Twisting Induced by Axial Loads (Bi-Moments)
T
Recall, for torsion of a beam we found at the wall
xx = E 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 look

For torsion of the I-beam shown below, determine the effective polar area moment of the cross section, the maximum shearing stress ( neglecting stress concentrations) and the warping displacement of the center lines. All distances are center line distance

Combined bending of unsymmetrical beams
Internal forces and moments are defined positive as shown
z y Mz
My Mx =T Vz
Vy
x
Looking down the negative y -axis
Vz
My
O
Vz +
dVz dx dx
My + dM y dx dx
dx +
M
O
=0
dM y M y + Vz dx M y + dx = 0 dx
dM y dx
= Vz
L

Determine the effective area moment for torsion of the two-cell box
beam shown below. The thickness is t all around except for the
vertical center section, whose thickness is t/2. All distances shown
are measured from the center lines.
t
T
2L
L
t/2
L
T =

Euler-Bernoulli Bending Theory (Pure Bending Moment)
A
z
D
dw
C dx
M
neutral axis
M
x
B
ux
uz = w(x) = vertical deflection of the neutral axis
z
u x = z ( x )
If the plane AB remains perpendicular to CD
ux = z
dw
dx
dw
dx
=
dw
dx
ux = z
dw
dx
u x
d 2w
= z

If T = 6000 N-mm, G = 80 GPa, compute the angle of twist/unit length and
the maximum shearing stresses (neglecting stress concentrations) in the
closed section and the fin
2 mm
2 mm
1 mm
2 mm
1 mm
20 mm
15 mm
19 mm
Center line distances
18.5
19.5
15.5
19.

Restraint of Warping (open , thin sections)
z
warping constrained
y
warping free to occur
x
T
d
ux =
dx
if d/dx is no longer a constant, this will cause a normal stress to develop
xx = Eexx = E
d 2
= E 2
dx
ux
x
this normal stress must not produce any a

Determination of the location of the shear center through use of the
sectorial area function
Vz
r
ds
P
P
Vy
q
If the shear forces Vy and Vz act through the shear center, P, and hence
produce only bending of the beam, then the shear flow, q, that produces

Torsion of thin, open sections
shear stress distribution torque,T a a a
max
a
b
max =
J eff
t (t < b)
Tt J eff
T = GJ eff
twist/ unit length
1 = bt 3 3
Torsion of a Thin Closed Section (single cell) = area contained within the centerline of the cross s

Torsion of a Thin Rectangular Section
z
ds = dy
n y = 0, nz = 1
1 d 2
=
( y + z2 )
n 2 ds
1 d
=
( y2 + z2 ) = y
z 2 dy
ds = dz
n y = 1, nz = 0
n y = 1, nz = 0
ds = dz
1 d 2
=
( y + z2 )
n 2 ds
1 d
=
( y2 + z2 ) = z
y 2 dz
=z
y
y
ds = dy
1 d 2
=
( y +

Torsion of Bars with Circular Cross Sections A Review
The twisting of a bar of a circular cross section by an axial torque is normally considered
in strength of materials courses, as shown in Fig. 1. This problem can be analyzed by
assuming that each plan

Shear stresses in the bending of thin, unsymmetrical sections
ds
z
s
Mz
dx
y
t(s)
My
Vy
Vz
x
consider an element of a thin beam, as shown
xx
q =t
q + dq
ds
dx
dA
F
x
d xx
xx +
dx
dx
=0
d xx
dx dA xx dA = 0
( q + dq ) dx qdx + xx +
dx
d
dq = xx dA
dx
If

For the C-section shown below, using the initial point I for the integration, determine the distance, e, to the shear center and the constant, 0 , using the conditions
dA = 0 y dA = 0 z dA = 0
z
tf
h
S e
I y
tw
tf
b
A
s2
h/2
b
B
for AC
= 0 + es1
for AB