Torsion of a Thin Rectangular Section
z
ds = − dy
n y = 0, nz = 1
∂ψ 1 d 2
=
y + z2 )
(
∂n 2 ds
∂ψ 1 d
=
y2 + z2 ) = − y
(
∂z 2 − dy
ds = dz
n y = −1, nz = 0
n y = 1, nz = 0
ds = − dz
∂ψ 1 d 2
=
y + z2 )
(
∂n 2 ds
∂ψ 1 d
=
y2 + z2 ) = −z
(
−∂y 2 − dz
∂ψ
=z
∂y
y
ds = dy
∂ψ 1 d 2
y + z2 )
=
(
∂n 2 ds
∂ψ 1 d 2
y + z2 ) = y
=
(
−∂z 2 dy
n y = 0, nz = −1
∂ψ 1 d 2
y + z2 )
=
(
∂n 2 ds
∂ψ 1 d 2
y + z2 ) = z
=
(
∂y 2 dz
z
∂ 2ψ ∂ 2ψ
∇ψ = 2 + 2 =0
∂y
∂z
2
ψ = yz
∂ψ
=z
∂y
∂ψ
=z
∂y
y
t
⎛ ∂ψ
⎞
− z⎟
⎝ ∂y
⎠
⎛ ∂ψ
⎞
+ y⎟
σ xz = Gφ ′ ⎜
⎝ ∂z
⎠
σ xy = Gφ ′ ⎜
σ xz = 2Gφ ′ y =
T = GJ eff φ ′
2Ty
J eff
σ xy = 0
2
2
⎧⎪⎛
∂ψ ⎞ ⎛
∂ψ ⎞ ⎫⎪
J eff = ∫ ⎨⎜ z −
⎟ +⎜ y+
⎟ ⎬ dA
∂y ⎠ ⎝
∂z ⎠ ⎪⎭
A⎪
⎩⎝
bt 3
2
= 4∫ y dA = 4 I z =
bt 3
J eff =
3
A
3
σ xz
τ max
Tt
=
J eff
Out of plane warping
>> z= linspace(-5,5, 20);
>> y =linspace(-0.5,0.5,10);
>> [zz, yy] = meshgrid(z,y);
>> ux=zz.*yy;
>> mesh(zz,yy,ux)
>> axis equal
>> view (50,20)
no
warping
no warping
u x = φ ′ yz
The results we obtained for the torsion of a thin rectangle can also be
used, with some qualifications, for other thin open sections such as shown in the
figure below
For example, the effective area moments for the cross sections shown can be calculated as
1
(a ) J eff = bt 3
3
1
1
(b) Jeff = b1t13 + b2 t23
3
3
1
1
1
(c ) J eff = b1 t13 + b2 t23 + b3 t33
3
3
3
Also, the maximum shear stress formula can still be applied as
τ max
where
Ttmax
=
Jeff
t max
is the largest thickness of the cross section. However, this maximum shear stress occurs on the
outer edges of the thickest section and does not account for the stress concentrations that occur at reentrant corners such as those marked with a C in Fig. 1. At such locations, the stresses depend on the
local radius of curvature of the corner and may be considerably larger than the value predicted from Eq.
(1). Such stress concentrations can be taken into account by finding either numerically or experimentally
a stress concentration factor, K, for each re-entrant corner and then examining all high stress points and
choosing the one with the highest stress, i.e.
τ max
⎡ Tt ⎤
= ⎢K
⎥
J
⎣
eff ⎦ max
t
Stress Concentrations
open section
r
τ max τ
3
2.5
t
τ max
K=
τ
2
r
1.5
τ max τ
1
closed section
0.5
0
0.5
r /t
1.0
1.5
Centerline warping of thin open sections
u x = −φ ′ω
ω = ∫ r⊥ ds + ω0
sectorial area function
ds
r⊥
dΩ
T
O
d ω = r⊥ ds
1
r⊥ ds
2
ω = 2 ∫ d Ω = 2Ω + ω0
center of twist
dΩ =
ω+
since we are taking T as
positive counterclockwise,
ω is positive if the area is
swept out in a
counterclockwise manner
ω−
To locate the center of twist, O, we must have
∫ yω dA = 0
∫ zω dA = 0
To fix
ω0
y and z are measured from the centroid
of the cross section
we can specify
∫ ω dA = o
An ω satisfying all three of the above conditions is called a principal
sectorial area function, ωp
u x = −φ ′ ω p