Civil Engineering Reference
In-Depth Information
TABLE E8.12
Shear Stress (ksi)
Element of the Girder
z
=
0 (end)
z
=
L
/
2
=
108 in. (center)
Flange,
t
=
0.625 in.
7.45
0
Web,
t
=
0.375 in.
7.84
0
TABLE E8.13
Normal Stress (ksi)
Element of the Girder
z
=
0 (end)
z
=
L
/
2
=
108 in. (center)
Flange,
t
=
0.625 in.
0
22.27
Combined stresses (Tables E8.12 and E8.13)
:
Inthiscase,wheretherearenobearingstiffenerstotransferloadsbetween
flanges, local flexural normal stresses in the flange from the eccentric load
must also be considered and superimposed on the top flange stresses.
Example 8.4
Use the flexure analogy for torsion to find the shear and normal stresses due
to combined flexure and torsion of Example 8.3.
2.0
(
2.5
)
12.25
H
=
0.625
=
0.43 kips/ft,
−
0.43
(
216
)
2
(
12
)
8
M
H
=
=
209.0 kip-in.,
209.0
0.625
(
10
)
2
/
6
=
σ
bH
=
20.1 ksi.
, that reduce the normal lateral bending stress have
been developed as a corrective measure since the flexure analogy over-
estimates normal flange stresses due to warping. For the case of
L/a
Modification factors,
β
=
3.25
and uniform torsional moment,
t
,
β =
0.48 (Salmon and Johnson, 1980).
σ
bH
=
20.1
β =
20.1
(
0.48
)
=
9.65 ksi,
which is very close to the value of 9.81 ksi obtained in Example 8.3.
0.43
(
18
)
2
V
H
=
=
3.87 kips,
3
(
3.87
)
τ
bH
=
=
0.93 ksi.
2
[
10
(
0.625
)
]
