Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of

Q7  Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answers (1)

best_answer

Inner radii of each column, r1 = 30 cm = 0.3 m

Outer radii of each colum, r2 = 60 cm = 0.6 m

Mass of the structure, m = 50000 kg

Stress on each column is P

\\P=\frac{50000\times 9.8}{4\times \pi \times (0.6^{2}-0.3^{2})}\\ P=1.444\times 10^{5}Nm^{-2}

Youngs Modulus of steel is Y=2\times 10^{11}Nm^{-2}

\\Compressional\ strain=\frac{P}{Y} \\=\frac{1.444\times 10^{5}}{2\times 10^{11}}\\ =7.22\times 10^{-7}

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question