Euler-Bernoulli Bending Theory (Pure Bending Moment)
A
ψ
z
D
dw
C dx
M
neutral axis
M
x
B
ux
uz = w(x) = vertical deflection of the neutral axis
z
u x = − zψ ( x )
If the plane AB remains perpendicular to CD
ux = − z
dw
dx
dw
dx
ψ
ψ=
dw
dx
ux = − z
dw
dx
∂u x
d 2w
= −z 2
ε xx =
∂x
dx
If we assume
σ yy = σ zz = σ xy = σ yz = 0
The stress-strain relations give
ε xx =
1
⎡σ xx −ν (σ yy + σ zz ) ⎤
⎦
E⎣
d 2w
σ xx = − E z 2
dx
⎛ ∂u x ∂u z
+
∂x
⎝ ∂z
σ xz = G ⎜
⎞
⎛ dw dw ⎞
=
G
−
⎜
⎟=0
⎟
⎝ dx dx ⎠
⎠
z
d 2w
σ xx = − E z 2
dx
y
σ xx
M = − ∫ σ xx z dA
M
A
=E
xx
d w 2
z dA
dx 2 ∫A
dA = 0
A
∫ σ xx y dA = 0
A
x
A
2
d 2w
= EI 2
dx
∫σ
y
z
σ xx = −
Mz
I
∫ zdA = 0
neutral axis is at centroid
A
∫ y z dA = 0
A
cross-section must be symmetric
Engineering Beam Theory
z
x
qz
M ( x) z
σ
=
−
Let
xx
I
V ( x)Q ( z )
σ xz = −
I t ( z)
M
V
[ Note: we still have u x = − z
σ xz = 0
dM
2
= V ( x)
d
w
dx
M = EI 2
dx
dV
= qz ( x )
dx
dw ⎛
dw ⎞
=
ψ
⎜
⎟
dx ⎠
dx ⎝
(inconsistent) ]
d 4w
EI 4 = qz ( x )
dx
qz … applied force/unit length on beam in zdirection
so that
dM
= V ( x)
dx
dV
= qz ( x )
dx
How are these internal force and bending moment equilibrium relations
related to our local equilibrium equations?
∂σ xx ∂σ xy ∂σ xz
+
+
=0
∂x
∂y
∂z
∂σ xy
∂x
+
∂σ yy
∂y
+
∂σ yz
∂z
=0
∂σ xz ∂σ yz ∂σ zz
+
+
=0
∂x
∂y
∂z
∂σ xx ∂σ xy ∂σ xz
+
+
=0
∂x
∂y
∂z
multiply by z and integrate over the cross-section, A
∫z
A
⎛ ∂σ
∂σ xx
∂σ
dA + ∫ z ⎜ xy + xz
∂x
∂y
∂z
A ⎝
⎞
⎟ dA = 0
⎠
⎡∂
⎤
d
∂
z
σ
dA
+
z
σ
+
z
σ
(
)
(
)
xx
∫A ⎢⎣ ∂y xy ∂z xz ⎥⎦ dA − ∫A σ xz dA = 0
dx ∫A
or, equivalently
- V(x)
- M(x)
z
∂f
∫A ∂y dA = vC∫ f ny ds
n
nz
ny
A
C
y
∂f
∫A ∂z dA = vC∫ f nz ds
Gauss’ theorem (2-D)
−
dM
+ ∫ z ( n yσ xy + nzσ xz ) ds + V ( x ) = 0
dx v
C
Tx(
z
n)
n
y
Tx( ) = 0
n
dM
= V ( x)
dx
Now, consider
∂σ xz ∂σ yz ∂σ zz
+
+
=0
∂x
∂y
∂z
integrating over A
⎛ ∂σ yz ∂σ zz
d
σ xz dA + ∫ ⎜
+
∫
dx A
∂y
∂z
A⎝
-V(x)
⎞
⎟ dA = 0
⎠
−
dV
+ ∫ (σ yz n y + σ zz nz ) ds = 0
dx v
C
Tz( n )
n
Tz( n )
z
y
( )
T
z
v∫ ds = qz ( x )
n
C
applied force/unit length in zdirection
dV
= qz ( x )
dx
Last remaining equilibrium equation is:
∂σ xy
∂x
+
∂σ yy
∂y
Integrating over A gives
+
∂σ yz
∂z
=0
⎛ ∂σ yy ∂σ yz
d
dA
+
+
σ
⎜
xy
∫
∫
dx A
∂y
∂z
A⎝
⎞
⎟ dA = 0
⎠
Vy
dVy
dx
+ v∫ (σ yy n y + σ yz nz ) ds = 0
C
Ty( n )
( )
T
y
v∫ ds = q y ( x )
n
( n)
n
Ty
C
z
y
Vy
applied force/unit length in ydirection
dVy
dx
= −q y ( x )
which is identically satisfied if
Vy = 0, Ty( ) = 0
n
Timoshenko Beam Theory
A
ψ
z
D
dw
C dx
M
neutral axis
M
x
B
ψ
u x = − zψ ( x )
dψ
dx
⎛ ∂u ∂u
σ xz = G ⎜ x + z
∂x
⎝ ∂z
ψ ( x) ≠
dw
dx
dw
dx
σ xx = − E z
dw ⎞
⎞
⎛
=
−
ψ
+
G
x
(
)
⎜
⎟
⎟
dx ⎠
⎝
⎠
better than Euler/Bernoulli but still a constant across the
cross-section so introduce a form factor κ 2
σ xx = − E z
dψ
dx
dw ⎞
⎛
⎟
dx ⎠
⎝
= σ zz = σ xy = σ yz = 0
σ xz = κ 2G ⎜ −ψ ( x ) +
σ yy
For bending moment and shear force
M = − ∫ zσ xx dA = E
A
dψ
dx
2
z
∫ dA = EI
A
dψ
dx
dw ⎞
⎛
V = − ∫ σ xz dA = −κ 2G ⎜ −ψ +
⎟ ∫ dA
dx ⎠ A
⎝
A
dw ⎞
⎛
= −κ 2G ⎜ −ψ +
⎟ A = −σ xz A
dx ⎠
⎝
Timoshenko Beam theory
dψ
M = EI
dx
V ( x)
dw
−ψ ( x ) = − 2
dx
κ GA
Example:
Euler-Bernoulli Theory
d 2w
M = EI 2
dx
dw
ψ=
dx
P
P
x
M = - Px
dψ
= − Px
dx
Px 2
+ C1
EIψ = −
2
x
L
V=-P
EI
ψ ( L) = 0 ⇒ ψ =
This gives
σ xx =
Pxz
I
P ⎡⎣ L2 − x 2 ⎤⎦
rotation of the beam cross-section
2 EI
(same as ordinary beam theory)
dw
V
=− 2
+ψ
slope of neutral axis
κ GA
dx
P ( L2 − x 2 )
P
= 2
+
κ GA
2 EI
Integrating
P ( L2 x − x3 / 3)
Px
+
+ C2
w= 2
κ GA
2 EI
PL
PL3
+
+ C2 = 0
w( L) = 0 ⇒
2
κ GA 3EI
which gives
PL3
PL
C2 = w(0) = −
− 2
3EI κ GA
deflection due to:
− PL3
w ( 0) =
3EI
3E I ⎤
⎡
+
1
⎢⎣ κ 2G AL2 ⎥⎦
For a rectangular section of base b and height h
I
1 ⎛h⎞
=
⎜ ⎟
AL2 12 ⎝ L ⎠
2
bending
shear