---
title: "Moment area: slopes at A, B, deflection at D"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:moment-area-slopes-a-b-defl-d-2065"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-slopes-a-b-defl-d-2065
site: BCT Engineering
---

# Moment area: slopes at A, B, deflection at D

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2065 Chaitra, 8 marks
- Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=FFv2pErDeBU) — Civil Engineering Exam
- 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-slopes-a-b-defl-d-2065
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

Determine the slope at A and B and deflection at D of the beam loaded as shown in fig-3 using moment area method. Take EI as constant. [Figure: simply supported beam AB of 18 m (12 m + 6 m as dimensioned); 10 kN downward load at C, 9 m from A; D lies between A and C. Figure partly unclear.]

## Short answer

Slopes at A and B are 202.5/EI rad. Deflection at D (6 m from A) is 1035/EI downward. The tangent at midspan C is horizontal; first and second moment-area theorems give these values from the triangular M/EI diagram.

## Answer

**Assumptions:** simply supported beam $AB$ of span $L = 18$ m, $P = 10$ kN at $C$ (mid-span, 9 m from $A$); $D$ is taken 6 m from $A$ because its exact position is not clear in the figure. $EI$ is constant.

#### Reactions and BM diagram
$R_A = R_B = 5$ kN. Maximum BM at $C$: $M_C = 5\times 9 = 45$ kN m. At $D$: $M_D = 5\times 6 = 30$ kN m.

```text
  A o-----6-----D---3---C--------9--------o B
                        10 kN
  BMD: triangle, peak 45 kN m at C
```

#### Slope at A and B
The loading is symmetrical, so the tangent at $C$ is horizontal. By the first moment-area theorem, the slope at $A$ equals the area of the $M/EI$ diagram between $A$ and $C$:

$$
\theta_A = \frac{1}{EI}\times\frac12\times 9\times 45 = \frac{202.5}{EI}
$$

$$
\theta_B = \theta_A = \frac{202.5}{EI}\ \text{(opposite sense)}
$$

(Check: $PL^2/16EI = 10\times 18^2/16 = 202.5$.)

#### Deflection at D
By the second theorem, the deviation of $A$ from the tangent at $C$ is the moment of the $M/EI$ area $AC$ about $A$. Since the tangent at $C$ is horizontal and $A$ has no deflection, this is the deflection at $C$:

$$
y_C = \frac{1}{EI}\left[\frac12\times 9\times 45\right]\times\frac{2}{3}\times 9 = \frac{1215}{EI}
$$

Deviation of $D$ from the tangent at $C$ (moment about $D$ of the $M/EI$ area between $D$ and $C$; rectangle $30\times 3$ plus triangle $\frac12\times 3\times 15$):

$$
t_D = \frac{1}{EI}\left[30\times 3\times 1.5 + \tfrac12\times 3\times 15\times 2\right] = \frac{135+45}{EI} = \frac{180}{EI}
$$

$$
y_D = y_C - t_D = \frac{1215-180}{EI} = \frac{1035}{EI}
$$

(Check: $y = \dfrac{Px(3L^2-4x^2)}{48EI} = 1035/EI$.)

**Answer: $\theta_A = \theta_B = 202.5/EI$ kN m² (rad); $y_D = 1035/EI$ kN m³ downward (D at 6 m from A).**

## Check yourself

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

1. Why can the tangent at the mid-span point C be treated as horizontal in this beam?
   A. The beam and loading are symmetrical about C
   B. The beam is fixed at C
   C. The load P is zero at C
   D. The deflection at C is zero
   Answer: A. Symmetry of the beam and load makes the slope at C zero, so the tangent there is horizontal. (Key point: Tangent at C)

2. How is the slope at A found from the first moment-area theorem here?
   A. Moment of the M/EI area about A
   B. Area of the M/EI diagram between A and C
   C. Maximum M divided by EI
   D. Area of the M/EI diagram over the full span AB
   Answer: B. The change in slope between A and C equals the M/EI area, and the slope at C is zero. (Key point: First theorem)

3. What is the slope at A (and B) for P = 10 kN at mid-span of the 18 m beam?
   A. 101.25/EI
   B. 405/EI
   C. 202.5/EI
   D. 1215/EI
   Answer: C. theta = 1/2 x 9 x 45/EI = 202.5/EI, matching PL^2/16EI. (Key point: Slope value)

4. How is the deflection at D obtained?
   A. y_C plus the deviation of D from the tangent at C
   B. Area of M/EI between A and D
   C. Slope at A times 6 m
   D. y_C minus the deviation of D from the tangent at C
   Answer: D. The tangent at C is horizontal, so y_D = y_C - t_D = (1215 - 180)/EI = 1035/EI. (Key point: Deflection at D)

## Related in Theory of Structures I

- [Moment area: slopes at A, B, deflection at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-beam-60kn-12m) — Theory of Structures I, chapter 4
- [Moment area: slope and deflection under 50 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-12-ss-defl-4m-d) — Theory of Structures I, chapter 4
- [Moment area: overhang beam free-end deflection](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-overhang-10kn-m-2078) — Theory of Structures I, chapter 4
- [Moment area: B, C, maximum deflection](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-200-300kn) — Theory of Structures I, chapter 4
- [Moment area: slope and deflection at E and at 4 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-e-4m-a) — Theory of Structures I, chapter 4
- [Moment area: deflection at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-overhang-20kn-10kn-2071o) — Theory of Structures I, chapter 4

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