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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.
Stresses in Shafts and Shells
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
Stresses in Shafts and Shells
∴ 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
Consider a section X-X at a distance x from A,
[ ]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)
The governing equation for deflection is :
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
(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, 2A B
wlR R= =
Consider a section X-X at a distance x from A,
.2 2
wl xM x w x= −
14
Stresses in Shafts and Shells
2
2 2wl x w x
= − . . . (9.27)
Figure 9.8
The governing equation for deflection is :
2 2
2 2 2d y w l x w xEIdx
= − . . . (9.28)
Integrating the Eq. (9.28), we can get
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)
The constants C1 and C2 can be found from the boundary conditions.
The boundary conditions are :
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
(c) Deflection Diagram
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
Taking moments about A,
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)
Integrating the Eq. (9.41), we can get
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)
The constants C1 and C2 can be found from the boundary conditions.
The boundary conditions are :
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
(c) Deflection 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)
Taking moments about A,
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)
The governing equation for deflection is :
2
002
M xd yEI M Mldx
= = − + < > . . . (9.55)
Integrating the Eq. (9.55), we can get
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.
The boundary conditions are :
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)
The slope and deflection equation will be :
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
Stresses in Shafts and Shells
20 0 0 0 0
1 6 2 2 6 8 24M l M M l M l M llC l
l⎛ ⎞= − − = − =⎜ ⎟⎝ ⎠
The slope and deflection equation will be :
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
Taking moments about A,
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)
The governing equation for deflection is :
2
02
Md yEI Mldx
−= = x . . . (9.69)
Integrating the Eq. (9.69), we can get
2012
MdyEI xdx l
= − + C . . . (9.70)
3016
M2EIy x C x C
l= − + + . . . (9.71)
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) 0, 0x y= =
Applying the boundary condition (1) into the Eq. (9.71), C2 = 0.
Applying the BC (2) into the Eq. (9.71),
2
0 01 10
6 6M l MC l C= − + ∴ =
l
The slope and deflection equation will be :
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
Stresses in Shafts and Shells
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
(c) Deflection 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∑ = + =
Taking moments about A,
. . . (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
Consider a section X-X at a distance x from A,
0M M= − . . . (9.81)
The governing equation for deflection is
2
02d yEI Mdx
= = − M . . . (9.82)
Integrating the Eq. (9.82), we can get
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= =
Applying the BC (1) into the Eq. (9.84),
2 0C =
Applying the BC (2) into the Eq. (9.84),
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)
Stresses in Shafts and Shells
Slope at C, 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
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
(c) Deflection 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)
Consider a section X-X at a distance x from A,
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
Stresses in Shafts and Shells
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
Taking moments about A,
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
Stresses in Shafts and Shells
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
Taking moments about A,
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
Stresses in Shafts and Shells
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)
The governing equation for deflection is
2
2d yEI Mdx
= = = −W x . . . (9.93)
W
A
Figure 9.22
Integrating the Eq. (9.93), we can get
2
12dy W xEI Cdx
= − + . . . (9.94)
3
16W x
2EIy C x C= − + + . . . (9.95)
The constants C1 and C2 can be found from the boundary conditions.
The boundary conditions are :
At B, , dyx ldx
0= = (Fixed end) . . . (1)
At B, , 0x l y= = . . . (2)
Applying the BC (1) into the Eq. (9.94), we can get
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)
The slope and deflection equations will be :
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
(c) Deflection 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’.
Consider a section X-X at a distance x from left end.
Moment, 2lM W x⎡= − −⎢⎣ ⎦⎤⎥ (Hogging BM) . . . (9.104)
The governing equation for deflection is :
2
2 2d y lEI M W xdx
⎡ ⎤= = − −⎢ ⎥⎣ ⎦ . . . (9.105)
34
Stresses in Shafts and Shells
Figure 9.24
Integrating the Eq. (9.105), we can get
2
12 2dy W lEI xdx
⎡ ⎤ C= − − +⎢ ⎥⎣ ⎦ . . . (9.106)
3
16 2W l
2EIy x C x C⎡ ⎤= − − + +⎢ ⎥⎣ ⎦ . . . (9.107)
The constants C1 and C2 can be found from the boundary conditions.
The boundary conditions are :
At B, , 0dyx ldx
= = . . . (1)
At B, , 0x l y= = . . . (2)
Applying the BC (1) into the Eq. (9.106), we can get
22
102 2
W l ll C⎡ ⎤= − − +⎢ ⎥⎣ ⎦
∴ 2
1 8W lC = + . . . (9.108)
Applying BC (2) into the Eq. (9.107), we can get
3 2
206 2 8
W l W ll l⎛ ⎞= − − + × +⎜ ⎟⎝ ⎠
C
∴ 3
2548W lC −
= . . . (9.109)
The slope and deflection equations will be :
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
(c) Deflection Diagram
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)
The governing equation for deflection is :
2 2
2 2d y w xEI Mdx
= = − . . . (9.117)
A BC
X
Xx
w/unit length
l/2l/2
Figure 9.26
Integrating the Eq. (9.117), we can get
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)
Applying the BC (1) into the Eq. (9.118), we can get
3
106
wl C= − +
∴ 3
1 6wlC = + . . . (9.120)
Applying BC (2) into the Eq. (9.119), we can get
4 4
2024 6
wl wl C= − + +
∴ 4
2 8wlC = − . . . (9.121)
The slope and deflection equations will be :
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)
Deflection at A (x = 0),
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
(c) Deflection Diagram
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.
Consider a section X-X at a distance x from left end.
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
The governing equation for deflection is :
22
2 2 2d y w lEI M xdx
⎡ ⎤= = − −⎢ ⎥⎣ ⎦ . . . (9.129)
Integrating the Eq. (9.129), we can get
3
16 2dy w lEI xdx
⎡ ⎤= − − +⎢ ⎥⎣ ⎦C . . . (9.130)
4
1 224 2dy w lEIy x C x Cdx
⎡ ⎤= − − + +⎢ ⎥⎣ ⎦ . . . (9.131)
The constants C1 and C2 can be found from boundary conditions.
The boundary conditions are :
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
Stresses in Shafts and Shells
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)
The slope and deflection equations will be :
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)
Slope at C 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
3
48Cwl
EIθ = . . . (9.137) +
Deflection at A (x = 0),
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
(c) Deflection Diagram
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)
The governing equation for deflection is :
22 2
2 2 2 2d y w x w lEI M xdx
⎡ ⎤= = − + −⎢ ⎥⎣ ⎦ . . . (9.135)
Integrating the Eq. (9.135), we can get
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)
The constants C1 and C2 can be found from boundary conditions.
The boundary conditions are :
At B, , 0dyx ldx
= = . . . (1)
40
At B, , 0x l y= = . . . (2) Stresses in Shafts and Shells
Applying the boundary condition (1) into the Eq. (9.136), we can get
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)
The slope and deflection equations will be :
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)
Slope at A, (x = 0),
37
48AwlEI
θ = + . . . (9.142)
Slope at C 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
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)
Deflection at C, 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠
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
(c) Deflection Diagram
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
Stresses in Shafts and Shells
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)
The governing equation for deflection is :
2 3
2 6d y w xEI M
ldx= = − . . . (9.149)
Integrating the Eq. (9.149), we can get
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)
Applying the boundary condition (1) into the Eq. (9.150), we can get
3
1024wl C−
= +
∴ 3
1 24wlC = + . . . (9.152)
Applying the boundary condition (2) into the Eq. (9.151), we can get
4 4
20120 24wl wl C= − + +
∴ 3
2 24wlC = + . . . (9.153)
The slope and deflection equations will be :
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)
Deflection at C, 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
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
(c) Deflection Diagram
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
Stresses in Shafts and Shells
Moment, 00 .M M x= − . . . (9.160)
The governing equation for deflection is :
2
002
d yEI M Mdx
= = − x . . . (9.161)
Integrating the Eq. (9.161), we can get
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)
Applying the boundary condition (2) into the Eq. (9.163), we can get
2
00 20 .
2M l
M l l C= − + +
∴ 2
02 2
M lC = − . . . (9.165)
The slope and deflection equations will be :
0dy
0EI M x Mdx
= − + l . . . (9.166)
2 2
0 002 2
M x MEIy M l x= − + −l
. . . (9.167)
Slope at A, (x = 0),
0A
M lEI
θ = . . . (9.168)
Slope at C 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
0 02ClEI M Mθ = − + l
∴ 02CM l
EIθ = + . . . (9.169)
Deflection at A, (x = 0),
2
0A
M lyEI
= − . . . (9.170)
Deflection at C, 2lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
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
(c) Deflection 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
Stresses in Shafts and Shells
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)
Slope at A, (x = 0),
29
32Awl
EIθ = + . . . (11)
Slope at C, 4lx⎛ ⎞=⎜ ⎟
⎝ ⎠,
29
32Cwl
EIθ = + . . . (12)
Deflection at A, (x = 0),
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
Boundary conditions are :
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)
The boundary conditions are :
At B, 4 m, 0dyxdx
= = . . . (5)
At B, 4 m, 0x y= = . . . (6)
From Eqs. (3) and (5)
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× + × = × ∴ = ↑
From Eqs. (1) and (2),
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)
The boundary conditions are :
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
From Eqs. (6) and (8)
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)
From Eqs. (1) and (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)
The boundary conditions are :
At A, . . . (8) 0, 0x y= =
At B, . . . (9) 12 m, 0x y= =
From Eqs. (7) and (8),
C2 = 0 . . . (10)
From Eqs. (7) and (9),
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 :
Stresses in Shafts and Shells
(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
From Eqs. (1) and (2),
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)
Boundary conditions are :
At A, 2 m, 0x y= = . . . (8)
At B, 10 m, 0x y= = . . . (9)
From Eqs. (7) and (8),
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)
From Eqs. (1) and (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= =
From Eqs. (7) and (8),
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)
From Eqs. (1) and (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= =
From Eqs. (7) and (8),
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
From Eqs. (1) and (2),
98 kN ( )3AR = ↑ . . . (3)
o 298 4810 [ 1] [ 3]3 2
M x x x= + − − − . . . (4)
58
Stresses in Shafts and Shells
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)
The boundary conditions are :
At A, 0, 0x y= = . . . (8)
At B, 6 m, 0x y= = . . . (9)
From Eqs. (7) and (8),
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)
Boundary conditions are :
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=
From Eqs. (4) and (5),
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
From Eqs. (1) and (2),
. . . (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−
=
top related