Week 4 Challenge

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