Week 2 Challenge
- Plate size = 260 x 20mm ; Bolt size = 20mm; hole dia = 20+2= 22mm
a) Plate drilled hole
A = (b – nd) x t where { b – width of plate; n – no of bolts ; d – hole dia of bolt; t-thickness of plate}
= (260 – 4x22) x 20 = 3440 mm2
b) Plate : Zigzag bolting
A = (b –∑d +∑p2/4g) x t
minimum net section area along BCEG = 3648mm2
c) Angle (200 x 100 x 12 with M20 bolt)
Width of section b = 200+100 = 300mm; dia of hole = 20+2 = 22mm; t =12mm
Net area = {(200+100)-(3*22)+[50^2/4x90 + 50^2/4x130]} =2949mm2.
2. Design Of Sag Rod
Total Load = DL +LL = (60 +75) *9.8*8*1.2= 12700 N
Total load on sag rod =(12700/3 +100*9.8*8/3) sin15 = 1772 N
Tdn = 0.72 Fu An
An = Tdn/ 0.72 Fu = 1772 /(0.72 * 250) = 9.84
An = (pi/4)*d*d
d = (An*4/pi)^ 0.5 = ( 9.84*4 /3.14)^0.5 = 3.54mm
3. Determine section size for a given Tensile load
Design strength due to gross section yielding
Gross area = Tdg = 0.909 * Fy * Ag
500*10^3 = 0.909 * 350 * Ag
Ag = 1571mm2
Assume ISA 100 x 75 x 12 mm , Area = 1956mm2
Area of Section selected > Gross area ; 1956 >1571mm2
Design strength due to yielding = Tdg = 0.909 Fy Ag = 0.909 * 350 * 1956 = 622.3kN
Design Strength due to net section rupture
Tdn = 0.72 * fu * An = 0.72 *410*1956 = 577.4kN
Hence 622.3kN, 577.4kN > 500kN hence safe.