Civil Engineering Reference
In-Depth Information
Solution for maximum stress
. The shear centre (see Section 5.4.3) load of 6 kN
acting in the plane of the long leg of the angle has principal plane components of
6cos19.82
◦
=
5.645 kN parallel to the
z
axis, and
6sin19.82
◦
=
2.034 kN parallel to the
y
axis,
asindicatedinFigure5.41d.Theseloadcomponentscauseprincipalaxismoment
components at the support of
M
y
=−
5.645
×
1.5
=−
8.467 kNm about the
y
axis, and
M
z
=+
2.034
×
1.5
=+
3.052 kNm about the
z
axis.
The maximum elastic bending stress occurs at the point A (
y
1
A
=
15.1 mm,
z
1
A
=−
97.4 mm) shown in Figure 5.41d. The coordinates of this point can be
obtained by using equation 5.61 as
y
=
15.1cos19.82
◦
−
97.4sin19.82
◦
=−
18.8 mm, and
z
=−
15.1sin19.82
◦
−
97.4cos19.82
◦
=−
96.7 mm.
The maximum stress atAcan be obtained by using equation 5.12 as
σ
max
=
(
−
8.467
×
10
6
)
×
(
−
96.7
)
842.6
×
10
4
−
(
3.052
×
10
6
)
×
(
−
18.8
)
120.5
×
10
4
=
146.9 N/mm
2
Thestressesattheotherlegendsshouldbecheckedtoconfirmthatthemaximum
stress is atA.
Solution for maximum deflection
.Themaximumdeflectionoccursatthetipofthe
cantilever. Its components
v
and
w
can be calculated using
QL
3
/3
EI
.Thus
w
=
(
5.645
×
10
3
)
×
(
1500
)
3
/(
3
×
210,000
×
842.6
×
10
4
)
=
3.6 mm,
and
v
=
(
2.034
×
10
3
)
×
(
1500
)
3
/(
3
×
210,000
×
120.5
×
10
4
)
=
9.0 mm.
The resultant deflection can be obtained by vector addition as
δ
=
√
(
3.6
2
+
9.0
2
)
=
9.7 mm
Using non-principal plane properties
.Problemsofthistypearesometimesincor-
rectly analysed on the basis that the plane of loading can be assumed to be a
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