Civil Engineering Reference
In-Depth Information
Figure G6-6
Unbalanced load
and elevation for
Example 6-1.
Example 6-2
Unbalanced Snow Load, Wide
Gable Roof
Problem
Determine the unbalanced load for the structure in Example 4-3.
Solution
Because the roof has a slope of ½ on 12, unbalanced loads need to be con-
sidered. From Example 4-3, it has been determined that
p
s
is 25 lb/ft
2
and
p
g
is 30 lb/ft
2
.
Since
W
7.5 lb/
ft
2
, while the leeward side has a balanced load of
p
s
that is 25 lb/ft
2
plus a
rectangular surcharge. Since the eave-to-ridge distance
W
is 130 ft and
p
g
is
30 lb/ft
2
,
>
20 ft, the windward-side load is 0.3
p
s
=
0.3(25)
=
h
=
0 43
.
3
Wp
+− =
10
1 5
.
( .
0 43
)
3
130
30
+
10
−
1 5
.
=
3
.
97 ft
4
4
d
g
The unit weight of snow is given by Equation 7-3,
0.13(30 lb/ft
2
)
17.9 lb/ft
3
γ=
0.13
p
g
+
14
=
+
14
=
and the roof slope is 1 on 24 (
S
=
24). Hence, the intensity of the rectangular
surcharge is
h
γ
397
.
ft
(
179
24
.
lb/ft
3
)
d
=
=
14 5
.
lb/ft
2
S
