Civil Engineering Reference
In-Depth Information
in which
M
T
and
M
B
are the flange minor axis end restraining moments
M
T
=
1
/
2
EI
z
(
d
2
v
/
d
x
2
)
L
/
2
+
(
d
f
/
4
)
EI
z
(
d
2
φ/
d
x
2
)
L
/
2
M
B
=
1
/
2
EI
z
(
d
2
v
/
d
x
2
)
L
/
2
−
(
d
f
/
4
)
EI
z
(
d
2
φ/
d
x
2
)
L
/
2
,
and
R
2
and
R
4
arethedimensionlessminoraxisbendingandwarpingendrestraint
parameters.Ifthebeamissymmetricallyrestrained,thensimilarconditionsapply
at the end
x
=−
L
/
2.The other support conditions are
(
v
)
±
L
/
2
=
(φ)
±
L
/
2
=
0.
The particular case for which the minor axis and warping end restraints are
equal, so that
R
2
=
R
4
=
R
(6.51)
may be analysed. The differential equilibrium equations for a buckled position
v
,
φ
of the beam are
EI
z
d
2
v
d
x
2
=−
M
cr
φ
+
(
M
B
+
M
T
)
and
d
x
−
EI
w
d
3
φ
GI
t
d
φ
d
x
3
=
M
cr
d
v
d
x
.
These differential equations and the boundary conditions are satisfied by the
buckled shape
,
M
cr
φ
π
2
EI
z
/
k
cr
L
2
=
A
cos
π
x
k
cr
L
−
cos
π
v
=
2
k
cr
in which the effective length factor
k
cr
satisfies
1
−
R
=−
π
R
2
k
cr
cot
π
2
k
cr
.
(6.52)
The solutions of this equation are shown in Figure 6.18c.
6.14 Appendix - monosymmetric beams
When a monosymmetric beam (see Figure 6.27) is bent in its plane of symmetry
and twisted, the longitudinal bending stresses
σ
exert a torque which is similar
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