Lecture 4  2003
Pure Twist
pure twist around center of rotation D => neither axial (
σ
) nor bending forces (Mx, My) act on
section; as previously, D is fixed, but (for now) arbitrary point.
as before:
a) equilibrium of wall element:
d
q
+
d
σ
⋅
t
=
0
ds
dx
b) compatibility (shear strain)
d
u
+
d
v
=
γ
=
0
small deflections
ds
dx
c) tangential displacement (
δ
v)
in terms of
η
,
ζ
and
φ (
geometry)
δ
v
=
δη
⋅
cos
α
()
+
h
p
⋅
δφ
+
δζ
⋅
sin
α
δ
x
δ
x
δ
x
δ
x
N.B. h
p
=>h
D
from definition of problem
further assumptions:
1) preservation of cross section shape =>
ζ
=
ζ
(x);
η
=
η
(x)
φ
=
φ
(x)
2) shear though finite is small ~ 0 =>
d
u
=
−
d
v
ds
dx
3) Hooke's law holds
=>
σ
=
E
⋅
δ
u
axial stress
δ
x
 from equilibrium 
pure twist
⌠
⌠
⌠
⋅
σ
dA
=
N
x
τ⋅
h
p
dA
=
qh
p
ds
=
T
p
⌠
σ
dA
=
N
x
=
0
⌡
⌡
⌡
⌡
⌠
⌠
σ⋅
y
dA
=
M.z
⌠
dA
=
q
⋅
cos
α
⌠
⌡
cos
α
ds
=
V
y
y
dA
=
−
M
z
=
0
⌡
⌡
⌡
⌠
⌠
⌠
⌡
sin
α
ds
=
V
z
z
dA
=
M
y
=
0
z
dA
=
M
y
⌠
dA
=
q
⋅
sin
α
⌡
⌡
⌡
pure twist also => only
φ
is finite i.e. other displacements (and derivatives)
ζ
=
η
= 0 =>
δ
v
δη
+
δζ
⋅
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
using negligible shear assumption
d
u
=
−
d
v
=>
d
u
=
−
h
D
⋅
δφ
and integration along s =>
ds
dx
ds
δ
x
1
notes_13_pure_twist.mcd
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u
=
−
δφ
⋅
h
D
ds
+
u
0
x
δ
x
()
⌡
⋅
which showed u linear with y and z => plane sections plane.
previously
u
=
−η
'
⋅
Y
− ζ
'Z
+
u
0
x
⌠
here  only if h
D
is constant so it can come outside
h
D
1 ds

is u (longitudinal
⌡
displacement) linear. u is defined as warping displacement (function).
stress analysis can be made analogous for torsion and bending IF the integrand
h
D
⋅
ds
thought to
be a coordinate. calculation of stresses will involve statical moments, moments of inertia and
products of inertia which will be designated "sectorial" new coordinate =
Ω
Ω
wrt arbitrary origin and
ω
wrt normalized sectorial coordinate (as before like wrt center of area)
d
Ω
=
h
D
⋅
ds
=
d
ω
the warping function then becomes:
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 Spring '03
 DavidBurke
 Shear Stress, ........., σ, zd

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