8.1 Introduction
*beam deflections
- bending deformation: deflection due to bending moment = dominating deformation of beam
- shear deformation: deflection due to shearing force  negligible
*bending deformation of a beam = deformation of neutral surface
- curve of the neutral surface: y  y (x )  elastic curve or deflection curve
- deflection (y): distance between elastic curve and undeformed position (x-axis)
dy
 tan   
deflection angle ():
dx
 slope of the elastic curve  angle between x-axis and tangential line of elastic curve
*methods of deflection analysis
- methods using the geometry of elastic curve
- methods using strain energy
*differential equation for elastic curve:
 double integration method, moment-area method
d2y
M
: 2nd-order ordinary differential equation

2
dx
EI
8.2 Double Integration Method
8.2.1 Differential Equations for Elastic Curve
* curvature of elastic curve (Eq. 7.2.5):
- Use ds  d ,  
dy
,
dx
ds  dx
1
M
EI



1


d d d 2 y


ds dx dx 2
2
- cf (elementary calculus) :
1


d y
dx 2
  dy  2 
1    
  dx  
3/ 2
1 d2y
 dy

   1 

 dx 2
 dx

* differential equation for deflection curve
- Positive moment causes negative curvature: use negative sign in moment.
d2y
M

: 2nd-order ordinary differential equation

2
dx
EI
* solution of DE
- solution method: double integration
- determination of integration constants(2)  boundary conditions(BC) or compatibility conditions
* interpretation of solution
- deflection: positive y (+y)  downward
- deflection angle: positive  (+)  clockwise
8.2.2 Boundary and Compatibility Conditions
*physical meaning of derivatives
y : y = deflection
dy
  = deflection angle
y :
dx
d2y
y  : EI 2   M  bending moment
dx
d3y
dM
y  : EI 3  
 V  shearing force
dx
dx
d4y
dV
 w  load
y IV : EI 4  
dx
dx
*boundary conditions at supports
- simple support (hinge or roller): y  0
- fixed support: y  0 & y     0
*compatibility conditions: continuity of deflection or slope of elastic curve
- continuity on deflection and slope at any beam section: y x0  y x0
- continuity on deflection at internal hinge: y x0  y x0
&
dy
dy

dx x0 dx x0
8.2.3 Examples
(1) Cantilever beam – uniform load
 Identify BCs: fixed support at B ( x  L) : y ( L)  0 & y ( L)  0
 bending moment: M   12 wx 2 , 0  x  L
 DE for deflection curve: EIy    M  12 wx 2
 integration of DE: EIy '  16 wx3  C1
 Impose BC: y ( L)  0  C1   16 wL3
deflection angle: EIy '  16 wx 3  16 wL3 , or,
 integration & impose BC: EIy 
wL4
 deflection: y 
24 EI
1
24

dy
wL3

dx
6 EI
  x 3 
1    
  L  
wx 4  16 wL3 x  C2 [ y ( L)  0 )] C2   81 wL4
 x  4
x 
   4   3
 L  
 L 
 maximum deflection and deflection angle at free end, or x  0 :  max  
wL3
,
6 EI
ymax 
wL4
8EI
(2) Simple beam – uniform load
 Identify BCs: simple supports at A ( x  0) and B ( x  L) : y (0)  0 & y ( L)  0
 bending moment: M  12 w( Lx  x 2 ) , 0  x  L
 DE for deflection curve: EIy    M   12 w( Lx  x 2 )
 integration of DE: EIy '   12 w( 12 Lx 2  13 x 3  C1 ) ,
EIy   12 w( 16 Lx 3  121 x 4  C1 x  C2 )
 Impose BC: C2  0 , C1   121 L3
  x 3  x 2 
4   6   1
 L
  L 

4
3
wL4  x 
 x   x 

2
deflection: y 
    
 
24 EI  L 
 L   L 
deflection angle:  
dy
wL3

dx 24 EI
 maximum deflection angle at A ( x  0) and B ( x  L) :  max   A   B 
maximum deflection at mid-span, or x  12 L : ymax 
5wL4
wL4

384 EI 76.8EI
wL3
24 EI
(3) Cantilever beam – concentrated load
 Identify BCs:
fixed support at A ( x  0) :
y (0)  0 & y (0)  0
 bending moment:
 P( a  x ), 0  x  a
M 
, axL
 0
*piece-wise
function
continuous
 DE for deflection curve:
EIy    M
 P( a  x ), 0  x  a

, axL
 0
  1 P(a  x ) 2  C1 , 0  x  a
 integration of DE: EIy '   2
, axL
 C2
 Impose BC: y (0)  0  C1  12 Pa 2 , but C2 cannot be determined yet!!!
 1 P(a  x )3  12 Pa 2 x  C1 , 0  x  a
 integration again: EIy   6
, axL
 C2 x  C4
 Impose BC: y (0)  0  C3   16 Pa 3 , but C2 & C4 cannot be determined yet!!!
Use compatibility conditions to determine C2 & C4 :
y ( a  0)  y ( a  0)  C2  C1  12 Pa 2 ;
y ( a  0)  y ( a  0)  C4  C3   16 Pa 3


 P
 (a  x )2  a 2 , 0  x  a
dy  2 EI

 deflection angle:  
2
dx  Pa
, axL
 2 EI
 P
3
2
3
 6 EI ( a  x )  3a x  a , 0  x  a
deflection : y  
P

3a 2 x  a 3
, axL
 6 EI




 special case of a  L (a concentrated load at the free end)
2
2
3
dy PL2   x   x  
PL3   x   x  

y
- slope & deflection :  
 2      ,
3     
dx 2 EI   L   L  
6 EI   L   L  
PL3
PL2
- maximum values at free end, or x  L :  max 
, ymax 
3EI
2 EI
(4) Simple beam – concentrated load
 Identify BCs: simple supports at A ( x  0) and B ( x  L) : y (0)  0 & y ( L)  0
Compatibility conditions: point of concentrated load: y ( a  0)  y (a  0) , y (a  0)  y (a  0)
 Pb
x
, 0 xa

 bending moment: M   L
Pb
 x  P( x  a ), a  x  L
 L
*piece-wise continuous function
 Pb
x
, 0 xa

 DE for deflection curve: EIy    M   L
Pb

x  P( x  a ), a  x  L
 L
 Pb 2
x  C1
, 0 xa

 integration of DE: EIy '    2 L
Pb 2 P

x  ( x  a ) 2  C2 , a  x  L
 2L
2
 Compatibility condition: y ( a  0)  y (a  0)  C1  C2
 Pb 3
x  C1 x  C3
, 0 xa

 integration again: EIy    6 L
Pb 3 P

x  ( x  a ) 3  C1 x  C4 , a  x  L
6
 6L
 Compatibility condition: y (a  0)  y (a  0)  C3  C4
Impose BC: y (0)  0  C3  C4  0 , y ( L)  0  C2  C1 
Pb 2
( L  b2 )
6L
 Pb 2
 2 2
, 0 xa
dy  6 EIL L  b  3x

 deflection angle:  
Pb 2
P
dx 
x  a 2 , a  x  L

L  b 2  3x 2  
2 EI
 6 EIL
Pb

2
2
3
, 0 xa
 6 EIL L x  b x  x 
deflection : y  
Pb 2
P

x  a 3 , a  x  L

L x  b2 x  x 3  
6 EI
 6 EIL
 maximum values (in case of a  b )
- slope:  max   B  
- ymax 
Pb
L2  b2 3 / 2 @ x 
9 3EIL
* cf: y ( L / 2) 

Pab
(L  a) @ x  L
6 EIL
y ( 12 L)
ym a x
1
3
( L2  b2 )
Pb

3L2  4b 2 
48EI
( b  0 )  = 0.974 ;
or, 0.974    1
 special case of a  b  12 L (a concentrated load at the mid-span)
2
PL2 
x 
- slope & deflection 0  x  L :  
1  4   ,
16 EI 
 L  
1
2
- maximum values:  max   A   B 
PL2
@ x0&L ;
16 EI
3
PL3   x   x  
y
3   4  
48EI   L   L  
ymax 
PL3
@ x
48EI
1
2
L
(5) Simple beam – moment load
 Identify BCs: simple supports at A & B:
y ( 0)  0 & y ( L )  0
Compatibility conditions:
point of concentrated moment:
y ( a  0)  y ( a  0) ,
y ( a  0)  y ( a  0)
 bending moment:
 M0
, 0 xa
  L x
M 
M
 0 ( x  L) , a  x  L
 L
*piece-wise continuous function
 DE for deflection curve:
 M0
x
, 0 xa

EIy    M   L
M
 0 ( x  L) , a  x  L
 L
 integration of DE and use of BCs & CCs to get deflection angle and deflection:
 M0
2
2
2
, 0 xa
dy  6E I L3x  L  3b


dx  M 0 3( x  L) 2  L2  3a 2 , a  x  L
 6E I L
 M0 3
2
2
, 0 xa
 6 EIL x  ( L  3b ) x
y 
M
 0 ( x  L)3  ( L2  3a 2 )( x  L) , a  x  L
 6 EIL








 special case of a  0 & b  L [a concentrated moment at the support A ]
(6) Simple beam – member-end moment load: M 0  M A (clockwise)
- deflection (angle):
2

M AL   x 

3  1  1 ;
6 EI   L 

- support rotation:  A 
M AL
,
3EI
B  
M AL
6 EI
y
M A L2
6EI
 x  x  x

 L  L  1 L  2 


 
8.3 Moment-Area Method
8.3.1 Derivation of Moment-Area Theorems
d 
M
dx
EI
M
dx
A EI
    d  
= area of
B
M
-diagram
EI
dt  xd
B
 t   dt   xd   x
A
M
dx
EI
= First area-moment of
M
-diagram
EI
[Theorem 8.3.1 : The first moment-area theorem]
The angle between tangents drawn at A and B on the elastic curve
is equal to
the area of the corresponding portion of the bending moment diagram, divided by EI.
  BA   AB  AM
[Theorem 8.3.2 : The second moment-area theorem]
The vertical deviation of point B on the elastic curve away from a tangent drawn at A
is equal to
the moment of the area of the bending moment diagram with respect to B, divided by EI.
t  tBA  xB AM
8.3.2 Application of Moment-Area Theorems – Moment-area method
(1)
M
-diagram:
EI
BMD modified with EI
M
dx ,
x A EI
(2) Area and centroid: AM  
xB
xB  b
(3) Sketch of elastic curve
(4) Use of theorems
 BA   AB  AM (1st theorem);
 A   B   BA
 t AB  aAM , tBA  bAM (2nd theorem);  A (a  b)  ( B  tBA )   A ,
(5) Signs
 B (a  b)  ( A  t AB )   B
8.3.3 Examples
Cantilever beam: Use the fact that the tangent at fixed support is the beam axis itself.
(1) Cantilever beam – concentrated load at free end
 AB   B
1  PL 
PL2
L  
2  EL 
2 EL
[clockwise w.r.t l A  AB ]
 AB   
tBA   B
1  PL   2 
PL3
tBA   
L
L


 

2  EL   3 
3EL
[B is below l A  AB ]
(2) Cantilever beam – partial uniform load
Simple beam: Draw the tangents at both supports and use simple geometry.
(3) Simple beam – concentrated load
 the first moment-area theorem for  AB
 the second moment area theorem for tBA
 simple geometry: tBA  L A

t BA
L
from the first moment-area theorem)
A 
 deflection angle:    A   Ax (  Ax
 deflection: y  x A  t xA ( t xA from the second moment-area theorem)
8.4 Beams with Variable Section
M
due to variable EI
EI
* Either the double-integration method or the moment-area method may be used.
* piece-wise continuous function of
[<Example 8.4.1>] Cantilever of step sections
(i) double-integration method
 Px
d 2 y  EI ,
- DE:

dx 2  Px ,
EI
0 xa
axL
- BC & CC: y ( L)  0 , y ( L)  0
&
y ( a  0)  y ( a  0) ,
y ( a  0)  y ( a  0)
- integration of DE and application of BC and CC
2
 PL2   x  2

a

    (1   )   1 , 0  x  a
 L
dy  2EI   L 



2
2

dx  PL  x 
 1
, axL



 2EI  L 




 PL3

 6EI
y 
3
 PL
 6EI

2
3
  x 3

a x x
a
    3(1   )     3   2(  1)   2 , 0  x  a
 L  L  L
 L
  L 

3
 x 
x 
, axL
   3   2
 L  
 L 
(ii) moment-area method
8.5 Method of Superposition
[<Example 8.5.1>] Simple beam with equally-spaced concentrated loads
[<Example 8.5.2>] Cantilever with displacement constraint
[<Example 8.5.3>] Simple beam with half-span uniform load
[<Example 8.5.4>] Simple beam with partially-loaded uniform load
Table 8.5.1 Loads and Deflections of Various Beams
Loads on beams – Max moments – Max slope at ends – Deflections – Max deflections