Week 4 Challenge
Objective of this challenge is to:
a) understand the concept of design of beams and how to classify section based on shapes of beams.
b) How to calculate the effetive length of builtup sections for eg; rafter.
c) Calculation for the deflection of beam under flexural stiffness, based on type of loading.
d) Understand the concept of moment capacity for a given live load and dead load and know how to design purlin.
1. Section classification problem for listed beam shapes:
ISMB 500
As per Is800-2007 Table 46
Depth(d) =500mm
Width of flange(b)=180mm
Thickness of flange(tf)=17.2mm
Thickness of web(tb)=10.2mm
d/tw=500/10.2=49>84
Hence it is plastic.
ISMC 200
d=200mm
tf=11.4mm
tw=6.1mm
b=75mm
b/tw=200/6.1=32.78>43 Hence Plastic.
PG 1300 x 450 x 20 x 12
d=1300mm
tf=50mm
bf=450mm
tw=12mm
d/tw=1300/12=108.32~105-126. Hence it is semi-compact.
Double angle (Back to back) – 2 x ISA 75x75x6
d=75mm
tw=6mm
d/tw=75/6=12.5mm~10.5-15.5. Hence Semi-Compact.
RHS 150x100x10
Along X-X
d=100
t=10mm
d/t=10
Along Y-Y
d=150mm
t=10mm
d/t=150/10=15
In both axis d/t is less than 84. Hence it is plastic.
b=450-24mm=426mm
d=1000-40mm=960mm
td=12mm
tb=20mm
b/tb=426/20=21.3
d/tb=960/12=80 Hence Plastic.
2. Effective length problems:
Length of one side of rafter(L)=20/cos15=20.7m
As in the fig the base are pined and the top is restrained. But in the rafter it is lateral restraint at the top of flange.
Effective length along x axis=1.8L=1.4x20.7=28.98m
Effective length along y axis =2L=2x20.7=41.4m
3. Deflection of a beam problem:
EI=20 x 106 Nm2
Deflection at the center of beam=WL3/48EI+5WL3/384EI
=WL3/EI(1/48+5/384)
W=100+100+6x7=242N
Deflection=(242x7x7x7)/20x10^6(1/48=5/384)
=4.1x10^-3(0.034)
=0.038mtr.
EI=25 x 106 Nm2
W=2+5+1.5x5=14.5N
Deflection=4.5x10^3/48x25 x 106 [3x2/5-4(3/8)^3]
=0.1x106-6mtr
c) Given
UDL=10Kn/m
Deflection =5mm
We know that Deflection=5WL^4/384EI
EI=5WL^4/384x5=5x10x10x10^4/384x5
or EI=13.02x106 Nmsq
4. Moment capacity determination problem:
5. Purlin design problem
