---
title: "Shaft overhang: maximum deflection formula"
subject: "Theory of Structures I"
chapter: 4
key: "semester_4/theory_of_structures_i:shaft-max-deflection-pal2"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/shaft-max-deflection-pal2
site: BCT Engineering
---

# Shaft overhang: maximum deflection formula

Theory of Structures I · Chapter 4: Deflection of Beams

- Asked in: 2072 Asoj, 12 marks
- Video: [Method of superposition for beams explained (slope & deflection with tables)](https://www.youtube.com/watch?v=DdZc7kLa4iI) — Engineer4Free
- 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/shaft-max-deflection-pal2
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4

## Question

A uniform shaft ABC is simply supported in bearings A and B and overhanging to C. $AB = l$ and $BC = a$. When a transverse force P acts at C, show that the maximum deflection in the portion AB is $\frac{P a l^2}{9\sqrt{3}\,EI}$.

## Short answer

Using double integration, maximum deflection in AB occurs at x = l/√3 from A and equals Pal²/(9√3 EI) upwards.

## Answer

#### Given data
Shaft ABC, simply supported at A and B ($AB = l$), overhanging to C ($BC = a$). A transverse force $P$ acts at C. $EI$ constant. To show: maximum deflection in AB is $\dfrac{Pal^2}{9\sqrt3\,EI}$.

```text
 A ---------------- B ------- C
 ^        l         ^    a     | P
```

#### Reactions
Moments about B: $R_A\,l = -Pa$, so $R_A = -\dfrac{Pa}{l}$ (downward). $R_B = P\left(1 + \dfrac{a}{l}\right)$.

#### Bending moment in AB
Taking $x$ from A:

$$
M = -\frac{Pa}{l}\,x\quad(\text{hogging}),\qquad 0 \le x \le l
$$

#### Double integration
$$
EI\frac{d^2y}{dx^2} = M = -\frac{Pa}{l}x
$$

$$
EI\frac{dy}{dx} = -\frac{Pa}{2l}x^2 + C_1
$$

$$
EIy = -\frac{Pa}{6l}x^3 + C_1x + C_2
$$

Boundary conditions: $y = 0$ at $x = 0$ gives $C_2 = 0$; $y = 0$ at $x = l$ gives

$$
-\frac{Pal^2}{6} + C_1l = 0 \;\Rightarrow\; C_1 = \frac{Pal}{6}
$$

Therefore (positive y = upward)

$$
y = \frac{Pa}{6EIl}\left(l^2x - x^3\right)
$$

#### Position of maximum deflection
$$
\frac{dy}{dx} = 0 \;\Rightarrow\; l^2 - 3x^2 = 0 \;\Rightarrow\; x = \frac{l}{\sqrt3}
$$

#### Maximum deflection
$$
y_{max} = \frac{Pa}{6EIl}\left(\frac{l^3}{\sqrt3} - \frac{l^3}{3\sqrt3}\right) = \frac{Pa}{6EIl}\cdot\frac{2l^3}{3\sqrt3}
$$

$$
\boxed{y_{max} = \frac{Pal^2}{9\sqrt3\,EI}}
$$

The shaft bows **upward** in AB (opposite to the load at C), at a distance $l/\sqrt3 = 0.577\,l$ from A. This proves the required result.

## Check yourself

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

1. What is the reaction at A when P acts at the overhang C?
   A. Pa/l upward
   B. Pa/l downward
   C. P(1 + a/l) upward
   D. Zero
   Answer: B. Moments about B give RA l = -Pa, so RA is Pa/l acting downward. (Key point: Reaction at A)

2. What is the bending moment in AB at distance x from A?
   A. +Pax/l (sagging)
   B. -Pax/l (hogging)
   C. -Pa (constant)
   D. P(l - x)
   Answer: B. The downward reaction at A produces hogging moment M = -Pa x/l, so EI y'' = -Pax/l. (Key point: Bending moment in AB)

3. Which conditions fix the constants in the double integration?
   A. y = 0 and slope = 0 at A
   B. y = 0 at A and y = 0 at B
   C. Slope = 0 at B and y = 0 at C
   D. y = 0 at B only
   Answer: B. Both supports are rigid, giving C2 = 0 and C1 = Pal/6. (Key point: Boundary conditions)

4. Where is the maximum deflection in AB and in which direction?
   A. At l/2, downward
   B. At l/sqrt(3) from A, upward
   C. At l/sqrt(3) from A, downward
   D. At l/3 from A, upward
   Answer: B. dy/dx = 0 gives l^2 = 3x^2, x = l/sqrt3; the shaft bows upward, with y_max = Pal^2/(9 sqrt3 EI). (Key point: Location and result)

## Related in Theory of Structures I

- [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
- [Overhang beam deflection and slope at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Abeam-vw-overhang-c-10kn) — Theory of Structures I, chapter 3
- [Overhang beam ILD for shear force at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-sf-c-2071o) — Theory of Structures I, chapter 5
- [Strain energy and real work deflection, overhang beam](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Areal-work-strain-energy-overhang-5m) — Theory of Structures I, chapter 3
- [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
- [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

- [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2
- [Maximum length of pipe portion AB](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apipe-abc-max-length-ab) — Hydraulics, chapter 2
- [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7
- [Lateral vibration of a simply supported beam](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Abeam-lateral-vibration-simply-supported) — Theory of Mechanism and Machine II, chapter 10