Civil Engineering Reference
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to that which causes some short concentrically loaded compression members to
buckle torsionally (see Sections 3.7.5 and 3.11).The longitudinal bending force
σ
δ
A
=
M
y
z
I
y
δ
A
acting on an element
δ
A
of the cross-section rotates
a
0
d
φ/
d
x
(where
a
0
is the
distancetotheaxisoftwistthroughtheshearcentre
y
0
=
0,
z
0
),andsoitstransverse
component
σ
·
δ
A
·
a
0
d
φ/
d
x
exertsatorque
σ
·
δ
A
·
a
0
d
φ/
d
x
·
a
0
abouttheaxisof
twist.The total torque
T
M
exerted is
T
M
=
d
φ
d
x
a
0
M
y
z
I
y
d
A
A
in which
a
0
=
y
2
+
(
z
−
z
0
)
2
.
Thus
T
M
=
M
y
β
y
d
φ
d
x
,
(6.74)
in which the monosymmetry property
β
y
of the cross-section is given by
β
y
=
1
I
y
z
3
d
A
+
y
2
z
d
A
−
2
z
0
.
(6.75)
A
A
An explicit expression for
β
y
for a monosymmetric I-section is given in
Figure6.27, andthiscanalsobeusedfortee-sectionsbyputtingtheflangethick-
ness
t
1
or
t
2
equal to zero.Also given in Figure 6.27 is an explicit expression for
the warping section constant
I
w
of a monosymmetric I-section. For a tee-section,
I
w
is zero.
6.15 Worked examples
6.15.1 Example 1 - checking a beam supported at both ends
Problem
.The7.5mlong610
×
229UB125ofS275steelshowninFigure6.30is
simplysupportedatbothendswherelateraldeflections
v
areeffectivelyprevented
and twist rotations
φ
are partially restrained. Check the adequacy of the beam for
acentralconcentratedtopflangeloadcausedbyanunfactoreddeadloadof60kN
(which includes an allowance for self-weight) and an unfactored imposed load of
100 kN.
Design bending moment.
M
Ed
={
(
1.35
×
60
)
+
(
1.5
×
100
)
}×
7.5
/
4
=
433kNm.
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