Formwork for Concrete Structures by R.L.Peurifoy and G.D- By EasyEngineering.net

388 Chapter Thirteen
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density of 150 lb per cu ft will be (4/12 ft)(150 lb per cu ft) = 50 lb per
sq ft. Combining this dead load with the live load of 50 lb per sq ft,
the total load will be 100 lb per sq ft. For a total load of 100 lb per sq
ft, Table 4-13 indicates an allowable span length of 24 in. for ¾-in.
Class I Plyform with the face grain across supports. Therefore, the
joists will be spaced at 24 in. on centers.
Space the Ribs
Consider using 2 × 6 S4S joists. With joists spaced at 24 in., or 2 ft, on
centers, the uniformly distributed load on the joists will be 2 ft × 100 lb
per sq ft = 200 lb per lin ft. Table 4-17 indicates a maximum span of
73 in. for 2 × 6 S4S No. 2 grade Southern Pine lumber with an equivalent
uniform load of 200 lb per lin ft. For constructability use a span
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of 72 in., or 6 ft, between ribs.
Design the Ribs
Set the spacing of the shores under the ribs at 5 ft 0 in., measured
horizontally. The spans for the ribs near the crown will be 5 ft 0 in.,
whereas the spans near the outer edges of the roof will be slightly
greater. Consider each rib member to be a simple beam 5 ft, or 60 in.,
long with a maximum of three concentrated loads from the joists,
providing bending moment in the ribs. For the loading shown in
Figure 13-4, the bending moment at the center of the rib is calculated
as follows:
Load P from each joist = (100 lb per sq ft )(2 ft)(6 ft)
= 1,200 lb
Total load on each rib = 3(1,200 lb)
= 3,600 lb
Reaction at each end of rib = (3,600 lb)/2
= 1,800 lb
FIGURE 13-4 Loads on a rib member.
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