---
title: "Deflection at A: beam with load and moment"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:beam-deflection-point-a-moment-and"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-deflection-point-a-moment-and
site: BCT Engineering
---

# Deflection at A: beam with load and moment

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Lecture 011 - Slope and Deflecion Example Using Double Integration Method](https://www.youtube.com/watch?v=khf_cy_lFrU) — Pinoy Ipsum 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-deflection-point-a-moment-and
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Obtain deflection at point A of the following beam. [Figure: beam of length L with constant EI, fixed at B (right end) and free at A (left end); a point load P and a moment M act at A.]

## Short answer

Deflection at A is PL³/(3EI) + ML²/(2EI) downward.

## Answer

**Assumptions:** cantilever of length $L$, fixed at B, free end A carries a downward load $P$ and a couple $M$ whose sense also tends to bend the beam downward (same direction as $P$). $EI$ constant.

Take $x$ measured from the free end A and $y$ positive upward. Both $P$ and $M$ cause hogging, so the bending moment at $x$ gives:

$$
EI\frac{d^2y}{dx^2} = -(Px + M)
$$

Integrate:

$$
EI\frac{dy}{dx} = -\left(\frac{Px^2}{2} + Mx\right) + C_1
$$

At the fixed end B ($x = L$) the slope is zero:

$$
C_1 = \frac{PL^2}{2} + ML
$$

Integrate again:

$$
EI\,y = -\left(\frac{Px^3}{6} + \frac{Mx^2}{2}\right) + C_1 x + C_2
$$

At B ($x=L$), $y = 0$:

$$
C_2 = \frac{PL^3}{6} + \frac{ML^2}{2} - \frac{PL^3}{2} - ML^2 = -\frac{PL^3}{3} - \frac{ML^2}{2}
$$

At A ($x=0$), $EI\,y_A = C_2$:

$$
y_A = -\left(\frac{PL^3}{3EI} + \frac{ML^2}{2EI}\right)
$$

The negative sign means the deflection is downward.

(Superposition gives the same: $PL^3/3EI$ from the point load plus $ML^2/2EI$ from the couple.)

**Answer: deflection at A $= \dfrac{PL^3}{3EI} + \dfrac{ML^2}{2EI}$ downward.** If the couple acts in the opposite sense, subtract the second term.

## Check yourself

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

1. What is the bending moment equation used, with x measured from the free end A?
   A. EI y'' = -P(L - x)
   B. EI y'' = -Px^2/2
   C. EI y'' = -M only
   D. EI y'' = -(Px + M)
   Answer: D. Moment at x from the free end is the load times x plus the couple M. (Key point: Double integration start)

2. Which conditions give the constants?
   A. Slope and deflection are both zero at the fixed end B (x = L)
   B. Slope and deflection are zero at A
   C. Deflection zero at A and slope zero at B
   D. Moment zero at B
   Answer: A. B is fixed, so y = 0 and dy/dx = 0 there, which fixes C1 and C2. (Key point: Boundary conditions)

3. What is the deflection at A?
   A. PL^3/(3EI) + ML/(EI)
   B. PL^3/(3EI) + ML^2/(2EI)
   C. PL^3/(8EI) + ML^2/(2EI)
   D. PL^3/(3EI) - ML^2/(2EI) for a couple in the same sense as P
   Answer: B. Load gives PL^3/3EI and the couple gives ML^2/2EI; they add when the couple acts in the same sense as P. (Key point: Final result)

4. How does superposition confirm the result?
   A. Multiply the two deflections
   B. Use only the larger of the two
   C. Add the cantilever deflection due to P and that due to couple M
   D. Subtract always
   Answer: C. Deflections of linear systems add, giving PL^3/3EI + ML^2/2EI. (Key point: Superposition check)

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