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MECHANICS OF
MATERIALS
Third Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2002 The McGraw-Hill Companies, Inc. All rights reserved.
9Deflection of Beams
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MECHANICS OF MATERIALSThird
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Deflection of Beams
Deformation of a Beam Under Transverse
LoadingEquation of the Elastic Curve
Example 9.01
Example 9.02
Direct Determination of the Elastic CurveFrom the Load Distribution
Example 9.04
Sample 9.1
Using singularity functions todetermine the slope and deflection
of a beam
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Deformation of a Beam Under Transverse Loading
Relationship between bending moment and
curvature for pure bending remains valid for
general transverse loadings.
EI
xM )(1
EI
Px
1
Curvature varies linearly withx
At the free endA, x=0, A
A
,01
At the supportB,x=L,PL
EIB
B
,01
A Cantilever beam AB of length L subjected
to concentrated load P at the free end A. We
have
M(x) = -Px. Then it become
x
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Deformation of a Beam Under Transverse Loading Consider Overhanging beam AD support
concentrate load
Reactions atAand C (RA= 1kN an RC= 5 kN
Bending moment diagram
Curvature is zero at points where the bending
moment is zero, i.e., at each end and atE.
EIxM )(1
Between AE Beam is concave upwards where the
bending moment is positive and concave
downwards where it is negative at ED.
Maximum curvature occurs where the moment
magnitude is a maximum at the support C.
An equation for the beam shape or elastic curve
is required to determine maximum deflection
and slope.
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Deformation of a Beam Under Transverse Loading
From the information obtained its curvature,
we get fairly good idea of the shape of the
deformed beam.
The analysis and design of the beam require
more precise information on the deflectionand the slope of the beam at various points.
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Type 1. Equation of the Elastic Curve
From elementary calculus, simplified for beam
parameters,
2
2
232
2
2
1
1
dx
yd
dxdy
dx
yd
Substituting and integrating,
2
2
1
0
1 2
0 0
1
( )
( )
x
x x
d y
EI EI M xdx
dyEI EI M x dx C
dx
EI y dx M x dx C x C
The slope dy/dx = is
very small the
square negligible.
From EIxM )(1
Where dy/dx = tan
= (x)
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Type 1. Equation of the Elastic Curve
2100
CxCdxxMdxyEI
xx
Constants are determined from boundary
conditions
Three cases for statically determinant beams,
Simply supported beam0, 0A By y
Overhanging beam
0,0 BA yy
Cantilever beam0,0 AAy
More complicated loadings require multiple
integrals and application of requirement for
continuity of displacement and slope.
deflection and slope.
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Example 9.01
The cantilever beamABis of uniform cross
section and carries a load P at its free end A.
Determine the equation of the elastic curve
and the deflection and slope at A..
Using the free-body diagram of the portionAC
of the beam where Cis located at a distancex
fromend A, we find
M= -PxEI d2y/dx2= - Px
Integrating in x, we obtain
EI dy/dx = -Px2+ C1We now observe that at fixed end B we have
x=L and = dy/dx = 0 and we getC1= PL
2
Then EI dy/dx = -Px2+ PL2
Integrating both member we write
EIy = -1/6 Px3
+ PL2
x + C2
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Example 9.01
But at B we have x=L, y=0 and we have
0 = -1/6 PL3+ PL3+C2C2= -1/3 PL
3
Carrying the value of C2
EIy = -1/6 Px3+ PL2x -1/3 PL3
Or
y = P/6EI (-x3
+ 3L2
x 2L3
)
The deflection and slope at A are obtained by
letting x=0. We find
yA= -PL3/3EI
and
A= (dy/dx)A= PL
2
/2EI
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Example 9.02 The simply supported prismatic beam AB
carries a uniformly distributed load wper
unit length. Determine the equation of theelastic curve and the maximum deflection of
the beam
M = wLx wx2
EI d2
y/dx2
= - wx2
+ wLx
Integrating twice in x, we have
EI dy/dx = -1/6 wx3+ wLx2+ C1
EIy = -1/24 wx4+ 1/12 wLx3+ C1x + C2
Observing that y=0 at both ends of the beam,
we first let x = 0 and y=0 and obtain C2=0.
We then make x=L and y=0 in the same
equation and write
0 = -1/24 wx4
+ 1/12wLx3
+ C1LC1= 1/24wL
3
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Example 9.02 Carrying the value of C1and C2back we
obtain the equation of the elastic curve:
EIy = -1/24 wx4+ 1/12wLx31/24wL3x
Or y = w/(24EI) (-x4+ 2Lx3L3x)
We check that the slope of the beam is zero
for x = L/2 and the elastic curve has a max at
the midpoint C of the beam
yC= w/(24EI)(-L4/16 + 2LL3/8L3L/2)
= - 5wL4/384EI
The maximum deflection, the maximum
absolute value of the deflection is
/y/max= 5wL4/384EI
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Sample Problem 9.1
1.2m4.5mkN200
GPa200mm10300101360 46
aLP
ExIW
For portionABof the overhanging beam,
(a) derive the equation for the elastic curve,(b) determine the maximum deflection,
(c) evaluateymax.
SOLUTION:
Develop an expression for M(x)
and derive differential equation for
elastic curve.
Integrate differential equation twiceand apply boundary conditions to
obtain elastic curve.
Locate point of zero slope or point
of maximum deflection.
Evaluate corresponding maximum
deflection.
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Sample Problem 9.1
SOLUTION:
Develop an expression for M(x) and derivedifferential equation for elastic curve.
- Reactions:
L
aPR
L
PaR BA 1
- From the free-body diagram for sectionAD,
LxxL
aPM 0
xL
aP
dx
ydEI
2
2
- The differential equation for the elastic
curve,
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Sample Problem 9.1
PaLCLCLL
aPyLx
Cyx
6
1
6
10:0,at
0:0,0at
113
2
Integrate differential equation twice and apply
boundary conditions to obtain elastic curve.
213
12
6
1
2
1
CxCxL
aPyEI
CxL
aP
dx
dyEI
xL
aP
dx
ydEI
2
2
32
6 L
x
L
x
EI
PaLy
PaLxxL
aPyEI
Lx
EIPaL
dxdyPaLx
LaP
dxdyEI
6
1
6
1
3166
121
3
2
2
Substituting, a) equation ofthe elastic curve
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Sample Problem 9.1
Locate point of zero slope or pointof maximum deflection in portion AB
32
6 L
x
L
x
EI
PaLy
LL
xL
x
EI
PaL
dx
dym
m 577.03
316
02
Evaluate corresponding maximum
deflection.
32
max 577.0577.06
EI
PaLy
EI
PaLy
6
0642.02
max
46
2
max
m103002006
4.5m1.2m200kN0642.0
xGPay
mm5.2maxy
Evalution of ymax
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2. Direct Determination of the Elastic Curve From
the Load Distribution
The equation of elastic curve can be obtained
EI
xM
dx
yd )(2
2
Differentiating both member with respect to x3
3 1 ( )d y dM V xdx EI dx EI
And differentiating again
4
4
1 ( )d y dV w x
dx EI dx EI
We conclude that when a prismatic beam
supports a distributed load w(x) its elastic curve
is governed by the forth order linear differential
equation
EI
xw
dx
yd )(4
4
Where dM/dx =V
and dV/dx = -w
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2. Direct Determination of the Elastic Curve From
the Load Distribution
xwdx
ydEI
dx
Md
4
4
2
2
13
3
)( CdxxwxVdx
ydEI
xwdxydEI
dxMd 4
4
2
2
Equation for beam displacement becomes
xwdx
ydEI
dx
Md
4
4
2
2
432
2213
161 CxCxCxC
dxxwdxdxdxxyEI
Integrating four times yields
Constants are determined from boundary
conditions.
212
2
)( CxCdxxwdxxMdx
ydEI
32212/1)( CxCxCdxxwdxdxxEIdx
dyEI
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Example 9.04
13
3
CwxxVdx
ydEI
Since w = constant
wdx
ydEI
4
4
2122
2
2/1 CxCwxxMdx
ydEI
The boundary condition require that M=0 at both end
of the beam we first let x=0 and M=0 and obtain c2=0.
We then make x =L and M=0 and obtain C1= wL
Carrying the value of C1and C2EI d2y/dx2= - wx2+ wLx
EI dy/dx = -1/6 wx3+ wLx2 + C3EIy = -1/24 wx4+ 1/12 wLx3+ C3x + C4
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Example 9.04
But the bondary conditions also require that y=0 at both
end of the beam. Let x=0 and y=0, we obtain C4=0, let
x=L and y=0 in the same equation
EIy = -1/24 wx4+ 1/12 wLx3+ C3x + C4
0 = -1/24 wL4+ 1/12 wLL3+ C3LC3= -1/24 wL
3
We obtain the equation of elastic curve
y = w/24EI (-x4+ 2Lx3L3x)
The value of the maximum deflection is obtained by
making x =L/2.
We have
/y/max = 5wL4/384EI
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Type 3. Using singularity functionLet us consider the beam and loading and draw the free
Body diagram of the beam. We write
V(x) = PP x - L > 0
We have
M(x) = P x - P < x - L >
EI d2y/dx2= P x - P < x - L >
and integrating in x
EI= EI dy/dx = 3/8 Px2 P 2+ C1
EIy = 1/8 Px31/6 P 3 + C1x + C2
The constants C1and C2can be determined from. Letx=0, y=0 and we have C2=0 in Eq.
Let x =L, y=0
0 = 1/8 PL31/6 P (3/4 L)3+ C1L
C1 = -7PL2/128
EIy = 1/8 Px3
1/6 P 3
- 7pL2
/128 x
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Type 3. Using singularity function
For the beam and loading shown, determine
(a) the slope at end A,
(b) the deflection at point B,(c) the deflection at end D
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Type 3. Using singularity function
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Method of Superposition
Principle of Superposition:
Deformations of beams subjected to
combinations of loadings may be
obtained as the linear combination of
the deformations from the individualloadings
Procedure is facilitated by tables of
solutions for common types of
loadings and supports.
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Sample Problem 9.7
For the beam and loading
shown, determine the slope and
deflection at pointB.
SOLUTION:
Superpose the deformations due toLoading IandLoading
II as shown.
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Sample Problem 9.7
Loading I
EI
wLIB 6
3
EI
wLy IB 8
4
Loading II
EI
wLIIC 48
3
EI
wLy IIC 128
4
In beam segment CB, the bending
moment is zero and the elastic curve is a
straight line.
EI
wLIICIIB 48
3
EI
wLL
EI
wL
EI
wLy IIB 384
7
248128
434
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Sample Problem 9.7
EI
wLEI
wLIIBIBB 486
33
EI
wL
EI
wLyyy IIBIBB 384
7
8
44
EIwLB
487
3
EI
wLyB
384
41 4
Combine the two solutions,
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Application of Superposition to Statically
Indeterminate Beams
Method of superposition may be applied
to determine the reactions at the
supports of statically indeterminate
beams.
Designate one of the reactions as
redundant and eliminate or modify
the support.
Determine the beam deformation without
the redundant support.
Treat the redundant reaction as an
unknown load which, together with
the other loads, must produce
deformations compatible with the
original supports.
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Sample Problem 9.8
For the uniform beam and loadingshown, determine the reaction at
each support and the slope at endA.
SOLUTION:
Release the redundant support at B, and find deformation.
Apply reaction atBas an unknown load to force zero displacement atB.
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Sample Problem 9.8
Distributed Loading:
EI
wL
LLLLLEI
wy wB
4
334
01132.0
3
2
3
223
2
24
At pointB, Lx32
xLLxxEI
wy
wB334 2
24
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Sample Problem 9.8
Redundant Reaction Loading:
EIL
bPayax
3,At
22
EI
LR
L
LEIL
R
y
B
B
RB
3
22
01646.0
33
2
3
LbLa31
32 andFor
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Sample Problem 9.8
From statics,
wLRwLR CA 0413.0271.0
For compatibility with original supports,yB= 0
EI
LR
EI
wLyy BRBwB
34
01646.001132.00
wLRB 688.0
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Sample Problem 9.8
EI
wL
EI
wLwA
33
04167.024
EIwLLLLEILwLEILbLPbRA32
2
22
03398.0336
0688.06
EI
wL
EI
wLRAwAA
33
03398.004167.0 EI
wLA
3
00769.0
Slope at endA,