---
title: "Conjugate beam: vertical deflection and rotation at C"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:conjugate-udl-5knm-overhang-2071m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-udl-5knm-overhang-2071m
site: BCT Engineering
---

# Conjugate beam: vertical deflection and rotation at C

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2071 Magh, 8 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-udl-5knm-overhang-2071m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

Determine the vertical deflection and rotation at free end C of the overhanging beam ABC loaded as shown in figure below by using conjugate beam method. [Figure: beam ABC; AB = 5 m with 5 kN/m UDL; overhang BC = 2 m with 10 kN at C.]

## Short answer

Conjugate beam analysis with M/EI as load gives free end response. The deflection at C is 41.25/EI kN·m³ downward and the rotation is 27.29/EI rad clockwise.

## Answer

#### Given data
Overhanging beam ABC: $AB = 5$ m with UDL 5 kN/m, overhang $BC = 2$ m with 10 kN at C. $EI$ constant.

#### Step 1: Reactions
$R_{A} = 8.50$ kN, $R_{B} = 26.50$ kN (positive = upward; check: $\sum F_y = 35.00$ kN equals the total load 35.00 kN).

#### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces
In AB, $M = 8.5x - 2.5x^2$. This is the free (simply supported) parabola of the UDL plus a hogging triangle caused by the overhang moment $M_B = -10\times2 = -20$ kNm. In BC, $M = 10x - 70$.

| Piece | Area of $M/EI$ | $\bar x$ from A (m) |
|---|---|---|
| Parabola on AB (peak $wL^2/8 = 15.625$ kNm) | $\tfrac23(5)(15.625) = +52.083$ | 2.500 |
| Triangle on AB (0 at A to $-20$ at B) | $\tfrac12(5)(-20) = -50.000$ | 3.333 |
| Triangle on BC ($-20$ at B to 0 at C) | $\tfrac12(2)(-20) = -20.000$ | 5.667 |

(Areas are in units of $1/EI$; the net area on AB is $+2.083$.)

#### Step 3: Conjugate beam
The real beam has a hinge at A, a roller at B and a free end at C, so the conjugate beam has a simple support at A, an internal hinge at B and a fixed end at C. 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 span AB about the hinge B (conjugate moment is zero there):

$$
R'_A\times5 = (52.083)(2.500) + (-50.000)(1.667) = 130.208 - 83.333 = 46.875
$$

$$
R'_A = \frac{9.375}{EI}\ \text{(slope at A, clockwise)}
$$

**At C (free end) ($x = 7$ m):**

$$
V' = R'_A - \sum A = 9.375 - [(52.083) + (-50.000) + (-20.000)] = \frac{27.292}{EI}\ \text{(slope, clockwise)}
$$

$$
M' = R'_A\,x - \sum A(x-\bar x) = 65.625 - [(52.083)(4.500) + (-50.000)(3.667) + (-20.000)(1.333)] = 65.625 - 24.375 = \frac{41.250}{EI}\ \text{(deflection, downward)}
$$

**Answer: deflection at the free end C $= \dfrac{41.25}{EI}$ downward; rotation at C $= \dfrac{27.29}{EI}$ rad (clockwise)** (kNm$^3$/EI and kNm$^2$/EI).

## Check yourself

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

1. What are the reactions at A and B?
   A. RA = 26.5 kN, RB = 8.5 kN
   B. RA = 8.5 kN, RB = 26.5 kN
   C. RA = 12.5 kN, RB = 22.5 kN
   D. RA = 17.5 kN, RB = 17.5 kN
   Answer: B. Moments about A: RB x 5 = 5x5x2.5 + 10x7 = 132.5, so RB = 26.5 kN and RA = 8.5 kN. (Key point: Reactions)

2. What conjugate supports correspond to the real beam ABC (hinge A, roller B, free end C)?
   A. Hinge at A, internal hinge at B, fixed at C
   B. Roller at A, fixed at B, free at C
   C. Fixed at A, hinge at B, roller at C
   D. Hinge at A, roller at B, hinge at C
   Answer: A. These are the standard conjugate substitutions for ends and interior supports. (Key point: Hinge/fixed in conjugate)

3. How is the rotation at C obtained and what is it?
   A. Conjugate moment at C, 41.25/EI
   B. Conjugate shear at C, 27.29/EI clockwise
   C. Conjugate shear at A, 9.375/EI
   D. Conjugate moment at B, zero
   Answer: B. Rotation is the conjugate shear: 9.375 - (2.08 - 20) = 27.29/EI. (Key point: Rotation at C)

4. What is the deflection at C?
   A. 41.25/EI downward
   B. 27.29/EI downward
   C. 41.25/EI upward
   D. 65.63/EI downward
   Answer: A. Conjugate moment at C is 65.625 - 24.375 = 41.25/EI, downward. (Key point: Deflection at C)

## Related in Theory of Structures I

- [Conjugate beam: overhanging beam free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-overhang-300-150) — Theory of Structures I, chapter 4
- [Conjugate beam: slope and deflection at free end C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-2knm-10kn-5kn-2068) — Theory of Structures I, chapter 4, asked 2×
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