Civil Engineering Reference
In-Depth Information
Detailed design of a web plate
:
V
0.35
F
y
≥
486
0.35
(
50
)
≥
27.8 in.
2
(
OK, 53.13 in.
2
provided
)
.
A
w
≥
Flexural buckling
:
4.18
E
4.18
29,000
h
t
w
=
85
0.625
=
136
≤
f
c
≤
22.6
≤
150.
Therefore,nolongitudinalstiffenersarerequiredforwebflexuralbuckling
stability. This was also shown in the calculation relating to the compression
flange vertical buckling above.
Shear buckling
:
2.12
t
w
E
2.12
(
0.625
)
29,000
50
h
=
85
≥
F
y
≥
≥
31.9 in.
Therefore, transverse web stiffeners are required.
Combined bending and shear
:
0.75
F
y
9814
(
12
)
4469
1.05
f
v
F
y
f
b
=
=
26.3
≤
−
0.75
50 ksi
1.05
486/53.13
50
≤
−
=
27.9.
However,
f
b
=
27.5 ksi OK.
This interaction criteria does not generally require checking, and in the
case where
f
v
/(
0.35
F
y
)
=
0.55
F
y
=
by the web plate is approximately
((
53.1
/
6
)/(
53.1
/
6
15.0 %
of the total moment, combined bending and shear need not be considered.
Flange-to-web connection
:
Top weld
:
+
(
20
)(
2.5
)))
100
=
VQ
f
I
486
(
50.0
)(
43.75
)
223,444
2
1.80
W
S
w
2
2
1.80
(
40
)
3
(
12
)
2
=
q
+
=
+
4.62
2
2.00
2
=
+
=
5.04 k/in.
0.5
√
2
(
5.04
)/(
0.35
(
50
))
≥
For a CJP weld, the weld size must be
≥
0.20 in. OK
since the web thickness is 0.625 in.
Bottom weld
:
Δ
VQ
f
I
(
310
)(
50.0
)(
43.75
)
201,100
Δ
q
=
=
=
3.37 k/in.
