Civil Engineering Reference
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and equal to the applied torque
T
. In this case, a particular integral is
p
(
x
)
=
Tx
GI
t
,
while the boundary conditions require that
0
=
A
1
+
A
2
+
A
3
0
=
T
GI
t
+
A
1
a
−
A
2
,
a
0
=
A
1
a
2
e
L
/
a
+
A
2
a
2
e
−
L
/
a
so that
e
(
x
−
2
L
)/
a
−
e
−
x
/
a
+
1
−
e
−
2
L
/
a
φ
=
Tx
Ta
GI
t
(
1
+
e
−
2
L
/
a
)
GI
t
−
.
At the fixed end, the uniform torque is zero because d
φ
/d
x
=
0. Because of
this,thetorque
T
x
thereisresistedsolelybythewarpingtorque,andthemaximum
warping shear stresses occur at this point. The value of d
2
φ
/d
x
2
is greatest at the
fixedend,andthereforethevalueofthewarpingnormalstressisalsogreatestthere.
Thevalueofd
φ
/d
x
isgreatestattheloadedendofthecantilever,andthereforethe
valueoftheshearstressduetouniformtorsionisalsogreatestthere.Thewarping
torque at the loaded end is
2
e
−
L
/
a
1
+
e
−
2
L
/
a
,
T
wL
=
T
which decreases from
T
to zero as
L
/
a
increases from 0 to
∞
. Thus the warp-
ing shear stresses at the loaded end are not zero, but are less than those at the
fixed end.
10.9 Worked examples
10.9.1 Example 1 - approximations for the serviceability
twist rotation
Problem.
ThecantilevershowninFigure10.28is5mlongandhastheproperties
shown in Figure 10.37. If the serviceability end torque
T
is 3 kNm, determine
approximate values of the serviceability end twist rotation either by
(a) assuming that the cantilever is in uniform torsion (
EI
w
≡
0), or
(b) assuming that the cantilever is in warping torsion (
GI
t
≡
0), or
(c) using the approximation of equation 10.56.
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