---
title: "Conjugate beam: slope at A, deflection at C"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:conjugate-slope-a-defl-c-30-20"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-slope-a-defl-c-30-20
site: BCT Engineering
---

# Conjugate beam: slope at A, deflection at C

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2078 Chaitra, 10 marks
- Video: [Conjugate Beam Method Slope and Deflection - Introduction and Step by Step Example](https://www.youtube.com/watch?v=QsNvwRQVkBE) — 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/conjugate-slope-a-defl-c-30-20
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

Find the slope at A and deflection at C, using conjugate beam method for the given beam. Take EI to be constant. [Figure: beam ABC; A hinge; 30 kN load at 3 m from A; B roller 3 m beyond that load; overhang BC = 2 m with 20 kN at C.]

## Short answer

With the conjugate beam method, the conjugate reaction at A gives the slope at A as 27.5/EI clockwise. The deflection at the free end C is 78.33/EI downward.

## Answer

#### Given data
Beam ABC: A hinge, 30 kN at 3 m from A, B roller 3 m beyond the load (at 6 m), overhang $BC = 2$ m with 20 kN at C. $EI$ constant.

#### Step 1: Reactions
$R_{A} = 8.33$ kN, $R_{B} = 41.67$ kN (positive = upward; check: $\sum F_y = 50.00$ kN equals the total load 50.00 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 3 | $\frac{25 x}{3}$ | EI | 37.50 | 2.000 |
| 3 to 6 | $90 - \frac{65 x}{3}$ | EI | -22.50 | 6.667 |
| 6 to 8 | $20 x - 160$ | EI | -40.00 | 6.667 |

(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. In the portion 3 to 6 m the moment changes sign, so its net area is small and negative and its "centroid" falls outside the portion; it is used only to give the correct first moment of that portion.)

#### Step 3: Conjugate beam
The real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

#### Step 4: Conjugate reaction at A (slope at A)
Taking moments of the conjugate beam about the support at $x = 6$ m (conjugate moment is zero there):

$$
R'_A\times6 = (37.50)(4.000) + (-22.50)(-0.667) = 165.00
$$

$$
R'_A = \frac{27.500}{EI}\ \text{(slope at A, clockwise)}
$$

**At the 30 kN load ($x = 3$ m):**

$$
V' = R'_A - \sum A = 27.500 - [(37.50)] = \frac{-10.000}{EI}\ \text{(slope, anticlockwise)}
$$

$$
M' = R'_A\,x - \sum A(x-\bar x) = 82.500 - [(37.50)(1.000)] = \frac{45.000}{EI}\ \text{(deflection, downward)}
$$

**At C (free end) ($x = 8$ m):**

$$
V' = R'_A - \sum A = 27.500 - [(37.50) + (-22.50) + (-40.00)] = \frac{52.500}{EI}\ \text{(slope, clockwise)}
$$

$$
M' = R'_A\,x - \sum A(x-\bar x) = 220.000 - [(37.50)(6.000) + (-22.50)(1.333) + (-40.00)(1.333)] = \frac{78.333}{EI}\ \text{(deflection, downward)}
$$

**Answer: slope at A $= \dfrac{27.5}{EI}$ (clockwise); deflection at C $= \dfrac{78.33}{EI}$ 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. For this beam (hinge at A, roller at B, free end C) what are the conjugate supports?
   A. Hinge A, roller B, free C
   B. Fixed A, internal hinge B, free C
   C. Hinge A, internal hinge B, fixed C
   D. Hinge A, fixed B, hinge C
   Answer: C. A remains a hinge, interior support B becomes an internal hinge and free end C becomes fixed. (Key point: Conjugate beam supports)

2. How is R'_A (the slope at A) determined?
   A. Sum of all M/EI areas
   B. Conjugate moment about B is zero for the part AB, so R'_A x 6 = 165 giving 27.5/EI
   C. Divide the 30 kN load by EI
   D. Take moments about C
   Answer: B. The conjugate moment at the internal hinge B is zero, which gives R'_A x 6 = 165, so theta_A = 27.5/EI clockwise. (Key point: Slope at A)

3. With R'_A = 27.5/EI and a triangle area 37.5/EI to the left of the load, what is the slope under the load?
   A. 27.5 + 37.5 = 65/EI clockwise
   B. 27.5 - 37.5 = -10/EI, i.e. 10/EI anticlockwise
   C. 37.5/EI clockwise
   D. 27.5/EI clockwise
   Answer: B. Slope is conjugate shear, V' = R'_A - area to the left = 27.5 - 37.5 = -10/EI (anticlockwise). (Key point: Slope under the 30 kN load)

4. How is the deflection at C obtained and what is the result?
   A. Conjugate shear at C: 52.5/EI
   B. Conjugate moment at the fixed end C: 78.33/EI downward
   C. R'_A x 8 = 220/EI
   D. Conjugate moment at C: 78.33/EI upward
   Answer: B. Deflection is the conjugate moment at the fixed end: 27.5x8 minus the moments of the areas = 78.33/EI, positive so downward. The shear 52.5/EI is the slope. (Key point: Deflection at the free end)

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