Week 4 Challenge

1. Section classification problem for listed beam shapes

Calculate section classification of following beam shapes

Here Fy = 210 N/mm2

Section of the following shape can be classified based on the table 2 of IS 800 : 2007

• ISMB 500

Ans

b= 90 mm and tf= 17.2mm

Hence b/tf= 90/17.2= 5.23

From table 2 for rolled section the value for plastic class 1 section - 9.4*sqrt(250/210)=10.26

The value we obtained is less than 10.26. Hence this section falls under plastic class 1 sections.

• ISMC 200

  Ans

  b= 75 mm and tf= 11.4mm

  Hence b/tf= 75/11.4= 6.58

  From table 2 for rolled section the value for plastic class 1 section - 9.4*sqrt(250/210)=10.26

  The value we obtained is less than 10.26. Hence this section falls under plastic class 1 sections.

  • PG 1300 x 450 x 20 x 12

Ans

b= (450/2)-6= 219 mm and tf= 50mm

Hence b/tf= 75/11.4= 4.38

From table 2 for welded section the value for plastic class 1 section - 8.4*sqrt(250/210)=9.17

The value we obtained is less than 9.17. Hence this section falls under plastic class 1 sections.

• Double angle (Back to back) – 2 x ISA 75x75x6

Ans

b= 150 mm and tf= 6mm

Hence b/tf= 150/6= 25

From table 2 for welded section the value for semi compact class 03 section - 13.6*sqrt(250/210)=14.84

The value we obtained is greater than 14.84. Hence this section falls under semi compact class 03 section.

• RHS - 150 x 100 x10

Ans

b= 100-10-10=80 mm and t= 10mm

Hence b/tf= 80/10= 8

From table 2 for rolled section the value for plastic class 1 section - 9.4*sqrt(250/210)=10.26

The value we obtained is less than 9.17. Hence this section falls under plastic class 1 sections.

• Builtup box – 1000 x 450 x 20 x 12

b= 450-12-12= 426 mm and tf= 20mm

Hence b/tf= 426/20= 21.3

From table 2 for welded section the value for semi compact class 03 section - 13.6*sqrt(250/210)=14.84

The value we obtained is greater than 14.84. Hence this section falls under semi compact class 03 sections.

2. Effective length problems

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.

Ans)

The length of the rafter = L= Sqrt(20^2 + 5.36^2) = 20.71 m

Since the rafter is is connected to purlin it is rotationally fully restrained, hence the effective length of the rafter is 0.75*20.71= 15.53 m.

3. Deflection of a beam problem

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

^Ans)

The deflection at centre of the beam is calculated by calculating the following equation

= (WL^3/48EI)[(3a1/L)-4(a1/L)^3]+(WL^3/48EI)[(3a2/L)-4(a2/L)^3]+(5WL^3)/384EI

= (100*7^3/48*20x10^6) [(3*2/7)-4(2/7)^3]+(100*7^3/48*20x10^6) [(3*5/7)-4(5/7)^3]+(5*6*7^3)/384*20x10^6

=2.73x10^-5 +2.44x10^-5+ 1.34x10^-6

=5.30x10^-5 m

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

^Ans

The deflection at centre of the beam is calculated by calculating the following equation

= (WL^3/48EI)[(3a1/L)-4(a1/L)^3]+(WL^3/48EI)+(5WL^3)/384EI

= (2*5^3/48*25000) [(3*1.5/5)-4(1.5/7)^3]+(5*5^3/48*25000) +(5*1.5*5^3)/384*25000

=1.65x10^-4+ 5.21x10^-4+9.77x10^-5

=7.84x10^-4 m

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

Ans

L= 10 m, W= 10Kn/m, EI=?, Deflection at centre = 5 mm

The equation for deflection at centre for UDL = (5WL^3)/384EI

EI=(5WL^3)/384*deflection at centre

EI= (5*10*10^3)/384*(5/1000)= 26041.67 Knm2

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

Ans

Calculate Bending moment

Total bending moment = BM using Live load+ BM using Dead load +BM using Point load.

Here W for live load = 6*0.3= 1.8Kn/m

W for dead load is = 5*.3= 1.5Kn/m

BM using live load = WL^2/8 = 1.8*6^2/8= 8.1Knm

BM using Dead load= WL^2/8 = 1.5*6^2/8= 6.75Knm

BM using Point load.= WL/4 = 100*6/4 =150Knm

Total bending moment of the section is = 8.1+6.75+150= 164.85 Knm

Total factored moment 150*1.5= 247.28Knm

Now let's calculate section modulus

Zp=M-gama m0 /fy

= 247.28*1000*1000*1.1/210

= 1295276.19 mm3

From table 2.1 of IS 808

The Section modulus of ISMB450 is 1350000 mm3 which is greater than calculated section modulus. Hence Considering ISMB 450 beam as for moment capacity determination.

Now let's calculate design shear strengh

Vmax =1.5*( Shear force due to live load + shear force due to dead load + shear force due to concentrated load)

Shear force due to live load= (8.1*6/2)=24.3 Kn

Shear force due to dead load = (6.75-*6/2)=20.25Kn

Shear force due to concentrated load = 100/2 = 50Kn

Vmax = 1.5(24.3+20.25+50)=141.83Kn

The design shear strength of a section Vd= Vn/1.1 = 296.1/1.1 =269.18Kn

Vn = Vp = 296.1

Vp=Av*fy/3 = 4230*210/3 = 296100/1000= 296.10Kn

Av= h*tw= 450*9.4= 4230 mm2

Here Vd > Vmax hence the design section is safe.

Let's now calculate limiting width to thickness ratio

b= 75mm and tf = 9.4 mm

b/tf = 75/9.4 = 7.98

From table 2 for rolled section the value for plastic class 1 section - 9.4*sqrt(250/210)=10.26

The value we obtained says that the section is plastic hence the section is having satisfactory design capacity.

Check for deflection

Total deflection = 1.5* (Deflection due to live load + deflection due to dead load + defelection due to concentrated load)

Deflection due to live load= (5WL^3)/384EI= (5*1800*6^3/384*25x10^6) = 2.03x10^-4 m

Deflection due to dead load= (5WL^3)/384EI= (5*1500*6^3/384*25x10^6) = 1.69x10^-4 m

Deflection due to concentrated load= (WL^3/48EI)=100*1000*6/48*25x10^6 = 5x10^-4 m

Total deflection = 1.5*((2.03+1.69+5)x10^-4) =1.31x10^-3

From table 6 of IS 800 Maximum deflection for Simple beam is to be Span/300 = 6/300 =0.02 m

The design deflection is less than Maximum deflection hence the section is satisfactory

Check for web buckling

Buckling capacity Vcr= Buckling Coeffecient * Area of the web * fy

To know buckling coefficient we need to know the buckling class of the section from Table 10 of IS 800

From table 10 we get to know the buckling class b as the buckling takes place at Y axis the the tf of the section is 17.4 mm which is less than 40 mm

So from Table 7 of IS 800 the buckling coefficient is 0.34

the area of the web = (450-17.4-17.4)*9.4=3902.88 mm2

Hence Vcr = 0.34*3902.88*210=278.62Kn

Herer Vcr > V max hence buckling capacity of the section is satisfactory.

As per cl.8.7.4 of IS 800 :2007 Page No.67

Fw">Fw= ((b1+n2)×">×tw ×">×fy/γmo">γmo

where b1">b1=length of stiff bearing =50 mm

n2">n2= length of load dispersion

= (50+95.75)x9.4x210/1.1

261555 N = 261.56KN > V ma

section is safe in bearing

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

^Ans