---
title: "Point of contraflexure; w-V-M relation"
subject: "Strength of Materials"
chapter: 1
key: "semester_3/strength_of_materials:contraflexure-load-shear-moment-relation"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/contraflexure-load-shear-moment-relation
site: BCT Engineering
---

# Point of contraflexure; w-V-M relation

Strength of Materials · Chapter 1: Axial Forces, Shearing Forces and Bending Moments

- Asked in: 2076 Chaitra, 2+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/contraflexure-load-shear-moment-relation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Define point of contraflexure. Derive the relationship between rate of loading, Shear force and Bending moment.

## Short answer

A point of contraflexure is where bending moment changes sign, passing through zero. For a beam element, dV/dx = -w and dM/dx = V, so d²M/dx² = -w. The rate of shear force change equals negative load intensity; the rate of bending moment change equals shear force.

## Answer

#### Point of contraflexure
A point of contraflexure (inflexion point) is a point on a beam where the bending moment changes sign (passes through zero), so the beam's curvature changes from sagging to hogging or vice versa. The bending moment there is zero, but a zero moment does not always mean contraflexure: the moment must also change sign.

#### Relation between loading rate, shear force and bending moment
Take an element of length $dx$ of a beam on which the load intensity is $w$ per unit length, acting downward. $V$ and $M$ act on the left face; $V + dV$ and $M + dM$ act on the right face.

```text
          w
   ↓ ↓ ↓ ↓ ↓ ↓
  M ┌─────────┐ M + dM
 V ↑│         │↓ V + dV
    └─────────┘
        dx
```

**Vertical equilibrium**
$$
V - w\,dx - (V + dV) = 0 \;\Rightarrow\; \frac{dV}{dx} = -w
$$

**Moment equilibrium** about the right face (the term $w\,dx^2/2$ is negligible):
$$
M + V\,dx - w\,dx\cdot\frac{dx}{2} - (M + dM) = 0 \;\Rightarrow\; \frac{dM}{dx} = V
$$

Combining:
$$
\frac{d^2M}{dx^2} = \frac{dV}{dx} = -w
$$

**Meaning:** the rate of change of shear force equals the negative of the load intensity, and the rate of change of bending moment equals the shear force. The moment is a maximum where $V = 0$. A point of contraflexure occurs where $M = 0$ and $M$ changes sign, i.e. where the curvature of the elastic curve ($1/R = M/EI$) changes sign. Where there is no load near that point ($w = 0$), it is simply where the straight-line BMD crosses the axis.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A point of contraflexure is a point where:
   A. The shear force is maximum
   B. The load is zero
   C. The bending moment changes sign and passes through zero
   D. The deflection is maximum
   Answer: C. The curvature changes between sagging and hogging, so M passes through zero and reverses sign. (Key point: Definition of contraflexure)

2. Is every point of zero bending moment a point of contraflexure?
   A. Yes, always
   B. No, the moment must also change sign there
   C. Yes, if the shear is zero
   D. No, only at supports
   Answer: B. A point where M merely touches zero, such as an internal hinge or a free end, is not contraflexure unless the sign changes. (Key point: Zero moment is not enough)

3. Vertical equilibrium of an element dx with downward load w gives:
   A. dV/dx = w
   B. dM/dx = w
   C. dV/dx = -w, and with dM/dx = V, d^2M/dx^2 = -w
   D. dV/dx = 0
   Answer: C. V - w dx - (V + dV) = 0 gives dV/dx = -w, and combining with dM/dx = V gives d^2M/dx^2 = -w. (Key point: Derivation result)

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