6.0
ELASTIC DEFLECTION OF BEAMS
6.1
Introduction
6.2
Double-Integration Method
6.3
Examples
6.4
Moment Area Method
6.5
Examples
Introduction
P
x
P
y
Elastic curve
The deflection is measured from the original neutral axis to the neutral axis of the
deformed beam.
The displacement y is defined as the deflection of the beam.
It may be necessary to determine the deflection y for every value of x along the
beam. This relation may be written in the form of an equation which is frequently
called the equation of the deflection curve (or elastic curve) of the beam
Importance of Beam Deflections
A designer should be able to determine deflections, i.e.
In building codes ymax <=Lbeam/300
Analyzing statically indeterminate beams involve the use of various deformation relationships.
Methods of Determining Beam Deflections
a) Double-Integration Method
b) Moment-Area Method
c) Elastic Energy Methods
d) Method of singularity functions
Double-Integration Method
The deflection curve of the bent beam is
d2y
EI 2  M
dx
In order to obtain y, above equation needs to be integrated twice.
y
r
Radius of
curvature
x
y
EI
M
r
1 M


  (Curvature )
r EI
An expression for the curvature at any point along the curve representing the
deformed beam is readily available from differential calculus. The exact formula
for the curvature is 2

d y
dx 2
  dy  2 
1    
  dx  
3
2
dy
is small
dx
d2y
  2
dx
d2y
EI 2  M
dx
The Integration Procedure
Integrating once yields to slope dy/dx at any point in the beam.
Integrating twice yields to deflection y for any value of x.
The bending moment M must be expressed as a function of the coordinate x
before the integration
Differential equation is 2nd order, the solution must contain two constants of
integration. They must be evaluated at known deflection and slope points
(i.e. at a simple support deflection is zero, at a built in support both slope
and deflection are zero)
Sign Convention
Positive Bending
Negative Bending
Assumptions and Limitations
Deflections caused by shearing action negligibly small compared to bending
Deflections are small compared to the cross-sectional dimensions of the beam
All portions of the beam are acting in the elastic range
Beam is straight prior to the application of loads
y
Examples
L
x
x
PL
P
P
M  PL  Px
d2y
EI 2  M
dx
d2y
EI 2   PL  Px
@x
dx
dy
x2
EI
  PLx  P  c1
Integrating once
dx
2
2

dy
0
 0  EI 0   PL0  P
 c1  c1  0
@x=0
dx
2
2
3
PLx
x
Integrating twice
EIy  
 P  c2
2
6 3
PL 2
@ x = 0 y  0  EI 0   
0  P 0  c2  c2  0
2
6
PLx 2
x3
EIy  
P
2
6
@ x = L y = ymax
EIymax
 max
PL L2
L3
PL3
PL3

P 
 ymax  
2
6
6
3EI
PL3

3EI
y
W N per unit length
x
x
WL2
2 WL
L
d2y
W
2
EI 2   L  x 
dx
2
@x
W
2
M   L  x 
2
2
d y
EI 2  M
dx
dy W L  x 
EI

 c1
dx 2
3
3
Integrating once
dy
W L  0 
WL3
 0  EI 0 
 c1  c1  
dx
2
3
6
3
@x=0

dy W
WL3
3
EI
 L  x  
dx 6
6
W L  x  WL3
EIy  

x  c2
6
4
6
4
Integrating twice
W L  0  WL3
WL4
0  c2  c2 
y  0  EI 0   

6
4
6
24
4
@x=0
W
WL3
WL4
4
EIy   L  x  
x
24
6
24
Max. occurs @ x = L
EIymax
W L4 WL4
WL4
WL4



 ymax  
6
24
8
8EI
 max
WL4

8 EI
y
Example
x
x
L
WL
2
WL
2
WL
x
M
x  Wx
2
2
2
d y WL
x2
EI 2 
x W
dx
2
2
dy WL x 2 W x 3
EI


 c1
Integrating
dx
2 2 2 3
L
dy
@ x
0
Since the beam is symmetric
dx
2
3 2
L
L
 
 
WL3
L
WL  2  W  2 
@ x
EI 0 

 c1  c1  
24
2
2
2
2 3

dy WL 2 W 3 WL3
EI

x  x 
dx
4
6
24
Integrating
WL x 3 W x 4 WL3
EIy 


x  c2
4 3 6 4
24
WL 0  W 0  WL3
@ x = 0 y = 0  EI 0  
0  c2


4 3
6 4
24
3
4
WL 3 W 4 WL3
 EIy 
x 
x 
x
12
24
24
Max. occurs @ x = L /2
EIymax
5WL4

384
 max
5WL4

384 EI
 c2  0
y
Example
x
P
x
P
2
L/2
L/2
P
2
L
P
M x
2
2
2
d y P
L
EI 2  x
for 0  x 
dx
2
2
dy P x 2
EI

 c1
Integrating
dx 2 2
L
dy
@ x
0
Since the beam is symmetric
2
dx
2
L
 
PL2
L
P 2
c1  
@ x
EI 0 
 c1 
16
2
2 2
for 0  x 

dy P 2 PL2
EI
 x 
dx 4
16
P x 3 PL2
EIy 

x  c2
4 3 16
Integrating
P 0  PL2
0  c2
 EI 0  

4 3
16
3
@x=0 y=0
P 3 PL2
 EIy  x 
x
12
16
Max. occurs @ x = L /2
EIymax
PL 3

48
 max
PL3

48 EI
 c2  0
Moment-Area Method
First Moment –Area Theorem
r
The first moment are theorem
states that: The angle between the
tangents at A and B is equal to the
area of the bending moment diagram
between these two points, divided by
the product EI.
d
A
B
ds

B
dx
M

dx
EI
A
d

x
M
The second moment area theorem states that: The vertical distance of point B on a deflection
curve from the tangent drawn to the curve at A is equal to the moment with respect to the vertical
through B of the area of the bending diagram between A and B, divided by the product EI.
M
EI
r
Mx
dx
A EI
M

ds  rd  r 
M
d 
dx integratin g will give
EI
ds
d
B
d 
B
EI
ds
M
   d   dx
EI
A
it is small lateral deflection s replace ds with dx
Mx
xd 
dx
EI
B

Mx
dx
EI
A

The Moment Area Procedure
1. The reactions of the beam are determined
2. An approximate deflection curve is drawn. This curve must be consistent with
the known conditions at the supports, such as zero slope or zero deflection
3. The bending moment diagram is drawn for the beam. Construct M/EI diagram
4. Convenient points A and B are selected and a tangent is drawn to the assumed
deflection curve at one of these points, say A
5. The deflection of point B from the tangent at A is then calculated by the second
moment area theorem
Comparison of Moment Area and Double Integration Methods
If the deflection of only a single point of a beam is desired, the moment-area
method is usually more convenient than the double integration method.
If the equation of the deflection curve of the entire beam is desired the
double integration method is preferable.
Assumptions and Limitations
Same assumptions as Double Integration Method holds.
Examples
PL
A
L
Tangent at A

?
P
P
B
Tangent at B
M
PL
L
PL
 2L 
EI    PL   
2
3
 3 
L
EI    PL 
2
3
PL 3

3EI
PL2

2 EI
A
W N per unit length
Tangent A
=?
WL2
2 WL
1 WL2
A
L
3 2
3
x L
4
L
B
x
WL2
2
L  W 2  3 
WL4
EI     L   L   
3  2  4 
8
WL 4

8EI
Example
a
P
P
a
=?
A
P
L
P
Tangent A
Pa
L
a
2
a
2
2

 P 3
L
La
La
a
L
 L a
 Paa 2

   a
EI   Pa  a    a  
a  Pa 
4
4
2 3
2 3
8
2
 4 2

PaL2 Pa3 PL3  3a 4a 3 



  3 
8
6
24  L
L 
PL3

24 EI
 3a 4a 3 
  3 
L 
L