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Design of purlin given 10 span=4m 2)spacing =1.72m 3)$\theta=304^{\circ}$
1 Answer
written 5.7 years ago by |
load calculation
D.L/---weight of GI sheet= 150Pa
self weight at purlin=100 ,pa
Total DL=250Pa=0.25Kpa
plan Area=1.73$\times cos 30^{\circ}$=1.5m
D.L on Plan area =0.25$\times $ 1.5=0.375kN/M
L.L--L.L on plan =0.4 $\times$1.5=0.6kN/m
W.L--W.L on sloping Area=0.723$\times $1.73=-1.31 kN
Load Normal to slope Zz
Wdz=0.375$\times cos30^{\circ}$
= 0.325 kN/m
WLz=0.6$\times 10530^{\circ}$
=0.52KN/m
Total load=0.845kN/m
Load parallel to slope Y-Y
WDy=0.375$\times sin 30^{\circ}$
=0.19kN/m
WLy=0.6$\times sin30^{\circ}$
=0.3 kN/m