Unit 9

5

Deflection of BeamsUNIT 9 DEFLECTION OF BEAMS

Structure 9.1 Introduction

Objectives

9.2 Different Methods 9.2.1 Governing Equation for Deflections 9.2.2 Sign Conventions 9.2.3 The Boundary Conditions

9.3 Deflections of Simply Supported Beams (SS Beams) 9.3.1 SS Beams with Central Point Load 9.3.2 SS Beams with a Point Load anywhere on Span 9.3.3 SS Beams with UDL 9.3.4 SS Beams with Triangular Load 9.3.5 SS Beams with a Couple 9.3.6 SS Beams with an End Couple 9.3.7 SS Beans with Equal End Couples 9.3.8 Examples

9.4 Deflection of Cantilever Beams 9.4.1 Cantilever Beams with Single Concentrated Load at Free End 9.4.2 Cantilever Beams with a Central Point Load 9.4.3 Cantilever Beams with a UDL 9.4.4 Cantilever Beam with a UDL on Some Portion 9.4.5 Cantilever Beams with a Triangular Load 9.4.6 Cantilever Beams with an End Couple 9.4.7 Examples

9.5 Deflection of Overhang Beams

9.6 Application of Deflections of Beams

9.7 Summary

9.8 Answers to SAQs

9.1 INTRODUCTION

Deflection is defined as the vertical displacement of a point on a loaded beam.

Slope is defined as the angle between the deflected central line and original central line of the beam.

Illustration

From Figure 9.1,

yc = Deflection at point c.

The beam is replaced by central line of neutral plane. The bent form of this central line is called elastic curve of beam. The expression for this elastic curve is in the form of y = f (x) where y is the deflection at x.

θA = Slope at A,

θB = Slope at B, B

ACB = Original beam, and

AC′B = Deflected beam.

6

Stresses in Shafts and Shells

Figure 9.1

The deflections of beams are very significant in the design of structures. The excessive deflection cause cracks in walls, cracks in ceilings, create a feeling of lack of safety and affect geometry, shape and appearance. In machine parts, such as shafts may result in undesirable interference between mating parts such as gears. Hence, the maximum deflection is minimized in the design.

Objectives After studying this unit, you should be able to

• conceptualise the deflections,

• calculate the deflection and slope of simply supported beams,

• calculate the deflection and slope of cantilevers, and

• calculate the deflection and slope of other determinate beams.

9.2 DIFFERENT METHODS

The following methods are used to calculate the slopes and deflections :

(a) Double Integration Method

This is most suitable when concentrated or Udl over entire length is acting on the beam.

(b) Macaulay’s Method

This method is most conveniently used when a concentrated load is acting on a beam at some point other than mid-point.

(c) Moment Area Method

This is conveniently used when the area of BMD can be easily calculated and beam cross-section changes after certain length.

(d) Conjugate Beam Method

This is most suitable for beams with varying moment of inertia.

(e) Strain Energy Method

(f) Virtual Work Method

(g) Unit Load Method

9.2.1 Governing Equation for Deflections Let us consider a beam represented by its elastic curve as shown in Figure 9.2, subjected to some arbitrary loading.

Figure 9.2

The ACDB is the elastic curve of the beam. The original central line was along x-axis.

θB

A B C

θAC1 yc

A BC D

dx

x

7

Deflection of BeamsLet us consider a small length CD = ds of the deflected beam, where horizontal distance between C and D is dx.

dθ

D

B θ

dy

y

A

x

C

ds

dx

Figure 9.3

The normal drawn to the curve at C and D will meet at centre of curvature.

Let ds = length of the portion CD,

R = radius of the curvature,

θ = angle made by tangent at D with x-axis, and

dθ = angle made by radius at C with radius at D.

From Figure 9.2

ds = R d θ

[since ]ds dxR ds dxd d

= = ≈θ θ

∴ 1 dR dx

θ= . . . (9.1)

tan dydx

θ =

For small angle, tan θ = θ

∴ dydx

θ = . . . (9.2)

dydx

is defined as the slope of the curve. The curve in case of deflected beam is the shape

of the central line of the beam.

Differentiating the Eq. (9.2), w.r.t., x, we can get

2

2d ddx dxθ=

y . . . (9.3)

From Eqs. (9.1) and (9.3),

2

21 d yR dx= . . . (9.4)

From the theory of pure bending, we know,

M EI R= . . . (9.5)

From Eqs. (9.4) and (9.5),

2

2EI dM EIR dx

= =y . . . (9.6)

8

This is the equation for elastic curve of beam and governing equation for deflections.

Note that y and dydx

can be found by integration once and twice, respectively. The

condition for integration is that function M = f (x) must be continuous.


9.2.2 Sign Conventions

(a) x is +ve towards right.

(b) y is +ve upwards.

(c) Anticlockwise slopes are +ve.

(d) Sagging BM are +ve.

9.2.3 The Boundary Conditions

Boundary conditions are the known values of deflection and slope at specified values of

x. The y and dydx

are invariably known at the supports. dydx′

may also be ascertained from

symmetry of bent shape of beam central line. For example,

(a) At simple supports, 0, 0dyydx

= ≠ .

(b) At fixed supports, 0, 0dyydx

= = .

(c) At middle of SS beam loaded either by a single central or Udl over entire

length 0, 0dyydx

≠ = .

9.3 DEFLECTIONS OF SIMPLY SUPPORTED BEAMS (SS BEAMS)

Simply supported beam is supported on one hinged support and other roller support. The loads may be point loads or concentrated or distributed. Distributed loads may be uniformly distributed or distributed in any other manner. Triangular distribution is common, Udl, triangular loads, etc.

9.3.1 SS Beams with Central Point Load

2lAC CB= =

Because of symmetry,

2A B

WR R= =

Boundary conditions are the known values of deflection and slope at specified values of x. The Boundary Condition (BC) for this problem are :

(a) At x = 0, y = 0.

(b) At x = l, y = 0

(c) At , 02l dyx

dx= =

9

Deflection of Beams W

W/2 W/2 x

Figure 9.4

Consider a section X-X at a distance x from A,

2AlM R x W x⎛ ⎞= − −⎜ ⎟

⎝ ⎠

2 2

W lx W x⎛ ⎞= − −⎜ ⎟⎝ ⎠

. . . (9.7)

The governing equation for deflection is :

2

2d yEI Mdx

=

or 2

2 2 2 2d y W l lEI x W x xdx

⎛ ⎞ ⎛= − − < <⎜ ⎟ ⎜⎝ ⎠ ⎝

l ⎞⎟⎠

. . . (9.8)

Integrating the above equation, we can get

22

12 2 2 2dy W x W lEI xdx

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

C . . . (9.9)

C1 is constant for integration.

Apply B.C 3, i.e. at , 02l dyx

dx= =

2

102 8

W l C= +

2

1 16W lC −

=

Integrating the above equation, we can get

33

1 2.4 3 6 2

W x W lEIy x C x C⎛ ⎞= − − + +⎜ ⎟⎝ ⎠

. . . (9.10)

∴ 33 2

24 3 6 2 16W x W l W l xEIy x C⎛ ⎞= − − − +⎜ ⎟

⎝ ⎠

with B.C 2, y = 0 at x = l

3 3 320

12 48 16W W Wl l l= − − + C

3

21 1 1

4 3 12 4W l C⎛ ⎞− +⎜ ⎟

⎝ ⎠= −

or C2 = 0

Here is the constant of integration.

Applying the boundary condition (2) into the Eq. (9.10).

Applying the BC (2) into the Eq. (9.11), we can get

2 2

1 10 .4 4 16

W l W lC C −= + =

10


∴ The slope and deflection will be :

22 2

4 2 2 16 2dy W x W l W l lEI xdx

⎛ ⎞ ⎛ x l ⎞= − − − < <⎜ ⎟ ⎜⎝ ⎠ ⎝

⎟⎠

. . . (9.11)

33 2

12 6 2 16W x W l W lEIy x x⎛ ⎞= − − −⎜ ⎟

⎝ ⎠ . . . (9.12)

The variation of slope is parabolic. The variation of deflection is cubic parabola . The maximum slopes are at A and B. The maximum deflection is at C. Slope at A, i.e. at x = 0, from Eq. (9.11)

2

16AW l

EI−

θ =

Slope at B, x = l, from Eq. (9.11),

2

16BW l

EIθ = +

Deflection at C, 2lx = , from Eq. (9.12),

3

48CW ly

EI= −

The slope and deflection diagrams are shown in Figure 9.5.

A B

W

C

(a) Beam Parabola

A

θA

B

θB

(b) Slope Diagram

yCA

Cubic Parabola

B

(c) Deflection Diagram

Figure 9.5

9.3.2 SS Beams with a Point Load Anywhere on Span AB = l, AC = a, CB = b, a + b = l

0y∑ = A BR R W+ = . . . (9.13)

Taking moments about A,

orB BW AC R AB Wa R l= × = ×

∴ ( )BW aR

l= ↑ . . . (9.14)

11

Deflection of BeamsFrom Eqs. (9.13) and (9.14)

( )AW bR

l= ↑ . . . (9.15)

BA

l

a x

lw

Figure 9.6


[ ]AM R x W x a= − −

Note that if x < a, M = RA x, i.e. second term is not applicable.

or, [ ]W bM x W x al

= − − . . . (9.16)


2

2 [ ]d y W bEI M x W xldx

= = − − a . . . (9.17)

Integrating the Eq. (9.17), we can get

2

21[ ]

2 2dy W b x WEI xdx l

= − − +a C . . . (9.18)

3

31 2. [ ]

2 3 6W b x WEIy x a C x C

l= − − + + . . . (9.19)

The constants C1 and C2 can be found from the boundary conditions.

The boundary conditions are :

at A, . . . (1) 0, 0x y= =

at B, . . . (2) , 0x l y= =

Applying BC (1) to Eq. (9.19) and noting that W (x – a) is not applicable if x < a, as is required for BC (1) or BC at A.

1 20 (0) (0)6

W b C Cl

= + +

∴ C2 = 0

Applying BC (2) into the Eq. (9.19), C2 = 0.

2 310

6 6W b Wl b C= − + l

2 21 ( )

6W bC b l

l= −

The slope and deflection will be given by

2 2 2[ ] [2 2 6

dy W b W Wb 2 ]EI x x a bdx l l

= − − + − l . . . (9.20)

3 3 2[ ] [ ]6 6 6

W b W W b 2EIy x x a b l xl l

= − − + − . . . (9.21)

12

Stresses in Shafts and Shells Slope at A, (at x = 0), and noting that [

2W ]x a− is not applicable

2 2[ ]6A

A

dy W b l bdx EI l

⎛ ⎞ = θ = − −⎜ ⎟⎝ ⎠

2 2[( ) ] ( 2 )6 6W b W aba b a a bEI l EI l

= − + − = − + . . . (9.22)

Slope at B, (at x = l),

2 2 2. [ ] [2 2 6B

B

dy W b W W bl l a bdx l l

⎛ ⎞ θ = − − + −⎜ ⎟⎝ ⎠

2 ]l

2 2[3 3 ]6W b l bl b lEI l

= − + 2−

2 2[2 ( ) 3 ( ) ]6W b a b b a b bEI l

= + − + +

2(2 ) (2 )6 6W b Waba ab a bEI l EI l

= + = +

or (2 )6BW ab a b

EI lθ = + . . . (9.23)

Deflection at centre 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

3 3

2 2. [6 2 6 2 6C

W b l W l W b lEI y a b ll l

⎛ ⎞ ⎡ ⎤= − − + −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦]

2

2

2 2( )48 12

W bl W b b l= + −

2 2[ 4 448

W b l b l= × + − 2 ]

2 2[4 3 ]48

W b b l= × −

∴ 2 2[4 3 ]48CW by b

EI= × − l

2 2[3 4 ]48W b l b

EI= − × − . . . (9.24)

For deflection under the load yw again 3[6

W ]x a− vanishes. Hence, by putting x = a

3 2 2[ ]6wW by a b aEI l

= + − l a

2 2 2 2[ 26W ab a b a b ab

EI l= + − − − ]

or 2 2

3wW a by

EI l= . . . (9.25)

13

Deflection of BeamsFor maximum deflection, 0dy

dx= .

From Eq. (9.20),

2 2 20 [0 ] [2 2 6

W b W W b 2 ]x a bl l

= − − + − l

2 2. [2 6

W b W b 2 ]x l bl l

= −

∴ 2 2

3l bx −

=

Substituting the value of x into the Eq. (9.21), we can get

2 2 2max [ ]

6W b xEIy x b l

l= + −

2 2 2 2

2 2

6 3 3W b l b l b b l

l⎡ ⎤− −⎢ ⎥⎢ ⎥⎣ ⎦

= + −

3

2 2 2( )9 3W b

= − l bl

−

∴ 3

2 2 2max ( )

9 3W by l

EI l= − − b . . . (9.26)

The slope and deflection diagram are shown in Figure 9.7.

A B

Cx

(a) Beam

θB

θA

A B

x

C

(b) Slope Diagram

Elastic curve

A BD

yWymax yc

x


Figure 9.7

9.3.3 SS Beams with UDL Figure 9.8 shows a simply supported beam of span l and loaded by a udl of w per unit length.

Because of symmetry, 2A B

wlR R= =


.2 2

wl xM x w x= −

14


2

2 2wl x w x

= − . . . (9.27)

Figure 9.8


2 2

2 2 2d y w l x w xEIdx

= − . . . (9.28)


2 3

14 6dy w l x w xEIdx

C= − + . . . (9.29)

3 4

112 24w l x w x

2EIy C x C= − + + . . . (9.30)



at A, 0, 0x y= = . . . (1)

at B, , 0x l y= = . . . (2)

at C, , 02l dyx

dx= = . . . (3)

Applying the BC (1) to Eq. (9.30), C2 = 0.

Applying the BC (2) to the Eq. (9.30),

4 4

1 1012 24 24wl wl wlC l C= − + ∴ =

3 . . . (9.31)

The slope and deflection equations will be,

2 3

4 6 2dy w l x w x wlEIdx

= − −3

4 . . . (9.32)

3 4 3

12 24 24w l x w x wlEIy x= − − . . . (9.33)

Slope at A, x = 0

3

24AA

dy wldx EI

⎛ ⎞ = θ = −⎜ ⎟⎝ ⎠

. . . (9.34)

Slope at B, 2lx =

3 3 3

4 6 24 2BB

dy wl wl wl wlEI EIdx

⎛ ⎞ = θ = − − =⎜ ⎟⎝ ⎠

2

4

∴ 3

24Bwl

EIθ = . . . (9.35) +

A C

X

X

w/unit length

l

x

B x

15

Deflection of BeamsBecause of symmetry, the maximum deflection occurs at mid-span, i.e. at C,

2lx⎛ ⎞=⎜ ⎟

⎝ ⎠.

From Eq. (9.33),

4 4

max 96 24 16 24 2wl wl wlEIy = − −

× ×

4

4

45[4 1 8]384 384wl wl−

= − − =

∴ 4

max5

384wlyEI

−= . . . (9.36)

A B

w/Unit length

Cl/2 l/2

(a) Beam

(b) Slope Diagram

Θ

θB

C θA

+

3° Curve

4° Curve

A BC yc = ymax


Figure 9.9

9.3.4 SS Beams with Triangular Load Figure 9.10 shows a simply supported beam with distributed load which uniformly increases from 0 at A (x = 0) to w/unit length at B (x = l). Thus, to load diagram appears

as a triangle. Apparently the total load = area of the triangle 12

wl= and at any

section X-X at a distance x from A the rate of loading will be w xl

, so that the load of

shaded triangle 1 12 2

w x w xxl l

= . This load will act at a distance of =23x from A

(centroid of triangles). The sum of the reactions (RA + RB) will be equal to the load of

entire triangle, i.e.

B

2wl .

0y∑ =

or, 12 2A B

wlR R l w+ = × × = . . . (9.37)

Xl

w

BA

Xx

16

Stresses in Shafts and Shells Figure 9.10 : SS Beam with Triangular Load


22 3 B

wl l R l× = ×

( )3B

wlR = ↑ . . . (9.38)

From Eqs. (9.37) and (9.38),

( )6A

wlR = ↑ . . . (9.39)

Consider a section X-X at distance x from A,

Intensity of load wlx

⎛ ⎞= ⎜ ⎟⎝ ⎠

1 . .2 3A

wl xM R x xx

⎛ ⎞= − ⎜ ⎟⎝ ⎠

3

6 6wl x w x

l= − . . . (9.40)

The governing equation for deflection is

2 3

2 6 6d y wl x w xEI M

ldx= = − . . . (9.41)


2 4

112 24dy wl x w xEI Cdx l

= − + . . . (9.42)

3 5

1 236 120wl x w xEIy C x C

l= − + + . . . (9.43)



at A, 0, 0x y= = . . . (1)

at B, , 0x l y= = . . . (2)

Applying the BC (1) to the Eq. (9.43), we can get C2 = 0.

Applying the BC (2),

4 4

1036 120

wl wl C l= − +

4 3

17[3 10]

360 360wl wlC

l−

= − = . . . (9.44)

The slope and deflection equation will be :

2 4

3712 24 360

dy wl x w xEIdx l

= − − wl . . . (9.45)

17

Deflection of Beams

3 537

36 120 360wl x w xEIy wl x

l= − − . . . (9.46)

Slope at A, (x = 0)

37

360AA

dy wldx EI

−⎛ ⎞ = θ =⎜ ⎟⎝ ⎠

. . . (9.47)

Slope at B, (x = l)

3 3

3712 24 360B

B

dy wl wlEI EIdx

⎛ ⎞ = θ = − −⎜ ⎟⎝ ⎠

wl

3 3

[30 15 7]360 45wl wl

= − − =

∴ 3

45Bwl

EIθ = . . . (9.48)

Slope at mid-span, 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠.

3 3

3748 24 16 360C

C

dy wl wlEI EIdx

⎛ ⎞ = θ = − −⎜ ⎟ ×⎝ ⎠wl

3 367[30 15 7 16]

24 16 15 5760wl wl

= − − × = −× ×

∴ 367

5760Cwl

EIθ = − . . . (9.49)

Deflection at centre, 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠.

4 4

4736 8 12 32 360 2Cwl wlEI y wl= − −× × ×

4 471[40 3 7 16]

630 32 11520wl wl

= − − × = −×

. . . (9.50)

For maximum deflection,

0dydx

=

From Eq. (9.42),

2 4

37012 24 360

wl x w x wll

= − −

⇒ 2 2 4 40 30 15 7l x x l= − −

⇒ 4 2 2 472 015

x l x l− + =

∴ 0.5193x l=

From Eq. (9.43),

3 5max (0.5193 ) (0.5193 )

36 120wl wEI y l l

l= − −

18

Stresses in Shafts and Shells 37 (0.5193 )

360 2wl l−

×

4 4

[14 0.11 3.64]360 153wl wl−

= − − =

∴ 4

max 154wly = − . . . (9.51)

(a) Beam

(b) Slope Diagram


Figure 9.11

9.3.5 SS Beams with a Couple A simply supported beam AB of span l on which a couple of moment M0 acts at C is shown in Figure 9.12.

Figure 9.12

Since sum of all forces in y direction is equal to zero,

0A BR R+ = . . . (9.52)


0 BM R l= ×

0 ( )BMRl

= ↑ . . . (9.53)

From Eqs. (9.52) and (9.53),

W

B C

A

l/2 l/2

4° Curve

Θ

θB

D

+

θA

B

5° Curve

θC

X = 0.5193 l A

yyc

C DA B

max

A B MO

X

Xa

MOb/l

x

C

l

l

MOa/l

19

Deflection of Beams 0 ( )A

MR

l= − ↓

Consider a section X-X at a distance x from A and write expression for BM at X-X. At this point you must note two important features of expression for BM at X-X. If X-X lies between A and C, i.e. 0 < x ≤ a the BM is only due to RA. If X-X is between C and B, i.e. a < x < b, then BM is due to RA and M0. So we write the second BM because the first is included but take care that M0 is not considered if integration is in the region 0 < x ≤ a and x is taken as (x – a) if integration is in the region a < x ≤ l and x = l. Thus, in two cases we can apply BC at x = 0 and x = l. For specific presentation in the expression for M, M0 is placed in < > meaning that < M0 > to be ignored in 0 < x ≤ a and to be considered in a < x ≤ b with x replaced by (x – a). The procedure is known as Macaulays method and has already been used in Section 9.3.2.

0AM R x M= + < >

00

M xM

l= − + < > . . . (9.54)


2

002

M xd yEI M Mldx

= = − + < > . . . (9.55)


2

00 1[ ]

2M xdyEI M x

dx l= − + < > − +a C . . . (9.56)

3

20 01[ ]

6 2M x M

2EIy x a C x Cl

= − + − + + . . . (9.57)

The constants of integration C1 and C2 can be found from the boundary conditions.


at A, . . . (1) 0, 0x y= =

at B, . . . (2) , 0x l y= =

Applying the boundary condition (1) to the Eq. (9.57), with M0 ignored, C2 = 0.

Applying the BC (2) into the Eq. (9.57),

2

20 010 (

6 2M l M l a C l= − + − +)

20 01 ( )

6 2M l M

C ll

= + − a . . . (9.58)


2

20 00 [ ] (

2 6M x M l Mdy 0 )

2EI M x a

dx l l= − + < > − + − −l a . . . (9.59)

3

2 20 0 0 0[ ] ( )6 2 6 2

M x M M l MEIy x a x l a x

l l= − + − + − − . . . (9.60)

If the moment M0 is applied at the centre,

2la =

20


20 0 0 0 0

1 6 2 2 6 8 24M l M M l M l M llC l

l⎛ ⎞= − − = − =⎜ ⎟⎝ ⎠


2

0 002 2 24

M xdy lEI M xdx l

⎛ ⎞= − + − +⎜ ⎟⎝ ⎠

M l . . . (9.61)

23

0 0 06 2 2 24

M x M M llEIy x xl

⎛ ⎞= − + − +⎜ ⎟⎝ ⎠

. . . (9.62)

Slope at A, (x = 0)

024AM l

EIθ = + . . . (9.63)

Slope at B, (x = l)

0 0 02 2 24 24B

0M l M l M l M lEI θ = − + + =

024BM l

EIθ = . . . (9.64)

Slope at C 2lx⎛ ⎞

⎜ ⎟ , =⎝ ⎠

0 08 24 12C

0M l M l M lEI θ = − + = −

∴ 012CM l

EIθ = − . . . (9.65)

Deflection at C 2lx⎛ =⎜

⎝ ⎠⎞⎟ , y = 0

To get new slope,

2

0 02 24

M x Ml

+ =l

12 2 3l lx = =

9.3.6 SS Beams with an End Couple . . . (9.66) 0 so thaty AF R R∑ = + = 0B


0BR l M× =

∴ 0 ( )BMRl

= + ↑ . . . (9.67)

A

B

X

Xl

xMO

M BO B

BDM

21

Deflection of Beams

Figure 9.13 : SS Beam with End Couple

From Eqs. (9.66) and (9.67),

0 ( )AM

Rl

= − ↓

Consider a section X-X at a distance from x from A,

0.AM

M R x xl

= = − . . . (9.68)


2

02

Md yEI Mldx

−= = x . . . (9.69)


2012

MdyEI xdx l

= − + C . . . (9.70)

3016

M2EIy x C x C

l= − + + . . . (9.71)



at A, . . . (1) 0, 0x y= =

at B, . . . (2) 0, 0x y= =

Applying the boundary condition (1) into the Eq. (9.71), C2 = 0.


2

0 01 10

6 6M l MC l C= − + ∴ =

l


20 02 6M M ldyEI x

dx l= − + . . . (9.72)

30 06 6M M l x

EIy xl

= − + . . . (9.73)

Slope at A, (x = 0)

06AM l

EIθ = + . . . (9.74)

Slope at B, (x = l)

0 02 6 3B

0M l M l M lEI θ = − + = −

∴ 03BM l

EIθ = − . . . (9.75)

Slope at C, 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

22


20 0 0

8 2 6 8 6C0M M l M l M llEI ⎛ ⎞θ = − + = − +⎜ ⎟

⎝ ⎠

∴ 024CM l

EIθ = + . . . (9.76)

Deflection at C, 2lx⎛ ⎞

⎜ ⎟ , =⎝ ⎠

3

0 06 2 6 2M M ll lEIy

l⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2

0 0 048 12 16

2M l M l M l= − + =

∴ 2

016CM l

yEI

= + . . . (9.77)

A

(a) Beam

(b) Slope Diagram


Figure 9.14

For maximum deflection, the slope is zero.

2

0 0 02 6

M x M ll

− + =

2

2

3lx =

0.5773lx l= =

The maximum deflection,

3

0 0max . .

6 63 3 3M M ll lEIy

l= − +

2 2

0 0[ 3 ]18 3 9 3M l Ml l

+ = += −

∴ 2

0max 9 3

M lyEI

= + . . . (9.78)

B

D

MO

C

2° Curve

+

θB

D

Θ

θA θC

0.577l

A CB

3° Curve

A BC

+ymaxyc

D

23

Deflection of Beams

0B

9.3.7 SS Beams with Equal End Couples . . . (9.79) 0 so thaty AF R R∑ = + =


. . . (9.80) 0 0 0B BM M R l R− = × ∴ =

From Eqs. (9.79) and (9.80),

AR = 0

Figure 9.15 : SS Beam with Equal End Couples


0M M= − . . . (9.81)


2

02d yEI Mdx

= = − M . . . (9.82)


0 1dyEI M xdx

= − + C . . . (9.83)

2

01 22

M xEIy C x C= − + + . . . (9.84)

The constants C1 and C2 can be found from the boundary conditions. The boundary conditions are : at A, . . . (1) 0, 0x y= =

at B, . . . (2) , 0x l y= =


2 0C =


2

0 01 10

2 2M l MC l C= − + ∴ =

l . . . (9.85)

The slope and deflection equations will be :

00 2

M ldyEI M xdx

= − + . . . (9.86)

2

0 02 2

M x M lEIy x= − + . . . (9.87)

Slope at A, (x = 0)

02AM l

EIθ = + . . . (9.88)

Slope at B, (x = l)

0 00 2 2B

M l MEI M lθ = − + = −l

X

A

B

X

C

MO

x MO

24

∴ 02BM l

EIθ = − . . . (9.89)



⎝ ⎠,

0 02 2C

M l M lEI θ = − +

∴ 0Cθ = . . . (9.90)

Deflection at C, 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

2 2

0 02 2 2 2 8C

0M M l M ll lEIy ⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∴ 2

08C

M ly

EI= + . . . (9.91)

CA

(a) Beam

(b) Slope Diagram


Figure 9.16

9.3.8 Examples Example 9.1

A simply supported beam of span l is subjected to two concentrated loads at one-third span from two supports. Find the maximum slope and maximum deflection EI is constant.

Solution

By symmetry, A BR R W= = . . . (1)


2.3 3l lM W x W x W x⎡ ⎤ ⎡= − − − − ⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ . . . (2)

B

MOMO

C

θA

θB

1° Curve (Straight line)

Θ

1° Curve (Straight line)

B +A

A+

A

B yc = ymax

C

B l/3 l/3 l/3

W W

X

Elastic curve

BM

X

25

Deflection of Beams

Figure 9.17

The equation for deflection is :

2

22

3 3d y l lEI M W x W x W xdx

⎡ ⎤ ⎡= = = − − − − ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

. . . (3)

Integrating the Eq. (3),

2 22

12

2 2 3 2 3dy W x W l W lEI x xdx

⎡ ⎤ ⎡ ⎤= − − − − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦C . . . (4)

3 33

12

6 6 3 6 3W x W l W l

2EIy x x C x C⎡ ⎤ ⎡ ⎤= − − − − + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ . . . (5)

The boundary conditions : at A, 20, 0 0x y C= = ∴ =

It should be understood that the Eqs. (3), (4) and (5) pertain to the region 23lx > .

Hence second and third terms vanish when BC at x = 0 is used. at B, , 0x l y= =

3 33

120

6 6 3 6 3W l W l W l C l⎛ ⎞ ⎛ ⎞= − − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3 2

18 11

6 27 27W l W lC ⎡ ⎤= − − − =⎢ ⎥⎣ ⎦ 9

. . . (6)

∴ 2 22 22

2 2 3 2 3 9dy W x W l W l W lEI x xdx

⎡ ⎤ ⎡ ⎤= − − − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Apparently since the problem is symmetric the maximum deflection occurs in the centre. 1 2 3C Cy y y+ = C

3 1 2 3A A Aθ + θ = θ = θ B

Deflection under the load, 3lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

3 2

6 3 9 3DW l W l lEIy ⎛ ⎞= −⎜ ⎟

⎝ ⎠×

3 31 51

27 6 27 6W l W l−⎛ ⎞= − =⎜ ⎟ ×⎝ ⎠

35

162DW lyEI

−= . . . (7)

At A, (x = 0),

2

9AW l

EIθ = − . . . (8)

At B (x = l),

26


2 2 24. .2 2 9 2 9 9B

W l W l W l W lEI θ = − − −2

2 2

[9 4 1 2]18 9

W l W l= − − − = +

∴ 2

9BW l

EIθ = + . . . (9)

For maximum deflection, slope is zero.

22 2

02 2 3 9

W x W l W lx⎡ ⎤= − − −⎢ ⎥⎣ ⎦

Again note that maximum deflection will occur between the loads which is easily ascertained from symmetry. However, to prove this Eq. (5) is used and since

23lx < between the loads, the third term vanishes.

⇒ 2

2 20 9 9 23lx x l⎛ ⎞= − − −⎜ ⎟

⎝ ⎠

2

2 2 29 9 29 3l l x 2x x l

⎛ ⎞= − + − −⎜ ⎟⎜ ⎟

⎝ ⎠

2 26 2l l x= − + − l

26 3l x l=

2lx = . . . (10)

3 2

3max 6 6 3 9

W W l WlEIy x x x⎛ ⎞= − − −⎜ ⎟⎝ ⎠

Now put 2lx =

3 3 3

max 6 2 6 2 3 18W l W l l WlEIy ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

3 2 32 1.

6 2 3 9 2 6 216 3 8W l l W l l Wl⎛ ⎞ ⎛= − − − = − + −⎜ ⎟ ⎜

⎝ ⎠ ⎝

1 1 ⎞⎟⎠

3 31 1 1 72 1 27

6 8 36 3 6 216W l Wl + −⎡ ⎤ ⎛= − − = − ⎜ ⎟⎢ ⎥

⎞⎣ ⎦ ⎝ ⎠

3

3 23[36 8 96]36 8 6 648

W l Wl ⎛ ⎞= − − = − ⎜ ⎟× × ⎝ ⎠

∴ 3

max23648

WlyEI

= . . . (11)

Example 9.2 A simply supported beam of span 6 m is subjected to Udl of 24 kN/m for a length of 2 m from left support. Find the deflection at the centre, maximum deflection and slopes at the ends and at the centre. Take EI = 20 × 106 N-m2.

Solution

27

Deflection of Beams . . . (1) 0, so that 24 2 48 kNy A BF R R∑ = + = × =

A BC

6m

24 kN/m

2m D

Figure 9.18


24 2 1 6BR× × = ×

. . . (2) 8 kN ( )BR = ↑

. . . (3) 48 8 40 kN ( )AR = − = ↑

Apply the Udl over the portion DB downwards and upwards,

A B

24 kN/m

2mD

4m24 kN/m

Figure 9.19

( 240 24 24 ( 2)2 2x xM x x x )−

= − × + −

Note that the third term vanishes if x < 2 m.

. . . (4) 2 240 12 12 ( 2)x x x= − + −

2

2 22 40 12 12 ( 2)d yEI x x x

dx= − + − . . . (5)

2 3 3

140 12 12 ( 2)

2 3 3dy x x xEI Cdx

−= − + +

2 3 3120 4 4 ( 2)x x x C+ − += − . . . (6)

2

4 41 2

20 ( 2)3xEIy x x C x C= − + − + + . . . (7)

Here again note that the third term vanishes for x < 2 m.

at A, 20, 0 0x y C= = ∴ =

at B, 6 m, 0x y= =

3

4 41

20 60 6 (6 2)3

C×= − + − + 6×

116 16 20020 12 36 6

6 3C ×

= − × + × − = −

∴ 2 3 3 20020 4 4 ( 2)3

dyEI x x xdx

= − + − − . . . (8)

The third term vanishes.

Slope at A, (x = 0),

28


33

6200 200 10 1 rad 3.33 10 rad

3 3003 20 10A EI−− − ×

θ = = = − = − ×× ×

Slope at B, (x = 6 m),

2 3 3 200200 6 4 6 4 (6 2)3BEI θ = × − × + − −

3

36

136 136 10 2.27 10 radian3 3 20 10B EI

−×θ = = = + ×

× ×

Slope at C, (x = 3 m), i.e. x > 2 m

2 3 3 20020 3 4 3 4 (3 2)3CEI θ = × − × + − −

320 0.47 10 radians3C EI

−θ = = ×

3

4 420 200( 2)3 3xEIy x x x= − + − − . . . (9)

Deflection at centre, (x = 3 m),

3 4 420 2003 3 (3 2)3 3CEIy 3= × − + − − ×

3 3

6100 100 10 10 5 mm

20 10CyEI

− − × ×= = = −

×

For maximum deflection,

0dydx

=

2 3 3 2000 20 4 4 ( 2)3

x x x= − + − −

2 3 3 2 20020 4 4 32 24 483

x x x x x= − + − − + −

2 2964 483

x x= − + −

∴ 2 7412 03

x x− + =

2.63 mx = , x > 2m

3 4 4max

20 2002.63 2.63 (2.63 2) 2.63 101.73 3

EIy = × − + − − × = −

∴ max 5.087 mm 5.1 mmy = − −;

Example 9.3 A beam of span 4 m is subject to a point load of 20 kN at 1 m from the left support and a Udl of 10 kN/m over a length of 2 m from the right support. Determine :

(a) Slope at the ends. (b) Slope at the centre. (c) Deflection under the load. (d) Deflection at the centre.

29

Deflection of Beams(e) Maximum deflection.

Take EI = 20 × 106 N-m2.

Solution

. . . (1) 0, so that 20 10 2 40 kNy A BF R R∑ = + = + × =

A B

10 kN/m

D

x

1m 1m C 2m

20 kN

Figure 9.20


4 20 1 10 2 3 80BR × = × + × × =

. . . (2) 20 kNBR =

20 kNAR =

[ 220 20 [ 1] 10 [ 2]2

xM x x x ]−= − − − −

. . . (3) 220 20 [ 1] 5 [ 2]x x x= − − − −

2

2d yEI Mdx

=

. . . (4) 220 20 [ 1] 5 [ 2]x x x= − − − −

2 2 31

510 10 [ 1] [ 2]3

dyEI x x xdx

= − − − − + C . . . (5)

3

3 41 2

10 10 5[ 1] [ 2]3 3 12xEIy x x C x C= − − − − + + . . . (6)

at A, 20, 0, 0x y C= = =

at B, 4 m, 0x y= =

3

3 41

10 4 10 50 (4 1) (4 2)3 3 12

C×= − − − − + 4×

= −

C 1 29.17

2 2 3510 10 [ 1] [ 2] 29.173

dyEI x x xdx

= − − − − −

(a) Slope at A, (x = 0),

3

36

29.17 29.17 10 1.46 10 radians20 10A EI

−− − ×θ = = = − ×

×

(b) Slope at B, (x = 4 m),

2 2 3510 4 10 (4 1) (4 2) 29.17 27.53BEI θ = × − − − − − = +

θ = 31.38 10 radiansB−+ ×

(c) Slope at centre, (x = 2 m),

30


2 210 2 10 (2 1) 29.17CEI θ = × − − −

30.04 10 radiansC−θ = + ×

Deflection under the load :

3

3 410 10 5[ 1] [ 2] 29.173 3 12xEIy x x x= − − − − −

At x = 1 m,

10 29.173DEIy = −

3 3

625.84 10 10 1.29 mm

20 10DEIy − × ×= = −

×

(d) Deflection at the centre :

x = 2 m

3

310 2 10 (2 1) 29.17 23 3CEIy ×

= − − − ×

1.75 mmCy = −

(e) Maximum deflection : Let the maximum deflection b/w D and C (x < 2 m).

0dydx

=

2 210 10 ( 1) 29.17 0x x− − − =

2 210 10 10 20 29.17 0x x x− − + − =

1.96 m 2 mx = <

3 3max

10 10(1.96) (1.96) 29.17 1.96 353 3

EIy = − − × = −

∴ max 1.7501 mmy = −

Example 9.4

A beam of span 8 m is loaded with UDL of 10 kN/m over the middle half portion. Find the maximum deflection. EI is constant.

Solution

By symmetry, 10 4 20 kN2A BR R ×

= = =

Figure 9.21

2 210 ( 2) 10 ( 6)20

2 2x xM x − −

= − −

. . . (1) 220 5 ( 2) 5( 6)x x x= − − − − 2

A B

10 kN/m

x2m 4m C 2m

x

31

Deflection of Beams

2

2d yEI Mdx

=

. . . (2) 220 5 ( 2) 5( 6)x x x= − − − − 2

2 3

31

20 5 ( 2) 5 ( 6)2 3 3

dy x xEIdx

−= − − − +x C . . . (3)

3

4 41 2

10 5 5( 2) ( 6)3 12 12xEIy x x C x C= − − − − + + . . . (4)

at A, 20, 0, 0x y C= = =

at B, 8 m, 0x y= =

3

4 41

10 8 5 50 (8 2) (8 6)3 12 12

C×= − − − − + 8×

1 145C = −

2 3 35 510 ( 2) ( 6) 1453 3

dyEI x x xdx

= − − − − −

3

4 410 5 5( 2) ( 6) 1453 12 12xEIy x x x= − − − − −

The maximum deflection at centre, since the beam is symmetrical and symmetrically loaded.

∴ 3

4 4max

10 4 5 5(4 2) (4 6) 145 43 12 12

EIy ×= − − − − − ×

max11203

yEI

−=

SAQ 1 (a) A simply supported beam of span 5 m carries a concentrated load of

100 kN at a distance 2 m from the left support. Determine the deflection at mid span and the maximum deflection. Take EI = 20 × 106 N-m2.

(b) A simply supported beam of span 6 m is subjected to an end couple of 30 kN-m. Determine the maximum deflection. EI = 30 × 106 N-m2.

(c) A simply supported beam of span 6 m carries a Udl of 48 kN/m for a length 3 m from the right support and a clockwise moment of 20 kN-m is applied at a distance of 1 m from left support. Determine the deflection at a distance of 1 m from left end.

9.4 DEFLECTION OF CANTILEVER BEAMS

Cantilever beams are fixed at one end and the other end is free. Since the x-axis is taken from left to right, the free end is taken at the left for convenience.

32

9.4.1 Cantilever Beams with Single Concentrated Load at Free End Stresses in Shafts and Shells

Consider a cantilever beam loaded with a point load ‘W’ at free end.

Consider a section X-X at a distance x from left end.

.M W x= − (Hogging BM) . . . (9.92)


2

2d yEI Mdx

= = = −W x . . . (9.93)

W

A

Figure 9.22


2

12dy W xEI Cdx

= − + . . . (9.94)

3

16W x

2EIy C x C= − + + . . . (9.95)



At B, , dyx ldx

0= = (Fixed end) . . . (1)

At B, , 0x l y= = . . . (2)


2

102

W l C= − +

∴ 2

1 2W lC = + . . . (9.96)

Applying BC (2) into the Eq. (9.95), we can get

3 2

206 2

W l W l l C= − + × +

3

2 3W lC = − . . . (9.97)


2 2

2 2dy W x W lEIdx

= − + . . . (9.98)

3 2

6 2W x W l x W lEIy = − + −

3

3 . . . (9.99)

The maximum slope occurs at x = 0.

Slope at free end, 2

2AW l

EIθ = + . . . (9.100)

lx

XX

X

B

33

Deflection of BeamsThe maximum deflection occurs at x = 0.

3

max 3W ly

EI= − . . . (9.101)

Slope at centre 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

2 2

2 2 2CW l W lEI ⎛ ⎞θ = − +⎜ ⎟

⎝ ⎠

23

8AW lEI

θ = + . . . (9.102)

Deflection at centre 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

3 2 3

6 2 2 2 3CW l W l l W lEIy ⎛ ⎞= − + × −⎜ ⎟

⎝ ⎠

3 35[ 1 12 16]

48 48W l W l−

= − − + − =

∴ 35

48CW ly

EI= − . . . (9.103)

W

A l/2

C

l/2 B

(a) Cantilever

A

θA

C B

+ θC

2° Curve

(b) Slope Diagram


yA

B

Θ

C

yc

A

Figure 9.23

9.4.2 Cantilever Beams with a Central Point Load Consider a cantilever beam loaded with a point load ‘W’ at the centre ‘C’.


Moment, 2lM W x⎡= − −⎢⎣ ⎦⎤⎥ (Hogging BM) . . . (9.104)


2

2 2d y lEI M W xdx

⎡ ⎤= = − −⎢ ⎥⎣ ⎦ . . . (9.105)

34


Figure 9.24


2

12 2dy W lEI xdx

⎡ ⎤ C= − − +⎢ ⎥⎣ ⎦ . . . (9.106)

3

16 2W l

2EIy x C x C⎡ ⎤= − − + +⎢ ⎥⎣ ⎦ . . . (9.107)



At B, , 0dyx ldx

= = . . . (1)

At B, , 0x l y= = . . . (2)


22

102 2

W l ll C⎡ ⎤= − − +⎢ ⎥⎣ ⎦

∴ 2

1 8W lC = + . . . (9.108)


3 2

206 2 8

W l W ll l⎛ ⎞= − − + × +⎜ ⎟⎝ ⎠

C

∴ 3

2548W lC −

= . . . (9.109)


2 2

2 2 8dy W l W lEI xdx

⎡ ⎤= − − +⎢ ⎥⎣ ⎦ . . . (9.110)

3 2 35

6 2 8 48W l W l WEIy x⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

l . . . (9.111)

Slope at (x = 0),

2

8AW l

EIθ = + . . . (9.112)

Slope at C 2lx⎛ ⎞

⎜ ⎟ , =⎝ ⎠

2

8CW l

EIθ = + . . . (9.113)

Deflection at A (x = 0),

W

A Cx

XX

X

l/2 l/2 B

35

Deflection of Beams

3548AW ly

EI= − . . . (9.114)

Deflection at C 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

2 35.

8 2 48 24CW l l W l W lEIy = + − = −

3

∴ 2

24CW ly

EI= − . . . (9.115)

W

A l/2 l/2 BC

(a) Cantilever Beam with a Central Point Load

A

θA

C B

+ θC

2° Curve

(b) Slope Diagram

2° Curve

B C

yc

A

yA

St. line


Figure 9.25

9.4.3 Cantilever Beams with a UDL Consider a cantilever beam loaded with a Udl of w/unit length on total length. Consider a section X-X at a distance x from left end.

Moment, 2

2 2x w xM w x= − = − (Hogging BM) . . . (9.116)


2 2

2 2d y w xEI Mdx

= = − . . . (9.117)

A BC

X

Xx

w/unit length

l/2l/2

Figure 9.26


3

16dy w xEIdx

= − + C . . . (9.118)

4

1 224w xEIy C x C= − + + . . . (9.119)

The constants C1 and C2 can be found from boundary conditions.

36

The boundary conditions are : Stresses in Shafts and Shells

At B, , 0dyx ldx

= = . . . (1)

At B, , 0x l y= = . . . (2)


3

106

wl C= − +

∴ 3

1 6wlC = + . . . (9.120)


4 4

2024 6

wl wl C= − + +

∴ 4

2 8wlC = − . . . (9.121)


3 3

6 6dy w x wlEIdx

= − + . . . (9.122)

3 4

24 6 8w wl x wlEIy = − + − . . . (9.123)

Slope at (x = 0), 3

6Awl

EIθ = + . . . (9.124)

Slope at C 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

3 3 7

48 6 48Cwl wl wlEI θ = − + = +

3

∴ 37

48CwlEI

θ = . . . (9.125)


4

8Awly

EI= − . . . (9.126)

Deflection at C 2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

4 3 4

24 2 6 2 8Cw l wl l wlEIy ⎛ ⎞ ⎛ ⎞= − + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4 417( 1 32 48)

24 16 384wl wl−

= − + − =×

∴ 417

384CwlyEI

−= . . . (9.127)

B

C

w/unit length

l/2 (a) Cantilever Beam with Udl

Al/2

θBAB

+ θBC B

3° Curve

37

Deflection of Beams

(b) Slope Diagram

4° Curve

B CyC

A

yA


Figure 9.27

9.4.4 Cantilever Beam with a UDL on Some Portion Case I

Udl on the right half portion :

Consider a cantilever beam loaded with a Udl of w/unit length on portion CB as shown in Figure 9.28.


Moment, 22 2

lxlM w x

⎛ ⎞−⎜ ⎟⎛ ⎞ ⎝ ⎠= − −⎜ ⎟⎝ ⎠

2

2 2w lx⎛ ⎞= − −⎜ ⎟⎝ ⎠

(Hogging BM) . . . (9.128)

A B

w/unit length

x

C

X

X

l/2l/2

Figure 9.28


22

2 2 2d y w lEI M xdx

⎡ ⎤= = − −⎢ ⎥⎣ ⎦ . . . (9.129)


3

16 2dy w lEI xdx

⎡ ⎤= − − +⎢ ⎥⎣ ⎦C . . . (9.130)

4

1 224 2dy w lEIy x C x Cdx

⎡ ⎤= − − + +⎢ ⎥⎣ ⎦ . . . (9.131)



At B, , 0dyx ldx

= = . . . (1)

At B, . . . (2) , 0x l y= =

Applying the boundary condition (1) into the Eq. (9.130), we can get

38


3

106 2w ll C⎡ ⎤= − − +⎢ ⎥⎣ ⎦

∴ 3

1 48wlC = . . . (9.132)

Applying boundary condition (2) into the Eq. (9.131), we can get

4 3

2024 2 48w l wll l⎛ ⎞= − − + × +⎜ ⎟

⎝ ⎠C

4

2[ 1 8]24 16

wl C= − + +×

∴ 4

27384

wlC −= . . . (9.133)


3 3

6 2 4dy w l wlEI xdx

⎡ ⎤= − − +⎢ ⎥⎣ ⎦ 8 . . . (9.134)

4 3 47

24 2 48 384w l wl wlEIy x x⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

. . . (9.135)

Slope at (x = 0),

3

48Awl

EIθ = + . . . (9.136)


⎝ ⎠,

3

48Cwl

EIθ = . . . (9.137) +


47

384AwlyEI

−= . . . (9.138)

Deflection at C 2lx⎛ ⎞

⎜ ⎟ , =⎝ ⎠

3 47

48 2 384Cwl l wlEIy = + × −

4 43(4 7)

384 384wl wl−

= − =

∴ 43

384CwlyEI

= − . . . (9.139)

B

w/unit length

AC

l/2 l/2

(a) Cantilever Beam with Udl on the Right Half

39

Deflection of Beams

A

θA

C B

+ θC

3° Curve

(b) Slope Diagram

4° Curve

B C yc

A

yA

St. line


Figure 9.29

Case II

Consider a cantilever beam loaded with a Udl of w/unit length on portion AC as shown in Figure 9.30(a).

The Udl is added in both directions on the portion CB as shown in Figure 9.30(b).

A B

w/unit length

C l/2 l/2

D

(a)

A B

w/unit length

x l/2

C

l/2

(b)

Figure 9.30

Consider a section X-X at a distance x from left end,

Moment, . .2 2 2 2x w lM w x x x⎛ ⎞ ⎛ ⎞= − + − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

l

22

2 2 2w x w lx⎡= − + −⎢⎣ ⎦

⎤⎥ (Hogging BM) . . . (9.134)


22 2

2 2 2 2d y w x w lEI M xdx

⎡ ⎤= = − + −⎢ ⎥⎣ ⎦ . . . (9.135)


33

16 6 2dy w x w lEI xdx

⎡ ⎤= − + − +⎢ ⎥⎣ ⎦C . . . (9.136)

42

1 224 24 2w x w lEIy x C x C⎡ ⎤= − + − + +⎢ ⎥⎣ ⎦

. . . (9.137)



At B, , 0dyx ldx

= = . . . (1)

40

At B, , 0x l y= = . . . (2) Stresses in Shafts and Shells


33

106 6 2

wl w ll C⎛ ⎞= − + − +⎜ ⎟⎝ ⎠

∴ 3 3

11 71

6 8 4wl wlC ⎡ ⎤= − =⎢ ⎥⎣ ⎦ 8

. . . (9.138)

Applying the boundary condition (2) and value of C1 into the Eq. (9.137), we can get

44 4

270

24 24 2 48wl w l wll C⎛ ⎞= − + − + +⎜ ⎟

⎝ ⎠

∴ 4 4

21 71 (16 1 56)

24 16 2 24 16wl wlC ⎡ ⎤= − − = − −⎢ ⎥ ×⎣ ⎦

439

24 16wl

= −×

4

239384

wlC −= . . . (9.139)


33 37

6 6 2 48dy w x w l wlEI xdx

⎡ ⎤= − + − +⎢ ⎥⎣ ⎦ . . . (9.140)

44 37 39

24 24 2 48 384w x w l wl wlEIy x x⎡ ⎤= − + − + −⎢ ⎥⎣ ⎦

4 . . . (9.141)


37

48AwlEI

θ = + . . . (9.142)


⎝ ⎠,

3 3 37 6

6 2 48 48Cw l wl wlEI

EI E⎛ ⎞θ = − + = +⎜ ⎟⎝ ⎠ I

. . . (9.143)

Deflection at D 34lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

3 3 33 3 7

6 4 6 4 2 48Dw l w l l wlEI ⎛ ⎞ ⎡ ⎤θ = − + − +⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

3 327 7

24 16 24 16 48wl wl wl−

= + +× ×

3

3 330 5[ 27 1 56]

24 16 384 64wl wl wl

= − + + = = +×

3

41

Deflection of Beams∴

3564D

wlEI

θ = + . . . (9.144)

Deflection at A, (x = 0)

439

384Cwly

EI= − . . . (9.145)


⎝ ⎠

4 3 47 39

24 2 48 2 384Cw l wl l wlEI y ⎛ ⎞= − + × −⎜ ⎟

⎝ ⎠

4 412[ 1 28 39]

384 384wl wl−

= − + − =

∴ 412

384CwlyEI

−= . . . (9.146)

Deflection at D, 34lx⎛ ⎞=⎜ ⎟

⎝ ⎠

4 4 3 43 3 7 3 39

24 4 24 4 2 48 4 384Dw l w l l wl l wlEI y ⎛ ⎞ ⎡ ⎤ ⎛ ⎞= − + − + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎝ ⎠

4

[ 18 1 21 32 39 16]24 256

wl= − + + × −

××

4

432 224 256 384

wl wl−= = −

×

4 4

192Dwl wly

EI= . . . (9.147)

w/unit length

A BC l/2 l/4

D

l/4

(a) Cantilever Beams with Udl on Left Portion

A

θA

C B

θC

3° Curve

θD

D

(b) Slope Diagram

yC

yAӨ


Figure 9.31

9.4.5 Cantilever Beams with a Triangular Load Consider a cantilever beam loaded with a triangular load as shown in Figure 9.32.

w/unit length

BA l

x

xx

42


Figure 9.32

Consider a section X-X at a distance x from A as shown in Figure 9.32.

Intensity of loading .w xl

=

Moment, 31

2 3w xM x xl l

−⎛ ⎞ ⎛ ⎞= − × × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 6

w x . . . (9.148)


2 3

2 6d y w xEI M

ldx= = − . . . (9.149)


4

124dy w xEIdx l

C= − + . . . (9.150)

5

1 2120w xEIy C x C

l= − + + . . . (9.151)

The constants C1 and C2 can be found from boundary conditions. The boundary conditions are :

At B, , 0dyx ldx

= = . . . (1)

At B, , 0x l y= = . . . (2)


3

1024wl C−

= +

∴ 3

1 24wlC = + . . . (9.152)


4 4

20120 24wl wl C= − + +

∴ 3

2 24wlC = + . . . (9.153)


4 3

24 24dy w x wlEIdx l

= − + . . . (9.154)

5 3

120 24 30w x wl wlEIy x

l= − + −

3 . . . (9.155)

Slope at A, (x = 0), 3

24Awl

EIθ = + . . . (9.156)

43

Deflection of BeamsSlope at C

2lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

4 3

24 2 24Cw l wlEI

l⎛ ⎞θ = − +⎜ ⎟⎝ ⎠

3 31 15124 16 24 16

wl wl−⎡ ⎤= + = +⎢ ⎥ ×⎣ ⎦

∴ 315

384Cwl

EIθ = + . . . (9.157)

Deflection at A, (x = 0),

4

30Awly

EI= − . . . (9.158)


⎝ ⎠,

4 4

120 48 30Cwl wl wlEI y −

= + −4

4 4[4 10 16]

480 240wl wl− −

= + − =

∴ 4

240Cwly

EI−

= . . . (9.159)

BA C l/2l/2

w/unit length

(a) Cantilever Beams with a Triangular Load

+ θA

4° Curve

θC

A B C

(b) Slope Diagram

yC

yA Ө

A C


Figure 9.33

9.4.6 Cantilever Beams with an End Couple Consider a cantilever beam loaded with an end couple M0 as shown in Figure 9.34.

A B

X

C

MO X

x l

Figure 9.34

Consider a section X-X at a distance x from A as shown in Figure 9.34.

44


Moment, 00 .M M x= − . . . (9.160)


2

002

d yEI M Mdx

= = − x . . . (9.161)


00

dy1EI M x

dx= − + C . . . (9.162)

2

0 12x

2EIy M C x C= − + + . . . (9.163)

The constants C1 and C2 can be found from boundary conditions. The boundary conditions are :

At B, , 0dyx ldx

= = . . . (1)

At B, , 0x l y= = . . . (2)

Applying the boundary condition (1) into the Eq. (9.162), we can get 0 10 M l C= − +

∴ 1 0C M l= . . . (9.164)


2

00 20 .

2M l

M l l C= − + +

∴ 2

02 2

M lC = − . . . (9.165)


0dy

0EI M x Mdx

= − + l . . . (9.166)

2 2

0 002 2

M x MEIy M l x= − + −l

. . . (9.167)


0A

M lEI

θ = . . . (9.168)


⎝ ⎠,

0 02ClEI M Mθ = − + l

∴ 02CM l

EIθ = + . . . (9.169)


2

0A

M lyEI

= − . . . (9.170)


⎝ ⎠,

2 2

0 00 .

2 2 2 2CM M ll lEI y M l⎛ ⎞= − + −⎜ ⎟

⎝ ⎠

45

Deflection of Beams

20

8CM l

yEI

= − . . . (9.171)

A B

CMO

l/2 l/2

(a) Cantilevers with an End Couple

θA

B

St. line

C A

θC

(b) Slope Diagram


yC

yAӨ

A C B

Figure 9.35

9.4.7 Examples Example 9.5

A cantilever beam of length l carries a point load W at a distance 4l from free end.

Determine the slopes at free end and under load and the maximum deflection and deflection under the load.

Solution

4lM w x⎡ ⎤= − −⎢ ⎥⎣ ⎦

. . . (1)

2

2 4d y lEI w xdx

⎡ ⎤= − −⎢ ⎥⎣ ⎦ . . . (2)

2

12 4dy w lEI xdx

⎡ ⎤= − − +⎢ ⎥⎣ ⎦C . . . (3)

3

16 4w l

2EIy x C x C⎡ ⎤= − − + +⎢ ⎥⎣ ⎦ . . . (4)

A B

l/4

l

Figure 9.36

Boundary conditions are :

At B, , 0dyx ldx

= = . . . (5)

At B, . . . (6) , 0x l y= =

From Eqs. (3) and (5),

2

102 4w ll C⎡ ⎤= − − +⎢ ⎥⎣ ⎦

46


2

19

32wlC = + . . . (7)

From Eqs. (4), (6) and (7)

3 3

290

6 4 32w l wll C⎡ ⎤= − − + +⎢ ⎥⎣ ⎦

3

2 [ 108 27]6 64

wlC = − +×

3 381 27

6 64 128wl wl−

= − =×

. . . (8)

The Eqs. (3) and (4) will be :

∴ 2 29

2 4 32dy w l wlEI xdx

⎡ ⎤= − − +⎢ ⎥⎣ ⎦ . . . (9)

3 2 39 27

6 4 32 128w l wl wlEIy x x⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

. . . (10)


29

32Awl

EIθ = + . . . (11)


⎝ ⎠,

29

32Cwl

EIθ = + . . . (12)


327

128AwlyEI

= − . . . (13)

Deflection under load at C, 4lx⎛ ⎞=⎜ ⎟

⎝ ⎠,

2 39 27 18.

32 4 128 128Cwl l wl wlEI y −

= − =3

∴ 318

128CwlyEI

−=

Example 9.6

Find the slope and deflection at the free end of a cantilever shown in Figure 9.37. Take EI = 200 × 106 N-m2.

Solution

( 330 60 [ 2] 24 ( 3)2

xM x x x )−= − − − − −

. . . (1) 230 60 [ 2] 12 [ 3]x x x= − − − − −

2

2d yEI Mdx

=

47

Deflection of Beams . . . (2) 230 60 [ 2] 12 [ 3]x x x= − − − − −

2 2 3115 30 [ 2] 4 [ 3]dyEI x x x

dx= − − − − − + C . . . (3)

3 3 41 25 10 [ 2] [ 3]EIy x x x C x C= − − − − − + + . . . (4)

A BC 2 m D

30 kN 60 kN

1 m 2 m

24 kN/m

Figure 9.37


At B, 5 m, 0dyxdx

= = . . . (5)

At B, . . . (6) 5 m, 0x y= =

From Eqs. (3) and (5)

2 2 310 15 5 30 (5 2) 4 (5 3) C= − × − − − − +

∴ . . . (7) 1 677C =

From Eqs. (4), (6) and (7)

3 3 420 5 5 10 (5 2) (5 3) 677 5 C= − × − − − − + × +

∴ . . . (8) 2 2474C = −

2 2 315 30 [ 2] 4 [ 3] 677dyEI x x xdx

= − − − − − + . . . (9)

. . . (10) 3 3 415 10 [ 2] [ 3] 677 2474EIy x x x x= − − − − − + −

At free end, i.e. at A, x = 0,

From Eq. (9)

3

36

677 677 10 3385 10 radians200 10A EI

−×θ = + = = ×

×

From Eq. (10)

3 3

62474 2474 10 10 12.37 mm

200 10AyEI

− − × ×= = = −

×

= − 12.37 mm ( )↓

Example 9.7

Find the slope and deflection at the free end of the cantilever shown in Figure 9.38. Take EI = 200 × 106 N-m2.

Solution

Apply Udl in both the upward and downward directions in the portion CB.

BA

24 kN/m

C 2mD1m 1m

30 kN

Figure 9.38

48

Stresses in Shafts and She

lls

Figure 9.39

( 1)24 . 24 ( 1) 30 [ 2]2 2x xM x x x−

= − + − − −

2 212 12 [ 1] 30 [ 2x x x ]= − + − − − . . . (1)

2

2 22 12 12 [ 1] 30 [ 2d yEI M x x x

dx]= = − + − − − . . . (2)

3 3 214 4 [ 1] 15 [ 2]dyEI x x x

dxC= − + − − − + . . . (3)

4 4 31 2[ 1] 5 [ 2]EIy x x x C x C= − + − − − + + . . . (4)


At B, 4 m, 0dyxdx

= = . . . (5)

At B, 4 m, 0x y= = . . . (6)


3 3 310 4 4 4 (4 1) 15 (4 2) C= − × + − − − +

∴ 1 208C = . . . (7)

From Eqs. (4), (6) and (7)

4 4 320 4 (4 1) 5 (4 2) 208 4 C= − + − − − + × +

∴ 2 617C = − . . . (8)

∴ 3 3 24 4 [ 1] 15 [ 2] 20dyEI x x xdx

= − + − − − + 8

7

. . . (9)

. . . (10) 4 4 3[ 1] 5 [ 2] 208 61EIy x x x x= − + − − − + −

At free end A, x = 0,

From Eq. (9)

3

36

208 208 10 1.04 10 radians200 10A EI

−×θ = + = + = + ×

×

From Eq. (10)

3 3

6617 617 10 10 3.085 mm ( )

200 10AyEI

− − × ×= = = −

×↓

B A

SAQ 2 Find the displacement of free end of a cantilever shown in Figure 9.40. Take EI = 40 × 106 N-m2.

24 kN/m 30 kN

X

Xx

C 24 kN/m D1m

20 kN 24 kN/m

2mA B

2m

49

Deflection of Beams

Figure 9.40

9.5 DEFLECTION OF OVERHANG BEAMS

The overhang beams have overhangs on one side or on both sides. This type of beams are generally supported on rollers and hinged supports (Figure 9.41).

(a) Single Overhang Beam

A B

(b) Double Overhang Beam

Figure 9.41

For this types of beams the deflection are generally calculated using Macaulay’s Method.

Example 9.8

Determine the deflection under the loads as shown in Figure 9.42.

1m C 1m 1mA D

X

X

80kN

x

B20 kN

Figure 9.42

Solution

. . . (1) 0, so that 80 20 100 kNy A BF R R∑ = + = + =

Σ M about A = 0.

80 . . . (2) 1 20 3 2 70 kN ( )A BR R× + × = × ∴ = ↑


30 kN ( )AR = ↑

. . . (3) 30 80 ( 1) 70 ( 2)M x x x= − − + −

2

2 30 80 [ 1] 70 [ 2]d yEI M x x xdx

= = − − + − . . . (4)

2 2 2115 40 [ 1] 35 [ 2]dyEI x x x

dx= − − + − + C . . . (5)

3 3 31 2

40 355 [ 1] [ 2]3 3

EIy x x x C x C= − − + − + + . . . (6)


At A, . . . (7) 0, 0x y= =

50

At B, 2 m, 0x y= = . . . (8) Stresses in Shafts and Shells

From Eq. (6) and (7), C2 = 0


3 31

400 5 2 [2 1] 23

C= × − − + ×

1403

C = −

∴ 3 3 340 35 405 [ 1] [ 2]3 3 3

EIy x x x= − − + − − x . . . (7)

Deflection at C,

x = 1 m

3 40 255 1 13 3CEIy −

= × − × =

∴ 253Cy

EI−

= . . . (8)

Deflection at D,

x = 3 m

3 3 340 35 405 3 (3 1) (3 2) 3 03 3 3DEIy = × − − + − − × =

0Dy =

Example 9.9

For the beam shown in Figure 9.43, determine the deflection at free end and the maximum deflection.

Figure 9.43

Solution

48 6 60 348 kNA BR R+ = × + = . . . (1)

Σ M about A = 0.

48 6 (2 3) 60 16 12BR× × + + × = ×

∴ 200 kN ( )BR = ↑ . . . (2)


148 kN ( )AR = ↑ . . . (3)

Apply the Udl in downward and upward directions in the portion EC (Figure 9.44).

E B4m C

48kN/m

4m

60 kN

A D

2m 6m

B

48 kN/m

AD

148 kN200 kN

C E

60 kN X

Xx8m

51

Deflection of BeamsFigure 9.44

48 48148 [ 2] [ 2] [ 8] [ 8] 200 [ 12]2 2

M x x x x x x= − − − + − − + −

. . . (4) 2 2148 24 [ 2] 24 [ 8] 200 [ 12]x x x x= − − + − + −

2

2d yEI Mdx

=

. . . (5) 2 2148 24 [ 2] 24 [ 8] 200 [ 12]x x x x= − − + − + −

2 3 2 2174 8 [ 2] 8 [ 8] 100 [ 12]dyEI x x x x

dx= − − + − + − + C . . . (6)

3 4 4 31

74 1002 [ 2] 2 [ 8] [ 12]3 3 2EIy x x x x C x C− − + − + − + + . . . (7)


At A, . . . (8) 0, 0x y= =

At B, . . . (9) 12 m, 0x y= =


C2 = 0 . . . (10)


3 4 41

740 12 2 [12 2] 2 [12 8] 123

C= × − − + − + ×

∴ . . . (11) 1 1928C = −

∴ 3 4 4 374 1002 [ 2] 2 [ 8] [ 12] 19283 3

EIy x x x x x= − − + − + − −

Deflection at free end, x = 16 m

3 4 4 374 10016 2 [16 2] 2 [16 8] [16 12] 1928 16 36803 3CEIy = × − − + − + − − × =

∴ 3680Cy

EI= +

Maximum Deflection : It occurs between D and E.

0dydx

=

2 30 74 8 [ 2] 1928x x= − − −

2 3 274 8 64 48 96 1928x x x x= − + + − −

3 28 122 96 1928x x x= − + +

By trial and error, x = 5.7 m

3 4max

74 5.7 2 [5.7 2] 1928 5.7 6796.33

EIy = × − − − × = −

∴ max6796.3y

EI= −

Example 9.10

52

A double overhanging beam of 12 m length rests symmetrically on supports, 8 m apart. A load of 80 kN acts at one free end and a load of 40 kN acts at other free end. Determine the deflection :


(a) at free ends, and

(b) at mid-span.

Solution

. . . (1) 0, so that 80 40 120 kNy A BF R R∑ = + = + =

Σ M about A = 0,

⇒ 8 40 10 80 2BR × = × − ×

30 kN ( )BR = ↑ . . . (2)

Figure 9.45


90 kN ( )AR = ↑ . . . (3)

80 90 [ 2] 30 [ 10]M x x x= − + − + − . . . (4)

2

2 80 90 [ 2] 30 [ 10]d yEI M x x xdx

= = − + − + − . . . (5)

2 2 2140 45 [ 2] 15 [ 10]dyEI x x x

dxC= − + − + − + . . . (6)

2 3 31

40 15 [ 2] 5 [ 10]3 2EIy x x x C x C= − + − + − + + . . . (7)


At A, 2 m, 0x y= = . . . (8)

At B, 10 m, 0x y= = . . . (9)


32

400 23

C−= × +

∴ 2320

3C = + . . . (10)

From Eqs. (7), (8) and (9),

3 31

40 3200 = × 10 15 [10 2] 103 3

C−+ − + × +

11664

3C = + . . . (11)

3 3 340 1664 32015 [ 2] 5 [ 10]3 3

EIy x x x x= − + − + − + +3

. . . (12)

D

X

X

40 kN

BC

80 kN

2m

A

2m E

8m

53

Deflection of BeamsDeflection at free ends :

At D, x = 0

3203Dy

EI= + . . . (13)

At E, x = 12 m,

3 3 340 1664 32012 15 [12 2] 5 [12 10] 123 3EEIy = − × + − + − + × +

3

∴ 37123Ey

EI= −

Deflection at mid-span :

x = 6 m

3 340 1664 3206 15 [6 2] 63 3CEIy = − × + − + × +

3

45443Cy

EI= −

SAQ 3 Determine the deflection at C and D of the bean shown in Figure 9.46.

A

BC

12 kN /m

2m 2m

40 kN 40 kN

2m D

Figure 9.46

9.6 APPLICATION OF DEFLECTIONS OF BEAMS

The calculation of deflection of beams is useful in the analysis of structures and also in the design of structures, and machine parts.

(a) In the design of structures, the check for deflection can be done.

(b) In the analysis of propped cantilever, propped beams, fixed beams, the deflection calculations are used.

9.7 SUMMARY

(a) Deflection is the vertical displacement of a loaded beam. (b) Slope is the rotation of the loaded beam,

(c) 2

2d yEI Mdx

=

1dyEI Mdx

= +∫ C

1 2EIy M C x C= + +∫∫

54

Stresses in Shafts and Shells (d)

2

16AwlEI

−θ =

3

48CwlyEI

−=

A B

W

C l/2 l/2

Figure 9.47

(e) ( 26Awa b a bEI l

−θ = + )

(2 )6Bwb a bEI l

θ = + +

2 2[3 4 ]48C

wby lEI

−= − b

32 2 2

max ( )9 3

wby lEI l

−= − b

A B

W

Ca b

Figure 9.48

(f) 3

24AwlEI

−θ =

45384C

wlyEI

= −

A B

w/unit length

C Figure 9.49

(g) 2

2Bwl

EIθ =

3

3BwlyEI

−=

w

Figure 9.50

(h) 2

6Bwl

EIθ =

4

8BwlyEI

−=

A B

w/unit length

A B

55

Deflection of Beams Figure 9.51

9.8 ANSWERS TO SAQs

SAQ 1

(a) 6 220 10 N-mEI = ×100 kN

3m2m A B

x

X

X

Figure 9.52

0 100 kNA By R R∑ = ⇒ + = . . . (1)

Σ M about A = 0

⇒ 5 100 2BR × = ×

∴ 40 kN ( )BR = ↑ . . . (2)


60 kN ( )AR = ↑ . . . (3)

60 100 [ 2]M x x= − − . . . (4)

2

2 60 100 [ 2]d yEI M x xdx

= = − − . . . (5)

2 2130 50 [ 2]dyEI x x

dx= − − + C . . . (6)

3 31 2

5010 [ 2]3

EIy x x C x C= − − + + . . . (7)

The boundary conditions are : At A, . . . (8) 0, 0x y= =

At B, . . . (9) 5 m, 0x y= =


2 0C = . . . (10)

From Eqs. (7), (8) and (9)

3 31

500 10 5 [5 2] 53

C= × − − + ×

∴ . . . (11) 1 160C = −

2 230 50 [ 2] 160dyEI x xdx

= − − − . . . (12)

3 35010 [ 2] 1603

EIy x x x= − − − . . . (13)

Deflection at C, x = 2 m,

310 2 160 2CEIy = × − ×

56

Stresses in Shafts and Shells ∴

3 3

6240 240 10 10 12 mm

20 10CyEI

− − × ×= = = −

×

Maximum Deflection It occurs between C and B.

0dydx

=

2 230 50 [ 2] 160 0x x− − − =

2 230 50 200 200 160 0x x x− − + − =

220 200 360 0x x− + − =

⇒ 2 10 18 0x x− + =

∴ 2.35 mx =

Deflection at Mid-span 2.35 mx =

3 35010 2.5 [2.5 2] 160 2.53DEIy = × − − − ×

12.29 mmDy = −

3 3max

5010 2.35 [2.35 2] 160 2.353

EIy = × − − − ×

max 12.35 mmy = −

(b) 6 230 10 N-mEI = ×

Figure 9.53

0 0y AF R RB∑ = ⇒ + = . . . (1)

Σ M about A = 0

⇒ 6 30BR × =

∴ 5 kN ( )BR = ↓ . . . (2)


5 kN ( )AR = − ↓ . . . (3)

o5 30M x x+= − . . . (4)

2

o2 5 30d yEI x

dx= − + x . . . (5)

2

15 30

2dy xEI xdx

C= − + + . . . (6)

3

21

5 153x

2EIy x C x C= − + + + . . . (7)

A B

x

30 kN-m

6m

57

Deflection of BeamsThe boundary conditions are :

At A, . . . (8) 0, 0x y= =

At B, . . . (9) 6 m, 0x y= =


2 0C = . . . (10)

From Eqs. (7), (9) and (10)

3

21

5 60 15 66

C× 6= − + × + ×

∴ 1 60C = − . . . (11)

25 30 60

2dy xEI xdx

= − + − . . . (12)

3

25 15 606xEIy x x= − + − . . . (13)

For maximum deflection, 0dydx

= ,

∴ 25 30 60 02

x x− + − =

or 2 12 24 0x x− + =

∴ 2.54 mx =

3 2max

5 (2.54) 15 (2.54) 60 2.54 69.286

EIy = − + − × = −

max 2.31 mmy = −

(c) 0yF∑ =

∴ 48 3 144 kNA BR R+ = × = . . . (1)

or 6 20 48 3 (3 1.5)BR × = + × × +

∴ 334 kN ( )3BR = ↑ . . . (2)

x

A B

48 kN /m

x

CD

x20 kN -mm 1m 2m 3m

Figure 9.54


98 kN ( )3AR = ↑ . . . (3)

o 298 4810 [ 1] [ 3]3 2

M x x x= + − − − . . . (4)

58


2o

298 4810 [ 1] [ 3]3 2

d yEI x x xdx

= + − − − 2 . . . (5)

2 31

98 4810 [ 1] [ 3]3 6

dyEI x x xdx

C= + − − − + . . . (6)

3 2 41

49 5 [ 1] 2 [ 3]9 2EIy x x x C x C= + − − − + + . . . (7)


At A, 0, 0x y= = . . . (8)

At B, 6 m, 0x y= = . . . (9)


C2 0= . . . (10)

From Eqs. (7), (8) and (10),

3 2 41

490 6 5 (6 1) 2 (6 3)9

C 6= × + − − − + ×

∴ 11139

6C = −

Deflection at D, x = 1 m,

349 11391 19 6DEIy = × − ×

184.4Dy

EI−

=

SAQ 2

6 240 10 N-mEI = ×

Figure 9.55

22420 [ 2]2

M x x= − − − . . . (1)

2

22 20 12 [ 2]d yEI x x

dx− − − . . . (2)

2 3110 4 [ 2]dyEI x x

dxC− − − + . . . (3)

3

41 2

10 [ 2]3xEIy x C x C= − − − + + . . . (4)


A

x20 kN

B

24 kN/m

2m

x

C 3m

x

59

Deflection of BeamsAt A, 4 m, 0dyx

dx= =

=

. . . (5)

At B, . . . (6) 4 m, 0x y=


2 310 10 4 4 (4 2) C= − × − − +

∴ . . . (7) 1 192C = +

From Eqs, (4), (6) and (7),

3

22

10 40 [4 2] 1923

C− ×= − − + × 4 +

∴ 21616

3C = −

3410 1616[ 2] 192

3 3xEIy x x= − − − + − . . . (8)

Deflection at free end, x = 0

16163Ay

EI= −

SAQ 3

0yF∑ = . . . (1) 12 6 2 40 152 kNA BR R+ = × + + =

0AM∑ =

⇒ 4 12 6 3 40 2 40 6BR × = × × + × + ×

∴ . . . (2) 134 kN ( )BR = ↑

A

12 kN/m

x

C

40 kN

B

X

D

X40 kN

2m 2m 2m

Figure 9.56


. . . (3) 18 kN ( )AR = ↑

21218 40 [ 2] 134 [ 4]

2xM x x x= − − − + − . . . (4)

2

22 18 6 40 [ 2] 134 [ 4]d yEI x x x x

dx= − − − + − . . . (5)

2 3 2 219 2 20 [ 2] 67 [ 4]dyEI x x x x

dx= − − − + − + C . . . (6)

4

3 3 31

20 673 [ 2] [ 4]3 3 3x

2EIy x x x C x C= − − − + − + + . . . (7)

The boundary conditions are : At A, . . . (8) 0, 0x y= =

At B, . . . (9) 4 m, 0x y= =

60

From Eqs. (7) and (8), Stresses in Shafts and Shells 2 0C = . . . (10)

From Eqs, (7), (9) and (10),

4

3 31

4 200 3 4 [4 2] 42 3

C= × − − − + ×

∴ 18

3C −

= . . . (11)

43 320 67 83 [ 2] [ 4]

2 3 2 3x 3EIy x x x x= − − − + − − . . . (12)

At x = 2 m,

3 24 83 2 22 3CEIy = × − − ×

323Cy

EI= +

At x = 6 m,

3 3 364 20 67 83 6 [6 2] [6 4] 62 3 3 3DEIy −

= × − − − + − ×

7923Dy

EI−

=

Unit 9

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Transcript of Unit 9