Week 4 Challenge
1. a) ISMB 500
b/tf = 180/172 = 10.46 < 10.6, hence section is compact.
b) ISMC 200
b/tf = 75/11.4 = 6.57 < 9.4, hence section is plastic.
c) PG 1300X450X50X12
b/tf = 450/50 = 9 < 9.4 ,hence section is plastic.
d) 2 x ISA 75 x 75 x 6
b/tf = d/t = 75/6 = 12.5 < 15.7>10.5
Hence section is semi compact
e) RHS 150* 100 *10
d/tw = 150/10 = 15 <54 Section is plastic.
f) Builtup box 1000 x 450 x 20 x 12
d/tw = 1000/12 = 83.33 <84 , Section is plastic
2) Effective length of top chord 1= 0.85L1
Effective length of top chord 2 = 0.85L2
Effective length of vertical member = 0.85L3
Effective length of inclined member = 0.75L4
length of rafter = 20.7m
Effective length of rafter = 0.85L = 0.85 * 20.7 = 17.6m
3) a) 
y = WL^3/48EI {(3a1/L) - 4(a1/L)^3} + WL^3/48EI {(3a2/L) - 4(a2/L)^3} + 5WL^3/(384 EI)
= {100*7^3/(48*20*10^6)}[3*2/7 - 4(2/7)^3] + {100*7^3/(48*20*10^6)}[3*5/7 - 4(5/7)^3] +( 5 x 6 x7^3/(384*20*10^6) = 0.0531mm
b) 
y = WL^3/48EI {(3a1/L) - 4(a1/L)^3} + WL^3/48EI {(3a2/L) - 4(a2/L)^3} + 5WL^3/(384 EI)
= {2000*1.5^3/(48*25*10^6)}[3*1.5/5 - 4(1.5/5)^3] + {2000*2.5^3/(48*25*10^6)}[3*2.5/5 - 4(2.5/5)^3] +( 5 x 1500 x5^3/(384*25*10^6) = 0.78mm
c) y = 5mm
y = 5wl^3/(384 EI)
EI = 5 * 10*1000*10^3/(384x5x10^3)
EI = 26.04 x 10^6N/mm2
4. Dead load = 5kN/m2
live load = 6kN/m2
Total load = 5+6 = 11kN/m2
UDL = 11 x 3 = 33kN/m
max bending moment = 33*6^2/8 + 100*6/4 = 298.5kNm
Plastic modulus = Zp = M x 'gammam0'/fy = 298.5 x 10^6 x 1.1/250 = 1.31 x 10^6mm3
Assume ISMB 450 Zp = 1533.36 mm3
Overall depth = 450mm
width of flange = 150mm
thickeness of flange tf = 17.4mm
Depth of web d = h-2(tf+R) = 450 - 2tf = 450-35 = 415mm
thkness of web = 9.4 , MI about major axis = 30390.8cm^4
Section Classification
b/tf = 150/17.4 = 8.62 <9.4
d/tf = 415/9.4 = 44.14 <84 hence plastic section .
Design of shear force
Shear force = 100 + 33*6 =198+100=298/2= 149kN
Design of shear strength Vd = fy/'gammam0' x `root`3 x h x tw
=250/(1.1 * 'root'3) * 450 * 9.4 = 555.04 kN
0.6 Vd = 333kN ; Vd< 0.6 Vd
Check for design capacity
d/tw = 44.14 < 67
Md= Bb x Zb x fy/ 'gammam0' < 1.2 x Ze x fy /1.1
= 1 * 1533.36 *250/1.1 =340.5kNm < 1.2 x 1350.7 x 250/1.1 = 368.37kNm
`Check for deflection = 'delta' udl + 'delta' pl`
= 5*33*6000^3*10^3/(384*2*10^5*30390.8x 10^4)+[100x(6000)^3*10^3/48*2*10^5*30390.8x10^4) = 0.89mm
Allowable max deflection = L/300=6000/300 = 20mm
5. Purlin Design
Span of truss = 20m
Spacing of Truss = 4m c/c
Roof Slope = 20 degrees
Spacing of purlin = 2m
Wind suction = 1* 2 = 2kN/m
Live Load = 0.75 * 2 =1.5kN/m
Dead Load = 1.4 * 2= 2.8kN/m
Wdz = 2.8 cos20 = 2.63kN/m
Wlz = 1.5 cos20 = 1.41 kN/m
Ww2 = -2kn/m
Wdy = 2.8 x Sin20 = 0.95
Wly = 1.5 x Sin20 = 0.513
Factored load combination (z direction) = 1.2Wl + 1.2DL + 1.2LL
= 1.2 x 2 + 1.2 x 2.63 + 1.2x 1.41= 7.248 kN/m
Provide ISMC 150 d= 150mm , ry = 22mm ; width b = 75mm
thickness of flange = tf= 9mm , thickness of web tw=5.7mm
Elastic section modulus Zez= 105 x 10^3 mm3
Elastic section modulus Zey= 19.5 x 10^3 mm3
plastic section modulus Zez= 119.82 x 10^3 mm3
MOment of inertia = 788 x 10^4mm^4
Section Classification
b/tf = 75 /9 = 8.33 < 9.4
d/tw =[ 150-2(9)]/5.7 = 19.65 <42 hence section is plastic
Design capacity of the section
Mdz = Zpz x fy/ 'gammam0' = (119.82x10^3 x 250/1.1 x 10^6) = 27.23kNm
< 1.2 x Zez fy/ 'gammam0' = [1.2 x 105 x 10^3 x 250/1.1 x 10^6) = 28.63kNm
`py = {2tf bf^2/4}` + (h-2tf)tw^2/4
`nbsp; = (2 x 9 x 75^2/4) + {(150-2*9)*5.7^2/4` = 26384.6mm3
Mdy = (Zpy x fy /'gammam0') = 26384.6 x 250/ (1.1 x 10^6) = 6kNm < 1.5Zeyfy/'gammam0'= 1.5 x 19.5x 10^3x250/1.1 x 10^6 = 6.6kNm , hence safe
Check for deflection = 5/384 * wL^4/EI = 5/384 * 7*4000^4/2*10^5 *788*10^4 = 8.17mm <22.22, hence safe
Member strength [Mz/Mdz] + [My/Mdy] < 1
[18.7/27.33] + [1.097/6] < 1 ; 0.868 <1 hence safe.