Giáo trình Strength of Materials - Chapter 8: Beam deflection

CHAPTER 8: BEAM DEFLECTION 
8.1 Deflection of a beam under transverse loading 
8.2 Equation of the elastic curve 
8.3 Direct determination of the elastic curve from the loading 
8.4 Statically indeterminate beams 
8.5 Examples 
8.6 Statically indeterminate beams 
8.1 DEFLECTION OF A BEAM UNDER TRANSVERSE 
LOADING 
• Relationship between bending moment and 
curvature for pure bending remains valid for 
general transverse loadings. 
EI
xM )(1


• Cantilever beam subjected to concentrated 
load at the free end, 
EI
Px


1
• Curvature varies linearly with x 
• At the free end A,  A
A
 ρ
ρ
,0
1
• At the support B, 
PL
EI
B
B
 

 ,0
1
8.1 DEFLECTION OF A BEAM UNDER TRANSVERSE 
LOADING 
• Overhanging beam 
• Reactions at A and C 
• Bending moment diagram 
• Curvature is zero at points where the bending 
moment is zero, i.e., at each end and at E. 
EI
xM )(1


• Beam is concave upwards where the bending 
moment is positive and concave downwards 
where it is negative. 
• Maximum curvature occurs where the moment 
magnitude is a maximum. 
• An equation for the beam shape or elastic curve 
is required to determine maximum deflection 
and slope. 
8.2 EQUATION OF THE ELASTIC CURVE 
• From elementary calculus, simplified for beam 
parameters, 
2
2
23
2
2
2
1
1
dx
yd
dx
dy
dx
yd


















• Substituting and integrating, 
 
 
  21
00
1
0
2
21
CxCdxxMdxyEI
CdxxM
dx
dy
EIEI
xM
dx
yd
EIEI
xx
x






8.2 EQUATION OF THE ELASTIC CURVE 
  21
00
CxCdxxMdxyEI
xx
 
• Constants are determined from boundary 
conditions 
• Three cases for statically determinant beams, 
– Simply supported beam 
0,0  BA yy
– Overhanging beam 
0,0  BA yy
– Cantilever beam 
0,0  AAy 
• More complicated loadings require multiple 
integrals and application of requirement for 
continuity of displacement and slope. 
8.3 DIRECT DETERMINATION OF THE ELASTIC CURVE 
FROM THE LOAD DISTRIBUTION 
• For a beam subjected to a distributed load, 
   xw
dx
dV
dx
Md
xV
dx
dM

2
2
• Equation for beam displacement becomes 
 xw
dx
yd
EI
dx
Md

4
4
2
2
   
43
2
22
13
16
1 CxCxCxC
dxxwdxdxdxxyEI

 
• Integrating four times yields 
• Constants are determined from boundary 
conditions. 
8.4 STATICALLY INDETERMINATE BEAMS 
• Consider beam with fixed support at A and roller 
support at B. 
• From free-body diagram, note that there are four 
unknown reaction components. 
  21
00
CxCdxxMdxyEI
xx
 
• Also have the beam deflection equation, 
 which introduces two unknowns but provides 
three additional equations from the boundary 
conditions: 
0,At 00,0At  yLxyx 
• Conditions for static equilibrium yield 
000  Ayx MFF
 The beam is statically indeterminate. 
8.5 EXAMPLES 
EXAMPLE 9.01 
ft 4ft15kips50
psi1029in7236814 64


aLP
EIW
For portion AB of the overhanging beam, 
(a) derive the equation for the elastic curve, 
(b) determine the maximum deflection, 
(c) evaluate ymax. 
SOLUTION: 
•Develop an expression for M(x) and 
derive differential equation for elastic 
curve. 
•Integrate differential equation twice 
and apply boundary conditions to 
obtain elastic curve. 
•Locate point of zero slope or point of 
maximum deflection. 
•Evaluate corresponding maximum 
deflection. 
8.5 EXAMPLES 
SOLUTION: 
•Develop an expression for M(x) and derive 
differential equation for elastic curve. 
-Reactions: 







L
a
PR
L
Pa
R BA 1
-From the free-body diagram for section AD, 
 Lxx
L
a
PM  0
x
L
a
P
dx
yd
EI 
2
2
-The differential equation for the elastic 
curve, 
EXAMPLE 9.01 
8.5 EXAMPLES 
EXAMPLE 9.01 
•Locate point of zero slope or point of 
maximum deflection. 















32
6 L
x
L
x
EI
PaL
y
L
L
x
L
x
EI
PaL
dx
dy
m
m 577.0
3
31
6
0
2
















•Evaluate corresponding maximum 
deflection. 
  3
2
max 577.0577.0
6

EI
PaL
y
EI
PaL
y
6
0642.0
2
max 
   
  46
2
max
in723psi10296
in180in48kips50
0642.0

y
in238.0max y
8.5 EXAMPLES 
EXAMPLE 9.02 
For the uniform beam, determine the 
reaction at A, derive the equation for 
the elastic curve, and determine the 
slope at A. (Note that the beam is 
statically indeterminate to the first 
degree) 
SOLUTION: 
•Develop the differential equation for the 
elastic curve (will be functionally 
dependent on the reaction at A). 
•Integrate twice and apply boundary 
conditions to solve for reaction at A and 
to obtain the elastic curve. 
•Evaluate the slope at A. 
8.5 EXAMPLES 
EXAMPLE 9.02 
•Consider moment acting at section D, 
L
xw
xRM
M
x
L
xw
xR
M
A
A
D
6
0
32
1
0
3
0
2
0












L
xw
xRM
dx
yd
EI A
6
3
0
2
2

•The differential equation for the elastic curve, 
8.5 EXAMPLES 
EXAMPLE 9.02 
L
xw
xRM
dx
yd
EI A
6
3
0
2
2

•Integrate twice 
21
5
03
1
4
02
1206
1
242
1
CxC
L
xw
xRyEI
C
L
xw
xREI
dx
dy
EI
A
A

 
•Apply boundary conditions: 
0
1206
1
:0,at 
0
242
1
:0,at 
0:0,0at 
21
4
03
1
3
02
2



CLC
Lw
LRyLx
C
Lw
LRLx
Cyx
A
A
•Solve for reaction at A 
0
30
1
3
1 4
0
3  LwLRA  LwRA 0
10
1
8.5 EXAMPLES 
EXAMPLE 9.02 
xLw
L
xw
xLwyEI 











 30
5
03
0
120
1
12010
1
6
1
 xLxLx
EIL
w
y 43250 2
120

•Substitute for C1, C2, and RA in the elastic 
curve equation, 
 42240 65
120
LxLx
EIL
w
dx
dy

EI
Lw
A
120
3
0
•Differentiate once to find the slope, 
at x = 0, 
8.6 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 individual 
loadings 
•Procedure is facilitated by tables of 
solutions for common types of 
loadings and supports. 
8.6 METHOD OF SUPERPOSITION 
For the beam and loading shown, 
determine the slope and deflection at 
point B. 
SOLUTION: 
Superpose the deformations due to Loading I and Loading II as shown. 
EXAMPLE 9.03 
8.6 METHOD OF SUPERPOSITION 
EXAMPLE 9.03 
Loading I 
 
EI
wL
IB 6
3
  
EI
wL
y IB 8
4

Loading II 
 
EI
wL
IIC 48
3
  
EI
wL
y IIC 128
4

In beam segment CB, the bending moment is 
zero and the elastic curve is a straight line. 
   
EI
wL
IICIIB 48
3
 
 
EI
wLL
EI
wL
EI
wL
y IIB 384
7
248128
434







8.6 METHOD OF SUPERPOSITION 
EXAMPLE 9.03 
   
EI
wL
EI
wL
IIBIBB 486
33
 
   
EI
wL
EI
wL
yyy IIBIBB 384
7
8
44

EI
wL
B
48
7 3

EI
wL
yB
384
41 4

Combine the two solutions, 
8.6 METHOD OF SUPERPOSITION 
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. 
8.6 METHOD OF SUPERPOSITION 
EXAMPLE 9.04 
For the uniform beam and loading shown, 
determine the reaction at each support and 
the slope at end A. 
SOLUTION: 
•Release the “redundant” support at B, and find deformation. 
•Apply reaction at B as an unknown load to force zero displacement at B. 
8.6 METHOD OF SUPERPOSITION 
EXAMPLE 9.04 
•Distributed Loading: 
 
EI
wL
LLLLL
EI
w
y wB
4
3
34
01132.0
3
2
3
2
2
3
2
24




























•Redundant Reaction Loading: 
 
EI
LRL
L
EIL
R
y BBRB
322
01646.0
33
2
3













•For compatibility with original supports, yB = 0 
   
EI
LR
EI
wL
yy BRBwB
34
01646.001132.00 
 wLRB 688.0
•From statics, 
 wLRwLR CA 0413.0271.0
8.6 METHOD OF SUPERPOSITION 
EXAMPLE 9.04 
 
EI
wL
EI
wL
wA
33
04167.0
24

 
EI
wLL
L
L
EIL
wL
RA
32
2 03398.0
336
0688.0






















   
EI
wL
EI
wL
RAwAA
33
03398.004167.0  
EI
wL
A
3
00769.0
Slope at end A,