Beam Deflection 1 1

Topics: Integral, Derivative, Differential equation Pages: 10 (482 words) Published: March 25, 2015
Beam Deflection
By Touhid Ahamed

Introduction
• In this chapter rigidity of the beam will be considered
• Design of beam (specially steel beam) base on strength
consideration and deflection evaluation

Introduction
Different Techniques for determining beam deflection
• Double integration method
• Area moment method
• Conjugate-beam method
• Superposition method
• Virtual work method

Double Integration Method
The edge view of the neutral surface of a deflected beam is
called the elastic curve

1 M ( x)


EI

ρ

Double Integration Method
• From elementary calculus, simplified
for beam parameters,
d2y

2
2
1
d
y
dx


2
 
2 3 2
dx
 dy 
1    
  dx  

• Substituting and integrating,
1
d2y
EI EI 2 M  x 

dx
x

dy
EI  EI
 M  x  dx  C1
dx 
0

x

x

EI y dx M  x  dx  C1x  C2
0

0

Double Integration Method
Boundary conditions for statically determinate beam

x

x

0

0

EI y dx M  x  dx  C1 x  C2

Solved Problem

SOLUTION:
• Develop an expression for
M(x) and derive differential
equation for elastic curve.
W 14 68

I 723 in 4

P 50 kips L 15 ft

E 29 106 psi
a 4 ft

For portion AB of the overhanging
beam, (a) derive the equation for
the elastic curve, (b) determine
the maximum deflection,
(c) evaluate ymax.

• 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.

Solved Problem
SOLUTION:
• Develop an expression for M(x) and
derive differential equation for
elastic curve.
- Reactions
:
Pa
 a
RA 

L

RB P1   
 L



- From the free-body diagram for
section AD,
a
M  P

L

x

 0  x  L

- The differential equation for the
elastic curve,
EI

d2y

a


P
x
2
L
dx

Solved Problem
• Integrate differential equation twice
and apply boundary conditions to
obtain
dy elastic
1 a 2 curve.
EI

dx



2

P

L

x  C1

1 a 3
P x  C1x  C2
6 L
at x 0, y 0 : C2 0
EI y 

EI

d2y

a


P
x
2
L
dx

Substituting,
dy
1 a
1
EI
 P x 2  PaL
dx
2 L
6
EI y 

1 a 3 1
P x  PaLx
6 L
6

at x L, y 0 : 0 
dy PaL 

1 
dx 6 EI 

 x
3 
 L

2




3
PaL2  x  x  
y
   
6 EI  L  L  

1 a 3
1
P L  C1L C1  PaL
6 L
6

Solved Problem
• Locate point of zero slope or
point of maximum
deflection.
2
dy
PaL
x 
0 
1  3 m  
dx
6 EI 
 L  

3
PaL2  x  x  
y
   
6 EI  L  L  

xm 

L
0.577 L
3

• Evaluate corresponding
maximum deflection.
PaL2
ymax 
0.577   0.577  3
6 EI





PaL2
ymax 0.0642
6 EI
ymax


50 kips  48 in 180 in  2
0.0642





6 29 106 psi 723 in 4



ymax 0.238 in

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