Restraint of Warping (open , thin sections)
z
warping constrained
y
warping free to occur
x
T
dφ
ux = − ω
dx
if dφ/dx is no longer a constant, this will cause a normal stress to develop
σ xx = Eexx = E
d 2φ
= −E ω 2
dx
∂ux
∂x
this normal stress must not produce any axial force or bending moments
∫ σ xx dA = Eφ ′′ ∫ ω dA = 0
A
A
∫ yσ xx dA = Eφ ′′ ∫ yω dA = 0
A
A
∫ zσ xx dA = E φ ′′ ∫ zω dA = 0
A
A
which says that the sectorial area function must be a principal sectorial
area function and we have
d 2φ
σ xx = −E ω p 2
dx
σ xx = 0
Note: these stresses are self-equilibrated but
they do not rapidly decay from the ends ( thus,
they violate Saint Venant's principle)
However, a changing axial stress will require that additional shear
flows be developed (recall how we obtained the VQ/It expression for
shear stress in bending)
∑F
x
=0
⎛
⎛
∂q ⎞
∂σ xx ⎞
⎜ σ xx +
dx ⎟ tds − σ xx tds + ⎜ q(s) + ds⎟ dx − q(s)dx = 0
⎝
⎠
⎝
∂x
∂s ⎠
∂σ
∂q
= −t xx
∂s
∂x
d 3φ
d 2β
= E 3 t ωp = E 2 t ωp
dx
dx
β = φ′ =
dφ
dx
integrating gives
d βs
q(s) = E 2 ∫ ω p tds
dx 0
2
d β
= E 2 ∫ ω p dA
dx
2
which is the shear flow due to restraint of warping. The torque
associated with this shear flow is
s =s f
Tq = ∫ r⊥ q ds
s =0
sf
= ∫ q dω p
0
Thus, the total torque is
T = TSV + Tq
sf
= GJeff β + ∫ q dω p
0
Integration by parts lets us express this in the form
⎛ ∂q ⎞
T = GJeff β + qω p s= 0 − ∫ ω p ⎜ ⎟ ds
⎝ ∂s ⎠
0
s= s f
sf
d 2β
= GJeff β − E 2 ∫ ω 2p dA
dx A
and defining
Jω = ∫ ω 2p dA
A
gives
where
d2 β
T
2
2
−
k
β
=
−
k
dx 2
GJeff
GJeff
k =
EJω
2
differential equation
for β(x)
If the torque T = constant, solution is
β = C sinh(kx) + D cosh(kx) +
T
GJeff
C, D are constants. We need to find these from the boundary conditions
Fixed end:
β =0
Unconstrained end:
( ux = 0 )
dβ
=0
dx
( σxx = 0 )
z
y
x
L
β x= 0 = 0
dβ
dx
which gives
⇒ D=−
=0 ⇒ C=
x= L
β=
T
T
GJeff
T
tanh (kL)
GJeff
T
[tanh(kL)sinh(kx ) − cosh(kx) + 1]
GJ eff
β=
dφ
T
=
⎡⎣ tanh ( kL ) sinh ( kx ) − cosh ( kx ) + 1⎤⎦
dx GJ eff
To get the twist we must integrate, using the additional boundary condition
φ(0) = 0. We obtain at x = L
L
φ (L) = ∫ β dx =
0
TL
GJ eff
⎡ tanh (kL) ⎤
⎢⎣1 − kL ⎦⎥
The axial stress in the bar is
σ xx = −E ω p
dβ
Tk
= −E ω p
[tanh(kL)cosh(kx) − sinh(kx)]
dx
GJ eff
which is a maximum at the fixed end
σ xx x = 0
Tk
= − Eω p
tanh(kL)
GJeff
varies along the section
Problem: Explain how
d2 β
T
2
2
−
k
β
=
−
k
dx 2
GJeff
can be used to solve this problem for φ = φ(x) :
T1
T2
x
a
T = T1 − T2
0<x<a
b
T
β = β1 = A sinh ( kx ) + B cosh ( kx ) +
a<x<a+b
T1 − T2
GJ eff
T = −T2
β = β 2 = C sinh ( kx ) + D cosh ( kx ) −
T2
GJ eff
To find A, B, C, D
β1 ( 0 ) = 0
( ux = 0)
d β2 ( a + b )
=0
dx
(σxx = 0)
solve for A,B,C,D
β1 ( a ) = β 2 ( a )
(continuity of ux )
d β1 ( a ) d β 2 ( a )
=
dx
dx
(continuity of σxx )
in 0 < x < a, integrate to get twist
φ = φ1 = ∫ β1dx + C1
in a < x < a + b integrate to get twist
φ = φ2 = ∫ β 2 dx + C2
Find C1 , C2 from
φ1 ( 0 ) = 0
φ1 ( a ) = φ2 ( a )