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

# Relation between w, V and M

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

- Asked in: 2080 Bhadra, 2 marks; 2068 Chaitra, 6 marks (asked 2 times)
- Video: [Relation Between Loading Intensity |Shear Force and Bending Moment | Strength of Materials #zafarsir](https://www.youtube.com/watch?v=pp-Pgy2TcLM) — Impact Academy Official
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/load-shear-moment-relation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-1

## Question

Show the relation between loading intensity 'w', shear force 'V' and bending moment 'M'.

## Short answer

dW/dx = -w, dM/dx = V, thus d²M/dx² = -w. The slope of the shear diagram equals minus the load intensity, and the slope of the moment diagram equals the shear force. The area under the load diagram equals the change in shear, and the area under the shear diagram equals the change in moment.

## Answer

For a beam carrying a distributed load of intensity $w$ (taken positive downward), the shear force $V$ and the bending moment $M$ at a section $x$ are related by

$$
\frac{dV}{dx} = -w,\qquad \frac{dM}{dx} = V,\qquad \frac{d^2M}{dx^2} = -w
$$

#### Derivation
Consider a small element of length $dx$ with load $w$ (acting down). Shear $V$ and moment $M$ act on the left face; $V + dV$ and $M + dM$ act on the right face.

```text
        w (per unit length)
   ↓ ↓ ↓ ↓ ↓ ↓ ↓
  M ┌───────────┐ M + dM
 V ↑│           │↓ V + dV
    └───────────┘
        dx
```

Vertical equilibrium: $V - w\,dx - (V + dV) = 0 \Rightarrow \dfrac{dV}{dx} = -w$.

Moments about the right face (ignoring the second-order term $w\,dx^2/2$): $M + V\,dx - M - dM = 0 \Rightarrow \dfrac{dM}{dx} = V$.

#### Integrated forms between sections 1 and 2
$$
V_2 - V_1 = -\int_{x_1}^{x_2} w\,dx\;(\text{= minus the area of the load diagram}),\qquad
M_2 - M_1 = \int_{x_1}^{x_2} V\,dx\;(\text{= area of the shear force diagram})
$$

#### Consequences
- The slope of the SFD at any point equals minus the load intensity there; the slope of the BMD equals the shear force.
- The bending moment is a maximum or minimum where $V = 0$ (or where $V$ changes sign).
- No load: the SFD is horizontal and the BMD is linear. UDL: the SFD is linear and the BMD is parabolic. Triangular load: the SFD is parabolic and the BMD is cubic.
- A point load causes a sudden jump in $V$ equal to the load; a couple causes a jump in $M$ equal to the couple, while the SFD is unaffected.

## Check yourself

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

1. For a downward load intensity w, what is the relation between w and V?
   A. dV/dx = w
   B. dV/dx = -w
   C. dM/dx = -w
   D. V = w x M
   Answer: B. Vertical equilibrium of the element gives V - w dx - (V + dV) = 0, so dV/dx = -w. (Key point: Load-shear relation)

2. What does the slope of the bending moment diagram at a section equal?
   A. The load intensity w
   B. Minus the load intensity
   C. The shear force V at that section
   D. The area of the load diagram
   Answer: C. Moment equilibrium of the element gives dM/dx = V. (Key point: Shear-moment relation)

3. Where does the bending moment have a maximum or minimum?
   A. Where w is largest
   B. At the supports always
   C. Where the slope of the SFD is zero
   D. Where V = 0 and changes sign
   Answer: D. Since dM/dx = V, M has a stationary value where V = 0, and a sign change of V makes it a maximum or minimum. (Key point: Maximum moment)

4. Under a UDL, what are the shapes of the SFD and BMD?
   A. SFD linear, BMD parabolic
   B. SFD constant, BMD linear
   C. SFD parabolic, BMD cubic
   D. SFD linear, BMD linear
   Answer: A. A UDL makes V vary linearly and M parabolically, since each is the integral of the one before. (Key point: Diagram shapes)

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