Week 4 Challenge

Answer- 1

Aim – To calculate section classification of beam

As per IS 800-2007 Table – 2 and steel table here classified of given beams-

1. ISMB -500

b/t_f = 180/17.2 = 10.46 is less than 10.5 so it is compact section

2. ISMB – 200

b/t_f = 75/11.4 = 6.57 is less than 9.4 so it is plastic section

3. Plate Girder –

b/t_f = 450/50 = 9 is less than 9.4 so it is plastic section

4. Double angle 2 x ISA 75x75x6

d/t_f = 75/6 = 12.5 is les than 15.7 so section is semi compact.

5. RHS – 150x100x10

d/t_w = 150/10 = 15 is less than 15.7 so it is semi compact section.

6. Built Box –

d/t_w = 1000/12 = 83.35 is less than 84 so it is plastic section

Answer -2

Aim - To calculate Effective Length

Effective Length –

Effective Length Top Chord 1 = 0.85 L1

Effective Length Top chord 2 = 0.85 L2

Effective Length Vertical member = 0.85 L3

Effective Length Inclined member = 0.75 L4

Effective Length of rafter = 0.85 L ( L =20.7)

= 0.85 x 20.7 = 17.6 m

Answer -3

Aim- To Calculate Deflection at center of beam

Given EI = 20 x 10⁶ Nm²

^{Calculation as per attachement-}

Given EI = 25 x 10⁶ Nm²

^{Calculation as per attachement-}

• 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.

^{Given - Span - 10 m, W = 10 KN/m (Udl), Deflection y = 5mm}

^{Calculation as per attachement-}

^{Answer -4}

^{Aim - Design of Purlin member}

^{Procedure -}

1. Calculate all dead load and live load and also acting load normal to sheet and parllel to sheet , wind load
2. Calculate moment of acting load
3. Calculate required plastic modulus with assuming width and depth and according this choose section
4. Check section limit and width
5. Calculate moment of select section and check
6. Calculation deflection and check.

Calculation and check in this attachment -