Civil Engineering Reference
In-Depth Information
TABLE E8.4
Flexural Normal Stress,
σ
b
(ksi)
Element of the Girder
z
=
0 (end)
z
=
L
/
2
=
108 in. (center)
Top Flange,
t
=
0.75 in.
0
−
5.58
32
−
0.75
30.5
2
(
0.44
)
30.5
4
191.8 in.
3
,
Q
web
=[
(
12
)(
0.75
)
]
+
=
2
17.5
(
135.5
)
5430
(
0.75
)
=
τ
b flange
(
0
)
=
0.58 ksi,
17.5
(
191.8
)
5430
(
0.44
)
=
τ
bweb
(
0
)
=
1.40 ksi,
Combined stresses (
Table E8.6)
:
Both pure,
τ
w
, shear stresses due to torsion are relatively
small, but warping normal stress (18.0 ksi) is large.
τ
t
, and warping,
Example 8.2
Use the flexure analogy (
Figure E8.2)
for torsion to find the shear and normal
stresses due to combined flexure and torsion of Example 8.1.
35
(
6
)
32
H
=
0.75
=
6.72 kips,
−
6.72
(
216
)
4
M
H
=
=
362.9 kip-in.,
362.9
σ
bH
=
=
20.2 ksi.
0.75
(
12
)
2
/
6
[
]
TABLE E8.5
Flexural Shear Stress,
τ
b
(ksi)
Element of the Girder
z
=
0 (end)
z
=
L
/
2
=
108 in. (center)
Flange,
t
=
0.75 in.
0.58
0.58
Web,
t
=
0.44 in.
1.40
1.40
TABLE E8.6
Shear Stress (ksi)
Element of the Girder
z
=
0 (end)
z
=
L
/
2
=
108 in. (center)
Flange,
t
=
0.75 in.
3.97
1.14
Web,
t
=
0.44 in.
3.10
1.40
