Civil Engineering Reference
In-Depth Information
is shown in Figure 4.16b, where the nodes are 1 and 2, the axis of the cylinder is
the
x
axis, and twisting moments are positive according to the right-hand rule.
From elementary strength of materials, it is well known that the angle of twist per
unit length of a uniform, elastic circular cylinder subjected to torque
T
is given by
T
JG
=
(4.70)
where
J
is polar moment of inertia of the cross-sectional area and
G
is the shear
modulus of the material. As the angle of twist per unit length is constant, the total
angle of twist of the element can be expressed in terms of the nodal rotations and
twisting moments as
TL
JG
x
2
−
x
1
=
(4.71)
or
JG
L
T
=
(
x
2
−
x
1
)
=
k
T
(
x
2
−
x
1
)
(4.72)
Comparison of Equation 4.72 with Equation 2.2 for a linearly elastic spring and
consideration of the equilibrium condition
M
x
1
+
M
x
2
=
0
lead directly to the
element equilibrium equations:
JG
L
1
x
1
x
2
M
x
1
M
x
2
−
1
=
(4.73)
−
11
so the torsional stiffness matrix is
1
JG
L
−
1
[
k
torsion
]
=
(4.74)
−
11
While this development is, strictly speaking, applicable only to a circular cross
section, an equivalent torsional stiffness
J
eq
G
/
L
is known for many common
structural cross sections and can be obtained from standard structural tables or
strength of materials texts.
Adding the torsional characteristics to the general beam element, the element
equations become
u
1
u
2
v
1
z
1
v
2
z
2
w
1
y
1
w
2
y
2
x
1
x
2
f
x
1
f
x
2
f
y
1
M
z
2
f
y
2
M
z
2
f
z
1
M
y
1
f
z
2
M
y
2
M
x
1
M
x
2
[
k
axial
]
[0]
[0]
[0]
[0]
[
k
bending
]
xy
[0]
[0]
=
(4.75)
[0]
[0]
[
k
bending
]
xz
[0]
[0]
[0]
[0]
[
k
torsion
]
