Civil Engineering Reference
In-Depth Information
from which
S
y
I
z
×
10
6
q
s
,0
=−
×
1
.
04
Taking moments about the mid-point of the wall AD we have
2
150
0
800
−
S
y
z
s
=
786
q
OB
d
s
A
+
294
q
BA
d
s
B
(vi)
0
Noting that
q
OB
=
q
b,OB
+
q
s
,0
and
q
BA
=
q
b,BA
+
q
s
,0
we rewrite Eq. (vi) as
150
2
S
y
I
z
4
s
2
A
−
10
6
)d
s
A
S
y
z
s
=
786(
+
1
.
4
×
0
800
15
16
s
B
−
10
6
)d
s
B
+
294(
+
1500
s
B
+
0
.
95
×
(vii)
0
Integrating Eq. (vii) and eliminating
S
y
gives
z
s
=
282 mm
.
PROBLEMS
P.10.1
A cantilever has the inverted T-section shown in Fig. P.10.1. It carries a vertical
shear load of 4 kN in a downward direction. Determine the distribution of vertical
shear stress in its cross-section.
0
.
004(44
2
y
2
)N
/
mm
2
,
0
.
004(26
2
y
2
)N
/
mm
2
.
Ans.
In web:
τ
=
−
in flange:
τ
=
−
10 mm
60 mm
10 mm
40 mm
F
IGURE
P.10.1
P.10.2
An I-section beam having the cross-sectional dimensions shown in Fig. P.10.2
carries a vertical shear load of 80 kN. Calculate and sketch the distribution of vertical
shear stress across the beam section and determine the percentage of the total shear
load carried by the web.
