Lecture 5  2003
Twist closed sections
As this development would be almost identical to that of the open section, some of the
development is simply repeated (copied) from the open section development.
pure twist around center of rotation D => neither axial (
σ
) nor bending forces (Mx, My) act on
section
 from equilibrium 
pure twist
⌠
⌠
⌠
⋅
σ
dA
=
N
x
τ⋅
h
p
dA
=
q h
p
ds
=
T
p
⌠
σ
dA
=
0
⌡
⌡
⌡
⌡
⌠
⌠
σ⋅
y
dA
=
M.z
⌠
(
)
dA
=
q
⋅
cos
α
⌠
⌡
τ⋅
cos
α
(
)
ds
=
V
y
σ⋅
y
dA
=
0
⌡
⌡
⌡
⌠
⌠
⌠
⌡
τ⋅
sin
α
(
)
ds
=
V
z
σ⋅
z
dA
=
0
σ⋅
z
dA
=
M
y
⌠
(
)
dA
=
q
⋅
sin
α
⌡
⌡
⌡
a) equilibrium of wall element:
pure twist => .
ξ
=
η
= 0 =>
δ
v
δψ
(
)
+
δη
⋅
sin
α
=
δ
x
⋅
cos
α
δ
x
(
)
+
h
p
⋅
δφ
becomes
δ
v
=
h
D
⋅
δφ
δ
x
δ
x
δ
x
δ
x
b) compatibility (shear strain)
d
u
+
d
v
=
γ
ds
dx
here is first change. we cannot set
γ
= 0 as we did in the open problem
=>
d
u
=
γ
⋅
−
d
v
=>
d
u
=
τ
⋅ −
h
D
⋅
δφ
⌠
s
τ
δφ
⌠
ds
dx
ds
G
δ
x
u
=
ds
−
⋅
h
D
ds
+
u
0
x
and integration along s =>
G
δ
x
( )
⌡
0
⌡
⌠
δ
x
( )
as
γ
is small => = 0
for open sections
u
=
−
δφ
⋅
h
D
ds
+
u
0
x
⌡
other assumptions: section shape remains etc. same
1
notes_14_twist_closed.mcd
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δ
x
δ
x
M
x
⌠
s
τ
ds
See: Torsion of ThinWalled, Noncircular Closed Shafts; Shames Section
G
14.5 particularly; equations 14.17, 14.18 and 14.21 (Bredt's formula)
⌡
0
also: Hughes 6.1.19, 6.1.21, 6.1.22 and section 6.1
δφ
G J
δφ
⋅
⋅
⋅
M
x
=
2 q
⋅
A
q
:=
2 A
and
......
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 Spring '03
 DavidBurke
 Shear Stress, Shear strength, hD ds

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