Question

A beam is subjected to a triangular distributed load whose value at right end of the beam is w=8.1 kN/m. Draw the free- body diagram of the beam and determine the vertical reaction at A (in kN). Sign: Upward is positive A B 30 m

Answer 1

Sol: Information provided in the question:

A simply supported beam AB:

Such that, support at Point A = Hinge

Support at Point B = Roller

Span of beam = 30 m

Load on beam is triangular distributed load with peak height w = 8.1 KN/m (At right support)

W = 8.1 KN/m A B 30 m

=> Now we draw the free body Diagram of given Beam:

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. A free-body diagram is a special example of the vector diagrams .

=> For this we can analyze that there are two reaction at support A for Hinge support: Va and Ha in vertical and Horizontal direction respectively:

=> At Point B we have Roller support which is applying only one reaction at in vertical direction that is Vb and in horizontal direction it is free to move so (Hb = 0) here

NOW THE FREE BODY DIAGRAM FOR BEAM (A-B)

' z8,1KN/m Ha 8 ܒܐ Vo Va Va Bvo 6 H

=> Now we calculate reactions at end support:

$\sum F_{horizontal} = 0$

Ha = 0 KN

$\sum F_{vertical} = 0$

Va + Vb = (1/2)x8.1x30 = 121.5 KN

Taking Sum of Moment at A is zero:
$\sum M_{A} = 0$

Vb x 30 = (1/2)x8.1x30 x ( 2/3 x30)

Vb x 30 = 81 x 30

Vb = 81 KN

Then Va = 121.5 - Vb = 121.5 - 81 = 40.5 KN

Hence Vertical Reaction at Point A is Va = 40.5 KN