Question

Determine the reactions at supports of simply supported beam of 6m span carrying increasing load of 1500N/m to 4500N/m from one end to other end.

Answer 1

Since Beam is simply supported i.e., at point A and point B only point load is acting. First change UDL and UVL in to point load. 

As shown in fig(b). Let ∑H & ∑V is the sum of horizontal and vertical component of the Resultant forces, The supported beam is in equilibrium, hence resultant force is zero. Draw the FBD of the beam as shown in fig(b), 

Divided the diagram ACBE in to two parts A triangle CDE and a rectangle ABCE. 

Point load of Triangle CDE =1/2 × CD × DE = 1/2 × 6 × (4.5 –1.5)= 9KN 

act at a distance 1/3 of CD (i.e., 2.0m )from point D 

Point load of Rectangle ABCD = AB × AC = 6 × 1.5 = 9KN 

act at a distance 1/2 of AB (i.e., 3m )from point B 

Now apply condition of equilibrium:

∑H = 0;

R_AH = 0

∑V = 0;

R_A –1500 × 6 – 3000 × 3 + R_B = 0

R_A + R_B = 18000 N  .........(i)

Now taking moment about point ‘A’

– R_B × 6 + 9000 × 3 9000 × 4 = 0

R_B = 10500 Nm

Putting the value of R_B in equation (i)

R_A = 7500 Nm.