---
title: "Strain energy method for frame displacement"
subject: "Theory of Structures I"
chapter: 2
key: "semester_4/theory_of_structures_i:strain-energy-frame-500kn"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/strain-energy-frame-500kn
site: BCT Engineering
---

# Strain energy method for frame displacement

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

- Asked in: 2066 Kartik, 10 marks
- Video: [Strain Energy Method - Analysis of Frames - Problem No 1](https://www.youtube.com/watch?v=jIyET3Nm7aU) — Stan Academy
- Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/strain-energy-frame-500kn
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2

## Question

Use the strain energy method for the given frame to calculate the vertical displacement of the point with load 500 kN. The frame is made of steel rod of diameter 100 mm. Take $E = 2\times10^5\ \text{N/mm}^2$. [Figure: frame fixed at joint 0; member 0-1 horizontal, 3 m; member 1-2 vertical, 1 m; member 2-3 horizontal, 2 m, with 500 kN acting downward at joint 3; a steel bar of diameter 100 mm runs from joint 0 to the lower member.]

## Short answer

Castigliano’s theorem including bending and axial strain energy gives vertical displacement ≈ 23.26 m downward (bending integral 45.667 m³, bending dominates; axial contributes only 0.32 mm). Load is very heavy for the 100 mm rod.

## Answer

#### Given data and assumptions
Frame fixed at joint 0, all members of solid circular section $d = 100$ mm, $E = 2\times10^5$ N/mm$^2$ ($= 2\times10^8$ kN/m$^2$). Taking the figure as a cantilever frame with horizontal member 0-1 (3 m), vertical member 1-2 (1 m) and horizontal member 2-3 (2 m), the 500 kN load $P$ acts downward at joint 3. Shear energy is neglected (small for slender members; $G$ is not given). Vertical displacement of joint 3 is found by Castigliano's theorem: $\delta_3 = \partial U/\partial P$.

$$
I = \frac{\pi d^4}{64} = 4.9087e-06\ \text{m}^4,\quad EI = 981.7\ \text{kNm}^2,\quad A = 0.00785\ \text{m}^2,\quad AE = 1.5708e+06\ \text{kN}
$$

#### Internal forces (s measured from joint 3)
| Member | Length | Bending moment $M$ | Axial force $N$ |
|---|---|---|---|
| 3-2 (horizontal) | 2 m | $Ps$ | 0 |
| 2-1 (vertical) | 1 m | $2P$ (constant) | $P$ (compression) |
| 1-0 (horizontal) | 3 m | $P(2+s)$ | 0 |

#### Strain energy and displacement
$$
U = \int\frac{M^2}{2EI}dx + \int\frac{N^2}{2AE}dx
$$

$$
\delta_3 = \frac{\partial U}{\partial P} = \frac{P}{EI}\left[\int_0^2 s^2ds + (2)^2(1) + \int_0^3 (2+s)^2 ds\right] + \frac{PL_{2-1}}{AE}
$$

$$
\int_0^2 s^2ds = 2.667,\qquad 2^2\times1 = 4,\qquad \int_0^3(2+s)^2ds = \frac{5^3-2^3}{3} = 39
$$

$$
\delta_{bending} = \frac{500\times(2.667+4+39)}{981.7} = \frac{500\times45.667}{981.7} = 23.258\ \text{m}
$$

$$
\delta_{axial} = \frac{500\times1}{1.5708e+06} = 0.318\ \text{mm}
$$

**Answer: vertical displacement of the loaded joint $\approx 23.26$ m downward** (bending governs; axial part is only 0.32 mm). The very large value shows that the given load is extremely heavy for a 100 mm rod; the small-deflection theory is used as asked.

## Check yourself

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

1. The vertical displacement of the loaded joint is obtained by:
   A. Setting delta = P L / (A E) for the whole frame
   B. Castigliano's theorem, delta = dU/dP, with U from bending and axial energy
   C. Equating the load to the shear energy
   D. Maxwell's reciprocal theorem on the vertical member only
   Answer: B. U is written for bending and axial force and differentiated with respect to the load P at joint 3. (Key point: Method used)

2. In the vertical member 2-1, the bending moment is:
   A. Zero
   B. Ps varying from 0 to 2P
   C. A constant 2P and axial force P
   D. P(2 + s)
   Answer: C. The load acts at a 2 m horizontal lever arm from the vertical member, so M = 2P is constant, and the member carries axial compression P. (Key point: Moment in vertical member)

3. The bending term sum of integrals of (dM/dP) M used is 2.667 + 4 + 39. Where does the 39 come from?
   A. Integral of s^2 from 0 to 2
   B. Integral of (2+s)^2 from 0 to 3 for member 1-0
   C. 2^2 x 1 for the vertical member
   D. Axial force energy
   Answer: B. In member 1-0, M = P(2+s) over 3 m, so the integral of (2+s)^2 = (5^3 - 2^3)/3 = 39. (Key point: Bending integral)

4. Which part dominates the vertical displacement?
   A. Axial part, 23.26 m
   B. Bending part, about 23.26 m, with axial only 0.32 mm
   C. Both equal
   D. Shear part
   Answer: B. Bending gives 500 x 45.667/981.7 = 23.26 m while the axial part PL/AE is only 0.318 mm, so bending governs. (Key point: Bending versus axial)

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