Formwork Homework #2
Design the formwork for a wall 8-ft high Use 1” Plyform Assume 2 x 4 studs and 2 x 4 double wales, and braces Limit deflection to l/360 Tie capacity 3000# Rate of placement = 6 lb/ft2 Temperature = 80 oF Design load for forms: Since Rate of placement = 6 lb/ft2 < 7 lb/ft2 P = 150 + 9000 R/T = 150 + (9000)(6)/80 = 825 lb/ft2 Checking for limiting pressure: P… = 150h = (150)(8) = 1200 lb/ft < 2000 lb/ft2 Since P < P… P = 825 lb/ft2 Material properties:
2
Lumber E = 1.6 x 106 x 0.97 = 1.55 x 106 psi Fb = 1400 x 1.25 x 0.86 = 1505 psi Fc = 975 x 1.25 x 0.70 = 853 psi Fc⊥ = 565 x 1.25 x 0.67 = 473 psi Fv = 90 x 1.25 x 0.97 = 109 psi Ft = 825 x 1.25 x 0.84 = 866 psi Plyform I = 0.427 S = 0.737 IB/Q= 8.871 E = 1.65 x 106 Fb = 1930 Fv = 80 Find maximum spacing of studs = maximum span of plyform Assume 12-in wide strip of plywood, 3 spans
Design load: w = 1 x 825 = 825 lb/ft Bending: lb= 4.46d (Fb b/w)1/2 =10.95[(1930)(0.737)/825]1/2 = 14.4 in. Shear: ls= 20 Fv (IB/Q)/w + 2d = 20 [80 (8.871)/825] + 2(1) = 19.2 in. Deflection: ld = 1.69 (E I /w)1/3 = 1.69 [(1.65 x 106)(0.427)/825]1/3 = 16 in. Bending controls. Maximum stud spacing is 14.4 inches. Use Plywood-96” long. Spacing 96 96 = = 6.85" . Use 7 spacings. = 13.7" . Use 14” for design. 14 7
Find maximum spacing of wales (double 2x4’s) = maximum span of studs. Assume 3 spans.
Design load: P = (14/12) x 825 = 962.5 lb/ft Bending: lb = (4.46)(3.5)[(1505)(1.5)/962.5]1/2 =23.9” Shear: ls = [(13.3)(109)(1.5)(3.5)/962.5] + (2)(3.5) = 14.9” Deflection: ld = (1.69) [1.55 x 106(5.36)/962.5]1/3 = 34.7”
Shear governs. Maximum span is 14.9 inches. For 8’ wall, 96”/14 = 6.44. So use 7 spacings of wales. 96/7 = 13.67, so use 14” for design. Find maximum spacing of ties = maximum span of wales. Assume 3 spans, double wales.
Design load: w = 825 x (14/12) = 962.5 lb/ft Bending: lb = (4.46)(3.5)[1505(3.0)/962.5]1/2 = 33.8” Shear:
ls = 13.3[(109)(3)(3.5)/962.5] + (2)(3.5) = 22.8” Deflection: ld = 1.69 [1.55 x 106(5.359)(2)/962.5]1/3 = 43.7”
Check maximum spacing of ties based on tie strength: l = (3000/962.5) x 12 = 37.4”
max
Shear governs. Maximum tie spacing is 22.8”. For a modular value, use tie spacing of 19.2”. (8 x 12 = 96”, 96/4 = 24”, 96/5 = 19.2” – use). Check crushing of studs against wales. Wale spacing is 14 inches. Force of stud against wales at point of support: P =(14/12) x 962.5 = 1122.9 lb Unit stress at point of contact: F = 1122.9/(2 x 1.5 x 1.5) = 249.5 psi < 473 psi O.K.
c⊥
Design lateral bracing: P=qGCp Cp = 0.80 G = 0.85 – Exposure condition C & D q = 0.00256KzKztV2I Kz = 1.03 …………….. Table 6-3 Kzt = 1.0 V = 90 I=1.0 q = 0.00256(1.03)(1)(1)(90)2(1) = 21.35 PSF P = 21.35(0.85)(0.80) = 14.5 PSF Use 15 PSF
15 PSF x 1 = 15 PLF
R1 6' 8'
R2 2'
Assume 1’ width R2 = 15 x (2 + 6/2) = 75 PLF
l = 7.81 x 12 = 93.72”
l 93.72 = = 62.48 > 50 d 1.5
6' 7.81
Brace at mid point l = 31.24 d So long column So
5'
Fc′ =
0.30 E (l / d ) 2
0.30(1.55 × 106 ) = 31.242 = 476 PSI Pmax = 476(1.5)(3.5) = 2501lb ⎛ 5 ⎞ PH = 2501⎜ ⎟ ⎝ 7.81 ⎠ = 1601lb Spacing = 1601lb 75 PLF = 21 ft.