Shear Force and Bending Moment Diagrams

Why it matters

Shear force and bending moment diagrams are essential tools in civil engineering for analyzing the internal forces within beams. These diagrams help engineers design safe and efficient structures by understanding how loads affect the beam's strength and stability.

Key ideas

• Shear Force (SF): The force that causes one part of a material to slide past another. It is perpendicular to the axis of the beam.
• Bending Moment (BM): The tendency of a force to cause rotation about a point or axis. It is the product of the force and the distance from the point of interest.
• Sign Conventions: Use a consistent cut-face convention: here positive bending is sagging, with dM/dx = V and dV/dx = -w for downward distributed load w. External force signs alone do not define internal moment signs.
• Types of Loads: Concentrated loads, uniformly distributed loads (UDL), and varying loads affect the shear force and bending moment differently.
• Support Types: Simply supported, cantilever, and fixed beams have different reactions and internal force distributions.
• Equilibrium Equations: Used to calculate reactions at supports and internal forces.

Formulas

• SF = V
  • Where V is the shear force (N).
• BM = M
  • Where M is the bending moment (Nm).
• M = F·d
  • Where F is the force (N) and d is the perpendicular distance from the point of interest (m).

Worked example

Given: A simply supported beam of length 6 m with a point load of 10 kN at the center.

1. Calculate reactions at supports:

   • By symmetry, reactions at both supports are equal.
   • R1 + R2 = 10 kN
   • R1 = R2 = 5 kN

2. Shear Force Diagram (SFD):

   • At x = 0, SF = 5 kN
   • At x = 3 m, SF = 5 kN - 10 kN = -5 kN
   • At x = 6 m, SF = 0

3. Bending Moment Diagram (BMD):

   • At x = 0, BM = 0
   • At x = 3 m, BM = 5 kN * 3 m = 15 kNm
   • At x = 6 m, BM = 0

Final Answer: Maximum bending moment is 15 kNm at the center.

Common mistakes

• Incorrect sign conventions for shear forces and bending moments.
• Miscalculating reactions at supports, especially in indeterminate structures.
• Forgetting to consider the effects of distributed loads.
• Not drawing the diagrams to scale, leading to misinterpretation.

For GATE CE

• Questions often involve calculating shear forces and bending moments for various loading conditions.
• Practice problems with different types of beams and loads to understand the influence on internal forces.
• Be familiar with drawing and interpreting shear force and bending moment diagrams.

Quick check

1. What is the shear force at the midpoint of a simply supported beam with a central point load?
2. How does a uniformly distributed load affect the bending moment diagram?
3. What is the bending moment at the supports of a simply supported beam?

Answers: 1. Shear jumps from +P/2 just left to -P/2 just right of the central point load; it is not a unique zero value at the load, 2. It creates a parabolic curve, 3. Zero.