---
title: "Strain energy deflection at A, spring support"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:strain-energy-beam-spring-deflection-a"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/strain-energy-beam-spring-deflection-a
site: BCT Engineering
---

# Strain energy deflection at A, spring support

Theory of Structures I · Chapter 2: Analysis by the Strain Energy Method

- Asked in: 2081 Chaitra, 6 marks
- Video: [Deflection of Beam with Spring Support | IStructE Structural Behaviour Course](https://www.youtube.com/watch?v=Uy02gf-Y9nM) — Dr Jawed Qureshi
- 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/strain-energy-beam-spring-deflection-a
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

Using strain energy method, determine the deflection at point A for the beam shown in the figure below. Here, $k$ stands for stiffness of spring. [Figure: beam ABC with EI constant; free end A carries a 20 kN downward load; support B (hinge) is 1.5 m from A; end C is 3 m from B and is supported on a spring of stiffness $k = EI/9$.]

## Short answer

Using Castigliano’s theorem with beam bending and spring energy (k = EI/9), deflection at A is δ_A = 5.625P/EI; with P=20 kN, δ_A = 112.5/EI m downward (EI in kNm²).

## Answer

#### Given and assumptions
Beam ABC, $EI$ constant: overhang $AB = 1.5$ m, span $BC = 3$ m. Load $P = 20$ kN (down) at the free end A, hinge at B, and C supported on a spring of stiffness $k = EI/9$ (kN/m, with $EI$ in kNm$^2$). The deflection of A is found by Castigliano's theorem, $\delta_A = \partial U/\partial P$, using the total strain energy of beam bending plus the spring.

#### Reactions
Take moments about B:

$$
R_C\times3 = P\times1.5 \;\Rightarrow\; R_C = 0.5P = 10\ \text{kN (downward, spring in tension)}
$$

$$
R_B = P + 0.5P = 1.5P = 30\ \text{kN (upward)}
$$

#### Bending moments (x from A, y from C)
- $AB$: $M = Px$, $0 \le x \le 1.5$
- $BC$: $M = R_C\,y = 0.5Py$, $0 \le y \le 3$

#### Strain energy
$$
U = \int_0^{1.5}\frac{(Px)^2}{2EI}dx + \int_0^{3}\frac{(0.5Py)^2}{2EI}dy + \frac{R_C^2}{2k}
$$

$$
U = \frac{P^2}{2EI}\left[\frac{1.5^3}{3} + 0.25\times\frac{3^3}{3}\right] + \frac{(0.5P)^2\times 9}{2EI}
= \frac{P^2}{2EI}\left[1.125 + 2.25\right] + \frac{1.125P^2}{EI}
$$

$$
U = \frac{1.6875P^2 + 1.125P^2}{EI} = \frac{2.8125\,P^2}{EI}
$$

#### Deflection at A
$$
\delta_A = \frac{\partial U}{\partial P} = \frac{5.625\,P}{EI} = \frac{5.625\times20}{EI}
$$

Parts: beam bending $= 67.5/EI$; spring (rotation of the beam about B caused by the spring's extension) $= 45/EI$.

**Answer: $\delta_A = \dfrac{112.5}{EI}$ downward (in metres when $EI$ is in kNm$^2$).**

## Check yourself

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

1. How is the deflection at A found from the total strain energy U?
   A. delta_A = U / P
   B. delta_A = dU/dP
   C. delta_A = d^2U/dP^2
   D. delta_A = 2U / P^2
   Answer: B. Castigliano's theorem: the deflection under a load equals the partial derivative of the total strain energy with respect to that load. (Key point: Castigliano theorem)

2. Taking moments about B for the 3 m span with load P at 1.5 m overhang, the spring reaction R_C is:
   A. 2P
   B. 0.5P (downward, spring in tension)
   C. 1.5P upward
   D. P
   Answer: B. R_C x 3 = P x 1.5 gives R_C = 0.5P = 10 kN, acting downward so the spring is in tension. (Key point: Spring reaction)

3. Which terms make up the total strain energy U?
   A. Beam bending in AB only
   B. Beam bending in AB and BC only
   C. Bending in AB and BC plus spring energy R_C^2/(2k)
   D. Only spring energy R_C^2/(2k)
   Answer: C. The spring stores energy R_C^2/(2k) in addition to the bending energy of both beam portions. (Key point: Spring energy included)

4. With U = 2.8125 P^2/EI and P = 20 kN, the deflection at A is:
   A. 56.25/EI
   B. 112.5/EI
   C. 225/EI
   D. 67.5/EI
   Answer: B. dU/dP = 5.625P/EI = 5.625 x 20/EI = 112.5/EI downward (67.5 from bending plus 45 from the spring). (Key point: Final deflection)

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