Determine the resultant force and specify where it acts on t

Determine the resultant force and specify where it acts on the beam measured from A.

Solution

>> First Let\'s divide the whole distributed load into two parts ;

1). A triangular part, upto A ( for 3 m) , Resultant = F1

2). Another Triangular part, from A to B , (for 6 m) , Resultant = F2

As, Resultamt force is equal to area under distributed load.

>> As, F1 = (1/2)*6*3 = 9 kN

It acts at 3/3 = 1 m from A , in left side

>> and, F2 = (1/2)*6*6 = 18 kN

It acts at 6/3 = 2 m from A in right side

>> Let, F = Resultant force of whole system and is acting at distance \"x\" from A in right side

As, F = F1 + F2 = 9 + 8 = 27 kN

>> Now, Moment about A = F*x = 2*F2 - 1*F1

=> 27*x = 2*18 - 1*9 = 27

=> x = 1 m

>> So, Resultant force = 27 kN and is acting at 1 m from A, in right side (towards B)....