A timber beam of rectangular section is to support a load of 20KN uniformly distributed over a span of 3.6m when beam is simply supported. If the depth is to be twice the breadth, and the stress in timber is not exceed 7N/mm2, find the dimensions of the cross section. How could you modify the dimensions with 20KN of concentrated load is present at centre with same breadth and depth ratio.

 A timber beam of rectangular section is to support a load of 20KN uniformly distributed over a span of 3.6m when beam is simply supported. If the depth is to be twice the breadth, and the stress in timber is not exceed 7N/mm², find the dimensions of the cross section. How could you modify the dimensions with 20KN of concentrated load is present at centre with same breadth and depth ratio.

Solution

Step 1: case 1: when simply supported beam of length 3.6m carries udl of 20KN and depth is twice the width

We know that W = w L

                            = 20 X 1000X3.6

                            = 5.56N

Moment = WL/8

       M    = 5.56 X 1000X 3.6 /8

       M    = 2499.75 N-mm

Step 2: Calculation of cross sectional dimensions of the beam

σ = 7N/mm²

M /I= σ / y

2499.75/(bd³/12) = 7/(d/2)

b = 8.12mm

d = 2b = 16.24mm

Step 3 :Case 2: when simply supported beam of length 3.6m carries point load of 20KN and depth is twice the width

Moment = WL/4

       M    = 20 X 10⁶X 3.6 /4

       M    = 18X 10⁶ N-mm

Step 4: Calculation of cross sectional dimensions of the beam

σ = 7N/mm²

M /I= σ / y

18X 10⁶ /(bd³/12) = 7/(d/2)

b = 156.82mm

d = 2b = 313.65mm