Week 4 Challenge

1.

ISMB 500 : - b/tf = 180/17.2 = 10.46 < 10.6 hence section is compact/

ISMC 200 : - b/tf = 7.5/11.4 =6.57 <9.4 , hence section is plastic.

PG 1300x450x50x12 :- b/tf =450/50 = 9< 9.4 , hence is plastic.

2TSA 75x75x6 :- b/tf = d/t = 75/6 =12.5 < 15.7, section is semi compact.

RHS 150x100x10 :- d/tw = 150/10= 15< 15.4 , section is plastic

Built up box 1000x450x20x12

d/tw =1000/12 = 83.35 < 84 , hence section is plastic.

2.

Truss is given in which top chord, vertical member and inclined members have been with their length have been marked with effective length are to be determined

Procedure( Solution)

Top chords with length L1&L2">L1&L2L1&L2 are in compression which is continous so Effective length will be as follows-

L1">L1L1 (Effective) = 0.85 L1">L1L1

L2">L2L2(Effective) = 0.85 L2">L2L2 ( 0.7 L to 1.0 L )

L3">L3L3(Effective) = 1.0 L3 ( Vertical member in tension)

L4">L4L4(Effective) = 0.85 L4">L4L4 ( Discontinous compression member)

Result :- It is concluded that the Member having lengthL1">L1L1 , L2">L2L2 is under compression are of effctive length 0.85 L1">L1L1, and 0.85 L2">L2L2 , effective length of L3">L3L3 is 1.0L3">L3L3and that of inclined member with length L4">L4L4 is 0.85L4">L4

3.

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5.

Design channel purlin, considering sag rod at mid span for the roof truss having following data

• Span of truss : 20 m
• Spacing of Truss : 4 m c/c
• Roof slope : `theta`=20 degrees
• Spacing of purlin : 2 m
• Wind suction : 1 kN/m²
• Live Load : 0.75 kN/m² ,Dead Load 1.4 kN/m2

^{total DL +LL = 1.4 +0.75 =2.15 KN/m2 = 2.15 x 2 = 4.. KN/m2}

^{Factoral load = w = 4.5 x1.5 = 6.45 KN/m2}

^{Component of the load acting normal to the sheathing = w cos`theta` = 4.3cos20 = 4.22KN/m2}

^{Component of the load parallel to the sheathing =w sin`theta` =4.3 sin 20 = 1.53KN/m2}

^{Factored wind load = 1.5x1x2 = 3Kn/m2}

^{total load moment = 6.06 +3 =9.06KN/m2}

^{Total load parallel = 2.20 Kn/m}

^{Mzz = wl2/8 = 9.06*4^2/ 8 =18.12KN}

^{Myy=wl2/8=2.20x4^2 /8 =4.4KN}