THEORY OF STRUCTURES II

Compiler: CA Hollenbach

Moderator: IG Watts

Revisor: BD Ikotun

Department of Civil and Chemical Engineering

University of South Africa, Florida

Only study guide for

TST271Z

2

THEORY OF STRUCTURES II

(TST271Z)

THEORY OF STRUCTURES II

STUDY GUIDE

(TST271Z)

Compiled by: CA HOLLENBACH PrEng BSc BSc (Eng) FIStructE MSAICE TED

Moderator: IG WATTS NDT(CIVIL) MSAICET

Revised by: BD IKOTUN BEng (CIVIL) MSc (Eng)

3

COPYRIGHT DECLARATION

In terms of Copyright Act, 98 of 1978 no part of this material may be produced, published, redistributed, screened or used in any form without prior written permission from Unisa. When materials have been used from other sources permission must be obtained directly from the original source.

4

CHAPTER CONTENTS PAGE

CHAPTER 1 STRESSES AND STRAINS 6

CHAPTER 2 SECTIONAL PROPERTIES 52

CHAPTER 3 SIMPLE BEAMS 86

CHAPTER 4 THEORY OF ELASTIC BENDING 130

CHAPTER 5 MOMENT-AREA METHOD OF ANALYSIS 158

CHAPTER 6 MATHEMATICAL METHOD OF ANALYSIS 197

CHAPTER 7 ELASTIC REINFORCED CONCRETE THEORY 231

CHAPTER 8 STATICALLY DETERMINATE PIN-JOINTED

FRAMES 260

CHAPTER 9 IMPACT LOADING 298

CHAPTER 10 ANALYSIS OF STRESSES 313

5

TST271Z CHAPTER 1

STRESSES AND STRAINS

SPECIFIC OBJECTIVES

By the end of this chapter, you should be able to do the following:

Obtain the mechanical properties of a material given the load-elongation data obtained from a tension test.

Define factor of safety. Determine the stresses and strains in bars that are connected in series and/or parallel,

which are subjected to axial loads and/or changes in temperature.

6

TST271Z CHAPTER 1

STRESSES AND STRAINS

CONTENTS PAGE

1.1 INTRODUCTION .............................................................................................................. 7

1.2 STRESS ............................................................................................................................... 8

1.3 DIRECT STRESS AND STRAIN; YOUNGS MODULUS ........................................... 9

1.4 STRESS-STRAIN CURVE FOR MILD STEEL ........................................................... 13

1.5 WORKING STRESS, STRESS FACTOR AND LOAD FACTOR .............................. 16

1.6 POISSONS RATIO ......................................................................................................... 20

1.7 BARS IN SERIES AXIALLY LOADED ........................................................................ 27

1.8 COMPOSITE BARS ........................................................................................................ 30

1.9 TEMPERATURE STRESS IN COMPOSITE BARS ................................................... 35

1.10 BENDING STRESS ....................................................................................................... 41

1.11 SHEAR STRESS ............................................................................................................ 42

1.12 TORSIONAL STRESS .................................................................................................. 45

1.13 VOLUMETRIC STRESS AND STRAIN AND BULK MODULUS .......................... 46

1.14 RELATIONSHIP BETWEEN THE ELASTIC CONSTANTS ................................. 48

1.15 HARDNESS TEST ......................................................................................................... 49

1.16 TUTORIAL .................................................................................................................... 50

7

STRESSES AND STRAINS

1.1 INTRODUCTION

This chapter deals with the effects of forces. A force can be defined as that which changes or

tends to change the position of rest or motion of a body. Force is a vector quantity, i.e. it has

magnitude as well as direction. It can be represented graphically by a straight line drawn to

scale, with an arrowhead indicating the direction of the force.

The unit of force is the Newton which is defined as a force that will accelerate 1 kg mass 1

metre per second2 (i.e. 1 metre per second per second). The acceleration due to gravity

equals 9.81 m/sec2. Thus 1 kg mass exerts a force of 9.81 Newtons.

4 cm

4 kN Scale: 1 cm = 1 kN

Graphical representation of a force of 4 kN acting horizontally from left or right

The moment of a force F about a point A is equal to the force multiplied by the shortest

distance between point A and the line of action of the force.

Moment of F about A = Fa Nm or MA = Fa Nm

A

F (N)

8

1.2 STRESS

Stress is defined as force per unit area.

= FA

The basic unit of stress is the Pascal: 1 Pa = 1 N/m2

Stress may be due to any of the following:

1) Direct or axial force

2) Bending

3) Shear Force

4) Torsion

9

1.3 DIRECT STRESS AND STRAIN: YOUNGS MODULUS

Direct stress may be either tensile or compressive

Tensile stress tFA

= Compressive stress CFA

=

Where F = direct axial force (either tensile or compressive)

A = cross sectional area

Figure 1.1: Direct stress

Stress causes deformation of the body in which it occurs. Direct stress causes -:

a) Lengthening if it is tensile

b) Shortening if it is compressive

Strain is defined as change of length per unit length:

= ll

Where l = original length

l = change of length

Youngs modulus E (or Modulus of direct elasticity)

According to Hookes Law:

FF F F

10

Strain stress

Hence, E x strain = stress (E = Constant)

i.e. E = stressstrain

E stressstrain

FAll

Forcearea

changein lengthoriginal length

= = = =

E is termed Youngs Modulus and since strain is dimensionless, E is usually expressed in

kN/mm2 i.e. in GPa.

The above relationship holds good for any elastic material within its elastic range.

Values of E for different materials are found experimentally. The following are typical

values for E:

Structural or mild steel: E = 210 kN/mm2 = 210 GPa

Concrete: E = 14-28 kN/mm2 = 14-28 GPa

Structural grade timber: E = 6-16.5 kN/mm2 = 6-16.5 GPa

Note : 1 GPa = 1 000 000 000 Pa

= 1 000 000 000 N/m2

= 1 000 N/mm2

= 1 kN/mm2

11

Example 1.3.1

A bar of sectional area 1 250 mm2 and 2 m in length extended 0.4 mm when an axial load of

52.5 kN was applied. Calculate Youngs modulus for the material of the bar.

SOLUTION

( )GPaormmkNmmN

mmNmmN

ll

AF

strainstressE

210210210000

20004.0

125010005.52

2

2

22

==

=

=

===

Example 1.3.2

A short timber post of rectangular section has one side of its section twice the other. When

the post is loaded axially with 9.8 kN it contracts 0.119 mm per metre length. If E for the

timber = 8.4 GPa calculate the cross sectional dimensions of the post.

SOLUTION

Let x be smaller side

Then 2x = larger side

Area of section = 2x2 mm2

12

mmx

mmx

mmNxmmN

strainstressE

strain

mmNx

mmNx

NareaLoadstress

70

4902000119.08400

4900000119.0

4900

8400

000119.01000

119.0

49002

9800

22

22

2

22

22

=

=

=

=

=

==

===

Dimensions of post are 140 mm 70mm

13

1.4 STRESS-STRAIN CURVE FOR MILD STEEL

If the changes in length of a test specimen as a result of different loads are measured, the

stress-strain curve can be plotted. For mild steel a tensile load is usually applied, the curve

for a compressive load being virtually the same up to the point when the change in cross

section of the test specimen becomes substantial.

The ends of the test specimen are gripped in the jaws of a special testing machine in which

measurable tensile loads can be applied. A strain gauge is attached over a length l on the test

specimen, and read for various increasing loads until the specimen breaks.

The stress-strain curve can then be plotted. For mild steel it has the form as shown in fig. 1.2.

Figure 1.2 Stress-strain curve for mild steel

From 0 to the elastic limit A is a straight line, and Hookes law, which states that stress is

proportional to strain applies. If the load is removed before the elastic limit is reached, the

test specimen will revert to its original dimensions.

After the elastic limit has been passed, the strain is semi-plastic and on removal of the load,

some extension will remain, called the permanent set.

E

limit of proportionality

(elastic limit)

Semi-plastic range Upper yield stress

Plastic rangeElastic range

A B

General extension Local extension

E = =

tan

strain ll

=

lower yield stress y

Ultimate stress ult

C D

F

0

14

On increasing the load further, the yield stress is reached at point B, where a sudden

elongation from C to D and a rapid drop in stress from B to C takes place. Point B is called

the upper yield point and C the lower yield point. Beyond D, strain hardening occurs, and

stress again increases with strain, and reaches a maximum value, known as the ultimate stress

ult at point E, where the phenomenon of necking occurs. This is a rapid reduction in cross-

sectional area at some weak point in the test specimen. From E to F there is a reduction in

nominal stress (i.e. areaoriginalload ) until fracture occurs at F.

For mild steel and other ductile materials:

Ultimate tensile strength = max imum load

original cross sectional area

Nominal fracture stress = load at fractureoriginal cross sectional area

True fracture stress = load at fracturefinal cross sectional area at fracture

Example 1.4.1

A tensile test is carried out on a mild steel bar of 20 mm diameter. The bar yields under a

load of 80 kN. It reaches a maximum load of 150 kN and finally breaks at a load of 70 kN.

The diameter at the fracture was measured as 10.2 mm.

Calculate: (a) The tensile stress at yield point

(b) The ultimate tensile stress

(c) The nominal fracture stress

(d) The true fracture stress

15

SOLUTION

(a) Nominal tensile stress at yield point

MPa

mmN

areationalcrossoriginalAwhereAF

mmdA

yy

6.2542.314

1080

sec

2.3144

)20(4

23

00

222

0

=

=

==

===

(b) Ultimate tensile stress

MPaxA

Fult 4.4772.314

10150 3

0

max===

(c) Nominal fracture stress

MPaxAFf

fn 8.2222.3141070 3

0

===

(d) True fracture stress

MPax

AF

mmA

f

fft

f

6.85672.811070

72.814

)2.10(

3

22

===

==

16

1.5 WORKING STRESS, STRESS FACTOR AND LOAD FACTOR

The stresses that are present in a component of a structure under normal working conditions

are called the working stresses (w). The ratio of the yield stress of the material of the

component to the working stress is the stress factor against yielding.

Stress factor against yielding =

y

w

Previously this ratio was called the factor of safety. Modern literature prefers the term stress

factor since this defines more precisely that working stress is compared with yield stress.

In practical problems, working stresses can only be estimated approximately in stress

calculations. For this reason the stress factor may give little indication of the safety of a

component. A more realistic estimate of safety can be made by finding the extent to which

the applied or working loads may be increased before collapse or fracture occurs. A working

load F to which a load factor n is applied becomes a factored load nF.

Example 1.5.1

(a) A mass of 200 kg is suspended from a metal rod of 5 mm diameter and 14 m long.

Find the change of length of the rod if E for the metal is 180 GPa.

(b) If the yield stress of the metal is 380 MPa, find the largest mass that the rod can

support if a stress factor of 2.5 is applied.

SOLUTION

(a)

( )mm

mmAEFll

lAFl

ll

AF

E

8.7

1010180005.04

1481.9200 392

=

==

==

Change in length = +7.8 mm

17

(b)

( )kgm

Nx

AF

MPa

factorstress

yw

w

y

3.3058.9

2985

005.04

10152

1525.2

3805.2

5.2

26

==

=

=

===

==

Mass = 305.3 kg

Example 1.5.2

A strut 2 m long and 20 mm diameter is subjected to a compressive force of 40 kN. Find the stress and the change in length if E = 210 GPa.

SOLUTION

( )

( ) mmEl

AF

AEFll

lAFl

llAFE

MPaStressmN

mNAF

39

6

26

2

2

3

1010210

210127

12710127

02.04

1040

=

==

==

==

==

l = 1.21 mm (shortening)

18

Example 1.5.3

A tie of 25 mm diameter and 12 m long is stressed to 140 MPa. The total extension at this

stress is 9.8 mm. Find E for the tie and the total load.

SOLUTION

( )N

NAF

GPaEmNE

lstrain

68722

025.04

10140

44.1710008166.0

10140

0008166.0120098.0

26

26

=

==

=

==

===

l

Load = 68.7 kN

Note:

In paragraph 1.4 the stress-strain curve for mild steel, which is a ductile material is described.

In brittle materials, such as high tensile steel, there is no marked yield point. The stress-strain

graph is a straight line until the limit of proportionality is reached. Thereafter it curves

upward until fracture occurs at ultimate stress.

Figure 1.3

Fracture at ult

stress limit of proportionality

strain

19

Whereas the stress factor on yielding is y/w for mild steel, it is more convenient to use the

term proof stress for high tensile steel.

The proof stress is found by drawing the line parallel to the linear-elastic line at the

appropriate proof strain, as illustrated in figure 1.4 below.

Figure 1.4

strain

stress 0,2% proof stress

0,1 % proof stress

0

20

1.6 POISSONS RATIO

When a body is subjected to tension (or compression) the stretching (or contraction) in the

direction of the force is accompanied by a smaller contraction (or expansion) at right angles

to the force.

Figure 1.5

Let = applied stress

Let = strain in direction of applied stress

Then: lateral strain = m

where 1m

= Poissons ratio

The value of is found experimentally and for metals it is approximately 0,3.

Note Although we might well apply a positive or negative sign to 1m

, it is accepted usage

not to give a sign, but to state increase or decrease.

l

Stretching contraction

contraction expansion

21

Example 1.6.1

A strut 1.6 m long has a cross section of 3 cm 5 cm. If a compressive force of 220 kN acts

along the long axis of the strut, find the change in the longitudinal and lateral dimensions.

Poissons ratio = 1/3 and E = 200 GPa.

SOLUTION

( )

)(0073.0

1003.031

106.117.13)

:dim31

106.117.1

1

17.1

101020005.003.0

6,110220

33

3

39

3

increasemm

mmsidecma

ensioninchangeLateral

mll

mstrainLateral

decreasemml

mm

EAFll

ll

AF

E

=

=

=

==

=

=

=

==

)(0122.0

0073.0355)

increasemm

mmsidecmb

=

=

22

Example 1.6.2

A cube of 12 cm sides and made of material for which E = 80 GPa, is subjected to a

compressive force of 2000 kN in the x direction, a tensile force of 1500 kN in the y direction

and no force in the z direction. Calculate the change of dimensions in each direction if

Poissons ratio = 9.21 . Also calculate the new dimensions.

SOLUTION

0

2.1041012.012.0

101500

9.1381012.012.0

102000

36

63

=

=

=

=

=

z

y

x

MPa

MPa

Consider the force in the x direction. Fx = 2000 kN compression

z

x

1500 kN

y 1500 kN

2000 2000

23

( )

( )( )

( )( )

( )increasemmlmll

increaseml

l

increasemmlmmll

increaseml

ldecreasemmlmm

lldecrease

EAF

ll

zzzz

x

z

zz

yyyy

x

y

yy

x

xxx

x

x

x

xx

072.01209.21074.1

9.21074.1

072.01209.21074.1

9.21074.1

209.01201074.1

1074.1108012.012.0

102000

3

3

3

3

3

3

9

3

=

==

===

=

==

===

==

=

=

===

Consider the force in the y direction. Fy = 1500 kN tension

( )( )

( )

( )decreasemmlmmllml

l

decreasemmlmmll

mll

increasemmlmmllincrease

EAF

ll

zzzz

y

z

zz

xxxx

y

x

xx

yyyy

y

y

y

yy

0554.01209.21034.1

9.2103.1

054.01209.2103.1

9.2103.1

156.0120103.11030.1

108012.012.0101500

3

3

3

3

3

3

9

3

=

==

===

=

==

===

===

=

===

Total changes of dimension

x direction : -0.209 0.054 = -0.263 mm

y direction : +0.072 + 0.156 = +0.228 mm

z direction : +0.072 0.054 = +0.018 mm

24

New dimensions

x direction : 120 0.263 = 119.737 mm

y direction : 120 + 0.228 = 120.228 mm

z direction : 120 + 0.018 = 120.018 mm

Example 1.6.2 in reverse

If we know the strains in three mutually perpendicular directions, as well as the values for

Youngs modulus E and Poissons ratio 1m

for the material, we can find the stresses in the

three directions (and hence the stress in any other direction, if required).

Let stresses x y and z occur in three mutually perpendicular directions x, y and z. The

strains x, y and z resulting in each direction will be made up of the following:

(a) Longitudinal strain due to stress in the given direction; plus

(b) Lateral strains due to the stresses in each of the other two directions.

Let tensile stress be positive, compressive stress negative. Let increase in dimension be

positive, decrease negative.

The following equations will hold good:

In x direction

xx y z

x xy z

E mE mE

ie Em m

=

=

In y direction

yy x z

yx

yz

E mE mE

ie Em m

=

= +

1

2

25

In z direction

zz x y

zx y

z

E mE mE

ie Em m

=

= +

If x, y and z are known, and also E and 1m

, then the three equations above can be solved for

x, y and z.

Now again consider example 1.6.2 (in reverse). Suppose the strains are now given and E and 1/m are known, and it is required to determine the stresses in the three directions and hence the

loads.

Let us first determine the total strains in the three directions:

( )3

33 10188.20

9.2103.11074.1

=+

=++=mm

zyxtx

( )33

3

109.10103.19.2

1074.1

+=+++

=++=mm

zy

xty

( )3

33

10152.009.2103.1

9.21074.1

+=+

+=++= z

yxtz mm

Using these total strains we can now find the stresses by substituting in equations ,

and above :

80 109 (-2.188 10-3) 9.29.2zy

x

+=

80 109 (+1.9 10-3) 9.29.2z

yx

+=

80 109 (+0.152 10-3) z

+=9.29.2

3

1 2

3

1a

2a

3a

26

We now have three equations with three unknowns, which on solving give:

x = -139.1 MPa

y = +104 MPa

z = 0

These are the same stresses we had at the start.

27

1.7 BARS IN SERIES AXIALLY LOADED

Consider three bars in series axially loaded, as shown in fig 1.6(a) forming member A B

(a)

(b)

Figure 1.6

The member as a whole is in equilibrium: P = 0

i.e -P1 - P2 - P3 + P4 + P5 = 0

Each part of the member must be in equilibrium.

Considering free body diagram to left (or right) of MM

FM = P1 (= -P2 -P3 + P4 +P5)

B

N O

RA M

P1 x P4 P5 P3P2

RM

O N

P3P2P4

A B

P5 P1

FMFM

R O N M M

O RNMM

28

Similarly: FN = -P1 -P2 (= P3 - P4 - P5)

F0 = -P1 - P2 - P3 (= - P4 - P5)

FR = -P1 - P2 - P3 +P4 (= - P5)

The total change in the length of the member equals to the algebraic sum of the change in

length of each part.

Example 1.7.1

Two round mild steel bars AB and CD are connected by a square copper bar BC. The

lengths, diameters and sizes are shown in the sketch. The compound bar is subjected to a

tensile axial force P. If the total elongation of the compound bar is 1.2 mm, find (a) the force

P and (b) the forces in the bars. Youngs modulus for mild steel is 200 GPa and for copper

100 GPa.

SOLUTION

The applied force P is transmitted throughout the length AD

FAB = FBC = FCD = P

Also: LAB + LBC + LCD = 1.2 mm

20mm square

DCB

A

0.4m0.8m0.6m

P P

15mm diameter 15mm diameter

m.s.copperm.s.

29

(a)

( ) ( ) ( )( )

kN

kNP

P

P

EAPl

EAPl

EAPl

AEFllei

lAFl

llAFE

CDCD

CD

BCBC

BC

ABAB

AB

8.24

103.48

100012.0

0012.032.112098.1610

0012.010200015.0

4

4.01010002.0

8.0

102000150.04

6.0

0012.0

..

39

9

9292

92

=

=

=++

=

+

+

=++

===

(b) ( )

MPaCDAB 3.14010015.0

4

108.24 62

3

=

==

( ) MPaBC 6210020.0108.24 6

2

3

=

=

30

1.8 COMPOSITE BARS

A composite bar is made up of two or more different materials. They are connected in such a

way that the change of length under load is the same for each constituent material.

(a) Under tension (b) Under compression

Figure 1.7 Composite bars

Because the constituent members of a composite bar under load remain the same length:

strain = ll

is the same for all constituent members

But strain =

=

EEstress

If a composite bar is made up of materials A, B, C ...

then ====C

C

B

B

A

A

EEEll

If FA, FB, FC.... are the forces carried by A B C ....

and AA, AB, AC .... are the cross sectional areas of A, B, C .....

l l

F

F F

l l

F

31

thenFA

FA

FA

FA E

FA E

FA E

andF F F F

AA

AB

A

BC

C

C

A

A A

B

B B

C

C C

Total A B C

= = =

= =

= + + +

The above two equations can be solved to find the portion of the total load carried by each

material of which the composite bar is made.

Example 1.8.1

A strut is made of three strips of metal glued together. One strip is steel with E = 210 GPa

and cross section 8 cm 2 cm; the second strip is an aluminium alloy with E = 70 GPa and

cross section 8 cm 3 cm; and the third strip is bronze with E = 110 GPa and cross section 8

cm x 1 cm. If the whole strut is subjected to a compressive force of 0.3 MN, find the force

carried by each strip and the stress in each. Also find the change in length of the strut if it is

0.6 m long.

SOLUTION

8 cm

2 cm

3 cm

1 cm Bronze B

Steel S

Aluminium 0.3MN 0.3MN

0.6 m

32

88168336..

1011001.008.0107003.008.01021002.008.0 999

BAS

BAs

BB

B

AA

A

SS

S

FFFei

FFFEA

FEA

FEA

F

==

=

=

===

From the above:

168 FS = 336 FA

88 FS = 336 FB

Also FS + FA + FB = 0.3 103 kN

On solving the three equations with three unknowns we get the following:

MPaeikNF SS 4.1061002.008.0103.170..3.170 6

3

=

==

MPaeikNF AA 5.351003.008.0102.85..2.85 6

3

=

==

MPaeikNF BB 6.551001.008.0105.44..5.44 6

3

=

==

Check : FS + FA + FB = 300 kN

Change in length The strain is the same for S, A and B.

Using steel:

mm

Ell

Ell

39

6

1010210

6.0104.106

==

==

Change in length l = 0.304 mm

1

2

3

33

Example 1.8.2

The figure shows the cross section of a short reinforced concrete column. Calculate the stress

in the concrete and the stress in the steel if an axial load of 735.5 kN is applied to the column.

Assume that the bond between the steel and the concrete is sufficient to prevent slip.

Given: E for steel = 210 GPa

E for concrete = 14 GPa

420 steel bars as reinforcement

SOLUTION

( )( ) 232 102566.102.044 mAs ==

Strain = ll

is the same for the steel and the concrete

S

S

C

CE E=

S C C= =21014

15

Also FS + FC = Ftotal

S AS + C AC = 735.5 103 N

Substitute S = 15C in :

15C 1.257 10-3 + C 106.74 10-3 = 735.5 103

1

2

2

33-

2-3C

m 10 106.74

m 10 1.2566 - 0.3 0.36=A

=

360 mm

300 mm

34

C (125.6 10-3) = 735.5 103 N/m2

C = 6 106 N/m2 C = 6 MPa

From S = 15C S = 90MPa

Note The factor ES/EC (= m) is used in the modular ratio method of reinforced concrete

design which is dealt with in greater detail in Chapter 7.

1

35

1.9 TEMPERATURE STRESS IN COMPOSITE BARS

In composite bars made up of materials with different rates of thermal expansion, internal

stresses are set up by temperature changes.

Consider a simple composite bar consisting of two members - a solid round bar B contained

inside a circular tube T. The coefficients of linear thermal expansion are B and T

respectively.

If the ends of the bar and tube are attached rigidly to each other, longitudinal stresses are set

up by a change of temperature.

If the bars are not attached, each bar will extend freely:

Bar B will extend B Lo t

and Tube T will extend T L0 t

where t = increase in temperature

L0 = original length of B and T.

If the members are attached to each other, the one with the higher coefficient of expansion

will be compressed by a force F, while the one with the lower coefficient of expansion will be

extended by an equal force F.

The two forces must be equal and opposite in order to maintain equilibrium of internal forces.

36

Original member Temperature raised tC

Cross sectional areas: Members free Members

BAR B: A(B) to expand attached

Tube T: A(T) separately to each other

Figure 1.8

( ) ( )( ) ( )

( )

El l

l lE

Also F A

l lE

and l lE

But l l l t l tl

El

El t

BB o

BT

T o

T

B T B o T o

B o

B

T o

To B T

= = =

=

= =

+ =

+ =

Also : Compressive force in bar B = tensile force in tube T

B AB = T AT

We now have two equations with two unknowns B and T, which can be solved.

Note The original length l0 is immaterial as it cancels out in equation .

1

1

2

l (B)

T l0t

B l0 t

l (T)Bar B

Tube T

B > T

l0

37

Example 1.9.1

An aluminium rod 2.2 cm diameter is threaded at the ends, and passes through a steel tube 2.5

cm internal diameter and wall thickness 0.3 cm. Both are heated to a temperature of 140C,

when the nuts on the rod are lightly screwed onto the ends of the tube. Calculate the stress in

the rod and in the tube when the common temperature has fallen to 20C.

Given: ES = 200 GN/m2 S = 1.2 10-5 per C

EA = 70 GN/m2 A = 2.3 10-5 per C

SOLUTION

At 140C Temperature falls 140C to 20C

AA = /4 (22)2 mm2 Members free Nut

AS = /4 (302 - 252) mm2 to contract screwed on

separately

( ) ( )

( ) ( )( )

( )( )

26

4

599

000

00

00

101848720

101.11207020720

102.13.2120102001070

mN

tlE

lE

ltltlllBut

Ell

Ell

SA

SA

SA

SAS

S

A

A

SASA

S

SS

A

AA

=+

=+

=

+

=+

=+

==

1

Al0t

Sl0t (l)S

(l)A A

S

38

( ) ( )

SA

SA

SASSAA

ei

eiAAAlso

694.022

2531..

25314

224

.

2

22

222.

=

=

=

=

Substituting in :

20 (0.694) S + 7 S = 1848 106 N/m2

20.88 S = 1848 106 N/m2

S = 88.5 MPa

and A = 61.4 MPa

Example 1.9.2

A bar of brass 25 mm diameter is enclosed in a steel tube 50 mm external diameter and 25

mm internal diameter. The bar and tube are both initially 1 m long and are rigidly fastened at

both ends. Find the stresses in the two materials if the temperature rises from 15C to 95C.

If the composite bar is then subjected to an axial tensile load of 50 kN, find the resulting

stresses and the increase in length from the original state.

Given: ES = 200 GN/m2 EB = 100 GN/m2

S = 11.6 10-6 / C B = 18.7 10-6 / C

1

2

39

SOLUTION

Brass bar

Steel tube

Original state Temperature raised 80C

Free Rigidly connected at ends

( ) ( )( ) ( )

( )26

2699

0000

00

00

106.1132

106.117.18801020010100

mN

mN

tltlE

lE

ltltlllBut

Ell

Ell

Ell

llE

SB

SB

SBS

S

B

B

SBSB

S

SS

B

BB

=+

=

+

=+

=+

==

===

Also : compressive force in brass = tensile force in steel

B AB = S AS

B (/4 252) = S (/4) [502 - 252]

Giving B = 3S

Substitute in :

6S + S = 113.6 106 N/m2

S = 16.23 MPa

B = 48.69 MPa

l0Bt (l)B

(l)S l0st

1

1

2

40

The composite bar is now subjected to an axial tensile load of 50 kN.

Total force = force in brass + force in steel

50 x 103 = B AB + S AS

i.e ( ) ( ) ( )[ ] 3222 1050025.0050.04025.04 =+ SB giving B + 3S = 101.86 106 N/m2

Also: extension of bronze = extension of steel.

B

B

S

S

B S

BS

lE

lE

0 0

100 200

2

=

=

=

Substitute in :

S2

+ 3S = 101.56 106 MPa

giving S = 29.02 MPa

B = 14.51 MPa

Resultant tension in steel = 29.02 + 16.23 = 45.25 MPa

Resultant tension in brass = 14.51 + 48.69 = 63.2 MPa

1

1

2

41

1.10 BENDING STRESS

Consider a member AB subjected to bending due to an applied bending moment M, as shown

in fig. 1.9. The top of the beam reduces in length and is, therefore, in compression. The

bottom of the beam increases in length and is, therefore, in tension. At a certain plane

between the top and bottom fibres, the length remains the same. This plane is called the

neutral axis, where the stress is zero.

Maximum C

ompres

n.a n.a B

tension

(a) Elevation (b) Cross (c) Stress

section diagram

Figure 1.9 Member subject to bending

From the stress diagram in the figure it can be seen that the bending stress is a maximum in

compression at the top of the cross section. It gradually reduces to zero at the neutral axis,

and then again gradually increases to a maximum tensile stress at the bottom.

Bending stress is dealt with in more detail in Chapter 4

n.a

M

n.a

M

compression A B

maximum t

42

1.11 SHEAR STRESS

A shear force consists of two equal parallel forces acting in opposite directions, i.e. not in the

same line.

Average shear stress = shear force

area resisting shear force

= F/A

Consider a rectangular block of material PQRS acted upon by a shear force couple F l as

shown in fig.1.10 (a).

F S S1

strained shape

1

(a) (b)

Figure 1.10

Let the deformation in the direction of F be .

Then shear strain = deformation

leverarm l

Since is small, tan =

=l

Modulus of rigidity G (or shearing modulus)

A

1 1

FQ P Q

R S

P

R

l

R1

43

If a body is subjected to a shear stress within its elastic range, the shear strain is directly

proportional to the shear stress.

Hence G x shear stress = shear strain (G = constant)

i.e G F Al l

= =

Note: Compare Young's modulus E (paragraph 1.3)

Complementary shear stress

If shear stress acts on planes PQ and RS of a body as shown in figure 1.10 (b), a clockwise

moment ( x PQ x t) x Q R, where t = thickness of the body, will result. Since the body is in

equilibrium, there must be an equal and opposite moment acting on the body.

( PQ t) Q R = (1 Q R t) P Q

= 1

Hence for a shear stress in a plane of a body there always exists an equal shear stress in a

perpendicular plane. This shear stress is called the complimentary shear stress.

Rivets in single and double shear planes

single shear plane two shear planes

(a) Lap joint (b) Butt joint with cover plates

Figure 1.11

F F F F

44

The rivet in the lap joint above is in single shear, having only one shear plane. If diameter of

rivet = d, then

=

F

d4

2

The rivets in the butt joint above are in double shear, having two shear planes each. If

diameter of rivets = d, then

=

F

d24

2

Note Shear stress will be more fully investigated in STRUCTURAL ANALYSIS II

45

1.12 TORSIONAL STRESS

L

A B

(a) Elevation F (b) View X F

Figure 1.12 Torsion

Consider a solid circular shaft AB fixed at A and with a pulley of radius A rigidly attached at

B. Twisting moment (or torque) on shaft: T = FA.

It can be shown that, subject to certain assumptions, the following relationship holds good:

TJ r

GL

= =

where T = twisting moment or torque

J = polar moment of inertia (see paragraph 2.8)

= shear stress

r = distance from axis of shaft in a plane perpendicular to it

G = shearing modulus (or modulus of rigidity)

= relative angle of twist of the two ends of the shaft

L = length of shaft

The above relationship is analogous to the basic formula MI y

ER

= =

for elastic bending

(see Chapter 4).

We shall not deal further with torsional stress in this course.

a A B

B A

46

1.13 VOLUMETRIC STRESS AND STRAIN AND BULK MODULUS

When a body is immersed in a liquid, it is subjected to the same pressure on all its faces.

This also occurs on a soil sample at depth.

This pressure creates a volumetric stress. This causes a strain on the body, called the

volumetric strain.

Volumetric stressvolumetric strain

=

v

v

= Constant

This constant is called the bulk modulus K.

Volumetric strain

Consider a square bar acted on by a tensile axial force as in figure 1.13.

a

Figure 1.13

If x is the longitudinal strain, then the lateral strain is

y xm=

1 where 1m

is Poisson's ratio.

Volume of bar before stretching V0 = a2 L0

After straining the volume is V = (a - ya)2 (L0 + x L0)

i.e V = a2L0 (1 - y)2 (1 + x)

= V0 (1 - y)2 (1 + x)

L0 xL0

a

ya

47

If x and y are small quantities compared to unity, we may ignore squares and products of x

and y and we may write

(1 - y)2 (1 + x) = (1 + x - 2 y)

V = V0 (1+ x - 2 y)

Volumetric strain is defined as the ratio of the change of volume to the original volume

and is, therefore,

V V

V x y

= 0

0

2

If y xm=

1 , then the volumetric strain is: V VV mx

=

0

0

1 2

48

1.14 RELATIONSHIP BETWEEN THE ELASTIC CONSTANTS

where E = Youngs modulus or modulus of direct elasticity

G = Modulus of rigidity or shear modulus

K = bulk modulus

1/m = Poissons ratio (also referred to as )

49

1.15 HARDNESS TEST

Hardness represents the resistance of a material against in indentation. There are various

methods of determining the hardness number k.

Brinell method

A hardened steel ball is pressed into the surface under a specified load which is held on for a

fixed period and then removed. A permanent impression is left in the surface and the

Brinell number is defined as the ratio of the applied load in kg to the spherical area of the

impression in mm2.

Other tests

There are various other tests, e.g. the Vickers Pyramid Diamond method, the Firth

Hardometer, the Rockwell hardness tester, the Shore scleroscope method and the Knoop

hardness test.

It has been found that there is an approximate linear relation between ultimate strength and

hardness number :

Ultimate tensile strength (N/mm2) = k Hardness number

For mild steel and using the Brinell method, k = 3.5.

50

1.16 TUTORIAL

1. A compound bar consists of a steel core, 15 mm diameter, within an alloy bar 25 mm

square. If the coefficient of thermal expansion for both metals is the same, and equal

to 12 10-6 per C, find the stress in each if the positions of the two ends are fixed in

position and the temperature rises 60 above that at which the bar is unstressed. Also

find the total force in the bar. ES = 200 GPa, Ealloy = 90 GPa

[Ans : S = 144 MPa; alloy = 64.8 MPa; F = 54.5 kN ]

2. A rod with a diameter of 25 mm and a length of 500 mm is subjected to an axial force

of 50 kN which causes an elongation of 0.25 mm. Determine (a) the stress in the rod;

(b) the strain; and (c) the modulus of elasticity of the material.

[Ans : 101.86 MPa; 50 x 10-6; 203.7 GPa]

3. A composite rod of total length 200 mm consists of a steel rod 120 mm long and 10

mm in diameter, which is rigidly attached to the end of a brass rod 80 mm long and

20m mm in diameter. The rod is used as a tie in a link mechanism and the strain in the

brass rod is limited to 0.53 x 10-3. Given that the total extension of the composite rod

must not exceed 0.1624 mm and E for the steel is 200 GPa, calculate the (a) strain in

the steel rod; (b) load carried by the steel rod; (c) load carried by the brass rod; and (d)

modulus of elasticity for the brass.

[Ans : 1 x 10-3; 15.71KN; 15.71 KN; 94.35 GPa]

4. A steel rim must be shrunk on a wheel 3 m in diameter without exceeding a stress of

77.25MPa in the rim. Calculate (a) the inside diameter of the rim; and (b) the least

temperature increase for the rim to fit the wheel ( = 11 x10-6/oC; E = 200GPa)

[Ans : 2.9988 m; 35.12oC]

5. An aluminium rod 20 mm in diameter is screwed at the ends and passes through a 25

mm bore steel tube 3 mm thick. Rigid washers and nuts are then fitted to the screwed

ends of the aluminium rod and the whole system heated to 140oC, when the nuts on

the rod are lightly tightened to take up any slack. Calculate the stress in the rod and in

51

the tube when the assembly has cooled to 20oC (EST = 200GPa; EAL = 70GPa; ST =

12 X 10-6/oC; AL = 23 X 10-6/oC).

[Ans : AL = 65.23 MPa (T); ST = 77.625 MPa (C)]

52

TST271Z CHAPTER 2

SECTIONAL PROPERTIES

SPECIFIC OBJECTIVES

By the end of this chapter, you should be able to do the following:

Determine the position of the centroid of a built up section.

Explain what the second moment area of a section and the neutral axis of a section are.

Determine the second moment of area about the horizontal and vertical axes passing through the centroid of the section.

Determine the section modulus of the cross section of a beam.

53

TST271Z CHAPTER 2

SECTIONAL PROPERTIES

CONTENTS PAGE

2.1 INTRODUCTION ............................................................................................................ 54

2.2 CROSS SECTIONAL AREA .......................................................................................... 55

2.3 CENTROID ...................................................................................................................... 56

2.4 SECOND MOMENT OF AREA (MOMENT OF INERTIA I).................................... 61

2.5 PARALLEL AXES THEOREM.....................................................................................64

2.6 RADIUS OF GYRATION................................................................................................73

2.7 SECTION MODULUS (ELASTIC) ............................................................................... 74

2.8 PERPENDICULAR AXES THEOREM........................................................................75

2.9 PROPERTIES OF PLANE AREAS ............................................................................... 77

2.10 SECTION MODULUS (PLASTIC) .............................................................................. 79

2.11 TUTORIAL .................................................................................................................... 84

54

2. SECTIONAL PROPERTIES

2.1 INTRODUCTION

By virtue of their shapes alone, various sections have the following properties:

(1) Cross sectional area

(2) Position of the centroid

(3) Second moment of area

(4) Radius of gyration

(5) Section modulus (elastic)

(6) Section modulus (plastic)

55

2.2 CROSS SECTIONAL AREA

A table of properties, which include cross sectional areas of some common sections, is given

in paragraph 2.9. A comprehensive table is given in the South Africa Steel Construction

Handbook (Red Book).

The cross sectional areas of structural steel sections are given in the Red Book. Three

examples are given below:

I-beam 460 140 46 I A = 5,90 103 mm2

Channel 140 60 x 16 [ A = 2,04 103 mm2

Unequal angle 75 50 6 A = 0,719 103 mm2

56

2.3 CENTROID

The centre of gravity (C.G.) of a body is the point through which the weight of the body acts,

for all positions of the body.

The C.G. of a body is not necessarily inside the body itself.

The cross section of a structural section is a plane figure without mass (and hence it cannot

have weight) and the term centroid is used. An axis through the centroid is called a

centroidal axis, and two mutually perpendicular centroidal axes intersect at the centroid of the

section.

To determine the centroid of a compound cross section made up of different parts, proceed as

follows:

(1) Divide the cross section into its different parts, whose areas and positions of centroids

are known.

(2) Assume the area of each part to act as a force through its centroid.

(3) Assume the total area to act through the centroid of the total cross section at a distance

y from the top or bottom of the section for a horizontal centroidal axis. Let the total

area be A and the areas of the parts be a1 a2 ....... distant y1 y2 .......... from the top or

bottom of the section.

(4) Take moments of all areas about the top or bottom of the cross section.

Then: A y = a1 y1 + a2 y2 +.........

Solve for y

(5) Follow the same steps for a vertical centroidal axis, taking moments about left-hand

or right-hand edges.

57

Example 2.3.1

Find the position of the centroid of the following figure.

SOLUTION

a1 = 20 150 = 3 000

a2 = 80 30 = 2 400

A = 5400

Moments about bottom edge :

5400 y = 3 000 75 + 2 400 15

= 225 000 + 36 000

= 261 000

y = 48.33

Moments about left-hand edge:

5400 x = 3 000 10 + 2 400 60

= 30 000 + 144 000

= 174 000

20

100

a2 a1

x

58

x = 32.22

Example 2.3.2

Find the position of the centroid of the following figure

SOLUTION

a1 = 1/2 (40 50) = 1 000

a2 = 60 50 = 3 000

A = 1 000 + 3 000 = 4 000

Moments about bottom edge:

4 000 y = 1 000 50/3 + 3 000 25

= 16666.67 + 75 000

= 9 1666.67

y = 22.92

Moments about right-hand edge:

4 000 x = 1 000 (60 + 40/3) + 3 000 30

= 73333.33 + 90 000

a1

x

60

a1

100

59

= 163 333.33

x = 40.83

Example 2.3.3

Find the position of the centroid of the compound girder shown in the following figure.

Notes:

(1) Due to symmetry, the centroid of course lies on the Y-Y axis.

(2) The addition of the single plate renders the compound sections un-symmetrical about

the X-X axis, and the position of x must be calculated as before.

From the South Africa Steel Construction Handbook (Red Book), the area of 254 x152 x59 I

section is 7.57 103 mm2

SOLUTION

254 152 59 I. A = 7.57 103 mm2 = 7 570 mm2

Y

Y

254 152 59 I

Centroid of I

X X

A A

200 15 plate

60

200 15 pl. A = = 3 000 mm2

Total A = = 10 570 mm2

Moments about AA:

10 570 x = 7 570 142 + 3 000 7.5

= 1 074 940 + 22 500

= 1 097 440

x = 103.8 mm

61

2.4 SECOND MOMENT OF AREA (MOMENT OF INERTIA I)

In Fig. 2.1 below, A is an element of area, distant y from any axis XX

Figure 2.1

Total area A = A

First moment of area about XX = yA

I = Second moment of area about XX = y2A

or Ixx = y dA2 for a continuous area.

Example 2.4.1

Rectangular section d x b

GG = centroidal axis

Find IGG

X X

Area A Element of area A

y

y

y

GG

A

b

62

SOLUTION

Consider an element of area A = by, distant y from GG

IGG for this element = A.y2

= by.y2

I for complete sectionGG =

=

=

=

=

=

by dy

by

b d d

b d

bd

I bd

d

d

d

d

GG

2

2

2

3

2

2

3 3

3

3

3

3

3 8 8

3 4

12

12

Example 2.4.2

Rectangular section b d.

GG = centroidal axis

Find IGG

d

G G b

63

SOLUTION

Using the same method as for Example 2.4.1 above, we find that

I dbGG =3

12

If the sections in examples 2.4.1 and 2.4.2 are compared, with d = 2b, we get the following

result:

( ) ( )

44

33

61

32

21212

121

bIbI

bbIbbI

GGGG

GGGG

==

==

It can be seen that the moment of inertia of the upright beam is 4 x the moment of inertia of

the flat beam, but they have equal cross-sectional areas. This is an important result, as will be

seen later, when we deal with resistance to bending and the deflection of beams.

G G

b

2b G G

2b

b

Areas are equal = 2b2

64

2.5 PARALLEL AXES THEOREM

Figure 2.2

The parallel axes theorem states:

IQQ = IGG + Ae2

where IGG = moment of inertia about centroidal axis G-G

QQ GG

e = distance between the axes.

Proof

Since the area would balance about G-G if it had mass:

(+ y) A = (- y) A

yA = 0

IQQ = (e +y)2 A

= ( e2 + 2 ey + y2) A

= e2 A + 2ey A + y2 A

= e2 A + 2e yA + y2 A

A

G

Gy G

Q e

Q

65

= e2 A + 0 + IGG

IQQ = IGG + Ae2

Example 2.5.1

Find IBB, IQQ and IYY of the rectangular cross section d b

SOLUTION

(1) IBB

Method 1 Method 2

From first principles Using parallel axes theorem

A b y

I y bdy

by

bd

BB

d

d

=

=

=

=

120

13

03

3

3

3

4121

2121

3

33

23

2

bd

bdbd

dbdbd

AeII GGBB

=

+=

+=

+=

Y

G

B B

b

d2

d2

Q Q

G

y1

Y

d y

e

66

(2) IQQ

I I Ae

i e I bd bde

QQ GG

QQ

= +

= +

2

32

12. .

(3) IYY

IGG = ( )( )1123width depth

In this case width = d

depth = b

IYY = 112 b3d

Example 2.5.2

Find Ixx of the I section shown below

SOLUTION

Method 1

Subtract the moment of inertia of

the shaded portion from the

moment of inertia of the

rectangular D B

Method 2

Divide I-section into three rectangules : the two

flanges and the web.

Top flange:

d

s

B

XX

G1 t

G1

D

t

G2 G2

e

e

67

I I Ae

Bt Bt D t

XX G G= +

= +

1 1

2

3 2

12 2 2

Width of shaded portion = B-s =b

(say)

Depth of shaded portion = D-2t =

d (say) Section is symmetrical

Bottom flange :

I I Ae

Bt Bt D t

XX G G= +

= +

2 2

2

3 2

12 2 2

Centroid of shaded portion

Coincides with centroid of D B

= I BD bdXX3 3

12 12

Web :

( )

( )

I s D t

Total I Bt Bt D t s D t

XX

XX

=

= +

+

112

2

212 2 2

112

2

3

3 23

Example 2.5.3

Find IXX the I section shown

below.

X X 10

50 All dimensions in mm

68

SOLUTION

Method 1

( )( ) ( )( )

4

4

4

433

33

246

24667.17067.416

84121105

121

121

121

cmI

cmcm

cmI

bdBDI

XX

XX

XX

=

=

=

=

=

Method 2

( )

( )( ) ( )( )[ ]

( )( )

4

4

4

43

4

4

423

2467.423.203

67.42

81121

:3.203

25.10142.02

5.415151212

:22

cmcmITotal

cm

cmI

webcm

cm

cmI

flangeswebIflangesII

XX

XX

XX

XXXXXX

=

+=

=

=

=

+=

+=

+=

Example 2.5.4

Find the moment of inertia about the centroidal axis XX of the T-section below.

100

B

G2

A

A2 G2

X X

A1

20 All dimensions in mm

G1 G1

69

SOLUTION

For a rectangle, I-section etc. the position of the centroid is known without calculation. For a

T-section or other asymmetrical section it is necessary first to calculate the position of the

centroid.

Find position of centroidal axis XX

Take moments about AB

A y = A1 y 1 + A2 y 2

(14 2 + 10 1) y = 14 2 8 + 10 1 0.5

38 y = 224 + 5

y = 22938

cm

y = 6.03 cm

Web: Ixx = IGG + Ae2

= 1/12 (2) (14)3 + (14 x 2) (8 6.03)2 cm4

= 457.3 + 108.7 cm4

= 566 cm4

Flange: Ixx = IGG + Ae2

= 1/12 (10) (1)3 + (10 1) (6.03 0.5)2

= 0.8 + 305.8 cm4

= 306.6 cm4

Total Ixx = 566 + 306.6

70

Ixx = 872.6 cm4

Example 2.5.5

Find Ixx and Iyy of an H-section 152 152 37 with a plate 180 mm x 15 mm on each flange.

SOLUTION

IXX

H-section Ixx = 2 210 cm4 (Red Book)

2 plates Ixx = 2 [IGG + Ae2]

= 2 [1/12 (18) (1.5)3 + (18 1.5) (8.09 + 0.75)2]

= 2 (5.06 + 2 109.93) cm4

= 4 230 cm4

Total Ixx = 6 440 cm4

IYY

H-section Iyy = 706 cm4 (Red Book)

2 plates Iyy = 2 (1/12 db3)

All dimensions in mm

Y180 15 plate

X X 152 152 37 H

Y 180 15 plate

71

= 43

6185.1 cm

= 1 458 cm4

Total Iyy = 2 164 cm4

Example 2.5.6

Find Ixx and Iyy for the built-up section shown

SOLUTION

IXX

2 flange pls Ixx = 2 [1/12 (40) (2)3 + (40 x 2) (11)2] = 19 413 cm4

2 web pls Ixx = 2 [1/12 (2) (20)3] = 2 667 cm4

4 Ls Ixx = 4 [34.9 + 11.1 (10 -1.85)2] = 3 089 cm4

Total Ixx = 25 169 cm4

*18,5

Y 400 20 pl.

200 20 pl. 200 20 pl.

X X

2015020

105 105

60 60 10 angle sections (4)A = 11.1 cm2 *

400 x 20 pl

*18,5 All dimensions in mm

* = from the Red Book

72

IYY

2 flange pls Iyy = 2 [1/12 (2) (40)3] = 21 333 cm4

2 web pls Iyy = 2 [1/12 (20) (2)3] + (20 x 2) (8.5)2] = 5 807 cm4

4 Ls Iyy = 4 [34.9 + 11.1 (7.5 + 2 + 1.85)2] = 5 859 cm4

Total Iyy = 32 999 cm4

73

2.6 RADIUS OF GYRATION

The radius of gyration of a section is that distance from the centroid of the section which

would give the same moment of inertia if the whole area were concentrated at that distance.

Figure 2.3

If radius of gyration = r, then

I = Ar2

i.e r = I/A

This property will be used later in column calculations.

A r X X

Area A

74

2.7 SECTION MODULUS (ELASTIC)

The elastic section modulus Ze of a cross section is defined as

Ze =Iy

where I = Moment of inertia of the cross section about centroidal axis

y = The extreme fibre distance from the centroidal axis.

This property will be used later in bending stress calculation.

75

2.8 PEPENDICULAR AXES THEOREM

The perpendicular axes theorem states that

Izz = IXX + IYY

where XX, YY and ZZ are three mutually perpendicular axes.

Figure 2.4

Consider an elementary area A that is distant:

y from axis XX

x from axis YY

r from axis ZZ

Ixx + Iyy = y2A + x2A

= (x2 + y2) A

= r2A

= Izz

Z X

yr

x A r2 = x2 + y2

Y

76

Izz = Ixx + Iyy

Izz is called the polar moment of inertia and is denoted by J. This property is used in torsion

calculations.

77

2.9 PROPERTIES OF PLANE AREAS

TRIANGLE

A bd

c d

I bd

Z bd

r d

e

=

=

=

=

=

2

3

36

24

18

3

2

CIRCLE

A d R

c d

I d R

Z d R

r d R

e

= =

=

= =

= =

= =

22

4 4

3 2

4

2

64 4

32 4

4 2

RECTANGLE

A bd

c d

I bd

Z bd

r d

e

=

=

=

=

=

2

12

6

12

3

2

R

GG c d

G G

b

cd

GG

b

d c

78

HOLLOW CIRCLE

( )

( )

( )

Ad d

c d

Id d

Zd d

d

rd d

e

=

=

=

=

=

212

414

414

212

4

2

64

32

4

I - SECTION

A bd b d

C d

I bd b d

Z

bd b d

d

r IA

e

=

=

=

=

=

1 1

31 1

3

31 1

3

2

12 12

12 12

2

d

d1 G

c

G

b12

d d1 GG

c

b1 = b - t

t

b

79

2.10 SECTION MODULUS (PLASTIC)

The last sectional property we will deal with in this chapter is the plastic section modulus Zp, also called the first moment of area.

For plastic section modulus the neutral axis must be in such a position that the area above the

plastic neutral axis equals the area below it.

Consider the following T-section.

Figure 2.5

The first step is to find the position of the equal-area axis.

15a = (300 20 + 430 15)

which gives a = 415 mm

Then Zp(xx) = First Moment of area about XX

= 300 20 25 + 15 (15) 2

2 + 2

415)(15 2

= 1 443 400 mm3

Example 2.10.1

The figure below shows a built-up section. Calculate the following with MM as reference

point.

20 300

= 415a

15

430xx

15

80

a. The position of the centroid of the section.

b. The second moment of area about horizontal and vertical axis (Ixx and Iyy).

c. The radius of gyration, rx and ry.

d. The elastic section modulus.

e. The plastic section modulus.

SOLUTION Since YY is an axis of symmetry, its position is known, x = 150mm.

( ) ( )

44

333

44

232323

2

2

67.116557.11655666667.7291663327500112500000

0507012/101103012/103005012/1

7.292667.292661161690587567.1429166125482524750063037503125000

)5.45115(3500705012/1)5.4565(33003011012/1)2545(150005030012/1

)(

5.4521800

40250021450037500021800

)115(3500)65(3300)25(1500021800350033001500

507030110)50300(

cmmmxxxxxxI

cmmmxxxxxxI

AeIIb

mmy

mmxxxA

YY

XX

GGXX

==++=

+++++=

==+++++=

+++++=

+=

=

++=

++=

=++

++=

81

(c)

mmA

Ir

mmA

Ir

YYy

XXx

12.7321800

7.116556666

64.3621800

67.29266116

===

===

(d)

3)( 36.6432115.45

67.29266116 mmyIZ

top

XXtopex ===

(e)

Let us take the distance that cut the figure into two equal areas to be P from MM

3744956

27534594611665.177019335.197980

)67.133035)(7050()67.1315)(30110(33.36267.13)67.13300(165.18)33.36300(

33.36300

10900

3002

218003002/1

mmZ

Z

xxxxZ

mmP

P

xPA

p

p

p

=

+++=

+++++

++=

==

=

=

82

Example 2.10.2

Calculate Zp(xx) for the section shown in example 2.7.1

SOLUTION

Find equal-area axis

Area A = 3 4578 mm2

A mm2

345782

17280 2= =

17289 = (500 25) + (901 2) + (yp - 25) 12

= 14302 + 12yp - 300

giving yp = 273.92 mm

Calculate first moment of area Zp(xx):

500 25 pl. (500 25) (273.92 12.5) = 3.268 106 mm3

2/90 65 L's (901 2) (273.92 - 25 27.9) = 0.397 106 mm3

12 1280 pl. above XX .12(273.92 12.5) 2

2592.273 = 0.372 106 mm3

500

2/65506 LS

121280 pl

2/90656 LS

1280

50

2590

12

12

equal area axis x x

300

273,92

1031,08

yp

83

12 1280 pl. below XX (1031.08 12) 2

08.1031 = 6.379 106 mm3

300 12 pl. (300 12) (1031.08 + 6) = 3.733 106 mm3

2/65 50 L's (2 658) (12.9 + 12 + 1031.08) = 1.4 106 mm3

Zp(xx) = 15.549 106 mm3

84

2.11 TUTORIAL

Calculate the position of the centroids and the second moment of areas about a horizontal and

vertical axis through the centroid for the following built-up sections. Use the section tables

for standard sections (all dimensions are in millimetres).

1.

[Ans : Ix = 13.928 x 106 mm; Iy = 16.85 x 106 mm4]

2.

[Ans : Ix = 193.41 x 106 mm4; Iy = 8.718 x 106 mm4]

80

80

40

204020

40

160

50

80

120

15 200

90

85

3.

[Ans: x = 262.55 mm from the left hand side; Ix = 1090.6 x 106 mm4; Iy = 1285.4 x 106 mm4]

400

30

340

30

300

260

30

40 40

86

TST271Z CHAPTER 3

SIMPLE BEAMS

SPECIFIC OBJECTIVES

By the end of this chapter, you should be able to do the following:

Define shear force and bending moment and obtain these values at any given point on a beam.

Explain the relationship between load, shear force and bending moment.

Plot the shear force and bending moment diagrams for statically determinate beams.

Determine the maximum shear force and bending moment for a statically determinate beam.

Define and obtain the point of contraflexure on a beam.

87

CHAPTER 3

TST271Z SIMPLE BEAMS

CONTENTS PAGE

3.1 INTRODUCTION............................................................................................................ 88

3.2 CALCULATION OF REACTIONS .............................................................................. 90

3.3 SHEAR FORCE ............................................................................................................... 93

3.4 SHEAR FORCE DIAGRAM .......................................................................................... 96

3.5 BENDING MOMENT ..................................................................................................... 99

3.6 RELATIONSHIP BETWEEN LOAD, SHEAR FORCE AND BENDING

MOMENT ....................................................................................................................... 103

3.7 POINT OF CONTRAFLEXURE.................................................................................107

3.8 BENDING MOMENT DIAGRAM .............................................................................. 109

3.9 TUTORIAL.....................................................................................................................128

88

SIMPLE BEAMS

3.1 INTRODUCTION

Simple beams are statically determinate beams, i.e. all reactions can be found by considering

the three basic conditions for static equilibrium:

V = 0

H = 0

M = 0

Note: If V 0, the beam would move up or down

If H 0, the beam would move to the left or right

If M 0, the beam would rotate clockwise or anti-clockwise

Simple beams may be:

(1) Simply supported beams

(a) With no overhang (b) With overhang at one end (c) With overhang both ends

(2) Cantilevers

(a) Fixed in position and direction at one end

(b) Free at the other end

Figure 3.1

w/m W

v

W W W1 W2

w/m w/m

VL VLVL VR VR VR

89

Loading may be point loads W, W1.... etc., a uniformly distributed load (u.d.l.) of w per

meter run OR a linear varying load of w per meter run.

90

3.2 CALCULATION OF REACTIONS

The reactions are found by considering the three basic conditions for static equilibrium:

H = 0, V = 0, M = 0

Example 3.2.1: Simply supported (s.s) beam, no overhang

SOLUTION

MB = 0 : 5 VA - 800 = 0 VA = 160 kN

MA = 0 : 5 VB - 1200 = 0 VB = 240 kN

Check : V = 0 : VA + VB = 400 kN

ALTERNATIVELY

VA = 160 kN as calculated above

V = 0 : VA + VB = 400

VB = 400 - 60

VB = 240 kN

Note: VA and VB are the reactions to the external loading system.

BA 5 m

400 kN

3 m 2 m

VB VA

91

Example 3.2.2: Simply supported beam with overhang

SOLUTION

MC = 0 : 6VB - 120 8 - 70 4.5 - 70 1.5 - 50 8 4 = 0

6VB = 2 980

VB = 496.667 kN

MB = 0 : 6VC + 120 2 - 70 1.5 - 70 4.5 - 50 8 2 = 0

6VC = 980

VC = 163.333 kN

Check: V = 0:

VB + VC - 120 - 70 - 70 - 50 8 = 0

496.667 + 163.333 - 660 = 0

ALTERNATELY: VB = 496.667 kN as calculated above

V = 0: 496.667 + VC = 120 + 70 + 70 + 400

VC = 163.333 kN

Note: VA and VB are the reactions to the external loading system.

1.5m1.5 3m 2m C

50 kN/m 70 kN 70 kN120 kN

A B

VC VB 6 m

92

Example 3.2.3 Cantilever

SOLUTION

V = 0: VA = 100 + 5 + 20 3 kN VA = 210 kN

M = 0: MA = 100 3 + 50 6 + 20 3 4.5 MA = 870 kNm

Note: VA and MA are the reactions to the external loading system.

MA 3 m

VA

3 m

50kN 100 kN 20 kN/m

6 m

93

3.3 SHEAR FORCE

Shear force is the internal force which occurs in a beam, resisting the external loading

system. It acts perpendicular to the longitudinal axis of the beam, and is transmitted from

point to point along the beam.

A beam as a whole must satisfy the three basic conditions for static equilibrium. Similarly

any portion of a beam must satisfy the same conditions. A portion of a beam showing the

external loading system as well as the internal forces is called a free body diagram.

Consider the free body diagram of portion LX of beam LR as shown in figure 3.2.

Figure 3.2

The internal force at X resisting the external loading system = shear force S.

For equilibrium V = 0

S = VL - W - wx

Ww/unit length

RL

X

VRVL

x

W w/unit length

L X Free body diagram of LX of beam

x S

VL

94

Definition

The shear force (SF) at any section of a beam is the resultant vertical force of all the forces

acting on one side of the section.

Sign Convention

Shear force up on left and down on right = +

Shear force down on left and up on right = -

Example 3.3.1

Calculate the SF at A and at 2m from A for the beam shown.

SOLUTION

Find reactions

MB = 0 : 5VA = 400 2 + 300 2.5 3.75

VA = 722.5 kN

MA = 0 : 5VB = 400 3 + 300 2.5 1.25

VB = 427.5 kN

Check : V = 0 : 722.5 + 427.5 - 400 - 300 2.5 = 0

SF immediately to the left of A = 0, Since there are no forces to the left of A (see definition

of SF)

400 kN 300 kN/m

A 2 m B 2.5 m

VB VA 5 m

95

SF immediately to the right of A :

V = 0 : SA = VA

SA = + 722.5 kN

(SF is positive since force is up on left, down on right)

SF at 2 m from A

V = 0: S = 722.5 - 2 300

S = 122.5 kN

300 kN/m

A SA = 722.5 kN

VA 722.5kN

Free body

diagram at A

Free body diagram of first 2 m of beam

300 kN/m

A

S2m = 122.5 kNVA = 722.5 kN

2m

96

3.4 SHEAR FORCE DIAGRAM

A shear force diagram is a scale representation of the shear force at any point along the beam.

If the SF at various points is calculated and plotted, the SF diagram will result.

For u.d.l.s and point loads the S.F. diagram will consist of straight lines so that the only

shear forces that need to be calculated and plotted are those at point loads and at the ends of

u.d.l.s.

Notes

(1) At point loads the SF changes from just left of the load to just right of the load by an

amount equal to the load.

(2) For the purpose of the SF diagram the reactions at the supports are taken as loads.

Example 3.4.1

Calculate the shear forces and draw the SF diagram for the beam shown below:

SOLUTION

VA and VB have been calculated in example 3.3.1

SA = 722.5 kN

SC = 722.5 - 300 2.5 = -27.5 kN

SD (L) = 722.5 - 300 2.5 = -27.5 kN

400 kN

BDA C

2.5 m 2 m

5 m

VB = 427.5 kNVA = 722.5 kN

300kN/m

97

SD (R) = 722.5 - 300 2.5 - 400 = -427.5 kN

SB = 722.5 - 300 2.5 - 400 = -427.5 kN

SF Diagram

Example 3.4.2

Draw the SF diagram for the beam below

SOLUTION

From inspection VB = VC = 120 + 70 + 5 50 = 440 kN

SA = - 120 kN

SB (L) = - 120 - 2 50 = -220 kN

SB (R) = - 220 + 440 = +220 kN

722.5 kN

+

- 427.5 kN

2m 2.5 m 27.5

5 m

0.5

1.5m 1.5m3m

120 kN70 kN 70 kN 120 kN

VC = 440 VB = 440

50 kN/m

2 m6 m 2 m

98

SE (L) = + 220 - 50 1,5 = +145 kN

SE (R) = 145 - 70 = +75 kN

SF (L) = + 75 - 3 50 = -75 kN

SF (R) = - 75 - 70 = -145 kN

SC (L) = - 145 1.5 50 = -220 kN

SC (R) = - 220 + 440 = +220 kN

SD (L) = + 220 - 2 50 = +120 kN

SD(R) = + 120 - 120 = 0

SF Diagram

Values in kN

145 120

75

75

DCA E FB

145120

220220

220220

99

3.5 BENDING MOMENT

Bending moment (BM) is the internal moment which occurs in a beam, resisting the moments

of the external loading system. It acts in the vertical plane through the longitudinal axis of

the beam, and is transmitted from point to point along the beam.

Consider the free body diagram of portion LX of beam LR as shown in fig 3.3.

Figure 3.3

Internal moment at X resisting moments of external loads = Mx

For equilibrium M = 0

( )M V x W x a wxX L= 2

2

W

L R

Xx

w/unit length

a

VR VL

x

w/unit length

Free body diagram of LX of beam LRa X

W

MxL

VL

100

Definition

The bending moment M at a point in a beam equals the sum of the moments about the point

considered of all external forces to the left (or to the right) of the point.

Sign convention

Sagging BM is taken as positive

Hogging BM is taken as negative

Example 3.5.1 Point load only

Find the BM at any distance x from A in the beam shown in the figure

SOLUTION

From inspection

The expressions for BM differ for values of x < a and x > a

VWa

lB=

( )V

W l alA

=

( )l

alWVA

=

( )l

alWVA

=

W

BA

x2x1

a l-a

l

101

x1 < a

x2 > a

Example 3.5.2 Point load + u.d.l.

Find the bending moment at any distance x from A in the beam shown in the figure

SOLUTION

Find VA and VB

MB = 0 : 9VA = 2 000 6 + 400 9 4.5

VA = 3133.33 kN

MA = 0 : 9VB = 2 000 3 + 400 9 4.5

VB = 2466.67 kN

Check : VA + VB = 3133.33 + 2466.67 = 5 600 kN

( ) ( )axWl

xalWM x

= 22

( )1xl

alWM x

=

400 kN/m

2000 kN

A B

x 3 m 6 m

9 m

VBVA

102

Example 3.5.3 Cantilever

SOLUTION

Reactions : VB = W

MB = Wl

Considering moments of forces to left of C about C:

Mx = -Wx

OR

Considering moments of forces to right of C about C:

MX = -MB + W (l-x)

= -Wl + Wl - Wx

= -Wx (which, of course, is the same as above)

Note : There is hogging in the beam, - BM

x < 3 m

kNmxx

xxM x2

2

20033.31332

40033.3133

=

=

x > 3 m

( )( ) kNmxxx

xxxM x2

2

2003200033.31332

4003200033.3133

=

=

W

C B MB

x l-xA

l

VB = W

103

3.6 RELATIONSHIP BETWEEN LOAD, SHEAR FORCE AND BENDING

MOMENT

Consider an element AB, length x, of a beam carrying a load of w/unit length.

Figure 3.4

For equilibrium of element x: V = 0

= + +

=

S S S w x

Sx

w

in the limit wdxdS

x=

lim

0

i.e S wdx= Also for equilibrium of element x: M = 0

Taking moments about A:

( ) ( )M M M W x x S S x + + + + = .2

0

Sx - M = 0 (neglecting 2nd order of smallness)

in the limit

x

WxS+S

M+MM+M

S

M

BAM

S S+S

104

==

SdxMei

Sdx

dMx

..

lim0

These are important results and the following deductions can be made:

(1) BM is a maximum when dMdx

= 0

But dMdx

S=

Thus BM is a maximum at point where the shear force = 0

(2) If w can be expressed in terms of x, then expressions for shear force S and bending

moment M can be obtained by integration.

(3) If the load is uniform and w is constant then: -

S wdx wx A= = + (a linear function)

M Sdx wx Ax B= = + + 22 (a parabolic function)

Example 3.6.1

Find the point where the maximum bending moment occurs in the beam below, and calculate

its value.

2000 kN

C B A 3m

9 m VB = 2466.67 kN VA = 3133.33 kN

400kN/m

105

SOLUTION

VA and VB were calculated in Example 3.5.2.

The S.F. diagram can now be drawn :

SA = 3133.33 kN

Sc (L) = 3133.33 - 3 400 = 1933.67 kN

Sc (R) = 3133.33 - 3 400 - 2000 = -66.67 kN

SB = -66.67 - 6 400 = -2466.67 kN

S.F. diagram

(values in kN)

Position of maximum BM is where .F. = 0 i.e. at point load

Mmax = 3133.33 3 - 400 3 1.5

= 7600 kNm

3m 66.67

1933.673133.33

2466.67

106

ALTERNATIVELY USING CALCULUS

x < 3m

anomalyaniswhich

xgivesthis

Mimumfor

xdx

dM

xxM

x

x

x

3400

33.3133max0

40033.3133

240033.3133

2

==

=

=

x > 3m

( )

anomalyanalsoiswhich

xgivesthis

Mimumfor

xdx

dMxxxM

x

x

x

3400

33.1133max0

200040033.3133

3200020033.3133 2

==

=

=

Mmaxcan only occur at x=3

107

3.7 POINT OF CONTRAFLEXURE

A point of contraflexure is a point where the curvature of a beam under loading changes from

concave upwards to concave downwards or vice versa. Thus for an infinitely small length the

beam is straight and BM here = 0.

Figure 3.5

To find the point/s of contraflexure, an expression for the bending moment at any point x

must be developed, equated to zero and solved for x.

Example 3.7.1

Find the points of contraflexure in the beam shown below. Sketch the deflected shape.

SOLUTION

From inspection VB = VC = 5w

( )

mandmgivingxxxei

wwxwxei

urecontraflexofspoforwxxwM x

84.116.8,01510..

05.752

..

int02

5.15

2

2

2

==+

=+

==

w/mDB A C

7m 1.5m1.5m

x

VC = 5wVB = 5w

point of contraflexure BM = 0

108

Deflected shape

109

3.8 BENDING MOMENT DIAGRAM

A diagram representing bending moments at any point along the beam can be drawn similarly

as for shear force diagrams.

You should be familiar with the following standard cases of bending moment diagrams.

(1) Simply supported beam with central point load

BM Diagram

(2) Simply supported beam with u.d.l over whole span

M wl x wx

This is a parabolic functiondMdx

wl wx

for M

Solving x l

M wl l wl l

wl

x

x

x

=

=

=

=

=

=

2 2

20

2

2 2 2 4

8

2

2

max

:

max

wl2/8 +

wl/2 wl/2l

w/unit length

wl/4

W

l/2 l/2

W/2 W/2l

+

110

(3) Simply supported beam with two equal point loads equi-distant from supports

BM Diagram

(4) Simply supported beam with single non-central point load

BM Diagram

W(l-a) Wa

Wa(l-a)

W

l-a a

l

+

WW

a a

l WW

Wa +

111

(5) Simply supported beam with u.d.l over whole span as well as central point load

BM Diagram

(6) Cantilever with point load at end

BM Diagram

W Wl

W l

Wl -

W w/unit run

l/2 l/2

l W+wl 2 2

W+wl 2 2

+ Wl+wl2

4 8

112

(7) Cantilever with u.d.l. over whole length

BM Diagram

Figure 3.6 (1) to (7)

Example 3.8.1

Calculate the reactions and draw the shear force and bending moment diagrams for the

cantilever beam below. Calculate the value of the shear force and bending moment at K.

SOLUTION

Reactions : V = 0 : VA = 30 + 40 + 20 + 8 4 = 122 kN

M = 0 : MA - 20 2 - 40 4 - 30 5 - 8 4 2 = 0

w/unit length

wl

l

-

wl 2

2

wl 2

2

0.5

30kN 40kN 20 kN 8kN/m

A MA

VA 1m 2m

5m

1.5m

KB C D

113

MA = -414 kNm

Shear forces Bending moments

SA = 122 kN MA = -414 kNm

SB(L) = 122 - 8 2 = 106 kN MB = -414 + 122 x 2 - 8 2 1 = -186 kNm

SB(R) = 106 - 20 = 86 kN MK = -30 2.5 - 40 1.5 - 8 1.5 0.75 =-1444kNm

SK = 86 0.5 8 = 82 kN MC = -30 1 = -30 kNm

SC(L) = 82 - 8 1.5 = 70 kN MD = 0

SC(R) = 70 - 40 = 30 kN

SD(L) = 30 kN

SD(R) = 0

Shear force Diagram (kN)

-

Bending Moment Diagram (kNm)

7030

106 86

DA B

30144

186

414

122

K C

82

A B K C D

+

114

Note

You will have noticed that positive bending moments are shown below the beam line and

negative bending moments above, contrary to normal practice in mathematics. This

convention will be adhered to as far as bending moment diagrams are concerned, as the

deflected shape of the beam can be more easily visualised.

Example 3.8.2

Find the equations for the shear force and the bending moment for the beam with uniformly

increasing u.d.l. shown below and sketch the SF and BM diagrams showing values and

positions at which they occur.

SOLUTION

Find reactions

M V

V kN

B A

A

= = +

=

0 6 200 6 3 150 62

13

6

750

:

M V

V kN

Check

A B

B

= = +

=

+ = +

0 6 200 6 3 150 62

23

6

900

900 750 200 6 150 62

:

:

Equation for SF:

6 mVA VB

x

150 350 kN/m

200 kN/m

115

2

2

5.122007500:0

5.122007502

1506

200

xxSFor

xx

xxxVS Ax

=

=

=

=

Solving gives x = +3.12m or -19.12m

Ignore the -19.12 m value

SF Diagram

Equation for BM

M V x x x x x

x x x

x A=

=

2002 6

1502 3

750 100 256

2

23

.

Mx is a maximum at SF = 0 i.e at x = 3.12 m

ALTERNATIVELY : dMdx

x x forx = =750 200 756

02

maximum BM. Solve for x

900 kN

3.12 m

750 kN

116

( ) ( )kNm

M

1240612.32512.310012.3750

32

max

=

=

BM Diagram

Example 3.8.3

Sketch the shear force and bending moment diagrams for the beam shown below and

calculate the maximum values and the positions at which they occur. Sketch the deflected

shape.

SOLUTION

Find reactions

MB = 0 : 10VA = 150 12 4 - 2 500 2

VA = 220 kN

MA = 0 : 10VB = 150 12 6 + 2 500 12

VB = 4 080 kN

Check : 220 + 4 080 = 150 12 + 2 500

2m

150 kN/m 2 500 kN

CBA

VBVA

10 m

3.12 m

1240 kNm

117

Sketch SF diagram

SA = 220 kN

SB(L) = 220 - 10 150 = -1280

SB(R) = -1 280 + 4 080 = 2 800

SC(L) = 2 800 - 2 150 = 2 500

Calculate x (position where SF = 0)

220 - 150x = 0 x = 2215

m

ALTERNATIVELY : Mx = 220x - 150

2

2x

dMdx

x = 220 - 150x = 0 for maximum BM

Solve for x

Sketch BM diagram

BM Diagram (values in kNm)

SF Diagram

Values in kN

1200

+

-

28002500

220

x

2215m

161 13

CBD A

5300

118

Point of contraflexure

Contraflexure occurs where M = 0 i.e. at poin