Twisting Induced by Axial Loads (Bi-Moments)
T
Recall, for torsion of a beam we found at the wall
σ xx = − Eω p
Tk
tan ( kl )
GJ eff
Since the stress is proportional to the principal sectorial area function,
for the I-beam this normal stress distribution looks like:
σ xx = 0
T
Thus, we see that the stresses in the flanges produce a selfequilibrated set of moments, called bi-moments:
T
If torsion (twisting) can generate bi-moments which are self-equilibrated
axial stresses, then axial stress distributions that generate bi-moments
should induce twisting
Thus, consider a thin, open section subjected to axial loads that
generate both axial extension (or shortening) and twisting:
σ xx
x
Then the axial displacement, by superposition is
u x = −ω p ( y, z ) β ( x ) + uo ( x )
If we assume σxx is the only significant normal stress, then
σ xx = E
du x
du
dβ
= − Eω p
+E 0
dx
dx
dx
Let P = the axial load. Then
P = ∫ σ xx dA = − E
A
=0
du0
dβ
dA
EA
ω
+
p
dx ∫A
dx
so we have
du0
P
=
dx AE
just our usual axial load relationship
Also note that
∫ yσ xx dA = − E
A
dβ
dx
∫ yω p dA + E
A
du0
ydA = 0
dx ∫A
if y is measured from the centroid since then
∫ yω dA = o
p
A
∫ ydA = 0
A
Similarly
∫ zσ xx dA = − E
A
du0
dβ
z
dA
E
zdA = 0
ω
+
p
dx ∫A
dx ∫A
if z is also measured from the centroid.
Thus, these axial stresses do not generate bending, only extension and
twisting
Now, define the bi-moment, M Γ , as
M Γ = ∫ σ xxω p dA
A
=0
Then
M Γ = −E
du0
dβ
2
ω
dA
+
e
ω p dA
p
∫
∫
dx A
dx A
Jω
M Γ = − E Jω
σ xx = − Eω p
and our stress
becomes
dβ
dx
σ xx =
M Γω p ( y, z )
Jω
du
dβ
+E 0
dx
dx
P
+
A
Note: if the stress distribution is purely uniform (constant) over the cross
section then no twisting will be induced since
M Γ = ∫ σ xxω p dA = σ xx ∫ ω p dA = 0
A
A
However, other axial stress distributions that generate a bi-moment will
induce twisting (and extension)
If σxx generates a bi-moment, how do we find the twisting, β(x), this
stress generates?
Answer: consider the relationship
MΓ
dβ
= −
dx
EJω
as a boundary condition on the end where the load is applied
z
For example:
y
σ xx
L
d 2β
2
−
k
β =0
2
dx
boundary conditions
β ( 0) = 0
−M
dβ
( L) = Γ
dx
E Jω
(since T = 0)
x
Consider a specific case of the cross-section shown below loaded by a
concentrated axial force, P = 100 kN, acting through the centroid, O.
Determine the twist, φ(x), and the stress distribution at the wall.
100 mm
z
O
P = 100 kN
O
y
5 mm all
around
L=3m
100 mm
Note: O is also the shear center
E = 200 GPa = 2 × 105 N / mm 2
G / E = 0.36
200 mm
s
B
d
Take initial integration point at A with ω = ω0
A
s
for AB
ω = ω0 + hs / 2
h/2
for BC
O
for CD
h/2
d
ω = ω0 + hd / 2
ω = (ω0 + hd / 2 ) − hs / 2
C
D
s
Setting
∫ ω dA = 0
A
gives
d
h
d
0
0
0
t ∫ (ω0 + hs / 2 ) ds + t ∫ (ω0 + hd / 2 ) ds + t ∫ (ω0 + hd / 2 − hs / 2 ) ds = 0
− hd ( h + d ) − ( 200 )(100 ) ( 200 + 100 )
=
ω0 =
2 ( h + 2d )
2
( 200 + 200 )
= −7.5 × 103 mm 2
principal sectorial area function, ωp
2.5x103 mm2
-7.5x103 mm2
2.5x103 mm2
1
1
J eff = 2 × d t 3 + h t 3 = 0.17 × 105 mm 4
3
3
-7.5x103 mm2
J ω = ∫ ω p2 dA = 2.08 × 1010 mm6
A
k=
G J eff
E Jω
= 5.4 × 10−4 mm −1
Since T = 0
d 2β
2
−
k
β =0
2
dx
β ( x ) = A sinh ( kx ) + B cosh ( kx )
Boundary conditions
ux ( 0 ) = 0 → β ( 0 ) = 0 → B = 0
d β ( L ) −M Γ ( L )
=
dx
E Jω
−M Γ ( L )
Ak cosh ( kL ) =
E Jω
A=
−M Γ ( L )
E k J ω cosh ( kL )
M Γ ( L ) = ∫ σ xx ( y, z ) ω p ( y, z ) dA
A
= ω p ( 0, 0 )
∫
A( 0,0 )
σ xx dA
= Pω p ( 0, 0 ) = 2.5 × 103 P
N − mm
−M Γ ( L )
A=
E k J ω cosh ( kL )
−2.5 ×103 P
=
= −4.23 × 10−5 mm −1
EkJ ω cosh ( kL )
Thus,
β ( x ) = −4.23 × 10−5 sinh ( 5.4 × 10−4 x ) mm −1
Integrating
β=
dφ
dx
Boundary condition
φ ( x) =
A
cosh ( kx ) + C
k
φ ( 0) = 0 → C = − A / k
A
φ ( x ) = ⎡⎣cosh ( kx ) − 1⎤⎦ rad
k
= −0.08 ⎡⎣ cosh ( 5.4 × 10−4 x ) − 1⎤⎦ rad
at the free end
φ ( L ) = −0.13 rad
At the fixed wall
dβ
( 0 ) E Jω
dx
= 95 ×106 N − mm 2
M Γ ( 0) = −
P M Γω p
σ xx = +
A
Jω
N-mm2
mm2
95 ×106 ) ω p
(
105
=
+
( 400 )( 5) 2.08 ×1010 mm6
15.7 MPa
= 50 + 4.57 × 10−3 ω p
61.4 MPa
15.7 MPa
MPa