Lesson 2.3: Shear Force & Bending Moment Diagrams

Shear Force & Bending Moment analysis is a core topic in Strength of Materials. GATE and PSU exams often test beam reactions, shear force (SF), bending moment (BM), and diagram construction for various supports and loads.

🔹 1. Basic Concepts

• Shear Force (SF): Internal force acting perpendicular to the cross-section of a beam.

$$V=\text{Sum of vertical forces on one side of section}$$

• Bending Moment (BM): Internal moment causing bending of the beam.

$$M=\text{Sum of moments about the section}$$

• Positive sign conventions:

  • SF: Upwards on left, downwards on right

  • BM: Sagging (+), Hogging (−)

🔹 2. Types of Supports & Reactions

• Simply Supported Beam: Pin + Roller → Vertical reactions only

• Cantilever Beam: Fixed at one end → Vertical, Horizontal, Moment reactions

• Overhanging Beam: Extensions beyond supports → Extra reactions

• Propped Cantilever / Fixed Beam: Mixed boundary conditions

Applications:

• Bridges, building beams, crane structures

🔹 3. Relationships Between Load, SF & BM

• Differential Relations:

dVdx=−w(x),dMdx=V(x)\frac{dV}{dx} = -w(x), \quad \frac{dM}{dx} = V(x)dxdV​=−w(x),dxdM​=V(x)

Where $w(x)$ = distributed load

• Key Points:

  • SF = 0 → BM maximum/minimum

  • BM = 0 → SF changes sign

🔹 4. Steps to Draw SF & BM Diagrams

1. Calculate reactions using equilibrium equations (∑Fy=0,∑M=0\sum F_y = 0, \sum M = 0∑Fy​=0,∑M=0)

2. Cut sections at points of interest (just left/right of loads/supports)

3. Calculate shear force for each section

4. Calculate bending moment for each section

5. Plot SF & BM diagrams

6. Identify maximum BM → design section

🔹 5. Common Loading Cases

🔹 6. Solved Examples (PYQ Style)

Example 1 (GATE ME 2018):
Simply supported beam, span 6 m, point load 12 kN at midspan. Draw SF & BM diagrams.

👉 Solution:

• Reactions: R1 = R2 = 6 kN

• SF left of load = +6 kN, right = -6 kN

• BM maximum at midspan: Mmax=6∗3=18 kNmM_{\text{max}} = 6*3 = 18 \text{ kNm}Mmax​=6∗3=18 kNm

• Diagrams: SF → triangular, BM → parabolic

Example 2 (PSU Exam):
Cantilever beam, length 4 m, end load 10 kN. Draw SF & BM diagrams.

👉 Solution:

• SF constant along length = -10 kN

• BM maximum at fixed end = 10*4 = 40 kNm

• Diagrams: SF → horizontal line, BM → linear decreasing

🔹 7. Practice Exercises

1. Simply supported beam 8 m, UDL 5 kN/m over entire span. Find reactions & draw SF/BM diagrams.

2. Cantilever beam 3 m, point load 6 kN at free end. Find maximum bending moment.

3. Beam with overhang 2 m beyond support, UDL 4 kN/m. Draw SF/BM diagrams.

4. Identify location of maximum BM for simply supported beam with central point load.

🔹 8. Summary

• Shear Force (SF): Vertical internal force → affects shear stress

• Bending Moment (BM): Moment causing bending → affects bending stress

• Relationships: dM/dx = V, dV/dx = -w(x)

• Diagrams: Triangular/linear for SF, Parabolic/linear for BM depending on load

• Applications: Beam design, structural analysis, mechanical systems