Week 4 Challenge

ANSWER NO 1

AIM

1. Section classification problem for listed beam shapes

Calculate section classification of following beam shapes

• ISMB 500
• ISMC 200
  • PG 1300 x 450 x 20 x 12

• Double angle (Back to back) – 2 x ISA 75x75x6
• RHS - 150 x 100 x10
• Builtup box – 1000 x 450 x 20 x 12

ANSWER

ANSWER NO 2

AIM:

Calculate effective length of rafter madeup of Builtup I section about major (x-x) and minor axis (y-y) having purlin spaced at 1.2 m.

ANSWER NO 3

AIM:

3. Deflection of a beam problem

• Calculate deflection at the centre of the beam shown below. EI = 20 x 10⁶ Nm²

^SOULTION

Deflection of the member above is calculated as follows:

Deflection = (5wl^3/384EI)+(wl^3)/48EI*((3a1)/l-4(a1/l)^3)+(wl^3)/48EI*((3a2)/l-4(a2/l)^3)

=(5*6*7^3)/(384*20*10^6)+(100*7^3)/(48*20*10^6)(3*2)/7–4(2/7)^3)+ 100*7^3)(3*5)/7 – 4*(5/7)^3)

=1.339*10^-6+2.7*10^-5+2.43*10^-5m

=0.0527mm

• Calculate deflection at the centre of the beam shown below. EI = 25 x 10⁶ Nm²

^SOULTON

Deflection of the member above is calculated as follows:

Deflection = (5wl^3/384EI)+(wl^3)/48EI*((3a1)/l-4(a1/l)^3)+(wl^3)/48EI*((3a2)/l-4(a2/l)^3)

=(5*1500*5^3)/(384*25*10^6)+(2000*1.5^3)/(48*25*10^6)(3*1.5)/5–4(1.5/5)^3)+ (5000*5^3)/(48*25*10^6)(3*2.5)/5 – 4*(2.5/5)^3)

=9.76*10^-5+1.65*10^-4+5.208*10^-4

=0.783mm

3) A simply supported beam is 10 m long with udl of 10 kN/m Calculate the Flexural stiffness (EI) of the beam which limits the deflection to 5 mm at the middle

SOULTION

Deflection for simply supported beam is given by = (5wl^3)/(384EI)

EI = (5wl^3)/(384*deflection)

EI = (5*10*1000*10^3)/(384*5*1000)

=26.04N/m2

ANSWER NO 4

AIM:

4. Moment capacity determination problem

Select a standard hot rolled section for the beam which is simply supported of length 6 m spaced at 3 m c/c, carrying a floor Dead load of 5 kN/m² (including Self weight) & Live load of 6 kN/m² and a concentrated load of 100 kN at mid span of beam. The beam is laterally restrained and assumes bearing length of 50 mm

Tip steps:

• Calculate maximum bending moment & required plastic modulus
• Select suitable trial section
• Calculate design shear force and design shear strength
• Check design capacity of the section
• Check for deflection, web buckling, bearing

SOULTION

Ans: Given: Dead load = 5kN/m2

Live load = 6kN/m2

Total load = 5+6 = 11kN/m2

The beam is spaced at 3m c/c thus the load = 11*3 = 33kN/m

The maximum bending moment = BM due to UDL+BM due to concentrated load

=(wl2)/8+(wl)/4

=(33*6*6)/8+(100*6)/4

=298.5kNm

Plastic modulus Zp(required) = (M*γm0)/fy

=298.5*10^6*1.1/250

=1313.4*10^3m3

Let us adopt ISMB 500 with section modulus of 1808.7*10^3m3

Properties of ISMB 500

Depth (d) = 500mm

Width of the flange (b) = 180mm

Thickness of flange (tf) = 17.2mm

Thickness of web (tw) = 10.2mm

Depth of the web (h) = 500-2*17.2 = 465.6mm

Section classification

b/tf = 180/17.2 = 10.46<10.5

Thus the section is compact section

Calculation of shear force

Shear force = 100+33*6 = 298kN

Reaction = 298/2 = 149kN

Design shear strength Vd = (fy*d*tw)/(γm0√3)

= (250*500*10.2)/(1.1*√3)

=669.2kN

0.6Vd = 0.6*669.2 = 401.52kN

149<401.52kN

Design capacity Md = βp*Zb*fy/γm0<1.2*Ze*fy/1.1

Βp = 1 for compact section

=(1*1313.4*250)/1.1<1.2*1808.7*250/1.1

= 298.5kN-m<493.28kN-m

Hence it is ok

Check for deflection

Max deflection = 5wl^3/384EI+wl^3/(48EI)

=(5*33*10^3*6000^3)/(384*2*10^5*45218.3*10^4)+(100*10^3*3000^3)/(48*2*10^5*45218.3*10^4)

=1.026+0.62 = 1.64mm

Deflection due to load

= span/300 = 6000/300 = 20mm

1.64<20mm

Hence it is ok

Check for web buckling

Assume stiffener plate of width b1 = 50mm

Ieff of web =(b1*tw^3)/12

= (50*10.2^3)/12

= 4421.7mm

Area of the web = 50*10.2 = 510mm2

Radius of gyration = sqrt(I/A)

= √(4421.7/510)

r= 2.9mm

Effective length of the web = 0.7*depth of the web

=0.7*465.6

d(effective)= 325.92mm

λ = d(effective)/r

=325.92/2.9 = 112.38

For λ and fy 250 from IS 800-2007 we get

Fcd= 83.7Mpa

Check for web buckling

n1 = 500/2 = 250mm

b1+n1 = 50+250 = 300mm

Fwb = (b1+n1) *tw*fcd

= 300*10.2*83.7

= 256.12kN<669.2kN

Hence it is safe

Check for web crippling

F(crippling) = (b1+n2)*t*fy/γm0

n2=2.5d1 = 2.5*465.6 = 1164mm

F(crippling) = (50+625)*10.2*250/1.1

=569.12<669.2kN

Hence it is safe.

ANSWER NO 5

AIM:

5. Purlin design problem

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 : 20 degrees
• Spacing of purlin : 2 m
• Wind suction : 1 kN/m²
• Live Load : 0.75 kN/m² ,Dead Load 1.4 kN/m²