---
title: "Cost comparison of materials for a tie rod"
subject: "Machine Design I"
chapter: 2
key: "semester_6/machine_design_i:material-cost-tie-rod"
source: https://www.bctengineer.com/q/semester_6/machine_design_i/material-cost-tie-rod
site: BCT Engineering
---

# Cost comparison of materials for a tie rod

Machine Design I · Chapter 2: Materials Selection

- Asked in: year not given, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/machine_design_i/material-cost-tie-rod
- Whole chapter: https://www.bctengineer.com/md/semester_6/machine_design_i/chapter-2

## Question

A tie rod of length 1 m must carry a steady tensile load of 50 kN with a factor of safety of 2 based on yield strength. Compare the following materials on the basis of mass and material cost (assumed prices): mild steel (yield strength 250 MPa, density 7850 kg/m^3, Rs 120/kg); alloy steel 4140 (650 MPa, 7850 kg/m^3, Rs 180/kg); aluminium alloy 6061-T6 (270 MPa, 2700 kg/m^3, Rs 450/kg); titanium alloy Ti-6Al-4V (880 MPa, 4430 kg/m^3, Rs 3000/kg). Which material is cheapest and which is lightest? Comment on the economics of material selection.

## Short answer

Required area A = (load × safety factor) / yield strength. Mass = density × A × length, cost = mass × price. Cheapest: alloy steel 4140 (Rs 217, 1.21 kg); lightest: Ti-6Al-4V (0.50 kg, Rs 1510).

## Answer

The tie rod is in simple tension, so the required area is $A = \dfrac{P \times FS}{S_y}$, mass $m = \rho A L$ and cost $= m \times$ price.

Given: $P = 50\ \text{kN}$, $FS = 2$, $L = 1\ \text{m}$. Design load $= 50 \times 2 = 100\ \text{kN}$.

#### Area, mass and cost

For mild steel:

$$
\begin{aligned} A &= \frac{100\,000}{250} = 400\ \text{mm}^2 \\ m &= 7850 \times 400\times10^{-6} \times 1 = 3.14\ \text{kg} \\ \text{Cost} &= 3.14 \times 120 = \text{Rs } 377 \end{aligned}
$$

The other materials are worked out in the same way:

| Material | $S_y$ (MPa) | A (mm$^2$) | Mass (kg) | Price (Rs/kg) | Cost (Rs) |
|---|---|---|---|---|---|
| Mild steel | 250 | 400.0 | 3.14 | 120 | 377 |
| Alloy steel 4140 | 650 | 153.8 | 1.21 | 180 | 217 |
| Al 6061-T6 | 270 | 370.4 | 1.00 | 450 | 450 |
| Ti-6Al-4V | 880 | 113.6 | 0.50 | 3000 | 1510 |

#### Comment

- **Cheapest:** alloy steel 4140 (about Rs 217), although its price per kg is higher than mild steel, because its higher strength needs much less material.
- **Lightest:** titanium alloy (0.50 kg), but it costs about seven times as much as 4140, so it is justified only where weight is critical (aircraft, aerospace).
- Price per kg alone is misleading; the cost should be compared on the basis of the **function performed** (cost per unit strength, or cost of the finished part). Machining, joining, corrosion protection, availability and the cost of a heavier structure supporting the part must also be included.
- For stiffness-limited or buckling-limited parts the comparison would use modulus, and the ranking can change.

**Answer: Cheapest = alloy steel 4140 (Rs 217, 1.21 kg); lightest = Ti-6Al-4V (0.50 kg, Rs 1510).**

## Check yourself

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

1. How is the required area of the tie rod found?
   A. A = P x FS / Sy
   B. A = P / (FS x Sy)
   C. A = P x Sy / FS
   D. A = P / Sy without factor of safety
   Answer: A. Design load P x FS divided by yield strength. (Key point: Area formula)

2. What is the area for mild steel (P=50 kN, FS=2, Sy=250 MPa)?
   A. 200 mm^2
   B. 400 mm^2
   C. 800 mm^2
   D. 153.8 mm^2
   Answer: B. 100 000 N / 250 MPa = 400 mm^2. (Key point: Mild steel area)

3. Which material gives the lowest material cost?
   A. Mild steel (Rs 377)
   B. Aluminium 6061 (Rs 450)
   C. Alloy steel 4140 (about Rs 217)
   D. Titanium (Rs 1510)
   Answer: C. Higher strength needs much less material even though the price per kg is higher than mild steel. (Key point: Cheapest material)

4. Why is price per kg alone misleading?
   A. Price per kg is always proportional to cost of the part
   B. Mass does not affect cost
   C. Only density matters
   D. Cost should be compared on the function performed, since stronger materials need less mass
   Answer: D. Cost per unit of function and finished-part cost must be compared. (Key point: Economics)

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