---
title: "Deflection at B, C and slope at A, D"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:beam-deflection-bc-slope-ad"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/beam-deflection-bc-slope-ad
site: BCT Engineering
---

# Deflection at B, C and slope at A, D

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2080 Bhadra, 8 marks
- Video: [Deflection of beams 10 //Macaulay’s method//simply supported beam with two/2 point loads](https://www.youtube.com/watch?v=pHsuuNCktuQ) — Mech PM
- 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-bc-slope-ad
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Determine deflection at points B and C and slope at points A and D. [Figure: beam A-B-C-D-E with EI constant; hinge support at A and roller support at D; 5 kN point load at B; 1 kN/m UDL over C to D (3 m); 2 kNm couple at the overhanging end E; lengths A-B 1 m, B-C 1 m, C-D 3 m, D-E 2 m.]

## Short answer

Downward deflection at B: 6.66/EI m, at C: 9.65/EI m. Slope at A: 7.41/EI rad clockwise; at D: 4.34/EI rad anticlockwise (EI in kN·m²).

## Answer

**Reading of the figure:** $AB = 1$ m, $BC = 1$ m, $CD = 3$ m, $DE = 2$ m; hinge at A (x = 0) and roller at D (x = 5 m). Loads: 5 kN at B (x = 1), UDL 1 kN/m over CD (x = 2 to 5), and a 2 kN·m couple at the free end E (taken clockwise). $EI$ is constant. Deflection is taken positive upward.

#### Reactions
Moments about A (anticlockwise positive):

$$
R_D(5) - 5(1) - 3(3.5) - 2 = 0 \Rightarrow R_D = 3.5\ \text{kN}, \qquad R_A = 5 + 3 - 3.5 = 4.5\ \text{kN}
$$

#### Macaulay's method
The UDL ends at D, so it is continued to the end and cancelled by an equal upward load (+1 kN/m beyond D). Bending moment (sagging positive) at a section $x$ from A:

$$
EI\frac{d^2y}{dx^2} = 4.5x - 5\langle x-1\rangle - \frac{1}{2}\langle x-2\rangle^2 + 3.5\langle x-5\rangle + \frac{1}{2}\langle x-5\rangle^2 + 2\langle x-7\rangle^0
$$

Integrating:

$$
EI\,\theta = 2.25x^2 - 2.5\langle x-1\rangle^2 - \frac{1}{6}\langle x-2\rangle^3 + 1.75\langle x-5\rangle^2 + \frac{1}{6}\langle x-5\rangle^3 + 2\langle x-7\rangle + C_1
$$

$$
EI\,y = 0.75x^3 - \frac{5}{6}\langle x-1\rangle^3 - \frac{1}{24}\langle x-2\rangle^4 + \frac{7}{12}\langle x-5\rangle^3 + \frac{1}{24}\langle x-5\rangle^4 + \langle x-7\rangle^2 + C_1x + C_2
$$

#### Constants
- At A, $y = 0$ at $x = 0$: $C_2 = 0$.
- At D, $y = 0$ at $x = 5$: $0.75(125) - \frac{5}{6}(64) - \frac{1}{24}(81) + 5C_1 = 0 \Rightarrow C_1 = -7.4083$

#### Slopes at A and D
$$
EI\,\theta_A = C_1 = -7.408\ \text{kN·m}^2 \quad\Rightarrow\quad \theta_A = \frac{7.408}{EI}\ \text{rad (clockwise)}
$$

$$
EI\,\theta_D = 2.25(25) - 2.5(16) - \frac{27}{6} + C_1 = 4.342\ \text{kN·m}^2 \quad\Rightarrow\quad \theta_D = \frac{4.342}{EI}\ \text{rad (anticlockwise)}
$$

#### Deflections at B and C
$$
EI\,y_B = 0.75(1) + C_1(1) = -6.658\ \text{kN·m}^3 \Rightarrow y_B = \frac{6.658}{EI}\ \text{downward}
$$

$$
EI\,y_C = 0.75(8) - \frac{5}{6}(1) + C_1(2) = -9.650\ \text{kN·m}^3 \Rightarrow y_C = \frac{9.650}{EI}\ \text{downward}
$$

**Answer (with $EI$ in kN·m²):** $y_B = 6.66/EI$ and $y_C = 9.65/EI$ (downward); $\theta_A = 7.41/EI$ (clockwise) and $\theta_D = 4.34/EI$ (anticlockwise).

## Check yourself

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

1. How is R_D found?
   A. Moments about A equal to zero
   B. Using Macaulay constants
   C. Sum of horizontal forces
   D. Setting slope zero at D
   Answer: A. Taking moments about A gives R_D = 3.5 kN, and R_A = 4.5 kN. (Key point: Reaction at D)

2. In Macaulay's method, how is a UDL that ends at D handled?
   A. Doubled
   B. Replaced by a point load
   C. Ignored beyond D
   D. Extended to the end and cancelled by an equal upward UDL
   Answer: D. Macaulay terms must continue to the end of the beam, so the load is extended and the extra cancelled. (Key point: UDL not reaching the end)

3. Which conditions determine C1 and C2?
   A. y = 0 at E
   B. y = 0 at B and C
   C. y = 0 at A and at D
   D. slope zero at A and D
   Answer: C. Both supports have zero deflection: C2 = 0, then C1 = -7.408. (Key point: Boundary conditions)

4. Since EI theta_A = C1 = -7.408 kN·m^2, what is the slope at A?
   A. Clockwise 7.41/EI
   B. 4.34/EI clockwise
   C. Anticlockwise 7.41/EI
   D. Zero
   Answer: A. Slope at A equals C1, negative, meaning clockwise 7.41/EI. (Key point: Slope at A)

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