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

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Design of Wood Members for Formwork 151
Live load of workers and tools = 50.0 lb per sq ft
Design load = 143.0 lb per sq ft
Plywood Decking to Resist Vertical Load
Because the type and thickness of plywood have already been
chosen, it is necessary to determine the allowable span length of
the plywood, that is, the distance between joists, such that the
bending stress, shear stress, and deflection of the plywood will be
within allowable limits. Table 4-9 provides the section properties
for the 7 ⁄8-in.-thick plywood selected for the design. The plywood
will be placed with the face grain across the supporting joists.
Therefore, the properties for stress applied parallel to face grain
will apply. The physical properties for a 1-ft-wide strip of the 7 ⁄8-in.-
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thick plywood installed with the face grain across supports are as
follows:
Approximate weight = 2.6 lb per sq ft
Effective thickness for shear, t = 0.586 in.
Cross-sectional area, A = 2.942 in. 2
Moment of inertia, I = 0.278 in. 4
Effective section modulus, S e
= 0.515 in. 3
Rolling shear constant, Ib/Q = 8.05 in. 2
Table 4-10 indicates the allowable stresses for the plywood of
Group II species and S-2 stress rating for normal duration of load.
However, values for stresses are permitted to be increased by 25% for
short-duration load conditions, less than 7 days, as discussed in
Chapter 4 (Table 4-4). The 25% increase does not apply to the modulus
of elasticity. For this design, it is assumed that a wet condition will
exist. Following are the allowable stresses that will be used for the
design:
Allowable bending stress, F b
= 1.25(820) = 1,025 lb per sq in.
Allowable rolling shear stress, F s
= 1.25(44) = 55 lb per sq in.
Modulus of elasticity, E = 1,300,000 lb per sq in.
Bending Stress in Plywood Decking
Because the plywood will be installed over multiple supports, three
or more, Eq. (5-41) can be used to determine the allowable span length
of the plywood based on bending.
From Eq. (5-41), l b
= [120F b
S e
/w b
] 1/2
= [120(1,025)(0.515)/(143)] 1/2
= 20.0 in.
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