---
title: "Moment area: slope and deflection at E and at 4 m"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:moment-area-e-4m-a"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-e-4m-a
site: BCT Engineering
---

# Moment area: slope and deflection at E and at 4 m

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2076 Bhadra, 12 marks
- Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=jkwPzvWsqIY) — AF Math & Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-e-4m-a
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

Using moment-area theorems, calculate the slope and deflection at end E of the beam shown in the figure. Also calculate the deflection at 4 m from the support A. [Figure: beam A to E; A hinge, B roller; AC = 2 m, CD = 4 m with 10 kN/m UDL on CD, DB = 2 m, overhang BE = 1 m with 15 kN at E.]

## Short answer

Reactions: 18.12 kN at A, 36.88 kN at B. Using moment-area theorems and the M/EI diagram, slope at free end E is 99.17/EI anticlockwise and deflection is 101.67/EI upward. Deflection at 4 m from A is 320/EI downward.

## Answer

#### Given data
Beam A to E: A hinge at 0, C at 2 m, D at 6 m, B roller at 8 m, E free end at 9 m. UDL 10 kN/m on CD (from 2 m to 6 m), 15 kN at E. $EI$ constant.

```text
          10 kN/m
 A --- C =========== D --- B -- v E
 ^  2 m      4 m       2 m  ^ 1 m 15 kN
```

#### Step 1: Reactions
$R_{A} = 18.12$ kN, $R_{B} = 36.88$ kN (positive = upward; check: $\sum F_y = 55.00$ kN equals the total load 55.00 kN).

#### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces
| Portion (m) | $M$ (kNm) | stiffness | Area of $M/EI$ | $\bar x$ from A (m) |
|---|---|---|---|---|
| 0 to 2 | $\frac{145 x}{8}$ | EI | 36.25 | 1.333 |
| 2 to 6 | $- 5 x^{2} + \frac{305 x}{8} - 20$ | EI | 183.33 | 3.945 |
| 6 to 8 | $160 - \frac{175 x}{8}$ | EI | 13.75 | 5.939 |
| 8 to 9 | $15 x - 135$ | EI | -7.50 | 8.333 |

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

#### Step 3: Tangent at A
Deviation of the support at $x = 8$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$
t = (36.25)(6.667) + (183.33)(4.055) + (13.75)(2.061) = \frac{1013.33}{EI}
$$

That support does not deflect, so $\theta_A = -t/8$:

$$
\theta_A = \frac{-126.667}{EI}\ \text{(clockwise)}
$$

**At the point 4 m from A ($x = 4$ m):**

$$
\theta = \theta_A + \sum A = -126.667 + [(36.25) + (95.42)] = \frac{5.000}{EI}\ \text{(anticlockwise)}
$$

$$
y = \theta_A\,x + t_{P/A} = -506.667 + [(36.25)(2.667) + (95.42)(0.943)] = \frac{-320.000}{EI}\ \text{(downward)}
$$

**At E (free end) ($x = 9$ m):**

$$
\theta = \theta_A + \sum A = -126.667 + [(36.25) + (183.33) + (13.75) + (-7.50)] = \frac{99.167}{EI}\ \text{(anticlockwise)}
$$

$$
y = \theta_A\,x + t_{P/A} = -1140.000 + [(36.25)(7.667) + (183.33)(5.055) + (13.75)(3.061) + (-7.50)(0.667)] = \frac{101.667}{EI}\ \text{(upward)}
$$

**Answer:** at E: slope $= \dfrac{99.17}{EI}$ (anticlockwise) and deflection $= \dfrac{101.67}{EI}$ upward; at 4 m from A: deflection $= \dfrac{320}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI).

## Check yourself

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

1. How is theta_A = 126.667/EI obtained?
   A. Total area of M/EI divided by 8
   B. Area of triangle AC divided by 2
   C. Deviation of the roller B from the tangent at A (t = 1013.33/EI) divided by 8 m
   D. R_A divided by EI
   Answer: C. B stays at the level of A, so theta_A = t/8 = 1013.33/8 = 126.67/EI (clockwise). (Key point: Slope at A from tangent deviation)

2. For the 1 m overhang with 15 kN at E, how is its M/EI diagram described?
   A. Sagging, 15/EI at E
   B. Zero at B, hogging towards E
   C. Constant 15/EI
   D. Hogging, 15/EI at B reducing to 0 at E (area -7.5/EI)
   Answer: D. M = 15x - 135 gives -15 kNm at B and zero at E, a hogging triangle of area 0.5x1x15 = 7.5, taken negative. (Key point: Overhang moment)

3. At x = 4 m, theta = -126.667 + 36.25 + 95.42 = 5/EI. What does this show?
   A. Slope is still clockwise and large
   B. Beam has zero deflection there
   C. The slope is now small and anticlockwise, so the beam has almost reached its lowest point
   D. The M/EI areas cancel the deflection
   Answer: C. Slope = theta_A plus the area up to the section. The result of +5/EI is nearly zero, so the deflection (320/EI downward) is close to its maximum. (Key point: Slope at 4 m from A)

4. What are the slope and deflection at E?
   A. 99.17/EI anticlockwise and 101.67/EI upward
   B. 99.17/EI clockwise and 101.67/EI downward
   C. 126.67/EI clockwise and 320/EI downward
   D. 7.5/EI and 101.67/EI downward
   Answer: A. Slope = theta_A + total areas = 99.17/EI (anticlockwise). Deflection = theta_A x 9 plus the areas' moments = +101.67/EI, upward. (Key point: Results at free end E)

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