---
title: "Conjugate beam: stepped rigidity, central load"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:ss-beam-stepped-flexural-rigidity"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/ss-beam-stepped-flexural-rigidity
site: BCT Engineering
---

# Conjugate beam: stepped rigidity, central load

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2072 Magh, 12 marks
- Video: [CONJUGATE BEAM METHOD FOR SLOPE & DEFLECTION - SIMPLY SUPPORTED BEAM WITH UDL & POINT LOADS](https://www.youtube.com/watch?v=Dhx6x7GKqzU) — Structural Funda by:- KheNiSiR
- 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/ss-beam-stepped-flexural-rigidity
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

A simply supported beam carries a point load W at mid span L. The middle one third portion of length has flexural rigidity 2EI and the rest two third portion has flexural rigidity EI. Determine the maximum deflection and slope at supports. Use conjugate beam method.

## Short answer

By conjugate beam method: slope at each support = 13WL²/(288EI), maximum mid-span deflection = 35WL³/(2592EI) downward.

## Answer

#### Given data
Simply supported beam, span $L$, central load $W$. The middle third ($L/3$) has rigidity $2EI$; the two outer thirds have $EI$. By symmetry the reactions are $W/2$ each and the maximum deflection is at mid-span.

```text
             W
 A --- EI ---|--- 2EI ---|--- EI --- B
 ^    L/3         L/3         L/3     ^
```

#### Step 1: Bending moment and $M/EI$ diagram (left half)
$M = \dfrac{Wx}{2}$ for $0 \le x \le \dfrac{L}{2}$. $M/EI$ ordinates:

- At $x = L/3$: in the outer third $M/EI = \dfrac{WL}{6EI}$; in the middle third $M/(2EI) = \dfrac{WL}{12EI}$ (a sudden drop).
- At mid-span: $M/(2EI) = \dfrac{WL}{8EI}$.

#### Step 2: Conjugate beam
The conjugate beam is simply supported, loaded with the $M/EI$ diagram. Loads (left half, in units of $W L^2/EI$ with centroids from A):

| Part | Area | Centroid from A |
|---|---|---|
| Triangle, $0$ to $L/3$ (peak $WL/6EI$) | $\tfrac12\cdot\tfrac{L}{3}\cdot\tfrac{WL}{6EI} = \dfrac{WL^2}{36EI}$ | $\dfrac{2L}{9}$ |
| Rectangle, $L/3$ to $L/2$ (height $WL/12EI$) | $\tfrac{L}{6}\cdot\tfrac{WL}{12EI} = \dfrac{WL^2}{72EI}$ | $\dfrac{5L}{12}$ |
| Triangle on top (height $\tfrac{WL}{24EI}$) | $\tfrac12\cdot\tfrac{L}{6}\cdot\tfrac{WL}{24EI} = \dfrac{WL^2}{288EI}$ | $\dfrac{4L}{9}$ |

Sum of the half-loads $= \dfrac{WL^2}{EI}\left(\dfrac{8 + 4 + 1}{288}\right) = \dfrac{13\,WL^2}{288\,EI}$.

#### Step 3: Slope at the supports
The conjugate reaction equals half the total conjugate load (symmetry):

$$
\theta_A = \theta_B = R'_A = \frac{13\,WL^2}{288\,EI} = 0.0451\,\frac{WL^2}{EI}
$$

#### Step 4: Maximum deflection (conjugate moment at mid-span)
$$
y_{max} = R'_A\frac{L}{2} - \frac{WL^2}{36EI}\left(\frac{L}{2}-\frac{2L}{9}\right) - \frac{WL^2}{72EI}\left(\frac{L}{2}-\frac{5L}{12}\right) - \frac{WL^2}{288EI}\left(\frac{L}{2}-\frac{4L}{9}\right)
$$

$$
y_{max} = \frac{WL^3}{EI}\left[\frac{13}{576} - \frac{5}{648} - \frac{1}{864} - \frac{1}{5184}\right] = \frac{WL^3}{EI}\cdot\frac{117 - 40 - 6 - 1}{5184} = \frac{70}{5184}\frac{WL^3}{EI}
$$

$$
y_{max} = \frac{35\,WL^3}{2592\,EI} = 0.0135\,\frac{WL^3}{EI}
$$

**Answer: slope at each support $= \dfrac{13WL^2}{288EI}$; maximum (mid-span) deflection $= \dfrac{35WL^3}{2592EI}$ downward.** For comparison, a uniform beam of rigidity $EI$ has $\dfrac{WL^3}{48EI} = 0.0208\dfrac{WL^3}{EI}$, so the stiffer middle reduces the deflection by about 35%.

## Check yourself

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

1. At x = L/3 where the rigidity changes from EI to 2EI, what happens to the M/EI ordinate?
   A. It is continuous
   B. It drops suddenly from WL/6EI to WL/12EI
   C. It doubles
   D. It becomes zero
   Answer: B. M = WL/6 is the same on both sides, but dividing by 2EI instead of EI halves the ordinate. (Key point: M/EI jump)

2. Why can the slope at the supports be found as half the total conjugate load?
   A. The conjugate reactions are equal because the load and rigidity are symmetric
   B. Because slope equals the load at midspan
   C. Because the conjugate beam is a cantilever
   D. Because deflection is zero at midspan
   Answer: A. Symmetric loading and rigidity give a symmetric conjugate load, so each reaction is half the total. (Key point: Use of symmetry)

3. What is the slope at each support?
   A. WL^2/16EI
   B. 13WL^2/288EI
   C. 35WL^3/2592EI
   D. WL^2/48EI
   Answer: B. Half-load sum is 13WL^2/(288EI), equal to R'A. (Key point: Slope at supports)

4. Where is the maximum deflection and what is its value?
   A. Mid-span, 35WL^3/(2592EI)
   B. At L/3, 13WL^3/(288EI)
   C. Mid-span, WL^3/(48EI)
   D. At the supports, 13WL^2/(288EI)
   Answer: A. By symmetry it is at midspan, equal to the conjugate moment there, 35WL^3/2592EI (about 35% less than the uniform-EI value). (Key point: Maximum deflection)

## Related in Theory of Structures I

- [Conjugate beam: stepped symmetric beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-sym-beam-3portions) — 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
- [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4
- [Conjugate beam: slope and deflection at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-slope-defl-c-120kn-3ei) — Theory of Structures I, chapter 4
- [Beam mid-span deflection with temperature gradient](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-midspan-defl-temperature) — Theory of Structures I, chapter 3
- [ILD and BM at mid span for triangular load](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aild-triangular-load-mid-span-c) — Theory of Structures I, chapter 5

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