What is Shear Force?

Shear force is the internal force within a structural element, such as a beam, that acts parallel to the cross-section of the element. It represents the sum of all forces acting on one side of a section that tend to cause the section to slide or shear relative to the other side. Shear force is typically measured in units of force, such as newtons (N) or pounds (lb), and is critical in structural engineering to ensure the stability and strength of structures.

How to Calculate Shear Force?

To calculate shear force in a beam, follow these steps:

• Identify all external forces: Determine all applied loads (point loads, distributed loads, moments) and support reactions (from supports like pins or rollers).
• Draw a free-body diagram: Represent the beam and all forces acting on it.
• Choose a section: Select a point along the beam where you want to calculate the shear force.
• Sum forces on one side: For the chosen section, sum all vertical forces (upward and downward) on either the left or right side of the section. The shear force is the algebraic sum of these forces.
  • Use the sign convention: Typically, upward forces are positive, and downward forces are negative.
• Account for distributed loads: If a distributed load (e.g., w N/m) exists, its effect is calculated as the load intensity multiplied by the length over which it acts up to the section.
• Shear force formula:

For a section at position $x$, the shear force $V$ is given by: $$V = \sum (\text{Forces on one side of the section})$$ Ensure to include reactions and loads up to the section.

• Ensure to include reactions and loads up to the section.

Sample Calculation of How to Find Shear Force?

Problem: A simply supported beam of length 4 m has a point load of 10 kN at 1 m from the left support (A) and a uniformly distributed load (UDL) of 5 kN/m over the entire length. Calculate the shear force at 2 m from the left support.

Step-by-Step Solution:

• Step 1: Calculate support reactions:

Total UDL = 5 kN/m × 4 m = 20 kN (acting at the center, 2 m from A).Take moments about A to find reaction at B ($R_B$RB​): $$\sum M_A = 0: \quad R_B \times 4 – 10 \times 1 – 20 \times 2 = 0$$ $$4R_B – 10 – 40 = 0 \implies 4R_B = 50 \implies R_B = 12.5 \, \text{kN}$$

Sum vertical forces to find reaction at A

A ($R_A$): $$R_A + R_B – 10 – 20 = 0 \implies R_A + 12.5 – 10 – 20 = 0 \implies R_A = 17.5 \, \text{kN}$$

Step 2: Section at 2 m from A:

• Consider the left side of the section.
• Forces acting: Reaction at A (17.5 kN upward), point load at 1 m (10 kN downward), and UDL from 0 to 2 m (5 kN/m × 2 m = 10 kN downward).
• Shear force

$$V = R_A – \text{Point load} – \text{UDL up to 2 m} = 17.5 – 10 – 10 = -2.5 \, \text{kN}$$

• Result: The shear force at 2 m from the left support is -2.5 kN (negative indicates downward shear).

FAQs about Shear Force