Civil Thinking › Beam Analysis Problems Loaded Beam under Distributed Loading

Problem Statement

Beam Subjected to Distributed Loading

A prismatic beam of span 3 m is subjected to the following loading and support conditions.

Supports:
• Support at x = 0 m
• Support at x = 3 m

Distributed Loads:
• 0.953 kN/m acting from x = 0 to 3 m

Required:
• Determine the support reactions
• Draw the Shear Force Diagram (SFD)
• Draw the Bending Moment Diagram (BMD)

Equilibrium equation ∑M = 0 Expanded form − W1·1.5 + R2y·3 = 0 Substitution − 2.859 kN×1.5 m + 1.43 kN×3 m = 0 Solved reaction

Equilibrium equation ∑Fy = 0 Expanded form − W1 + R1y + R2y = 0 Substitution − 2.859 kN + 1.43 kN + 1.43 kN = 0 Solved reaction

ΣFy check − 2.859 kN + 1.43 kN + 1.43 kN = 0 = 0 ΣM check − 2.859 kN×1.5 m + 1.43 kN×3 m = 0 = 0

Position (m)	Shear (kN)	Remark
0	1.4295	Start of beam
1.5	0	Zero shear location
3	-1.4295	End of beam
3	-1.4295	Just before point load / reaction
3	0	Just after point load / reaction

s = 0
(1.4295) × 0 = 0
ΣM = 0

s = 1.5
(1.4295) × 1.5 = -2.14425
1.0721 (UDL resultant)
ΣM = -1.0721

s = 3
(1.4295) × 3 = -4.2885
(1.4295) × 0 = 0
4.2885 (UDL resultant)
ΣM = 0