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

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Forms for Footings 201
Therefore, the maximum spacing of studs will be 12 in. center to center,
or 1 ft.
Determine the maximum spacing of form ties based on the horizontal
concrete pressure where the two panels are joined: the 2 × 4 top
plate of the bottom panel and the bottom plate of the top panel. Thus,
two plates will resist the pressure along this row of form ties. When
the forms are filled with concrete, the depth at this section will be 2 ft
from the top of the form. The maximum pressure from Eq. (3-1) will
be 2 ft × 150 lb per cu ft = 300 lb per sq ft. Consider a horizontal strip
12 in. above and 12 in. below this section. The lateral pressure on the
form will be 2 ft × 300 lb per sq ft = 600 lb per lin ft.
The uniformly distributed load of 600 lb per lin ft is resisted by
2 plates. Check the adequacy of plates for bending, shear and deflection.
The physical properties of 2 × 4 lumber can be obtained from
Table 4-1 as follows:
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Area, A = 2(5.25) = 10.5 in. 2
Section modulus, S = 2(3.06) = 6.12 in. 3
Moment of inertia, I = 2(5.36) = 10.72 in. 4
Referring to Tables 4-3, 4-3a, and 4-4 for 2 × 4 S4S No. 2 grade
Douglas Fir-Larch lumber adjusted for size and short-load duration
the allowable stresses will be:
Bending stress, F b
= C F
× C D
× (reference design value for bending)
= 1.5(1.25)(900 lb per sq in.)
= 1,687 lb per sq in.
Shear stress, F v
= C D
× (reference design value for shear)
= 1.25(180 lb per sq in.)
= 225 lb per sq in.
Modulus of elasticity, E = 1,600,000 lb per sq in.
The allowable span lengths for bending, shear, and deflection in
the double 2 × 4 plates can be calculated as follows:
Bending in the double plates:
Eq. (5-34), l b
= [120F b
S/w] 1/2
= [120(1,687)(6.12)/600] 1/2
= 45.4 in.
Shear in the double plates:
Eq. (5-36), l v
= 192F v
bd/15w + 2d
= 192(225)(10.5)/15(600) + 2(3.5)
= 57.4 in.
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