---
title: "Position of 2t force for zero net elongation"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:bar-force-2t-zero-elongation"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-force-2t-zero-elongation
site: BCT Engineering
---

# Position of 2t force for zero net elongation

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2078 Bhadra, 8 marks
- Video: [Problem on Principle of superposition |Simple Stresses & Strains | Strength of Materials | MOM | MOS](https://www.youtube.com/watch?v=OZSiLSB9l2M) — Sampurna Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-force-2t-zero-elongation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

At what distance 'x' from the fixed end of the uniform bar should the '2t' force be applied in order that the net overall change in length of the bar will be zero? [Figure: bar fixed at the left end, length 5 m in total split into 1 m, 2 m and 2 m; forces of 1.5 t acting to the right at 1 m and at 3 m from the fixed end, and a force of 1.5 t acting to the left at the free end.]

## Short answer

Without the 2t force, net elongation is -1.5/AE. A tensile 2t at distance x adds 2x/AE, so -1.5+2x=0 gives x=0.75 m from the fixed end.

## Answer

Let the bar have uniform area $A$ and modulus $E$. The bar is fixed at the left end, and the loads are (positions from the fixed end): 1.5 t (→) at 1 m, 1.5 t (→) at 3 m, 1.5 t (←) at the free end (5 m). The 2 t force is applied at distance $x$ from the fixed end; to cancel a net shortening it must act away from the wall (→).

#### Force in each portion (tension +ve) without the 2 t force
Take the sum of the forces to the right of the section:

| Portion | Length (m) | Axial force (t) |
|---|---|---|
| 0 – 1 m | 1 | $1.5 + 1.5 - 1.5 = +1.5$ |
| 1 – 3 m | 2 | $1.5 - 1.5 = 0$ |
| 3 – 5 m | 2 | $-1.5$ |

$$
\delta_{without} = \frac{1}{AE}\left[1.5(1) + 0(2) + (-1.5)(2)\right] = \frac{-1.5}{AE}\ \text{(t·m)}
$$

The bar is net shortened, so a tensile 2 t force must lengthen it by $1.5/AE$.

#### Effect of the 2 t force at distance $x$
It adds $+2$ t of tension in every portion between the wall and $x$, so it adds $2x/AE$ to the length:

$$
-1.5 + 2x = 0 \Rightarrow x = 0.75\ \text{m}
$$

Since $x = 0.75 < 1$ m, the force lies in the first portion, as assumed.

Check: portions become 0 – 0.75 m: $+3.5$ t; 0.75 – 1 m: $+1.5$ t; 1 – 3 m: 0; 3 – 5 m: $-1.5$ t. Sum $= 3.5(0.75) + 1.5(0.25) + 0 - 1.5(2) = 2.625 + 0.375 - 3 = 0$.

**Answer: $x = 0.75$ m from the fixed end, with the 2 t force acting away from the fixed end.**

## Check yourself

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

1. What is the axial force in the 1-3 m portion without the 2 t force?
   A. Zero
   B. +1.5 t
   C. -1.5 t
   D. +3 t
   Answer: A. Forces right of the section: 1.5 (right) and 1.5 (left) cancel, giving 0. (Key point: Section method)

2. What is the bar's change in length without the 2 t force?
   A. Net elongation of 1.5/AE
   B. Net shortening of 1.5/AE (t.m)
   C. Zero
   D. Net shortening of 3/AE
   Answer: B. Sum N L = 1.5(1) + 0 - 1.5(2) = -1.5, so the bar is shortened. (Key point: Net change)

3. How does the 2 t force at distance x change the length?
   A. It adds 2 x 5/AE
   B. It subtracts 2x/AE
   C. It adds 2x/AE by adding 2 t tension between the wall and x
   D. It adds 2(5 - x)/AE
   Answer: C. The force increases tension by 2 t only in the portion from the wall to x. (Key point: Effect of 2 t force)

4. What is x?
   A. 0.75 m
   B. 1.5 m
   C. 0.5 m
   D. 0.75 cm from the free end
   Answer: A. -1.5 + 2x = 0 gives x = 0.75 m, which lies within the first 1 m portion. (Key point: Result)

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