Shear Force and Bending Moment

Why it matters

Understanding shear force and bending moment is essential for designing safe and efficient structural elements like beams and bridges. These concepts help engineers determine how loads affect structures, ensuring they can withstand applied forces without failure.

Key ideas

• Shear Force (SF): The internal force parallel to the cross-section of a structural element. It results from external loads, reactions, or both.
• Bending Moment (BM): The internal moment that causes a beam to bend. It is the result of external loads, reactions, or both, acting at a distance from a point.
• Sign Convention: In this lesson, x increases from left to right, sagging bending moment is positive, and positive shear is the net upward transverse force on the portion to the left of the cut. With downward distributed load w taken positive, dV/dx = −w and dM/dx = V. Internal actions on opposite cut faces have opposite directions.
• Shear Force Diagram (SFD): A graphical representation showing how shear force varies along the length of a beam.
• Bending Moment Diagram (BMD): A graphical representation showing how bending moment varies along the length of a beam.
• Types of Loads: Concentrated loads, uniformly distributed loads (UDL), and varying distributed loads.
• Types of Supports: Simply supported, fixed, and cantilever.

Diagram rules

On a smooth segment, dV/dx = −w. Thus V(x) − V(x₀) = −∫w dx and M(x) − M(x₀) = ∫V dx. Include the integration constants or known boundary values.

A downward point load P creates a jump ΔV = −P. Without an applied couple, M remains continuous across that point, but its slope changes. A concentrated couple causes a jump in M. Point loads do not always create a maximum moment; extrema require checking shear signs and boundaries.

Formulas

• V = dM/dx
  • V: Shear force (N)
  • M: Bending moment (Nm)
  • x: Distance along the beam (m)
• M = ∫V dx
  • M: Bending moment (Nm)
  • V: Shear force (N)
  • x: Distance along the beam (m)

Worked example

Given: A simply supported beam of length 6 m carries a uniformly distributed load of 2 kN/m over its entire length.

1. Calculate reactions at supports.

   • Total load = 2 kN/m × 6 m = 12 kN
   • Reactions at supports (R1 and R2) are equal due to symmetry.
   • R1 = R2 = 12 kN / 2 = 6 kN

2. Draw the Shear Force Diagram (SFD).

   • Just right of A (x = 0⁺), V = +6 kN
   • Just left of B (x = 6⁻ m), V = +6 kN - 12 kN = -6 kN; the 6 kN upward reaction then returns shear to zero. Throughout the span, V(x) = 6 − 2x kN.

3. Draw the Bending Moment Diagram (BMD).

   • At A (x = 0), M = 0
   • At C (x = 3 m), M = 6 kN × 3 m - 2 kN/m × (3 m)^2 / 2 = 9 kNm
   • At B (x = 6 m), M = 0

Final Answer: Maximum bending moment is 9 kNm.

Common mistakes

• Incorrect sign convention for shear forces and bending moments.
• Miscalculating reactions at supports, especially in asymmetrical loading.
• Forgetting to integrate shear force to find bending moment.

For GATE ME

Questions often involve calculating shear forces and bending moments for different loading conditions and support types. Practice drawing SFDs and BMDs accurately and quickly.

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 point load affect the bending moment diagram?
3. What is the relationship between shear force and bending moment?

Answers: 1. For central downward point load P, the shear is +P/2 immediately to its left and −P/2 immediately to its right. It jumps at the load; assigning zero there hides the discontinuity. 2. A point force changes the slope of the bending-moment diagram while moment stays continuous unless a couple is also applied. 3. dM/dx = V with the stated convention.