---
title: "Conjugate beam: deflection and slope at E and B"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:conjugate-fixed-link-30-50"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-fixed-link-30-50
site: BCT Engineering
---

# Conjugate beam: deflection and slope at E and B

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2075 Baisakh, 12 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-fixed-link-30-50
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

For the beam shown in figure below, find the deflection and slope at E and B. Take $I = 6\times10^7\ \text{mm}^4$ and $E = 200\ \text{kN/mm}^2$. [Figure: beam fixed at A; segment AB (I) of 5 m connected by a link at B to a second segment (2I) with a support at D; 30 kN load at 3 m from the link end and 2 m before D; free end E, 4 m beyond D, carries a 50 kN load. Details of this figure are partly illegible.]

## Short answer

Splitting at the link, the cantilever AB (EI = 12 000 kN·m²) gives tip B upward deflection 1166.67/EI (0.0972 m) and slope 350/EI anticlockwise. For segment BE, conjugate beam yields right‑side slope at B = 62.33/EI anticlockwise; at E, deflection = 1104/EI (0.092 m) downward and slope = 342.67/EI clockwise.

## Answer

#### Given data and assumptions
(The figure is partly illegible; this is the reading used.) Beam fixed at A. Segment AB: 5 m, rigidity $I$, ending at a **link (internal hinge)** at B. From B the beam (rigidity $2I$) continues with a 30 kN load 3 m from B, a roller support D at 5 m from B and a free end E at 9 m from B carrying 50 kN. $I = 6\times10^7$ mm$^4 = 6\times10^{-5}$ m$^4$, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, so $EI = 12\,000$ kNm$^2$.

```text
 A=====|B      30 kN          50 kN
 fixed |link ----v---- D(roller) ------v E
  I 5 m  |    3 m    2 m ^      4 m (2I)
```

#### Step 1: Split at the link
The link carries shear but no moment. The part BE rests on the roller D and on the link at B (which is carried by the cantilever AB). Moments about D (BE alone):

$$
R_B\times5 = 30\times2 - 50\times4 \;\Rightarrow\; R_B = -28\ \text{kN},\qquad R_D = 80 + 28 = 108\ \text{kN}
$$

The support force on BE at B is 28 kN downward, so BE pulls the tip of the cantilever AB with an **upward** force of 28 kN.

#### Step 2: Cantilever AB (rigidity $EI$)
Upward tip load 28 kN, length 5 m:

$$
y_B = \frac{28\times5^3}{3EI} = \frac{1166.67}{EI}\ \text{(upward)},\qquad \theta_{B,left} = \frac{28\times5^2}{2EI} = \frac{350}{EI}\ \text{(anticlockwise)}
$$

#### Step 3: Segment BE by the conjugate beam
First treat BE as a beam with $y = 0$ at B and D (rigidity $2EI$, $x$ measured from B):

**(a) Reactions of BE**
$R_{B} = -28.00$ kN, $R_{D} = 108.00$ kN (positive = upward; check: $\sum F_y = 80.00$ kN equals the total load 80.00 kN).

**(b) Bending moment of BE, as $M/EI$ pieces**
| Portion (m) | $M$ (kNm) | stiffness | Area of $M/EI$ | $\bar x$ from B (m) |
|---|---|---|---|---|
| 0 to 3 | $- 28 x$ | 2EI | -63.00 | 2.000 |
| 3 to 5 | $90 - 58 x$ | 2EI | -142.00 | 4.136 |
| 5 to 9 | $50 x - 450$ | 2EI | -200.00 | 6.333 |

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

**(c) Conjugate beam**
The real segment BE has a support at B (provided by the link), a roller at D and a free end at E, so the conjugate beam has a pin at B, an internal hinge at D and a fixed end at E. It carries the $M/EI$ diagram of part (b) as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

**(d) Conjugate reaction at B (slope at B)**
Taking moments of the conjugate beam about the support at $x = 5$ m (conjugate moment is zero there):

$$
R'_B\times5 = (-63.00)(3.000) + (-142.00)(0.864) = -311.67
$$

$$
R'_B = \frac{-62.333}{EI}\ \text{(slope at B, anticlockwise)}
$$

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

$$
V' = R'_B - \sum A = -62.333 - [(-63.00)] = \frac{0.667}{EI}\ \text{(slope, clockwise)}
$$

$$
M' = R'_B\,x - \sum A(x-\bar x) = -187.000 - [(-63.00)(1.000)] = \frac{-124.000}{EI}\ \text{(deflection, upward)}
$$

**At D ($x = 5$ m):**

$$
V' = R'_B - \sum A = -62.333 - [(-63.00) + (-142.00)] = \frac{142.667}{EI}\ \text{(slope, clockwise)}
$$

$$
M' = R'_B\,x - \sum A(x-\bar x) = -311.667 - [(-63.00)(3.000) + (-142.00)(0.864)] = \frac{0.000}{EI}\ \text{(deflection, zero)}
$$

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

$$
V' = R'_B - \sum A = -62.333 - [(-63.00) + (-142.00) + (-200.00)] = \frac{342.667}{EI}\ \text{(slope, clockwise)}
$$

$$
M' = R'_B\,x - \sum A(x-\bar x) = -561.000 - [(-63.00)(7.000) + (-142.00)(4.864) + (-200.00)(2.667)] = \frac{1104.000}{EI}\ \text{(deflection, downward)}
$$

#### Step 4: Add the movement of B
Point B actually moves up by $1166.67/EI$ while D does not move, so BE also rotates as a rigid body clockwise by $1166.67/(5EI) = 233.33/EI$. The rigid-body deflection at distance $x$ from B is $1166.67\,(1 - x/5)/EI$ upward, which becomes negative (downward) beyond D. Adding this to the results of Step 3:

| Point | Deflection | Slope |
|---|---|---|
| B (link, left side = AB tip) | $\dfrac{1166.67}{EI}$ upward | $\dfrac{350}{EI}$ anticlockwise |
| B (right of the link) | same as left | $\dfrac{171}{EI}$ clockwise |
| E (free end) | $\dfrac{2037.33}{EI}$ downward | $\dfrac{576}{EI}$ clockwise |

(At E: Step 3 gives $1104/EI$ downward and $342.67/EI$ clockwise; the rigid-body rotation adds $0.8\times1166.67/EI = 933.33/EI$ downward and $233.33/EI$ clockwise: $1104 + 933.33 = 2037.33$ and $342.67 + 233.33 = 576$. At B, right of the link: Step 3 gives $62.33/EI$ anticlockwise, and $62.33 - 233.33 = -171$, i.e. $171/EI$ clockwise.)

#### Numerical values ($EI = 12\,000$ kNm$^2$)
- Deflection at B $= 97.2$ mm upward.
- Slope at B: left $= 0.02917$ rad (anticlockwise), right $= 0.01425$ rad (clockwise).
- Deflection at E $= 169.8$ mm downward; slope at E $= 0.0480$ rad (clockwise).

**Answer:** $\Delta_B = 97.2$ mm upward; $\Delta_E = 169.8$ mm downward; $\theta_E = 0.0480$ rad clockwise. At the link the slope jumps from $0.0292$ rad (anticlockwise, left) to $0.0143$ rad (clockwise, right).

## Check yourself

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

1. How is the beam split at the link at B?
   A. The link carries moment but no shear
   B. The link carries shear but no moment, so BE is treated separately and loads the tip of cantilever AB
   C. The link is a fixed support
   D. The link carries both shear and moment equally
   Answer: B. A link (internal hinge) transmits shear only. BE rests on the roller D and the link, and the shear at B is the load on cantilever AB. (Key point: Role of the link)

2. Moments about D for BE give R_B = -28 kN. What is the consequence for cantilever AB?
   A. AB carries 28 kN downward at its tip
   B. AB carries 108 kN at its tip
   C. AB carries no load
   D. AB is pulled upward at its tip by 28 kN
   Answer: D. R_B is negative, meaning the support force on BE at B acts downward, so BE pulls the tip of AB upward by 28 kN. (Key point: Force at the link)

3. How is the deflection of B on cantilever AB found?
   A. 28x5^2/(2EI)
   B. 28x5^3/(3EI) = 1166.67/EI upward
   C. 28x5/EI
   D. 28x5^3/(8EI)
   Answer: B. A cantilever with a tip load P deflects PL^3/(3EI) = 28x125/3 = 1166.67/EI, here upward. The slope is PL^2/(2EI) = 350/EI. (Key point: Deflection of the cantilever tip)

4. Why must a rigid-body rotation be added to the conjugate results for BE?
   A. Because EI changes along BE
   B. Because D deflects
   C. Because the E load is large
   D. B actually moves up by 1166.67/EI while D stays fixed, so BE rotates clockwise by 1166.67/(5EI) = 233.33/EI
   Answer: D. The conjugate analysis assumed zero deflection at B and D. Adding the rigid rotation about D gives at E 1104 + 933.33 = 2037.33/EI downward and 342.67 + 233.33 = 576/EI clockwise. (Key point: Rigid-body rotation of BE)

## Related in Theory of Structures I

- [Conjugate beam: slope and deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-beam-24kn-10knm) — 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-15-3kn-2076b) — Theory of Structures I, chapter 4, asked 2×
- [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×
- [Conjugate beam: slope and deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-slope-defl-free-end-2073) — Theory of Structures I, chapter 4
- [Conjugate beam: free end of overhang, UDL and point load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-10kn-4m-3m-2071b) — Theory of Structures I, chapter 4
- [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

## Related in other subjects

- [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3
- [Slope deflection: beam with 50 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-50kn-30kn-bmd) — Theory of Structures II, chapter 3
- [Slope deflection: continuous beam, 90 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-15knm-90kn-30kn) — Theory of Structures II, chapter 3
- [Cantilever: slope and deflection at free end](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acantilever-slope-deflection-50kn-40kn) — Strength of Materials, chapter 7