6.5 narrow rectangular cross section
• If section is long and narrow soap bubble will be
almost independent of long coordinate
The Prandtl soap-film analogy
• Soap bubble equation
p
∂ z ∂ z
=
−
+
2
2
S
∂y
∂x
2
2
• By comparing to the
torsion equation get
φ
z
2Gθ S
=
z
, φ=
p / S 2Gθ
p
Taking advantage of analogy
• Soap film deflection
2
⎡
⎛ y ⎞ ⎤
z = z 0 ⎢1 − ⎜ ⎟ ⎥
⎝ h ⎠ ⎥⎦
⎢⎣
2 z0
∂2z ∂2z
+ 2 =− 2
2
∂x
∂y
h
• Get z0=ph2/2S and
2
⎡
⎛ y⎞ ⎤
2
φ = Gθ h ⎢1 − ⎜ ⎟ ⎥
⎢⎣ ⎝ h ⎠ ⎥⎦
Stresses and torsional constant
• Differentiate for stresses
σ zx =
∂φ
∂φ
= − 2 G θ y , σ zy = −
=0
∂y
∂x
τ max = 2Gθ h, for y = ± h
• Integrate for torque
b h
1
T = 2 ∫ ∫ φ dx dy = G θ (2b)(2h) 3 = G" J "θ
3
−b − h
1
" J " = (2b)(2h) 3
3
• When section gets narrower for constant
area, how do J and “J” change?
• Altogether
τ max =
3T
T
3T
2Th
θ
=
,
=
=
G (2b)(2h) 3 G" J "
(2b)(2h) 2 " J "
Cross sections made up of multiple
narrow rectangles
• You just straighten them up!
1
" J"= C
3
τ max
n
∑ (2b )(2h )
i =1
i
3
i
2Thmax
T
=
, θ =
" J"
G" J "
• If we have a rectangle, would we gain by halving
the thickness on half the length and increasing it
by 50% on the other half?
Example
• For the section shown calculate the maximum
shear stress and the angle of twist per unit
length when the member is subjected to torque
T = 100 in.lb. G=12Mpsi
Solution : Total length of the member, 2b = 2 x(1.125 - 0.05) + 1.5
0.05 ”
2b = 3.65 in and 2h = 0.05 in
100 in-lb
b
1.125 ”
= 73 > 10 ⇒ rectangula r approximat ion used.
h
3T
3(100)
=
= 32,876 ksi
2
2
1.5 ”
(2b)(2h) (3.65)(0.05)
G = 12 x 10 psi
3T
3(100)
θ=
=
= 0.0548 rad/in
3
6
3
G (2b)(2h)
12 x 10 (3.65)(0.05)
τ max =
6
Torsion of rectangular cross section
members
• Geometry:
φ = Gθ ( h − x
2
2
)−
32 G θ h 2
π3
∞
∑
( − 1)
n = 1, 3 , 5 ,...
• How fast does this converge?
( n −1 )
2
nπ x
nπ y
cosh
2h
2h
nπ b
3
n cosh
2h
cos
Bringing to nondimensional form
• Stiffness and stress
" J " = k1 (2h) 3 (2b)
1 ⎡ 192 ⎛ h ⎞
k1 = ⎢1 − 5 ⎜ ⎟
3⎣ π ⎝b⎠
∞
∑
n = 1, 3 , 5 , ....
nπ b ⎤
1
tanh
⎥
5
n
2h ⎦
k2
8
= 1− 2 ∑
π n = 1,3,5,.....
k1
τ max = 2Gθ hk2 / k1
⎛
⎞
⎜
⎟
1
⎜
⎟
π
n
b
⎜ n2 cosh
⎟
2h ⎠
⎝
Torsional Parameters for Rectangular Cross Sections
b/h
1
1.5
2
2.5
3
4
6
10
∞
k1
0.141
0.196
0.229
0.249
0.263
0.281
0.299
0.312
0.333
k2
0.208
0.231
0.246
0.256
0.267
0.282
0.299
0.312
0.333
Hollow thin-wall torsion members
and multiply connected cross sections
• Hollow sections much more efficient than open
ones
• Compare a hollow cylinder of radius R and
thickness t to same cross section with a slit cut
open
1
J cut = (2π r )t 3
3
2
J pipe
r
⎛ ⎞
= 3⎜ ⎟
J cut
⎝t⎠
J pipe = 2π r 3t
Reading assignment
Section 6.7: Question: For a hollow cross section do we gain
by reducing the thickness on half the perimeter and
increasing it on the other half? Why?
Source: www.library.veryhelpful.co.uk/ Page11.htm