Shear Force and Bending Moment Diagrams

Why it matters

Understanding shear force and bending moment diagrams is crucial for mechanical engineers as they help in analyzing and designing beams and structural elements. These diagrams provide insights into how beams will react under different loading conditions, ensuring safety and efficiency in structures like bridges, buildings, and machinery.

Key ideas

• Shear Force (SF): The internal force parallel to the cross-section of a beam. It results from external loads, reactions at supports, or changes in the beam's geometry.
• Bending Moment (BM): The internal moment that causes the beam to bend. It is the result of external loads and reactions causing rotation about a point.
• Sign Conventions: Use sagging moment positive, x from left to right, and shear equal to the net upward force on the left portion. With downward distributed load w positive, dV/dx = −w and dM/dx = V.
• Types of Loads: Concentrated loads, distributed loads, and moments can all affect shear force and bending moments.
• Diagrams: SF and BM diagrams graphically represent how shear force and bending moment vary along the length of the beam.

Jumps and integration

A downward point force P causes a shear jump −P. Moment remains continuous there unless a concentrated couple is also present. A couple causes a moment jump. Use M(x) − M(x₀) = ∫V dx, retaining the boundary value. In the example, shear is +5 kN for 0 < x < 3 m and −5 kN for 3 < x < 6 m.

Formulas

• V = dM/dx
  • V: Shear force (N)
  • M: Bending moment (N·m)
  • x: Distance along the beam (m)
• M = ∫V dx
  • M: Bending moment (N·m)
  • V: Shear force (N)
  • x: Distance along the beam (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.
   • R_A = R_B = 10 kN / 2 = 5 kN

2. Draw the Shear Force Diagram (SFD).

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

3. Draw the Bending Moment Diagram (BMD).

   • At x = 0, M = 0
   • At x = 3 m, M = 5 kN * 3 m = 15 kN·m
   • At x = 6 m, M = 0

Final Answer: Maximum bending moment is 15 kN·m at the center.

Common mistakes

• Incorrect sign conventions leading to wrong diagrams.
• Miscalculating reactions at supports.
• Forgetting to consider all types of loads, including distributed loads.

For GATE ME

Questions often involve calculating shear forces and bending moments for various loading conditions. Practice drawing accurate diagrams and understanding the effects of different loads on beams.

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. It jumps from +P/2 immediately left of the central downward load to −P/2 immediately right; there is no single continuous value at the load. 2. A constant nonzero UDL produces a parabolic bending-moment segment. 3. Zero at a simple end support with no applied end couple.