Double Integration Method

UNIT 9

DEFLECTION OF BEAMS

Deflection of Beams

Structure 9.1

Introduction Objectives

9.2

9.3

9.4

Different Methods 9.2.1

Governing Equation for Deflections

9.2.2

Sign Conventions

9.2.3

The Boundary Conditions

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

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 y = f ( x x) where y is the deflection at x.

θ A = Slope at A, θ B = Slope at B, B

ACB = Original beam, and AC ′ B ′ = Deflected beam.

5

Stresses in Shafts and Shells

C A

B θ B

θ A

C 1

y c c

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. A

C

D

B

x

dx

Figure 9.2

6

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

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

Deflection of Beams

y

A θ d θ ds

C

θ

B

x D

dy

dx

Figure 9.3

The normal drawn to the curve at C and D 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 d θ = angle made by radius at C with radius at D.

From Figure 9.2 ds = R d θ ds

R =

∴

1 R

=

dθ

dx d θ

[since ds ≈ dx]

d θ

=

. . . (9.1)

dx dy

tan θ =

dx

For small angle, tan θ = θ

∴

dy

θ=

. . . (9.2)

dx

dy

is defined as the slope of the curve. The curve in case of deflected beam is the shape dx of the central line of the beam. Differentiating the Eq. (9.2), w.r.t., x, we can get dθ dx

2

d y

=

dx

. . . (9.3)

2

From Eqs. (9.1) and (9.3),

1 R

2

d y

=

. . . (9.4)

dx 2

From the theory of pure bending, we know, M I

=

E

. . . (9.5)

R

From Eqs. (9.4) and (9.5), M =

EI R

= EI

2

d y dx

2

. . . (9.6)

7


This is the equation for elastic curve of beam and governing equation for deflections. dy Note that y and can be found by integration once and twice, respectively. The dx condition for integration is that function M = f ( x 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 dy dy′ x. The y and are invariably known at the supports. may also be ascertained from dx dx symmetry of bent shape of beam central line. For example, dy

(a)

At simple supports, y = 0,

(b)

At fixed supports, y = 0,

(c)

At middle of SS beam loaded either by a single central or Udl over entire dy = 0. length y ≠ 0, dx

dx

dy dx

≠ 0.

=0.

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 AC = CB =

l

2

Because of symmetry, R A = RB =

W

2

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

8

(a)

At x = 0, y = 0.

(b)

At x = l, y = 0

(c)

At x =

l

,

dy

2 dx

=0

Deflection of Beams

W

W/2

W/2

x

Figure 9.4

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

⎛ ⎝

l⎞ ⎟ 2⎠

⎛ ⎝

l⎞

M = R A x − W ⎜ x − W

=

x − W ⎜ x −

2

. . . (9.7)

⎟

2⎠

The governing equation for deflection is : 2

EI

d y dx

= M

2

2

or

EI

d y dx

2

⎛ ⎝

W

=

x −W ⎜x −

2

l⎞ ⎟ 2⎠

⎛ l ⎞ ⎜2 < x < l⎟ ⎝ ⎠

. . . (9.8)

Integrating the above equation, we can get 2

2

W ⎛ l⎞ EI = − ⎜ x − ⎟ + C 1 dx 2 2 2 ⎝ 2⎠ dy

W x

. . . (9.9)

C 1 is constant for integration. l

Apply B.C 3, i.e. at x = 0=

dy

,

=0

2 dx

W l

2

2 8

+ C 1

− W l2

C 1 =

16

Integrating the above equation, we can get 3

3

W ⎛ l⎞ − EIy = . x − ⎟ + C1 x + C 2 ⎜ 4 3 6 ⎝ 2⎠ W

∴

EIy =

x

W x3

4 3

−

W ⎛

l⎞

3

W l2 x

⎜ x − 2 ⎟ − 16 ⎠

6 ⎝

. . . (9.10)

+ C 2

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

0=

W

12

l3 −

W

48

l3 −

W

16

l 3 + C 2

1 1⎞ W l3 ⎛ 1 = − − ⎜ ⎟ + C 2 4 ⎝ 3 12 4 ⎠ or

C 2 = 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 0=

W

4

.

l2

4

+ C1 C 1 =

− W l2 16 9


∴ The slope and deflection will be : dy

EI

dx

=

Wx

2

2

2

W ⎛ l⎞ Wl x− ⎟ − − ⎜ 2 ⎝ 2⎠ 16

4

3

3

⎛ l ⎞ ⎜ 2 < x < l ⎟ ⎝ ⎠

2

W ⎛ l⎞ Wl − EIy = x− ⎟ − x ⎜ 12 6 ⎝ 2⎠ 16 Wx

. . . (9.11)

. . . (9.12)

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

θ A =

− W l2 16 EI

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

θ B = + Deflection at C , x =

l

2

W l2

16 EI

, from Eq. (9.12), yC = −

Wl

3

48 EI

The slope and deflection diagrams are shown in Figure 9.5. W A

B

C

(a) Beam Parabola θ B

B

A θ A

(b) Slope Diagram A

B

y C C

Cubic Parabola (c) Deflection Diagram Figure 9.5

9.3.2 SS Beams with a Point Load Anywhere on Span AB = l,

∑ y = 0

AC = a,

CB = b,

a+b=l

R A + RB = W

. . . (9.13)

Taking moments about A, W × AC = R B × AB or Wa = RB l

∴ 10

R B =

Wa l

(↑)

. . . (9.14)

Deflection of Beams

From Eqs. (9.13) and (9.14) Wb

R A =

(↑)

l

. . . (9.15)

x

w

a

l

B

A l

Figure 9.6

Consider a section X - X X at a distance x from A, M = R A x − W [ x − a]

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

M =

Wb

x − W [ x − a]

l

. . . (9.16)


EI

d y dx

2

Wb

=M =

l

x − W [ x − a]

. . . (9.17)

W [ x − a]2 + C 1 2

. . . (9.18)

Integrating the Eq. (9.17), we can get EI

dy

=

dx

EIy =

W b x2

2

l

Wb

2l

−

.

x

3

−

3

W [ x − a ]3 + C1 x + C 2 6

. . . (9.19)

The constants C 1 and C 2 can be found from the boundary conditions. The boundary conditions are : at A,

x = 0, y = 0

. . . (1)

at B,

x = l , y = 0

. . . (2)

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

0=

∴

Wb

6l

(0) + C1 (0) + C 2

C 2 = 0

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

0=

Wb

6

C1 =

2

l −

Wb

6l

W

6

3

b + C 1 l

(b 2 − l 2 )

The slope and deflection will be given by EI

dy dx

EIy =

=

Wb

2l

Wb

6l

3

2

x −

x −

W

6

W

2

[ x − a]2 +

[ x − a]3 +

Wb

. . . (9.20)

[b2 − l 2 ] x

. . . (9.21)

6l

Wb

6l

[b2 − l 2 ]

11


W

Slope at A, (at x = 0), and noting that

2

[ x − a] is not applicable

Wb ⎛ dy ⎞ 2 2 ⎜ dx ⎟ = θ A = − 6 EI l [l − b ] ⎝ ⎠ A Wb

=−

6 EI l

[(a + b)2 − a 2 ] = −

W ab

6 EI l

(a + 2b)

. . . (9.22)

Slope at B, (at x = l), Wb 2 W Wb 2 ⎛ dy ⎞ 2 2 ⎜ dx ⎟ θ B = 2l . l − 2 [l − a] + 6l [b − l ] ⎝ ⎠ B

= = = θ B =

or

⎛ ⎝

Deflection at centre ⎜ x =

Wb

6 EI l Wb

6 EI l Wb

6 EI l

[3l 2 − 3b l + b2 − l 2 ] [ 2 (a + b )2 − 3 b ( a + b) + b 2 ] (2 a 2 + ab) =

W ab

Wa b

6 EI l

( 2a + b )

(2 a + b)

6 EI l

. . . (9.23)

l⎞ ⎟, 2⎠ 3

3

Wb ⎛l ⎞ W ⎡l Wb 2 ⎤ 2 l EI yC = .⎜ ⎟ − a [ b l ] − + − ⎥⎦ 6l ⎝ 2 ⎠ 6 ⎢⎣ 2 6l 2

=

∴

W bl2

+

48

=

Wb

=

Wb

yC =

Wb

48 48

× [4b 2 − 3l 2 ] × [4b2 − 3l 2 ]

Wb

48 EI

× [3l 2 − 4b2 ]

For deflection under the load y yw again yw =

=

or

12

yw =

Wb

6 EI l W ab

6 EI l

12

(b 2 − l 2 )

× [l 2 + 4b2 − 4l 2 ]

48 EI

=−

Wb

W

6

. . . (9.24)

[ x − a]3 vanishes. Hence, by putting x = a

[ a 3 + b2 a − l 2 a ] [ a 2 + b 2 − a 2 − b 2 − 2ab]

W a2 b2

3 EI l

. . . (9.25)

For maximum deflection,

dy dx

Deflection of Beams

= 0.

From Eq. (9.20), Wb

0=

2l

Wb

2l

∴

x 2 −

. x 2 =

x =

W

[0 − a]2 +

2

Wb

Wb

6l

[b2 − l 2 ]

[l 2 − b2 ]

6l

2 2 l −b

3

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

=

W bx

6l

[ x 2 + b2 − l 2 ] 2 2 2 2 l − b ⎡l − b

Wb

6l

=−

Wb

2

9 3l

∴

ymax = −

⎢ ⎣⎢

3

Wb

(l −

2

9 3 EI l

3

⎤ + b2 − l 2 ⎥ ⎦⎥

3 2 2 b )

(l −

3 2 2 b )

. . . (9.26)

The slope and deflection diagram are shown in Figure 9.7. B

A

x

C

(a) Beam θ B

A

x

θ A

C

B

(b) Slope Diagram D

A

y max max y c c

Elastic curve

B

x

y W W

(c) Deflection Diagram 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,

R A = RB =

wl

2

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

wl

2

x − wx .

x

2 13


wl x

=

2

2

wx

−

. . . (9.27)

2 w/unit length X

x A

B

C l

x

X

Figure 9.8

The governing equation for deflection is : EI

d2y dx

=

2

wl x

2

−

w x2

. . . (9.28)

2

Integrating the Eq. (9.28), we can get dy

EI

=

dx

w l x2

4

w l x3

EIy =

12

−

−

w x3

+ C 1

. . . (9.29)

+ C1 x + C 2

. . . (9.30)

6

w x4

24


x = 0,

y=0

. . . (1)

at B,

x = l , y = 0

. . . (2)

at C ,

x =

l

2

dy

,

=0

dx

. . . (3)

Applying the BC (1) to Eq. (9.30), C 2 = 0. Applying the BC (2) to the Eq. (9.30), 0=

wl

4

12

−

wl

4

24 24

+ C1 l ∴ C 1 =

wl

3

24

. . . (9.31)

The slope and deflection equations will be, EI

dy dx

EIy =

=

w l x2

4

wl x

12

3

−

−

w x3

wx

6 4

24

−

− wl

wl3

24

. . . (9.32)

3

24

x

. . . (9.33)

Slope at A, x = 0 wl3 ⎛ dy ⎞ = θ = − A ⎜ dx ⎟ 24 EI ⎝ ⎠ A

Slope at B, x =

. . . (9.34)

l

2 wl3 wl3 wl3 wl2 ⎛ dy ⎞ EI ⎜ ⎟ = EI θ B = 4 − 6 − 24 = 24 ⎝ dx ⎠ B

∴ 14

θ B = +

w l3

24 EI

. . . (9.35)

⎛ ⎝

Because of symmetry, the maximum deflection occurs at mid-span, i.e. at C , ⎜ x =

l⎞ ⎟. 2⎠

Deflection of Beams

From Eq. (9.33), EIymax =

=

∴

ymax =

wl4

96 wl 4

384

−

wl4

−

24 × 16

[4 − 1 − 8] =

wl4

24 × 2

−5 384

wl4

− 5 wl4

. . . (9.36)

384 EI w/Unit length

B

A l /2 /2

/2 l/2 C

(a) Beam 3° Curve θ B

+ θ A

Θ

C

(b) Slope Diagram A

B

C yc = ymax

4° Curve


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 x = 0) to w/unit length at B ( x x = l). Thus, to load diagram appears 1 as a triangle. Apparently the total load = area of the triangle = w l and at any 2 wx section X - X , so that the load of X at a distance x from A the rate of loading will be l 1 wx 1 wx 2 x x = shaded triangle = . This load will act at a distance of from A 2 l 2 l 3 (centroid of triangles). The sum of the reactions ( R A + R B) will be equal to the load of wl entire triangle, i.e. . 2 B

∑ y = 0 or,

R A + RB =

1 2

×l × w=

wl

. . . (9.37)

2 X

x

w A l

B X

15

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

Taking moments about A,

×

2l

R B =

wl

wl

2

= RB × l

3

3

(↑)

. . . (9.38)

(↑)

. . . (9.39)

From Eqs. (9.37) and (9.38), wl

R A =

6

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

⎛ wl ⎞ ⎟ ⎝ x ⎠

Intensity of load = ⎜ M = R A x −

=

wl x

6

−

1 ⎛ wl ⎜ 2 ⎝ x

x ⎞ . x . ⎟ 3 ⎠

w x3

. . . (9.40)

6l

The governing equation for deflection is 2

EI

d y dx

= M =

2

wl x

6

−

3

wx

. . . (9.41)

6l


dy dx

=

wl x

−

12

w l x3

EIy =

2

36

−

wx

4

+ C 1

. . . (9.42)

+ C1 x + C 2

. . . (9.43)

24l

w x5

120l


x = 0,

y=0

. . . (1)

at B,

x = l , y = 0

. . . (2)

Applying the BC (1) to the Eq. (9.43), we can get C 2 = 0. Applying the BC (2), 0=

wl4

36

C 1 =

wl

− 4

360 l

wl4

120

+ C1 l

[3 − 10] =

− 7 w l3

. . . (9.44)

360

The slope and deflection equation will be : EI

16

dy dx

=

w l x2

12

−

w x4

24l

−

7 360

wl3

. . . (9.45)

EIy =

wl x

3

wx

−

36

5

120 l

−

7

Deflection of Beams

3

360

. . . (9.46)

wl x

Slope at A, ( x x = 0)

− 7 w l3 ⎛ dy ⎞ ⎜ dx ⎟ = θ A = 360 EI ⎝ ⎠ A

. . . (9.47)

Slope at B, ( x x = l) 3

3

wl wl 7 ⎛ dy ⎞ EI ⎜ = EI θ B = − − w l3 ⎟ 12 24 360 ⎝ dx ⎠ B wl

=

∴

θ B =

⎛ ⎝

Slope at mid-span, ⎜ x =

3

360

[30 − 15 − 7] =

wl

3

45

wl3

. . . (9.48)

45 EI

l⎞ ⎟. 2⎠

wl3 wl3 7 ⎛ dy ⎞ 3 = EI θ = − − wl C ⎟ 48 24 × 16 360 ⎝ dx ⎠C

EI ⎜

wl3

=

∴

24 × 16 × 15

θC = −

⎛ ⎝

[30 − 15 − 7 × 16] = −

67 w l 3

67 w l 3 5760 . . . (9.49)

5760 EI

l⎞ ⎟. 2⎠

Deflection at centre, ⎜ x =

EI y C =

=

wl

4

36 × 8

−

wl4

63 0 × 32

wl

4

12 × 32

−

7 360 × 2

wl

[40 − 3 − 7 × 16] = −

4

71 w l 4 11520

. . . (9.50)

For maximum deflection, dy dx

=0

From Eq. (9.42), 0=

w l x2

12

−

w x4

24 l

−

7 360

wl

⇒

0 = 30 l 2 x 2 − 15 x 4 − 7 l 4

⇒

x − 2 l x +

∴

x = 0.5193 l

4

2

2

7 15

3

4

l =0

From Eq. (9.43), EI y max =

wl

36

(0.5193 l )3 −

w

120 l

− (0.5193l )5 17


− =

∴

7

3

360 × 2 wl4

360

[14 − 0.11 − 3.64] =

wl

ymax = −

w l (0.5 (0.519 193 3l )

− wl4 153

4

. . . (9.51)

154 W

A

B

C

2 l/ 2

l/ 2 2

(a) Beam 4° Curve B +

X = 0.5193 l

θ B

A D

θ C C

Θ

θ A

(b) Slope Diagram C A

D B

y max

y c c

5° Curve


9.3.5 SS Beams with a Couple A simply supported beam AB of span l on which a couple of moment M 0 acts at C is shown in Figure 9.12. X x MO A

C a

B

l

X l

M Ob/ l M Oa/ l

Figure 9.12

Since sum of all forces in y direction is equal to zero, R A + RB = 0

. . . (9.52)

Taking moments about A, M 0 = RB × l

R B =

M 0 l

From Eqs. (9.52) and (9.53), 18

(↑)

. . . (9.53)

R A = −

M 0

Deflection of Beams

(↓)

l

Consider a section X - X X at a distance x from A and write expression for BM at X - X X . At this point you must note two important features of expression for BM at X - X X . If X X - X X lies between A and C , i.e. 0 < x ≤ a the BM is only due to R A. If X X - X X is between C and B B, i.e. a < x < b, then BM is due to R A and M M 0. So we write the second BM because the first is included but take care that M 0 is not considered if integration is in the region 0 < x ≤ a and x x is taken as ( x x – a) if integration is in the region a < x ≤ l and x x = l. Thus, in two cases we can apply BC at x = 0 and x x = l. For specific presentation in the expression for M , M 0 is placed in < > meaning that < M 0 > 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. M = R A x + < M 0 >

=−

M 0 x l

+ < M 0 >

. . . (9.54)


EI

d y dx

=M =−

2

M 0 x

+ < M 0 >

. . . (9.55)

+ < M 0 > [ x − a] + C 1

. . . (9.56)

l


dy dx

=−

EIy = −

2

M 0 x

2l

M 0 x

3

M 0

+

6l

2

[ x − a ]2 + C1 x + C 2

. . . (9.57)

The constants of integration C 1 and C 2 can be found from the boundary conditions. The boundary conditions are : at A,

x = 0, y = 0

. . . (1)

at B,

x = l , y = 0

. . . (2)

Applying the boundary condition (1) to the Eq. (9.57), with M 0 ignored, C 2 = 0. Applying the BC (2) into the Eq. (9.57),

0=− C1 =

M 0 l 2

+

6

M 0 l

+

6

M 0

2

M 0

2l

(l − a)2 + C1 l

(l − a ) 2

. . . (9.58)


dy dx

=−

EIy = −

M 0 x

2

2l

M 0 x

6l

3

+

+ < M 0 > [ x − a] + M0

2

[ x − a]2 +

M0 l

6

M0 l

6

−

M 0

x−

M 0

2l

2l

(l − a) 2

. . . (9.59)

( l − a) 2 x

. . . (9.60)

If the moment M 0 is applied at the centre, a=

l

2 19


C1 =

2

M ⎛ M l M l M l l⎞ − 0 ⎜l − ⎟ = 0 − 0 = 0 2l ⎝ 2⎠ 6 8 24

M 0 l

6

The slope and deflection equation will be : dy

EI

dx

=−

M 0 x

l ⎞ M l ⎛ + M0 ⎜ x − ⎟ + 0 2⎠ 24 ⎝

2l

M 0 x

EIy = −

2

3

6l

. . . (9.61)

2

M ⎛ M l l⎞ + 0 ⎜x − ⎟ + 0 x 2 ⎝ 2⎠ 24

. . . (9.62)


θ A = +

M 0 l

. . . (9.63)

24 EI

Slope at B, ( x x = l) EI θ B = −

θ B = ⎛ ⎝

Slope at C ⎜ x =

M 0 l

2

+

M0 l

2

+

M0 l

24

=

M0 l

24

M 0 l

. . . (9.64)

24 EI

l⎞ ⎟, 2⎠ EI θC = −

θC = −

∴ ⎛ ⎝

Deflection at C ⎜ x =

M 0 l

8

+

M0 l

24

=−

M0 l

12

M 0 l

. . . (9.65)

12 EI

l ⎞

⎟ , y = 0

2⎠

To get new slope,

+

M 0 x

2

2l

x =

l

=

M 0 l

=

l

12

24

2 3

9.3.6 SS Beams with an End Couple ∑ F y = 0 so that RA + R B = 0

. . . (9.66)

Taking moments about A, R B × l = M 0

∴

R B = +

M 0 l

(↑)

. . . (9.67) X

x A

B

MO

l

X

20

BDM

MO B

B

Deflection of Beams Figure 9.13 : SS Beam with End Couple

From Eqs. (9.66) and (9.67), R A = −

M 0

(↓)

l

Consider a section X - X X at a distance from x from A, M = R A . x = −

M 0

x

l

. . . (9.68)


EI

d y dx

2

= M =

− M 0

x

. . . (9.69)

x + C 1

. . . (9.70)

x + C1 x + C 2

. . . (9.71)

l


dy dx

M 0

=−

EIy = −

2l

M 0

2

3

6l


x = 0, y = 0

. . . (1)

at B,

x = 0, y = 0

. . . (2)

Applying the boundary condition (1) into the Eq. (9.71), C 2 = 0. Applying the BC (2) into the Eq. (9.71),

0=−

M 0 l

2

+ C1 l ∴ C 1 =

6

M 0 l

6


dy dx

=−

EIy = −

M 0

2l

M 0

2 x +

x3 +

6l

M0 l

. . . (9.72)

6

M0 l x

. . . (9.73)

6


θ A = +

M 0 l

. . . (9.74)

6 EI

Slope at B, ( x x = l) EI θ B = −

M 0 l

θ B = −

M 0 l

∴ ⎛ ⎝

Slope at C , ⎜ x =

2 3 EI

+

M0 l

6

=−

M0 l

3 . . . (9.75)

l⎞ ⎟, 2⎠

21


EI θC = −

⎛ ⎝

Deflection at C , ⎜ x =

2

M0 l

⎜ ⎟ + 8 ⎝2⎠

M0 l

=−

6

+

8

M0 l

6

M 0 l

θC = +

∴

M 0 ⎛ l ⎞

. . . (9.76)

24 EI

l⎞ ⎟, 2⎠ 3

M ⎛ l ⎞ M l ⎛l⎞ EIy = − 0 ⎜ ⎟ + 0 ⎜ ⎟ 6l ⎝ 2 ⎠ 6 ⎝2⎠

∴

=−

M 0 l

yC = +

M 0 l

2

M0 l

+

48

2

M0 l

=

12

2

16

2

. . . (9.77)

16 EI

A

B C

MO

D

(a) Beam 2° Curve θ A

+

D

θ C C

A

B

C

Θ

θ B

0.577l

(b) Slope Diagram 3° Curve y max max +

y c c

A

C

B

D


For maximum deflection, the slope is zero.

−

M 0 x 2

2l

2

x =

l

+

M0 l

6

=0

2

3

x =

l

3

= 0.577 l

The maximum deflection, EIymax = −

=

M 0

6l

M 0 l

2

18 3

∴ 22

ymax = +

.

l3

3 3

+

M0 l

6

[− l + 3 ] = +

M 0 l

.

l

3

M 0 l

2

9 3

2

9 3 EI

. . . (9.78)

Deflection of Beams

9.3.7 SS Beams with Equal End Couples ∑ F y = 0 so that RA + R B = 0

. . . (9.79)

Taking moments about A, M 0 − M 0 = R B × l ∴ RB = 0

. . . (9.80)

From Eqs. (9.79) and (9.80), R A = 0 X C A

B MO x

X

MO

Figure 9.15 : SS Beam with Equal End Couples

Consider a section X - X X at a distance x from A, M = − M 0

. . . (9.81)

The governing equation for deflection is 2

EI

d y dx

2

= M = − M 0

. . . (9.82)


dy dx

= − M 0 x + C 1

EIy = −

M 0 x 2

2

. . . (9.83)

+ C1 x + C 2

. . . (9.84)


x = 0, y = 0

. . . (1)

at B,

x = l , y = 0

. . . (2)

Applying the BC (1) into the Eq. (9.84), C 2 = 0

Applying the BC (2) into the Eq. (9.84),

0=−

M 0 l 2

2

+ C1 l ∴ C 1 =

M 0 l

2

. . . (9.85)

The slope and deflection equations will be : EI

dy dx

= − M0 x +

EIy = −

M 0 x

2

2

+

M 0 l

. . . (9.86)

2

M0 l

2

x

. . . (9.87)


θ A = +

M 0 l

. . . (9.88)

2 EI

Slope at B, ( x x = l) EI θ B = − M 0 l +

M 0 l

2

=−

M 0 l

2 23


θ B = −

M 0 l

EI θC = −

M 0 l

∴ ⎛ ⎝


. . . (9.89)

2 EI

l⎞ ⎟, 2⎠

∴

2

+

M0 l

2

θC = 0

⎛ ⎝


. . . (9.90)

l⎞

⎟,

2⎠

2

M 0 ⎛ l ⎞ M0 l ⎛ l ⎞ M0 l2 EIyC = − ⎜ ⎟ + 2 ⎜2⎟ + 8 2 ⎝2⎠ ⎝ ⎠ yC = +

∴

M 0 l

2

. . . (9.91)

8 EI C

A

B MO

MO

(a) Beam 1° Curve (Straight line) θ A

+

B

A

C

Θ

θ B

(b) Slope Diagram y c c = y max max

+

A

B


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 R A = RB = W

By symmetry,

. . . (1)

X at a distance x from A, Consider a section X - X l⎤ 2l ⎤ ⎡ ⎡ M = W . x − W ⎢ x − ⎥ − W ⎢ x − ⎥ 3⎦ 3⎦ ⎣ ⎣ W

C

X

W

A

B /3 l /3

/3 l /3

l /3 /3

X

24

Elastic curve

. . . (2)

Deflection of Beams

Figure 9.17

The equation for deflection is : EI =

d2y

l⎤ ⎡ M W x W x = = − − −W ⎢ 3 ⎥⎦ dx 2 ⎣

2l ⎤ ⎡ ⎢⎣ x − 3 ⎥⎦

. . . (3)

Integrating the Eq. (3), EI

dy

=

dx

Wx

3

−

6

2

W ⎡ l⎤ W ⎡ 2l ⎤ x x − − − − + C 1 2 ⎢⎣ 3 ⎥⎦ 2 ⎢⎣ 3 ⎥⎦

2

Wx

EIy =

2

2

W ⎡

l⎤

3

W ⎡

2l ⎤

. . . (4)

3

x− ⎥ − x − ⎥ + C1 x + C 2 6 ⎢⎣ 3⎦ 6 ⎢⎣ 3⎦

. . . (5)

The boundary conditions : at A,

x = 0, y = 0 ∴ C 2 = 0

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

2l 3

.

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

x = l , y = 0

0=

Wl

3

6

3

3

W ⎛ 2l ⎞ W ⎛l⎞ − − ⎜ ⎟ ⎜ ⎟ + C1 l 6 ⎝ 3⎠ 6 ⎝3⎠

3 8 1 ⎤ W l2 Wl ⎡ 1− C 1 = − − = 6 ⎢⎣ 27 27 ⎥⎦ 9

∴

EI

dy

=

dx

Wx

2

2

2

. . . (6) 2

W ⎡ l⎤ W ⎡ 2l ⎤ Wl − x− ⎥ − x− ⎥ − ⎢ ⎢ 2 ⎣ 3⎦ 2 ⎣ 3⎦ 9

2

Apparently since the problem is symmetric the maximum deflection occurs in the centre. y1C + y2 C = y3C

θ1 A + θ 2 A = θ3 A = θ3 B

⎛ ⎝

Deflection under the load, ⎜ x = 3

EIy D

l⎞

⎟,

3⎠

2

W ⎛l⎞ Wl l = − × ⎜ ⎟ 6 ⎝ 3⎠ 9 3 3

3

Wl ⎛1 ⎞ − 5W l = − 1⎟ = ⎜ 27 ⎝ 6 27 × 6 ⎠ y D =

− 5 W l3 162 EI

. . . (7)

At A, ( x x = 0),

θ A = −

W l2

9 EI

. . . (8)

At B ( x x = l), 25


Wl

EI θ B =

−

2 W l2

=

∴

2

18

θ B = +

W

2

4l 2

.

9

−

W

2

.

l

[9 − 4 − 1 − 2] = +

2

9

−

Wl

2

9

W l2

9

W l2

. . . (9)

9 EI

For maximum deflection, slope is zero.

0=

Wx

2

2

2

W ⎡ l⎤ Wl − − − x 2 ⎢⎣ 3 ⎥⎦ 9

2

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 2l x < between the loads, the third term vanishes. 3 2

l⎞ ⎛ 0 = 9 x − 9 ⎜ x − ⎟ − 2l 2 3⎠ ⎝ 2

⇒

⎛ 2 l 2 2l x ⎞ 2 = 9 x − 9 ⎜ x + − ⎟⎟ − 2l ⎜ 9 3 ⎠ ⎝ 2

= − l 2 + 6 l x − 2l 2 6l x = 3l 2 x =

l

. . . (10)

2

EIy max

Now put x =

3

W ⎛ l⎞ Wl Wl x − x− ⎟ − = ⎜ 6 6 ⎝ 3⎠ 9 W

3

2

x

l

2 3

EIy max

3

3

W ⎛l⎞ W ⎛l l⎞ Wl = − − ⎟ − ⎜ ⎟ ⎜ 6 ⎝2⎠ 6 ⎝ 2 3⎠ 18 3

2

3

W ⎛l 2l ⎞ Wl l Wl ⎛ 1 1 1 ⎞ . =− =− − ⎟ − + − ⎟ ⎜ ⎜ 6 ⎝2 3⎠ 9 2 6 ⎝ 216 3 8 ⎠

1 1⎤ W l3 ⎡1 Wl 3 ⎛ 72 + 1 − 27 ⎞ = − − =− ⎜ ⎟ 6 ⎢⎣ 8 36 36 3 ⎥⎦ 6 ⎝ 216 ⎠

=

∴

y max =

W l3

36 × 8 × 6 23 Wl 3 648 EI

⎛ 23 ⎞ ⎟ ⎝ 648 ⎠

[36 − 8 − 96] = − Wl 3 ⎜

. . . (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

26

∑ F y = 0, so that RA + RB = 24 × 2 = 48 kN

. . . (1)

Deflection of Beams

24 kN/m C A

B 2m

D 6m

Figure 9.18

Taking moments about A, 24 × 2 × 1 = R B × 6

R B = 8 kN kN (↑)

. . . (2)

R A = 48 − 8 = 40 kN (↑ )

. . . (3)

Apply the Udl over the portion DB downwards and upwards, 24 kN/m A

B D

2m

24 kN/m

4m

Figure 9.19

M = 40 x − 24 x ×

x

2

+ 24 ( x − 2)

( x − 2) 2

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

= 40 x − 12 x 2 + 12 ( x − 2)2 2

d y

EI

dx

EI

= 40 x − 12 x 2 + 12 ( x − 2)2

2

dy dx

. . . (4)

=

40 x 2 2

−

12 x 3 3

+

. . . (5)

12 ( x − 2) 3 3

+ C 1

= 20 x 2 − 4 x 3 + 4 ( x − 2)3 + C 1 EIy =

20 x 2 3

− x 4 + ( x − 2) 4 + C1 x + C 2

. . . (6) . . . (7)

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

x = 0, y = 0 ∴ C 2 = 0

at B,

x = 6 m, 0=

y=0

20 × 63 3

(6 − 2)4 + C 1 × 6 − 64 + (6

C 1 = − 20 × 12 + 36 × 6 −

∴

EI

dy dx

16 × 16 6

=−

= 20 x2 − 4 x3 + 4 ( x − 2)3 −

200 3

200 3

. . . (8)

The third term vanishes. x = 0), Slope at A, ( x

27


θ A =

− 200

1 − 200 × 103 rad = − 3.33 × 10− 3 rad = − 6 300 3 × 20 ×10

=

3 EI

Slope at B, ( x x = 6 m), 2

3

3

EI θ B = 200 × 6 − 4 × 6 + 4 (6 − 2) −

θ B =

136 3 EI

=

136 × 103 6

3 × 20 ×10

200 3

= + 2.27 × 10− 3 radian

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

3

3

EI θC = 20 × 3 − 4 × 3 + 4 (3 − 2) −

θC = EIy =

20 3 EI

20 x 3

3

= 0.47 × 10− 3 radians

− x 4 + ( x − 2) 4 −

3

200

200 3

. . . (9)

x

Deflection at centre, ( x = 3 m), EIyC =

20

× 33 − 34 + (3 − 2)4 −

3

200 3

×3

− 100 × 103 × 103 yC = = = − 5 mm EI 20 × 106 − 100

For maximum deflection, dy dx

=0

0 = 20 2 0 x 2 − 4 x3 + 4 ( x − 2)3 −

200 3

24 x2 + 48 x − = 20 x 2 − 4 x3 + 4 x3 − 32 − 24

= − 4 x 2 + 48 x − ∴

2

x − 12 x +

74 3

200 3

296 3

=0

2.63 m , x > 2m x = 2.63 EIymax =

∴

20 3

× 2.633 − 2.634 + (2.63 − 2)4 −

200 3

× 2.63 = − 101.7

m m ; − 5. 1 m m ymax = − 5.087 mm

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 :

28

(a)

Slope at the ends.

(b)

Slope at the centre.

(c)

Deflection under the load.

(d)

Deflection at the centre.

(e)

Deflection of Beams

Maximum deflection.

Take EI = 20 × 106 N-m2. Solution

∑ F y = 0, so that RA + RB = 20 + 10 × 2 = 40 kN 20 kN

. . . (1)

10 kN/m

A

B

1m

D

1m

2m

C

x

Figure 9.20

Taking moments about A, R B × 4 = 20 × 1 + 10 × 2 × 3 = 80 R B = 20 kN

. . . (2)

R A = 20 kN

M = 20 x − 20 [ x − 1] − 10 [ x − 2]

[ x − 2] 2

= 20 x − 20 [ x − 1] − 5 [ x − 2]2 2

d y

EI

dx

. . . (3)

= M

2

= 20 x − 20 [ x − 1] − 5 [ x − 2]2 EI

dy dx

= 10 x2 − 10 [ x − 1]2 −

EIy =

10 x 3 3

−

10 3

[ x − 1]3 −

at A,

x = 0, y = 0, C 2 = 0

at B,

x = 4 m, y = 0

0=

10 × 43 3

−

10 3

5 3

[ x − 2]3 + C 1

5 12

(4 − 1) 3 −

. . . (4)

[ x − 2]4 + C1 x + C 2

5 12

. . . (5)

. . . (6)

(4 − 2) 4 + C 1 × 4

C 1 = − 29.17

EI

(a)

dy dx

5 3

[ x − 2]3 − 29.17

Slope at A, ( x x = 0),

θ A = (b)

= 10 x2 − 10 [ x − 1]2 −

− 29.17 EI

=

− 29.17 × 103 = − 1.46 × 10− 3 radians 6 20 × 10

Slope at B, ( x x = 4 m), EI θ B = 10 × 42 − 10 (4 − 1)2 −

5 3

(4 − 2)3 − 29.17 = + 27.5

θ B = + 1.38 × 10 − 3 radians (c)

Slope at centre, ( x x = 2 m), 29


2 2 EI θC = 10 × 2 − 10 ( 2 − 1) − 29.17

θC = + 0.04 × 10− 3 radians Deflection under the load : EIy =

10 x 3

−

3

10

[ x − 1]3 −

3

5 12

[ x − 2]4 − 29.17 x

At x = 1 m, EIy D =

10

− 29.17

3

− 25.84 × 103 × 103 = − 1.29 EIy D = 1.29 mm 20 × 106 (d)

Deflection at the centre : x = 2 m EIyC =

10 × 23 3

−

10 3

(2 − 1)3 − 29.17 × 2

1.75 mm yC = − 1.75

(e)

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

=0

10 x 2 − 10 ( x − 1)2 − 29.17 = 0 10 x 2 − 10 x 2 − 10 + 20 x − 29.17 = 0 x = 1.96 m < 2 m EIymax =

∴

10 3

(1.96)3 −

10 3

(1.96)3 − 29.17 × 1.96 = − 35

1.7501 1 mm ymax = − 1.750

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

R A = RB =

By symmetry,

10 × 4 2

= 20 kN

10 kN/m x A

B

C 4m

2m

2m

x

Figure 9.21

M = 20 x −

10 ( x − 2) 2 2

−

10 ( x − 6) 2 2

= 20 x − 5 ( x − 2)2 − 5 ( x − 6)2

30

. . . (1)

EI

d2y dx

2

Deflection of Beams

= M = 20 x − 5 ( x − 2)2 − 5 ( x − 6)2

EI

dy dx

=

EIy =

20 x 2 2 10 x 3 3

− −

5 ( x − 2) 3 3 5 12

3

( x − 2) 4 −

at A,

x = 0, y = 0, C 2 = 0

at B,

x = 8 m, 0=

5

−

. . . (2)

( x − 6)3 + C 1

5 12

( x − 6) 4 + C1 x + C 2

. . . (3) . . . (4)

y=0

10 × 83 3

−

5 12

(8 − 2)4 −

5 12

(8 − 6)4 + C 1 × 8

C 1 = − 145

EI

dy dx

= 10 x2 −

EIy =

10 x 3 3

−

5 3

( x − 2)3 −

5 12

5 3

( x − 2) 4 −

( x − 6)3 − 145 5

12

( x − 6) 4 − 145 x

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

∴

EIymax =

ymax =

10 × 43 3

−

5 12

( 4 − 2) 4 −

5 12

( 4 − 6)4 − 145 × 4

− 1120 3 EI

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.

31


9.4.1 Cantilever Beams with Single Concentrated Load Load at Free End Consider a cantilever beam loaded with a point load ‘ W ’ at free end. Consider a section X - X X at a distance x from left end. M = − W . x

(Hogging BM)

. . . (9.92)

The governing equation for deflection is EI =

2

d y

= M = − W x

dx 2

. . . (9.93)

W X B

A

l

x

X

Figure 9.22


dy

W x2

=−

dx

EIy = −

+ C 1

2

W x3

. . . (9.94)

+ C1 x + C 2

6

. . . (9.95)

The constants C 1 and C 2 can be found from the boundary conditions. The boundary conditions are : dy

At B,

x = l ,

=0

At B,

x = l , y = 0

dx

(Fixed end)

. . . (1) . . . (2)

Applying the BC (1) into the Eq. (9.94), we can get 0=−

∴

Wl

2

2

C 1 = +

Wl

+ C 1 2

. . . (9.96)

2

Applying BC (2) into the Eq. (9.95), we can get 0=−

W l3

C 2 = −

W l2

+

6

2

× l + C 2

W l3

. . . (9.97)

3


dy dx

=−

EIy = −

W x2

2

W x3

6

+

+

W l2

. . . (9.98)

2

W l2 x

2

−

W l3

3

. . . (9.99)

The maximum slope occurs at x = 0. Slope at free end, 32

θ A = +

W l2

2 EI

. . . (9.100)

Deflection of Beams

The maximum deflection occurs at x = 0. ymax = −

⎛ ⎝

Slope at centre ⎜ x =

W l3

. . . (9.101)

3 EI

l⎞

⎟,

2⎠

2

W ⎛l⎞ Wl EI θC = − + ⎜ ⎟ 2 ⎝2⎠ 2 3W l 2

θ A = +

⎛ ⎝

Deflection at centre ⎜ x =

. . . (9.102)

8 EI

l⎞ ⎟, 2⎠

EIyC = −

=−

∴

2

yC = −

W ⎛l⎞

3

W l2

W l3

l

⎜ ⎟ + 2 ×2− 3

6 ⎝2⎠ Wl

3

48

[− 1 + 12 − 16] =

− 5 W l3 48

5 W l3

. . . (9.103)

48 EI

W

C A

/2 l /2

B

/2 l /2

(a) Cantilever 2° Curve θ A

+

θ C C

A

B

C

(b) Slope Diagram C y A

B

Θ

y c c

A


9.4.2 Cantilever Beams with a Central Point Load Consider a cantilever beam loaded with a point load ‘ W ’ at the centre ‘C ’. ’. Consider a section X - X X at a distance x from left end.

⎡ ⎣

Moment, M = − W ⎢ x −

l⎤ 2 ⎥⎦

(Hogging BM)

. . . (9.104)


d2y

l⎤ ⎡ = = − − M W x ⎢⎣ 2 ⎥⎦ dx 2

. . . (9.105) 33


W A

X

C

l /2 /2 x

l /2 /2

B

X

Figure 9.24


dy

=−

dx

W ⎡

l⎤

2

x − ⎥ + C 1 2 ⎢⎣ 2⎦

. . . (9.106)

3

W ⎡ l⎤ EIy = − x − + C1 x + C 2 6 ⎢⎣ 2 ⎥⎦

. . . (9.107)

The constants C 1 and C 2 can be found from the boundary conditions. The boundary conditions are : dy

At B,

x = l ,

At B,

x = l , y = 0

dx

=0

. . . (1) . . . (2)

Applying the BC (1) into the Eq. (9.106), we can get 2

2

l⎤ ⎡ 0=− l − ⎥ + C 1 ⎢ 2 ⎣ 2⎦ Wl

∴

C 1 = +

W l2

. . . (9.108)

8

Applying BC (2) into the Eq. (9.107), we can get 3

2

W ⎛ l⎞ Wl × l + C 2 0=− l− ⎟ + ⎜ 6 ⎝ 2⎠ 8

∴

C 2 =

− 5 W l3

. . . (9.109)

48

The slope and deflection equations will be : 2

W ⎡ l⎤ Wl EI x− ⎥ + =− ⎢ dx 2 ⎣ 2⎦ 8 dy

EIy = −

W ⎡

l⎤

3

x− ⎥ + 6 ⎢⎣ 2⎦

W l2

8

2

−

. . . (9.110)

5 W l 3 48

. . . (9.111)

Slope at ( x = 0),

θ A = +

⎛ ⎝

Slope at C ⎜ x =

2

8 EI

. . . (9.112)

l⎞ ⎟, 2⎠

θC = + Deflection at A ( x x = 0), 34

Wl

W l2

8 EI

. . . (9.113)

⎛ ⎝


. . . (9.114)

48 EI

l⎞ ⎟, 2⎠

EIyC = +

∴

Deflection of Beams

5 W l3

y A = −

yC = −

Wl

2

.

8

l

−

2

5 W l3 48

=−

Wl

3

24

W l2

. . . (9.115)

24 EI W

A

l /2 /2

B

l/2

C

(a) Cantilever Beam with a Central Point Load 2° Curve +

θ A

θ C C

A

B

C

(b) Slope Diagram C

A y A

B

y c c St. line

2° Curve


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 X at a distance x from left end. x

Moment, M = − w x

2

=−

w x2

2

(Hogging BM)

. . . (9.116)


d y

EI

dx

2

= M = −

wx

2

. . . (9.117)

2

w/unit length X A

C

/2 l /2 x

B /2 l/2 X

Figure 9.26


dy dx

=−

EIy = −

wx

6

w x4

24

3

+ C 1

+ C1 x + C 2

. . . (9.118) . . . (9.119)

The constants C 1 and C 2 can be found from boundary conditions. 35


The boundary conditions are : dy

At B,

x = l ,

=0

At B,

x = l , y = 0

dx

. . . (1) . . . (2)

Applying the BC (1) into the Eq. (9.118), we can get

∴

3

wl

0=−

+ C 1

6

wl 3

C 1 = +

. . . (9.120)

6

Applying BC (2) into the Eq. (9.119), we can get wl4

0=−

∴

wl4

+

24

+ C 2

6

wl4

C 2 = −

. . . (9.121)

8

The slope and deflection equations will be : dy

EI

=−

dx

θ A = +

⎛ ⎝

Slope at C ⎜ x =

3

6

w

EIy = −

Slope at ( x = 0),

wx

+

3

. . . (9.122)

6

wl3 x

+

24

wl

6

−

wl4

. . . (9.123)

8

w l3

. . . (9.124)

6 EI

l⎞

⎟,

2⎠

EI θC = −

∴

wl

3

+

48

wl

6

3

=+

7 w l3 48

7 w l3

θC =

. . . (9.125)

48 EI

Deflection at A ( x x = 0), y A = −

⎛ ⎝


wl

4

. . . (9.126)

8 EI

l⎞

⎟,

2⎠

4

3

w ⎛l⎞ wl ⎛ l ⎞ wl EIyC = − + ⎜ ⎟ ⎜ ⎟− 8 24 ⎝ 2 ⎠ 6 ⎝ 2⎠

=

∴

yC =

wl4

24 × 16

( − 1 + 32 − 48) =

4

− 17 w l 4 384

−17 wl 4

. . . (9.127)

384 EI w/unit length

A

B

/2 l /2

C

/2 l /2

(a) Cantilever Beam with Udl

36

3° Curve θ A B

B

+

θ C B

B

Deflection of Beams

(b) Slope Diagram C

A

B

y C C

y A

4° Curve


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

Moment,

l⎞ ⎛ x − ⎟ ⎜ l⎞⎝ 2⎠ ⎛ M = − w ⎜ x − ⎟ 2⎠ 2 ⎝

w⎛ l⎞ = − ⎜ x − ⎟ 2 ⎝ 2⎠

2

(Hogging BM) w/unit length

X A

B

C

l /2 /2

x

. . . (9.128)

/2 l/2 X

Figure 9.28


EI

d y dx 2

w⎡ l⎤ = M = − ⎢x − ⎥ 2 ⎣ 2⎦

2

. . . (9.129)


EIy

dy dx dy dx

w⎡

=−

l⎤

3

x − ⎥ + C 1 6 ⎢⎣ 2⎦

=−

w ⎡

l⎤

. . . (9.130)

4

x − ⎥ + C1 x + C 2 24 ⎢⎣ 2⎦

. . . (9.131)

The constants C 1 and C 2 can be found from boundary conditions. The boundary conditions are : dy

At B,

x = l ,

At B,

x = l , y = 0

dx

=0

. . . (1) . . . (2)

Applying the boundary condition (1) into the Eq. (9.130), we can get 37


w⎡

0=−

∴

3

l − ⎥ + C 1 6 ⎢⎣ 2⎦

wl

C 1 =

l⎤

3

. . . (9.132)

48

Applying boundary condition (2) into the Eq. (9.131), we can get 4

3

w ⎛ l⎞ wl 0=− l− ⎟ + × l + C 2 ⎜ 24 ⎝ 2⎠ 48 wl

=

∴

C 2 =

4

24 × 16

[ − 1 + 8] + C 2

− 7 wl 4

. . . (9.133)

384

The slope and deflection equations will be : 3

3

w⎡ l⎤ wl EI = − ⎢x − ⎥ + dx 6 ⎣ 2⎦ 48 dy

4

. . . (9.134)

3

4

w ⎡ l⎤ wl 7 wl EIy = − x x − + − 24 ⎢⎣ 2 ⎥⎦ 48 384

. . . (9.135)

Slope at ( x = 0), wl3

θ A = +

⎛ ⎝

Slope at C ⎜ x =

. . . (9.136)

48 EI

l⎞

⎟,

2⎠

wl

θC = +

3

. . . (9.137)

48 EI

Deflection at A ( x x = 0), y A =

⎛ ⎝


∴

. . . (9.138)

384 EI

l⎞ ⎟, 2⎠

EIyC = +

=

− 7 wl4

wl

48

wl4

384

yC = −

3

3

×

l

2

−

7 wl4

( 4 − 7) = wl

384

− 3 wl4 384

4

. . . (9.139)

384 EI w/unit length

A /2 l /2

C

B /2 l /2

(a) Cantilever Beam with Udl on the Right Half

38

Deflection of Beams

3° Curve θ A

+

θ C C

A

B

C

(b) Slope Diagram C

A

B

y c c y A

4° Curve

St. line


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). w/unit length D A

B

C

/2 l /2

l /2 /2

(a) w/unit length C A

B l /2 /2

/2 l /2 x

(b) Figure 9.30

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

M = − w . x . wx

=−

2

x

2

w⎛ l⎞⎛ l⎞ x − ⎟ ⎜x − ⎟ ⎜ 2 ⎝ 2⎠ ⎝ 2⎠

+

w⎡ l⎤ + ⎢ x − ⎥ 2 ⎣ 2⎦

2

2

(Hogging BM)

. . . (9.134)


EI

d y dx

2

=M =−

wx

2

2

w⎡ l⎤ + ⎢x − ⎥ 2 ⎣ 2⎦

2

. . . (9.135)


dy dx

=−

wx

6 2

3

3

w⎡ l⎤ + ⎢ x − ⎥ + C 1 6 ⎣ 2⎦ 4

w ⎡ l⎤ EIy = − x − ⎥ + C1 x + C 2 + ⎢ 24 24 ⎣ 2⎦ wx

. . . (9.136)

. . . (9.137)

The constants C 1 and C 2 can be found from boundary conditions. The boundary conditions are : At B,

x = l ,

dy dx

=0

. . . (1) 39


x = l ,

At B,

y=0

. . . (2)

Applying the boundary condition (1) into the Eq. (9.136), we can get wl

0=−

3

+

6

3

w⎛

l⎞

⎜ l − 2 ⎟ + C 1 ⎠

6 ⎝

wl3 ⎡ 1 ⎤ 7 w l3 1− ⎥ = C 1 = 6 ⎢⎣ 8⎦ 48

∴

. . . (9.138)

Applying the boundary condition (2) and value of C 1 into the Eq. (9.137), we can get 4

4

w ⎛ l⎞ 7 wl 0=− l + − + ⎜ ⎟ 24 24 ⎝ 2⎠ 48 wl

4

4

+ C 2

4

1 7⎤ wl ⎡ 1 (16 − 1 − 56) − − = C 2 = 24 ⎢⎣ 16 2 ⎥⎦ 24 × 16 wl

∴

=−

39 w l 4 24 × 16

− 39 w l 4

C 2 =

. . . (9.139)

384


dy

=−

dx

wx

w ⎡ l⎤ 7 wl x− ⎥ + + ⎢ 6 ⎣ 2⎦ 48

6

w x4

EIy = −

3

3

+

24

w ⎡

l⎤

4

x− ⎥ + 24 ⎢⎣ 2⎦

7 wl3 48

3

. . . (9.140)

x−

39 w l 4 384

. . . (9.141)

Slope at A, ( x x = 0), 7 w l3

θ A = +

⎛ ⎝

Slope at C ⎜ x =

. . . (9.142)

48 EI

l⎞

⎟,

2⎠

3

EI θC

7 wl3 6 wl3 w⎛l⎞ =− ⎜ ⎟ + =+ 6 ⎝2⎠ 48 EI 48 E I

⎛ ⎝

Deflection at D ⎜ x =

3l ⎞

⎟,

4⎠

3

EI θ D = −

=

=

40

. . . (9.143)

w ⎛ 3l ⎞

⎜

w ⎡ 3l

l⎤

3

7 w l3

⎟ + 6 ⎢ 4 − 2 ⎥ + 48 ⎣ ⎦

6 ⎝ 4⎠

− 27 w l 3 wl3 7 w l3 + + 24 × 16 24 × 16 48 wl

3

24 × 16

[− 27 + 1 + 56] =

30 w l 3 384

=+

5 w l3 64

∴

θ D = +

Deflection of Beams

5 w l3

. . . (9.144)

64 EI

Deflection at A, ( x x = 0) yC = −

39 w l 4

⎛ ⎝


l⎞ ⎟ 2⎠

EI y C = −

=

∴

yC =

⎛ ⎝

. . . (9.145)

384 EI

Deflection at D, ⎜ x =

w ⎛l⎞

4

7 wl3

wl4

[ − 1 + 28 − 39] =

384

− 12 w l 4 384

− 12 w l 4

. . . (9.146)

384 EI

3l ⎞

⎟

4⎠ 4

EI y D

39 w l 4

l

⎜ ⎟ + 48 × 2 − 384 24 ⎝ 2 ⎠

4

3

w ⎛ 3l ⎞ w ⎡ 3l l⎤ 7 w l ⎛ 3 l ⎞ 39 w l =− + − + ⎜ ⎟ ⎜ ⎟ − 384 24 ⎝ 4 ⎠ 24 ⎢⎣ 4 2 ⎥⎦ 48 ⎝ 4 ⎠ wl4

=

24 × 256

4

[ − 18 + 1 + 21 × 32 − 39 × 16]

2 − 32 w l 4 4 =− wl 24 × 256 384

=

wl4 wl4

y D =

. . . (9.147)

192 EI w/unit length D

A

2 l/ 2

C

B

/4 l /4

/4 l /4

(a) Cantilever Beams with Udl on Left Portion 3° Curve

θ A

θ C C

A

θ D

D

C

B

(b) Slope Diagram

y A

Ө

yC


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

w/unit length

41 A

B x

l


Figure 9.32

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

Intensity of loading =

. x

l

⎛1 w ⎞ ⎛ x ⎞ − w x3 M = − ⎜ × x × x⎟ ⎜ ⎟ = 6l ⎝2 l ⎠ ⎝3⎠

Moment,

. . . (9.148)


d2y

= M = −

dx 2

w x3

6l

. . . (9.149)


dy dx

wx

=−

+ C 1

24 l

w x5

EIy = −

4

. . . (9.150)

+ C1 x + C 2

120 l

. . . (9.151)


At B,

x = l ,

=0

At B,

x = l , y = 0

dx

. . . (1) . . . (2)

Applying the boundary condition (1) into the Eq. (9.150), we can get 0=

∴

− w l3

+ C 1

24

C 1 = +

wl 3

. . . (9.152)

24

Applying the boundary condition (2) into the Eq. (9.151), we can get 0=−

∴

C 2 = +

wl4

120

+

wl4

24

+ C 2

wl3

. . . (9.153)

24


dy dx

=−

EIy = −

w x4

24 l

wx

5

120 l

+

+

wl3

wl

. . . (9.154)

24 3

24

x−

wl

3

30

. . . (9.155)


θ A = +

42

w l3

24 EI

. . . (9.156)

⎛ ⎝

Slope at C ⎜ x =

Deflection of Beams

l⎞ ⎟, 2⎠ 4

w ⎛l⎞ wl3 EI θC = − + ⎜ ⎟ 24 l ⎝ 2 ⎠ 24

=

∴

w l3 ⎡ − 1

15 w l 3 ⎤ + = + 1 ⎥⎦ 24 ⎢⎣ 16 24 × 16 16 15 w l 3

θC = +

. . . (9.157)

384 EI

Deflection at A, ( x x = 0), y A = −

⎛ ⎝

Deflection at C , ⎜ x = EI yC =

=

∴

yC =

wl4

. . . (9.158)

30 EI

l⎞

⎟,

2⎠

− wl4 120

− wl4 480

+

wl

4

48

−

wl

4

30

− wl4

[ 4 + 10 − 16] =

240

− wl4

. . . (9.159)

240 EI w/unit length A

B C

/2 l /2

l/ 2 2

(a) Cantilever Beams with a Triangular Load 4° Curve

+

θ A

θ C C

A

B C

(b) Slope Diagram C

A yC y A

Ө


9.4.6 Cantilever Beams with an End Couple Consider a cantilever beam loaded with an end couple M 0 as shown in Figure 9.34. MO

X B

A l

x

C X

Figure 9.34

Consider a section X - X X at a distance x from A as shown in Figure 9.34. 43


0 M = − M 0 . x

Moment,

. . . (9.160)


d2y dx

= M = − M 0 x0

2

. . . (9.161)


dy

= − M 0 x0 + C 1

dx

x

EIy = − M 0

2

. . . (9.162)

+ C1 x + C 2

2

. . . (9.163)


At B,

x = l ,

At B,

x = l , y = 0

dx

=0

. . . (1) . . . (2)

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

∴

C1 = M 0 l

. . . (9.164)

Applying the boundary condition (2) into the Eq. (9.163), we can get

0=−

M 0 l 2

C 2 = −

∴

+ M 0 l . l + C 2

2 M 0 l

2

. . . (9.165)

2


dy dx

= − M 0 x + M 0 l

EIy = −

M 0 x

2

+ M0 l x −

2

. . . (9.166) M 0 l

2

. . . (9.167)

2


θ A = ⎛ ⎝

Slope at C ⎜ x =

M 0 l

. . . (9.168)

EI

l⎞ ⎟, 2⎠

EI θC = − M 0

∴

θC = +

l

2

+ M 0 l

M 0 l

. . . (9.169)

2 EI

Deflection at A, ( x x = 0), y A = −

⎛ ⎝


M 0 l

. . . (9.170)

EI

l⎞

⎟,

2⎠

EI y C = −

44

2

M 0 ⎛ l ⎞

2

l

⎜ ⎟ + M0 l . 2 −

2 ⎝2⎠

M0 l

2

2

yC = −

M 0 l

Deflection of Beams

2

. . . (9.171)

8 EI MO

C B

A

2 l/ 2

/2 l /2

(a) Cantilevers with an End Couple St. line

θ A θC B

C

A

(b) Slope Diagram C

A Ө

y A

B

y C C


9.4.7 Examples Example 9.5

A cantilever beam of length l carries a point load W at a distance

l

from free end. 4 Determine the slopes at free end and under load and the maximum deflection and deflection under the load. Solution l⎤ ⎡ M = − w ⎢ x − ⎥ 4⎦ ⎣

. . . (1)

2

EI

l⎤ ⎡ = − − w x ⎢ 2 4 ⎥⎦ dx ⎣

d y

. . . (2)

2

w⎡ l⎤ EI = − ⎢ x − ⎥ + C 1 dx 2 ⎣ 4⎦ dy

EIy = −

w⎡

l⎤

. . . (3)

3

x − ⎥ + C1 x + C 2 6 ⎢⎣ 4⎦

. . . (4)

l/ 4

B

A l

Figure 9.36

Boundary conditions are : dy

At B,

x = l ,

At B,

x = l , y = 0

dx

=0

. . . (5) . . . (6)

From Eqs. (3) and (5),

0=−

w⎡

l⎤

2

l − ⎥ + C 1 2 ⎢⎣ 4⎦

45


C 1 = +

9 wl2

. . . (7)

32

From Eqs. (4), (6) and (7) 3

3

w⎡ l⎤ 9 wl − + + C 2 0=− l 6 ⎢⎣ 4 ⎥⎦ 32 w l3

C 2 =

[− 108 + 27]

6 × 64

81 w l 3

=−

6 × 64

− 27 w l 3

=

. . . (8)

128

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

∴

2

w ⎡ l⎤ 9 wl = − ⎢x − ⎥ + EI dx 2 ⎣ 4⎦ 32 dy

3

w⎡ l⎤ 9 wl EIy = − x − + 6 ⎢⎣ 4 ⎥⎦ 32

2

2

. . . (9)

x−

27 w l 3

. . . (10)

128


θ A = +

⎛ ⎝


9 wl2

. . . (11)

32 EI

l⎞

⎟,

4⎠

θC = +

9 wl2

. . . (12)

32 EI

x = 0), Deflection at A, ( x y A = −

27 w l 3

. . . (13)

128 EI

⎛ ⎝

Deflection under load at C , ⎜ x = EI y C =

∴

yC =

9 wl2 32

.

l

4

−

l⎞ ⎟, 4⎠

27 w l 3 128

=

− 18 w l 3 128

− 18 w l 3 128 EI

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 M = − 30 x − 60 [ x − 2] − 24 ( x − 3)

= − 30 x − 60 [ x − 2] − 12 [ x − 3]2 EI

46

d2y dx

2

= M

( x − 3) 2 . . . (1)

= − 30 x − 60 [ x − 2] − 12 [x − 3]2 EI

dy dx

. . . (2)

= − 15 x 2 − 30 [ x − 2]2 − 4 [ x − 3]3 + C 1

3 3 4 EIy = − 5 x − 10 [ x − 2] − [ x − 3] + C1 x + C 2

60 kN

30 kN A

2m

C

Deflection of Beams

. . . (3) . . . (4)

24 kN/m 2m

1m D

B

Figure 9.37

Boundary conditions are : dy

At B,

x = 5 m,

=0

At B,

x = 5 m, y = 0

dx

. . . (5) . . . (6)

From Eqs. (3) and (5) 0 = − 15 × 52 − 30 (5 − 2)2 − 4 (5 − 3)3 + C 1

∴

C 1 = 677

. . . (7)

From Eqs. (4), (6) and (7) 0 = − 5 × 53 − 10 (5 − 2)3 − (5 − 3)4 + 677 × 5 + C 2

∴

C 2 = − 2474

EI

dy dx

. . . (8)

= − 15 x2 − 30 [ x − 2]2 − 4 [ x − 3]3 + 677

EIy = − 15 x 3 − 10 [ x − 2]3 − [ x − 3]4 + 677 x − 2474

. . . (9) . . . (10)

At free end, i.e. at A, x = 0, From Eq. (9)

θ A = +

677 EI

=

677 × 103 6

200 × 10

= 3385 × 10− 3 radians

From Eq. (10) y A =

− 2474 EI

− 2474 × 103 × 103 12.37 7 mm = = − 12.3 200 × 106

= − 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. 24 kN/m A

1m

C

1m

30 kN D

2m

B

Figure 9.38

47

Stresses in Shafts and Shells Shells 30 kN

24 kN/m

X

A

B 1m

C

D

x

24 kN/m X

Figure 9.39

M = − 24 x .

x

2

+ 24 ( x − 1)

( x − 1) 2

− 30 [ x − 2]

= − 12 x 2 + 12 [ x − 1]2 − 30 [ x − 2] 2

EI

d y dx

= M = − 12 x2 + 12 [ x − 1]2 − 30 [ x − 2]

2

dy

. . . (1) . . . (2)

= − 4 x3 + 4 [ x − 1]3 − 15 [ x − 2]2 + C 1

. . . (3)

EIy = − x 4 + [ x − 1]4 − 5 [ x − 2]3 + C1 x + C 2

. . . (4)

EI

dx

The boundary conditions are : dy

At B,

x = 4 m,

At B,

x = 4 m, y = 0

dx

=0

. . . (5) . . . (6)

From Eqs. (3) and (5) 0 = − 4 × 43 + 4 ( 4 − 1)3 − 15 ( 4 − 2)3 + C 1

∴

C 1 = 208

. . . (7)

From Eqs. (4), (6) and (7) 0 = − 44 + ( 4 − 1)4 − 5 ( 4 − 2)3 + 208 × 4 + C 2

∴

C 2 = − 617

∴

EI

dy dx

. . . (8)

= − 4 x3 + 4 [ x − 1]3 − 15 [ x − 2]2 + 208

EIy = − x 4 + [ x − 1]4 − 5 [ x − 2]3 + 208 x − 617

At free end A A, x = 0, From Eq. (9)

θ A = +

208 EI

=+

208 × 103 6

200 × 10

= + 1.04 × 10− 3 radians

From Eq. (10)

− 617 × 103 × 103 y A = = = − 3.085 mm ( ↓ ) EI 200 × 106 − 617

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

48

A 2m

20 kN 2m

B

. . . (9) . . . (10)

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

B

(a) Single Overhang Beam

(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. 80kN

X

B A

1m

C

1m

D

1m

x

20 kN

X

Figure 9.42

Solution

∑ F y = 0, so that RA + RB = 80 + 20 = 100 kN

. . . (1)

Σ M about A = 0. 80 × 1 + 20 20 × 3 = R A × 2 ∴ RB = 70 kN (↑ )

. . . (2)

From Eqs. (1) and (2), R A = 30 kN kN ( ↑ ) M = 30 x − 80 ( x − 1) + 70 ( x − 2) 2

EI

EI

d y dx

dy dx

. . . (3)

= M = 30 x − 80 [ x − 1] + 70 [ x − 2]

. . . (4)

= 15 x2 − 40 [ x − 1]2 + 35 [ x − 2]2 + C 1

. . . (5)

2

3

EIy = 5 x −

40 3

[ x − 1] 1]3 +

35 3

[ x − 2]3 + C1 x + C 2

. . . (6)

The boundary conditions are : At A,

x = 0, y = 0

. . . (7) 49


At B,

x = 2 m, y = 0

. . . (8)

From Eq. (6) and (7), C 2 = 0 From Eqs. (6) and (8)

0 = 5 × 23 − C 1 = −

∴

40 3

[ 2 − 1]3 + C 1 × 2

40 3

EIy = 5 x3 −

40 3

[ x − 1] 1]3 +

35 3

[ x − 2] 2 ]3 −

40 3

x

. . . (7)

Deflection at C , x = 1 m EIyC = 5 × 13 −

∴

yC =

40 3

×1 =

− 25 3

− 25

. . . (8)

3 EI

Deflection at D, x = 3 m

EIy D = 5 × 33 −

40 3

(3 − 1)3 +

35 3

(3 − 2)3 −

40

× 3= 0

3

y D = 0

Example 9.9

For the beam shown in Figure 9.43, determine the deflection at free end and the maximum deflection. 48kN/m A

D

60 kN B

E

2m

4m

6m

C

4m

Figure 9.43

Solution R A + RB = 48 × 6 + 60 = 348 kN

. . . (1)

Σ M about A = 0. 48 × 6 × ( 2 + 3) + 60 × 1 6 = R B × 12

∴

R B = 200 kN (↑ )

. . . (2)


k N (↑) R A = 148 kN

. . . (3)

Apply the Udl in downward and upward directions in the portion EC (Figure 9.44). 60 kN

48 kN/m X A E

D 8m 148 kN

50

x

C B 200 kN

X

Deflection of Beams

Figure 9.44

48

M = 148 x −

2

[ x − 2] [ x − 2] +

48 2

[ x − 8] [ x − 8] + 200 [ x − 12]

= 148 x − 24 [ x − 2]2 + 24 [ x − 8]2 + 200 [x − 12] 2

d y

EI

dx 2

. . . (4)

= M

= 148 x − 24 [ x − 2]2 + 24 [ x − 8]2 + 200 [x − 12] dy

EI

= 74 x 2 − 8 [ x − 2]3 + 8 [ x − 8]2 + 100 [ x − 12]2 + C 1

dx

EIy

. . . (5)

74 3

3

4

4

x − 2 [ x − 2] + 2 [ x − 8] +

100 3

. . . (6)

[ x − 12]3 + C1 x + C 2

. . . (7)


x = 0, y = 0

. . . (8)

At B,

x = 12 m, y = 0

. . . (9)

From Eqs. (7) and (8), C 2 = 0

. . . (10)


0=

∴

74 3

× 123 − 2 [12 − 2]4 + 2 [12 − 8]4 + C 1 × 12

C 1 = − 1928

EIy =

∴

74 3

. . . (11)

3

4

4

x − 2 [ x − 2] + 2 [ x − 8] +

100 3

[ x − 12]3 − 1928 x

Deflection at free end, x = 16 m EIyC =

∴

74 3

× 163 − 2 [16 − 2]4 + 2 [16 − 8]4 + yC = +

100 3

[16 − 12]3 − 1928 × 16 = 3680

3680 EI

Maximum Deflection : It occurs between D and E E . dy dx

=0

0 = 74 x 2 − 8 [ x − 2]3 − 1928

= 74 x 2 − 8 x 3 + 64 + 48 x 2 − 96 x − 1928 = 8 x3 − 122 x 2 + 96 x + 1928 By trial and error, x = 5.7 m EIymax =

∴

74 3

ymax = −

× 5.73 − 2 [5.7 − 2]4 − 1928 × 5.7 = − 6796.3

6796.3 EI

Example 9.10

51


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

∑ F y = 0, so that RA + RB = 80 + 40 = 120 kN

. . . (1)

Σ M about A = 0, ⇒

R B × 8 = 40 × 10 − 80 × 2

R B = 30 kN kN ( ↑ )

. . . (2) X

80 kN

40 kN A

D

B

2m

C 8m

2m

E

X

Figure 9.45

From Eqs. (1) and (2), R A = 90 kN kN ( ↑ )

. . . (3)

M = − 80 x + 90 [ x − 2] + 30 [ x − 10]

. . . (4)

2

EI

EI

d y dx

= M = − 80 x + 90 [ x − 2] + 30 [ x − 10]

. . . (5)

= − 40 x2 + 45 [ x − 2]2 + 15 [ x − 10]2 + C 1

. . . (6)

2

dy dx

40

EIy = −

x2 + 15 [ x − 2]3 + 5 [ x − 10]3 + C1 x + C 2

3

. . . (7)

Boundary conditions are : At A,

x = 2 m, y = 0

. . . (8)

At B,

x = 10 m, y = 0

. . . (9)


0=

∴

− 40 3

× 23 + C 2

320

C 2 = +

. . . (10)

3

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

0=

− 40 3

C 1 = + EIy = −

52

40 3

× 103 + 15 [10 − 2]3 + C 1 × 10 +

320 3

1664

. . . (11)

3 3

3

3

x + 15 [ x − 2] + 5 [ x − 10] +

1664 3

x+

320 3

. . . (12)

Deflection of Beams

Deflection at free ends : At D,

x = 0 y D = +

At E ,

320

. . . (13)

3 EI

x = 12 m,

EIy E = −

40 3

× 123 + 15 [12 − 2]3 + 5 [12 − 10]3 +

y E = −

∴

1664 3

× 12 +

320 3

3712 3 EI

Deflection at mid-span : x = 6 m EIyC = − yC = −

40 3

× 63 + 15 [6 − 2] 2 ]3 +

1664 3

×6+

320 3

4544 3 EI

SAQ 3 Determine the deflection at C and D D of the bean shown in Figure 9.46. 12 kN /m

40 kN

40 kN

A 2m

D C

2m

B

2m

Figure 9.46

9.6 APPLICATION OF DEFLECTIONS OF BEAMS 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, 2

(c)

EI

EI

d y dx dy dx

EIy =

2

=

= M

∫

M + C 1

∫∫ M + C

1

x + C 2

53


(d)

θ A = yC =

− wl2 16 EI

− wl3 48 EI W A

B C

l/2

l/2

Figure 9.47

(e)

θ A =

− wa b 6 EI l

θ B = + yC =

wb

6 EI l

− wb 48 EI

ymax =

(a + 2b) (2 a + b)

[3 l 2 − 4 b2 ]

− wb

2

(l −

9 3 EI l

3 2 2 b )

W A

B a

C

b

Figure 9.48

(f)

θ A =

− w l3 24 EI

yC = −

wl4

5

384 EI w/unit length A

B C

Figure 9.49

(g)

θ B = y B =

wl 2

2 EI

− w l3 3 EI w A

B

Figure 9.50

(h)

θ B = y B =

wl2

6 EI

− wl4 8 EI w/unit length

54

A

B

Deflection of Beams Figure 9.51

9.8 ANSWERS TO SAQs SAQ 1

(a)

EI = 20 × 106 N-m 2 100 kN

X

2m

3m

A

B x X

Figure 9.52

∑ y = 0 ⇒ R A + RB = 100 kN

. . . (1)

Σ M about A = 0 ⇒

R B × 5 = 100 × 2

∴

R B = 40 kN k N ( ↑)

. . . (2)

From Eqs. (1) and (2), R A = 60 kN k N ( ↑)

. . . (3)

M = 60 x − 100 [ x − 2]

. . . (4)

2

EI

EI

d y dx

2

dy dx

= M = 60 x − 100 [ x − 2]

= 30 x 2 − 50 [ x − 2]2 + C 1 3

EIy = 10 x −

50 3

[ x − 2]3 + C1 x + C 2

. . . (5) . . . (6) . . . (7)


x = 0,

y=0

. . . (8)

At B,

x = 5 m, y = 0

. . . (9)

From Eqs. (7) and (8), C 2 = 0

. . . (10)

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

0 = 10 × 53 −

∴

50 3

[5 − 2]3 + C 1 × 5

C 1 = − 160

EI

dy dx

. . . (11)

= 30 x2 − 50 [ x − 2]2 − 160 3

EIy = 10 x −

50 3

[ x − 2]3 − 160 x

. . . (12) . . . (13)

Deflection at C , x = 2 m, EIyC = 10 × 23 − 160 × 2

55


yC =

∴

− 240 EI

=

− 240 × 103 × 103 = − 12 mm 20 × 106

Maximum Deflection

It occurs between C and B B. dy dx

=0

30 x 2 − 50 [ x − 2]2 − 160 = 0 30 x 2 − 50 x 2 − 200 + 200 x − 160 = 0

− 20 x 2 + 200 x − 360 = 0

⇒

2 x − 10 x + 18 = 0

∴

x = 2.35 m

Deflection at Mid-span x = 2.35 m 3

EIy D = 10 × 2.5 −

50 3

[2.5 − 2]3 − 160 × 2.5

12.29 9 mm y D = − 12.2 3

EIymax = 10 × 2.35 −

50 3

[2.35 − 2]3 − 160 × 2.35

12.35 5 mm ymax = − 12.3

(b)

EI = 30 × 106 N-m 2 30 kN-m

A

B 6m

x

Figure 9.53

∑ F y = 0 ⇒ RA + R B = 0

. . . (1)

Σ M about A = 0 ⇒

R B × 6 = 30

∴

R B = 5 kN kN (↓)

. . . (2)

From Eqs. (1) and (2), R A = − 5 kN kN (↓) o M = − 5 x + 30 x

2

EI

EI

d y dx dy dx

2

=−

EIy = −

56

= − 5 x + 30 xo 5 x2

5 x 3 3

2

+ 30 x + C 1

+ 15 x 2 + C1 x + C 2

. . . (3) . . . (4) . . . (5)

. . . (6)

. . . (7)

Deflection of Beams


x = 0,

y=0

. . . (8)

At B,

x = 6 m, y = 0

. . . (9)

From Eqs. (7) and (8), C 2 = 0

. . . (10)

From Eqs. (7), (9) and (10) 5 × 63

0=−

∴

+ 15 × 62 + C 1 × 6

6

C 1 = − 60 EI

dy dx

. . . (11) 5 x2

=−

EIy = −

+ 30 x − 60

. . . (12)

+ 15 x 2 − 60 x

. . . (13)

2

5 x 3 6

dy

For maximum deflection,

dx

= 0,

5

− x 2 + 30 x − 60 = 0

∴

2

or

2 x − 12 x + 24 = 0

∴

x = 2.54 m EIymax = −

5

(2.54)3 + 15 (2 (2.54)2 − 60 × 2.54 = − 69.28

6

2.31 mm ymax = − 2.31

(c)

∑ F y = 0 ∴

kN R A + RB = 48 × 3 = 144 kN

or

R B × 6 = 20 + 48 × 3 × (3 + 1.5)

∴

R B =

334 3

. . . (1)

kN ( ↑ )

. . . (2) x 48 kN /m

A

D 1m 2m 20 kN -mm x

C

B 3m x

Figure 9.54

From Eqs. (1) and (2), R A =

98

M =

98

3

3

kN ( ↑ ) o x + 10 [ x − 1] −

. . . (3)

48 2

[ x − 3]2

. . . (4)

57

2


EI

EI

d y dx

2

dy

3

49

EIy =

x + 10 [ x − 1]o −

3

98

=

dx

98

=

48 2

48

2

x + 10 [ x − 1] −

6

[ x − 3]2

[ x − 3]3 + C 1

x3 + 5 [ x − 1]2 − 2 [ x − 3]4 + C1 x + C 2

9

. . . (5)

. . . (6)

. . . (7)


x = 0,

y=0

. . . (8)

At B,

x = 6 m,

y=0

. . . (9)

From Eqs. (7) and (8), C 2 = 0

. . . (10)

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

0=

49 9

C 1 = −

∴

× 63 + 5 (6 − 1)2 − 2 (6 − 3)4 + C 1 × 6 1139 6

Deflection at D, x = 1 m, EIy D = y D =

49 9

× 13 −

1139 6

×1

− 184.4 EI

SAQ 2 6 2 EI = 40 × 10 N-m

x 20 kN A

24 kN/m C

2m

B 3m

x

x

Figure 9.55

M = − 20 x − 2

EI

EI

d y dx dy dx

2

2

[ x − 2]2

− 20 x − 12 [ x − 2]2

− 10 x 2 − 4 [ x − 2]3 + C 1

EIy = −

10 x 3 3

Boundary conditions are :

58

24

− [ x − 2]4 + C1 x + C 2

. . . (1)

. . . (2)

. . . (3)

. . . (4)

At A,

x = 4 m,

At B,

x = 4 m,

dy

Deflection of Beams

=0

. . . (5)

y=0

. . . (6)

dx

From Eqs. (4) and (5), 0 = − 10 × 42 − 4 ( 4 − 2)3 + C 1

∴

C 1 = + 192

. . . (7)

From Eqs, (4), (6) and (7), 0=

∴

− 10 × 43 3

C 2 = −

− [ 4 − 2]2 + 192 × 4 + C 2

1616 3

EIy = −

10 x 3

− [ x − 2]4 + 192 x −

3

1616

. . . (8)

3

Deflection at free end, x = 0 y A = −

1616 3 EI

SAQ 3

∑ F y = 0

R A + RB = 12 × 6 + 2 + 40 = 152 kN

. . . (1)

∑ M A = 0

⇒

R B × 4 = 12 × 6 × 3 + 40 × 2 + 40 × 6

∴

R B = 134 kN (↑)

. . . (2)

40 kN

12 kN/m

40 kN X

A C

2m

2m

B

2m

x

D

X

Figure 9.56

From Eqs. (1) and (2), R A = 18 kN k N ( ↑) M = 18 x − 2

EI EI

d y dx dy dx

2

. . . (3)

12 x 2 2

− 40 [ x − 2] + 134 [ x − 4]

= 18 x − 6 x2 − 40 [ x − 2] + 134 [ x − 4]

= 9 x2 − 2 x3 − 20 [ x − 2]2 + 67 [ x − 4]2 + C 1 3

EIy = 3 x −

x

4

3

−

20 3

[ x − 2]3 +

67 3

[ x − 4]3 + C1 x + C 2

. . . (4) . . . (5) . . . (6) . . . (7)


x = 0, y = 0

. . . (8)

At B,

x = 4 m, y = 0

. . . (9)

59


From Eqs. (7) and (8), C 2 = 0

. . . (10)

From Eqs, (7), (9) and (10), 3

0 =3×4 −

∴

C 1 =

44 2

−

20 3

[4 − 2]3 + C 1 × 4

−8

. . . (11)

3 3

EIy = 3 x −

x

4

2

−

20 3

[ x − 2]3 +

67 2

[ x − 4]3 −

8 3

. . . (12)

x

At x = 2 m, 3

EIyC = 3 × 2 − yC = +

24 2

−

8

−

20

3

×2

32 3 EI

At x = 6 m, 3

EIy D = 3 × 6 − y D =

60

− 792 3 EI

64 2

3

[6 − 2]3 +

67 3

[6 − 4]3

−8 3

×6

Double Integration Method

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