---
title: "Position of 3.5 tonne load on cross piece"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:two-rods-steel-copper-position-of"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/two-rods-steel-copper-position-of
site: BCT Engineering
---

# Position of 3.5 tonne load on cross piece

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2068 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/two-rods-steel-copper-position-of
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

Two vertical rods of steel and copper are rigidly fixed with the ceiling at their ends at 100cm apart. Each rod is 3m long and 25mm diameter. A horizontal cross piece connects the lower ends of the rods. Where should a load of 3.5 tonnes be placed on the cross piece so that it remains horizontal after being loaded. Take $E_s = 2\times10^6\ \text{kg/cm}^2$, $E_c = 1.0\times10^6\ \text{kg/cm}^2$.

## Short answer

For equal extension, steel carries twice the load of copper due to double E. With total 3500 kgf, steel takes 2333.3 kgf, copper 1166.7 kgf. Moment equilibrium about steel rod gives distance x = (1166.7 × 100) / 3500 = 33.33 cm from steel rod.

## Answer

**Idea:** the cross piece stays horizontal only if both rods stretch equally. The rods have the same area and length, so the load carried by each rod must be proportional to its $E$.

Let the steel rod carry $P_s$ and the copper rod carry $P_c$, with $P_s + P_c = 3500$ kgf. Equal extension:

$$
\frac{P_sL}{AE_s} = \frac{P_cL}{AE_c} \Rightarrow \frac{P_s}{P_c} = \frac{E_s}{E_c} = \frac{2\times10^6}{1\times10^6} = 2
$$

$$
P_s = \frac{2}{3}(3500) = 2333.3\ \text{kgf}, \qquad P_c = \frac{1}{3}(3500) = 1166.7\ \text{kgf}
$$

(With $A = \dfrac{\pi}{4}(2.5)^2 = 4.909$ cm², the extension of each rod is $\dfrac{2333.3\times300}{4.909\times2\times10^6} = 0.0713$ cm, the same for both rods.)

#### Position of the load
Let the load act at distance $x$ from the steel rod. Taking moments about the steel rod (rods 100 cm apart):

$$
3500\,x = P_c(100) = 1166.7\times100 \Rightarrow x = 33.33\ \text{cm}
$$

```text
   steel            copper
    |<----- 100 cm ---->|
    |                    |
  ==|====================|==  cross piece
    |  ^                 |
    |<-x->  3.5 t load   |
```

**Answer:** the load should be placed **33.33 cm from the steel rod** (66.67 cm from the copper rod).

## Check yourself

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

1. What must be true for the cross piece to stay horizontal?
   A. Both rods stretch equally
   B. Both rods carry equal load
   C. Both rods have equal stress
   D. The load acts at the centre
   Answer: A. Equal extension is required for horizontality. (Key point: Condition)

2. Why is P_s / P_c = 2?
   A. Steel is longer
   B. Same A and L, so P is proportional to E
   C. Steel is thicker
   D. Copper is stronger
   Answer: B. P L/(A E) equal gives P_s/P_c = E_s/E_c = 2. (Key point: Load ratio)

3. How is x found?
   A. Moments about the steel rod: 3500 x = P_c x 100
   B. 3500 x = P_s x 100
   C. x = 50 cm
   D. x = P_c / 3500 x 50
   Answer: A. x = 1166.7 x 100/3500 = 33.33 cm. (Key point: Position)

4. Where should the load be placed?
   A. 66.67 cm from the steel rod
   B. 33.33 cm from the steel rod
   C. 33.33 cm from the copper rod
   D. At the mid-point
   Answer: B. It must be nearer the stiffer steel rod which takes more load. (Key point: Result)

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