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

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390 Chapter Thirteen

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Because the allowable shear stress F v

of 218 lb per sq in. is larger

than the applied shear stress of 194.5 lb per sq in., the 2 × 10 S4S size

rib is satisfactory in shear. However, due to required curvature the rib

members must be sawn from 2 × 12 size lumber. Therefore the stresses

must be checked for a 2 × 12, and the net depth of the member must

be checked to ensure it is adequate.

The maximum depth that will be removed from the ends of a rib

member can be determined by applying Eq. (13-3). For the required

curve, R is approximately 24.5 ft and x is 2.5 ft, measured from the

midpoint of a member. Let the depth to be removed equal y.

y=R−

R −x

ww.EasyEngineering.n

2 2

= 24.5 – ( 24. 5) − ( 2. 5)

= 0.128 ft, or 1.54 in.

2 2

The loss in depth at the ends of a 2 × 12 rib member will be 1.54 in.,

which will leave a net depth of approximately 11.25 in. – 1.54 in. = 9.7 in.

Because the 9.7 in. is greater than the 9.25 in. for a 2 × 10, it is satisfactory.

Because the 2 × 10 will be cut from a 2 × 12, the allowable bending

stress for a 2 × 12 must be checked against the section modulus for a

2 × 10. The allowable bending stress from Tables 4-2 and 4-4 will be:

Allowable bending stress, F b

= C D

× (reference bending stress)

= 1.25(975)

= 1,218 lb per sq in.

The actual bending stress can be calculated from Eq. (5-7) as follows:

Actual bending stress, f b

= M/S

= 25,200 in.-lb/21.39 in 3

= 1,178 lb per sq in.

The allowable bending stress of 1,218 lb per sq in. is greater than

the applied bending stress of 1,178 lb per sq in. Therefore cutting the

curvature from 2 × 12 to make a 2 × 10 rib is satisfactory.

Determine the Load on the Shores

The circular shell roof formwork system in Figure 13-2 will be spaced

at 5 ft in the horizontal direction. The spacing of shores in the 30-ft

direction will be 6 ft on center. Therefore the area supported by each

shore will be 5 ft × 6 ft = 30 sq ft. The load will be:

P = (100 lb per sq ft)(30 sq ft)

= 3,000 lb

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