In both beamVIII.3-1
theory, only stress
resultants
Timoshenko
Beams
3D
problems (1)
 1D problems !!
(sum over cross section area) are considered.
Timoshenko beam theory
x3
q
x3
x1
Elementary beam theory
(Euler-Bernoulli beam theory)
x2
4. A plane normal to the beam axis
in the undeformed state remains
symmetric axis
normal in the deformed state.
Assume: r b  0
t1  t2  0, t3 x1, x2 , x3   t3 x1, x2 , x3 
1. A plane normal to the beam axis in the undeformed state
remains plane in the deformed state.
2. All the points on a normal cross-sectional plane have the
same transverse displacement.  no thickness stretch
3. There is no stretch along the beam axis.
 u1  x3   x1 
 u1  x3 w' x1  neglect shear


u

??
deformation!!
2
 u3  wx1 
 u3  wx1 
8- 0
Institute of Applied Mechanics
 u1  x3   x1 
t1  t2  0, t3 x1, x2 , x3   t3 x1, x2 , x3 

t2  12n1   22n2   32n3
 u3  wx1 
E ij  1    ij   kk ij
u2  ??
prismatic beam: n = 0:   =
VIII.3-2 Timoshenko Beams (2)
x3
x3
q
x1
strain field:
11  x3 ,1
13  12   w,1 
x2
symmetric axis
 33  0
other  ij  ??
stress field:
11  E11    22   33 
 Ex3  ,1    22   33 
 13  2 13     w,1 
8- 1
1
equations of equilibrium:
 11,1   21,2   31,3  0
 12,1   22, 2   32,3  0
  dA
A
 13,1   23, 2   33,3  0
0    12,1   22, 2   32,3 dA
A
d
 12dA    22n2   32n3  ds

C
d x1 A
B.C.
d
Ok!!


d
A

t
d
s
12
2


C
d x1 A

geometry, loading: symmetric w.r.t.
x3-axis  12 odd function of x2
 dA  0of Applied Mechanics
Institute
A
12
 11,1   21,2   31,3  0
 12,1   22, 2   32,3  0
VIII.3-3
 13,1   23, 2   33,3  0
t1  t2  0, t3 x1, x2 , x3   t3 x1, x2 , x3 
t3  13n(3)
1   23n2   33n3
Timoshenko Beams
t1  11n1   21n2   31n3
0    13,1   23, 2   33,3 dA
A
d
 13dA    23n2   33n3  ds

C
d x1 A
d

 13dA   t3 ds

A
C
d x1

dM

  x3  21n2   31n3 ds    31dA
d x1 C
A

dM
dM
  x3t1 ds  Q 
Q
d x1 C
d x1
px1    t3 ds  d M  Q  0
C
 d x
1
dQ
prismatic beam: n1 = 0:   = 
 p  x1   0
 d x1
0   11,1   21,2   31,3  x3 dA
dQ

 p  x1 
dx1
A


d

 11x3dA    21x3 , 2   31x3 ,3   31 dA

A
d x1 A
8- 2
Q    13dA
Summary:
A
  2     w,1 dA
A
M   x3 11dA
A
  x3 Ex3 ,1    22   33 dA
A
 EI,1     22   33 x3dA
A
Institute of Applied Mechanics
Q   2     w,1 dA
d M
 d x  Q  0
1
dQ

 p  x1   0
 d x1
A
x dA (4)
 EI      Beams
VIII.3-4MTimoshenko
,1
A
22
33
3
Approximations:
1. Neglect    22   33 x3dA
A
2.
Replace  by 2 
 : shear factor, a correction factor
 is used to adjust the approximate theory to agree with the
3D theory.
When  = 0.3,  = 0.850 for rectangular cross-section and
0.886 for circular cross-section.
 dw

d  d 
 EI
   2  A
    0
dx1  dx1 
 dx1

8- 3

d  2  dw
    p  0
  A
dx1 
 dx1

Timoshenko beam equation
Institute of Applied Mechanics
d M
 d x  Q  0
1
dQ

 p  x1   0
 d x1
Q   2     w,1 dA
 dw

d  d 
2
 EI
    A
    0
dx1  dx1 
 dx1


d  2  dw
    p  0
  A
dx1 
 dx1

A
M  EI ,1
VIII.3-5 Remarks
1. Euler-Bernoulli beam theory neglects shear deformation
 2  
   dw dx1
M   EI
2
dw
dx12
d  d2w 
 EI 2 
Q
dx1  dx1 
d2  d2w 
 EI 2   px1 
2 
dx1  dx1 
2. The Timoshenko beam theory accounts for flexural as well as shear
deformation. While the Euler-Bernoulli beam theory accounts
only for flexural deformation.
3. Two B.C.s are required at both ends


8- 4
either w or Q
either dw/dx1 or M
Institute of Applied Mechanics
d M
 d x  Q  0
1
dQ

 p  x1   0
 d x1
Q   2     w,1 dA
A
M  EI ,1
VIII.3-6
Example (1)
px   t ds
1
q
x3
x1
L
cross-sectional area A
moment of inertia
I
correction factor
2
B.C.s:
 dw

d  d 
2
 EI
    A
    0
dx1  dx1 
 dx1


d  2  dw
    p  0
  A
dx1 
 dx1

w(0)  w( L)  0
M (0)  M ( L)  0

C
3

d  2  dw
    p  0
  A
dx1 
 d x1

EI    2  Aw   '  EI   p  0
p
x1  c1
EI
by M (0)  0
p 2
 
x1  c1 x1  c2
2 EI
  
p 3 c1 2
x1  x1  c2 x1  c3
6 EI
2
 dw

d  d 
pL w( x )  ??
2
 EI
    A
    0 from M ( L)  0
c1  
1
dx1  d x1 
d
x
2 EI
 1

8- 5

Institute of Applied Mechanics
p
pL
x1 
B.C.s:
EI
2 EI
w(0)  w( L)  0
p 3 pL 2

x1 
x1  c3
6 EI
4 EI
  
 dw

d  d 
 EI
   2  A
    0
dx1  d x1 
 d x1

VIII.3-7 Example (2)
pL 
p 3 pL 2
 p
 

2
EI 
x1 



A
w

x

x

c


1
1
3  0
2 EI 
6 EI
4 EI
 EI


p 3 pL 2


2 px1  pL  
w  2
x1 
x1  c3 
2  A
4 EI
 6EI

1
by w(0)  0
pL 3
 p 4


w 2
px  pLx1   
x1 
x1  c3 x1   c4
2  A
12 EI
 24 EI

1
from w( L)  0
2
1
pL3
c3  
24 EI
p 4
pL 3 pL3
 p 2 pL 
 w 2
x1  
x1 
x1 
x1
 x1 
  A 2
2  24 EI
12 EI
24 EI
1
8- 6
Institute of Applied Mechanics
p 3 pL 2
pL3

x1 
x1 
6 EI
4 EI
24 EI
p 4
pL 3
pL3
 p 2 pL 
w 2
x1  
x1 
x1 
x1
 x1 
  A 2
2  24 EI
12 EI
24 EI
1
VIII.3-8 Example (3)
p 2 pL
px1
x1  L 
M  x1 
x1 
2
2
2
Q
8- 7
dM
d  px1




x

L
1

d x1
d x1  2
L

 p x1  
2

M  EI ,1
Q   2     w,1 dA
A
d M
 d x  Q  0
1
dQ

 p  x1   0
 d x1
Institute of Applied Mechanics