Megson ISM

ISM for

Structural and StressAnalysisSecond Edition

by

Dr. T.H.G. Megson

CD1

4/13 US

(2005)

Solutions Manual

S o l u t i o n s t o C h a p t e r 2 P r o b l e m s

S.2.1

(a) Vectors representing the 10 and 15 kN forces are drawn to a suitable scale as shownin Fig. S.2.1. Parallel vectors AC and BC are then drawn to intersect at C. Theresultant is the vector OC which is 21.8 kN at an angle of 23.4◦ to the 15 kN force.

B

R

C

A15 kN

10 kN

60°u

O FIGURE S.2.1

(b) From Eq. (2.1) and Fig. S.2.1

R2 = 152 + 102 + 2 × 15 × 10 cos 60◦

which gives

R = 21.8 kN

Also, from Eq. (2.2)

tan θ = 10 sin 60◦

15 + 10 cos 60◦

so that

θ = 23.4◦.

3

4 • Solutions Manual

S.2.2

(a) The vectors do not have to be drawn in any particular order. Fig. S.2.2 shows thevector diagram with the vector representing the 10 kN force drawn first.

12 kN

8 kN

10 kN20 kN Ru

FIGURE S.2.2

The resultant R is then equal to 8.6 kN and makes an angle of 23.9◦ to the negativedirection of the 10 kN force.

(b) Resolving forces in the positive x direction

Fx = 10 + 8 cos 60◦ − 12 cos 30◦ − 20 cos 55◦ = −7.9 kN

Then, resolving forces in the positive y direction

Fy = 8 cos 30◦ + 12 cos 60◦ − 20 cos 35◦ = −3.5 kN

The resultant R is given by

R2 = (−7.9)2 + (−3.5)2

so that

R = 8.6 kN

Also

tan θ = 3.57.9

which gives

θ = 23.9◦.

Solutions to Chapter 2 Problems • 5

S.2.3 Initially the forces are resolved into vertical and horizontal components as shownin Fig. S.2.3.

FIGURE S.2.3

20.0 kN

30°

30°

45°

35.4 kN50 kN

40 kN 80 kN

69.3 kN

35.4 kN

40 kN34.6 kN

(�1, 1.25)

(0, 0.5)

(1.25, 0.25)

(1.0, 1.6)

O

y

Ry

Rx

x

60 kN

x

y

Then

Rx = 69.3 + 35.4 − 20.0 = 84.7 kN

Now taking moments about the x axis

Rx y = 35.4 × 0.5 − 20.0 × 1.25 + 69.3 × 1.6

which gives

y = 1.22 m

Also, from Fig. S.2.3

Ry = 60 + 40 + 34.6 − 35.4 = 99.2 kN

Now taking moments about the y axis

Ry x = 40.0 × 1.0 + 60.0 × 1.25 − 34.6 × 1.0

so that

x = 0.81 m

The resultant R is then given by

R2 = 99.22 + 84.72

from which

R = 130.4 kN

Finally

θ = tan−1 99.284.7

= 49.5◦.


S.2.4

(a) In Fig. S.2.4(a) the inclined loads have been resolved into vertical and horizontalcomponents. The vertical loads will generate vertical reactions at the supportsA and B while the horizontal components of the loads will produce a horizontalreaction at A only since B is a roller support.

3 kN

4 m 6 m 5 m 5 m

7 kN 8 kN

5.7 kN6.1 kN

3.5 kN 5.7 kN

A B

RBRA,V

RA,H

60° 45°

FIGURE S.2.4(a)

Taking moments about B

RA,V × 20 − 3 × 16 − 6.1 × 10 − 5.7 × 5 = 0

which gives

RA,V = 6.9 kN

Now resolving vertically

RB,V + RA,V − 3 − 6.1 − 5.7 = 0

so that

RB,V = 7.9 kN

Finally, resolving horizontally

RA,H − 3.5 − 5.7 = 0

so that

RA,H = 9.2 kN

Note that all reactions are positive in sign so that their directions are thoseindicated in Fig. S.2.4(a).

(b) The loads on the cantilever beam will produce a vertical reaction and a momentreaction at A as shown in Fig. S.2.4(b).

Resolving vertically

RA − 15 − 5 × 10 = 0

which gives

RA = 65 kN


MA

RA

10 m

15 kN

5 kN/m

A B

FIGURE S.2.4(b)

Taking moments about A

MA − 15 × 10 − 5 × 10 × 5 = 0

from which

MA = 400 kN m

Again the signs of the reactions are positive so that they are in the directionsshown.

(c) In Fig. S.2.4(c) there are horizontal and vertical reactions at A and a verticalreaction at B.

RA,H

A

B

RA,V RB

10 kN

2 m 4 m 2 m 2 m

5 m

15 kN5 kN/m

20 kN

FIGURE S.2.4(c)

By inspection (or by resolving horizontally)

RA,H = 20 kN


RB × 8 + 20 × 5 − 5 × 2 × 9 − 15 × 6 − 10 × 2 = 0

which gives

RB = 12.5 kN

Finally, resolving vertically

RA,V + RB − 10 − 15 − 5 × 2 = 0


so that

RA,V = 22.5 kN.

(d) The loading on the beam will produce vertical reactions only at the supports asshown in Fig. S.2.4(d).

A B

RA RB

75 kN/m8 kN/m

9 m3 m

FIGURE S.2.4(d)


RA × 12 + 75 − 8 × 12 × 6 = 0

Hence

RA = 41.8 kN


RB + RA − 8 × 12 = 0

so that

RB = 54.2 kN.

S.2.5

(a) The loading on the truss shown in Fig. P.2.5(a) produces only vertical reactions atthe support points A and B; suppose these reactions are RA and RB respectivelyand that they act vertically upwards. Then, taking moments about B

RA × 10 − 5 × 16 − 10 × 14 − 15 × 12 − 15 × 10 − 5 × 8 + 5 × 4 = 0

which gives

RA = 57 kN (upwards)


RB + RA − 5 − 10 − 15 − 15 − 5 − 5 = 0

from which

RB = −2 kN (downwards).


(b) The angle of the truss is tan−1(4/10) = 21.8◦. The loads on the rafters are sym-metrically arranged and may be replaced by single loads as shown in Fig. S.2.5.These, in turn, may be resolved into horizontal and vertical components and willproduce vertical reactions at A and B and a horizontal reaction at A.

FIGURE S.2.5

2000 N 8000 N

7427.9 N

1857.0 NRA,V

RA,H RB

20 m

4 m742.7 N

2970.9 N21.8°

21.8°

21.8°


RA,V × 20 + 742.7 × 2 − 1857.0 × 15 + 2970.9 × 2 + 7427.9 × 5 = 0

which gives

RA,V = −835.6 N (downwards).


RB + RA,V − 1857.0 + 7427.9 = 0

from which

RB = −4735.3 N (downwards).


RA,H − 742.7 − 2970.9 = 0

so that

RA,H = 3713.6 N.



S.3.1 Fig. S.3.1(a) shows the mast with two of each set of cables; the other two cablesin each set are in a plane perpendicular to the plane of the paper.

FIGURE S.3.1 (a)

AB

C

D

E

20 m

40 m

15 m

25 m

35 m

20 m

uB

uD

uC

(b)

22.5kN

119.6 kN

211.4 kN

314.9 kN

247.4 kN

166.4 kN

74.6 kN

Then,

θB = tan−1(

2035

)= 29.7◦, θC = tan−1

(2025

)= 38.7◦, θD = tan−1

(2015

)= 53.1◦.

The normal force at any section of the mast will be compressive and is the sum of theself-weight and the vertical component of the tension in the cables. Furthermore theself-weight will vary linearly with distance from the top of the mast. Therefore, ata section immediately above B,

N = 5 × 4.5 = 22.5 kN.

At a section immediately below B,

N = 22.5 + 4 × 15 cos 29.7◦ = 74.6 kN.

At a section immediately above C,

N = 74.6 + 10 × 4.5 = 119.6 kN.

At a section immediately below C,

N = 119.6 + 4 × 15 cos 38.7◦ = 166.4 kN.

At a section immediately above D,

N = 166.4 + 10 × 4.5 = 211.4 kN.

At a section immediately below D,

N = 211.4 + 4 × 15 cos 53.1◦ = 247.4 kN.

Finally, at a section immediately above E,

N = 247.4 + 15 × 4.5 = 314.9 kN.

The distribution of compressive force in the mast is shown in Fig. S.3.1(b).


S.3.2 The beam support reactions have been calculated in S.2.4(a) and are as shownin Fig. S.3.2(a); the bays of the beam have been relettered as shown in Fig. P.3.2.

(a)

(b)

(c)

(d)

A

3 kN7 kN 8 kN

5.7 kN

5.7 kN

6.1 kN

6.9 kN 7.9 kN

9.2 kN

3.5 kNB

C D E

5.7 kN

9.2 kN 9.2 kN

�ve

A B C D E

6.9 kN3.9 kN

2.2 kN

7.9 kN

�ve

�ve

A B C

D E

27.6 kN m

51.0 kN m

40.0 kN m

�ve

A B C D E

FIGURE S.3.2

Normal force

The normal force at any section of the beam between A and C is constant and givenby NAC = 9.2 kN (the vertical 3 kN load has no effect on the normal force).

Then

NCD = 9.2 − 3.5 = 5.7 kN

and

NDE = 9.2 − 3.5 − 5.7 = 0

Note that NDE could have been found directly by considering forces to the right ofany section between D and E. The complete distribution of normal force in shown inFig. S.3.2(b).

Shear force

The shear force in each bay of the beam is constant since only concentrated loads areinvolved.


At any section between A and B,

SAB = −RA,V = −6.9 kN.

At any section between B and C,

SBC = −6.9 + 3 = −3.9 kN.

At any section between C and D,

SCD = −6.9 + 3 + 6.1 = 2.2 kN.

Finally, at any section between D and E,

SDE = +RE = 7.9 kN.

The complete shear force distribution is shown in Fig. S.3.2(c).

Bending moment

Since only concentrated loads are present it is only necessary to calculate values ofbending moment at the load points. Note that MA = ME = 0.

At B,

MB = 6.9 × 4 = 27.6 kN m.

At C,

MC = 6.9 × 10 − 3 × 6 = 51.0 kN m.

At D,

MD = 6.9 × 15 − 3 × 11 − 6.1 × 5 = 40 kN m.

Alternatively, MD = 7.9 × 5 = 39.5 kN m; the difference in the two values is due torounding off errors. The complete distribution is shown in Fig. S.3.2(d).

S.3.3 There will be vertical and horizontal reactions at E and a vertical reaction atB as shown in Fig. S.3.3(a). The inclined 10 kN load will have vertical and horizon-tal components of 8 and 6 kN respectively, the latter acting to the right. Resolvinghorizontally, RE,H = 6 kN. Now taking moments about E

RB × 10 − 2 × 8 × 11 − 8 × 3 = 0

which gives

RB = 20 kN


RE,V + RB − 2 × 8 − 8 = 0


so that

RE,V = 4 kN

A

(a)

B

5 m

x

3 m 4 m

10 kN

6 kN

2 kN/m

RB RE,V

RE,H

8 kN

3 m

C D

E

(b)

�ve

6 kN 6 kN

A B C D E

(c)

�ve

10 kN

10 kN

�ve

�ve

4 kN

4 kN 4 kN

4 kN

A B

C D

E

(d)

25 kN m

12 kN m

�ve

�ve4 kN m

A

B C

D E

FIGURE S.3.3

Normal force

The normal force at all sections of the beam between A and D is zero since thereis no horizontal reaction at B and no horizontal forces between A and D. In CD,NCD = RE,H = 6 kN (compressive and therefore negative); the distribution is shownin Fig. S.3.3(b).

Shear force

We note that over the length of the uniformly distributed load the shear force will varylinearly with distance from, say, A. Then, at any section between A and B, a distancex from A, the shear force is given by

SAB = +2x


Therefore, when x = 0, SAB = 0 and when x = 5 m (i.e. at section just to the left of B),SAB = 10 kN. Also, at any section between B and C a distance x from A

SBC = +2x − RB = 2x − 20

Therefore, when x = 5 m (i.e. at a section just to the right of B), SBC = −10 kN andwhen x = 8 m (i.e. at a section just to the left of C), SBC = −4 kN.

Between C and D the shear force is constant and SCD = 2 × 8 − 20 = −4 kN. Finally,between D and E, SDE = 2 × 8 − 20 + 8 = +4 kN (or, alternatively, SDE = +RE,V =+4 kN). The complete distribution is shown in Fig. S.3.3(c).

Bending moment

At any section between A and B the bending moment is given by

MAB = −2x( x

2

)= −x2

Then, when x = 0, MAB = 0 and when x = 5 m, MAB = −25 kN m. Note that thedistribution is parabolic and that when x = 0, (dMAB/dx) = 0.

At any section between B and C

MBC = −x2 + RB(x − 5) = −x2 + 20(x − 5)

When x = 5 m, MBC = −25 kN m and when x = 8 m, MBC = −4 kN m. The distributionof bending moment between B and C is parabolic but has no turning value betweenB and C.

At D the bending moment is most simply given by MD = RE,V × 3 = +12 kN m; thecomplete distribution is shown in Fig. S.3.3(d).

S.3.4 By inspection, the vertical reactions at A and B are each equal to W as shownin Fig. S.3.4(a).

Shear force

The shear force in AB is equal to −W while that in BC = −W + W = 0. Also the shearforce in CD is equal to +W and the complete distribution is shown in Fig. S.3.4(b).

A B C D

W

W

W

W

L4

L2

L4

(a)FIGURE S.3.4(a)


�ve

�ve

A B C

W

W

D

(b)

(c)

�ve

A DB C

WL4

WL4 FIGURES S.3.4(b) and (c)

Bending moment

From symmetry, MA = MD = 0 and MB = MC = WL/4 giving the distribution shown inFig. S.3.4(c).

S.3.5 The support reactions for the beam have been calculated in S.2.4(b). However,in this case, if forces and moments to the right of any section are considered, thecalculation of the support reactions is unnecessary.

65 kN

15 kN�ve

A

(b)

B

(c)

400 kN m

�ve

A B

5 kN/m15 kN

10 mx

A B

(a)

FIGURE S.3.5


Shear force

At any section a distance x, say, from B the shear force is given by

SAB = −15 − 5x

Then, when x = 0, SAB = −15 kN and when x = 10 m, SAB = −65 kN; the distributionis linear as shown in Fig. S.3.5(b).

Bending moment

The bending moment is given by

MAB = −15x − 5x( x

2

)= −15x − 5x2

2

When x = 0, MAB = 0 and when x = 10 m, MAB = −400 kN m. The distribution, shownin Fig. S.3.5(c), is parabolic and does not have a turning value between A and B.

S.3.6 Only vertical reactions are present at the support points. Referring toFig. S.3.6(a) and taking moments about C

A B

RB RC

x

1 kN/m 5 kN

5 m 5 m10 m

C D

(a)

5 kN

5 kN

�ve�ve

�ve�ve

5 kN

6.25 kN

3.75 kN

A

(b)

B C D

5.5 kN m

�ve

3.75 m12.5 kN m

25 kN m(c)

A B C D

FIGURE S.3.6


RB × 10 − 1 × 15 × 7.5 + 5 × 5 = 0

from which

RB = 8.75 kN


RC + RB − 1 × 15 − 5 = 0

so that

RC = 11.25 kN

Shear force

The shear force at any section between A and B a distance x from A is given by

SAB = +1x

At A, where x = 0, SAB = 0 and at B where x = 5 m, SAB = +5 kN.

In BC the shear force at any section a distance x from A is given by

SBC = +1x − RB = x − 8.75 (i)

Then, when x = 5 m, SBC = −3.75 kN and when x = 15 m, SBC = 6.25 kN.

In CD the shear force is constant and given by

SCD = −5 kN (considering forces to the right of any section)

The complete distribution is shown in Fig. S.3.6(b).

Bending moment

At any section between A and B the bending moment is given by

MAB = −1x( x

2

)= −x2

2Therefore, when x = 0, MAB = 0 and when x = 5 m, MAB = −12.5 kN m. The distribu-tion is parabolic and when x = 0, (dMAB/dx) = 0.

In BC

MBC =(

−x2

2

)+ RB(x − 5) = −0.5x2 + 8.75(x − 5) (ii)

When x = 5 m, MBC = −12.5 kN m and when x = 15 m, MBC = −25 kN m. The distri-bution is parabolic and has a turning value when SBC = 0 (see Eq. (3.4)) and, fromEq. (i), this occurs at x = 8.75 m. Alternatively, but lengthier, Eq. (ii) could be dif-ferentiated with respect to x and the result equated to zero. When x = 8.75 m, fromEq. (ii), MBC = −5.5 kN m.

In CD the bending moment distribution is linear, is zero at D and −25 kN m at C. Thecomplete distribution is shown in Fig. S.3.6(c).


S.3.7 Referring to Fig. S.3.7(a) and taking moments about C

(a)

(b)

(c)

A

A

A

B

B

B

C

C

C

D

D

D

3 m 3 m

4.4 kN

2.9 m

7.4 kN

1.5 kN

16.8 kN m

0.9 kN m

5.6 kN

�ve

�ve

�ve

�ve

�ve

1.5 m

10 kN1 kN/m

xRA RC

FIGURE S.3.7

RA × 6 − 10 × 3 − 1 × 4.5 × 0.75 = 0

from which

RA = 5.6 kN


RC + RA − 10 − 1 × 4.5 = 0

i.e.

RC = 8.9 kN

Shear force

In AB the shear force is constant and equal to −RA = −5.6 kN.

At any section in BC a distance x from A

SBC = −RA + 10 + 1(x − 3) = 1.4 + x (i)

Therefore, when x = 3 m, SBC = 4.4 kN and when x = 6 m, SBC = 7.4 kN.


In CD the shear force varies linearly from zero at D to −1 × 1.5 = −1.5 kN at C. Thecomplete distribution is shown in Fig. S.3.7(b).

Bending moment

The bending moment in AB varies linearly from zero at A to RA × 3 = 5.6 × 3 =16.8 kN m at B.

In BC

MBC = 5.6x − 10(x − 3) − 1(x − 3)2

2= −1.4x − 0.5x2 + 25.5 (ii)

so that when x = 3, MBC = 16.8 kN m and when x = 6 m, MBC = −0.9 kN m. Note thatthe bending moment at C, by considering the overhang CD, should be equal to−1 × 1.52/2 = −1.125 kN m; the discrepancy is due to rounding off errors. We seefrom Eq. (i) that there is no turning value of bending moment in BC and that theslope of the bending moment diagram at D is zero. Also, the value of x at whichMBC = 0 is obtained by setting Eq. (ii) equal to zero and solving. This gives x = 5.9 mso that MBC = 0 at a distance of 2.9 m from B. The complete distribution is shown inFig. S.3.7(c).

S.3.8 The vertical reaction at A is given by (taking moments about B)

RA × 10 − w × 10 × 5 + 10 × 2 = 0

which gives

RA = 5w − 2

The position of the maximum sagging bending moment in AB is most easily foundby determining the shear force distribution in AB. Then, at any section a distance xfrom A

SAB = −(5w − 2) + wx (i)

From Eq. (i), SAB = 0 when x = (5w − 2)/w. This corresponds to a turning value, i.e.a maximum value, of bending moment (see Eq. (3.4)). Therefore, for x = 10/3 m,w = 1.2 kN/m. The maximum value of bending moment in AB is then

MAB(max) = (5 × 1.2 − 2) × 103

− 1.2 ×(

103

)2

2= 6.7 kN m.

S.3.9 Suppose that the vertical reaction at A is RA, then, taking moments about B

RAL − nw(

L2

)(5L4

)− wL

(L2

)= 0

which gives

RA = wL(5n + 4)8

(i)


The shear force at any section of the beam between A and B a distance x from A isgiven by

SAB = −RA + nwL2

+ wx = −wL5n + 4

8+ nw

L2

+ wx = −nwL8

− wL2

+ wx (ii)

The shear force is zero at a position of maximum bending moment and in this caseoccurs at L/3 from the right hand support, i.e. when x = 2L/3. Then, from Eq. (ii)

n = 43

A point of contraflexure occurs at a section where the bending moment changes sign,i.e. where the bending moment is zero. At a section of the beam a distance x from Athe bending moment is given by

MAB = RAx − nw(

L2

)(L4

+ x)

− wx2

2(iii)

Substituting in Eq. (iii) for RA from Eq. (i) and the calculated value of n and equatingto zero gives a quadratic equation in x whose factors are L/3 and −L. Clearly therequired value is L/3.

S.3.10 Referring to Fig. S.3.10(a) the support reaction at A is given by, takingmoments about B

RA × 20 − 5 × 20 × 10 + 20 × 5 = 0

from which

RA = 45 kN

Resolving forces vertically

RB + RA − 5 × 20 − 20 = 0

which gives

RB = 75 kN

Shear force

The shear force at any section of the beam between A and B a distance x from A isgiven by

SAB = −RA + 5x = −45 + 5x (i)

When x = 0, SAB = −45 kN and when x = 20 m, SAB = 55 kN.

In BC, considering forces to the right of any section, SBC = −20 kN and the completedistribution is shown in Fig. S.3.10(b).


(a)

(b)

A

202.5 kN m

B C

B�ve

�ve

�ve �ve

�ve

C

55 kN

20 kN

20 kN

RBRA

100 kN m18 m

45 kN

9 m

5 m

5 kN/m

B C

A

20 mx

A

(c) FIGURE S.3.10

Bending moment

The bending moment in AB will be a maximum when SAB = 0, i.e. when, from Eq. (i),x = 9 m. Also

MAB = 45x − 5x2

2(ii)

so that

MAB(max) = 202.5 kN m

The bending moment distribution is parabolic in AB and when x = 0, MAB = 0 andwhen x = 20 m, MAB = −100 kN m.

In BC the bending moment distribution is linear and varies from zero at C to−100 kN m at B.

Finally, from Eq. (ii), MAB changes sign, i.e. there is a point of contraflexure, atx = 18 m (Fig. S.3.10(c)).


S.3.11 The beam and its loading are shown in Fig. S.3.11.

FIGURE S.3.11

AC

x

D

120 kN (total)

100 kN (total)

B

3.5 m2.5 m2 m

RBRA


RA × 8 − 100 × 4 − 120 × 4.75 = 0

which gives

RA = 121.25 kN

Suppose that the maximum bending moment occurs in the bay CD. The shear forcein CD, at any section a distance x from A is given by

SCD = −121.25 + 100x8

+ 120(x − 2)2.5

= −217.25 + 60.5x (i)

For the bending moment to be a maximum SCD = 0 so that, from Eq. (i), x = 3.6 m.This value of x is within CD so that the assumption that the bending moment is amaximum in CD is correct. Then

MAB(max) = 121.25 × 3.6 −(

1008

)3.62

2−(

1202.5

)(3.6 − 2)2

2= 294.1 kN m

At mid-span x = 4 m, so that

MAB = 121.25 × 4 −(

1008

)42

2−(

1202.5

)(4 − 2)2

2= 289 kN m.

S.3.12 The beam and its loading are shown in Fig. S.3.12(a)


RA × 6 − 12

× 6 × 2 × 2 = 0

from which

RA = 2 kN


�ve

�ve

�ve

A

A

(a)

(b)

(c)

2 kN

B

B

4.62 kN m

3.46 m

6 m

w

4 kN

RA RB

2 kN m

A B

x

FIGURE S.3.12

Shear force

The shear force at any section a distance x from A is given by

SAB = −2 +(

12

)xw

where w = (1/3)x from similar triangles. Then

SAB = −2 + x2

6(i)

When x = 0, SAB = −2 kN and when x = 6 m, SAB = 4 kN. Examination of Eq. (i) showsthat SAB = 0 when x = 3.46 m and that (dSAB/dx) = 0 when x = 0. The distribution isshown in Fig. S.3.12(b).

Bending moment

The bending moment is given by

MAB = 2x −(

12

)xw( x

3

)= 2x − x3

18(ii)

From Eq. (ii) MAB = 0 when x = 0 and x = 6 m. Also, from Eq. (i), MAB is a maximumwhen x = 3.46 m. Then, from Eq. (ii)

MAB(max) = 4.62 kN m.

and the distribution is as shown in Fig. S.3.12(c).


S.3.13 The arrangement is shown in Fig. S.3.13.

Self-weight w

Sling

AB

L

xa

FIGURE S.3.13

The same argument applies to this problem as that in Ex. 3.11 in that the optimumposition for the sling is such that the maximum sagging bending moment in AB willbe numerically equal to the hogging bending moment at B. Then, taking momentsabout B

RA(L − a) − wL(

L2

− a)

= 0

from which

RA = wL

(L2 − a

)L − a

The shear force at a section a distance x from A in AB is given by

SAB = −RA + wx = −wL

(L2 − a

)L − a

+ wx (i)

At the position of maximum bending moment in AB the shear force is zero giving

x = L

(L2 − a

)L − a

Then

MAB(max) = RAL

(L2 − a

)L − a

− w2

L

(L2 − a

)L − a

2

(ii)

The bending moment at B due to the cantilever overhang is −wa2/2 which is numer-ically equal to the maximum value of MAB. Therefore substituting for RA in Eq. (ii)and adding the two values of bending moment (their sum is zero) we obtain

2a2 − 4aL + L2 = 0


Solving gives

a = 0.29L

Alternatively, as in Ex. 3.11, the relationship of Eq. (3.7) could have been used. Then,the area of the shear force diagram between A and the point of maximum saggingbending moment is equal to minus half the area of the shear force diagram betweenthis point and B.

S.3.14 The truss of Fig. P.3.14 is treated in exactly the same way as though it were asimply supported beam. The support reactions are calculated by taking moments andresolving forces and are as shown in Fig. S.3.14(a).

BA C D

150 kN

560 kN m480 kN m

140 kN

140 kN

10 kN

�ve

�ve

�ve� ve

(a)

(b)

(c)

50 kN

60 kN

A

A

B

B

C

C

D

D

60 kN

FIGURE S.3.14

Shear force

Between A and B the shear force is constant and equal to −60 kN. Between B and Cthe shear force is equal to −60 + 50 = −10 kN and between C and D the shear forceis equal to +140 kN; the complete distribution is shown in Fig. S.3.14(b).


Bending moment

The bending moment at the supports is zero. At B the bending moment is equalto +60 × 8 = 480 kN m while at C the bending moment is +140 × 4 = 560 kN m; thedistribution is shown in Fig. S.3.14(c).

S.3.15 The support reactions have been previously calculated in S.2.5(a) and are asshown in Fig. S.3.15(a).

FIGURE S.3.15

10 kN

30 kN

5 kN

A

A

A

B

B

B

C

C

C

D

D

D

E

E

E

F

F

F

G

G

G

H

H

H

15 kN

15 kN

5 kN

5 kN

15 kN 5 kN 5 kN

2 kN57 kN

7 kN12 kN

(a)

(b)

(c)

�ve

�ve

�ve

100 kNm

40 kNm

20 kNm10 kNm

76 kNm

Shear force

In AB the shear force is equal to +5 kN.

In BC the shear force is equal to +5 + 10 = +15 kN.


In CD the shear force is equal to +5 + 10 + 15 = +30 kN.

In DE the shear force is equal to +5 + 10 + 15 + 15 − 57 = −12 kN.

In EF the shear force is equal to +5 + 10 + 15 + 15 − 57 + 5 = −7 kN or, more simply,considering forces to the right of any section, −5 − 2 = −7 kN.

In FG the shear force is equal to −5 kN while in GH the shear force is zero. Thecomplete distribution is shown in Fig. S.3.15(b).

Bending moment

Since only concentrated loads are involved it is only necessary to calculate values ofbending moment at the load points. Then

MA = MG = MH = 0

MB = −5 × 2 = −10 kN m

MC = −5 × 4 − 10 × 2 = −40 kN m

MD = −5 × 6 − 10 × 4 − 15 × 2 = −100 kN m

ME = −5 × 12 − 2 × 8 = −76 kN m

MF = −5 × 4 = −20 kN m

The complete distribution is shown in Fig. S.3.15(c).

S.3.16 In this problem it is unnecessary to calculate the support reactions. Also, theportion CB of the cantilever is subjected to shear and bending while the portion BAis subjected to shear, bending and torsion as shown in Fig. S.3.16.

Consider CB

The shear force in CB is constant and equal to +3 kN (if viewed in the direction BA).The bending moment is zero at C and varies linearly to −3 × 2 = −6 kN m at B. Thetorque in CB is everywhere zero.

Consider AB

The shear force in AB is constant and equal to −3 kN. The bending moment varieslinearly from zero at B to −3 × 3 = −9 kN m at A. The torque is constant and equal to6 kN m.


FIGURE S.3.16

A

3 kN

C2 m

B

3 m

3 kN

6 kN m Equivalentloading on BA

S.3.17 From Fig. P.3.17 the torque in CB is constant and equal to −300 N m. In BAthe torque is also constant and equal to −300 − 100 = −400 N m.

S.3.18 Referring to Fig. S.3.18(a)

(b) A B C

1500 Nm

1000 Nm�ve

(a)

AB C

500 Nmx

1 Nm/mm

FIGURE S.3.18

the torque in CB at any section a distance x from C is given by

TCB = 1x (T is in N m when x is in mm)

Then, when x = 0, TCB = 0 and when x = 1 m, TCB = 1000 N m.

In BA the torque is constant and equal to 1 × 1000 + 500 = 1500 N m.



S.3.19 From symmetry the support reactions are equal and are each 2400 N m. Then,referring to Fig. S.3.19(a)

FIGURE S.3.19

400 N m

400 N m

400 N m

400 N m

2 N m/mm

0.5 m 0.5 m1.0 m

A

A

(b)

(a)

x

B

B

C

C

D

D

2400 N m

1400 N m

1400 N m

�ve�ve

�ve �ve

1000 N m

1000 N m

2400 N m

the torque at any section a distance x from D is given by

TDC = 400 + 2x (T is in N m when x is in mm)

Therefore, when x = 0, TDC = 400 N m and when x = 0.5 m, TDC = 1400 N m.

In CB the torque is given by

TCB = 400 + 2x − 2400 = 2x − 2000

and when x = 0.5 m, TCB = −1000 N m. Note that TCB = 0 when x = 1.0 m.

The remaining distribution follows from antisymmetry and is shown in Fig. S.3.19(b).The maximum value of torque is 1400 N m and occurs at C and B.


S.4.1 Initially the support reactions are calculated. Only vertical reactions are presentso that, referring to Fig. S.4.1(a), and taking moments about E

RB × 18 − 30 × 24 + 60 × 6 = 0

i.e.

RB = 20 kN (upwards)


Now resolving vertically,

RE + RB − 30 − 60 = 0

which gives

RE = 70 kN (upwards)

A

30 kN 60 kNRB RE

B

G H J K

C

5 � 6 m

8 m

u

D E F

FIGURE S.4.1(a)

All members are assumed to be in tension as shown in Fig. S.4.1(a) and for simplicitythe forces in the members are designated by their joint letters. Also the diagonals ofthe truss are each 10 m long so that cos θ = 0.6 and sin θ = 0.8.

By inspection

HC = 0, BG = −20 kN and EK = −70 kN

Joint A:

Resolving vertically,

AG sin θ − 30 = 0

i.e.

AG = +37.5 kN

Resolving horizontally,

AB + AG cos θ = 0

i.e.

AB = −37.5 × 0.6 = −22.5 kN

Joint B:


BC − BA = 0


i.e.

BC = BA = −22.5 kN

Joint G:


GC sin θ + GB + GA sin θ = 0

i.e.

GC × 0.8 − 20 + 37.5 × 0.8 = 0

GC = −12.5 kN


GH + GC cos θ − GA cos θ = 0

i.e.

GH − 12.5 × 0.6 − 37.5 × 0.6 = 0

GH = 30 kN

Joint H:

By inspection (or resolving horizontally)

HJ = HG = 30 kN

Joint C:


CJ sin θ + CG sin θ = 0

i.e.

CJ × 0.8 − 12.5 × 0.8 = 0

CJ = 12.5 kN

Joint J:


JD + JC sin θ = 0

i.e.

JD + 12.5 × 0.8 = 0

JD = −10 kN



JK − JH − JC cos θ = 0

i.e.

JK − 30 − 12.5 × 0.6 = 0

JK = 37.5 kN

Joint D:


DK sin θ + DJ = 0

i.e.

DK × 0.8 − 10 = 0

DK = 12.5 kN


DE + DK cos θ − DC = 0

i.e.

DE + 12.5 × 0.6 + 37.5 = 0

DE = −45 kN

Joint F:


FK sin θ − 60 = 0

i.e.

FK × 0.8 − 60 = 0

FK = 75 kN


FE + FK cos θ = 0

i.e.

FE + 75 × 0.6 = 0

FE = −45 kN


To check the forces in the members JK and DE take a section cutting JK, KD and DEas shown in Fig. S.4.1(b).

FIGURE S.4.1(b)and (c)

KKJ

EDE F DE

D

DJ

KJK

FKD

60 kN 60 kNRE � 70 kN RE � 70 kN

E

(b) (c)

Taking moments about K,

ED × 8 − 60 × 6 = 0

ED = 45 kN

Taking moments about D,

KJ × 8 − 60 × 12 + 70 × 6 = 0

KJ = 37.5 kN

To check the force in the member JD take a section cutting JK, JD and CD as shownin Fig. S.4.1(c).


DJ + 70 − 60 = 0

DJ = −10 kN

S.4.2 Referring to Fig. S.4.2 the reactions at A and B are each 15 kN from symmetry.

C

15 kN

10 kN

10 kN

10 kN

15 kN

4 mE

4 � 2m

G

J

H

A P F

60�u

B

30�

FIGURE S.4.2


Also sin θ = 4/8 = 0.5 so that θ = 30◦, the remaining angles follow. By inspection (orby resolving vertically) BJ = −15 kN. Further, by inspection, FB = 0. All members areassumed to be in tension.

Joint A:


AC sin θ + 15 = 0

i.e.

AC × 0.5 + 15 = 0

AC = −30 kN


AP + AC cos θ = 0

i.e.

AP − 30 × 0.866 = 0

AP = 26.0 kN

Joint C:

Resolving parallel to CP,

CP + 10 cos 30◦ = 0

i.e.

CP = −8.7 kN

Resolving parallel to CE,

CE − CA − 10 sin 30◦ = 0

i.e.

CE + 30 − 10 × 0.5 = 0

CE = −25 kN

Joint P:


PE cos 30◦ + PC cos 30◦ = 0


i.e.

PE = −PC = 8.7 kN


PF + PE cos 60◦ − PC cos 60◦ − PA = 0

i.e.

PF + 8.7 × 0.5 + 8.7 × 0.5 − 26.0 = 0

PF = 17.3 kN

Joint F:


FE cos 30◦ + FH cos 30◦ = 0

i.e.

FE = −FH


FH cos 60◦ − FE cos 60◦ − FP = 0 (FB = 0)

FH × 0.5 + FH × 0.5 − 17.3 = 0

FH = −FE = 17.3 kN

Joint G:

Resolving parallel to GH,

GH + 10 cos 30◦ = 0

i.e.

GH = −8.7 kN

Resolving parallel to GE,

GJ − GE − 10 cos 60◦ = 0

i.e.

GJ − GE − 10 × 0.5 = 0 (i)

Joint H:

Resolving perpendicularly to HJ,

HE cos 30◦ + HG cos 30◦ = 0


i.e.

HE = −HG = 8.7 kN

Resolving parallel to HJ,

HJ − HF + HG cos 60◦ − HE cos 60◦ = 0

i.e.

HJ − 17.3 − 8.7 × 0.5 − 8.7 × 0.5 = 0

HJ = 26.0 kN

Joint J:


JG cos 60◦ + JH cos 30◦ + JB = 0

i.e.

JG × 0.5 + 26.0 × 0.866 − 15 = 0

JG = −15.0 kN

Substituting for JG in Eq. (i) gives

GE = −20.0 kN

S.4.3 From Fig. P.4.3, by inspection, there will be a horizontal reaction of 4 kN actingto the right at A. Taking moments about B,

RA,V × 12 − 40 × 10 − 40 × 8 = 0

i.e.

RA,V = 60 kN


RB + 60 − 40 − 40 = 0

i.e.

RB = 20 kN

First take a section cutting the members EG, EH and FH as shown in Fig. S.4.3(a).Note that tan θ = 1.5/2 so that θ = 36.9◦


FIGURE S.4.3

EGEH

FHH

2 m

1.5 m20 kN

G B

u

4 kN 4 kN

40 kN

60 kN

A C EC

EF

FH

F

(a) (b)


EH sin 36.9◦ + 20 = 0

i.e.

EH = −33.3 kN

Taking moments about H,

EG × 1.5 + 20 × 6 = 0

i.e.

EG = −80 kN


FH + EH cos 36.9◦ + EG = 0

i.e.

FH − 33.3 cos 36.9◦ − 80 = 0

FH = 106.6 kN

Now take a section cutting EC, EF and FH as shown in Fig. S.4.3(b)


EF − 40 + 60 = 0

EF = −20 kN

S.4.4 There will be horizontal and vertical reactions at A and a vertical reaction at B.Then, resolving horizontally,

RA,H − 2 × 6 cos 30◦ = 0


i.e.

RA,H = 10.4 kN

Taking moments about B,

RA,V × 12 − 5 × 36 × 6 − 6 × 3 = 0

i.e.

RA,V = 91.5 kN

From symmetry at K, the forces in the members KE and KG are equal. Then, assum-ing they are tensile and resolving vertically,

2 × KE cos 30◦ + 36 = 0

i.e.

KE = −20.8 kN (=KG)

Knowing KE (and KG) we can now take a section cutting through KG, EG, EF andDF as shown in Fig. S.4.4.

10.4 kN

36 kN 36 kN

36 kN

91.5 kN

AD

DFEF

K

ECKG

EG

FIGURE S.4.4


EF cos 30◦ + KG cos 30◦ + 3 × 36 − 91.5 = 0

i.e.

EF × 0.866 − 20.8 × 0.866 + 108 − 91.5 = 0

EF = 1.7 kN

Taking moments about E,

DF × 1.5 tan 60◦ − KG × 1.5 tan 60◦ − 36 × 1.5 + 36 × 3 − 91.5 × 4.5

+ 10.4 × 1.5 tan 60◦ = 0


i.e.

DF × 2.6 + 20.8 × 2.6 − 54 + 108 − 411.8 + 10.4 × 2.6 = 0

DF = 106.4 kN


EG + EF cos 60◦ + KG cos 60◦ + DF + 10.4 = 0

i.e.

EG + 1.7 × 0.5 − 20.8 × 0.5 + 106.4 + 10.4 = 0

EG = −107.3 kN

S.4.5 The equation of the parabola which passes through the upper chord pointsis y = kx2. When x = 9 m, y = 7 m so that 7 = k × 81 which gives k = 0.086. At A,when x = 3 m, y = 0.086×9 = 0.77 m so that BA = 7−0.77 = 6.23 m and when x = 6 m,y = 0.086 × 36 = 3.1 m and CD = 7 − 3.1 = 3.9 m.

From symmetry, the vertical reactions at the supports are both equal to 3.5 kN. Thetruss is now cut through AD, BD and BC as shown in Fig. S.4.5.

u1 D

DA

DB

CBC

3.9 m

u2

2 kN 3.5 kN FIGURE S.4.5

From Fig. S.4.5, tan θ1 = (6.23 − 3.9)/3 = 0.78 so that θ1 = 37.8◦. Also tan θ2 = 3.9/3 =1.3, i.e. θ2 = 52.4◦. Taking moments about D,

CB × 3.9 − 3.5 × 3 = 0

i.e.

CB = 2.7 kN



DA sin θ1 − DB sin θ2 − 2 + 3.5 = 0

i.e.

0.61 DA − 0.79 DB + 1.5 = 0

DA − 1.3 DB + 2.46 = 0 (i)

Now resolving horizontally,

DA cos θ1 + DB cos θ2 + CB = 0

i.e.

0.79 DA + 0.61 DB + 2.7 = 0

or

DA + 0.77 DB + 3.42 = 0 (ii)

Eq. (i) − Eq. (ii)

−2.07 DB − 0.96 = 0

DB = −0.5 kN

From Eq. (i) or Eq. (ii)

DA = −3.1 kN

S.4.6 In Fig. S.4.6 the horizontal through A meets the diagonal DE at I, the lengthAI = 3 m.

B (3, 1)

T

J

D (1, 1)

y

H (3, 0) G (2, 0)

7.5 kN 5 kN

F (1, 0) x

C (2, 1)

I 1 m

E (0, 0)

45�

A(3.5, 0.5)

0.5 m 1 m 1 m 1 mFIGURE S.4.6



T × JA − 7.5 × 1.5 − 5 × 3.5 = 0

i.e.

T × 3 sin 45◦ − 11.25 − 17.5 = 0

T = 13.6 kN

The coordinates of each joint, referred to the xy axes shown in Fig. S.4.6, are nowcalculated and inserted in Fig. S.4.6.

Joint E:

x; tEF(xF − xE) + tED(xD − xE) = 0

tEF(1 − 0) + tED(1 − 0) = 0

i.e.

tEF + tED = 0 (i)

y; tEF(yF − yE) + tED(yD − yE) − 5 = 0

tEF(0 − 0) + tED(1 − 0) − 5 = 0

i.e.

tED = 5

Substituting in Eq. (i),

tEF = −5

Joint F:

x; tDE(xE − xD) + tDC(xC − xD) + tDF(xF − xD) + T cos 45◦ = 0

tDE(0 − 1) + tDC(2 − 1) + tDF(1 − 1) + 13.6 cos 45◦ = 0

i.e.

tDC − tDE + 9.6 = 0

Substituting for tDE from the above,

tDC = −4.6

y; tDE(yE − yD) + tDC(yC − yD) + tDF(yF − yD) + T sin 45◦ = 0

tDE(0 − 1) + tDC(1 − 1) + tDF(0 − 1) + 13.6 sin 45◦ = 0


i.e.

−tDE − tDF + 9.6 = 0

Substituting for tDE from the above

tDF = 4.6

Now, instead of writing down the equations in terms of x and y initially, we shall insertthe coordinates directly.

Joint F:

x; tFE(0 − 1) + tFD(1 − 1) + tFC(2 − 1) + tFG(2 − 1) = 0

−tFE + tFC + tFG = 0 (ii)

y; tFE(0 − 0) + tFD(1 − 0) + tFC(1 − 0) + tFG(0 − 0) = 0

tFD + tFC = 0

tFC = −tFD = −4.6

Then, from Eq. (ii),

tFG = −0.4

Joint C:

x; tCB(3 − 2) + tCF(1 − 2) + tCD(1 − 2) + tCG(2 − 2) = 0

tCB − tCF − tCD = 0

Substituting the values of tCF and tCD

tCB = −9.2

y; tCB(1 − 1) + tCF(0 − 1) + tCD(1 − 1) + tCG(0 − 1) = 0

−tCF − tCG = 0

tCG = 4.6

Joint G:

x; tGB(3 − 2) + tGH(3 − 2) + tGF(1 − 2) + tGC(2 − 2) = 0

tGB + tGH − tGF = 0 (iii)


y; tGB(1 − 0) + tGH(0 − 0) + tGF(0 − 0) + tGC(1 − 0) − 7.5 = 0

tGB + tGC − 7.5 = 0

Then

tGB = 2.9

Substituting in Eq. (iii) for tGB and tGF gives tGH = −3.3.

Joint B:

x; tBG(2 − 3) + tBA(3.5 − 3) + tBC(2 − 3) + tBH(3 − 3) = 0

−tBG + 0.5 tBA − tBC = 0 (iv)

Substituting in Eq. (iv) for tBG and tBC gives tBA = −12.6

y; tBG(0 − 1) + tBA(0.5 − 1) + tBC(1 − 1) + tBH(0 − 1) = 0

−tBG − 0.5 tBA − tBH = 0 (v)

Substituting in Eq. (v) for tBG and tBA gives tBH = 3.4

Joint H:

x; tHA(3.5 − 3) + tHB(3 − 3) + tHG(2 − 3) = 0

0.5 tHA − tHG = 0

tHA = −6.6

The tension coefficients are now multiplied by the length of each member to obtainthe force in each member:

Member Length (m) Tension coefficient Force (kN)

ED 1.414 5 7.1EF 1 −5 −5DC 1 −4.6 −4.6DF 1 4.6 4.6FG 1 −0.4 −0.4FC 1.414 −4.6 −6.5CB 1 −9.2 −9.2CG 1 4.6 4.6GB 1.414 2.9 4.1GH 1 −3.3 −3.3BA 0.707 −12.6 −8.9BH 1 3.4 3.4HA 0.707 −6.6 −4.7


S.4.7 Truss of P.4.1: The support reactions have been calculated in P.4.1 and are asshown in Fig. S.4.7(a); the spaces between the members and forces are now numbered.

2

7 8

965

1 4 3

70 kN20 kN30 kN 60 kN

10

11 12

A

G H J K

FB C D E

(a)

1

4

2

5

6 9 11

107, 8

(b) 123FIGURES S.4.7(a)

and (b)

Starting at joint A and moving in a clockwise sense round joint A the vector 12 isdrawn vertically downwards and is equal to 30 kN to a suitable scale. The vector 25 isdrawn parallel to AG and intersects the vector 51, which is horizontal, at 5. The vector25, equal to 37.5 kN, acts away from A so that AG is in tension. The vector 51, equalto 22.5 kN, acts towards A so that AB is in compression.

Now moving in a clockwise sense round joint B the vector 41 is drawn verticallyupwards to represent 20 kN. The vector 15 has already been drawn and the vector56 is vertical so that the point 6 is located by the intersection of this vector with thehorizontal vector 64. Then, BG = −20 kN (vector 56) and BC = −22.5 kN (vector 64).

At joint H the vector 72 is horizontal as is the vector 28. Further, the vector 87 isvertical but since 72 and 28 are both horizontal the vector 87 must be zero in lengthand the points 7 and 8 therefore coincide. From the force polygon HG = 30.0 kN (72)and HJ = 30.0 kN (28), HC = 0.


We now move to joint C where the point 9 is located by drawing the vectors 89, parallelto CJ, and 94 horizontally through 4. Then, CJ = 12.5 kN (89), CG = −12.5 kN (67)and CD = −37.5 (94).

The point 10 is found by drawing the vectors 2 10 (horizontal) and 9 10 (vertical).JK = 37.5 kN (2 10) and JD = −10.0 kN (9 10).

Point 11 is found by drawing the vectors 11 4 horizontally and 10 11 parallel to DK.Then, DK = 12.5 kN (10 11) and DE = −45 kN (11 4).

The point 3 is located by drawing the vector 23 vertically downwards and equal to60 kN. The point 12 is positioned at the intersection of the vertical through 11 andthe horizontal through 3. Then, KE = −70 kN (11 12), EF = −45 kN (12 3) and KF =75 kN (12 2).

Truss of P.4.2: The reactions have been calculated in S.4.2 and are shown inFig. S.4.7(c); The spaces between the loads and members are now numbered.

A B

C

E

G

H

FP

J

15 kN 15 kN

10 kN

10 kN

10 kN

(c)1

2

3

4 5

6

78

9

11

12

10

2

3

1, 12

4

6

7

10

11

9 5

8

(d) FIGURES S.4.7(c) and (d)

Starting at joint A the vector 12 is drawn vertically upwards to represent the 15 kNreaction (Fig. S.4.7(d)). The point 6 is located at the intersection of 16 (horizontal)and 26 (parallel to AC). Then, AC = −30 kN (26) and AP = 26 kN (61).

Moving to joint C the vector 23 is drawn to represent the 10 kN load and the point 7 isfound by drawing 37 (parallel to AC) and 76 (parallel to CP). CP = −8.7 kN (76) andCE = −25 kN (37).


At P, 78 is drawn parallel to PE and 81 drawn parallel to PF, hence the point 8. Then,PE = 8.7 kN (78) and PF = 17.3 kN (81).

We now cannot move to joints E, F, G or H since there are more than two unknownsat each of these joints. Therefore moving to joint B, the vector 51 is drawn verticallyupwards to represent the 15 kN reaction. Now 1 12 is horizontal and 12 5 is vertical sothat the points 1 and 12 must coincide; Then, FB = 0 (1 12) and BJ = −15 kN (12 5).The point 11 is found by drawing 11 5 parallel to GJ and 12 11 parallel to JH. This givesJH = 26 kN (12 11) and GJ = −15 kN (11 5). Considering joint G the point 4 is locatedby drawing the vector 45 vertically downwards to represent the 10 kN load. Then, point10 is the intersection of 10 4 (parallel to EG) and 11 10 (parallel to HG). This givesEG = −20 kN (10 4) and GH = −8.7 kN (11 10). Finally point 9 is the intersection of12 9 (parallel to HF) and 89 (parallel to EF) or the intersection of either of thesetwo with 10 9 (parallel to EH). Therefore, EH = 8.7 kN (9 10), EF = −17.3 kN (89),HF = 17.3 kN (12 9).

S.4.8 The x, y and z coordinates of each joint are, O(0, 0, 0), A(−3.5, −4, 9),B(6.5, −4, 9) and C(1, 8, 9). Then,

x; tOA(−3.5 − 0) + tOB(6.5 − 0) + tOC(1 − 0) + 5 = 0

i.e.

tOA − 1.86 tOB − 0.29 tOC − 1.43 = 0 (i)

y; tOA(−4 − 0) + tOB(−4 − 0) + tOC(8 − 0) + 40 = 0

i.e.

tOA + tOB − 2 tOC − 10 = 0 (ii)

z; tOA(9 − 0) + tOB(9 − 0) + tOC(9 − 0) = 0

i.e.

tOA + tOB + tOC = 0 (iii)

Eq. (ii) − Eq. (iii)

−3 tOC − 10 = 0

i.e.

tOC = −3.33

Eq. (i) − Eq. (ii)

−2.86 tOB + 1.71 tOC + 8.57 = 0


Substituting for tOC gives

tOB = 1.01

From Eq. (iii)

tOA = 2.32

The length of member OA is given by

LOA =√

(xA − xO)2 + (yA − yO)2 + (zA − zO)2

=√

(−3.5 − 0)2 + (−4 − 0)2 + (9 − 0)2 = 10.45 m

Similarly,

LOB = 11.8 m and LOC = 12.08 m

Then

OA = 2.32 × 10.45 = 24.2 kN, OB = 1.01 × 11.8 = 11.9 kN,

OC = −3.33 × 12.08 = −40.2 kN.

S.4.9 Take the origin of axes at A and assume the axes system shown in Fig. S.4.9.

A

x

z y

25 kN

25 kN

D

B

C

45�

45�

FIGURE S.4.9

The coordinates of each joint are as follows: A(0, 0, 0), B(−2.5, 0, 4), C(0, 3, 4) andD(2.5, 0, 4). Then,

x; tAB(−2.5 − 0) + tAC(0 − 0) + tAD(2.5 − 0) + 25 cos 45◦ − 25 cos 45◦ = 0

i.e.

tAB − tAD = 0 (i)


y; tAB(0 − 0) + tAC(3 − 0) + tAD(0 − 0) + 2 × 25 cos 45◦ = 0

i.e.

tAC = −11.79 (ii)

z; tAB(4 − 0) + tAC(4 − 0) + tAD(4 − 0) + 25 = 0

i.e.

tAB + tAC + tAD + 6.25 = 0 (iii)

Substituting in Eq. (iii) from Eqs (i) and (ii)

tAB = 2.77 = tAD

Now

LAB = LAD =√

42 + 2.52 = 4.72 m and LAC =√

42 + 32 = 5.0 m.

Then,

AB = 2.77 × 4.72 = 13.1 kN = AD, AC = −11.79 × 5 = −59.0 kN.

S.4.10 Referring to Fig. P.4.10 choose an origin of axes at E with the x axis parallel toEF, the y axis parallel to EB and the z axis vertical. The coordinates of the joints arethen E(0, 0, 0), B(0, 4, 0), C(0, 4, 3), F(3, 0, 0), A(3, 4, 0) and D(3, 4, 3).

Joint E:

x; tEB(0 − 0) + tEC(0 − 0) + tEF(3 − 0) + 3 = 0

i.e.

tEF = −1 (i)

y; tEB(4 − 0) + tEC(4 − 0) + tEF(0 − 0) = 0

i.e.

tEB + tEC = 0 (ii)

z; tEB(0 − 0) + tEC(3 − 0) + tEF(0 − 0) + 9 = 0

i.e.

tEC = −3

Therefore, from Eq. (ii),

tEB = 3


Joint F:

x; tFE(0 − 3) + tFB(0 − 3) + tFA(3 − 3) + tFD(3 − 3) = 0

i.e.

tFE + tFB = 0 (iii)

Therefore, from Eq. (i),

tFB = 1

y; tFE(0 − 0) + tFB(4 − 0) + tFA(4 − 0) + tFD(4 − 0) = 0

i.e.

tFA + tFD + 1 = 0 (iv)

z; tFE(0 − 0) + tFB(0 − 0) + tFA(0 − 0) + tFD(3 − 0) + 6 = 0

i.e.

tFD = −2

Then, from Eq. (iv)

tFA = 1

From Fig. P.4.10, LEB = LFA = 4 m, LEF = 3 m and LEC = LFB = LFD = 5 m. Then

EF = −1 × 3 = −3 kN, EC = −3 × 5 = −15 kN, EB = 3 × 4 = 12 kN,

FB = 1 × 5 = 5 kN, FA = 1 × 4 = 4 kN, FD = −2 × 5 = −10 kN.

S.4.11 With the origin of axes at joint D as shown in Figs S.4.7(a) and (b) the coordi-nates of the joints are D(0, 0, 0), E(4, 0, 0), A(−2, 4, −4), B(6, 4, −4), C(2, −4, −4).Further, the analysis can only begin at joint D since there are more than three unknownsat every other joint.

FIGURE S.4.114 m

A, B C

D, E

z z

y 80 kN

4 m

4 m

2 m

A

40 kN D Ex

C B

2 m 2 m 2 m(a) (b)

Joint D:

x; tDE(4 − 0) + tDA(−2 − 0) + tDC(2 − 0) − 40 = 0


i.e.

2 tDE − tDA + tDC − 20 = 0 (i)

y; tDE(0 − 0) + tDA(4 − 0) + tDC(−4 − 0) = 0

i.e.

tDA − tDC = 0 (ii)

z; tDE(0 − 0) + tDA(−4 − 0) + tDC(−4 − 0) = 0

i.e.

tDA + tDC = 0 (iii)

Eqs (ii) and (iii) can only be satisfied if tDA = tDC = 0. Then, from Eq. (i) tDE = 10.

Joint E:

x; tED(0 − 4) + tEA(−2 − 4) + tEB(6 − 4) + tEC(2 − 4) = 0

i.e.

−40 − 6 tEA + 2 tEB − 2 tEC = 0 (iv)

y; tED(0 − 0) + tEA(4 − 0) + tEB(4 − 0) + tEC(−4 − 0) − 80 = 0

i.e.

4 tEA + 4 tEB − 4 tEC − 80 = 0 (v)

z; tED(0 − 0) + tEA(−4 − 0) + tEB(−4 − 0) + tEC(−4 − 0) = 0

i.e.

−4 tEA − 4 tEB − 4 tEC = 0 (vi)

Adding Eqs (v) and (vi) gives

−8 tEC − 80 = 0

i.e.

tEC = −10

Also, Eq. (v) − Eq. (vi) gives

tEA + tEB − 10 = 0 (vii)

Substituting in Eq. (iv) for tEC gives

−20 − 6 tEA + 2 tEB = 0 (viii)


2 × Eq. (vii) − Eq. (viii) gives

tEA = 0

Then, from Eq. (vii) tEB = 10

From Fig. S.4.11, LED = 4 m, LEB = LDA = LEC = LDC = √22 + 42 + 42 = 6 m. Then,

AD = 0, DC = 0, DE = 10 × 4 = 40 kN, AE = 0, CE = − 10 × 6 = − 60 kN, BE = 10 ×6 = 60 kN.


S.5.1 Referring to Fig. S.5.1 where H is the tension in the portion CD of the cable andtaking moments about A

RA,H

RA,V

A

B

H

C

ab

5 kN5 kN

1.5 m

FIGURE S.5.1

H × 1.5 − 5 × 5 − 5 × 2.5 = 0

i.e.

H = 25 kN = TCD

Resolving horizontally

RA,H − H = 0

i.e.

RA,H = 25 kN

From symmetry,

RA,V = 10 kN

The tension in AB is equal to the resultant of RA,V and RA,H,

i.e.

TAB =√

102 + 252 = 26.9 kN


From Fig. S.5.1,

tan α = 1025

= 0.4

Then

α = 21.8◦

Also the vertical distance of B below A is 2.5 tan α = 1.0 m. Then

tan β = 0.52.5

= 0.2

so that

β = 11.3◦

Now resolving horizontally at C

TCB cos 11.3◦ = TCD = 25 kN

i.e.

TCB = 25.5 kN

The tensions in the remaining parts of the cable follow from symmetry.

S.5.2 From Fig. S.5.2

CB

D

0.7 m

0.5 m

A

1 kN2 kN

RA,V

RA,H

RD,V

RD,H

a

g

FIGURE S.5.2

resolving vertically

RA,V + RD,V = 1 + 2 = 3 kN (i)

Taking moments about C

RD,H × 0.5 − RD,V × 2 = 0

i.e.

RD,H − 4 RD,V = 0 (ii)


RD,H × 0.7 + RD,V × 6 − 1 × 2 − 2 × 4 = 0


i.e.

RD,H + 8.57 RD,V − 14.29 = 0 (iii)

Eq. (ii) − Eq. (iii)

−12.57 RD,V + 14.29 = 0

i.e.

RD,V = 1.14 kN

From Eq. (ii)

RD,H = 4.56 kN = RA,H

From Eq. (i)

RA,V = 1.86 kN

Now

tan α = RA,V

RA,H= 1.86

4.56= 0.41

i.e.

α = 22.19◦

Therefore the sag of B relative to A is = 2 tan α = 0.81 m.

Further

γ = tan−1 (1.2 − 0.81)2

= 11.03◦

At A the resultant of RA,H and RA,V is equal to the tension TAB,

i.e.

TAB =√

4.562 + 1.862 = 4.9 kN

Also

TDC =√

4.562 + 1.142 = 4.7 kN

At B

TBC cos 11.03◦ − TBA cos 22.19◦ = 0

i.e.

TBC = 4.6 kN.

S.5.3 Suppose the horizontal and vertical components of reaction at A and E areRA,H, RA,V and RE,H, RE,V, respectively. Then, taking moments about B,

RA,V × 4 − RA,H × 2.6 = 0


i.e.

1.54 RA,V − RA,H = 0 (i)

Now taking moments about E

RA,V × 18 − RA,H × 0.5 − 3 × 14 − 5 × 9 − 4 × 4 = 0

i.e.

36 RA,V − RA,H − 206 = 0 (ii)

Eq. (i) − Eq. (ii)

−34.46 RA,V + 206 = 0

i.e.

RA,V = 5.98 kN

From Eq. (i)

RA,H = 9.21 kN

Then

TAB =√

5.982 + 9.212 = 10.98 kN

Suppose α is the angle AB makes with the horizontal, then

α = tan−1 2.64

= 33.02◦

Also let β be the angle BC makes with the horizontal. Then, resolving horizontallyat B

TBC cos β = TAB cos α = 10.98 cos 33.02◦ = 9.21 kN

Resolving vertically at B

TBC sin β + 3 = TAB sin α = 10.98 sin 33.02◦ = 5.98 kN

i.e.

TBC sin β = 2.98

Then

tan β = 2.989.21

= 0.324

i.e.

β = 17.93◦

and

TBC = 9.68 kN


The sag of C below A is equal to 2.6 + 5 tan β = 4.22 m.

Let γ be the angle CD makes with the horizontal. Then, resolving horizontally at C,

TCD cos γ = TCB cos β = 9.68 cos 17.93◦ = 9.21 kN


TCD sin γ = 5 − TCB sin β = 2.02

Then

γ = tan−1 2.029.21

= 12.37◦

and

TCD = 9.21cos γ

= 9.431 kN

The sag of D below A is equal to 4.22 − 5 tan γ = 3.13 m.


RE,H = RA,H = 9.21 kN

and resolving vertically

RE,V = 12 − RA,V = 6.02 kN

Then

TED =√

6.022 + 9.212 = 11.0 kN.

S.5.4 Consider half the cable as shown in Fig. S.5.4, it is immaterial which half.

40 m

D

B

HC

10 kN/m

RB,H

RB,V

FIGURE S.5.4


H × D − 10 × 402

2= 0

i.e.

H = 8000D



RB,H − H = 0

i.e.

RB,H = 8000D


RB,V = 10 × 40 = 4000 kN

Now

Tmax =√

R2B,V + R2

B,H =√

4002 +(

8000D

)2

But

Tmax = 1000 kN =√

4002 +(

8000D

)2

which gives

D = 8.73 m.

S.5.5 Consider half the cable shown in Fig. S.5.5; the cable is symmetrical about themid-span point.

H C D

B

d

10 kN

RB,H

RB,V

17 m 34 m

36 N/m

3 m

FIGURE S.5.5


RB,V = 0.036 × 51 + 10 = 11.84 kN


RB,H × 3 − 11.84 × 51 + 10 × 17 + 0.036 × 512

2= 0

which gives

RB,H = 129.0 kN

Then

Tmax =√

11.842 + 129.02 = 129.5 kN


Now taking moments about D

11.84 × 34 − 129.0 d − 0.036 × 342

2= 0

i.e.

d = 2.96 m.

S.5.6 The cable is symmetrical about the mid-span point. Then, referring to Fig. S.5.6and taking moments about B

H × 8 − 24 × 402

2= 0

24 kN/m

8 m

B

H

40 m

RB,V

RB,H

Tmax

a

C

FIGURE S.5.6

i.e.

H = 2400 kN = RB,H


RB,V = 24 × 40 = 960 kN

The angle the suspension cable makes with the horizontal at the top of each tower isgiven by

tan α = 9602400

= 0.4

i.e.

α = 21.8◦

Since the cable passes over frictionless pulleys and there is no bending moment at thebase of a tower the anchor cable must be inclined at the same angle to the horizontalas the suspension cable, i.e. 21.8◦.

Also

Tmax =√

24002 + 9602 = 2584.9 kN


The vertical force on a tower is caused by the tension in the suspension cable and thetension in the anchor cable, i.e.

Vertical force = 2 × 2584.9 sin 21.8◦ = 1919.9 kN.

S.5.7 The suspension cable is shown in Fig. S.5.7. Since the towers are of differ-ent height the lowest point in the cable will not be at mid-span. The position of thelowest point may be found by either using Eq. (5.17) directly or by working from firstprinciples. We shall adopt the latter approach.

A

C

B

25 kN/m

120 m

2.5 m

7.5 m

H

L1 L2

FIGURE S.5.7

Taking moments about B for CB

H × 10 = 25L22

2

i.e.

H = 5L22

4(i)

Now taking moments about A for CA

H × 7.5 = 25L21

2

i.e.

H = 5L21

3(ii)

Equating Eqs (i) and (ii) gives

L1 = 0.866L2

Then, since L1 + L2 = 120 m,

L2 = 64.31 m

From Eq. (i)

H = 5169.7 kN


H is equal to the horizontal component of the maximum tension in the suspensioncable which must occur at B since the span CB is greater than the span CA. Thevertical component of the maximum tension is equal to 25 × 64.31 = 1607.8 kN. Then

Tmax =√

5169.72 + 1607.82 = 5413.9 kN

We must now investigate the tension in the anchor cable since this will be differentto the tension in the suspension cable. There is no resultant horizontal force on thetop of a tower since the saddles are on rollers. Therefore

TAC cos 55◦ = H = 5169.7 kNi.e.

TAC = 9013.1 kN

The maximum tension in the cable is therefore 9013.1 kN and the maximum stress is

σmax = 9013.1 × 103(πD2

4

) = 240

Then

D = 218.7 mm

The vertical load on the tallest tower = 1607.8 + 9013.1 sin 55◦ = 8990.9 kN.

S.5.8 The central half of the cable has its cross-sectional area reduced to0.8 × 0.08 = 0.064 m2. Considering one half of the cable as shown in Fig. S.5.8 thecritical points are C where the tension is a maximum and B where the cross-sectionalarea is reduced.

A

B

C

40 m

Tmax

TB

100 m100 m

H

w kN/m

a

FIGURE S.5.8

Consider the portion AC and suppose that the load intensity is w kN/m. Takingmoments about C

H × 40 − w × 2002

2= 0

i.e.

H = 500 w


H is equal to the horizontal component of the maximum tension in the cable at C. Thevertical component is equal to 200w. Therefore

Tmax =√

(200w)2 + (500w)2 = 538.5w

The maximum allowable stress in the cable is 500 N/mm2. Therefore

538.5w = 500 × 103 × 0.08

which gives

w = 74.3 kN/m

The horizontal component of the tension in the cable at B is equal to H , the verticalcomponent is equal to 100w. Then

TB =√

(100w)2 + (500w)2 = 509.9w

Therefore

509.9w = 500 × 103 × 0.064

from which

w = 62.8 kN/m

The maximum allowable value of w is therefore 62.8 kN/m.

At the top of the towers the inclination of the cable to the horizontal is given by

α = tan−1 200w500w

= 21.8◦.

S.5.9 The suspension bridge is supported by two cables; the load/cable is therefore12.5 kN/m.

(a) Consider the right hand half of the cable shown in Fig. S.5.9.

12.5 kN/m

25 m

B

H

125 m

RB,V

RB,H

C

FIGURE S.5.9



H × 25 − 12.5 × 1252

2= 0

i.e.

H = 3906.25 kNResolving horizontally

RB,H = H = 3906.25 kNResolving vertically

RB,V = 12.5 × 125 = 1562.5 kNThen

Tmax =√

(3906.25)2 + (1562.5)2 = 4207.16 kN

Therefore the required area of cross-section of each cable is

A = 4207.16 × 103

800= 5259.0 mm2.

(b) (i) The load in the anchor cable is equal to 4207.16 kN and the overturning force isgiven by

Overturning force = RB,H − 4207.16 cos 45◦ = 3906.25 − 2974.9 = 931.3 kN.

(ii) The horizontal components of the maximum tension in the cable and the tensionin the anchor cable are equal and opposite since the cable passes over a saddleresting on rollers, i.e.

TAC cos 45◦ = 3906.25

so that

TAC = 5524.3 kN

There is zero overturning force on the tower.

S.5.10 As in P.5.7 we need, initially, to determine the position of the lowest point ofthe cable.

For CB, taking moments about B

H × 16 = 5L22

2

i.e.

H = 5L22

32(i)

Similarly, for CA and taking moments about A

H = 5L21

24(ii)


A

C

B

5 kN/m

80 m

12 m

16 m

H

L1 L2

RB,V

RB,H

FIGURE S.5.10

Equating Eqs (i) and (ii) gives

L1 = 0.866L2

Therefore, since L1 + L2 = 80,

L2 = 42.87 m

Then, from Eq. (i)

H = 287.16 kN = RB,H


RB,V = 5 × 42.87 = 214.35 kN

The maximum tension in the cable is given by

Tmax =√

(287.16)2 + (214.35)2 = 358.3 kN

The horizontal component of the tension in the anchor cable is equal to the horizontalcomponent of the maximum tension in the suspension cable since the cable passesover a saddle on rollers. Therefore

TAC cos 45◦ = 287.16

i.e.

TAC = 406.1 kN

The vertical thrust on the tallest tower is equal to the sum of the vertical componentsof the maximum tension in the suspension cable and the tension in the anchor cable,i.e.

Vertical thrust = 214.35 + 287.16 = 501.5 kN.



S.6.1 Referring to Fig. S.6.1 and taking moments about B

y10m

x

B

C

ARA,H

RA,V RB,V

RB,H

20 kN/m

FIGURE S.6.1

RA,V × 20 − 20 × 10 × 15 = 0

i.e.

RA,V = 150 kN

Now taking moments about C

RA,H × 10 − RA,V × 10 + 20 × 102

2= 0

i.e.

RA,H = 50 kN

With the origin of axes at the centre of the arch the equation of the arch is

x2 + y2 = 102

Therefore, when x = −5 m, y = √102 − 52 = 8.66 m.

The bending moment at this point is given by

M = RA,V × 5 − RA,H × 8.66 − 20 × 52

2= 150 × 5 − 50 × 8.66 − 250 = 67 kN m.

S.6.2 For the origin of axes shown in Fig. S.6.2 the equation of the arch is

x2 + y2 = 122

Then, when y = 4 m, x = 11.31 m and when y = 6 m, x = 10.39 m.


3 m 2 m

20 kND

A

RA,V

RB,V

RB,H

RA,H

8 m6 m

12m

y

x

C

f B

FIGURE S.6.2


RA,V × 21.7 − RA,H × 2 − 20 × 8.39 = 0

i.e.

10.85RA,V − RA,H − 83.9 = 0 (i)


RA,V × 11.31 − RA,H × 8 = 0

i.e.

1.41RA,V − RA,H = 0 (ii)

Eq. (i) − Eq. (ii)

9.44RA,V − 83.9 = 0

i.e.

RA,V = 8.89 kN

From Eq. (ii)

RA,H = 12.53 kN

The radius at D makes an angle with the vertical given by

φ = sin−1 312

= 14.48◦

Then

NF = 12.53 cos φ + 8.89 sin φ = 14.35 kN (compression)

and

SF = 8.89 cos φ − 12.53 sin φ = 5.48 kN

The vertical distance of D above A = (√

122 − 32) − 4 = 7.62 m. Then the bendingmoment at D is given by

BM = 8.89 × 8.31 − 12.53 × 7.62 = −21.6 kN m (hogging).


S.6.3 Take the origin of axes at the crown C of the arch as shown in Fig. S.6.3; theequation of the arch is then

y = k x2

When x = −7 m, y = 3 m so that k = 3/72 = 0.0612.

A

10 m 7 m

y

x

D

RB,V

RB,H

RA,V

RA,H

3 m

4 m

40 kN/m

B

C

FIGURE S.6.3

Therefore, when x = 10 m, y = 6.12 m (at A). Also, when x = −3 m, y = 0.55 m (D).

Taking moments about C for AC

RA,V × 10 − RA,H × 6.12 = 0

i.e.

RA,V − 0.612RA,H = 0 (i)

Now taking moments about B

RA,V × 17 − RA,H × 3.12 − 40 × 72

2= 0

i.e.

RA,V − 0.184RA,H − 57.65 = 0 (ii)

Then, Eq. (i) − Eq. (ii) gives −0.428RA,H + 57.65 = 0

i.e.

RA,H = 134.7 kN

From Eq. (i)

RA,V = 82.4 kN


The bending moment at D is then given by

MD = 82.4 × 13 − 134.7 × 5.57 − 40 × 32

2= 140.9 kN m (sagging).

S.6.4 For the parabolic part of the arch take the origin of axes at C as shown in Fig.S.6.4; the equation of this part of the arch is then y = kx2. When x = 9 m, y = 5 m sothat k = 0.0617. Then at D where x = 5 m, y = 1.54 m and the vertical height of Dabove A is 5 − 1.54 = 3.46 m.

C

y

B

RB,H

RB,V

RA,H

RA,V

A

D

x

4 m

9 m 3 m

5 m

18 kN/m30 kN/m

a

FIGURE S.6.4


RA,V × 9 − RA,H × 5 − 30 × 92

2= 0

i.e.

1.8RA,V − RA,H − 243 = 0 (i)


RA,V × 12 − 30 × 9 × 7.5 − 18 × 32

2= 0

i.e.

RA,V = 175.5 kN

Substituting in Eq. (i)

RA,H = 72.9 kN


The slope of the arch at D is dy/dx = 0.1234x. Therefore when x = 5 m, the slope =0.617 = tan α, i.e. α = 31.67◦. Then, at D

NF = 175.5 sin α + 72.9 cos α − 30 × 4 sin α = 91.2 kN (compression)

SF = 175.5 cos α − 72.9 sin α − 30 × 4 cos α = 9.0 kN

BM = 175.5 × 4 − 72.9 × 3.46 − 30 × 42

2= 209.8 kN m.

S.6.5 Suppose that the vertical and horizontal support reactions at B are RB,V andRB,H, respectively. Then, taking moments about A

RB,V × 12 − 10 × 7.5 − 10 × 9 − 10 × 10.5 = 0i.e.

RB,V = 22.5 kN


RB,H × 3 − 22.5 × 6 + 10 × 1.5 + 10 × 3 + 10 × 4.5 = 0

i.e.

RB,H = 15 kN

Normal force:

In BD

NF = 22.5 cos 45◦ + 15 cos 45◦ = 26.5 kN (compression)

In DE

NF = 26.5 − 10 cos 45◦ = 19.4 kN (compression)

In EFC

NF = 15 kN (compression)

Shear force:

In BD

SF = 22.5 sin 45◦ − 15 sin 45◦ = 5.3 kN

In DE

SF = 5.3 − 10 sin 45◦ = −1.77 kN

In EF

SF = 22.5 − 10 − 10 = 2.5 kN


In FC

SF = 2.5 − 10 = −7.5 kN

Bending moment:

At B

BM = 0

At D

BM = 22.5 × 1.5 − 15 × 1.5 = 11.25 kN m (sagging)

At E

BM = 22.5 × 3 − 15 × 3 − 10 × 1.5 = 7.5 kN m (sagging)

At F

BM = 22.5 × 4.5 − 15 × 3 − 10 × 3 − 10 × 1.5 = 11.25 kN m (sagging)

At C

BM = 0

All distributions are linear.

S.6.6 Referring to Fig. S.6.6 the calculated dimensions are as shown. Taking momentsabout D

RA,H

RD,V

RD,H

RA,V

A

B

C

D

5 kN

15 kN

10 kN

0.75 m

0.75 m

1.3 m

1.3 m

1.5 m

60°

30°

FIGURE S.6.6

RA,V × 2.05 − RA,H × 3.55 + 5 × 0.75 + 15 × 1.5 = 0


i.e.

RA,V − 1.73RA,H + 12.8 = 0 (i)


RA,V × 1.3 − RA,H × 0.75 = 0

i.e.

RA,V − 0.58RA,H = 0 (ii)

Then, Eq. (i) − Eq. (ii) gives

RA,H = 11.13 kN

Then, from Eq. (ii)

RA,V = 6.46 kN

Now resolving horizontally

RD,H = 11.13 − 15 = −3.87 kN (acting to the right)

and resolving vertically

RD,V = 6.46 + 5 + 10 = 21.46 kN

The bending moment at C is given by

MC = 3.87 × 1.5 = 5.81 kN m

Also the bending moment at D is zero. The bending moment varies linearly in DCand is drawn on the left hand side of the member.


S.7.1 In this problem there are two limiting criteria, one of stress and one of change inlength; we shall consider the maximum allowable stress criterion first. From Eq. (7.1)

σ(max) = 150 = 5000 × 103[(π4

)(3002 − d2)

] (i)

where d is the internal diameter of the column. Solving Eq. (i) gives

d = 218.1 mm


The shortening of the column must not exceed 2 mm. Therefore the strain in thecolumn, from Eq. (7.4), is limited to 2/(3 × 103) = 0.00067. Then, from Eqs (7.1)and (7.7)

0.00067 = 5000 × 103[200 000 × (

π4

)(3002 − d2)

] (ii)

Solving Eq. (ii) gives

d = 205.6 mm

The maximum allowable internal diameter of the column is therefore 205.6 mm.

S.7.2 The strain in the girder is given by Eq. (7.4), i.e.

ε = L − L0

L0= αT = 0.00005 × 30 = 0.0015

Therefore, from Eq. (7.7), the stress is given by

σ = 0.0015 × 180 000 = 270 N/mm2.

S.7.3 From Eq. (7.1)

σ = 150 = 10 000 × 103(πD2

4

)from which the required diameter of the column is D = 291.3 mm.

The shortening, δ, of the column is then, from Eqs (7.7) and (7.4), given by

δ = 150200 000

× 3 × 103 = 2.25 mm

From Eq. (7.12) the lateral strain is 0.3 × (150/200 000) = 0.000225. Then the increasein diameter is 0.000225 × 291.3 = 0.066 mm.

S.7.4 The temperature rise required to produce an extension of 1.5 mm is given by

T = L − L0

αL0= 1.5

0.000012 × 2 × 103 = 62.5◦

The effective strain in the member when it cools is αT = 0.000012 × 62.5 = 0.00075(or 1.5/2 × 103). The corresponding stress is then, from Eq. (7.7)

σ = 200 000 × 0.00075 = 150 N/mm2

Suppose that the section is of side a and b and suppose that the longitudinal strainis ε. The lateral strain is then, from Section 7.8, νε = 0.3ε. The sides of the section


are therefore reduced in length by 0.3εa and 0.3εb. The percentage change in cross-sectional area is therefore given by

%change = [ab − (a − 0.3εa)(b − 0.3εb)] × 100ab

This reduces to%change = 2 × 0.3εab × 100

ab= 2 × 0.3ε × 100

since ε2 is negligibly small. Then

%change = 2 × 0.3 × 0.00075 × 100 = 0.045%.

S.7.5 The required area of cross section is, from Eq. (7.1), given by

A = 100 × 103

155= 645.2 mm2

Reference to steel tables shows that two equal angles, 50 × 50 × 5 mm, each having an18 mm diameter bolt hole, have sufficient area of cross section.

S.7.6 The weight of the cable is = (π × 7.52/4) × 25 × 103 × 7850 × 9.81/109 = 85.05 Nand the total load on the cable is therefore = 85.05 + 5 × 103 = 5085.05 N.

The tensile stress in the cable at its point of support is then, from Eq. (7.1)

σ = 5085.05(π × 7.52

4

) = 115.1 N/mm2

Consider the cable shown in Fig. S.7.6. The weight of a length h is equal to ρAh whereρ is the density of the cable and A its area of cross section. The stress in the cable atthe section at the top of the length h is then ρh from Eq. (7.1).

h

dh L

FIGURE S.7.6

The extension of the elemental length δh due to the self-weight of the cable is then,from Eqs (7.4) and (7.7), given by

ext =(

ρhE

)δh


The extension of the complete cable due to self-weight is therefore

ext =∫ L

0

(ρhE

)dh = ρL2

2E= 7850 × 9.81(25 × 103)2

2 × 200 000 × 109 = 0.12 mm

The extension due to the load is, from Eq. (7.28), given by

ext (load) = 5 × 103 × 25 × 103(π × 7.52

4

)× 200 000

= 14.15 mm

The total extension is then 14.15 + 0.12 = 14.27 mm.

S.7.7 The concentrated loads applied to the chimney have previously been calculatedin S.3.1 and are 36.0 kN at a height of 15 m, 46.8 kN at a height of 25 m and 52.1 kN ata height of 35 m. The self-weight of the chimney is

SW = 40 × 0.15 × 2500 × 9.81 × 10−3 = 147.15 kN

The total force on the chimney base is then 147.15 + 36.0 + 46.8 + 52.1 = 282.05 kN

The maximum stress is then, from Eq. (7.1)

σ = 282.05 × 103

0.15 × 106 = 1.9 N/mm2

From S.7.6 the shortening due to the self-weight of the chimney is given by

shortening (SW) = 2500 × 9.81(40 × 103)2

2 × 20 000 × 109 = 0.98 mm

From Eq. (7.28) the shortening due to the loads is

shortening (loads) = (52.1 × 35 + 46.8 × 25 + 36 × 15) × 106

0.15 × 106 × 20 000= 1.18 mm

total shortening = 0.98 + 1.18 = 2.16 mm.

S.7.8 At any section a distance x from the top of the column the cross-sectional areais given by

A = a[

b2 + x(b1 − b2)h

]

Then, from Eq. (7.28), the shortening of an element δx is

δ� = Pδ xAE


The total shortening of the column is then

� =∫ h

0

PdxAE

Substituting for A

� =(

PaE

)∫ h

0

dx[b2 + x(b1−b2)

h

]i.e.

� =(

PaE

)[h

(b1 − b2)

]loge

[b2 + x(b1 − b2)

h

]h

0

from which

� =[

PhaE(b1 − b2)

]loge

(b1

b2

).

S.7.9 Assume that all the members are in tension. Then, using the method of joints(Section 4.6):

Joint C


CB cos 30◦ + 20 = 0i.e.

CB = −23.1 kNResolving horizontally

CD + CB cos 60◦ = 0i.e.

CD = 11.6 kN

Joint D

Resolving perpendicularly to DB

DA cos 30◦ − DC cos 30◦ = 0

i.e.

DA = DC = 11.6 kN

Resolving parallel to DB

DB + DC cos 60◦ + DA cos 60◦ = 0

i.e.

DB = −11.6 kN


Joint B


BA + BD cos 60◦ − BC cos 60◦ = 0

i.e.

BA = −5.1 kN

From Eqs (7.27) and (7.29)

20 × 103�

2=(

3 × 103

2 × 205 000

)[(23.12 + 11.62 + 5.12)

200+ (2 × 11.62)

100

]× 106

from which

� = 4.5 mm

Note that the negative signs indicating compression disappear in the above equation.

S.7.10 Assuming all members are in tension and, using the method of joints (note thatthe truss member forces, except CB and DA, may be obtained by inspection)

CB = −141.4 kN, CD = 100 kN, BA = −100 kN, BD = 100 kN,

DA = −141.4 kN , DE = 200 kN

Then, from Eqs (7.27) and (7.29)

100 × 103�

2= (141.42 × 2

√2 + 1002 × 2 + 1002 × 2 + 1002 × 2 + 141.42 × 2

√2

+ 2002 × 2) × 109

2 × 1200 × 205 000

i.e.

� = 10.3 mm.

S.7.11 Suppose that the loads in the bars are P1, P2 and P3. For vertical equilibriumof the system

P1 + P2 + P3 = P (i)

Taking moments about, say, bar 1

P2a + P32a − P3a2

= 0

i.e.

P2 + 2P3 = 3P2

(ii)


From Eq. (7.28) the extensions of the bars are P1L/AE, P2L/AE and P3L/AE where Ais the cross-sectional area of each bar and E is Young’s modulus. The displaced shapeof the system is shown in Fig. S.7.11.

a

P1 P2

P

P3

a2

a2

FIGURE S.7.11

Then, since the term L/AE is the same for each bar, the extension of each bar is directlyproportional to the load in the bar. The geometry of Fig. S.7.11 gives

P2 − P1

P3 − P1= 1

2(iii)

Rearranging Eq. (iii)

2P2 − P1 − P3 = 0 (iv)

Adding Eqs (i) and (iv)

P2 = P3

Substituting for P2 in Eq. (ii) gives

P3 = 7P12

Finally, from Eq. (i)

P1 = P12

.

S.7.12 Eqs (7.38) may be used directly to determine the stresses in the steel bar andalloy cylinder. The areas of cross section are

Ast = π × 202

4= 314.16 mm2, Aall = π(252 − 202)

4= 176.71 mm2

Then

σst = 50 × 103 × 200 000(314.16 × 200 000 + 176.71 × 70 000)

i.e.

σsteel = 132.9 N/mm2


Similarly

σall = 50 × 103 × 70 000(314.16 × 200 000 + 176.71 × 70 000)

i.e.

σalloy = 46.5 N/mm2

From Eq. (7.37) the shortening of the column is

δ = 50 × 103 × 200(314.16 × 200 000 + 176.71 × 70 000)

i.e.

δ = 0.13 mm

From Eq. (7.30) the strain energy stored in the column is

U =200

[(132.92×314.16

200 000

)+(

46.52×176.7170 000

)]2

i.e.

U = 3320.3 Nmm = 3.3 Nm.

S.7.13 This problem is very similar to P.7.12 and so the same equations may be useddirectly to obtain a solution. From Eqs (7.38)

σtim = 1000 × 103 × 15 000[(100 × 200)15 000 + (2 × 200 × 10)200 000]

i.e.

σtimber = 13.6 N/mm2

The allowable stress in the timber is 55/3 = 18.3 N/mm2 so that the timber size issatisfactory. Also

σst = 1000 × 103 × 200 000[(100 × 200)15 000 + (2 × 200 × 10)200 000]

i.e.

σsteel = 181.8 N/mm2

The allowable stress in the steel is 380/2 = 190 N/mm2 so that the size of the steel platesis satisfactory.


S.7.14 If the steel portion of the bar were disconnected from the aluminium part theseparate parts would take up the positions shown in Fig. S.7.14 where

δS = αSTL1, δA = αATL2 (i)

L1

dS

d

dA

AlSteel

L2

FIGURE S.7.14

Suppose that the connected parts of the bar take up the position shown so that thejunction of the two parts has suffered a displacement δ. Then

extension of steel = δS − δ

extension of aluminium = δA + δ

These extensions produce tensile stresses in the steel and aluminium which must beequal since their areas of cross section are the same. Therefore

σ = ES(δS − δ)L1

= EA(δA + δ)L2

(ii)

Rearranging Eq. (ii) gives

δ =(

ESδSL1

)−(

EAδAL2

)(

ESL1

)+(

EAL2

) (iii)

Now substituting for δ in the first of Eqs (ii) (or the second) and also for δS and δA

from Eq. (i) and rearranging gives

σ = T(αSL1 + αAL2)(L1ES

+ L2EA

) .

S.7.15 The cross-sectional areas of the steel tube and copper bar are, respectively,

Ast = π(362 − 302)4

= 311.02 mm2, Ac = π × 252

4= 490.9 mm2

Then, from Eqs (7.49)

σst = 80(0.000006 − 0.00001) × 490.9 × 100 000 × 200 000(490.9 × 100 000 + 311.02 × 200 000)


i.e.σsteel = −28.3 N/mm2

The stress in this case is negative so that, unlike the steel reinforcement in the concretecolumn of Fig. 7.28, the steel is in tension. This is logical since the coefficient of linearexpansion of copper is greater than that of steel. Again, from Eqs (7.49)

σc = 80(0.000006 − 0.00001) × 311.02 × 100 000 × 200 000(490.9 × 100 000 + 311.02 × 200 000)

i.e.σcopper = −17.9 N/mm2

The concrete in the column of Fig. 7.28 is in tension; here the negative answer indicatesthat the copper tube is in compression; again an expected result.

S.7.16 The first part of this question is identical in form to P.7.13. Therefore, we cansubstitute areas of cross section, etc. directly into Eqs (7.38).

Ast = π × 752

4= 4417.9 mm2, Aal = π(1002 − 752)

4= 3436.1 mm2

From Eqs (7.38)

σst = 106 × 200 000(4417.9 × 200 000 + 3436.1 × 80 000)

i.e.

σsteel = 172.6 N/mm2 (compression)

σal = 106 × 80 000(4417.9 × 200 000 + 3436.1 × 80 000)

i.e.

σaluminium = 69.1 N/mm2 (compression)

Due to the decrease in temperature in which no change in length is allowed the strainin the steel is αstT and the strain in the aluminium is αalT . Therefore, due to thedecrease in temperature

σst = Est αstT = 200 000 × 0.000012 × 150 = 360.0 N/mm2 (tension)

σal = Eal αalT = 80 000 × 0.000005 × 150 = 60.0 N/mm2 (tension)

The final stresses in the steel and aluminium are then

σsteel (total) = 360.0 − 172.6 = 187.4 N/mm2 (tension)

σaluminium (total) = 60.0 − 69.1 = −9.1 N/mm2 (compression).


S.7.17 The cross-sectional areas of the bolt and sleeve are

AB = π × 152

4= 176.7 mm2, AS = π(302 − 202)

4= 392.7 mm2

From equilibrium the compressive load in the sleeve must be equal to the tensile loadin the bolt, therefore the stresses in the bolt and sleeve due to the tightening of thenut are

σB = 10 × 103

176.7= 56.6 N/mm2 (tension)

σS = 10 × 103

392.7= 25.5 N/mm2 (compression)

When the external tensile load is applied the tensile stress in the bolt will be increasedwhile the compressive stress in the sleeve will be reduced. Equations (7.38) applyin which Young’s modulus is the same for the bolt and sleeve so that, due to the5 kN load

σB = 5 × 103

(176.7 + 392.7)= 8.8 N/mm2 (tension) = σS

The final stresses are then

σB = 56.6 + 8.8 = 65.4 N/mm2 (tension)

σS = 25.5 − 8.8 = 16.7 N/mm2 (compression).

S.7.18 The pressure of the water in the pipe is given by

p = 120 × 1000 × 9.81 × 10−3 = 1177.2 kN/m2

Therefore, from Eq. (7.63), the minimum wall thickness is given by

tmin = 1177.2 × 103 × 1 × 103

2 × 20 × 106 = 29.4 mm.

S.7.19 From Eq. (7.68) the required shell thickness is given by

0.8 treq = 0.75 × 3 × 103

4 × 80

i.e.

treq = 8.8 mm.



S.8.2 From Eq. (7.8) Young’s modulus E is equal to the slope of the stress–straincurve. Then, since stress = load/area and strain = extension/original length.

E = slope of the load-extension curve multiplied by (original length/area of crosssection).

From the results given the slope of the load-extension curve 402.6 kN/mm. Then

E 402.6 × 103 × 250(π×252

4

) 205 000 N/mm2

From Eq. (11.4) the modulus of rigidity is given by

G = TLθJ

Therefore the slope of the torque-angle of twist (in radians) graph multiplied by (L/J)is equal to G. From the results given the slope of the torque-angle of twist graph is12.38 kNm/rad. Therefore

G 12.38 × 106 × 250(π × 254

32

) 80 700 N/mm2

Having obtained E and G the value of Poisson’s ratio may be found from Eq. (7.21), i.e.

ν =(

E2G

)− 1 0.27

Finally, the bulk modulus K may be found using either of Eqs (7.22) or (7.23). FromEq. (7.22)

K E3(1 − 2ν)

148 500 N/mm2.

S.8.3 Suppose that the actual area of cross section of the material is A and that theoriginal area of cross section is Ao. Then, since the volume of the material does notchange during plastic deformation

AL = AoLo

where L and Lo are the actual and original lengths of the material respectively. Thestrain in the material is given by

ε = L − Lo

Lo= Ao

A− 1 (i)

from the above. Suppose that the material is subjected to an applied load P. Theactual stress is then given by σ = P/A while the nominal stress is given by σnom = P/Ao.Therefore, substituting in Eq. (i) for A/Ao

ε = σ

σnom− 1


Then

σnom(1 + ε) = σ = Cεn

or

σnom = Cε n

1 + ε(ii)

Differentiating Eq. (ii) with respect to ε and equating to zero gives

dσnom

dε= nC(1 + ε)ε n−1 − Cε n

(1 + ε)2 = 0

i.e.

n(1 + ε)ε n−1 − εn = 0

Rearranging gives

ε = n(1 − n)

.

S.8.4 Substituting in Eq. (8.1) from Table P.8.4

104

5 × 104 + 105

106 + 106

24 × 107 + 107

12 × 107 = 0.39 < 1

Therefore fatigue failure is not probable.


S.9.1 From symmetry the support reactions are equal and are each 5w kN. At eachsupport the bending moment is −w × 22/2 = −2w kN m. At mid-span the bendingmoment is 5w × 3 − w × 52/2 = 2.5w kN m which is therefore the maximum. Using themethod of Section 9.6

Iz = 200 × 3403

12− 185 × 3003

12= 2.39 × 108 mm4.

Then, substituting in Eq. (9.9)

150 = 2.5w × 106 × 1702.39 × 108

from which

w = 84.3 kN/m.

S.9.2 The maximum bending moment occurs at the built in end of the cantilever and isequal to 13 300 × 2.5 = 33 250 Nm. The second moment of area of the beam section is


230 × 3003/12 = 5.175 × 108 mm4. The maximum direct stress due to bending is then,from Eq. (9.9), given by

σmax = 33 250 × 103 × 1505.175 × 108 = 9.6 N/mm2.

S.9.3 The arrangement of the floor and joists is shown in Fig. S.9.3.

d m

d m

4 m

FIGURE S.9.3

If the joists are spaced a distance d m apart then each joist carries a uniformly dis-tributed load of intensity 16d kN/m. Since each joist is a simply supported beamthe maximum bending moment is, from Ex. 3.7, equal to 16d × 42/8 = 32d kN m.The second moment of area of each joist is, from Section 9.6, 110 × 3003/12 =2.475 × 108 mm4. Then, from Eq. (9.9)

7 = 32d × 106 × 1502.475 × 108

from which

d = 0.36 m.

S.9.4 The mast is shown in Fig. S.9.4.

At any section a distance h m from the top of the mast the diameter is given by

d = 100 +(

h15

)× 150

i.e.

d = 100 + 10h mm

The maximum direct stress due to bending at this section is, from Eq. (9.9)

σmax =Ph × 103 ×

(d2

)(

πd4

64

) = Ph × 103 × 32πd3 (i)


100 mm

h

15 m

250 mm FIGURE S.9.4

This maximum direct stress will be greatest when dσmax/dh = 0. Therefore, substitutingfor d in Eq. (i) and differentiating

0 =(

32 × 103Pπ

)(100 + 10h)3 − 3h(100 + 10h)2 × 10

(100 + 10h)6

i.e.

100 + 10h = 30h

which gives

h = 5 m

i.e. the mast will break at a distance of 5 m from the top.

The diameter at this section is then 150 mm and, from Eq. (9.9)

35 = P × 5 × 106 × 75(π × 1504

64

)from which

P = 2320 N.

S.9.5 With the two loads symmetrically positioned on the beam the support reactionsare equal and the maximum bending moment is equal to 20 × 1.5 = 30 kN m and isconstant between the loads. From Eq. (9.13) the required section modulus is given by

Z = 30 × 106

155= 193548.4 mm3.


A Universal Beam, 254 mm × 102 mm × 22 kg/m has a section modulus of 225.4 cm3,the least available. Now check to allow for its self-weight. The self-weight ofthe beam is equivalent to a uniformly distributed load along its complete length.From Ex. 3.7, the maximum bending moment due to self-weight is then equal to22 × 9.81 × 10−3 × 52/8 = 0.67 kN m. The total maximum bending moment is then30.0 + 0.67 = 30.67 kN m and the required section modulus is

Z = 30.67 × 106

155= 197871.0 mm3

The proposed Universal Beam is therefore adequate.

S.9.6 The area of cross section of the column is given by

A = 200 × 300 − 175 × 260 = 14 500 mm2

and its second moment of area about its z axis is, from Section 9.6

Iz = 200 × 3003

12− 175 × 2603

12= 1.94 × 108 mm4

The maximum compressive stress will occur at the outside of the flange where thecompressive stress due to the axial effect of P and the maximum compressive stressdue to the bending action of P coincide. Then, from Eq. (9.15) in which ey = 140 mmand ez = 0

−150 = −P14 500

+ −P × 1401.94 × 108 × 150

from which

P = 846.4 kN.

S.9.7 The cross-sectional area of the chimney is

A = 2 × 2 − 1.5 × 1.5 = 1.75 m2

and its second moment of area about the axis of bending is

I = 2 × 23

12− 1.5 × 1.53

12= 0.552 m4

The weight of the chimney is

W = 1.75 × 15 × 2000 × 9.81 × 10−3 = 515.03 kN

The wind pressure is equivalent to a uniformly distributed load through the height ofthe chimney of intensity

w = 750 × 2 × 10−3 = 1.5 kN/m

The maximum bending moment occurs at the base of the chimney and is

Mmax = 1.5 × 152

2= 168.75 kN m


The direct stress at A is then, from Eqs (7.1) and (9.9)

σA = −515.03 × 103

1.75 × 106 + 168.75 × 106 × 1 × 103

0.552 × 1012

i.e.

σA = −0.29 + 0.31 = 0.02 N/mm2 (tension)

and the direct stress at B is

σB = −0.29 − 0.31 = −0.6 N/mm2 (compression).

S.9.8 The area of cross section of the section is given by

A = 150 × 10 + 10 × 200 = 3500 mm2

For this problem the position of the z axis must be found since this will be the axisof bending. Suppose that the z axis is a distance y from the top of the flange. Takingmoments of area about the top of the flange

3500 y = 150 × 10 × 5 + 10 × 200 × 110

from which

y = 65 mm

The second moment of area about the z axis is, from Section 9.6

Iz = 150 × 653

3− 150 × 553

3+ 10 × 1453

3= 15.6 × 106 mm4

Note that this calculation uses the result for the second moment of area of a rectangleabout an axis through its base. Alternatively, the parallel axis theorem could have beenapplied to the flange and the leg separately but this approach would have been slightlylonger.

The maximum bending moment occurs at the built in end of the cantilever and isgiven by

Mmax = −5 × 22

2= −10 kN m (hogging)

Then, at the top of the flange the direct stress due to bending is

σ (bending) = 10 × 106 × 6515.6 × 106 = 41.7 N/mm2 (tension)

At the bottom of the leg the bending stress is

σ (bending) = 10 × 106 × (−145)15.6 × 106 = −92.9 N/mm2 (compression)


In addition there is a uniform direct stress caused by the axial load, i.e.

σ (axial) = −100 × 103

3500= −28.6 N/mm2 (compression)

Clearly the maximum direct stress occurs at the bottom of the leg and is

σmax = −92.9 − 28.6 = −121.5 N/mm2 (compression).

S.9.9 The beam section is antisymmetrical so that the position of the centroid of areais obvious by inspection, i.e. at the centre of the web, but the section will have a valuefor the product second moment of area Izy so that the direct stress distribution is givenby Eq. (9.31).

Iz = 2

(15 × 53

12+ 15 × 5 × 12.52

)+ 5 × 203

12= 27083.3 mm4

Iy = 2

(5 × 153

12+ 5 × 15 × 52

)+ 20 × 53

12= 6770.8 mm4

Izy = 5 × 15(5)(12.5) + 5 × 15(−5)(−12.5) = 9375.0 mm4

The coordinates of point A are z = 12.5 mm, y = 15 mm. The applied bendingmoments are

My = 55 Nm, Mz = 0

Also the denominator in both terms in Eq. (9.31) is

IzIy − I2zy = 27 083.3 × 6770.8 − 9375.02 = 9.548 × 107

Substituting in Eq. (9.31)

σx = −(

55 × 103 × 27 083.39.548 × 107

)12.5 −

(−55 × 103 × 9375.0

9.548 × 107

)15

i.e.

σx = −114 N/mm2 (compression).

S.9.10 Referring to Fig. S.9.10 the components of the bending moment are

Mz = −12 cos 30◦ = −10.4 kN m, My = −12 sin 30◦ = −6 kN m

The beam section is unsymmetrical so that the position of the centroid of area must befound before the second moments of area can be calculated. Suppose that the centroidof area is a distance y from the top of the top flange and a distance z from the left handface of the vertical leg. The area of cross section is

A = 140 × 10 + 200 × 10 + 40 × 10 = 3800 mm2


FIGURE S.9.10

150 mm

30°

y

J

AB

z G

F E

DC

H

10 mm

10 mm

10 mm

50 mm

200 mm z

y

Taking moments of areas about the top of the top flange

3800 y = 140 × 10 × 5 + 200 × 10 × 100 + 40 × 10 × 195

i.e.

y = 75 mm

Now taking moments about the left hand face of the vertical web

3800 z = 140 × 10 × 80 + 200 × 10 × 5 + 40 × 10 × 30

from which

z = 35.3 mm

The second moments of area are then

Iz = 140 × 103

12+ 140 × 10 × 702 + 10 × 2003

12+ 10 × 200 × 252

+ 40 × 103

12+ 40 × 10 × 1202

i.e.

Iz = 20.6 × 106 mm4

Iy = 10 × 1403

12+ 140 × 10 × 44.72 + 200 × 103

12+ 200 × 10 × 30.32

+ 10 × 403

12+ 10 × 40 × 5.32


i.e.

Iy = 7.0 × 106 mm4

Izy = 140 × 10(−44.7)(70) + 200 × 10(30.3)(−25) + 40 × 10(5.3)(−120)

i.e.

Izy = −6.2 × 106 mm4

The denominator in Eq. (9.31) is

IzIy − I2zy = (20.6 × 7.0 − 6.22) × 1012 = 105.8 × 1012

Now substituting in Eq. (9.31) for Mz, My, etc.

σx = −(−6 × 20.6 − 10.4 × 6.2

105.8

)z −

(−10.4 × 7.0 − 6 × 6.2105.8

)y

i.e.

σx = 1.78z + 1.04y (i)

It is not clear from Eq. (i) at which point the maximum will occur so that the stress atdifferent points in the cross section must be calculated.

At J, z = −114.7 mm, y = 65 mm and σx = −136.6 N/mm2 (this will be greater than thestress at A).

At B, z = 35.3 mm, y = 75 mm and σx = 140.8 N/mm2 (this will be greater than at H).

At D, z = −14.7 mm, y = −125 mm and σx = −156.2 N/mm2 (this will be greater thanat C, F or E).

Therefore the maximum direct stress is −156.2 N/mm2 (compression) and it occursat D.

S.9.11 The area of cross section is

A = 1 × 0.15 + 0.6 × 0.3 = 0.33 m2

The self-weight of the beam is then

SW = 0.33 × 2000 × 9.81 × 10−3 = 6.5 kN/m

The maximum bending moment occurs at mid-span and is given by

Mmax = 6.5 × 102

8+ 25 × 10

4= 143.75 kN m = Mz

Referring to Fig. S.9.11


1 m

0.15 m

0.75 m

0.3 m

y

zz G

y

FIGURE S.9.11

0.33 × 106y = (1 × 0.15 × 0.075 + 0.6 × 0.3 × 0.45) × 109

i.e.

y = 279.5 mm

0.33 × 106z = (1 × 0.15 × 0.5 + 0.6 × 0.3 × 0.15) × 109

i.e.

z = 309.1 mm

The second moments of area are then

Iz = 1000 × 1503

12+ 1000 × 150 × 204.52 + 300 × 6003

12+ 300 × 600 × 170.52

i.e.

Iz = 1.72 × 1010 mm4

Iy = 150 × 10003

12+ 150 × 1000 × 190.92 + 600 × 3003

12+ 600 × 300 × 159.12

i.e.

Iy = 2.38 × 1010 mm4

Izy = 150 × 1000(−190.9)(204.5) + 600 × 300(159.1)(−170.5)

i.e.

Izy = −1.07 × 1010 mm4

The denominator in Eq. (9.31) is then

IzIy − I2zy = 1.72 × 1010 × 2.38 × 1010 − (1.07 × 1010)2 = 2.95 × 1020

The direct stress is then, from Eq. (9.31)

σx = −(

143.75 × 1.072.95

)× 10−4z −

(143.75 × 2.38

2.95

)× 10−4y


i.e.

σx = −0.0046z − 0.0116y (i)

The coordinates of A are z = 9.1 mm, y = −470.5 mm, therefore

σx(A) = 5.4 N/mm2 (tension).

S.9.12 The maximum bending moment occurs at mid-span and is given by

Mmax = 100 × 42

8= 200 kN m = Mz

A

500 mm

300 mm

z

y

y

B

z

G

F D

C

100 mm

40 mm40 mm

50 mm

50 mm

FIGURE S.9.12

The area of cross section is given by

A = 300 × 500 − 220 × 400 + 100 × 50 = 67 000 mm2

Then,

67 000 y = (500 × 300 − 220 × 400) × 250 + 100 × 50 × 25

i.e.

y = 233.2 mm

and

67 000 z = (500 × 300 − 220 × 400) × 150 + 100 × 50 × 350

i.e.

z = 164.9 mm

The second moments of area are

Iz =(

300 × 5003

12+ 300 × 500 × 16.82

)−(

220 × 4003

12+ 220 × 400 × 16.82

)

+ 100 × 503

12+ 100 × 50 × 208.22 = 2.19 × 109 mm4


Iy =(

500 × 3003

12+ 500 × 300 × 15.12

)−(

400 × 2203

12+ 400 × 220 × 15.12

)

+ 50 × 1003

12+ 50 × 100 × 185.12 = 0.96 × 109 mm4

Izy = (500 × 300 − 400 × 220)(15.1)(16.8) + 100 × 50(−185.1)(−208.2)

= 0.21 × 109 mm4

The denominator in Eq. (9.31) is then equal to (2.19 × 0.96 − 0.212) × 1018

= 2.06 × 1018 and the direct stress is given by

σx = −(

−200 × 106 × 0.21 × 109

2.06 × 1018

)z −

(200 × 106 × 0.96 × 109

2.06 × 1018

)y

i.e.

σx = 0.0204z − 0.0932y (i)

The maximum stress will occur at a point whose coordinates are of opposite sign.By inspection the critical point is B where z = −135.1 mm and y = 266.8 mm. Then

σmax = −27.6 N/mm2.

S.9.13 The bending moments half way along the beam are

Mz = 800 × 1000 = 800 000 N mm, My = 400 × 1000 = 400 000 N mm

40 mm

y2.0 mm

2.0 mm

Gz

80 mm

1.0 mm

100 mm

y

z

FIGURE S.9.13

By inspection the centroid of area is midway between the flanges. Its distance z fromthe vertical web is given by

(40 × 2 + 100 × 2 + 80 × 1) z = 40 × 2 × 20 + 80 × 1 × 40


i.e.

z = 13.33 mm

The second moments of area of the cross section are calculated using the approxima-tions for thin-walled sections described in Section 9.6. Then

Iz = 40 × 2 × 502 + 80 × 1 × 502 + 2 × 1003

12= 5.67 × 105 mm4

Iy = 100 × 2 × 13.332 + 2 × 403

12+ 2 × 40 × 6.672 + 1 × 803

12+ 1 × 80 × 26.672

= 1.49 × 105 mm4

Izy = 40 × 2(−6.67)(50) + 80 × 1(−26.67)(−50) = 0.8 × 105 mm4

The denominator in Eq. (9.31) is then (5.67 × 1.49 − 0.82) × 1010 = 7.81 × 1010.

From Eq. (9.31)

σ = −(

400 000 × 5.67 × 105 − 800 000 × 0.8 × 105

7.81 × 1010

)z

−(

800 000 × 1.49 × 105 − 400 000 × 0.8 × 105

7.81 × 1010

)y

i.e.

σ = −2.08z − 1.12y

and at the point A where z = −66.67 mm, y = −50 mm

σ (A) = 194.7 N/mm2.

S.9.14 The bending moments at the built in end are

Mz = PL, My = 2P(

L2

)= PL

The centroid of area is at the centre of the enclosed part of the section and, using theapproximations for thin-walled sections described in Section 9.6

Iz = 2td(

d2

)2

+ 2td3

12= 2td3

3

Iy = 2td(

d4

)2

+ 2

[td3

12+ td

(d4

)2]

= 5td3

12

Izy = td(

d4

)(d2

)+ td

(−d4

)(−d2

)= td3

4


The denominator in Eq. (9.31) is equal to (td3)2[(2/3)(5/12) − (1/4)2] = 0.215(td3)2.

Substituting in Eq. (9.31) for Mz, My, etc.

σx = −PL(1.94z + 0.78y)td3

Then at A where z = 3d/4 and y = d/2

σ (A) = −1.85PLtd2

and at B where z = −d/4, y = d/2

σ (B) = 0.1PLtd2 .

S.9.15 Referring to Fig. S.9.15

N

z

a

u

A

y

G

6.4 mm

50mm

FIGURE S.9.15

Iz = 2∫ π

06.4(50 − 50 cos θ)2 × 50 dθ = 7.54 × 106 mm4

Iy = 2∫ π

06.4(50 sin θ)2 × 50 dθ = 2.51 × 106 mm4

Izy = 2∫ π

06.4(50 sin θ)(50 − 50 cos θ) × 50 dθ = 3.2 × 106 mm4

Since My = 0, Eq. (9.33) reduces to

tan α = − Izy

Iy= − 3.2

2.51= −1.275


i.e.

α = −51.9◦

which means that the neutral axis slopes downwards from left to right and makes anangle of 51.9◦ with the z axis.

The denominator in Eq. (9.31) is equal to (7.54 × 2.51 − 3.22) × 1012 = 8.69 × 1012

and since the bending moment is Mz = 3.5 kN m the direct stress distribution is givenby

σx = 1.29z − 1.01y (i)

Inspection of Eq. (i) shows that the direct stress will be a maximum when z = 0 andwhen y = ±100 mm. Then

σx( max ) = ±101.0 N/mm2.

S o l u t i o n s t o C h a p t e r 1 0 P r o b l e m s

S.10.1 The area of cross section is given by

A = 10 × 60 + 40 × 10 = 1000 mm2

(a) (b)

10 mm

G

A�

A�10 mm

60 mm

1.68 N/mm26.72 N/mm2

7.74 N/mm2

40 mm

y

z

y

y

FIGURE S.10.1

Referring to Fig. S.10.1(a)

1000 y = 10 × 60 × 40 + 40 × 10 × 5

from which

y = 26 mm

Then

Iz = 10 × 603

12+ 10 × 60 × 142 + 40 × 103

12+ 40 × 10 × 212 = 4.77 × 105 mm4


First consider the web.

The area A′ = 10(44 − y), b0 = 10 mm and y = (44 + y)/2 (i.e. the distance to thecentroid of A′) so that, from Eq. (10.4), the shear stress distribution is given by

τ =4 × 103 × 10(44 − y)

(44 + y

2

)10 × 4.77 × 105 = 0.004(442 − y2)

When y = 44 mm, τ = 0 and when y = −16 mm, τ = 6.72 N/mm2. The maximum valueof shear stress will occur when y = 0, i.e. τmax = 7.74 N/mm2.

Now consider the flange.

The area A′ = 40[−26 − (−y)], b0 = 40 mm and y = [−26 + (−y)]/2. Then, from Eq.(10.4)

τ =4 × 103 × 40(26 − y)

(26 + y

2

)40 × 4.77 × 105 = 0.004(262 − y2)

When y = −26 mm, τ = 0 and when y = −16 mm, τ = 1.68 N/mm2


S.10.2 The problem is identical to Ex. 10.2 except that numerical dimensions are given.Therefore, from the dimensions given in Fig. P.10.2

Iz = 150 × 4003

12− 135 × 3603

12= 2.75 × 108 mm4

Then, from Eq. (10.8)

τ (flange) =80 × 103

(4002

4 − y2)

2 × 2.75 × 108 = 1.45 × 10−4(40000 − y2)

When y = ±200 mm, τ (flange) = 0 and when y = ±180 mm, τ (flange) = 1.1 N/mm2

The shear stress distribution in the web is given by Eq. (10.11) and is

τ (web) =80 × 103

[150(4002 − 3602)

8 × 15 +(

36024 − y2

)2

]

2.75 × 108

i.e.

τ (web) = 15.77 − 1.45 × 10−4y2

When y = ±180 mm, τ (web) = 11.1 N/mm2. The maximum value of shear stress occursat y = 0 and is 15.77 N/mm2. The form of the distribution is identical to that shown inFig. 10.4(b).


The shear load carried by the web is given by

S(web) = 2∫ 180

015(15.77 − 1.45 × 10−4y2) dy = 76701.6 N = 76.7 kN

Therefore, the percentage of the total load carried by the web is given by

76.780

× 100 = 95.9%.

S.10.3 The area of cross section of the reinforced beam is given by

A = 400 × 40 + 2 × 200 × 30 + 25 × 540 = 41 500 mm2

FIGURE S.10.3

400 mm

40 mm 0.68

1.78

(b)(a)

N/mm2

1.3614.22

15.15

30 mm

30 mm

25 mm600 mm

200 mm

y

Gz

y

Referring to Fig. S.10.3(a) and taking moments of areas about the top edge of theplate

41 500 y = 400 × 40 × 20 + 200 × 30 × 55 + 25 × 540 × 340 + 200 × 30 × 625

i.e.

y = 216.6 mm

Then

Iz = 400 × 403

12+ 400 × 40 × 196.62 +

(200 × 6003

12− 175 × 5403

12

)

+ (200 × 600 − 175 × 540) × 123.42

i.e.

Iz = 2.31 × 109 mm4

In the plate, A′ = 400(216.6 − y), y = (216.6 + y)/2, b0 = 400 mm. Then from Eq. (10.4)

τ = 4.33 × 10−5(216.62 − y2)


When y = 216.6 mm, τ = 0 and when y = 176.6 mm, τ = 0.68 N/mm2.

In the flange of the I-beam, A′y = [400 × 40 × 196.6 + 200(176.6 − y)(176.6 + y)/2]and b0 = 200 mm. Substituting in Eq. (10.4)

τ = 4.33 × 10−7(6.26 × 106 − 100y2)

When y = 176.6 mm, τ = 1.36 N/mm2 and when y = 146.6 mm, τ = 1.78 N/mm2.

In the web of the I-beam, A′y = [400 × 40 × 196.6 + 200 × 30 × 161.6 + 25(146.6 − y)(146.6 + y)/2] and b0 = 25 mm.


τ = 3.46 × 10−6(4.38 × 106 − 12.5y2)

When y = 146.6 mm, τ = 14.22 N/mm2 and when y = 0, τ = 15.15 N/mm2.

The required distribution is shown in Fig. S.10.3(b).

The shear stress at the top of the flange is 1.36 N/mm2 so that the shear force/unitlength of beam is 1.36 × 200 = 272 N/mm or 272 kN/m.

S.10.4 From Ex. 10.1 the maximum shear stress in a rectangular section beam is givenby

τ (max) = 3Sy

2bd

The maximum shear force on a section of this beam is 90 kN, therefore

τ (max) = 3 × 90 × 103

2 × 150 × 300= 3 N/mm2.

S.10.5 The beam section is singly symmetrical so that the shear centre lies on the axisof symmetry, the z axis. The shear flow distribution is given by Eq. (10.22) which, sinceIzy = 0 and only Sy is applied, reduces to

qs = −Sy

Iz

∫ s

0ty ds (i)

where

Iz = 2∫ a

0ts2 sin2 α ds + 2

∫ a

0t(a sin α + s sin α)2 ds

In the first term on the right hand side of the expression for Iz, s is measured in thedirection CB from C and in the second term s is measured from B in the direction BA.


Then

Iz = 16a3t sin2 α

3

In AB,

y = a sin α + (a − sA) sin α = (2a − sA) sin α


qAB = −Sy

Iz

∫ s

0t(2a − sA) sin α ds

which gives

qAB =−3Sy

(2as − s2

A2

)16a3 sin α

When sA = 0, qAB = 0 and when sA = a, qAB = −9Sy/32a sin α.

In BC, y = (a − sB) sin α. Then, from Eq. (i)

qBC = −Sy

Iz

∫ s

0t(a − sB) sin α ds − 9Sy

32a sin α

i.e.

qBC =−3Sy

(32 + sB

a − s2B

2a2

)16a sin α

When sB = 0, qBC = −9Sy/32a sin α and when sB = a, qBC = −3Sy/8a sin α.


SyzS = −2∫ a

0qAB a sin 2α ds

Substituting for qAB from the above and integrating

SyzS = 3Sy cos α

4a2

[as2 − s3

A6

]a

0

which gives

zS = 5a cos α

8.

S.10.6 The shear centre is the point in a beam cross section through which shear loadsmust be applied for there to be no twisting of the section.

As in P.10.4, the z axis is an axis of symmetry so that the shear centre lies on this axis.Its position is found by applying a shear load Sy through the shear centre, determining


the shear flow distribution and then taking moments about some convenient point.Eq. (10.22) reduces to

qs = −Sy

Iz

∫ s

0ty ds (i)

in which, referring to Fig. S.10.6

2r

2r

t

z1

2 3

4O

u

zs

s1

Sy

S

s2

r

FIGURE S.10.6

Iz = 2

(tr3

3+ 2rtr2 +

∫ π2

0tr2 cos2 θ r dθ

)

i.e.

Iz = 6.22tr3

In the wall 12, y = s1. Therefore substituting in Eq. (i)

q12 = −Sy

Iz

∫ s

0ts1 ds = −Sy

Iz

ts21

2

Then

q2 = −Sy

Iz

tr2

2

In the wall 23, y = r, then

q23 = −(

Sy

Iz

)(∫ s

0tr ds + tr2

2

)

i.e.

q23 = −(

Sy

Iz

)(trs2 + tr2

2

)

and

q3 = −5Sy

Iz

tr2

2


In the wall 34, y = r cos θ , then

q34 = −Sy

Iz

(∫ θ

0tr2 cos θ dθ + 5tr2

2

)

i.e.

q34 = −Sy

Iztr2(

sin θ + 52

)

Taking moments about O

SyzS = −2

[∫ r

0q122r ds +

∫ 2r

0q23r ds +

∫ π2

0q34r2 dθ

]

The negative sign arises from the fact that the moment of the applied shear load is inthe opposite sense to the moments produced by the internal shear flows. Substitutingfor q12, q23 and q34 from the above

SyzS = Sy

Izt

[∫ r

0

(s212

)2r ds +

∫ 2r

0

(rs2 + r2

2

)r ds +

∫ π2

0r4(

sin θ + 52

)dθ

]

which gives

zS = 2.66r.

S.10.7 In this problem the axis of symmetry is the vertical y axis and the shear centrewill lie on this axis so that only its vertical position is required. Therefore we apply ahorizontal shear load Sz through the shear centre, S, as shown in Fig. S.10.7.

y

ys

S

4

3

21

O

u

Sz

s1s2

50 mm

100 mm 25 mm25 mm

50m

m

FIGURE S.10.7

The thickness of the section is constant and will not appear in the answer for the shearcentre position (see S.10.6), therefore assume the section has unit thickness.

Eq. (10.22), since Izy = 0, t = 1 and only Sz is applied, reduces to

qs = −Sz

Iy

∫ s

0z ds (i)


where

Iy = 253

12+ 25 × 62.52 + 50 × 502 +

∫ π2

0(50 cos θ)250 dθ

i.e.

Iy = 6.44 × 105 mm4

In the flange 12, z = 75 − s1 and

q12 = −Sz

Iy

∫ s

0(75 − s1) ds = −Sz

Iy

(75s1 − s2

12

)

and when s1 = 25 mm, q2 = −1562.5Sz/Iy

In the wall 23, z = 50 mm, then

q23 = −Sz

Iy

(∫ s

050 ds + 1562.5

)= −Sz

Iy(50s2 + 1562.5)

When s2 = 50 mm, q3 = −4062.5Sz/Iy

In the wall 34, z = 50 cos θ , therefore

q34 = −Sz

Iy

(∫ θ

050 cos θ 50 dθ + 4062.5

)= −Sz

Iy(2500 sin θ + 4062.5)

Now taking moments about O

SzyS = −2

(∫ 25

0q1250 ds1 −

∫ 50

0q2350 ds2 −

∫ π2

0q34502 dθ

)

Note that the moments due to the shear flows in the walls 23 and 34 are opposite insign to the moment produced by the shear flow in the wall 12. Substituting for q12, etc.gives

yS = 87.5 mm.

S.10.8 The section is singly symmetrical so that the shear centre S lies on the axis ofsymmetry, the z axis. Then, since only Sy is applied and Izy = 0, Eq. (10.22) reduces to

qs = −Sy

Iz

∫ s

0ty ds (i)

where

Iz = t0h3

12+ 2d

(3t02

)h2

4= t0h2

(h + 9d

12

)

In the wall AB, y = h/2, t = 2t0(1 − sA/2d). Substituting in Eq. (i) and integrating gives

qAB = −(

Sy

Iz

)t0h

(sA − s2

A4d

)


When sA = d, qB = −3Syt0hd/4Iz.

In the wall BC, y = h/2 − sB. Then

qBC = −Sy

Iz

[∫ s

0t0

(h2

− sB

)ds + 3t0hd

4

]

i.e.

qBC = − Sy

2Izt0

[hsB − s2

B + 3hd2

]

If moments are taken about the mid-point of the web qBC will not contribute to themoment equation, i.e.

SyzS = −2∫ d

0qAB

(h2

)ds

Substituting for qAB and Iz from the above gives

zS = 5d2

h + 9d.

S.10.9 The section is unsymmetrical so that both horizontal and vertical positions ofthe shear centre are unknown. To find the horizontal position apply a vertical shearload through the shear centre and then to find the vertical position apply a horizontalshear load through the shear centre. Initially the position of the centroid of area mustbe found so that the second moments of area can be determined.

The cross-sectional area of the flange DC is equal to the cross-sectional area of theflange AB so that the centroid of area is at the mid-depth of the section.


A = 10 × 50 + 10 × 100 + 5 × 100 = 2000 mm2

Referring to Fig. S.10.9 and taking moments of area about the web BC

100 mm

100 mm

10 mm10 mm

5 mm

50 mm

C

BA

yz

z G

D

ys

s

zs

Sy

SSz

FIGURE S.10.9


2000 z = 10 × 50 × 25 + 5 × 100 × 50

from which

z = 18.75 mm

Then

Iz = 10 × 5 × 502 + 5 × 100 × 502 + 10 × 1003

12= 3.33 × 106mm4

Iy = 10 × 503

12+ 10 × 50 × 6.252 + 5 × 1003

12+ 5 × 100 × 31.252

+ 10 × 100 × 18.752

= 1.38 × 106 mm4

Izy = 10 × 50(6.25)(50) + 5 × 100(31.25)(−50) = −0.625 × 106 mm4

The numerator in both terms on the right hand side of Eq. (10.22) is then

IzIy − I2zy = (3.33 × 1.38 − 0.6252) × 1012 = 4.205 × 1012

Since only the position of the shear centre is required it is not necessary to calculatethe shear flow distribution in the complete cross section; if moments are taken about,say, the corner C the shear flows in the walls BC and CD will not contribute to themoment equation so that only the shear flow distribution in the flange AB is needed.First consider the application of Sy. Equation (10.22) becomes

qs = Sy[Izy∫ s

0 tz ds − Iy∫ s

0 ty ds]

4.205 × 1012 (i)

In the flange AB, z = 81.25 − s and y = −50 mm. Therefore

qAB = Sy

[−0.149 × 10−6

∫ s

05(81.25 − s)ds − 0.328 × 10−6

∫ s

05(−50)ds

]

i.e.

qAB = Sy

[21.47 × 10−6s + 0.37 × 10−6s2

](ii)


SyzS =∫ 100

0qAB × 100 ds

Substituting for qAB from Eq. (ii) gives

zS = 23.1 mm

Now apply Sz through the shear centre. Eq. (10.22) becomes

qAB = Sz[−Iz

∫ s0 tz ds + Izy

∫ s0 ty ds

]4.205 × 1012 (iii)


The expressions for z and y in terms of s are identical to those above so that Eq. (iii)simplifies to

qAB = Sz

[−0.792 × 10−6

∫ s

05(81.25 − s)ds − 0.149 × 10−6

∫ s

05(−50)ds

]

i.e.

qAB = Sz

[−284.5 × 10−6s + 1.98 × 10−6s2

](iv)


SzyS = −∫ 100

0qAB × 100 ds

Again substituting for qAB from Eq. (iv) and carrying out the integration

yS = 76.3 mm.

S.10.10 The most straightforward approach for the solution of this problem is toreplace the applied shear load by a shear load through the shear centre, which coincideswith the centre of symmetry, together with a torque as shown in Fig. S.10.10. Theshear flows produced by the two separate loading systems are calculated separatelyand then superimposed. The torsion of thin-walled closed section beams is covered inSection 11.4.

30 kN

Cy

B

G S

sB

sA

Az D

EF

30°

2.6 � 106 N mm

FIGURE S.10.10

The torque is given by

T = 30 × 103 × 100 cos 30◦ = 2.6 × 106 N mm

The area enclosed by the mid-line of the section wall is given by

A = 100 × 2 × 100 cos 30◦ + 4 × 0.5 × 100 cos 30◦ × 100 sin 30◦ = 2.6 × 104 mm2

Then, from Eq. (11.20), the shear flow due to the torque is

qT = 2.6 × 106

2 × 2.6 × 104 = 50 N/mm


The shear flow distribution due to the shear load is given by Eq. (10.24) in whichSz = −30 × 103N and Sy = 0. Further, the section is doubly symmetrical so that Izy = 0.If an origin for s is taken at A on the horizontal axis of symmetry, i.e. the section is“cut” at A, then qs,0 = 0 and Eq. (10.24) reduces to

qs = −Sz

Iy

∫ s

0tz ds (i)

The thickness of the section walls is constant and, as we have seen previously, willdisappear from the expressions for shear flow distribution, therefore we assume thatthe walls are of unit thickness. Then

Iy = 2 × 1003

12+ 4

∫ 100

0(100 − sA sin 30◦)2ds = 2.5 × 106 mm4


qs = 0.012∫ s

0z ds

In the wall AB, z = 100 − sA sin 30◦ = 100 − 0.5sA. Then

qAB(shear) = 0.012(100sA − 0.25s2A) = 1.2sA − 0.003s2

A

When sA = 100 mm, qAB(shear) = 90 N/mm

In the wall BC, z = 50 − sB. Then

qBC(shear) = 0.012(50sB − 0.5s2B) + 90 = 0.6sB − 0.006s2

B + 90

When sB = 100 mm, qBC = 90 N/mm

In the wall CD, z = −50 − sC sin 30◦ = −50 − 0.5sC. Then

qCD(shear) = 0.012(−50sC−0.25s2C) + 90 = −0.6sC − 0.003s2

C + 90

The remaining shear flow distribution due to shear follows from symmetry.

Now superimposing the shear flows due to torsion on those due to shear,

qAB = 1.2sA − 0.003s2A + 50

qBC = 0.6sB − 0.006s2B + 140

qCD = −0.6sC − 0.003s2C + 140

In the remaining half of the section the shear flow distribution due to shear issymmetrical with that in ABCD but is then reduced by the shear flow due to torsion.

S.10.11 In Fig. P.10.11 BO is an axis of symmetry so take this as the z axis. Then Sy = Sand Izy = 0. Eq. (10.24) therefore reduces to

qs = − SIz

∫ s

0ty ds + qs,0


where

Iz = (2r)3 sin2 45◦

12+ 2

∫ π2

0(r sin θ)2r dθ = 0.62r3

in which the section is assumed to have unit wall thickness.

“Cutting” the section at O

qb,OA = − S0.62r3

∫ θ

0r sin θ r dθ = 1.61S( cos θ − 1)

r

and when θ = π/4, qb,OA = −0.47S/r.

Also

qb,AB = − S0.62r3

∫ s

0(r − s) sin 45◦ds − 0.47S

r

i.e.

qb,AB = −S(1.14rs − 0.57s2 + 0.47)r


Sr = 2∫ π

4

0qb,OAr r dθ + 2

(π r2

4

)qs,0

Substituting for qb,OA from the above

Sr = 2∫ π

4

01.61S( cos θ − 1) rdθ + π r2qs,0

2

from which

qs,0 = 0.8S/r

The complete shear flow distribution is then

qOA = S(1.61 cos θ − 0.81)r

qAB = S(0.57s2 − 1.14rs − 0.33)r

.

S.10.12 The shear centre lies on the vertical axis of symmetry therefore apply anhorizontal shear load Sz through the shear centre. Since Sy = 0 and Izy = 0 (the y axisis an axis of symmetry) Eq. (10.24) reduces to

qs = −Sz

Iy

∫ s

0tz ds + qs,0

where, referring to Fig. S.10.12

Iy = 23

12+ 2

∫ 3

0(s sin θ)2 ds


ys1

yss2

Sz

S

s

u

Gz

O 2

3

1

FIGURE S.10.12

assuming that the thickness is unity and s is measured from the point 3. Then

Iy = 2.67 m4

Now taking an origin for s at O

qb,O1 = − Sz

2.67

∫ s

0s1 ds = −0.19Szs2

1

and when s1 = 1 m, qb,O1 = −0.19Sz

qb,13 = −0.37Sz

∫ s

0(1 − s2 sin θ) ds − 0.19Sz = −0.37Sz(0.5 + s2 − 0.17s2

2)

In Eq. (10.30),∮

ds = 8 m. Then

qs,0 = 2 × 0.37Sz[0.5∫ 1

0 s21 ds + ∫ 3

0 (0.5 + s2 − 0.17s22)ds]

8= 0.44Sz

so that qO1 = Sz(0.44 − 0.19s21)

Take moments about 3

Sz(3 cos θ − yS) = 2∫ 1

0qO13 cos θ ds

from which

yS = 0.7 m.

S.10.13 Referring to Fig. P.10.13, the wall DB is 3 m long so that its cross-sectional area, 3 × 103 × 8 = 24 × 103 mm2, is equal to that of the wall EA,2 × 103 × 12 = 24 × 103 mm2. It follows that the centroid of area of the section liesmid-way between DB and EA on the vertical axis of symmetry. Also since Sy = 500 kN,Sz = 0 and Izy = 0, Eq. (10.24) reduces to

qs = −500 × 103

Iz

∫ s

0ty ds + qs,0 (i)


If the origin for s is taken on the axis of symmetry, say at O, then qs,0 is zero. Also

Iz = 3 × 103 × 8 × (0.43 × 103)2 + 2 × 103 × 12 × (0.43 × 103)2 + 2 × (1 × 103)3

× 10 × sin2 60◦/12

i.e.

Iz = 101.25 × 108 mm4

Eq. (i) then becomes

qs = −4.94 × 10−5∫ s

0ty ds

In the wall OA, y = −0.43 × 103 mm. Then

qOA = 4.94 × 10−5∫ s

012 × 0.43 × 103 ds = 0.25sA

and when sA = 1 × 103 mm, qOA = 250 N/mm

In the wall AB, y = −0.43 × 103 + sB cos 30◦. Then

qAB = −4.94 × 10−5∫ s

010(−0.43 × 103 + 0.866sB) ds + 250

i.e.

qAB = 0.21sB − 2.14 × 10−4s2B + 250

When sB = 1 × 103 mm, qAB = 246 N/mm

In the wall BC, y = 0.43 × 103 mm. Then

qBC = −4.94 × 10−5∫ s

08 × 0.43 × 103 ds + 246

i.e.

qBC = −0.17sC + 246

Note that at C where sC = 1.5 × 103 mm, qBC should equal zero; the discrepancy,−9 N/mm, is due to rounding off errors.

The maximum shear stress will occur in the wall AB (and ED) mid-way along itslength (this coincides with the neutral axis of the section) where sB = 500 mm. Thisgives, from Eq. (ii), qAB(max) = 301.5 N/mm so that the maximum shear stress is equalto 301.5/10 = 30.2 N/mm2.



S.11.1 From Eq. (11.4) the shear in the bar is given by

τ = TrJ

where J = πD4/32, i.e.

τ = 16TπD3

Therefore the shear stress varies directly with the applied torque but is inverselyproportional to the diameter cubed. It follows that while the torque in the bar wherethe diameter is 100 mm is four times that where the diameter is 50 mm the ratio of thediameters cubed is 8 so that the maximum shear stress will occur in the bar wherethe diameter is 50 mm. Then

τ (max) = 16 × 1 × 106

π × 503 = 40.7 N/mm2

Alternatively the stresses in both parts of the bar could be calculated and the maximumchosen.

Again from Eq. (11.4), the angle of twist is given by θ = TL/GJ . Therefore θ (freeend) = (4 × 106 × 200 × 32/70 000 × π × 1004 + 1 × 106 × 400 × 32/70 000 × π×504)/(180/π) = 0.6◦.

S.11.2 Referring to Fig. S.11.2 and using Eq. (11.8)

TA =(

0.52.0

)× 12 = 3 kN m

Therefore, from equilibrium

TC = 9 kN m

Maximum shear stress condition:

From Eqs (11.4)

Tmax

J= τmax(

D2

) (i)

A B C

TCTA

12 kN m

1.5 m 0.5 m FIGURE S.11.2


where D is the external diameter of the shaft and J = π(D4 − d4)/32 in which d is theinternal diameter of the shaft. Then, substituting in Eq. (i)

9 × 106[π(804−d4)

32

] = 150(802

)

which gives

d = 63.7 mm

Maximum angle of twist condition:

The maximum angle of twist will occur at B; then either the portion AB or the portionBC may be considered. Considering AB, from Eqs (11.4)

θB = TABLAB

GJ

i.e.

1.5 × π

180= 3 × 106 × 1.5 × 103

80 000 × π[

(804 − d4)32

]from which

d = 66.1 mm

The maximum allowable internal diameter is therefore 63.7 mm.

S.11.3 From equilibrium

TA + TD = 2500 Nm (i)

From compatibility

θB(AB) = θB(BC) + θC(CD)

Therefore, from Eqs (11.4) and since GJ = constant

TA × 2 = (TD − 1500) × 1 + TD × 1

i.e.

TA = TD − 750 (ii)

Solving Eqs (i) and (ii) gives

TA = 875 Nm, TD = 1625 Nm


Therefore, from Eqs (11.4)

τmax = 1625 × 103 × 25

π(

504

32

) = 66.2 N/mm2

The maximum angle of twist will occur at B. Then, considering AB

θB = 875 × 103 × 2 × 103

70 000 × π(

504

32

) = 0.0407 rad = 2.3◦.

S.11.4 From equilibrium

TA + TD = 1 (i)

and from compatibility

θB(AB) = θB(BC) + θC(CD)


TA × 1.0

π(

754

32

) = (TD − 1) × 0.5

π(

504

32

) + TD × 0.5

π(

504

32

)

i.e.

TA = 5.1 TD − 2.5 (ii)

Solving Eqs (i) and (ii)

TA = 0.43 kN m, TD = 0.57 kN m

Then

τmax = 0.57 × 106 × 25

π(

504

32

) = 23.2 N/mm2

0.57 kN m

A B C D�ve

�ve

0.43 kN m FIGURE S.11.4


The maximum angle of twist will occur at C. Then

θC = 0.57 × 106 × 0.5 × 103

70 000 × π(

504

32

) = 0.00664 rad = 0.38◦

The distribution of torque along the bar is shown in Fig. S.11.4.

S.11.5 The maximum torque in the box girder is 1 × 2 × 103 kN m. Then, from

Eq. (11.22) τmax = 1 × 2 × 103 × 106

2 × 1000 × 750 × 10= 133.3 N/mm2.

The rate of twist of the girder is given by Eq. (11.25) since G is constant. Also∮dst

= 2 × 100010

+ 2 × 75015

= 300

Eq. (11.25) then becomes

dθ

dx= T × 300

4 × (1000 × 750)2 × 70 000= 1.905 × 10−15 T

In AB, T = 2000 × 106 N mm. Therefore

θAB = 2000 × 106 × 1.905 × 10−15x + B = 3.81 × 10−6x + B

When x = 0, θAB = 0 so that B = 0. Also when x = 2000 mm

θAB = 0.00762 rad = 0.44◦

In BC, T = 1 × 106(4000 − x) so that

θBC = 1.905 × 10−9

(4000x − x2

2

)+ C

When x = 2000 mm, θBC = 0.00762 rad so that C = −0.00381. Then

θBC = 1.905 × 10−9

(4000x − x2

2

)− 0.00381

When x = 4000 mm, θBC = 0.01143 rad = 0.65◦

The distribution of angle of twist in degrees is shown in Fig. S.11.5.

ABC

0.65°0.44°

FIGURE S.11.5


S.11.6 The total torque applied to the beam is 20 × 4 × 103 Nm. From symmetry thereactive torques at A and D will be equal and are 40 × 103 Nm. Therefore,

TAB = 40 000 Nm

TBC = 40 000 − 20(x − 1000) = 60 000 − 20x Nm (x in mm)

Note that the torque distribution is antisymmetrical about the centre of the beam. Themaximum torque in the beam is therefore 40 000 Nm so that, from Eq. (11.22)

τmax = 40 000 × 103

2 × 200 × 350 × 4= 71.4 N/mm2

The rate of twist along the length of the beam is given by Eq. (11.25) in which∮= 2 × 200

4+ 2 × 350

6= 216.7

Thendθ

dx=[

216.74 × (200 × 350)2 × 70 000

]T = 15.79 × 10−14 T

In AB, TAB = 40 000 Nm so that

θAB = 6.32 × 10−6x + B

When x = 0, θAB = 0 so that B = 0 and when x = 1000 mm, θAB = 0.0063 rad (0.361◦)

In BC, TBC = 60 000 − 20x Nm. Then, from Eq. (11.25)

θBC = 15.79 × 10−14(60 000x − 10x2) × 103 + C

When x = 1000 mm, θBC = 0.0063 so that C = −0.0016. Then

θBC = 1.579 × 10−10(60 000x − 10x2) − 0.0016

At mid-span where x = 3000 mm, θBC = 0.0126 rad (0.722◦).

S.11.7 The torque is constant along the length of the beam and is 1 kN m. Also thethickness is constant round the beam section so that the shear stress will be a maximumwhere the area enclosed by the mid-line of the section wall is a minimum, i.e. at thefree end. Then

τmax = 1000 × 103

2 × 50 × 150 × 2= 33.3 N/mm2

The rate of twist is given by Eq. (11.25) in which∮

ds/t varies along the length of thebeam as does the area enclosed by the mid-line of the section wall. Then

∮dst

=2 × 50 +

(150 + 50x

2500

)2

= 125 + 0.01x


Also

A = 50(

150 + 50 x2500

)= 7500 + x

Then

dθ

dx=(

1 × 106

4 × 25 000

)(125 + 0.01x)(7500 + x)2

or

dθ

dx= 10

[(12 500 + x)

100(7500 + x)2

]

i.e.

dθ

dx= 0.1

[5000

(7500 + x)2 + 1(7500 + x)

]

Then

θ = 0.1[ −5000

(7500 + x)+ loge(7500 + x) + B

]

When x = 2500 mm, θ = 0 so that B = −6.41 and

θ = 0.1[ −5000

(7500 + x)+ loge(7500 + x) − 6.41

]rad

When x = 0, θ = 10.6◦ etc.

S.11.8 The maximum shear stress in the section is given by Eq. (11.31) in which, fromEq. (11.27)

J = 2 × 23(

20 + 15 + 25 + 253

)= 453.3 mm4

Then

τmax = 50 × 103 × 2453.3

= 220.6 N/mm2

From Eq. (11.31)

dθ

dx= T

GJ

i.e.

dθ

dx= 50 × 103

25 000 × 453.3= 0.0044 rad/mm.


S.11.9 From Fig. 10.16 the shear centre of the section lies at the point B. The shearloading may therefore be replaced by shear loads acting through B together with atorque as shown in Fig. S.11.9.

AsA

C

y2.0 mm

2000 N mm 2.5 mm500 N

sB

B

80 mm

60 mm

1000 N

G y

z

z

FIGURE S.11.9

The maximum shear stress due to the applied torque is given by Eq. (11.31) in which,from Eq. (11.27),

J = 60 × 23

3+ 80 × 2.53

3= 576.7 mm4

Then

τmax(torsion)AB = 2000 × 2.5576.7

= 8.67 N/mm2

τmax(torsion)BC = 2000 × 2.0576.7

= 6.94 N/mm2

The shear stress distribution due to shear is given by Eq. (10.22) in which Sz = −500 Nand Sy = −1000 N. Referring to Fig. S.11.9 the area of cross section is

A = 80 × 2.5 + 60 × 2.0 = 320 mm2

Then

320 z = 80 × 2.5 × 40

which gives

z = 25 mm

Also

320 y = 60 × 2.0 × 30

so that

y = 11.25 mm

Iz = 80 × 2.5 × 11.252 + 2 × 603

12+ 60 × 2 × 18.752 = 103 500 mm4

Iy = 60 × 2 × 252 + 2.5 × 803

12+ 80 × 2.5 × 152 = 226 667 mm4

Izy = 60 × 2(25)(18.75) + 80 × 2.5(−15)(−11.25) = 90 000 mm4


The denominator in Eq. (10.22) is

IzIy − I2zy = 103 500 × 226 667 − 90 0002 = 1.54 × 1010


qs =[

(−1000 × 90 000 + 500 × 103 500)1.54 × 1010

] ∫ s

0tzds

+[

(−500 × 90 000 + 1000 × 226 667)1.54 × 1010

] ∫ s

0ty ds

i.e.

qs = −0.0025∫ s

0tz ds + 0.0118

∫ s

0ty ds

Then

qAB = −0.0025∫ s

02.5(−55 + sA)ds + 0.0118

∫ s

02.5(−11.25)ds

i.e.

qAB = 0.0119 sA − 0.0031 s2A (i)

At B where sA = 80 mm, qAB = −18.9 N/mm.

Examination of Eq. (i) shows that qAB is zero at sA = 20.8 mm and has a turning valueat sA = 41.6 mm where qAB = − 4.9 N/mm. Therefore the maximum shear flow in AB

is −18.9 N/mm and the maximum shear stress is−18.9

2.5= −7.6 N/mm2.

Also

qBC = −0.0025∫ s

02 × 25 ds + 0.0118

∫ s

02(−11.25 + sB)ds − 18.9

i.e.

qBC = −0.3905 sB + 0.0118 s2B − 18.9 (ii)

From Eq. (ii), qBC is a maximum when sB = 16.5 mm, i.e. the maximum value of shearflow in BC is −22.1 N/mm and the maximum shear stress is −22.1/2 = −11.1 N/mm2.

Combining the torsional shear stresses with the shear stresses due to shear it can beseen that the maximum shear stress is 6.9 + 11.1 = 18.0 N/mm2 which occurs on theinside of BC at 16.5 mm from B.

S.11.10 The maximum shear stress in the section is given by Eq. (11.31) in which

J = 100 × 2.543

3+ 2 × 38 × 1.273

3+(

23

)∫ 50

0

[1.27 + 1.27

( s50

)]3ds = 854.2 mm4


in which s is measured from a lip/flange junction. Then

τmax = ±100 × 103 × 2.54854.2

= ±297.4 N/mm2

The rate of twist is given by Eq. (11.31) and is

dθ

dx= 100 × 103

26 700 × 854.2= 0.0044 rad/mm.


S.12.1 The second moments of area of the timber and steel are, respectively

It = 200 × 3003

12= 450 × 106 mm4

Is = 2 × 12 × 200 × 1562 = 117 × 106 mm4 (treating the plates as thin)


8 = 150M × 106

(450 + 20 × 117) × 106

which gives

M = 148.8 kN m

From Eq. (12.8)

110 = 162M × 106(117 + 450

20

)× 106

from which

M = 94.7 kN m

Therefore the allowable bending moment is 94.7 kN m.

S.12.2 The maximum bending moment is

Mmax = 46.5 × 3.52

8= 71.2 kN m (see Ex. 3.7)

Also, for the timber

It = 2

(100 × 753

12+ 100 × 75 × 112.52

)= 196.9 × 106mm4


and for the steel

Is = 2t × 3003

12= 4.5 × 106t mm4


8 = 71.2 × 106 × 150(196.9 + 20 × 4.5t) × 106

so that

t = 12.6 mm

From Eq. (12.8)

124 = 71.2 × 106 × 150(4.5t + 196.9

20

)× 106

which gives t = 17.0 mm.

Therefore the required thickness of the steel plates is 17 mm.

S.12.3 The maximum stress in the timber beam is given by

σt = Mt × 150(150 × 3003

12

) = 12 N/mm2

which gives

Mt = 27 kN m

Also, since σs = 20σt, the steel stress is limiting so that σt = 155/20 = 7.75 N/mm2.

300 mm

150 mm

12 mm

n

146.2 mm

FIGURE S.12.3

Referring to Fig. S.12.3

150(300 − n)2

2= 150n2

2+ 20 × 150 × 12(n + 6)

which gives

n = 80.7 mm


Now taking moments about the resultant of the compressive stress distribution in thetimber

M ×106 = 155×150×12(146.2+80.7+6)+(

7.752

)×150×80.7

(2 × 80.7

3+ 146.2

)

which gives

M = 74.4 kN m

Therefore, Ms/Mt = 74.4/27 = 2.76 so that the increase produced by the steelreinforcing plate is 176%.

S.12.4 From Eq. (12.13) for a critical section

140 = 8 × 15[

0.9d1

n− 1]

which gives

n = 0.415d1

Taking moments about the centroid of the steel reinforcement (see Fig. 12.6)

M =(

82

)× 0.5d1 × 0.415d1

(0.9d1 − 0.415d1

3

)

from which

M = 0.632d31

Then

0.632d31 = 16.8 × 4.52 × 106

8so that

d1 = 406.7 mm

Then, equating the tensile force in the steel to the compressive force in the concrete

140As =(

82

)× (0.5 × 406.7) × 0.415 × 406.7

which gives

As = 980.6 mm2.

S.12.5 The area of the compression steel is 3 × π × 252/4 = 1472.6 mm2 and of thetensile steel is 5 × π × 252/4 = 2454.4 mm2. Then, from Eq. (12.17)

300n2

2+ (16 − 1) × 1472.6(n − 25) = 16 × 2454.4(750 − n)


from which

n = 242.8 mm

Using the compatibility of strain condition, i.e. Eq. (12.13)

σs = 16σc(750 − 242.8)242.8

σs = 33.4σc

Therefore the limiting stress is the steel stress and σc = 125/33.4 = 3.7 N/mm2. Then,taking moments about the compressive steel

M × 106 = 125 × 2454.4(750 − 25) −(

3.72

)× 300 × 242.8

(242.8

3− 25

)

i.e.

M = 214.5 kN m.

S.12.6 Assume that the neutral axis is at the base of the flange. The maximum bendingmoment is

M = 42 × 62

8= 189 kN m

Then, taking moments about the resultant compressive force in the concrete flange

140As

(550 − 125

3

)= 189 × 106

i.e.

As = 2655.7 mm2

The tensile force in the steel is equal to the compressive force in the concrete, i.e.

140As =(

8.52

)b × 125

which gives

b = 699.9 mm

The required width of the flange is therefore 700 mm.

S.12.7 The maximum bending moment is

Mmax = 16.8 × 4.52

8= 42.5 kN m


Assume that the neutral axis is 0.45d1 from the top of the beam, i.e. M = Mu. Then

42.5 × 106 = 0.15 × 24 × 0.5d1 × (0.9d1)2

which gives

d1 = 307.8 mm

From Eq. (12.24)

42.5 × 106 = 0.65 × 280As × 0.9 × 307.8

from which

As = 843.0 mm2.

S.12.8 Assume that n = d1/2. Then, taking moments about the compression steel

M × 106 = 0.87 × 250 × 2454.4 × 725 − 0.4 × 22.5 × 300 × 750

(7504 − 25

2

)

i.e.

M = 222.5 kN m.

S.12.9 From S.12.6, Mmax = 189 kN m. Assuming the neutral axis to be at the base ofthe flange

189 × 106 = 0.87 × 280As

(550 − 125

2

)

from which

As = 1592 mm2

Then, since the compressive force in the concrete is equal to the tensile force in thesteel

0.4 × 25.5 × 125b = 0.87 × 280 × 1592

which gives

b = 304.2 mm

Say a flange width of 304 mm.

S.12.10 Equating the compressive force in the concrete to the tensile force in thesteel

0.4 × 30 × 2 × 103n = 0.87 × 350 × 9490

so that

n = 120.4 mm


Taking moments about the line of action of the resultant compressive force in theconcrete

M × 106 = 0.87 × 350 × 9490(405.7 − 87.5)

i.e.

M = 919.5 kN m.


S.13.1 By inspection the support reactions are each 32 kN. The simplest approach isto use Macauley’s method. The bending moment at a section in the bay of the beamfurthest from, say, the left hand end of the beam, the origin for x, is

M = −2x − 5x2

2+ 32[x − 2] + 32[x − 10]


EI

(d2v

dx2

)= −2x − 5x2

2+ 32[x − 2] + 32[x − 10]

Then

EI(

dv

dx

)= −x2 − 5x3

6+ 16[x − 2]2 + 16[x − 10]2 + C1

and

EIv = −x3

3− 5x4

24+(

163

)[x − 2]3 +

(163

)[x − 10]3 + C1x + C2

When x = 6 m, (dv/dx) = 0 from symmetry, i.e.

0 = −62 − 5 × 63

6+ 16[6 − 2]2 + C1

which gives

C1 = −40

When x = 2 m, v = 0 so that

0 = −23

3− 5 × 24

24− 40 × 2 + C2

from which

C2 = 86

Then

EIv = −x3

3− 5x4

24+(

163

)[x − 2]3 +

(163

)[x − 10]3 − 40x + 86


When x = 6 m,

EIv = −63

3− 5 × 64

24+(

163

)[6 − 2]3 − 40 × 6 + 86

which gives

v(mid-span) = −3.6 mm (downwards)

and when x = 0,

EIv = 86

so that

v(at end) = 2.0 mm (upwards).

S.13.2 Since there are no loads on the part CB of the beam and the implication inthe question is that the beam is weightless, the length CB will remain straight but willrotate about C. From Ex. 13.4 the mid-span deflection of the simply supported spanAC is 5wL4(1 − K)4/384EI (see Eq. (v)). Also, the slope of the beam at C is given byEq. (iii) and is wL3(1 − K)3/24EI. The deflection at B is therefore

wL3(1 − K)3KL/24EI.

The two deflections are numerically equal so that

5wL4(1 − K)4

384EI= wL3(1 − K)3KL

24EI

i.e. (516

)(1 − K) = K

from which

K = 0.24.

S.13.3 The beam is as shown in Fig. S.13.3.

30 kN/m

90 kN/m

RA RBx

6 m

BA

FIGURE S.13.3



RA × 6 − 30 × 62

2− 60

2× 6 × 6

3= 0

which gives

RA = 150 kN

The bending moment at any section a distance x from A is then

Mx = 150x − 30x2

2− (90 − 30)

( x6

) ( x2

) ( x3

)i.e.

Mx = 150x − 15x2 − 5x3

3Substituting in Eq. (13.3)

EI

(d2v

dx2

)= 150x − 15x2 − 5x3

3

EI(

dv

dx

)= 75x2 − 5x3 − 5x4

12+ C1

EIv = 25x3 − 5x4

4− x5

12+ C1x + C2

When x = 0, v = 0 so that C2 = 0 and when x = 6 m, v = 0. Then

0 = 25 × 63 − 5 × 64

4− 65

12+ 6C1

from which

C1 = −522

and the deflected shape of the beam is given by

EIv = 25x3 − 5x4

4− x5

12− 522x

The deflection at the mid-span point is then

EIvmid-span = 25 × 33 − 5 × 34

4− 35

12− 522 × 3 = −1012.5 kN m3units

Therefore

vmid-span = −1012.5 × 1012

120 × 106 × 206 000= −41.0 mm (downwards).

S.13.4 Take the origin of x at the free end of the cantilever. The load intensity at anysection a distance x from the free end is wx/L. The bending moment at this section isgiven by

Mx = −( x

2

) (wxL

) ( x3

)= −wx3

6L



EI

(d2v

dx2

)= −wx3

6L

EI(

dv

dx

)= −wx4

24L+ C1

EIv = −wx5

120L+ C1x + C2

When x = L, (dv/dx) = 0 so that C1 = wL3/24. When x = L, v = 0, i.e. C2 = −wL4/30.The deflected shape of the beam is then

EIv = −( w

120L

) (x5 − 5xL4 + 4L5

)

At the free end where x = 0

v = − wL4

30EI.

S.13.5 The uniformly distributed load is extended from D to F and an upward uni-formly distributed load of the same intensity applied over DF so that the overall loadingis unchanged (see Fig. S.13.5).

A

RA

1 m 2 m 2 m 1 m

D

F

1 kN/m

4 kN1 kN/m

6 kN

RF

CB

x

FIGURE S.13.5

The support reaction at A is given by

RA × 6 − 6 × 5 − 4 × 3 − 1 × 2 × 2 = 0

Then

RA = 7.7 kN

Using Macauley’s method, the bending moment in the bay DF is

M = 7.7x − 6[x − 1] − 4[x − 3] − 1[x − 3]2

2+ 1[x − 5]2

2



EI

(d2v

dx2

)= 7.7x − 6[x − 1] − 4[x − 3] − [x − 3]2

2+ [x − 5]2

2

EI(

dv

dx

)= 7.7x2

2− 3[x − 1]2 − 2[x − 3]2 − [x − 3]3

6− [x − 5]3

6+ C1

EIv = 7.7x3

6− [x − 1]3 − 2[x − 3]3

3− [x − 3]4

24− [x − 5]4

24+ C1x + C2

When x = 0, v = 0 so that C2 = 0. Also when x = 6 m, v = 0. Then

0 = 7.7 × 63

6− 53 − 2 × 33

3− 34

24− 14

24+ 6C1

which gives

C1 = −21.8

Guess that the maximum deflection lies between B and C. If this is the case the slopeof the beam will change sign from B to C.

At B,

EI(

dv

dx

)= 7.7 × 12

2− 21.8 which is clearly negative

At C,

EI(

dv

dx

)= 7.7 × 32

2− 3 × 22 − 21.8 = +0.85

The maximum deflection therefore occurs between B and C at a section of the beamwhere the slope is zero.

i.e.

0 = 7.7x2

2− 3[x − 1]2 − 21.8

Simplifying

x2 + 7.06x − 29.2 = 0

Solving

x = 2.9 m

The maximum deflection is then

EIvmax = 7.7 × 2.93

6− 1.93 − 21.8 × 2.9 = −38.8

i.e.

vmax = −38.8EI

(downwards).


S.13.6 Taking moments about B

RA × 30 − 4 × 20 + 20 = 0

which gives

RA = 2 kN

The bending moment in the bay CD is, using Macauley’s method

M = 2x − 4[x − 10] + 20[x − 20]0


EI

(d2v

dx2

)= 2x − 4[x − 10] + 20[x − 20]0

EI(

dv

dx

)= x2 − 2[x − 10]2 + 20[x − 20]1 + C1

EIv = x3

3− 2[x − 10]3

3+ 10[x − 20]2 + C1x + C2

When x = 0, v = 0 so that C2 = 0 and when x = 30 m, v = 0 so that C1 = −155.6

Guess that the maximum deflection occurs between B and C. If this is valid the slopeof the beam between B and C will be zero, i.e.

0 = x2 − 2[x − 10]2 − 155.6

or

x2 − 40x + 355.6 = 0

Solving

x = 13.3 m

which lies between B and C so that the ‘guess’ was correct. The maximum deflectionis then given by

EIvmax = 13.33

3− 2 × 3.33

3− 155.6 × 13.3

i.e.

vmax = −1309.2EI

m.

S.13.7 Taking moments about D

RA × 4 + 100 − 100 × 2 × 1 + 200 × 3 = 0

from which

RA = −125 N


RB − 125 − 100 × 2 − 200 = 0


Therefore

RB = 525 N

The bending moment at a section a distance x from A in the bay DF is given by

M = −125x + 100[x − 1]0 − 100[x − 2]2

2+ 525[x − 4] + 100[x − 4]2

2

in which the uniformly distributed load has been extended from D to F and an upwarduniformly distributed load of the same intensity applied from D to F.


EI

(d2v

dx2

)= −125x + 100[x − 1]0 − 50[x − 2]2 + 525[x − 4] + 50[x − 4]2

EI(

dv

dx

)= −125x2

2+ 100[x − 1]1 − 50[x − 2]3

3+ 525[x − 4]2

2+ 50[x − 4]3

3+ C1

EIv = −125x3

6+ 50[x − 1]2 − 50[x − 2]4

12+ 525[x − 4]3

6

+ 50[x − 4]4

12+ C1x + C2

When x = 0, v = 0 so that C2 = 0 and when x = 4 m, v = 0 which gives C1 = 237.5. Thedeflection curve of the beam is then

v = 1EI

(−125x3

6+ 50[x − 1]2 − 50[x − 2]4

12+ 525[x − 4]3

6+ 50[x − 4]4

12+ 237.5x

).

S.13.8 Referring to Fig. P.13.8 and choosing A as the origin for x, Eq. (13.10) becomes,for the portion AC of the beam

xC

(dv

dx

)C

− xA

(dv

dx

)A

− (vC − vA) =∫ C

A

(MxEI

)dx (i)

The coordinate of A is zero and the slope of the beam at C is zero from symmetry.Also the bending moment at any section between A and C is (W /2)x. Therefore Eq. (i)simplifies to

−vC =∫ C

A

(Wx2EI

)dx =

∫ L/4

0

(Wx2

2EI

)dx +

∫ L/2

L/4

(Wx2

4EI

)dx

i.e.

−vC =(

W2EI

)[

x3

3

]L/4

0

+(

12

)[x3

3

]L/2

L/4


which gives

vC = −3WL3

256EI.

S.13.9 The support reactions are each w0L/4 and the load intensity at any sectiona distance x from A in AB is 2w0x/L. The bending moment at this section is thenM = w0[(L/4)x − x3/3L]. The slope of the beam at B is zero from symmetry and thecoordinate of A is zero. Therefore Eq. (13.10) reduces to

−vB =(

1EI

)∫ L/2

0w0

[Lx2

4− x4

3L

]dx

i.e.

−vB =(w0

EI

)[Lx3

12− x5

15L

]L/2

0so that

vB = −w0L4

120EI.

S.13.10 Since the outer portions of the beam are rigid, EI is infinite and M/EIis zero over these portions. The M/EI diagram therefore has the form shown inFig. S.13.10.

A B C

WL8EI

WL8EI

WL4EI

Form of bending moment

diagram

L4

L4

L4

L4

FIGURE S.13.10

From Section 13.3 the deflection at B is equal to minus the moment of the area of theM/EI diagram between A and B about A. Then

vB = −(

L4

)(WL

8

)(3L8

)−(

12

)(L4

)(WL

8

)[(L4

)+(

23

)(L4

)]

i.e.

vB = −7WL4

384EI.


S.13.11 There are only vertical support reactions at A and B. Taking moments about B

RA × 2 + 20 × 1 = 0

i.e.

RA = −10 N

In AB the bending moments at any section a distance x from A are given by

Mz = RAx = −10x N mm, My = 0 (x in mm)

The deflected shape of the beam in the vertical plane is given by Eq. (13.15) and inthe horizontal plane by Eq. (13.14). The centroid of area of the beam section is at thecentre of the web and, with the usual axes system, the second moments of area are

Iz = 2

(15 × 53

12+ 15 × 5 × 12.52

)+ 5 × 203

12= 27 083.3 mm4

Iy = 2

(5 × 153

12+ 15 × 5 × 52

)+ 20 × 53

12= 6770.8 mm4

Izy = 15 × 5(5)(12.5) + 15 × 5(−5)(−12.5) = 9375.0 mm4

The denominator in Eqs (13.14) and (13.15) is then

IzIy − I2zy = 27083.3 × 6770.8 − 9375.02 = 95.48 × 106

Eq. (13.14) then reduces to

d2udx2 = 4.91 × 10−9x

Then

dudx

= 4.91 × 10−9x2

2+ C1

and

u = 4.91 × 10−9x3

6+ C1x + C2

When x = 0, u = 0 so that C2 = 0. When x = 2 × 103 mm, u = 0 so thatC1 = −3.27 × 10−3.

i.e.

u = 0.818 × 10−9x3 − 3.27 × 10−3x (i)

and when x = 1000 mm,

u = −2.45 mm (to the right)


From Eq. (13.15)

d2v

dx2 = −3.55 × 10−9x

dv

dx= −3.55 × 10−9x2

2+ C3

v = −3.55 × 10−9x3

6+ C3x + C4

When x = 0, v = 0 so that C4 = 0 and when x = 2000 mm, v = 0 so thatC3 = 2.37 × 10−3.

Therefore

v = −0.592 × 10−9x3 + 2.37 × 10−3x

When x = 1000 mm

v = 1.78 mm (upwards).

S.13.12 The components of the mid-span deflection may be determined from firstprinciples as in S.13.11 or deduced from the result for a symmetrical section sim-ply supported beam carrying a uniformly distributed load as in Eq. (v) of Ex. 13.4.Using the latter approach the horizontal component of the mid-span deflection is, bycomparing Eq. (13.3) and Eq. (13.14)

umid-span = −(

5wL4

384E

)(Iz

IzIy − I2zy

)(i)

Substituting the given values

umid-span = −47.8 mm (to the right)

Similarly

vmid-span =(

5wL4

384E

)(Izy

IzIy − I2zy

)(ii)

which gives

vmid-span = −21.2 mm (downwards)

The resultant deflection is then

Res.def. =√(47.82 + 21.22) = 52.3 mm at tan−1(21.2/47.8) = 23.9◦ below thehorizontal.


S.13.13 Again the problem may be solved by working from first principles or thesolution may be deduced from the symmetrical section beam case of Ex. 13.1, Eq. (v).The horizontal component of deflection of the free end of the cantilever with W actingalone is then

ufe =(

WL3

3E

)(Izy

IzIy − I2zy

)

and due to the tension T in the link acting alone

ufe = −(

TL3

3E

)(Iz

IzIy − I2zy

)

The sum of these two expressions is zero since there is no horizontal deflection.Therefore T = W (Izy/Iz).

With this value of T , My = −W (Izy/Iz)(L − x) and Mz = −W (L − x). Substituting inEq. (13.15)

d2v

dx2 =[−W (L − x)Iy + W

(IzyIz

)Izy(L − x)

]E(IzIy − I2

zy)= −W (L − x)

EIz

This expression is identical to Eq. (ii) in Ex. 13.1 so that we deduce that the verticaldeflection of the free end of the cantilever is −WL3/3EIz.

S.13.14 By inspection the centroid is 40 mm horizontally from the junction of the walls.

100 mm

5 mm60 mm

80 mm

5 mm

y

G

zy

FIGURE S.13.14

Referring to Fig. S.13.14 and taking moments of areas about the horizontal flange

(5 × 100 + 5 × 80) y = 5 × 100 × 30


i.e.

y = 16.7 mm

Then

Iz = 5 × 80 × 16.72 + 1003 × 5 ×(

60100

)2

12+ 100 × 5 × (13.3)2 = 0.35 × 106 mm4

Iy = 5 × 803

12+ 1003 × 5 ×

(80

100

)2

12= 0.48 × 106 mm4

Izy = 1003 × 5 ×(

60100

) (80100

)12

= 0.2 × 106 mm4

The beam is simply supported with a horizontal uniformly distributed load. The situa-tion is therefore identical to P.13.12 so that the components of deflection at mid-spanare given by Eqs (i) and (ii) in S.13.12. Substituting the appropriate values in theseequations

umid-span = −9.0 mm, vmid-span = 5.2 mm.

S.13.15 The moment-area method may be used to solve this problem. ComparingEq. (13.15) with Eq. (13.3) and examining the derivation of Eq. (13.10) it can be seenthat, for a beam of unsymmetrical section and subjected to bending moments Mz andMy, the equivalent expression to Eq. (13.10) is

xB

(dv

dx

)B

− xA

(dv

dx

)A

− (vB − vA) = 1E(IzIy − I2

zy)

∫ B

A(MzIy − MyIzy)x dx (i)

In P.13.15 there is no horizontal reaction at the simple support so that My = 0. In BAthe bending moment Mz = −W (L + x) + RBx where x is measured from B towards Aand RB is the vertical reaction at B. Since the origin for x is at B, the deflection of thebeam at B and at A is zero and the slope of the beam at A is zero Eq. (i) reduces to

0 =∫ A

B[ − W (L + x) + RBx] x dx =

∫ L

0(−WLx − Wx2 + RBx2) dx

which gives

RB = 5W2

.


S.13.16 Again the moment-area method may be used to solve this problem. Using thecomparison of Eq. (13.14) with Eq. (13.3) and from the derivation of Eq. (13.10)

xB

(dudx

)B

− xA

(dudx

)A

− (uB − uA) = 1E(IzIy − I2

zy)

∫ B

A(MyIz − MzIzy)x dx (i)

B

RB,V

RB,H

x 2L

L

A

C

W

FIGURE S.13.16

Referring to Fig. S.13.16 the x coordinate of B is zero, the slope of the beam at A iszero and the horizontal deflection at A and at B is zero. Also

Mz = −W (L + x) + RB,Vx and My = −RB,H x

Eq. (i) therefore reduces to

0 =∫ 2L

0[ − RB,HxIz + WIzy(L + x) − RB,VIzyx]x dx

which gives

0 = −4 RB,HIz + 7 WIzy − 4 RB,VIzy (ii)

Also, from Eq. (i) in S.13.15

0 = 4 RB,HIzy − 7 WIy + 4 RB,VIy (iii)

Solving Eqs (ii) and (iii) gives

RB,V = 7W4

, RB,H = 0.


S.13.17 Each half of the beam is subjected to a constant shear force of W /2 and deflectsby an amount vs. Therefore, from Eq. (13.22) and noting that, from Ex. 13.17, β = 6/5,the deflection at the mid-span point due to shear is

vs = 65G

∫ L/2

0

(W

2BD

)dx

which gives

vs = 3WL10GBD

.

S.13.18 The shear force at a section of the cantilever a distance x from its free end iswx. Substituting in Eq. (13.22)

vs = 65GBD

∫ L

0wx dx

i.e.

vs = 3wL2

5GBD.

S.13.19 The shear stress distribution has been calculated in Ex. 10.3 and is given byEq. (10.19), i.e.

τ =(

16W3πD2

)(1 − 4y2

D2

)

Now substituting in Eq. (13.20)

β =(

πD2

4W 2

)(16W3πD2

)2 ∫ D/2

−D/2

(1 − 4y2

D2

)2

b0 dy (i)

Changing to angular coordinates, (see Fig. 10.8)

y =(

D2

)sin θ , dy =

(D2

)cos θ dθ , b0 = D cos θ

Substituting in Eq. (i) and simplifying

β = 329π

∫ π/2

−π/2cos6 θ dθ

The value of the integral may be found using a reduction formula. Then

β = 109

The shear force at any section of the cantilever is W . Substituting in Eq. (13.22)

vs =(

109G

)β

∫ L

0

W(πD2

4

) dx


i.e.

vs = 40WL9πGD2 .

S.13.20 From Eq. (13.22) it can be seen that for a given beam and loading the deflectiondue to shear depends upon the value of β. Also, from Eq. (13.20) when a uniform shearstress is assumed, i.e. τ = Sy/A

β = AS2

y

∫ y,2

y,1

(Sy

A

)2

b0 dy

The integral in the above is equal to A so that

β = 1

For a rectangular section beam in which the shear stress distribution is given byEq. (10.4), β = 6/5 = 1.2 (see Ex. 13.17) so that the deflection in this case is 20%greater than for a shear stress distribution which is assumed to be uniform.

The deflected shape of the beam due to shear only is shown in Fig. S.13.20(a). Then,from Eq. (13.22)

vs = 1.270 000

∫ 1000

0

(500 × 103

200 × 400

)dx = 0.107 mm

FIGURE S.13.20

500 kN

(a)

A

B C

1 m 1 m

500 kN

(b)

AB

C

1 m 1 m

From Eq. (v) of Ex. 13.1 the deflection due to bending at B (Fig. S.13.20(b)) is

vb = 500 × 103 × 10003

3 × 200 000(

200×4003

12

) = 0.78 mm

From Eq. (iii) of Ex. 13.1 the slope of the beam at B is(dv

dx

)B

= 500 × 103 × 10002

2 × 200 000(

200×4003

12

) = 0.00117

Therefore the total deflection at C due to bending is 0.78 + 0.00117 × 1000 = 1.95 mmand the total deflection at C due to bending and shear isvtotal = 0.107 + 1.95 = 2.057 mm.


S.13.21 Release the beam at the mid-span point. The mid-span deflection of the result-ing simply supported beam is given by Eq. (v) of Ex. 13.4 and is 5w(2L)4/384EI.Suppose that the mid-span support reaction is R. The deflection of a simply supportedbeam carrying a central concentrated load is, from Eq. (v) of Ex. 13.5, R(2L)3/48EI.From the principle of superposition the two deflections must be equal but opposite indirection. Equating them gives R = 5wL/4.

From equilibrium and symmetry the reactions at the outside supports are

w(2L) − 5wL4

2= 3wL

8.

S.13.22 Release the beam at A and B. The slope of the beam at its supports due tothe concentrated load W is obtained from Eqs (xii) and (xiii) in Ex. 13.6. Then

EI(

dv

dx

)A

= Wab(L + b)6L

, EI(

dv

dx

)B

= Wab(L + a)6L

Now apply a moment MA at A to the released beam. The slope of the beam at A is,from Eq. (iv) of Ex. 13.10, MAL/3 and at B is MA/6. Now apply a moment MB at B.Again, from Eq. (iv) of Ex. 13.10 the slope of the beam at A is MB/6 and at B is MB/3.Since the final slope at each end of the beam due to the applied load and the endmoments is zero

MAL3EI

+ MAL6EI

= Wab(L + b)6LEI

and

MBL6EI

+ MBL3EI

= Wab(L + a)6LEI

Solving

MA = Wab2

L2 , MB = Wa2bL2

Then, from equilibrium

RA = Wb2(L + 2a)L3 , RB = Wa2(L + 2b)

L3 .

S.13.23 The moment-area method may be used to solve this problem. For the caseof the simply supported beam and taking the origin of axes at the left hand supportEq. (13.10) becomes

vmid-span =∫ L/4

0

Wx2

2(

EI2

) dx +

∫ L/2

L/4

[Wx2

2EI

]dx


which gives

vmid-span = 3WL3

128EI

When the beam is built in at both ends the slope of the beam at both ends and atmid-span is zero. Suppose that the fixed-end moments are M.Then, from Eq. (13.7)

0 =∫ L/4

0

(2

EI

)(Wx2

− M)

dx +∫ L/2

L/4

(1

EI

)(Wx2

− M)

dx

from which

M = 5WL48

.

S.13.24 Using the moment-area method and referring to Fig. S.13.24, Eq. (13.10),with the origin for x at B, becomes

A CB

RB

0.75 m 0.75 m

30 kN

16 kN/m

FIGURE S.13.24

vB = 0.12 × 10−3RB =(

1EI

)[∫ 0.75

0

(RBx − 16x2

2

)x dx

+∫ 1.5

0.75

(RBx − 16x2

2− 30(x − 0.75)

)x dx

]

i.e.

EI × 0.12 × 10−3RB =[

RBx3

3− 2x4

]1.5

0

− 30

[x3

3− 0.75x2

2

]1.5

0.75

which gives

RB = 23.4 kN.



S.14.1 The stress system is shown in Fig. S.14.1.

A C

40 N/mm2

40 N/mm2

50 N/mm2

35 N/mm2

B

u

FIGURE S.14.1

From Fig. S.14.1, σx = 50 N/mm2, σy = −35 N/mm2 and τxy = 40 N/mm2. Substitutingthese values in Eqs (14.8) and (14.9) gives

σI = (50 − 35)2

+√

(50 + 35)2 + 4 × 402

2= 65.9 N/mm2

σII = (50 − 35)2

−√

(50 + 35)2 + 4 × 402

2= −50.9 N/mm2

The angles the principal planes make with the plane on which σx acts is given byEq. (14.7), i.e.

tan 2θ = −2 × 40(50 + 35)

= −0.9411

which gives

θ = −21.6◦ and −111.6◦.

Referring to Fig. S.14.1 suppose that AB is the plane on which there is no direct stressalthough there will be a shear stress. Resolving forces in a direction perpendicular toAB

0 = 50 BC cos θ − 35 AC sin θ − 40 BC sin θ − 40 AC cos θ

Dividing through by BC cos θ gives

0 = 50 − 35 tan2θ − 80 tan θ

Solving this quadratic equation in tan θ gives

θ = 27.1◦ or 117.1◦.


S.14.2 The stress system is shown in Fig. S.14.2.

u

Plane A

139 N/mm2

108 N/mm2

62 N/mm2

Plane B

C

D

F

sn

τ

s

E

FIGURE S.14.2

(a) Resolving forces horizontally

139 CD − 108 CE cos θ − 62 CE sin θ = 0

Dividing through by CD

139 − 108 − 62 tan θ = 0

i.e.

tan θ = 0.5

and

θ = 26.6◦

(b) Resolving forces vertically

σED − 108 CE sin θ + 62 CE cos θ = 0

Dividing through by ED

σ = 108 − 62 cot θ

from which

σ = 108 − 62(

10.5

)= −16 N/mm2

(c) Resolving forces perpendicular to the plane FD

σnFD − 139 CD sin θ + 62 CF = 0

i.e.

σn = 139 − 62 cot θ = 139 − 124 = 15 N/mm2.


S.14.3 The required direct stress may be found directly by substituting the given valuesin Eq. (14.5) in which the angle θ = 30◦. Then

σn = 100 cos230◦ − 80 sin230◦ − 45 sin 60◦

i.e.

σn = 16.0 N/mm2

The principal stresses are given by Eqs (14.8) and (14.9). Substituting the given values

σI = (100 − 80)2

+√

(100 + 80)2 + 4 × 452

2= 110.6 N/mm2

σII = (100 − 80)2

−√

(100 + 80)2 + 4 × 452

2= −90.6 N/mm2

The angles the principal planes make with the plane on which σx acts is given byEq. (14.7),

i.e.

tan 2θ = −2 × 45(100 + 80)

= −0.5

from which

θ = −13.3◦ and −103.3◦

The graphical solution is shown in Fig. S.14.3.

FIGURE S.14.3

τ (N/mm2)

s (N/mm2)

sI � 110.6 N/mm2sII � �90.6 N/mm2

sn � 16 N/mm2

(�80, 45)

(100, �45)

�26.6°

�206.6°

Q2

Q1

C

60°0

2020 40 60 80 100

�20

�20�40�60�80

�40

�60

�80

40

60

80


S.14.4 The stress system acting on a triangular element formed by the plane AB andthe vertical and horizontal sides of the rectangular element is shown in Fig. S.14.4.

Aτ

sn

sy

sx

τxy

τxy a

C B

FIGURE S.14.4

Resolving forces perpendicular to AB

σnAB − σxAC sin α − σyBC cos α − τxyAC cos α − τxyBC sin α = 0

i.e.

σn = σx sin2α + σy cos2α + τxy sin 2α (i)

Resolving forces parallel to AB

τAB + σxAC cos α − σyBC sin α + τxyBC cos α − τxyAC sin α = 0

i.e.

τ = −σx sin α cos α + σy cos α sin α − τxy(cos2α − sin2α)

or

τ = −(

σx − σy

2

)sin 2α − τxycos 2α (ii)

(i) Substituting the given values in Eqs (i) and (ii) in turn gives

σn = 52.3 N/mm2, τ = 7.9 N/mm2

(ii) Similarly for this case

σn = −58.3 N/mm2, τ = 7.9 N/mm2.

S.14.5 In Fig. 14.11 the direct stress coordinate of the point Q1 is given by

σx = σI − P1B = σI − CB + CP1

In this expression CB is the radius of the circle which is equal to τmax and

CP1 =√

(CQ21 − P1Q2

1) =√

(τ 2max − τ 2

xy). Therefore

σx = σI − τmax +√

(τ 2max − τ 2

xy)

Similarly

σy = σI − τmax −√

(τ 2max − τ 2

xy).


S.14.6 The direct stress due to bending is given by Eq. (9.9) and is

σx = 100 × 103 × 25[π(

504 − 404

64

)] = 13.8 N/mm2

The shear stress due to the applied torque is obtained from Eq. (11.4) and is

τxy = 50 × 103 × 25[π(

504 − 404

32

)] = 3.45 N/mm2

Substituting these values in Eqs (14.8) and (14.9) and noting that σy = 0 gives

σI = 14.6 N/mm2 and σII = −0.8 N/mm2

The maximum shear stress can be found from either Eq. (14.11) or Eq. (14.12). Usingthe latter

τmax = 14.6 + 0.82

= 7.7 N/mm2

Finally, from Eq. (14.7)

tan 2θ = −2 × 3.4513.8

= −0.5

i.e.

θ = −13.3◦ and −103.3◦

The ratio of these strain gauge readings is constant and equal to Poisson’s ratio.

S.14.7 The internal pressure produces a longitudinal direct stress given by Eq. (7.62)and a circumferential direct stress given by Eq. (7.63). Then

σL = σx = 0.7 × 12004 × 1.2

= 175 N/mm2

σC = σy = 0.7 × 12002 × 1.2

= 350 N/mm2

The cylinder is thin-walled so that the maximum shear stress is given by Eq. (11.22).Then

τmax = τxy = 500 × 106

2(

π × 12002

4

)× 1.2

= 184.2 N/mm2

Substituting these values in Eqs (14.8) and (14.9) gives

σI = 466.4 N/mm2 and σII = 58.6 N/mm2


τmax =(

466.4 − 58.62

)= 203.9 N/mm2.


S.14.8 The stress system is two-dimensional but acts on a three-dimensional piece ofmaterial. The strains are therefore given by Eqs (7.14) in which σz = 0. Then

εx =(

83 − 0.3 × 65200 000

)= 3.18 × 10−4

εy =(

65 − 0.3 × 83200 000

)= 2.01 × 10−4

εz =(−0.3 × 83 − 0.3 × 65

200 000

)= −2.22 × 10−4

From Eq. (14.25)

γmax = (3.18 − 2.01) × 10−4 = 1.17 × 10−4

From Eq. (7.21)

G = 200 0002(1 + 0.3)

= 200 0002 × 1.3

= 76 923.1 N/mm2


τmax = 76 923.1 × 1.17 × 10−4 = 9.0 N/mm2

in the plane of the stresses and acting on planes at 45◦ to their directions.

S.14.9 Since the strain gauge is positioned on the neutral axis of the beam sectionit will record strains produced by the axial load and shear load but no strains due tobending.

Poisson’s ratio is given by (see Section 7.8)

ν = εc

εa= 300 × 10−6

1000 × 10−6 = 0.3

Substituting the given values of strain in Eqs (14.34) and (14.35) gives

εI = 1046.4 × 10−6, εII = −346.4 × 10−6

Then, substituting these values in Eqs (14.32) and (14.33) and using the calculatedvalue of Poisson’s ratio

σI = 207.1 N/mm2, σII = −7.1 N/mm2

Now adding Eqs (14.8) and (14.9) and noting that σy = 0

σI + σII = σx = 200 N/mm2 = P100 × 200

Note that σx could have been obtained directly from the strain gauge readings since

σx = εaE = 1000 × 10−6 × 200 000 = 200 N/mm2


Then

P = 4000 × 103 N = 4000 kN

Again from Eqs (14.8) and (14.9)

σI − σII =√

(200.02 + 4τ 2xy) = 214.2

from which

τxy = 38.3 N/mm2

The shear stress on the neutral axis of a rectangular section beam is 3Sy/2BD (seeFig. 10.3).

In this case Sy = 2 × 103w, therefore

38.3 = 3 × 2 × 103w2 × 100 × 200

from which

w = 255.3 N/mm = 255.3 kN/m.

S.14.10 Substituting the given values of strain in Eqs (14.34) and (14.35) gives

εI = 1201.5 × 10−6, εII = −501.5 × 10−6

Substituting these values in Eqs (14.32) and (14.33) gives

σI = 231.0 N/mm2, σII = −31.0 N/mm2

Adding Eqs (14.8) and (14.9)

σI + σII = σx = 200 N/mm2

This stress is produced by the bending moment M. Therefore, from Eq. (9.9)

200 = M × 50

2[(

2×1003

12

)+ 2 × 50 × 502

]from which

M = 3333 N m

Subtracting Eqs (14.8) and (14.9)

σI − σII = 262.0 =√

(2002 + 4τ 2xy)

which gives

τxy = 84.6 N/mm2

This shear stress is caused by the applied torque and since the beam is a thin-walledbox section the shear stress is given by Eq. (11.22), i.e.

84.6 = T2 × 50 × 100 × 2


from which

T = 1692 N m.

S.14.11 Since the strain gauge is placed on the neutral axis of the beam section it willnot record any strains due to bending. Therefore σx = 7 N/mm2. The principal strainsare calculated from Eqs (14.34) and (14.35) and are

εI = 94 × 10−6, εII = −271 × 10−6

Then, from Eqs (14.32) and (14.33)

σI = 1.29 N/mm2, σII = −8.15 N/mm2

Subtracting Eqs (14.8) and (14.9)

σI − σII =√

(72 + 4τ 2xy) = 9.44

from which

τxy = 3.27 N/mm2

The shear force in the section of the beam where the strain gauge is positioned is equalto W and the corresponding shear stress is 3W/2BD (see Fig. 10.3). Therefore

3.27 = 3W2 × 150 × 300

which gives

W = 98.1 kN.

S.14.12 The yield stress of the material is given by

σY = 70 00020 × 10

= 350 N/mm2

The principal stresses are given by Eqs (14.8) and (14.9), i.e.

σI,II = (σx − 70)2

± 12

{√[(σx + 70)2 + 4 × 602]

}

An examination of this expression shows that σII is negative so that for the Trescatheory Eq. (14.42) applies. Then

σI − σII =√

[(σx + 70)2 + 4 × 602] = 350

from which

σx = 259 N/mm2


The expressions for principal stress are cumbersome when inserted in the von Misescriterion for elastic failure, Eq. (14.55) for the two-dimensional case. Thereforesimplify them into the form

σI = A + B, σII = A − B

Substituting in Eq. (14.55) and simplifying gives

A2 + 3B2 = σ 2Y

Then [(σx − 70)

2

]2

+ 3

[(σx + 70)2 + 4 × 602

4

]= 3502

Solving the resulting quadratic equation gives

σx = 294 N/mm2.

S.14.13 The loading on the beam is equivalent to a central concentrated load of40 kN together with a torque of 40 kN m applied at mid-span. The maximum bendingmoment on the beam is (40 × 103 × 3 × 103)/4 = 30 × 106 N mm. The direct stressdue to bending is given by Eq. (9.9), i.e.

σx =30 × 106

(D2

)(

πD4

64

) = 306 × 106

D3

From symmetry the reaction torques are equal and of magnitude 20 × 106 N mm whichis therefore the maximum torque on the beam. Then, from Eq. (11.4)

τxy =20 × 106

(D2

)(

πD4

32

) = 102 × 106

D3

Now substituting in Eqs (14.8) and (14.9) for σx and τxy gives

σI = 336.9 × 106

D3 , σII = −30.9D3

Since σII is negative Eq. (14.42) for the Tresca theory applies. Then

(336.9 + 30.9) × 106

D3 = 145

which gives

D = 136 mm


Substituting now in the two-dimensional case for the von Mises theory, Eq. (14.55)

(336.92 + 30.92 + 336.9 × 30.9)

(106

D3

)2

= 1452

from which

D = 135 mm.

S.14.14 The direct stress due to bending is given by Eq. (9.9) and is

σx =M(

D2

)(

πD4

64

) = 32MπD3

The shear stress due to torsion is, from Eq. (11.4)

τxy =T(

D2

)(

πD4

32

) = 32MπD3

Substituting in Eqs (14.8) and (14.9) gives

σI = 32M(

1.62πD3

), σII = 32M

(−0.62πD3

)

Then, from the Tresca theory, Eq. (14.42)(32MπD3

)(1.62 + 0.62) = 150 × 106

which gives, since D = 150 mm,

M = 22.2 kN m

Therefore

T = 44.4 kN m

Now substituting in the failure criterion for the von Mises theory, Eq. (14.55)

(32MπD3

)2

(1.622 + 0.622 + 1.62 × 0.62) = (150 × 106)2

so that

M = 24.9 kN m

and

T = 49.8 kN m.


S.14.15 Let the unit stress be σ then the actual stresses are

σI = 3σ , σII = 2σ , σIII = −1.8σ

From the Tresca theory, since σIII is negative

σI − σIII = σY

i.e.

3σ + 1.8σ = 387

Then

σ = 80.6 N/mm2

and

σI = 241.8 N/mm2, σII = 161.2 N/mm2, σIII = −145.1 N/mm2

From the von Mises theory, Eq. (14.54)

(3σ − 2σ )2 + (2σ + 1.8σ )2 + (−1.8σ − 3σ )2 = 2 × 3872

from which

σ = 88.0 N/mm2

Then

σI = 264.0 N/mm2, σII = 176.0 N/mm2, σIII = −158.4 N/mm2.

S.14.16 The compressive load P is equivalent to an axial compressive load P togetherwith a bending moment P × 100 N mm (see Section 9.2). Therefore the axial stress isgiven by

σA = P200 × 400

= 1.25 × 10−5P N/mm2

and the maximum direct stress due to bending is

σB = ±P × 100 × 200(200 × 4003

12

) = ±1.88 × 10−5P N/mm2

Therefore the maximum compressive stress in the column section is

σ (max. comp) = (1.25 + 1.88) × 10−5P N/mm2

This will be the least algebraic value of principal stress so that Eq. (14.58) applies and

3.13 × 10−5P = 22


i.e.

P = 702.9 kN

The maximum tensile stress in the column is

σ (max. tens) = (1.88 − 1.25) × 10−5P N/mm2

From Eq. (14.57)

0.63 × 10−5P = 4

so that

P = 634.9 kN

The maximum allowable value of P is therefore 634.9 kN.


S.15.1 Give the beam at D a virtual displacement δD as shown in Fig. S.15.1. Thevirtual displacements of C and B are then, respectively, 3δD/4 and δD/2.

A B C D

RD

dDdD34

dD

2

L2

L4

L4

FIGURE S.15.1

The equation of virtual work is then

RDδD − 2W δD

2− W 3δD

4= 0

from which

RD = 1.75W

It follows that

RA = 1.25W .


S.15.2 The beam is given a virtual displacement δC at C as shown in Fig. S.15.2.

Ax

B C

RC

dCdC34dC

xL

3L4

L4

FIGURE S.15.2

The virtual work equation is then

RCδC − W 3δC

4−∫ L

0w( x

L

)δC d x = 0

from which

RC = 3W + 2wL4

so that

RA = W + 2wL4

.

S.15.3 The beam is given a virtual rotation θA at A as shown in Fig. S.15.3.

B CA

MA

uAL2uA

LuA

L2

L2

FIGURE S.15.3

The virtual work equation is then

MAθA − WLθA

2− 2WLθA = 0

from which

MA = 2.5WL

and

RA = 3W .

S.15.4 Give the beam virtual rotations α and β at A and B, respectively as shown inFig. S.15.4. Then, at C, (3L/4)α = (L/4)β so that β = 3α.

The relative rotation of AB and BC at C is (α + β) so that the equation of virtual workis MC(α + β) = ∫ 3L/4

0 wαx d x + ∫ L3L/4 w3α(L − x) d x


ABC

x

x(L�x)

b (L�x)

baax

3L4

L4

FIGURE S.15.4

i.e.

4MCα = wα

[∫ 3L/4

0x d x + 3

∫ L

3L/4(L − x) d x

]

from which

MC = 3wL2

32.

S.15.5 Suppose initially that the portion GCD of the truss is given a small virtualrotation about C so that G moves a horizontal distance δG and D a vertical distanceδD as shown in Fig. S.15.5(a).

FIGURE S.15.5

C D DD

20 kN

dG

dD

dGD

dCD

dCD cos 45°

dGD/cos 45°

20 kN

C CB

(a) (b) (c)

F G G G

Then, since CG = CD, δG = δD and the equation of virtual work is

FGδG = 20δD

so that

FG = +20 kN

The virtual displacement given to G corresponds to an extension of FG which, sincethe calculated value of FG is positive, indicates that FG is tensile.

Now suppose that GD is given a small virtual increase in length δGD as shown inFig. S.15.5(b). The vertical displacement of D is then δGD/cos 45◦ and the equation ofvirtual work is

GDδGD = 20δGD/ cos 45◦


from which

GD = +28.3 kN (tension)

Finally suppose that CD is given a small virtual extension δCD as shown in Fig. S.15.5(c).The corresponding extension of GD is δCD cos 45◦. Then the equation of virtual workis, since the 20 kN load does no work

CDδCD + GDδCD cos 45◦ = 0

Substituting for GD from the above gives

CD = −20 kN (compression).

S.15.6 First determine the deflection at the quarter-span point B. Then, referring toFig. S.15.6, the bending moment due to the actual loading at any section is given by

AB C

11

w

D

L4

L4

L2

FIGURE S.15.6

MA = wLx2

− wx2

2= w(Lx − x2)

2

and due to the unit load placed at B is

M1 = 3x4

in AB and M1 = (L − x)4

in BD

Then substituting in Eq. (iii) of Ex. 15.9

vB = w8EI

[∫ L/4

03(Lx2 − x3) d x +

∫ L

L/4(Lx − x2)(L − x) d x

]

which gives

vB = 57wL4

6144EI

For the deflection at the mid-span point the bending moment at any section due tothe actual loading is identical to the expression above. With the unit load applied at C

M1 = x2

in AC and M1 = (L − x)2

in CD


Substituting in Eq. (iii) of Ex. 15.9

vC = w4EI

[∫ L/2

0(Lx2 − x3) d x +

∫ L

L/2(Lx − x2)(L − x) d x

]

from which

vC = 5wL4

384EIwhich is identical to the result of Ex. 13.4.

S.15.7 Take the origin for x at the built in end A of the beam. Then, in AB the bendingmoment at any section due to the actual loading is given by

MA = −w(a − x)2

2and in BC, MA = 0

With a unit load applied vertically downwards at C

M1 = −1(L − x) at any section between A and C

Then, substituting these expressions in Eq. (iii) of Ex. 15.9 and noting that MA = 0between B and C

vC = w2EI

∫ a

0(a − x)2(L − x) d x

from which

vC = wa3(4L − a)24EI

which is identical to the result of Ex. 13.3.

S.15.8 Suppose the origin of x is at C. Then

MA = −Wx and M1 = −1x

From Eq. (iii) of Ex. (15.9)

vC = WEI

[∫ L/2

02x2 d x +

∫ L

L/2x2 d x

]

which gives

vC = 3WL3

8EI.

S.15.9 The calculation of deflections of trusses using the unit load method is illustratedin Ex. 15.7 in which the deflection at any joint is given by Eq. (viii). The method of


joints may be used to calculate the forces in the truss due to the applied load and theunit loads; the calculation is completed in tabular form below.

Member L(m) FA (kN) F1,V F1,H FAF1,VL FAF1,HL

AB 1 400 2.0 0 800 0BC 1.4 282.8 1.4 0 554.3 0CD 1 −200 −1.0 1 200 −200DE 1 −200 −1.0 1 200 −200BD 1 0 0 0 0 0BE 1.4 −282.8 −1.4 0 554.3 0∑ = 2308.6

∑ = −400

Therefore

δC,V = 2308.6 × 106

500 × 200 000= 23.1 mm (downwards)

δC,H = −400 × 106

500 × 200 000= −4.0 mm (to the left)

The resultant deflection is then

δ =√

(23.12 + 4.02) = 23.4 mm at tan−1(

4.023.1

)= 9.8◦ to the left of vertical.

S.15.10 The solution for this problem is completed in an identical manner to that forP.15.9 except that the cross-sectional area of each member depends on whether themember is in tension or compression and will therefore need to be included in thetabular solution as shown below.

Member L(m) FA (kN) F1,V F1,H A (mm2) FAF1,VL/A FAF1,HL/A

AB 1.4 282.8 1.4 0 750 0.74 0BC 1.4 −282.8 −1.4 0 1000 0.55 0AC 2 −200 −1 1 1000 0.40 −0.4BD 2 400 2 0 750 2.13 0CD 1.4 −159.1 −0.7 0 1000 0.16 0CF 2 −287.5 −1.5 1 1000 0.86 −0.58DG 2 175 1 0 750 0.47 0DF 1.4 159.1 0.7 0 750 0.21 0FG 1.4 −159.1 −0.7 0 1000 0.16 0FH 2 −62.5 −0.5 1 1000 0.06 −0.125GH 1.4 159.1 0.7 0 750 0.21 0∑ = 5.95

∑ = −1.105


Therefore, from Eq. (viii) of Ex. 15.7

δV = 5.95 × 106

200 000= 29.8 mm (downwards)

δH = −1.105 × 106

200 000= −5.5 mm (to the right)

The resultant deflection is then given by

δ =√

(29.82 + 5.52) = 30.3 mm at tan−1(

5.529.8

)= 10.5◦ to the right of vertical.

S.15.11 Fig. S.15.11 shows a plan view of the plate.

B

A

Dx

x

x

C FIGURE S.15.11

Suppose that the point of application of the load is at D, a distance x from each edgeof the plate. The unit load method may be used to determine the deflection of D but,initially, the forces FA must be calculated. From equlibrium

FA,A + FA,B + FA,C = 100 (i)

Taking moments about the edges of the plate in turn

4FA,A = 100x

FA,B × 4 sin A = FA,B × 4 × 0.6 = 100x

FA,C × 3 = 100x

Then

4FA,A = 2.4FA,B = 3FA,C

Therefore

FA,A = 0.6FA,B, FA,C = 0.8FA,B


Substituting in Eq. (i) gives

FA,B = 41.7 N

Then

FA,A = 25.0 N, FA,C = 33.4 N

Applying a unit load at D in the direction of the 100 N load

F1,A = 0.25, F1,B = 0.417, F1,C = 0.334

Then, substituting in Eq. (viii) of Ex. 15.7

δD = 1440(25 × 0.25 + 41.7 × 0.417 + 33.4 × 0.334)(π4

)× 12 × 196 000= 0.33 mm.

S.15.12 Suppose that the joints 2 and 7 have horizontal and vertical components ofdisplacement u2,v2, u7 and v7 respectively as shown in Fig. S.15.12.

3a

7

7�

a

2�

2

u2

�2

�7

u7 FIGURE S.15.12

The angle α which the member makes with the vertical is then given by

α = tan−1(

u7 − u2

3a + v7 − v2

)

which, since α is small and v7 and v2 are small compared with 3a may be written

α = u7 − u2

3a

The unit load method will be used to obtain the required deflections. Note that theparts of the truss 01234 and 56789 are identical.


Member L FA F1,2 F1,7 FAF1,2L FAF1,7L

27 3a 3P 0 0 0 087 5a 5P/3 0 5/3 0 125Pa/967 4a −4P/3 0 −4/3 0 64Pa/921 4a 4P −4/3 0 −64Pa/3 023 5a 0 5/3 0 0 026 5a −5P 0 0 0 038 3a 0 0 0 0 058 5a 0 0 0 0 098 5a 5P/3 0 5/3 0 125Pa/968 3a 0 0 0 0 016 3a 3P 0 0 0 056 4a −16P/3 0 −4/3 0 256Pa/913 3a 0 0 0 0 043 5a 0 5/3 0 0 093 (

√34)a 0 0 0 0 0

03 5a 0 0 0 0 015 5a −5P 0 0 0 010 4a 8P −4/3 0 −128Pa/3 0∑ = −192Pa/3

∑ = 570Pa/9

Therefore

u2 = −192Pa3AE

, u7 = 570Pa9AE

so that

α = 382P9AE

.

S.15.13 Using the unit load method, Eq. (iii) of Ex. 15.9 applies. Referring toFig. S.15.13

B

x

R

C

W

A

4R

u

FIGURE S.15.13

in CB

MA = W (R − R cos θ), M1 = −1 × R sin θ


in BA

MA = W 2R, M1 = 1x

Then

δC,H = 1EI

[∫ π

0−WR3(1 − cos θ) sin θ dθ +

∫ 4R

02WR x d x

]

which gives

δC,H = 14WR3

EI

where

I = π(1004 − 944)64

= 1.076 × 106 mm4

Substituting the values of I, W and R

δC,H = 53.3 mm.

S.15.14 In Eq. (15.42) t is the difference in temperature between the top and bottomsurfaces of the beam. In this case the temperature difference is (t2 − t1) so that Eq.(15.44) becomes

δ = −∫ L

0M1

[α(t2 − t1)

h

]d x

For a unit load applied at the free end of the cantilever and taking the origin of x atthe free end M1 = −1x so that

δ =∫ L

0x[α(t2 − t1)

h

]d x

from which

δ = α(t2 − t1)L2

2h.

S.15.15 The temperature at any section a distance x from the left hand support is givenby t = (x/L)t0. With unit load applied at the mid-span point of the beam Eq. (15.44)becomes

δ = −∫ L/2

0

(

12

)xα( x

L

)to

h

d x

δ = −αL2t048h

.


S.15.16 The frame is symmetrical about a vertical plane through its centre so that onlyhalf need be considered. Also, from symmetry the frame will act as though fixed at C(Fig. S.15.16).

B

30°

1HB

C

r

u

FIGURE S.15.16

If B is released the displacement at B due to the temperature rise is obtained fromEq. (15.44) in which

M1 = 1(r sin 30◦ + r sin θ) = r(0.5 + sin θ)

Then

δB,T =∫ π/2

−π/6r(0.5 + sin θ)

(2αT

d

)r dθ

which gives

δB,T = 3.83αTr2

d

The displacement at B due to the horizontal reaction HB at B is obtained using theunit load method. From Eq. (iii) of Ex. 15.9 in which

MA = −HBr(0.5 + sin θ) and M1 is given above

δB,H = −(

HB

EI

)∫ π/2

−π/6(0.5 + sin θ)2 r3dθ

from which

δB,H = −2.22HB r3

EI

The sum of the displacement of B due to the temperature rise and the displacementof B due to the support reaction is zero since B is a support point. Then

3.83αTr2

d− 2.22HBr3

EI= 0

from which

HB = 1.73EITα

dr


The maximum bending moment occurs at C and is HB × 1.5r = 2.6EITα/d. Then, fromEq. (9.9), the maximum direct stress is

σmax = 2.6EITα × 0.5dId

= 1.3ETα.

S.15.17 This problem may be solved using either complementary energy or potentialenergy. The strain energy U due to the bending of the beam is given by Eq. (9.21), i.e.U = ∫ L

0 (M2/2EI) dx

and the potential energy of the applied load is, from Ex. 15.11

V = −WvB

The total potential energy is then

TPE = U + V =∫ L

0

(M2

2EI

)d x − WvB

which, from Eq. (13.3) may be written

TPE = EI2

∫ L

0

(d2v

dx2

)2

d x − WvB

Substituting for d2v/dx2 from the given deflected shape function

TPE = EI2

∫ L

0

[(π2v1

L2

)sin(πx

L

)+(

9π2v3

L2

)sin(

3πxL

)]2

d x − WvB

or

TPE =(

π4EI2L4

)[v2

1

∫ L

0sin2

(πxL

)d x + 18v1v3

∫ L

0sin(πx

L

)sin(

3πxL

)d x

+81v23

∫ L

0sin2

(3πxL

)d x

]

Using the values of the integrals given

TPE =(

π4EI4L3

)(v2

1 + 81v23) − WvB

From the given deflected shape function, at mid-span when x = L/2

vB = v1 − v3

so that

TPE =(

π4EI4L3

)(v2

1 + 81v23) − W (v1 − v3)


Then, from the principle of the stationary value of the total potential energy

∂(TPE)∂v1

=(

π4EI2L3

)v1 − W = 0

∂(TPE)∂v3

=(

81π4EI2L3

)v3 + W = 0

Solving gives

v1 = 2WL3

π4EI, v3 = −2WL3

81π4EI

so that

vB =(

2WL3

π4EI

)(1 + 1

81

)

i.e.

vB = 0.02078WL3

EI

Note that Eq. (v) of Ex. 13.5 gives vB = 0.02083WL3/EI.

S.15.18 The method of solution used in S.15.17 may be used for this problem. Thestrain energy of the beam is, from Eqs (9.21) and (13.3)

U = 2(

EI2

)∫ L/4

0

(d2v

d x2

)2

dx + 2(

2EI2

)∫ L/2

L/4

(d2v

d x2

)2

d x

which gives, when the equation for the assumed displaced shape is substituted

U = 144EIv2m

L3

The total potential energy is then

TPE = 144EIv2m

L3 − Wvm

From the principle of the stationary value of the total potential energy

∂(TPE)∂vm

= 288EIvm

L3 − W = 0

from which

vm = 0.00347WL3

EI.


S.15.19 From Maxwell’s reciprocal theorem the deflection at A due to W at B is equalto the deflection at B due to W at A, i.e. δ2. What is now required is the deflection atB due to W at B.

Since the deflection at A with W at A and the spring removed is δ3 the load in thespring at A with W at B is (δ2/δ3)W which must equal the load in the spring at B withW at B. Therefore the resultant load at B with W at B is

W −(

δ2

δ3

)W = W

(1 − δ2

δ3

)(i)

The load W at A with the spring in place produces a deflection of δ1 at A. Thereforethe resultant load at A is (δ1/δ3)W so that, if the load in the spring at A with W at Ais F, then W − F = (δ1/δ3)W , i.e.

F = W(

1 − δ1

δ3

)(ii)

This then is the load at B with W at A and it produces a deflection δ2. Therefore, fromEqs (i) and (ii) the deflection at B due to W at B is

W

(1 − δ2

δ3

)W(

1 − δ1δ3

) δ2

and the extension of the spring with W at B is

(

1 − δ2δ3

)(

1 − δ1δ3

) δ2 − δ2

i.e.

δ2

[(δ1 − δ2)(δ3 − δ1)

].


AC

RA RB

0.36 m 0.72 m 0.72 m 0.6 m

F B D

2 kN

x

FIGURE S.15.20


the support reactions are each equal to 1 kN from symmetry. The slope of the beamat the supports is given by Eq. (iii) of Ex. 13.5 and is

dv

d x= ±2 × 103 × (1.44 × 103)2

16 × 240 × 108 = ±0.011

The deflection at C is then 0.011 × 0.36 × 103 = 3.96 mm and the deflection at D is0.011 × 0.6 × 103 = 6.6 mm.

From the reciprocal theorem the deflection at F due to a load of 3 kN atC = 3.96 × 3.0/2.0 = 5.94 mm and the deflection at F due to a load of 3 kN atD = 6.6 × 3.0/2.0 = 9.9 mm. Therefore the total deflection at F due to loads of 3 kNacting simultaneously at C and D is 5.94 + 9.9 = 15.84 mm.


S.16.1 The completely stiff structure for each of the structures shown in Fig. P.16.1 isshown in Fig. S.16.1.

(e)

(a)

(d)

(b) (c)

FIGURE S.16.1

The frames are all plane frames so that Eq. (16.3) for the degree of staticalindeterminacy applies.

(a) In this case M = 4, N = 4, r = 0 so that ns = 3. Also nk = 4 × 3 − 2 × 3 = 6.(b) M = 2, N = 2, r = 2 and ns = 1. nk = 2 × 3 − 2 × 2 = 2.(c) M = 3, N = 2, r = 4 and ns = 2. nk = 2 × 3 − 2 × 1 = 4.(d) M = 9, N = 8, r = 0 and ns = 6. nk = 8 × 3 − 3 × 3 = 15.(e) M = 6, N = 4, r = 7 and ns = 2. nk = 4 × 3 − 5 = 7.


S.16.2 The completely stiff structures are shown in Fig. S.16.2. Also, since thestructures are space frames Eq. (16.2) for the degree of statical indeterminacy applies.

FIGURE S.16.2 (b)(a) (c)

(a) M = 6, N = 6, r = 0 and ns = 6. nk = 6 × 6 − 2 × 6 = 24.(b) M = 18, N = 12, r = 0 and ns = 42. nk = 12 × 6 − 6 × 6 = 36.(c) M = 6, N = 4, r = 0 and ns = 18. nk = 4 × 6 − 3 × 6 = 6.

S.16.3 The method used to solve this problem is identical to the alternative solution ofEx. 16.5 except that there are two indeterminacies which may, but not necessarily, beselected as the support reactions at B and C. The beam is released at B and C and thedeflections at B and C calculated by any convenient method, say Macauley’s method.This gives

EIvC = −266.8 and EIvB = −109.3

A unit load is now applied at B and the deflections at B and C calculated again by, say,Macauley’s method. This gives

EIvC = −10.4 and EIvB = −4.6

A unit load is now applied at C and the deflections again calculated.

EIvC = −28.4 and EIvB = −10.4

Note that in the above the deflection at C due to unit load at B is equal to the deflectionat B due to unit load at C.

Since the resultant deflections at B and C are zero

−109.3 + 4.6RB + 10.4RC = 0

−266.8 + 10.4RB + 28.4RC = 0

Solving gives

RB = 14.7 kN, RC = 4.0 kN

Then from the vertical equilibrium of the beam

RA = 3.3 kN


and from the moment equilibrium of the beam about A

MA = 2.2 kN m.

S.16.4 For the beam of Fig. P.16.4 there are three indeterminacies. Suppose these arethe bending moment and vertical reaction at the end C and the vertical reaction atB; therefore release the beam at B and C. One of the releases is a moment so that inaddition to displacements an angle of rotation or slope of the beam is required. Againapplying Macauley’s method to the released beam

EIvB = −12 013.6, EIvC = −63 089.3, EI(

dv

dx

)C

= −2232.0

Now apply a unit load at B. Macauley’s method gives

EIvB = −576, EIvC = −2304, EI(

dv

dx

)C

= −72

With a unit load at C

EIvB = −2304, EIvC = −15 552, EI(

dv

dx

)C

= −648

Finally, apply a unit moment at C

EIvB = −72, EIvC = −648, EI(

dv

dx

)C

= −36

Since there is no deflection at B or C and no rotation at C

−12 013.1 + 576RB + 2304RC + 72MC = 0

−63 089.3 + 2304RB + 15 552RC + 648MC = 0

−2232.0 + 72RB + 648RC + 36MC = 0

Solving gives

MC = 19.0 kN m (hogging), RB = 9.0 kN, RC = 3.5 kN

Then, from equilibrium

RA = 3.5 kN, MA = 7 kN m (hogging).

S.16.5 The beam of Fig. P.16.5 has a degree of statical indeterminacy of 2. Releasethe beam at B and C. Then, using Macauley’s method

EIvB = −29 445.7, EIvC = −27 448.9

Applying unit load at B

EIvB = −985.9, EIvC = −808.5


Applying unit load at C

EIvB = −825.3, EIvC = −867.0

Then, since the deflections at B and C are zero

−29 445.7 + 985.9RB + 825.3RC = 0

−27 448.9 + 808.5RB + 867.0RC = 0

Solving gives

RB = 15.0 kN, RC = 17.8 kN

From moment equilibrium about D

RA = 4.3 kN, then RD = 15.9 kN.

S.16.6 The method illustrated in Ex. 16.8 will be used to solve this problem. Firstrelease the member CD. The forces F0 in the released structure are calculated andthen the forces F1 due to a unit load applied at C and D in the direction CD determined.The calculation is completed in tabular form as shown:

Member L (mm) A (mm2) F0(N) F1

F0F1LA

F21LA F (N)

AC 1385.6 30 50 −0.87 −2009.1 35.0 48.2CB 800 20 86.6 0.5 1732.0 10.0 87.6BD 1385.6 30 0 −0.87 0 35.0 −1.8CD 800 20 0 1.0 0 40.0 2.1AD 800 20 0 0.5 0 10.0 1.1∑ = −277.1

∑ = 130.0

From Eq. (ii) of Ex. 16.8

−277.1 + 130.0XCD = 0

from which

XCD = 2.1 N

Having obtained the force in CD the remaining forces are calculated by the methodof joints and are shown in the final column of the table.

S.16.7 In this problem the truss has a degree of statical indeterminacy equal to 2 sothat two releases are required. Release BG and GD and suppose that F0 are the forcesin the released structure and that F1 and F2 are the forces in the released structuredue to unit loads applied in the directions BG and GD, respectively. In the table belowthe length L is in m and the area of cross-section A in mm2.


Member L A F0 (kN) F1 F2

F0F1LA

F0F2LA

F21 LA

F22 LA

F1F2LA

AC 2.83 300 −70.7 1 0 −0.67 0 0.009 0 0CG 2 100 100 −0.707 −0.707 −1.4 −1.4 0.01 0.01 0.01CF 2.83 300 −70.7 0 1 0 −0.67 0 0.009 0AG 2 200 50 −0.707 0 −0.35 0 0.005 0 0GF 2 200 50 0 −0.707 0 −0.35 0 0.005 0AB 2 100 0 −0.707 0 0 0 0.01 0 0BG 2.83 300 0 1 0 0 0 0.009 0 0BC 2 200 0 −0.707 0 0 0 0.005 0 0CD 2 200 0 0 −0.707 0 0 0 0.005 0GD 2.83 300 0 0 1 0 0 0 0.009 0DF 2 100 0 0 −0.707 0 0 0 0.01 0

∑ = −2.42∑ = −2.42

∑ = 0.048∑ = 0.048

∑ = 0.01

Then

0.048XBG + 0.01XGD − 2.42 = 0

0.01XBG + 0.048XGD − 2.42 = 0

Solving

XBG = XGD = 41.8 kN

Then

AB = FD = −29.2 kN, BC = CD = −29.2 kN, AG = GF = 20.8 kN

BG = DG = 41.3 kN, AC = FC = −29.4 kN, CG = 41.6 kN.

S.16.8 In this problem it is the support system which is statically indeterminate.Therefore release the truss horizontally at F. Then

Member L (m) F0 (kN) F1 F0F1L F21L F (kN)

AB 5 −50 0.25 −62.5 0.3125 −32.48AD 5.6 0 −1.12 0 6.99 −78.31BC 5 −100 0.5 −250 1.25 −64.98BD 5.6 111.8 −0.56 −349.39 1.747 72.65CD 2.5 −100 0 0 0 −100.0CE 5 −100 0.5 −250 1.25 −64.98DE 5.6 111.8 −0.56 −349.39 1.747 72.65DF 5 0 −1 0 5.0 −70.06EF 2.5 −50 0.25 −31.25 0.1563 −32.49∑ = −1292.53

∑ = 18.45


Then

−1292.53 + 18.45RF,H = 0

from which

RF,H = 70.06 kN = RA,H

and

RA,V = 50 + 0.25 × 70.06 = 67.52 kN

so that, from equilibrium

RF,V = 32.48 kN

The actual forces F in the members of the truss are calculated as, for example for ABas follows

F = −50 + 0.25 × 70.06 = −32.48 kN

and are given in the final column of the above table.

S.16.9 Referring to Figs 16.9(a) and (b) it can be seen that the members 12,24 and 23 remain unloaded until P has moved through a horizontal distance0.25 cos α = 0.25 × (600/750) = 0.2 mm. Therefore until P has moved through a hor-izontal distance of 0.2 mm P is equilibrated solely by the forces in the members 13, 34and 41 which then form a triangular framework. The method of solution is to find thevalue of P which produces a horizontal displacement of 0.2 mm of joint 1. Using theunit load method:

Member L (mm) F0 (N) F1 F0F1L F (N)

13 750 1.25P 1.25 1171.9P 3683.814 450 −0.75P −0.75 253.1P −2210.343 600 0 0 0 0∑ = 1425.0P

Then

0.2 = 1425.0P300 × 70 000

which gives

P = 2947 N

The forces F are then as shown in the final column of the above table.

When P = 10 000 N additional forces are generated in these members correspondingto a load of P′ = 10 000 − 2947 = 7053 N which will now produce forces in the members


12, 24 and 23 of the frame. The solution is now completed using the unit load methodin which 24 is chosen as the released member.

Member L (mm) F0 (N) F1 F0F1L F21L F (N)

12 600 0 −0.8 0 384 2481.623 450 0 −0.6 0 162 1861.234 600 0 −0.8 0 384 2481.641 450 −0.75P′ −0.6 202.5P′ 162 −5638.913 750 1.25P′ 1 937.5P′ 750 9398.124 750 0 1 0 750 −3102.0∑ = 1140P′ ∑ = 2592

Then

1140 × 7053 + 2592X24 = 0

which gives

X24 = −3102.0 N

The final forces F in the frame are shown in the last column of the above table andare calculated, typically for member 41, as follows:

F41 = −0.6 × (−3102) − 0.75 × 7053 − 2210.3 = −5638.9 N.

S.16.10 The lengths of the members not given in Fig. P.16.10 are:

L12 = (9√

2)a, L13 = 15a, L14 = 13a, L24 = 5a

The force in the member 14 due to the temperature change is compressive and equalto 0.7 A. Also the change in length �14 of the member 14 due to a temperature changeT is L14αT = 13a × 2.4 × 10−6T . This must also be equal to the change in lengthproduced by the force in the member corresponding to the temperature rise. Supposethis force is X . The unit load method may be used to complete the solution in which,since X and the unit load are applied at the same points, in the same direction andno other loads are applied when only the temperature change is being considered,F0 = XF1 and Eq. (i) of Ex. 16.8 becomes

�14 = X∑(

F21 L

AE

)

The method of joints may be used to calculate the forces F1 in the members which areF14 = 1, F13 = −35/13, F12 = (16

√2)/13, F24 = −20/13, F23 = 28/13.


Substituting in the above equation for �14 and remembering that the cross-sectionalarea of member 12 is 1.414A

�14 = 29 532aX132AE

i.e.

13a × 24 × 10−6T = 29 532a(0.7A)132AE

from which

T = 5.5◦.

S.16.11 Release the truss at the support G. Then, using the unit load method

Member L (m) F1 F21L F (kN)

GF 1 2/√

3 4/3 1073.9GH 1 −1/

√3 1/3 −536.9

FH 1 −2/√

3 4/3 −1073.9FD 1 2/

√3 4/3 1073.9

JH 1 −3/√

3 9/3 −1610.8HD 1 2/

√3 4/3 1073.9

DC 1 4/√

3 16/3 2147.7CJ 1 2/

√3 4/3 1073.9

JA 1 −5/√

3 25/3 −2684.6AC 1 −2/

√3 4/3 −1073.9

JD 1 −2/√

3 4/3 −1073.9BC 0.5 6/

√3 18/3 3221.6∑ = 97/3

Suppose the vertical reaction at G is RV. Then

RV × 973AE

= 15 × 10−3

i.e.

RV = 15 × 10−3 × 3 × 2 × 109 × 10−3

97= 927.8 kN

The forces F in the members are then obtained using the method of joints and aregiven in the final column of the above table.

S.16.12 The internal force system in the framework and beam is statically determi-nate so that the unit load method may be used directly to determine the verticaldisplacement of D.


D

3wa

x2 x1

1.5w/unit length

C

B

E F G

A

RA,H

RA,V

RG,H

RG,V

3a

4a4a4a

FIGURE S.16.12

Referring to Fig. S.16.12 and taking moments about A

RG,H3a − 1.5w(8a)2

2− 3wa × 12a = 0

which gives

RG,H = 28wa

Therefore

RA,H = −28wa

From the vertical equilibrium of the support G, RG,V = 0 so that, resolving vertically

RA,V − 1.5w8a − 3wa = 0

i.e.

RA,V = 15wa

With unit vertical load at D

RG,H = 4, RA,H = −4, RA,V = 1, RG,V = 0

For the beam ABC in AB

MA = RA,Vx1 − 1.5wx21

2= 15wax1 − 0.75wx2

1, M1 = 1x1

and in BC

MA = −FCEx2 − FCF × 35

x2 − 1.5wx22

2

Substituting for FCE and FCF from the Table on P. 173

MA = 15wax2 − 0.75wx22, Similarly M1 = 1x2

Then∫L

(MAM1

EI

)dx = 16

Aa2E

[∫ 4a

0(15wax2

1 − 0.75wx31) dx +

∫ 4a

0(15wax2

2 − 0.75wx32) dx

]


i.e. ∫L

(MAM1

EI

)dx = 8704wa2

AE

Also

Member L A FA F1 FAF1L/A

AB 4a 4A 28wa 4 112wa2/ABC 4a 4A 28wa 4 112wa2/ACD 4a A 4wa 4/3 64wa2/3ADE 5a A −5wa −5/3 125wa2/3AEF 4a A −4wa −4/3 64wa2/3AFG 4a A −28wa −4 448wa2/ACE 3a A 3wa 1 9wa2/ACF 5a A −30wa −10/3 500wa2/ABF 3a A 18wa 2 108wa2/A∑ = 4120wa2/3A

Then

�D = wa2[8704 + (4120/3)]AE

i.e.

�D = 30 232wa2

3AE.

S.16.13 This problem may be solved using strain energy and Castigliano’s theorem(Eq. (15.36)). Suppose that the compressive load in the king post is R. Then there willbe tensile loads in the members AD and DC equal to (

√5)R/2. Further there will be

a resulting compressive load in the beam AC equal to R. Therefore, from Eq. (7.29)the strain energy due to the axial loads is given by

UA =R2[

4×103

5000 + 1×103

2000 + (5√

5)×103

2×200

]2E

= 14.63R2

EThe bending moment at any section a distance x from A in AB is given by

M = (50 × 103 − 0.5R)x

Therefore, from Eq. (9.21)

UB = 22EI

∫ 2000

0

[(50 × 103 − 0.5R)2

4

]x2 dx = (100 × 103 − R)2 × 43 × 109

96EI

The total strain energy is then

UA + UB = U = 14.63R2

E+ (100 × 103 − R)2 × 158.7

E


From Castigliano’s theorem

∂U∂R

= 0,

therefore

0 = 29.26R − 2(100 × 103 − R) × 158.7

from which

R = 91.6 kN.

S.16.14 Referring to Fig. S.16.14(a) and taking moments about D

x

w

ARA,H

RA,V

x x

D

RD,V

RD,H

L

B

2EI 2EI

CEI

2L3

FIGURE S.16.14a

RA,V

(2L3

)+(

wL2

)(2L3

)= 0

i.e.

RA,V = −wL2

Therefore

RD,V = wL2

For horizontal equilibrium

RA,H + wL2

= RD,H (i)

The total complementary energy of the frame is (see Eq. (i) of Ex. 15.9) given by

C =∫

L

∫ M

0dθdM +

∫ L

0w′�dx (ii)

Note that in Eq. (ii) the complementary energy due to the support reactions is zerosince the support displacements are all zero. Also w′ is the load intensity at any section


in AB a distance x from A where the horizontal displacement is �. From the principleof the stationary value of the total complementary energy and selecting the horizontalreaction at A as the variable

∂C∂RA,H

=∫

Ldθ

(∂M

∂RA,H

)= 0

or, from Eq. (9.19) ∫L

(MEI

)(∂M

∂RA,H

)dx = 0 (iii)

In AB

M = −RA,Hx − wx3

6L,

∂M∂RA,H

= −x

In BC

M = RA,Vx − RA,HL − wL2

6,

∂M∂RA,H

= −L

In DC

M = −RD,Hx = −(

RA,H + wL2

)x from Eq. (i),

∂M∂RA,H

= −x

Substituting in Eq. (iii)

∫ L

0

(1

2EI

)(−RA,Hx − wx3

6L

)(−x)dx+

∫ 2L/3

0

(1

EI

)(−wLx

2− RA,HL − wL2

6

)(−L)dx

+∫ L

0

(1

2EI

)(−RA,H − wL

2

)x(−x)dx = 0

from which

2RA,HL3 +(

2945

)wL4 = 0

so that

RA,H = −29wL90

Therefore, from Eq. (i)

RD,H = 8wL45

Then

MAB = 29wLx90

− wx3

6L

When x = 0, MAB = 0 and when x = L, MAB = 7wL2

45. Also

dMAB

dx= 0 = 29wL

90− 3wx2

6L


Therefore MAB is a maximum when x =(√

29/45)

L and MAB(max) = 0.173wL2.The bending moment distributions in BC and CD are linear and are shown inFig. S.16.14(b).

A

B

0.173wL2

C

D

2945 L

7wL2

45

7wL2

45

8wL2

45

8wL2

45

FIGURE S.16.14b

S.16.15 The frame of Fig. P.16.15 has three indeterminacies, the horizontal and ver-tical reactions at, say, the support D and the moment reaction at D as shown inFig. S.16.15(a); equally the reactions at the support A could be regarded as the inde-terminacies. The frame is therefore released at D and unit loads and a unit momentapplied as shown in Fig. S.16.15(b).

FIGURE S.16.15 (a) (b)

4 kN

2 kN 2EI

EI

A

C

D

10 m

RD,H

RD,V

MD

B

5 m 10 m

EI

A

C

D

B

I

I

I

It should be noted that the flexibility method of solution for this problem is lengthyand the most practicable approach is the moment distribution method which is usedto solve the same problem in Ex. 16.22.

The moments produced in the released frame are:

In DC,

M0 = 0, M1,H = 1x, M1,V = 0, M1,M = −1


In CF,

M0 = 0, M1,H = 1 × 10, M1,V = −1x, M1,M = −1

In FB,

M0 = 4(x − 10), M1,H = 1 × 10, M1,V = −1x, M1,M = −1

In BA,

M0 = 4 × 5 + 2x = 2(10 + x), M1,H = 1(10 − x), M1,V = −1 × 15, M1,M = −1

Then

∫ (M0M1,H

EI

)dx =

∫ 15

1040[

(x − 10)(2EI)

]dx +

∫ 10

020[

(10 + x)(10 − x)EI

]dx = 1583.3

EI∫ (M0M1,V

EI

)dx =

∫ 15

10−[

40(x − 10)x2EI

]dx +

∫ 10

0−[

15 × 2(10 + x)EI

]dx = −4833.3

EI∫ (M0M1,M

EI

)dx =

∫ 15

10−[

4(x − 10)2EI

]dx +

∫ 10

0−[

2(10 + x)EI

]dx = −325

EI

∫ (M21,H

EI

)dx =

∫ 10

0

(x2

EI

)dx +

∫ 15

0

(102

2EI

)dx +

∫ 10

0

[(10 − x)2

EI

]dx = 1416.7

EI

∫ (M21,V

EI

)dx =

∫ 15

0

(x2

2EI

)dx +

∫ 10

0

(152

EI

)dx = 2812.5

EI

∫ (M21,M

EI

)dx =

∫ 10

0

(12

EI

)dx +

∫ 15

0

(12

EI

)dx +

∫ 10

0

(12

EI

)dx = 27.5

EI

∫ (M1,VM1,H

EI

)dx =

∫ 15

0

(−10x2EI

)dx +

∫ 10

0

[−15(10 − x)EI

]dx = −1312.5

EI∫ (M1,HM1,M

EI

)dx =

∫ 10

0

(−xEI

)dx +

∫ 15

0

(−102EI

)dx +

∫ 10

0

[−(10 − x)EI

)dx = −175

EI∫ (M1,VM1,M

EI

)dx =

∫ 15

0

( x2EI

)dx +

∫ 10

0

(15EI

)dx = 206.3

EI

Then

1583.3 + 1416.7RD,H − 1312.5RD,V − 175MD = 0

−4833.3 − 1312.5RD,H + 2812.5RD,V + 206.3MD = 0

−325 − 175RD,H + 206.3RD,V + 27.5MD = 0

Solving gives RD,V = 1.89 kN, RD,H = 1.63 kN, MD = 8.02 kN m (anticlockwise)


Then, MCD = 8.28 kN m (anticlockwise), MAB = 3.63 kN m (anticlockwise),

MF = 10.62 kN m (sagging), MBC = 0.07 kN m (clockwise)

Compare these values with those given in Fig. 16.48.

S.16.16 This frame has two indeterminacies which may be chosen to be the supportreactions at D as shown in Fig. S.16.16(a). From Figs S.16.16(a) and (b)

FIGURE S.16.16 (a) (b)

20 kN/m

40 kN3 m

3 m

x x

x

6 m

RD,V

RD,H

4.5 m

F

D

CB

A

C

D

I

I

B

A

In DC,

M0 = 0, M1,H = 1(

67.5

)x = 0.8x, M1,V = −1

(4.57.5

)x = −0.6x

In CB,

M0 = 20 x2

2= 10 x2, M1,H = 1 × 6 = 6, M1,V = −1(4.5 + x)

In BF,

M0 = 20 × 6 × 3 = 360, M1,H = 1(6 − x), M1,V = −1 × 10.5 = −10.5

In FA,

M0 = 360 + 40(x − 3) = 40(6 + x), M1,H = 1(6 − x), M1,V = −10.5

Then∫ (M0M1,H

EI

)dx =

∫ 6

0

(60x2

EI

)dx +

∫ 3

0

[360(6 − x)

EI

]dx +

∫ 6

3

[40(36 − x2)

EI

]dx

= 10 980EI∫ (

M0M1,V

EI

)dx =

∫ 6

0

[−10x2(4.5 + x)

EI

]dx +

∫ 3

0

(−360 × 10.5EI

)dx

+∫ 6

3

[−40 × 10.5(6 + x)EI

]dx = −31 050

EI∫ (M21,H

EI

)dx =

∫ 7.5

0

(0.64x2

EI

)dx +

∫ 6

0

(62

EI

)dx +

∫ 6

0

[(6 − x)2

EI

]dx = 378

EI


∫ (M21,V

EI

)dx =

∫ 7.5

0

(0.36x2

EI

)dx +

∫ 6

0

[−6(4.5 + x)EI

]dx +

∫ 6

0

(10.52

EI

)dx

= 1067.6EI∫ (

M1,HM1,V

EI

)dx =

∫ 7.5

0

(−0.48x2

EI

)dx +

∫ 6

0

[−6(4.5 + x)EI

]dx

+∫ 6

0

[−10.5(6 − x)EI

]dx = −526.5

EI

Then10 980 + 378RD,H − 526.5RD,V = 0

−31 050 − 526.5RD,H + 1067.6RD,V = 0

Solving,

RD,V = 47.1 kN, RD,H = 36.6 kN

Then, MDC = 0, MCD = 7.5 kN m (anticlockwise), MBC = 84.8 kN m (anticlockwise)

MAB = 14.8 kN m (clockwise).

S.16.17

Referring to Fig. S.16.17 the half-span of the arch is given by L = 3.5 cos 30◦ = 3.03 m

R � 3.5m

L

PC

u

B

y

x

ARA,H

RA,V RB,V

RB,H

15 kN/m

FIGURE S.16.17

Also the ordinate of any point P is given by

y = R cos θ − R cos 60◦ = 3.5 cos θ − 1.75

and, taking moments about B

RA,V × 6.06 − 15 × 3.03 × 4.545 = 0

i.e.RA,V = 34.1 kN


so thatRB,V = 11.4 kN

Then, in AC,

M0 = 34.1x − 15x2

2

in which

x = 3.03 − 3.5 sin θ ,

i.e.

M0 = 34.47 + 39.73 sin θ − 91.88 sin2 θ

and in CB,

M0 = 34.1x − 15 × 3.03(x − 1.515),

i.e.

M0 = 34.47 + 39.73 sin θ

Then the numerator in Eq. (16.14) is (note that EI cancels)∫M0y ds =

∫ 0

π/3(34.47 + 39.73 sin θ − 91.88 sin2 θ)(3.5 cos θ − 1.75)3.5 dθ

+∫ −π/3

0(34.47 + 39.73 sin θ)(3.5 cos θ − 1.75)3.5 dθ = 462.1

Also ∫y2 ds =

∫ −π/3

π/3(3.5 cos θ − 1.75)23.5 dθ = −26.15

Then

RB,H = −462.126.15

= −17.7 kN (acting to the left)

and

RA,H = 17.7 kN (acting to the right)

The bending moment at the crown of the arch is given by

MC = 11.4 × 3.03 − 17.7 × (3.5 − 3.5 cos 60◦) = 3.6 kN m.

S.16.18 In this case the secant assumption applies so that the horizontal thrust at thearch supports is given by Eq. (16.15). Further, with the origin of axes at A the equationof the arch is given by

y =(

4hL2

)(Lx − x2)


where h is the rise of the arch and L its span.

From symmetry the vertical reactions at the supports are both 50 kN. Then in AD,M0 = 50x and in DF, M0 = 50x − 50(x − 10) = 500. The value of the numerator in Eq.(16.15) between F and B is identical to that between A and D. Therefore,

∫M0y dx = 2

∫ 10

050x

(4 × 3302

)(30x − x2) dx

+∫ 20

10500

(4 × 3302

)(30x − x2) dx = 24 451.7

and ∫y2dx =

∫ 30

0

(4 × 3302

)(30x − x2)2 dx = 144

Then

RA,H = RB,H = 24 451.7144

= 169.8 kN

Also

MD = MF = 50 × 10 − 169.8(

4 × 3302

)(30 × 10 − 102) = 47.2 kN m

and

MC = 50 × 15 − 169.8 × 3 − 50 × 5 = −9.4 kN m.

S.16.19 This has been demonstrated in the additional discussion to Ex. 16.14.

S.16.20 (i) P.16.3

From Table 16.6

MFAB = −12 × 1.6 × 0.82

2.42 = −2.13 kN m

MFBA = 12 × 1.62 × 0.8

2.42 = 4.27 kN m

MFBC = −10 × 1.2 × 0.82

2.02 = −1.92 kN m

MFCB = 10 × 1.22 × 0.8

2.02 = 2.88 kN m

Then, from Eqs (16.28), etc. and noting that θA = 0

MAB = −(

2EI2.4

)θB − 2.13 = −0.83EIθB − 2.13

MBA = −(

2EI2.4

)2θB + 4.27 = −1.67EIθB + 4.27


MBC = −(

2EI2.0

)(2θB + θC) − 1.92 = −2EIθB − EIθC − 1.92

MCB = −(

2EI2.0

)(2θC + θB) + 2.88 = −2EIθC − EIθB + 2.88

From internal equilibrium

MBA + MBC = 0 and MCB = 0

Substituting in these equations from the above gives

−3.67EIθB − EIθC + 2.35 = 0

−EIθB − EIθC + 1.44 = 0

Solving

EIθB = 0.29, EIθC = 1.29

Substituting these values back in the equations for bending moment gives

MAB = −2.37 kN m, MBA = 3.79 kN m, MBC = −3.79 kN m, MCB = 0

The support reactions produced by the applied loads are:

in AB, at A = 4 kN, at B = 8 kN

in BC, at B = 4 kN, at C = 6 kN

Due to the bending moments,

in AB, at A = 3.79 − 2.372.4

= 0.6 kN downwards and at B = 0.6 kN upwards

in BC, at B = 3.792.0

= 1.9 kN upwards and at C = 1.9 kN downwards

The final support reactions are then

at A = 3.4 kN, at B = 14.5 kN, at C = 4.1 kN.

(ii) P.16.4

From Table 16.6

MFAB = −MF

BA = −0.75 × 122

12= −9 kN m

MFBC = −MF

CB = −7 × 248

= −21 kN m

From Eqs (16.28), etc. in which θA = θC = 0

MAB = −(

EIθB

6

)− 9

MBA = −(

EIθB

3

)+ 9


MBC = −(

EIθB

6

)− 21

MCB = −(

EIθB

12

)+ 21

From internal equilibrium at B

MBA + MBC = 0

Substituting from the above gives EIθB = −24

Then

MAB = −5 kN m, MBA = −MBC = 17 kN m, MCB = 23 kN m

The support reactions due to the loads and the bending moments are then

at A = 3.5 kN, at B = 8.75 kN, at C = 3.75 kN.

S.16.21 The fixed end moments at A and B in BA and at C and D in CD are zero sinceno loads are applied within AB and CD. From Table 16.6

MBC = −4 × 5 × 102

152 = −8.89 kN m, MCB = 4 × 52 × 10152 = 4.44 kN m

From Eqs (16.28), etc. in which there will be a lateral movement, say v, of B and Cdue to sway in the frame and in which θA = θD = 0

MAB = −(

2EI10

)(θB − 3v

10

)= −0.2EIθB + 0.06EIv

MBA = −(

2EI10

)(2θB − 3v

10

)= −0.4EIB + 0.06EIv

MBC = −(

4EI15

)(2θB + θC) − 8.89 = −0.53EIθB − 0.27EIθC − 8.89

MCB = −(

4EI15

)(2θC + θB) + 4.44 = −0.53EIθC − 0.27EIθB + 4.44

MCD = −(

2EI10

)(2θC − 3v

10

)= −0.4EIθC + 0.06EIv

MDC =(

2EI10

)(θC − 3v

10

)= −0.2EIθC + 0.06EIv

Also from equilibrium

MBA + MBC = 0, MCB + MCD = 0

Substituting from the above and simplifying

−15.5EIθB − 4.5EIθC + EIv − 148.17 = 0

−4.5EIθB − 15.5EIθC + EIv + 74.0 = 0


Since there are three unknowns a further equation is required. This is obtained byconsidering the shear forces at A and D. From Eq. (16.31)

SAB = −(

6EI102

)θB +

(12EI103

)v

SDC = −(

6EI102

)θC +

(12EI103

)v

From equilibrium

SAB + SDC + 2 = 0

Substituting gives

−2.5EIθB − 2.5EIθC + EIv + 83.33 = 0

Solving gives

EIθB = −18.13, EIθC = +2.07, EIv = −123.5

Now substituting back in the expressions for bending moment

MAB = −3.78 kN m, MBA = −MBC = −0.16 kN m, MCB = −MCD = −8.24 kN m,

MDC = −7.82 kN m.

S.16.22 The distribution factors are obtained from Eq. (16.36) and are

DFBA =(

4EI12

)[(

4EI12

)+(

3EI24

)] = 0.73

Then

DFBC = 0.27

If C is considered to be simply supported

DFCB = 1 and DFCD = 0

The fixed end moments are found using Table 16.6 and are

MFAB = −MF

BA = −0.75 × 122

12= −9 kN m

MFBC = −MF

CB = −7 × 248

= −21 kN m

MFCD = −1 × 5 = −5 kN m


The moment distribution is carried out in Table S.16.22

TABLE S.16.22

DFs

FEMsBalance CCarry overBalance BCarry over

Final moments

A

0

�9

�7.3

�1.7

B

0.73

�9

0.27

�21

�8�5.4�14.6

�23.6 �23.6

C

1

�21�16

�5

0

�5

�5

D

0

0

The support reactions are calculated in the same way as in the slope–deflectionmethod of solution. Due to the loads the support reactions at A and B in AB areeach 4.5 kN. Due to the final end moments the support reactions in AB are, at A,(23.6 − 1.7)/12 = 1.8 kN downwards and at B, 1.8 kN upwards. The final reaction at Ais then 4.5 − 1.8 = 2.7 kN upwards with a moment of 1.7 kN m (hogging). Similarly thefinal reaction at B = 10.6 kN and at C = 3.7 kN.

S.16.23

The beam may be regarded as a two span beam with a cantilever overhang at the endof which a moment of 10 × 1 = 10 kN m is applied. Then, as in S.16.22, C will bebalanced and then regarded as pinned.

The distribution factors are

DFCD = 1, DFDC =(

3EI1.0

)[(

3EI1.0

)+(

4EI1.0

)] = 0.43, DFDE = 0.57

The fixed end moments are

MED = 0

MDE = −10 kN m

MDC = −MCD = 50 × 18

= 6.25 kN m

MCB = 10 kN m

The moment distribution is carried out in Table S.16.23.

The support reactions are calculated as before, e.g. at E, due to the applied momentthe reaction is = 10/1.0 = 10 kN upwards and due to the final end moments thereaction is (6.78 − 1.61)/1.0 = 5.17 kN downwards so that the total reaction at Eis 10 − 5.17 = 4.83 kN with a moment reaction of 1.6 kN m (hogging). Similarly thereactions at D and C are 17.0 and 28.2 kN, respectively.


TABLE S.16.23

DFs

FEMs

Balance C

Carry over

Balance D

Carry over

Final moments

C

0 1

�10 �6.25

�3.75

�10.0 �10.0

D

0.43

�6.25

�1.88

�2.41

�6.78

0.57

�10

�3.22

�6.78

E

0

0

1.61

1.61

S.16.24 The distribution factors are

DFBA =(

4 × 83.4 × 106

4

)/[(4 × 83.4 × 106

4

)+(

3 × 125.1 × 106

5

)]= 0.53

so that DFBC = 0.47. Note that E divides out of the above expression.

The fixed end moments due to the applied loads and the sinking support at B areobtained from Table 16.6 and are

MFAB = 6.25 × 42

12+ 6 × 207 000 × 103 × 83.4 × 106 × 10−12 × 0.01

42

= 8.3 + 64.8 = 73.1 kN m

MFBA = −8.3 + 64.8 = 56.5 kN m

MFBC = 50 × 5

8− 3 × 207 000 × 103 × 125.1 × 106 × 10−12 × 0.01

52

= 31.25 − 31.07 = 0.2 kN m

MFCB = −31.25 kN m


TABLE S.16.24

DFs

FEMs

Balance C

Carry over

Balance B

Carry over

Final moments

C

0.47

�31.25

�31.25

0

B

0.53

�0.2 �56.5

�15.6

�34.0 �38.3

�18.2 �18.2

A

�73.1

�19.2

�53.9


The support reactions then follow as before, i.e.

RA = 6.25 × 42

+ (53.9 + 18.2)4

= 30.5 kN with a moment reaction of 53.9 kN m

RB = 6.25 × 42

+ 502

− 18.25

− (53.9 + 18.2)4

= 15.8 kN

RC = 502

+ 18.25

= 28.6 kN


DFBA =(

3×2EI12

)[(

3×2EI12

)+(

3×3EI18

)+(

4EI16

)] = 0.4

DFBC =(

3×3EI18

)[(

3×2EI12

)+(

3×3EI18

)+(

4EI16

)] = 0.4

DFBD = 0.2


MAB = −MBA = −8 × 128

= −12 kN m

MBC = −MCB = −16 × 189

= −32 kN m

MDB = MBD = 0


TABLE S.16.25

Joint

Member

DFs

FEMs

Balance A,C

Carry over

Balance B

Carry over

Final moments

A

AB

1

�12

�12

0

0

BA

0.4

�12

�6

�12

�30

B

BC

0.4

�32

�16

�12

�36

BD

0.2

0

�6

�6

C

CB

1

�32

�32

0

0

D

DB

0

0

�3.0

�3

The final moments are as shown in the table.



DFBA = DFCD =(

3EI6

)[(

3EI6

)+(

4EI6

)] = 0.43

Then

DFBC = DFCB = 0.57


MFAB = −MF

BA = −25 × 68

= −18.75 kN m

MFBC = −50 × 2 × 42

62 = −44.4 kN m

MFCB = 50 × 22 × 4

62 = 22.2 kN m

MFCD = MF

DC = 0

The no-sway moments are now calculated in Table S.16.26(a).

TABLE S.16.26(a) No-sway case

DFs

FEMs

Balance A

Carry over

Balance

Carry over

Balance

Carry over

Balance

Carry over

Balance

Final no-sway moments

A

–

�18.75

�18.75

0

C

0.57

�22.2

�12.65

�4.64

�2.64

�1.8

�1.03

�0.38

�0.22

�12.48

0.43

0

�9.55

�2.0

�0.77

�0.16

�12.48

D

–

0

0

0.43

�18.75

�9.38

�7.0

�2.72

�0.57

�0.22

�38.64

B

0.57

�44.4

�9.27

�6.33

�3.61

�1.32

�0.75

�0.28

�38.64

�0.5

Suppose now that the frame is given an arbitrary sway δ. Then, since A and D arepinned supports

MFAB = MF

BC = MFCB = MF

DC = 0

and

MFBA = MF

CD = −3EIδ62


Suppose that δ = 100 × 62/3EI, Then

MFBA = MF

CD = −100 kN m

The arbitrary sway moments are now calculated in Table S.16.26(b).

TABLE S.16.26(b) Sway case

DFs

FEMs

Balance

Carry over

Balance

Carry over

Balance

Carry over

Balance

Final arbitrary swaymoments

A

–

0

0

C

0.57

�57

�28.5

�16.2

�8.1

�4.6

�2.3

�1.3

�66.8

0.43

�100

�43

�12.3

�3.5

�1.0

�66.8

0

D

–

0

0

0.43

�100

�43

�12.3

�3.5

�1.0

�66.8

0.57

0

�57

�28.5

�16.2

�8.1

�4.6

�2.3

�1.3

�66.8

B

The frame in this problem has the same form as the frame of Fig. 16.46 so that Eq.(16.46) may be used directly, i.e.

38.64 − 12.48 + k(−66.8 − 66.8) + 25 × 3 = 0

which gives

k = 0.76

The actual sway moments are then

MSBA = MS

CD = −0.76 × 66.8 = −50.8 kN m

Therefore the total end moments are

MAB = MDC = 0

MBA = 38.64 − 50.8 = −12.2 kN m

MCD = −12.48 − 50.8 = −63.3 kN m

The bending moment diagram for the frame is shown in Fig. S.16.26.


A D

B

53.643.7

12.3

12.3

63.3

63.3C

Bending moments in kN m.Drawn on tension sideof members

FIGURE S.16.26

S.16.27 In this problem the FEMs are caused by sway only. The distributionfactors are

DFBA =(

3×2EI5

)[(

3 × 2EI5

)+(

4×3EI5

)] = 13

DFBC = 23

DFCB =(

4×3EI5

)[(

4×3EI5

)+(

3×2EI5

)] = 23

DFCD = 13

Suppose that the frame is given an arbitrary sway as shown in Fig. S.16.27(a).

The corresponding sway FEMs are

MFSCD = −3 × 2EI × 5θ

4 × 5= −3θ

2

MFSBC = MFS

CB = 6 × 3EI × 3θ

4 × 5= 27θ

10

MFSBA = −3 × 2EI × θ

5= −6θ

5

Suppose that θ = 100 then the above moments become −150, 270 and −120,respectively. The moment distribution is carried out in Table S.16.27.


B

A

C

D

5u

u u

43u 4

3u

43u

43u

45u

45u5 m

10 kN

FIGURE S.16.27a

TABLE S.16.27

DFs

A

–

D

–

FEMs

Balance

Carry over

Balance

Carry over

Balance

Carry over

Balance

Final arbitrarysway moments

B

13

�120

�50

�13

�6

�2

�161

23

�270

�100

�40

�27

�17

�11

�5

�3

�161

C

23

�270

�80

�50

�33

�14

�9

�6

�4

�176

13

�150

�40

�17

�5

�2

�176

Using the principle of virtual work the sway equilibrium equation is

MBA(θ) + MBC

(−3θ

4

)+ MCB

(−3θ

4

)+ MCD

(5θ

4

)+ 10 × 5θ = 0

Substituting from the above table

k[7 × (−161) + 8 × (−176)] = −200

so that k = 0.0789

The final moments are then

MAB = MDC = 0, MBA = −MBC = 12.7 kN m, MCB = −MCD = 13.9 kN m.


The bending moment diagram is shown in Fig. S.16.27(b).

A

B C

D

13.913.9

12.7

12.7Bending moments in kN m.Drawn on tension sideof members

FIGURE S.16.27b


DFBA =(

3×2EI6

)[(

3×2EI6

)+(

4×2EI6

)] = 37

, DFBC = 47

DFCB =(

4×2EI6

)[(

4×2EI6

)+(

4EI3

)] = 12

, DFCD = 12

DFDC =(

4EI3

)[(

4EI3

)+(

3EI3

)+(

3EI3

)] = 410

Similarly

DFDF = 310

and DFDG = 310

The fixed end moments for the no-sway case are

MFBC = −MF

CB = −14 × 62

12= −42 kN m

The no-sway moment distribution is carried out in Table S.16.28(a).


TABLE S.16.28(a)

B C D

Member

DFs

FEMs × 10

Balance

Carry over

Balance

Carry over

Balance

Carry over

Balance

Carry over

Balance

Final no-swaymoments

BA37

�180

�45

�12.9

�5.5

�1.6

�24.5

BC47

�420

�240

�105

�60

�30

�17.1

�12.8

�7.3

�3.7

�2.1

�24.5

CB12

�420

�210

�120

�60

�30

�25.5

�8.6

�7.3

�3.7

�3.1

�27.6

CD12

�210

�60

�21

�25.5

�6

�7.3

�2.6

�3.2

�27.6

DC4

10

�105

�42

�30

�12

�12.8

�5.2

�3.7

�1.5

�9.1

DF3

10

�31.5

�9

�3.8

�1.1

�4.5

DG3

10

�31.5

�9

�3.8

�1.1

�4.5

Suppose that the member AB is given an arbitrary sway θ as shown in Fig. S.16.28.

40 kN

6 m

3 m

3 m

C

D

G

F

A

B

2u

2uu

u

FIGURE S.16.28

The arbitrary sway FEMs are then

AB :3 × 2EI × 6θ

62 ; CD, DC :6 × EI × 6θ

32

The ratio of these arbitrary sway moments is 1 : 4, therefore assume values of 70 and280 kN m. The arbitrary sway moment distribution is carried out in Table S.16.28(b).


TABLE S.16.28(b)

B D

MemberDFs

BA37

BC47

CB12

CD12

DC4

10

DF3

10

DG3

10

FEMs

Balance

Carry over

Balance

Carry over

Balance

Carry over

Balance

Final arbitrarysway moments

�70

�30

�30

�8.1

�3.6

�65

�40

�70

�40

�19

�10.9

�8.5

�4.9

�65.5

�140

�20

�38

�20

�17

�5.5

�4.7

�119.8

�280

�140

�56

�38

�14

�17

�3.8

�4.6

�119.8

�280

�112

�70

�28

�19

�7.6

�8.5

�3.5

�132.4

�84

�21

�5.7

�2.5

�66.2

�84

�21

�5.7

�2.5

�66.2

C

Referring to Fig. S.16.28 the sway equilibrium equation is

MBA(θ) + MCD(2θ) + MDC(2θ) + 40 × 6θ = 0

i.e.

MBA + 2(MCD + MDC) + 240 = 0

Then

24.5 + 2(−27.6 − 9.1) + k[65.5 + 2(119.8 + 132.4)] + 240 = 0

which gives

k = −0.335

The final end moments are then

MBA = −MBC = 24.5 − 0.335 × 65.5 = +2.6 kN m

Similarly

MCB = −MCD = 67.7 kN m, MDC = −53.5 kN m, MDF = 26.7 kN m,

MDG = 26.7 kN m.



S.17.1 Referring to Fig. P.17.1 and Fig. 17.3

Member 12 23 34 41 13Length L L L L

√2L

λ (cos θ) 1/√

2 −1/√

2 −1/√

2 1/√

2 0µ (sin θ) 1/

√2 1/

√2 −1/

√2 −1/

√2 1

The stiffness matrix for each member is obtained using Eq. (17.23). Thus

[K12] = AE2L

1 1 −1 −11 1 −1 −1

−1 −1 1 1−1 −1 1 1

[K23] = AE

2L

1 −1 −1 1−1 1 1 −1−1 1 1 −1

1 −1 −1 1

[K34] = AE2L

1 1 −1 −11 1 −1 −1

−1 −1 1 1−1 −1 1 1

[K41] = AE

2L

1 −1 −1 1−1 1 1 −1−1 1 1 −1

1 −1 −1 1

[K13] = AE√2L

0 0 0 00 1 0 −10 0 0 00 −1 0 1

The stiffness matrix for the complete framework is now assembled using the methoddescribed in the discussion of Eq. (17.14)

Fx,1

Fy,1

Fx,2

Fy,2

Fx,3

Fy,3

Fx,4

Fy,4

= AE2L

2 0 −1 −1 0 0 −1 10 2 + √

2 −1 −1 0 −√2 1 −1

−1 −1 2 0 −1 1 0 0−1 −1 0 2 1 −1 0 0

0 0 −1 1 2 0 −1 −10 −√

2 1 −1 0 2 + √2 −1 −1

−1 1 0 0 −1 −1 2 01 −1 0 0 −1 −1 0 2

u1 = 0v1

u2 = 0v2 = 0u3 = 0

v3

u4 = 0v4 = 0

(i)

In Eq. (i)

Fy,1 = −P, Fx,1 = Fx,3 = Fy,3 = 0

Then

Fy,1 = −P = AE2L

[(2 + √2)v1 − √

2v3] (ii)

Fy,3 = 0 = AE2L

[−√2v1 + (2 + √

2)v3] (iii)


From Eq. (iii)

v1 = (1 + √2)v3 (iv)

Substituting for v1 in Eq. (ii) gives

v3 = −0.293PLAE

Hence, from Eq. (iv)

v1 = −0.707PLAE

The forces in the members are obtained using Eq. (vi) of Ex. 17.1, thus

F12 = AE√2L

[1 1]

{0 − 0

0 + 0.707PL/AE

}= P

2= F14 from symmetry

F13 = AE√2L

[0 1]

{0 − 0

−0.293 PL/AE + 0.707 PL/AE

}= 0.293 P

F23 = AE√2L

[−1 1]

{0 − 0

−0.293 PL/AE − 0

}= −0.207 P = F43 from symmetry

The support reactions are Fx,2, Fy,2, Fx,4 and Fy,4. From Eq. (i)

Fx,2 = AE2L

(−v1 + v3) = 0.207P

Fy,2 = AE2L

(−v1 − v3) = 0.5P

Fx,4 = AE2L

(v1 − v3) = −0.207P

Fy,4 = AE2L

(−v1 − v3) = 0.5P.


Member 12 23 34 31 24Length l/

√3 l/

√3 l l l/

√3

λ (cos θ)√

3/2 0 1/2 −1/2√

3/2µ (sin θ) 1/2 1 −√

3/2 −√3/2 −1/2


From Eq. (17.23) the member stiffness matrices are

[K12] = AEl

3√

3/4 3/4 −3√

3/4 −3/43/4

√3/4 −3/4 −√

3/4−3

√3/4 −3/4 3

√3/4 3/4

−3/4 −√3/4 3/4

√3/4

[K23] = AEl

0 0 0 00

√3 0 −√

30 0 0 00 −√

3 0√

3

[K34] = AEl

1/4 −√3/4 −1/4

√3/4

−√3/4 3/4

√3/4 −3/4

−1/4√

3/4 1/4 −√3/4√

3/4 −3/4 −√3/4 3/4

[K31] = AEl

1/4√

3/4 −1/4 −√3/4√

3/4 3/4 −√3/4 −3/4

−1/4 −√3/4 1/4

√3/4

−√3/4 −3/4

√3/4 3/4

[K24] = AEl

3√

3/4 −3/4 −3√

3/4 3/4−3/4

√3/4 3/4 −√

3/4−3

√3/4 3/4 3

√3/4 −3/4

3/4 −√3/4 −3/4

√3/4

The stiffness matrix for the complete framework is now assembled as before and is

Fx,1

Fy,1

Fx,2

Fy,2

Fx,3

Fy,3

Fx,4

Fy,4

= AEl

1 + 3√

34

3 + √3

4 − 3√

34 − 3

4 − 14 −

√3

4 0 03 + √

34

3 + √3

4 − 34 −

√3

4 −√

34 − 3

4 0 0

− 3√

34 − 3

43√

32 0 0 0 − 3

√3

434

− 34 −

√3

4 0 3√

32 0 −√

3 34 −

√3

4

− 14 −

√3

4 0 0 12 0 − 1

4

√3

4

−√

34 − 3

4 0 −√3 0 3

2 + √3

√3

4 − 34

0 0 − 3√

34

34 − 1

4

√3

41 + 3

√3

4 − 3 + √3

4

0 0 34 −

√3

4

√3

4 − 34 − 3 + √

34

3 + √3

4

u1 = 0v1 = 0u2 = 0

v2

u3 = 0v3

u4 = 0v4 = 0

(i)

In Eq. (i)

Fx,2 = Fy,2 = 0, Fx,3 = 0, Fy,3 = −P, Fx,4 = −H


Then

Fy,2 = 0 = AEl

(3√

32

v2 − √3v3

)(ii)

and

Fy,3 = −P = AEl

[−√

3v2 +(

32

+ √3)

v3

](iii)

From Eq. (ii)

v2 = 23v3 (iv)

Now substituting for v2 in Eq. (iii)

− PlAE

= −2√

33

v3 + 32v3 + √

3v3

Hence

v3 = − 6Pl

(9 + 2√

3) AE

and, from Eq. (iv)

v2 = − 4Pl

(9 + 2√

3) AE

Also from Eq. (i)

Fx,4 = −H = AEl

(34v2 +

√3

4v3

)

Substituting for v2 and v3 gives

H = 0.449P.


Member 12 23 34 45 24Length l l l l lλ (cos θ) −1/2 1/2 −1/2 1/2 1µ (sin θ)

√3/2

√3/2

√3/2

√3/2 0


From Eq. (17.23) the member stiffness matrices are

[K12] = AEl

1/4 −√3/4 −1/4

√3/4

−√3/4 3/4

√3/4 −3/4

−1/4√

3/4 1/4 −√3/4√

3/4 −3/4 −√3/4 3/4

[K23] = AEl

1/4√

3/4 −1/4 −√3/4√

3/4 3/4 −√3/4 −3/4

−1/4 −√3/4 1/4

√3/4

−√3/4 −3/4

√3/4 3/4

[K34] = AEl

1/4 −√3/4 −1/4

√3/4

−√3/4 3/4

√3/4 −3/4

−1/4√

3/4 1/4 −√3/4√

3/4 −3/4 −√3/4 3/4

[K45] = AEl

1/4√

3/4 −1/4 −√3/4√

3/4 3/4 −√3/4 −3/4

−1/4 −√3/4 1/4

√3/4

−√3/4 −3/4

√3/4 3/4

[K24] = AEl

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

The stiffness matrix for the complete truss is now assembled and is

Fx,1

Fy,1

Fx,2

Fy,2

Fx,3

Fy,3

Fx,4

Fy,4

Fx,5

Fy,5

= AE4l

1 −√3 −1

√3 0 0 0 0 0 0

−√3 3

√3 −3 0 0 0 0 0 0

−1√

3 6 0 −1 −√3 −4 0 0 0√

3 −3 0 6 −√3 −3 0 0 0 0

0 0 −1 −√3 2 0 −1

√3 0 0

0 0 −√3 −3 0 6

√3 −3 0 0

0 0 −4 0 −1√

3 6 0 −1 −√3

0 0 0 0√

3 −3 0 6 −√3 −3

0 0 0 0 0 0 −1 −√3 1

√3

0 0 0 0 0 0 −√3 −3

√3 3

u1 = 0v1 = 0

u2

v2

u3 = 0v3 = 0

u4

v4

u5 = 0v5 = 0

(i)

In Eq. (i)

Fx,2 = Fy,2 = 0, Fx,4 = 0, Fy,4 = −P


Thus from Eq. (i)

Fx,2 = 0 = AE4l

(6u2 − 4u4) (ii)

Fy,2 = 0 = AE4l

(6v2) (iii)

Fx,4 = 0 = AE4l

(−4u2 + 6u4) (iv)

Fy,4 = −P = AE4l

(6v4) (v)

From Eq. (v)

v4 = − 2Pl3AE

From Eq. (iii)

v2 = 0

and from Eqs (ii) and (iv)

u2 = u4 = 0

Hence, from Eq. (vi) of Ex. 17.1

F24 = AEl

[1 0]

(0 − 0

−2 Pl/3 AE − 0

)

which gives

F24 = 0.

S.17.4 The uniformly distributed load on the member 26 is equivalent to concentratedloads of wl/4 at nodes 2 and 6 together with a concentrated load of wl/2 at node 4.Thus, referring to Fig. P.17.4 and Fig. 17.3

Member 12 23 24 46 56 67Length l l l/2 l/2 l lλ (cos θ) 0 −1/

√2 1 1 0 1/

√2

µ (sin θ) 1 1/√

2 0 0 1 1/√

2


From Eq. (17.34) and using the alternative form of Eq. (17.31)

[K12] = EIl3

12 SYM0 06 0 4

−12 0 −6 120 0 0 0 06 0 2 6 0 0

[K23] = EIl3

6 SYM6 6

6/√

2 6/√

2 46 6 −6/

√2 6

−6 −6 −6/√

2 6 66/

√2 6/

√2 2 6/

√2 −6/

√2 −4/

√2

[K24] = [K46] = EIl3

0 SYM0 960 −24 80 0 0 00 −96 24 0 960 −24 4 0 24 8

[K56] = EIl3

12 SYM0 06 0 4

−12 0 −6 120 0 0 0 06 0 2 6 0 0

[K67] = EIl3

6 SYM−6 6

6/√

2 −6/√

2 4−6 6 −6/

√2 6

6 −6 6/√

2 −6 66/

√2 −6/

√2 2 6/

√2 6/

√2 4/

√2

The member stiffness matrices are then assembled into a 21 × 21 symmetricalmatrix using the method previously described. The known modal displacements areu1 = v1 = θ1 = u5 = v5 = θ5 = u2 = u4 = u6 = θ3 = θ7 = 0 and the support reactions areobtained from {F} = [K ]{δ}. Having obtained the support reactions the internal shearforce and bending moment distributions in each member follow.

S.17.5 Referring to Fig. P.17.5, u2 = 0 from symmetry. Consider the members 23 and29. The forces acting on the member 23 are shown in Fig. S.17.5(a) in which F29 is the


force applied at 2 in the member 23 due to the axial force in the member 29. Supposethat the node 2 suffers a vertical displacement v2. The shortening in the member 29 isthen v2 cos θ and the corresponding strain is −(v2 cos θ)/l. Thus the compressive stressin 29 is −(Ev2 cos θ)/l and the corresponding compressive force is −(AEv2 cos θ)/l.

Thus

F29 = − (AEv2 cos2 θ)l

Now AE = 6√

2EI/L2, θ = 45◦ and l = √2 L. Hence

F29 = −3EIL3 v2

3

l�2

M2

Fy,3

M3

F29

2

9

u

u

P2 FIGURE S.17.5(a)

and

Fy,2 = −P2

− 3EIL3 v2 (i)

Further, from Eq. (11.26)

M3 = GJdθ

dz= −2 × 0.8EI

θ3

0.8L= −2EI

Lθ3 (ii)

From the alternative form of Eq. (17.31), for the member 23

Fy,2

M2/LFy,3

M3/L

= EIL3

12 −6 −12 −6−6 4 6 2

−12 6 12 6−6 2 6 4

v2

θ2L = 0v3 = 0θ3L

(iii)

Then, from Eqs (i) and (iii)

Fy,2 = −P2

− 3EIL3 v2 = 12EI

L3 v2 − 6EIL2 θ3


�ve

�ve�ve

321

�ve

�ve

P3

PL9

2PL9

PL9

(b)

(c) FIGURES S.17.5(b) and (c)

Hence

15v2 − 6θ3L = −PL3

2EI(iv)

From Eqs (ii) and (iii)

M3

L= −2EI

L2 θ3 = −6EIL3 v2 + 4EI

L2 θ3

which gives θ3 = v2/L.

Substituting for θ3 in Eq. (iv) gives

v2 = − PL3

18EI

Then

θ3 = − PL2

18 EIFrom Eq. (i)

Fy,2 = −P2

+ 3EIL3

PL3

18EI= −P

3and from Eq. (ii)

M3 = 2EIL

PL2

18EI= PL

9= −M1

Now, from Eq. (iii)

M2

L= −EI

L3 6v2 + 2EIL3 θ3L = 2P

9

Fy,3 = −12 EIL3 v2 + 6EI

L3 θ3L = P3


The force in the member 29 is F29/ cos θ = √2F29. Thus

F29 = F28 = √2

3EIL3

PL3

18EI=

√2P6

(tension)

The torques in the members 36 and 37 are given by M3/2, i.e.

M36 = M37 = PL18

The shear force and bending moment diagrams for the member 123 follow and areshown in Figs S.17.5(b) and (c), respectively.

S.17.6 The stiffness matrix for each element of the beam is obtained using the givenforce–displacement relationship, the complete stiffness matrix for the beam is thenobtained using the method previously described. This gives

Fx,1

M1/LFy,2

M2/LFy,3

M3/LFy,4

M4/L

= EIL3

24 −12 −24 −12−12 8 12 4−24 12 36 6 −12 −6−12 4 6 12 6 2

−12 6 36 −24 −24 −12−6 2 −6 12 12 4

−24 12 24 12−12 4 12 8

v1

θ1Lv2

θ2Lv3

θ3Lv4

θ4L

(i)

The ties FB, CH, EB and CG produce vertically upward forces F2 and F3 at B and C,respectively. These may be found using the method described in S.17.5 Thus

F2 = −a1E cos2 60◦

2L√3

+ a2E cos2 45◦√

2L

v2

But a1 = 384I/5√

3L2 and a2 = 192I/5√

2L2 so that

F2 = −96EI5L3 v2

Similarly

F3 = −96EI5L3 v3

Then

Fy,2 = −P − 96EI5L3 v2 and Fy,3 = −P − 96EI

5L3 v3

In Eq. (i), v1 = θ1 = v4 = θ4 = 0 and M2 = M3 = 0. Also, from symmetry, v2 = v3, andθ2 = −θ3. Then, from Eq. (i)

M2 = 0 = 6v2 + 12θ2L + 6v3 + 2θ3L


i.e.

12v2 + 10θ2L = 0

which gives

θ2 = − 65 L

v2

Also from Eq. (i)

Fy,2 = −P − 96EI5L3 v2 = EI

L3 (36v2 + 6θ2L − 12v3 − 6θ3L)

i.e.

−P − 96EI5L3 v2 = 48 EI

5L3 v2

whence

v2 = − 5PL3

144 EI= v3

and

θ2 = PL2

24EI= −θ3

The reactions at the ends of the beam now follow from the above values and Eq. (i).Thus

Fy,1 = EIL3 (−24v2 − 12θ2L) = P

3= Fy,4

M1 = EIL2 (12v2 + 4θ2L) = −PL

4= −M4.

Also

F2 = F3 = 96EI5L3

5PL3

144EI= 2P

3The forces on the beam are then as shown in Fig. S.17.6(a). The shear force andbending moment diagrams for the beam follow and are shown in Figs S.17.6(b) and(c), respectively.

The forces in the ties are obtained using Eq. (vi) of Ex.17.1 Thus

FBF = FCH = a1E2L√

3

[−1

2

√3

2

]{0 − 0v2 − 0

}

4321

P3 P � 2

3P �

P3

P �23

P �P3

PL4

P3

PL4

FIGURE S.17.6(a)


�ve

�ve�ve

4

4

3

2 3

21

1

�ve

�ve

P3

PL4

PL12

PL12

PL4

P3

(b)

(c) FIGURES S.17.6(b) and (c)

i.e.

FBF = FCH = 384EI√

3

5√

3 × 2L3

12

5PL3

144EI= 2

3P

and

FBE = FCG = a2E√2L

[− 1√

2

1√2

]{0 − 0v2 − 0

}

i.e.

FBE = FCG = 192EI

5√

2 × √2L3

1√2

5PL3

144EI=

√2P3

.

S.17.7 The forces acting on the member 123 are shown in Fig. S.17.7(a). The momentM2 arises from the torsion of the members 26 and 28 and, from Eq. (11.26) is given by

M2 = −2GJθ2

1.6 l= −EI

θ2

l(i)

FIGURE S.17.7(a)

1 2 3

l

M2 M3

Fy,2Fy,1P2

I2

Now using the alternative form of Eq. (17.31) for the member 12


Fy,1

M1/lFy,2

M2/l

= EIl3

12 −6 −12 −6−6 4 6 2

−12 6 12 6−6 2 6 4

v1

θ1Lv2

θ2L

(ii)

and for the member 23

Fy,2

M2/lFy,3

M3/l

= EIl3

96 −24 −96 −24−24 8 24 4−96 24 96 24−24 4 24 8

v2

θ2Lv3

θ3L

(iii)

Combining Eqs (ii) and (iii)

Fy,1

M1/lFy,2

M2/lFy,3

M3/l

= EIl3

12 −6 −12 −6 0 0−6 4 6 2 0 0

−12 6 108 −18 −96 −24−6 2 −18 12 24 4

0 0 −96 24 96 240 0 −24 4 24 8

v1

θ1lv2

θ2lv3

θ3l

(iv)

In Eq. (iv) v1 = v2 = 0 and θ3 = 0. Also M1 = 0 and Fy,3 = −P/2. Then from Eq. (iv)

M1

l= 0 = EI

l3(4θ1l + 2θ2l)

from which

θ1 = −θ2

2(v)

Also, from Eqs (i) and (iv)

M2

l= −EI

l2θ2 = EI

l3(2θ1l + 12θ2l + 24v3)

so that

13θ2l + 2θ1l + 24v3 = 0 (vi)

Finally from Eq. (iv)

Fy,3 = −P2

= EIl3

(24θ2l + 96v3)

which gives

v3 = − Pl3

192EI− θ2l

4(vii)

Substituting in Eq. (vi) for θ1 from Eq. (v) and v3 from Eq. (vii) gives

θ2 = Pl2

48EI


Then, from Eq. (v)

θ1 = − Pl2

96EI

and from Eq. (vii)

v3 = − Pl3

96EI

Now substituting for θ1, θ2 and v3 in Eq. (iv) gives Fy,1 = −P/16, Fy,2 = 9P/16,M2 = −Pl/48 (from Eq. (i)) and M3 = −Pl/6. Then the bending moment at 2 in 12 isFy,1l = −Pl/12 and the bending moment at 2 in 32 is −(P/2)(l/2) + M3 = −Pl/12. AlsoM3 = −Pl/6 so that the bending moment diagram for the member 123 is that shownin Fig. S.17.7(b).

13

2

�ve

�ve

Pl16

Pl12

Pl6 FIGURE S.17.7(b)

S.17.8

(a) The element is shown in Fig. S.17.8. The displacement functions for a triangularelement are given by Eq. (17.69). Thus

w1 = α1, v1 = α4

w2 = α1 + aα2, v2 = α4 + aα5

w3 = α1 + aα3, v3 = α4 + aα6

(i)

y

x

3 (0, a)

2 (a, 0)1 (0, 0) FIGURE S.17.8


From Eqs (i)

α1 = w1, α2 = w2 − w1

a, α3 = w3 − w1

a

α4 = v1, α5 = v2 − v1

a, α6 = v3 − v1

a

Hence in matrix form

α1

α2

α3

α4

α5

α6

=

1 0 0 0 0 0−1/a 0 1/a 0 0 0−1/a 0 0 0 1/a 0

0 1 0 0 0 00 −1/a 0 1/a 0 00 −1/a 0 0 0 1/a

w1

v1

w2

v2

w3

v3

which is of the form

{α} = [A−1]{δe}

Also, from Eq. 17.76

[C] =0 1 0 0 0 0

0 0 0 0 0 10 0 1 0 1 0

Hence

[B] = [C][A−1] =−1/a 0 1/a 0 0 0

0 −1/a 0 0 0 1/a−1/a −1/a 0 1/a 1/a 0

(b) From Eq. (17.81)

[Ke] =

−1/a 0 −1/a0 −1/a −1/a

1/a 0 00 0 1/a0 0 1/a0 1/a 0

E1 − ν2

1 ν 0

ν 1 00 0 1

2 (1 − ν)

×−1/a 0 1/a 0 0 0

0 −1/a 0 0 0 1/a−1/a −1/a 0 1/a 1/a 0

1

2a2t


which gives

[Ke] = Et4(1 − ν2)

3 − ν 1 + ν −2 −(1 − ν) −(1 − ν) −2ν

1 + ν 3 − ν −2ν −(1 − ν) −(1 − ν) −2−2 −2ν 2 0 0 2ν

−(1 − ν) −(1 − ν) 0 1 − ν 1 − ν 0−(1 − ν) −(1 − ν) 0 1 − ν 1 − ν 0

−2ν −2 −2ν 0 0 2

Continuity of displacement is only ensured at nodes, not along their edges.

S.17.9

(a) There are 6 degrees of freedom so that the displacement field must include sixcoefficients. Thus

w = α1 + α2x + α3y (i)

v = α4 + α5x + α6y (ii)

(b) From Eqs (i) and (ii) referring to Fig. S.17.9

w1 = α1 + α2 + α3, v1 = α4 + α5 + α6

w2 = α1 + 2α2 + α3, v2 = α4 + 2α5 + α6

w3 = α1 + 2α2 + 2α3, v3 = α4 + 2α5 + 2α6

2 (2, 1)

3 (2, 2)

1 (1, 1)

y

x FIGURE S.17.9

Thus

α2 = w2 − w1, α3 = w3 − w2, α1 = 2w1 − w3

α5 = v2 − v1, α6 = v3 − v2, α4 = 2v1 − v3

Therefore

α1

α2

α3

α4

α5

α6

=

2 0 0 0 −1 0−1 0 1 0 0 0

0 0 −1 0 1 00 2 0 0 0 −10 −1 0 1 0 00 2 0 0 0 −1

w1

v1

w2

v2

w3

v3

(iii)


which is of the form

{α} = [A−1]{δe}

From Eq. 17.76

[C] =0 1 0 0 0 0

0 0 0 0 0 10 0 1 0 1 0

Hence

[B] = [C][A−1] =−1 0 1 0 0 0

0 2 0 0 0 −10 −1 −1 1 1 0

(c) From Eq. (17.56)

{σ } = [D][B]{δe}

Thus, for plane stress problems (see Eq. (17.79))

[D][B] = E1 − ν2

1 ν 0

ν 1 00 0 1

2 (1 − ν)

−1 0 1 0 0 0

0 2 0 0 0 −10 −1 −1 1 1 0

i.e.

[D][B] = E1 − ν2

−1 2ν 1 0 0 −ν

−ν 2 ν 0 0 −10 − 1

2 (1 − ν) − 12 (1 − ν) 1

2 (1 − ν) 12 (1 − ν) 0

For plain strain problems (see Eq. (17.80))

[D][B] = E(1 − ν)(1 + ν)(1 − 2ν)

1 ν(1−ν) 0

ν(1−ν) 1 0

0 0 (1−2ν)2(1−ν)

×−1 0 1 0 0 0

0 2 0 0 0 −10 −1 −1 1 1 0

[D][B] = E(1 − ν)(1 + ν)(1 − 2ν)

−1 2ν1−ν

1 0 0 − ν1−ν

− ν1−ν

2 ν1−ν

0 0 −1

0 − 1−2ν2(1−ν) − 1−2ν

2(1−ν)1−2ν

2(1−ν)1−2ν

2(1−ν) 0


S.17.10

(a) The element is shown in Fig. S.17.10. There are 8 degrees of freedom so that adisplacement field must include eight coefficients. Therefore assume

w = α1 + α2x + α3y + α4xy (i)

v = α5 + α6x + α7y + α8xy (ii)

(b) From Eqs (17.75) and Eqs (i) and (ii)

εx = ∂w∂x

= α2 + α4y

y

4 (0, 2b) 3 (2a, 2b)

2 (2a, 0)1 (0, 0) x FIGURE S.17.10

εy = ∂v

∂y= α7 + α8x

γxy = ∂w∂y

+ ∂v

∂x= α3 + α4x + α6 + α8y

Thus since {ε} = [C]{α}

[C] =0 1 0 y 0 0 0 0

0 0 0 0 0 0 1 x0 0 1 x 0 1 0 y

(iii)

(c) From Eq. (iii)

[C]T =

0 0 01 0 00 0 1y 0 x0 0 00 0 10 1 x0 x y


and from Eq. (17.79)

[D] = E1 − ν2

1 ν 0

ν 1 00 0 1

2 (1 − ν)

Thus ∫vol

[C]T[D][C] dV =∫ 2a

0

∫ 2b

0[C]T[D][C]t dx dy (iv)

Substituting in Eq. (iv) for [C]T, [D] and [C] and multiplying out gives∫ 2a

0

∫ 2b

0[C]T[D][C]t dx dy

= Et1 − ν2

∫ 2a

0

∫ 2b

0

0 0 0 0 0 0 0 00 1 0 y 0 0 ν νx

0 012

(1 − ν)x2

(1 − ν) 012

(1 − ν) 0y2

(1 − ν)

0 yx2

(1 − ν) y2 + x2(1 − ν)2

0x2

(1 − ν) νy νxy + xy2

(1 − ν)0 0 0 0 0 0 0 0

0 012

(1 − ν)x2

(1 − ν) 012

(1 − ν) 0y2

(1 − ν)0 ν 0 νy 0 0 1 x

0 νxy2

(1 − ν) νxy + xy2

(1 − ν) 0y2

(1 − ν) x x2 + y2

2(1 − ν)

dx dy

= Et1 − ν2

0 0 0 0 0 0 0 00 4ab 0 4ab2 0 0 4abν 4ba2ν

0 0 2ab(1 − ν) 2a2b(1 − ν) 0 2ab(1 − ν) 0 2ab2(1 − ν)

0 4ab2 2a2b(1 − ν)83{2ab3 + a3b(1 − ν)} 0 2a2b(1 − ν) 4ab2ν 2a2b2(1 + ν)

0 0 0 0 0 0 0 00 0 2ab(1 − ν) 2a2b(1 − ν) 0 2ab(1 − ν) 0 2ab2(1 − ν)0 4abν 0 4ab2ν 0 0 4ab 4a2b

0 4a2bν 2ab2(1 − ν) 2a2b2(1 + ν) 0 2ab2(1 − ν) 4a2b83{2a3b + ab3(1 − ν)}

S.17.11 From the first of Eqs (17.83)

w1 = α1 − α2 − α3 + α4 = 0.1103 (i)

w2 = α1 + α2 − α3 − α4 = 0.3103 (ii)

w3 = α1 + α2 + α3 + α4 = 0.6103 (iii)

w4 = α1 − α2 + α3 − α4 = 0.1103 (iv)

Adding Eqs (i) and (ii)

w1 + w2 = 2α1 − 2α3 = 0.4103


i.e.α1 − α3 = 0.2

103 (v)

Adding Eqs (iii) and (iv)

w3 + w4 = 2α1 + 2α3 = 0.7103

i.e.α1 + α3 = 0.35

103 (vi)

Adding Eqs (v) and (vi)

α1 = 0.275103

Then from Eq. (v)

α3 = 0.075103

Now subtracting Eq. (ii) from Eq. (i)

w1 − w2 = −2α2 + 2α4 = − 0.2103

i.e.α2 − α4 = 0.1

103 (vii)

Subtracting Eq. (iv) from Eq. (iii)

w3 − w4 = 2α2 + 2α4 = 0.5103

i.e.α2 + α4 = 0.25

103 (viii)

Now adding Eqs (vii) and (viii)

2α2 = 0.35103

whenceα2 = 0.175

103

Then from Eq. (vii)

α4 = 0.075103

From the second of Eqs (17.83)

v1 = α5 − α6 − α7 + α8 = 0.1103 (ix)

v2 = α5 + α6 − α7 − α8 = 0.3103 (x)

v3 = α5 + α6 + α7 + α8 = 0.7103 (xi)

v4 = α5 − α6 + α7 − α8 = 0.5103 (xii)


Then, in a similar manner to the above

α5 = 0.4103

α7 = 0.2103

α6 = 0.1103

α8 = 0

Eqs (17.83) are now written

wi = (0.275 + 0.175x + 0.075y + 0.075xy) × 10−3

vi = (0.4 + 0.1x + 0.2y) × 10−3


εx = (0.175 + 0.075y) × 10−3

εy = 0.2 × 10−3

γxy = (0.075 + 0.075x + 0.1) × 10−3

= (0.175 + 0.075x) × 10−3

At the centre of the element x = y = 0. Then

εx = 0.175 × 10−3

εy = 0.2 × 10−3

γxy = 0.175 × 10−3

so that, from Eqs (17.79)

σx = 200 0001 − 0.32 (0.175 + 0.3 × 0.2) × 10−3 = 51.65 N/mm2

σy = 200 0001 − 0.32 (0.2 + 0.3 × 0.175) × 10−3 = 55.49 N/mm2

τxy = 200 0002(1 + 0.3)

× 0.175 × 10−3 = 13.46 N/mm2.

S.17.12 The displacement functions are

w = α1 + α2x + α3y

v = α4 + α5x + α6y


Then

w1 = α1 + 2α2 + 3α3

w2 = α1 + 3α2 + 3α3

w3 = α1 + 2.5α2 + 4α3

Extracting α1, α2 and α3 from the above gives

α1 = 4.5w1 − 0.5w2 − 3w3

α2 = w2 − w1

α3 = −0.5w1 − 0.5w2 + w3

Now substituting these expressions in the displacement function for w

w = (4.5 − x − 0.5y)w1 + (−0.5 + x − 0.5y)w2 + (−3 + y)w3

Similarly

v = (4.5 − x − 0.5y)v1 + (−0.5 + x − 0.5y)v2 + (−3 + y)v3

From Eqs (17.75)

εx = ∂w/∂x = −w1 + w2

εy = ∂v/∂y = −0.5v1 − 0.5v2 + v3

γxy = (∂w/∂y) + (∂v/∂x) = −0.5w1 − 0.5w2 + w3 − v1 + v2

Substituting for w1, w2, etc. gives

εx = 0.06 × 10−3, εy = 0.08 × 10−3, γxy = 0.17 × 10−3

Substituting these values in Eq. (17.80) gives

σx = 25.4 N/mm2, σy = 28.5 N/mm2, τxy = 13.1 N/mm2.

S.17.13 Assume displacement functions

w = α1 + α2x + α3y + α4xy

v = α5 + α6x + α7y + α8xy

Then

w1 = α1 − 2α2 − α3 + 2α4 = 0.001 × 10−3

w2 = α1 + 2α2 − α3 − 2α4 = 0.003 × 10−3

w3 = α1 + 2α2 + α3 + 2α4 = −0.003 × 10−3

w4 = α1 − 2α2 + α3 − 2α4 = 0


Solving these equations givesα1 = 0.00025 × 10−3

α2 = −0.000125 × 10−3

α3 = −0.00175 × 10−3

α4 = −0.000625 × 10−3

Similarlyα5 = −0.001 × 10−3

α6 = 0.00025 × 10−3

α7 = 0.002 × 10−3

α8 = −0.00025 × 10−3

Thenw = 0.00025 − 0.000125x − 0.00175y − 0.000625xy

v = −0.001 + 0.00025x + 0.002y − 0.00025xy

From Eqs (17.75)

εx = −0.000125 − 0.000625y

εy = 0.002 − 0.00025x

γxy = −0.0015 − 0.000625x − 0.00025y

At the centre of the element (x = 0, y = 0)

εx = −0.000125, εy = 0.002, γxy = −0.0015


σx = 104.4 N/mm2, σy = 431.3 N/mm2, τxy = −115.4 N/mm2 (see Ex. 17.4).

S.17.14 Assume displacement functions

w = α1 + α2x + α3y

v = α4 + α5x + α6y

Then

w1 = α1 + α2(0) + α3(0)

w2 = α1 + α2(2) + α3(0)

w3 = α1 + α2(0) + α3(3)

Solving

α1 = w1

α2 = w2 − w1

2

α3 = w3 − w1

3


Substituting in the displacement functions

w =[1 −

( x2

)−( y

3

)]w1 +

( x2

)w2 +

( y3

)w3

so that∂w∂x

= −(w1

2

)+(w2

2

)∂w∂y

= −(w1

3

)+(w3

3

)Similarly

∂v

∂x= −

(v1

2

)+(v2

2

)∂v

∂y= −

(v1

3

)+(v3

3

)Then the [B] matrix is given by

[B] =

∂w∂x∂v

∂y∂u∂y

+ ∂v

∂x

= 16

−3 0 3 0 0 0

0 −2 0 0 0 2−2 −3 0 3 2 0

w1

v1

w2

v2

w3

v3

Also the [D] matrix is

[D] =a b 0

b a 00 0 c

so that

[D][B] = 16

−3a −2b 3a 0 0 2b

−3b −2a 3b 0 0 2a−2c −3c 0 3c 2c 0

and

[B]T[D][B] = 136

9a + 4c 6b + 6c −9a −6c −4c −6b6b + 6c 4a + 9c −6b −9c −6c −4a

−9a −6b 9a 0 0 6b−6c −9c 0 9c 6c 0−4c −6c 0 6c 4c 0−6b −4a 6b 0 0 4a

The stiffness matrix for the element is

[K] =∫

area[B]T[D][B] t dx dy

from which

[K] = 3t[B]T[D][B].



S.18.1

Since the section is doubly symmetrical the elastic and plastic neutral axes coincide.Also the centroid of each semicircle is 0.424 r from the diameter. Then, from Eq. (18.6)

ZP = π r2 (2 × 0.424 r)2

= 1.33 r3

and, from Eq. (18.5)

MP = 1.33σY r3

The elastic section modulus is

Ze = Ir

= π r4

4r= 0.785r3

so that, from Eq. (18.7) the shape factor is given by

f = 1.33r3

0.785r3 = 1.69.


y

x

t

zd

b

y

FIGURE S.18.2

the plastic and elastic neutral axes coincide since the section is doubly symmetrical.Then

t(b + d)y = btd2

+ 2(

d2

)t(

d4

)which gives

y = d(2b + d)4(b + d)

From Eq. (18.6)

ZP =2(b + d)t

[2d(2b+d)

4(b+d)

]2


i.e.

ZP = td(2b + d)2

Therefore, from Eq. (18.5)

MP = σYtd(2b + d)2

The second moment of area of the beam section is given by

Iz = 2bt

(d2

4

)+ 2

(td3

12

)= td2 (3b + d)

6

so that

Ze = td(3b + d)

The shape factor is obtained from Eq. (18.7) and is

f = b + d2

b + d3

Substituting the given dimensions

f = 1.17.

S.18.3 In this case the beam section does not have a horizontal axis of symmetry sothat the elastic and plastic neutral axes do not coincide. In Fig. S.18.3

75 mm

15 mm

y

z

y

15 mm

300 mm

250 mm

Elastic N.A.

Plastic N.A.

15 mm

yP

FIGURE S.18.3

(75 × 15 + 270 × 15 + 250 × 15)y = 75 × 15 × 292.5 + 270 × 15 × 150 + 250 × 15 × 7.5

from which

y = 108.1 mm


Then

Iz = 75 × 153

12+ 75 × 15 × 184.42 + 15 × 2703

12+ 270 × 15 × 41.92

+ 250 × 153

12+ 250 × 15 × 100.62

i.e.

Iz = 108.0 × 106 mm4

Then

Ze = 108.0 × 106

191.9= 562847.8 mm3

Also, since the areas above and below the plastic neutral axis are equal

250 × 15 + 15yP = 75 × 15 + 15(270 − yP)

which gives

yP = 47.5 mm

Then ZP = 75×15×230.0+15×222.5×111.25+250×15×55.5+15×47.5 × 23.75

i.e.

ZP = 855093.8 mm3

Then

f = 855093.8562847.8

= 1.52

and

MP = 855093.8 × 300 × 10−6 = 256.5 kN m.

S.18.4 The two possible collapse mechanisms are shown in Figs S.18.4(a) and (b).

FIGURE S.18.4

2W

A B C

RC

W

D

2 m 2 m 2 m

(a)

A

B(b)

2WC

RC

D

W

In Fig. S.18.4(a) a plastic hinge occurs at C in CD. Then

W × 2 = MP = 256.5 kN m (from S.18.3)


i.e.

Wult = 128.3 kN

In Fig. S.18.4(b) plastic hinges develop at A, B and C in BC. Taking moments about A

MP = 2W × 2 − 4RC + W × 6 = 10W − 4RC

Also, taking moments about B

MP = 2RC − W × 4

Eliminating RC between these two equations gives

Wult = 1.5MP = 384.8 kN

Clearly the collapse mechanism shown in Fig. S.18.4(a) is the critical case and theminimum value of W is 128.3 kN with a plastic hinge at C in CD.

S.18.5 The collapse mechanism for the beam is shown in Fig. S.18.5.

W

L2

L2

FIGURE S.18.5

Considering half the span of the beam

2MP = wult

(L2

)(L4

)

i.e.

wult = 16MP

L2 .

S.18.6 There are three possible collapse mechanisms as shown in Figs S.18.6(a), (b)and (c).

In all three cases the support reaction at A is given by (taking moments about D)

RA × L = W(

2L3

)+ 2W

(L3

)− W

(L2

)

i.e.

RA = 5W6


(b)

A

B

C

D

EW

2W

W

(a)

W

AW

B

C

E

2W

RA

RA

D

L3

L3

L3

L2

(c)

A

W 2W

DW

E

CB

RA FIGURE S.18.6

Then for (a) (5W

6

)(L3

)= MP

i.e.

Wult = 18MP

5LFor (b) (

5W6

)(2L3

)− W

(L3

)= MP

i.e.

Wult = 9MP

2LFor (c)

W(

L2

)= MP

i.e.

Wult = 2MP

LThe collapse load of the beam is therefore 2MP/L with a plastic hinge at D.


S.18.7 The possible collapse mechanisms are shown in Figs S.18.7(a), (b) and (c).

(a)

B

C

D

WW

A

RD

C

WW

A

B

D

(b)RD

C

W

WA

B

D

(c)RD

L3

L3

L3

FIGURE S.18.7

Considering (a) and taking moments about A

MP = W(

L3

)+ W

(2L3

)− RDL = WL − RDL


MP = RD

(2L3

)− W

(L3

)

Eliminating RD from these two equations gives

Wult = 8MP

L

Now considering (b) and taking moments about A

MP = W(

L3

)+ W

(2L3

)− RDL = WL − RDL


MP = RD

(L3

)Eliminating RD gives

Wult = 4MP

L


Taking moments about B in (c)

MP = W(

L3

)− RD

(2L3

)


MP = RD

(L3

)

Then, eliminating RD gives

Wult = 9MP

LTherefore the ultimate value of W is 4MP/L with plastic hinges at A and C.

S.18.8 The ultimate load on the beam of P.18.7 is given by

Wult = 1.75 × 150 = 262.5 kN

Therefore, from S.18.7

MP = 262.5 × 64

= 393.75 kN m

Then

ZP = 393.75 × 106

300= 1 312 500 mm3

Steel tables give a UB of 406 mm × 152 mm × 67 kg/m as the most suitable section.

S.18.9 The beam mechanism is shown in Fig. S.18.9(a).

3 m

2M

M M

EB

C

D

(a)

30 kN 30 kN

3u

3 m 3 m

23u

23u

2u

u

FIGURE S.18.9(a)

Then, from the principle of virtual work

Mθ + 3Mθ + 0.5Mθ = 30 × 3θ + 30 × 3θ

2

which gives

M = 30 kN m


The sway mechanism is shown in Fig. S.18.9(b).

A

B E

F

u u

25 kN4u

FIGURE S.18.9(b)

Then

4Mθ = 25 × 4θ

i.e.

M = 25 kN m

For the combined mechanism there will be a hinge cancellation at B. Then

3Mθ + 0.5Mθ + 3Mθ = 30 × 3θ + 30 × 3θ

2+ 25 × 4θ

which gives

M = 36.2 kN m

The required moment of resistance is therefore 36.2 kN m.

S.18.10 There are three possible individual mechanisms as shown in Figs S.18.10(a),(b) and (c) in which (a) and (b) are beam mechanisms and (c) is a sway mechanism.

For (a)

W × 3θ + W × 3θ

2= 2Mθ + 2M × 3θ

2+ M × θ

2

Since M = 108 kN m

Wult = 132 kN

For (b)

W × 3θ + W × 6θ = 2Mθ + 2M × 3θ + M × 2θ

which gives

Wult = 120 kN

For (c)

0.75W × 6θ = 2Mθ + M × 3θ

2


2M

2M

3 m

(a)

3 m3 m

3uuB

C

D

W W

M

E

23u

23u

2u

(c)

9 m

0.75W

4 m

6 m

F

A

BE

u

6u 6u

23u

(b) 2M

2M

B

W

C

W

M

E

3u

2u

6u3uu

D

FIGURE S.18.10

from which

Wult = 84 kN

For the combined mechanisms, (a) + (c), with a hinge cancellation at B

9W θ = 2M × 3θ

2+ Mθ

2+ M × 3θ

2which gives

Wult = 60 kN

For the combined mechanisms, (b) + (c), with a hinge cancellation at B

13.5W θ = 2M × 3θ + M × 2θ + M × 3θ

2so that

Wult = 76 kN

Therefore the value of W which will just cause the frame to collapse is 60 kN.


S.18.11 For this frame there are three possible beam mechanisms as shown inFigs S.18.11(a), (b) and (c) and one sway mechanism, Fig. S.18.11(d).

(a)

3 m3 m3 m3 m

M

F

B

W W W

2M

2MD

C

uuE

2u3u

4u3

u3

(b)

W W W

FB

2M

2M

C

D

E

M

2u

3u3u6u

u u

(c)

B

CD

E

W W W

F

M2M

2M

3u 6u 9u3u

4u

u

(d)

A

B

G

FMW

u

9u 9u

12 m

6 m

2M

9 m

23u

FIGURE S.18.11(a–d)

For (a)

W (3θ + 2θ + θ) = M(

2θ + 8θ

3+ θ

3

)which gives

W = 0.833M


For (b)

W (3θ + 6θ + 3θ) = M(2θ + 4θ + θ)

i.e.W = 0.583M

For (c)W (3θ + 6θ + 9θ) = M(2θ + 8θ + 3θ)

so that

W = 0.722M

For (d)

W × 9θ = M(

2θ + 3θ

2

)i.e.

W = 0.389M

For (a) + (d) with a hinge cancellation at B

W (6θ + 9θ) = M(

8θ

3+ θ

3+ 3θ

2

)

so that

W = 0.3M

For (b) + (d) with a hinge cancellation at B

W (12θ + 9θ) = M(

4θ + θ + 3θ

2

)

i.e.

W = 0.309 M

For (c) + (d) with a hinge cancellation at B

W (18θ + 9θ) = M(

8θ + 3θ + 3θ

2

)

which gives

W = 0.463M

The minimum value of W to cause collapse is therefore 0.3 M with plastic hinges atC and F.

Referring to Fig. S.18.11(e)

6RG,H = M = W0.3

i.e.

RG,H = 0.56W


W

W W W

B

12 m

M

6 m

G

C D E F

9 m

ARA,H

RG,V

RG,H

RA,V(e) FIGURE S.18.11(e)


RA,H = W − 0.56W = 0.44W


RG,V × 12 + 0.56W × 3 = W (3 + 6 + 9 + 9)

i.e.

RG,V = 2.11W


RA,V = 3W − 2.11W = 0.89W .

S.18.12 The individual collapse mechanisms are shown in Figs 18.12(a), (b) and (c)where (a) and (b) are beam mechanisms and (c) is a sway mechanism.

For (a)

40(

3θ + 3θ

2

)= M

(θ + 2 × 3θ

2+ θ

2

)

which gives

M = 40 kN m

For (b)

40(3θ + 6θ) = M(θ + 2 × 3θ + 2θ)

from which

M = 40 kN m


3 m

2M

M M

CB

E

F

(a)

40 kN 40 kN

3u

3 m 3 m

23u

23u

2u

u

M

B

E

F2M

M

C

40 kN 40 kN

(b)3u

u 2u3u6u

20 kN

M

M

DA

6u

C

(c)

B

u

32u

u

32u

FIGURE S.18.12(a)–(c)

For (c)

20 × 6θ = M(

θ + 2θ

3+ 2θ

3+ θ

)i.e.

M = 36 kN m

For (a) + (c) with a hinge cancellation at B

5 × 40

(3θ + 3θ

2

)3

+ 120θ = 5 × M

(9θ2

)3

+ M(

10θ

3

)−5 × 2Mθ

3

so that

M = 56 kN m


For (b) + (c) with a hinge cancellation at B

5 × 40(3θ + 6θ)3

+ 120θ = 5 × 9Mθ

3+ 10Mθ

3−5 × 2Mθ

3

i.e.

M = 48 kN m

Therefore the required value of the plastic moment parameter is 56 kN m and plastichinges will occur at E and C.

Referring to Fig. S.18.12(d) and taking moments about A

40 kN

20 kN

40 kN

56 kN m

6 m

3 m 3 m 3 m 3 m 3 m(d)

A

RD,HRA,H

RD,VRA,V

D

CE F

B

FIGURE S.18.12(d)

RD,V × 15 − 40 × 9 − 40 × 6 − 20 × 6 = 0

so that

RD,V = 48 kN


RA,V + 48 − 40 − 40 = 0

i.e.

RA,V = 32 kN

Taking moments about C for CD

RD,H × 6 − 48 × 3 − 56 = 0

i.e.

RD,H = 33.3 kN


RA,H − 33.3 + 20 = 0

which gives

RA,H = 13.3 kN.


S.18.13 There are two possible collapse mechanisms, sway and gable, as shown inFigs S.18.13(a) and (b), respectively.

6u

9u

1.8u

2u

B C

E

D

A

5 kN

30 kN

(b)

u

5u

B

10 kN 5 kN

E

D

C

A

6 m 3 m

(a)

3 m

5 mu u

FIGURE S.18.13

For (a)

15 × 5θ = 4Mθ

so that

M = 18.75 kN m

For (b)

M(θ + 1.3 × 3θ + 3.8θ + 1.8θ) = 30 × 6θ + 5 × 9θ

which gives

M = 21.43 kN m

For (a) + (b) with a hinge cancellation at A

M(3θ + 9.5θ) = 75θ + 225θ

i.e

M = 24 kN m

Therefore the smallest value of the moment of resistance that can be used is 24 kN m.


S.18.14 In this case sway is prevented so that the possible collapse mechanisms areone of gable and two of rafter as shown in Figs S.18.14(a), (b) and (c).

u

u 3u

2u

PB

(c)C

A

P

(b)

B

u

u

3u 2u

AB

6 m 6 m

(a)

D

C

6u

4uP P

u

u

uu

P5

FIGURE S.18.14

For (a)

P(

3θ + 3θ + 4θ

5

)= 240 × 2θ + 200 × 2θ

which gives

P = 129.4 kN

For (b)

P × 3θ = 240 × 2θ + 240θ

i.e.

P = 240 kN

For (c)

P × 3θ = 240θ + 240 × 2θ + 200θ


from which

P = 306.7 kN

For (a) + 2 × (b) with a hinge cancellation at B

P(6.8θ + 2 × 3θ) = 200 × 2θ + 2 × 240θ + 240 × 2θ

so that

P = 106.3 kN

For (a) + 2 × (c) with a hinge cancellation at B

P(6.8θ + 2 × 3θ) = 200 × 2θ + 2 × 240θ + 240 × 2θ + 2 × 200θ

from which

P = 137.5 kN

Therefore the minimum critical value of P is 106.3 kN.



10 m

5 mB1

2.5B

12.5

A

12.5

A

12.5

45�

FIGURE S.19.1

the work absorbed in the yield lines is:

for the two panels A,

2m(

12.5

)× 10 = 8m

for the two panels B,

2 × 5(

12.5

)(0.4m + 0.6m) = 4m


The work done by the load is = 14[

5 × 5(

12

)+ 5 × 5

(13

)]= 291.7 kN m

Then

12m = 291.7

so that

m = 24.31 kN m/m.

S.19.2 Consider the slab shown in Fig. S.19.2(a).

FIGURE S.19.2

A

B

C13

13

14

10 m

3 m

4 m

(a)

A

B

C13

13

14

7 m 3 m

3 m

4 m

(b)

45� 45�

The work absorbed in the yield lines is:

for panel A,

m × 10(

13

)= 3.33m

for panel B,

2m × 10(

14

)= 5.0m

for panel C,

0.6m × 7(

13

)= 1.4m

The total work absorbed by the yield lines is therefore 9.73m.

The work done by the load = 10[

7 × 3 ×(

13

)+ 7 × 7

(12

)]= 315 kN m

Therefore

9.73m = 315

i.e.

m = 32.37 kN m/m

For the slab shown in Fig. S.19.2(b) the work absorbed in the yield lines is:

for panel A,

m × 7(

13

)= 2.33m


for panel B,

2m × 7(

14

)= 3.5m

for panel C,

0.6m × 7(

13

)= 1.4m

The total work absorbed in the yield lines is therefore 7.23m.


7 × 3(

13

)+ 7 × 4

(12

)+ 3 × 3

(12

)]= 255 kN m.

Therefore

7.23m = 255

i.e.

m = 35.27 kN m/m.


B

B

A

2.5 m

2.5 m

(12 kN/m2)

x

(8 kN/m2)

12.5

12.5

1x

7 m 3 m FIGURE S.19.3


for panel A,

0.4m × 5(

1x

)= 2m

x

for panel B,

2(m + m) × 10(

12.5

)= 16m

The total work absorbed by the yield lines is therefore m[16 + (2/x)].

The work done by the load is = 12[

5x(

13

)+ 5(7 − x)

(12

)]+ 8

[3 × 5

(12

)]

= 270 − 10x


Therefore

m(

16 + 2x

)= 270 − 10x

i.e.

m = 270 − 10x(16 + 2

x

)Differentiating the above with respect to x and simplifying gives

8x2 + 2x − 27 = 0

which gives

x = 1.716 m

Then

m = 14.73 kN m/m.


A

B

2.5 m

2.5 m

x 12.5

A

12.5

1x

8 m FIGURE S.19.4


for panels A,

2(20 + 16) × 8(

12.5

)= 230.4

for panel B,

(16 + 8) × 5(

1x

)= 120

x

The work done by the load = w[

(8 − x) × 5(

12

)+ 5x

(13

)]= w

[20 −

(2.5x

3

)]

Then

w(

20 − 2.5x3

)= 230.4 + 120

x


so that

w =(

230.4 + 120x

)[20 −

(2.5x

3

)]For a minimum value of w, dw/dx = 0. Differentiating the above with respect to x andsimplifying gives

x2 + 1.0417x − 12.5 = 0

which gives

x = 3.053 m

Substituting for x in the above expression for w gives

w = 15.45 kN/m2.


B

A A

C

11

B

11

12.5

12.5

1x

2.5 m 2.5 m1.0 m

(4�x) m

x

1.0 m

FIGURE S.19.5


for panels A,

2(m × 5 + 1.2m × 5)(

12.5

)= 8.8m

for panels B,

2(m × 2.5)(

11

)= 5.0m

for panel C,

(m × 6 + 1.2m × 6)(

1x

)= 13.2m

x


The total work absorbed by the yield lines is then = m[13.8 + (13.2/x)].


2 × 2.5 × 1(

13

)+ 2(4 − x) × 2.5

(12

)

+2 × 2.5x(

13

)+ 1x

(12

)+ 1(4 − x)

(11

)]

= 235 − 20x

Then

m(

13.8 + 13.2x

)= 235 − 20x

For x = 2.0 m, m = 9.56 kN m/m

For x = 2.5 m, m = 9.70 kN m/m

For x = 3.0 m, m = 9.62 kN m/m

i.e.

m = 9.70 kN m/m.


A

A

B

B

8 m

8 m

2.5 m

x

x

2.5 m

12.5

12.5

1x

1x

FIGURE S.19.6


for the panels A,

2[m(2.5 + x) + (1.4m × 8)](

12.5

)= m(10.96 + 0.8x)


for the panels B,

2(m × 2.5 + 1.4m × 2.5)(

1x

)= 12m

x

Therefore the total work absorbed in the yield lines = m[

10.96 + 0.8x +(

12x

)]

The work done by the applied load = 20[

2.52(

13

)+ 2.5x × 2

(13

)

+2 × 2.5(5.5 − x)(

12

)]

= 316.67 − 16.67x

Therefore

m[

10.96 + 0.8x + 12x

]= 316.67 − 16.67x

so that

m = 316.67 − 16.67x(10.96 + 0.8x + 12

x

)The maximum value of m is most easily found using the trial and error method. Then,

For x = 2.0 m, m = 15.27 kN m/m

For x = 2.5 m, m = 15.48 kN m/m

For x = 3.0 m, m = 15.36 kN m/m

Therefore the required value of the moment parameter is 15.48 kN m/m.


S.20.1

(a) A unit load is placed in different critical positions and the reaction calculated, i.e.

With unit load at C,

RAL − 1 × 5L4

= 0

which gives RA = 1.25

With unit load at A,

RA = 1

With unit load at B,

RA = 0


The RA influence line is shown in Fig. S.20.1(a).

1.25 1 �ve

C A B FIGURE S.20.1(a)

(b) With unit load at C, RA = 1.25 as in (a)


RA = 1


RA = 0

With unit load at D,

RAL + 1 × 0.25L = 0

which gives RA = −0.25

The influence line is shown in Fig. S.20.1(b).

1 �ve

�ve

1.25

0.25C A B

D

FIGURE S.20.1(b)

(c) With unit load at any point between A and B, RA = 1


RA = 0

The influence line is shown in Fig. S.20.1(c).

1

1

�ve

A B C FIGURE S.20.1(c)


S.20.2

(a) With unit load at D,

SC = −RA + 1 = −1.25 + 1 = −0.25


SC = −1 + 1 = 0

With unit load just to the left of C,

SC = −RA + 1 = −0.5 + 1 = +0.5

With unit load just to the right of C,

SC = 0.5 − 1 = −0.5


SC = −RA = 0

The influence line is shown in Fig. S.20.2(a)

�ve

�ve�veD A C

0.5

0.250.5

B

FIGURE S.20.2(a)

(b) With unit load at D,

SC = −RA + 1 = −1.25 + 1 = −0.25


SC = 0 (see (a))

With unit load just to the left of C,

SC = +0.5 (see (a))

The remaining part of the influence line follows from antisymmetry; the completeinfluence line is shown in Fig. S.20.2(b).

�ve 0.25�ve

�ve�veD A C

0.5

0.250.5

B

FIGURE S.20.2(b)


S.20.3

(a) With unit load at D,

MC = RA

(L2

)− 1

(3L4

)= −0.125L (RA from S.20.1(a))


MC = 0 since RA = 1


MC = RA

(L2

)= +0.25L


MC = 0 since RA = 0

The influence line for the bending moment at C is shown in Fig. S.20.3(a).

A

0.25L

0.125L

�ve

�ve C B

D

FIGURE S.20.3(a)

(b) The bending moment at C influence line may be determined from first principlesas in (a) or may be deduced from the influence line in (a) as shown in Fig. S.20.3(b).

0.125L 0.125L

0.25L

�ve �ve

�ve

A C B

D E

FIGURE S.20.3(b)

S.20.4 The shear force at C influence line and the bending moment at C influenceline are drawn as illustrated in S.20.2 and S.20.3 and are shown in Fig. S.20.4(a) and(b) respectively.

The maximum positive shear force will occur with the head of the load at C and is,

SC(max. pos) = 20 × 4 × 0.5 = 40 kN

The maximum negative shear force will occur with the tail of the load at C and is givenby

SC(max. neg) = 20(−4 × 0.5 + 1 × 0.125) = −37.5 kN


A

A

C

C

B D

0.5

10 m

1.5 1.5

2

4

0.5

�ve

�ve

�ve

�ve

B

2 m

8 m8 m

2 m

0.125 0.25

D

(a)

(b) FIGURE S.20.4

The maximum positive bending moment at C will occur with C dividing the load intotwo equal lengths. Then

MC(max. pos) = 20 × 2(1.5 + 4) × 52

= 550 kN m

The maximum negative bending moment at C will occur with the tail of the load at B.Then,

MC(max. neg) = −20 × 2 × 2 = −80 kN m.

S.20.5 The influence line for the bending moment at C is determined in an identicalmanner to that in S.20.3(b) and is shown in Fig. S.20.5; the ordinates at differentpositions 2 m apart have been calculated.

FIGURE S.20.5

2.5

0.75 0.52 m 2 m 2 m

0.5C

1.5

1.75 1.75

2.75 2.75

3.75

�ve

�ve

�ve

2.5

2 m 2 m 2 m 2 m 2 m

It is not clear which load of the group, when positioned at C, will give the maximumbending moment at C so that a trial and error method will be used. With the firstload at C

MC = 20 × 3.75 + 20 × 2.75 + 5 × 1.75 + 10 × 0.75 = 146.25 kN m


With the second load at C

MC = 20 × 2.75 + 20 × 3.75 + 5 × 2.75 + 10 × 1.75 = 161.25 kN m

With the third load at C

MC = 20 × 1.75 + 20 × 2.75 + 5 × 3.75 + 10 × 2.75 = 136.25 kN m

The maximum positive bending moment at C is therefore 161.3 kN m and occurs withthe second load at C.

Again it is not clear which position of the loads will give the maximum negative bendingmoment at C. With the first load at the end of the beam

MC = −20 × 2.5 − 20 × 1.5 − 5 × 0.5 + 10 × 0.5 = −77.5 kN m

With the second load at the end of the beam

MC = −20 × 2.5 − 5 × 1.5 − 10 × 0.5 = −62.5 kN m

Therefore the maximum negative bending moment at C is −77.5 kN m and occurs withthe first load at the end of the beam.

S.20.6 From the derivation of Eq. (20.18) the maximum bending moment occurs ata section of the beam under one of the loads such that the section and the centre ofgravity of the loads are positioned at equal distances either side of the mid-span of thebeam. The centre of gravity of the loads may be shown to be 4 m to the right of the lefthand load. Therefore positioning the inner 50 kN load 0.5 m to the right of mid-spanat D ensures that the above condition is satisfied; the maximum bending moment inthe beam then occurs under this load and the influence line and load positions are asshown in Fig. S.20.6(a).

BAD

2.61

4.99

2.89

4 m

10 m

50 kN 50 kN 30 kN

10 m

4 m0.5

m0.

5m

(a) FIGURE S.20.6(a)


The maximum bending moment in the beam is then

M(max) = 50 × 2.61 + 50 × 4.99 + 30 × 2.89 = 466.7 kN m

By inspection the maximum positive shear force will occur at B; the shear force at Binfluence line is shown in Fig. S.20.6(b).

A B

I

0.550.8

0.550.75

4 m5 m

4 m5 m

AB

I

(b)

(c) FIGURE S.20.6(b) and (c)

Then,

S(max. pos) = 30 × 1 + 50 × 0.8 + 50 × 0.55 = 97.5 kN

It may be easily verified that this value is greater than that when the inner 50 kN is at B.

Again by inspection, the maximum negative shear force will occur at A and in this casewill occur with the left hand 50 kN load at A. Then, from Fig. S.20.6(c)

S(max. neg) = 50(−1) + 50(−0.75) + 30(−0.55) = −104 kN.

S.20.7 The portion CE of the beam acts as a simply supported beam. Further, withunit load passing from left to right, it will cease to have any effect on the reactions,shear force and bending moment in the length AE as soon as it passes E.

RB IL:


RB = 0


RB = 1


RB × 4 − 1 × 5 = 0,

i.e.

RB = 1.25


With unit load at E etc.,

RB = 0

The IL is shown in Fig. S.20.7(a).

SK IL:

With unit load between A and B the influence line is the same as that for a simplysupported beam.

With unit load between B and C,

SK = −1 + RB

i.e.,

at B, SK = 0

at C, SK = +0.25

With unit load at E,

SK = 0

The IL is shown in Fig. S.20.7(b).

SD IL:

With unit load between A and D,

SD = 0

With unit load between D and C,

SD = 1


SD = 0

The IL is shown in Fig. S.20.7(c).

MK IL:

With the unit load between A and B the influence is the same as that for a simplysupported beam.


MK = RB × 2 − 1 × 3 = −0.5


MK = 0

The IL is shown in Fig. S.20.7(d).


MD IL:

With unit load between A and D,

MD = 0


MD = −1 × 0.5 = −0.5


MD = 0

The IL is shown in Fig. S.20.7(e).

A K B D C E

1 m2 m2 m 0.5 m0.5 m

1 1.25

(a)

0.5

0.5

0.25

�ve �ve

�ve

(b)

(c)

(d)

(e)

1

0.5

0.5

�ve

�ve

�ve

1

FIGURE S.20.7


S.20.8

RA IL:


RA = 1


RA = 0


RA = −13

The IL is shown in Fig. S.20.8(a).

RC IL:


RC = 0


RC × 20 − 1 × 15 = 0, i.e. RC = 34


RD × 15 − 1 × 20 = 0, i.e. RD = 1.33

Then,

RC × 20 − 1.33 × 15 = 0, i.e. RC = 1

The IL is shown in Fig. S.20.8(b).

ME IL:

ME = RC × 10 − RD × 5


RC = RD = 0 so that ME = 0


ME = 34

× 10 − 1 × 5 = 2.5


ME = 1 × 10 − 1.33 × 5 = 3.33

The IL is shown in Fig. S.20.8(c).


1

1

15 m 5 m

23

5 m

A D B

34

13

12

2 13

3

(a)

(b)

(c) FIGURE S.20.8

From the influence lines

RA(max) = 10 × 1 + 10(

23

)= 16.7 kN

RC(max) = 10(

34

)+ 10 × 1 = 17.5 kN

ME(max) = 10 × 2.5 + 10 × 3.33 = 58.3 kN m.

S.20.9 The influence lines for the shear force at sections spaced at 1 m intervals alongthe beam are shown in Fig. S.20.9(a).

ay b c d e

0.10.2 0.4 0.6

0.8

1

1

f

0.80.6

0.40.2

FIGURE S.20.9(a)


The maximum positive shear force at these sections is then:

With the head of the load at

b, S(max) = 1.2 × 0.1 × 1 = 0.12

c, S(max) = 1.2 × 0.3 × 1 = 0.36

d, S(max) = 1.2 × 0.5 × 1 = 0.60

e, S(max) = 1.2 × 0.7 × 1 = 0.84

f, S(max) = 1.2 × 0.9 × 1 = 1.08

y, S(max) = 1.2 × 0.05 × 0.5 = 0.03

The diagram of maximum positive shear force is then plotted to scale from thesevalues; the diagram of maximum negative shear force follows, both are shown in Fig.S.20.9(b). Superimposed on the diagrams of maximum shear force is the dead loadshear inverted. Then, from Fig. S.20.9(b) shear reversal will occur over the central1.3 m of the beam.

1.3 m

Shear reversalDead load

shear 1.25

1.25

FIGURE S.20.9(b)

S.20.10 The force in the member CD is given by −MF/4 (using the method of sections).The influence line for FCD is shown in Fig. S.20.10.

A C D E

0.625

1.25

0.625

FIGURE S.20.10

Then, with the loads in the given positions

FCD = −(20 × 0.625 + 10 × 1.25 + 5 × 0.625) = −28.1 kN.


S.20.11 From the method of sections FCG = MD/1.5, FHD = −MC/1.5, FFE sin α = RA

when unit load is to the right of F and FFE sin α = RB when unit load is to the left ofE; sin α = 1.5/2.5 = 0.6. The influence lines for the forces in CG, HD and FE are thenas shown in Figs S.20.11(a), (b) and (c), respectively.

(a)

2 m 3 m 3 m

1 1.67 1.67

� 2.678 � 8

16 � 1.5

(b)

11.67 1.75

� 2.56 � 10

16 � 1.5

(c)

1.25 �ve

1.67

0.29

0.21

FIGURE S.20.11

For the maximum force in CG place the right hand 70 kN load over the maximumordinate in the IL; the ordinates under the other loads are found by similar triangles.Then

FCG(max) = 15(

2.67 × 162

)+ 40 × 1 + 70 × 1.67 + 70 × 2.67 + 60 × 1.67 = 763 kN

For the maximum force in HD place the left hand 70 kN load over the maximumordinate in the IL. Then

FHD(max) = −[15(

2.5 × 162

)+ 40 × 1.67 + 70 × 2.5 + 70 × 1.75 + 60 × 1

]= −724 kN

For the maximum force in FE place the left hand 70 kN load over the 1.25 ordinate.Then

FFE(max) = 15(

0.21 × 2.292

)− 15

(1.25 × 13.71

2

)

+ 40 × 0.21 − 70 × 1.25 − 70 × 0.94 − 60 × 0.625

i.e.

FFE(max) = −307 kN.


S.20.12 When unit load is to the right of E, FCE may be found by taking moments aboutthe intersection of GC and ED produced and is then equal to 0.95RA. When unit loadis to the left of D, FCE may be found by taking moments about the intersection of GHand DE produced. Then, FCE = −2.085RB. The IL is then as shown in Fig. S.20.12(a).Further, FDE = MC/3 and the IL is as shown in Fig. S.20.12(b).

(a)

(b)

�ve

0.695 ED,C 0.475

2.085

2.38 m1.62 m

0.95

� 1.788 � 16

24 � 3

FIGURE S.20.12

Then

FCE(max. pos) = −30(

0.695 × 10.382

)+ 45

(0.475 × 13.62

2

)= 37.3 kN

FCE(max. neg) = −45(

0.695 × 10.382

)+ 30

(0.475 × 13.62

2

)= −65.3 kN

FDE = 45 × 24 × 1.782

= 961.2 kN.

S.20.13 The family of influence lines for the shear force in each panel of the truss isdrawn and is shown in Fig. S.20.13.

a b

ji

hg

fe

dc

rq

po

nm

lk

n1n2 n3 n4

1.0

1.0

12

34

56

78

9

FIGURE S.20.13

The central panel will require counterbracing.


For panel 4

S = 2.8 × area n3pa − 1.2(area n3fb − area n3pa)

= 4 × 3.375

(39

)2

− 1.2 × 5.625

(59

)2

= 0.38

Therefore panel 4 (and 6) require counterbracing.

For panel 3

S = 4 × 2.25

(29

)2

− 1.2 × 6.75

(69

)2

= −1.7

Therefore panel 3 and the remaining panels do not require counterbracing.

S.20.14 The influence lines are shown in Figs S.20.14(a), (b), (c) and (d).

FIGURE S.20.14

A

(a) RA

l

B D

C

(b) RB

A B

l

C D

(c) SE

l

A B E D

C

(d) ME l

A B

E

DC

S.20.15 Release the beam at A and apply a load W as shown in Fig. S.20.15.

wB

RB

A

x

2 m 2 m

C

FIGURE S.20.15

Then, RB = 2W .


Using Macauley’s method

EI

(d2v

dx2

)= Wx − 2W [x − 2]

EI(

dv

dx

)= Wx2

2− W [x − 2]2 + C1

EIv = Wx3

6− W [x − 2]3

3+ C1x + C2

The boundary conditions are: v = 0 when x = 2 m and 4 m. These give

C1 = −10W3

, C2 = 16W3

so that

v =(

WEI

){(x3

6

)− [x − 2]3

3−(

10x3

)+ 16

3

}

For the RA IL, v = 1 when x = 0

so that

1 =(

WEI

)(163

)

and

WEI

= 316

Then

RA =(

316

){(x3

6

)− [x − 2]3

3−(

10x3

)+ 16

3

}

When a udl of 30 kN/m covers the span AB

RA = 30(

316

)∫ 2

0

[(x3

6

)−(

10x3

)+ 16

3

]dx

i.e.

RA = 26.25 kN.



S.21.1 With the spring in position the forces acting on the column in its buckled stateare as shown in Fig. S.21.1.

4P

x�

L

y

kd

d

FIGURE S.21.1


EI

(d2v

dx2

)= 4P(δ − v) − kδ(L − x) (i)

The solution of Eq. (i) is

v = A cos µx + B sin µx + δ[4P + k(x − L)] (ii)

where

µ2 = 4PEI

When x = 0, v = 0 which gives

A = δ(kL − 4P)4P

Also, when x = 0, dv/dx = 0 from which

B = −δk4Pµ

Eq. (ii) then becomes

v =δ[(kL − 4P) cos µx − k( sin µx)

µ+ 4P + k(x − L)

]4P

(iii)

When x = L, v = δ, then

δ =δ[(kL − 4P) cos µL − k( sin µL)

µ+ 4P

]4P

from which

k = 4Pµ

(µL − tan µL).


S.21.2 Suppose that the buckling load of the column is P. Then from Eq. (21.1) andreferring to Fig. S.21.2,

P

P

�

x

D

C

EI

4EI

EI

Ay

B

L4

L4

L2

FIGURE S.21.2

in AB

EI

(d2v

dx2

)= −Pv (i)

and in BC

4EI

(d2v

dx2

)= −Pv (ii)

The solutions of Eqs (i) and (ii) are, respectively

vAB = A cos µx + B sin µx (iii)

vBC = C cos(µx

2

)+ D sin

(µx2

)(iv)

where

µ2 = PEI

When x = 0, vAB = 0 so that, from Eq. (iii), A = 0. Therefore

vAB = B sin µx (v)

When x = L/2, (dv/dx)BC = 0. Then, from Eq. (iv)

D = C tan(

µL4

)

and

vBC = C[

cos(µx

2

)+ tan

(µL4

)sin(µx

2

)](vi)

When x = L/4, vAB = vBC so that, from Eqs (v) and (vi)

B sin(

µL4

)= C sec

(µL4

)cos(

µL8

)(vii)


When x = L/4, (dv/dx)AB = (dv/dx)BC so that again, from Eqs (v) and (vi)

Bcos(

µL4

)= C

2sec(

µL4

)sin(

µL8

)(viii)

Dividing Eq. (vii) by Eq. (viii) gives

tan(

µL4

)tan(

µL8

)= 2

or (2 tan2 µL

8

)[1 − tan2

(µL8

)] = 2

from which

tan(

µL8

)= 0.707

i.e.µL8

= 0.615

i.e. (√PEI

)L8

= 0.615

so that

P = 24.2EIL2 .

S.21.3 The compressive load P will cause the column to be displaced from its initialcurved position to that shown in Fig. S.21.3.

y

P

P

L

�0

�x

FIGURE S.21.3


From Eq. (21.1) and noting that the bending moment at any point in the column isproportional to the change in curvature produced

EI

(d2v

dx2

)− EI

(d2v0

dx2

)= −Pv (i)

But

v0 = a(

4xL2

)(L − x)

Thend2v0

dx2 = −8aL2

and Eq. (i) becomes

d2v

dx2 +(

PEI

)v = −8a

L2 (ii)

The solution of Eq. (ii) is

v = A cos µx + B sin µx − 8a(µL)2 (iii)

where

µ2 = PEI

When x = 0, v = 0 so that

A = 8a(µL)2

Also, when x = L/2, (dv/dx) = 0 i.e.

B =8a tan

(µL2

)(µL)2

Eq. (iii) then becomes

v = 8a(µL)2 [ cos µx + tan

(µL2

)sin µx − 1] (iv)

The maximum bending moment occurs when v is a maximum at x = L/2,

i.e.

Mmax = −Pvmax

Therefore, from Eq. (iv)

Mmax =−8aP[sec

(µL2

)− 1]

(µL)2 .


S.21.4 The solution may be obtained directly from Eq. (21.40). Then

σmax =(

Pπdt

)1 +

1(

1 − PPCR

)(

δd2

)(

πd3t8πdt

)

i.e.

σmax =(

Pπdt

){1 +

[1

(1 − α)

](4δ

d

)}

S.21.5

(a) The Euler buckling load for a pin-ended column is given by Eq. (21.5). The secondmoment of area of a circular section column is πD4/64 which, in this case, is givenby I = π × 12.54/64 = 1198.4 mm4

The area of cross section is

A = π × 12.52

4= 122.72 mm2

(i)

P = π2 × 200 000 × 1198.4(500)2

i.e

P = 9462.2 N

Therefore the test result conforms to the Euler theory.

(ii)

P = π2 × 200 000 × 1198.4(200)2

i.e.

P = 59 137.5 N

Therefore the test result does not conform to the Euler theory.

(b) From Eq. (21.27) and considering the first test result

9800 = 122.72σs[1 + kL2(

1198.4122.72

)]

which simplifies to

79.86 + 2.05 × 106k = σs


Similarly, from the second test result

215.12 + 0.88 × 106k = σs

Solving gives

k = 1.16 × 10−4, σs = 317.2 N/mm2.

S.21.6 The second moment of area of the column is given by

I =π

[D4 −

(7D8

)4]

64= 0.0203D4


A =π

[D2 −

(7D8

)2]

4= 0.1841D2

Then

r2 = 0.0203D4

0.1841D2 = 0.11D2


300 × 3 × 103 = 0.1841D2[1 +

(1

7500

)(2.5×103)2

(0.11D2)

]which simplifies to

D4 − 14.85 × 103D2 − 0.125 × 103 = 0

Solving

D = 122 mm

Say

D = 128 mm, t = 8 mm.

S.21.7 The column will buckle about an axis parallel to its web. The second momentof area of the column is then given by

I = 2 × 8 × 1303

12+ 184 × 63

12= 2.93 × 106 mm4

The area of cross section is

A = 2 × 130 × 8 + 184 × 6 = 3184 mm2


Then

r2 = 2.93 × 106

3184= 920.2

so that

r = 30.3 mm

Then

Lr

= 2.5 × 103

30.3= 82.5

Also

σCR = π2 × 200 00082.52 (see Eq. (21.25))

i.e.

σCR = 290.0 N/mm2

Substituting these values etc. in Eq. (21.46) gives

σ = 180.9 N/mm2

Then the maximum load the column can withstand is given by

P = 180.9 × 3184 × 10−3 = 576 kN

The Euler buckling load is

PCR = 290 × 3184 × 10−3 = 923.4 kN

ThenP

PCR= 576

923.4= 0.62.

S.21.8 The bending moment at any section of the column is given by

M = PCRv = PCRkx(L − x)

Alsodv

dx= k(L − 2x)


U + V =(

P2CRk2

2E

)[(1I1

) ∫ a

0(Lx − x2)2dx +

(1I2

)∫ L−a

a(Lx − x2)2dx+

(1I1

)∫ L

L−a(Lx − x2)2dx

]−(

PCRk2

2

)∫ L

0(L − 2x)2dx


i.e.

U + V =(

PCRk2

2EI2

){(I2

I1− 1)[

L2a3

3− La4

2+ a5

5− L2(L − a)3

3

+L(L − a)4

2− (L − a)5

5

]+(

I2

I1

)(L5

30

)}− PCRk2L3

6


∂(U + V )∂k

= 0

Then, since I2 = 1.6I1 and a = 0.2L this gives

PCR = 14.96EI1

L2

Without the reinforcement

PCR = π2EI1

L2

The ratio of the two is 14.96/π2 = 1.52

Therefore the increase in strength is 52%.

S.21.9 Assume the equation of the deflected centre line of the column is

v =(

4δ

L2

)x2

in which δ is the deflection of the ends of the column relative to its centre and the originfor x is at the centre of the column. The second moment of area varies in accordancewith the relationship

I = I1

[1 − 1.6

( xL

)]The bending moment at any section of the column is given by

M = PCR(δ − v) = PCRδ

[1 − 4

(x2

L2

)]

Also, from the above

dv

dx=(

8δ

L2

)x


U + V =(

PCRδ2

EI1L3

)∫ L2

0

[(L2 − 4x2)2

L − 1.6x

]dx −

(64PCRδ2

L4

)∫ L2

0x2dx


i.e.

U + V = 0.3803P2CRδ2L

EI1− 8PCRδ2

3L


∂(U + V )∂δ

= 0.7606P2CRδL

EI1− 16PCRδ

3L= 0

which gives

PCR = 7.01EI1

L2

For a column of constant thickness and second moment of area I2

PCR = π2EI2

L2

For the columns to have the same buckling load

π2EI2

L2 = 7.01EI1

L2

so that

I2 = 0.7I1

Therefore, since the radii of gyration are the same, A2 = 0.7A1

The weight of the constant thickness column is equal to ρA2L = 0.7ρA1L.

The weight of the tapered column is equal to ρ × average thickness × L = ρ × 0.6A1Lso that the saving in weight is 0.1ρA1L, i.e.

saving in weight = 0.1 ρA1L0.7ρA1L

= 0.143

i.e. about 15%.

Megson ISM

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