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 effective length of builtup sections for ex.,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 (tw)= 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 it is 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)- 2xISA 75 x 75 x 6

d=75mm
tw= 6mm

d/tw = 75/6 = 12.5mm ~10.5-15.5 , Hence it is semi compact.

RHS 150 x 100 x 10

Along X-X
d= 100mm
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.

Builtup box - 1000 x 450 x 20 x 12

b= 450-24=426mm
d=1000-40= 960mm
td= 12mm
tb = 20mm

b/tb = 426/20 = 21.3
d/td =960/12 = 80 , Hence it is plastic.

2.Effective length problems:

Length of one side of rafter (L) = 20/cos15 = 20.7m

As in fig.the base are pinned and the top is restrained.But in the rafter it is lateral restraint at the top of flange.
Effective length along X axis = 1.4L = 1.4x20.7 = 28.98m
Effective length along Y axis = 2L = 2x20.7 = 41.4m

3.Deflection of a beam problem:

EI = 20x10^6Nm2
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) = 0.038m

EI = 25x10^6Nm2
W= 2+5+1.5x5 = 14.5N

Deflection = 4.5x10^3/48x25x10^6[3x2/5-4(3/8)^3] = 0.1x106-6m

c) Given
UDL = 10KN/m
Deflection = 5mm
We know that deflection = 5WL^4/384EI
EI = 5WL^4/384X5 = 5X10X10X10^4/384X5
or EI = 13.02x10^6Mm2

4.Moment capacity determination problem :




5.Purlin design problem :