Answer :

Given Data,


Mass of the big structure, M = 50,000 kg.


Inner radius of the column, r = 30 cm = 0.3 m


Outer radius of the column, R = 60 cm = 0.6 m


Young’s modulus of steel, Ys, from the standard chart is 2×1011 Pa


Total force exerted on all the columns, F = Mg = 5000kg × 9.8 N


Amount of force exerted on each column, F1 = (5000Kg× 9.8N)/4


F1 = 122500 N


Young’s modulus, Y = Stress/ Strain


Y = (/Strain


Strain = …………………………… (1)


Where, A = area of each column = π (R2-r2)


A = π ((0.6)2 - (0.3)2)


Substituting value of Area in equation (1), we get,


Strain = 122500 N/ π ((0.6)2-(0.3)2) × 2×1011 Pa


Strain = 7.22× 10-7


Hence, the compressional strain on each column is 7.22 × 10-7.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

What is the Bulk Physics - Exemplar

What is the YoungPhysics - Exemplar

Two identical solPhysics - Exemplar

A rectangular fraPhysics - Exemplar

A wire is suspendPhysics - Exemplar

A rod of length lPhysics - Exemplar

To what depth musPhysics - Exemplar

A mild steePhysics - Exemplar

A copper and a stPhysics - Exemplar

A spring is stretPhysics - Exemplar