A rod of steel is 20 m long at a temperature of 20°C. What will be the temperature stress produced when the temperature of the rod is raised to 70°C? The rod is permitted to expand by 6 mm. [Coefficient of linear expansion = 12 × 10-6 per °C, Modulus of elasticity. = 2 × 10N/mm2]

Task Id 1206 Daman (3)

This question was previously asked in
BHEL Engineer Trainee Mechanical 24 Aug 2023 Official Paper
View all BHEL Engineer Trainee Papers >
  1. 6 N/mm2
  2. 300 N/mm2
  3. 30 N/mm2
  4. 60 N/mm2

Answer (Detailed Solution Below)

Option 4 : 60 N/mm2
Free
BHEL Engineer Trainee Fluid Mechanics Mock Test
1.4 K Users
20 Questions 20 Marks 15 Mins

Detailed Solution

Download Solution PDF

Explanation:

The formula for thermal expansion is given by:

ΔLthermal = Linitial × α × ΔT

Where:

  • ΔLthermal = Thermal expansion (in meters)
  • Linitial = Initial length of the rod = 20 m = 20,000 mm
  • α = Coefficient of linear expansion = 12 × 10-6 per °C
  • ΔT = Temperature change = 70°C - 20°C = 50°C

Substituting the values, we get:

ΔLthermal = 20,000 mm × 12 × 10-6 per °C × 50°C

ΔLthermal = 20,000 × 0.000012 × 50

ΔLthermal = 12 mm

Step 2: Determine the actual expansion allowed.

According to the problem, the rod is permitted to expand by 6 mm.

Step 3: Compute the difference between the unconstrained expansion and the actual allowed expansion.

The difference in expansion, ΔLdifference, is given by:

ΔLdifference = ΔLthermal - ΔLallowed

Substituting the values, we get:

ΔLdifference = 12 mm - 6 mm

ΔLdifference = 6 mm

Step 4: Use the modulus of elasticity to find the stress produced due to this difference in expansion.

The formula for stress, σ, is given by:

σ = (E × ΔLdifference) / Linitial

Where:

  • σ = Stress (in N/mm2)
  • E = Modulus of elasticity = 2 × 105 N/mm2
  • ΔLdifference = Difference in expansion = 6 mm
  • Linitial = Initial length of the rod = 20,000 mm

Substituting the values, we get:

σ = (2 × 105 N/mm2 × 6 mm) / 20,000

σ = (1,200,000 N/mm2) / 20,000

σ = 60 N/mm2

Therefore, the temperature stress produced in the steel rod is 60 N/mm2.

Latest BHEL Engineer Trainee Updates

Last updated on Jul 8, 2025

-> The BHEL Cut Off 2025 has been uploaded on July 8, 2025 at the official website 

-> BHEL Engineer Trainee result has been released on July 8. 

-> BHEL Engineer Trainee answer key 2025 has been released at the official website. 

-> The BHEL Engineer Trainee Admit Card 2025 has been released on the official website.

->The BHEL Engineer Trainee Exam 2025 will be conducted on April 11th, 12th and 13th, 2025

-> BHEL Engineer Trainee 2025 Notification has been released on the official website.

-> A total of 150 Vacancies have been announced for various disciplines of Engineering like Mechanical, Electrical, Civil, etc.

-> Interested and eligible candidates can apply from 1st February 2025 to 28th February 2025.

-> The authorities has also released the BHEL Engineer Trainee Pattern 

-> The BHEL Engineer Trainee Selection Process is divided into two stages namely Written Test and Interview.

-> The selected candidates for the Engineer Trainee post will get a salary range between Rs. 60,000 - Rs. 1,80,000.

Get Free Access Now
Hot Links: online teen patti teen patti 51 bonus teen patti vip