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Beam Theory
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Engineering Beam Theory. Let. ( ). ( ) ( ). ( ) xx xz ... applied force/unit length on beam in z- direction. [ Note: we .... Timoshenko Beam Theory neutral axis dw dx x .
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
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