Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
2.2k views
in Physics by (63.7k points)

A compound shaft is made up of a steel rod of 50 mm diameter surrounded by a closely fitted brass tube. When a torque of 9 KN-m is applied on this shaft, its 60% is shared by the steel rod and the rest by brass tube. If shear modulus for steel is 85 GPa and for brass it is 45GPa. Calculate (a) The outside diameter of brass tube. (b) Maximum shear stress induced in steel and brass.

1 Answer

+2 votes
by (57.5k points)
selected by
 
Best answer

Diameter of steel ds = 50mm 

Total torque applied = 9KNm 

Torque sheared by steel rod Ts = 9 × 60/100 = 5.4KN-m 

Torque sheared by brass rod Tb = 9 × 40/100 = 3.6KN-m 

GS = 85GPa, Gb = 45GPa

Let; Outer diameter of brass = dob = ?, Inner diameter of brass = dib = 50mm, Maximum shear stress induced in steel = ιs Maximum shear stress induced in brass = ιb, Since shaft are parallel i.e.; θs = θb & T = Ts + Tb,Now applied θs = θb

TS.LS/GS.JS = Tb.Lb/Gb.Jb; LS = Lb 

We get; TS/GS.JS = Tb/Gb. Jb 

Or; Jb/JS = Tb.GS/TS.Gb 

Putting all the values 

{(π/32)(dob4 – 504)}/ {(π/32).504} = (3.6 × 85)/(5.4 × 45) 

dob = 61.29mm 

Again T/J = ι/R = TS/JS = ιS/RS 

ιS = TS.RS/JS = {(5.4 × 103)(25 × 10-3)}/{(π/32)(50 × 10–3)4

ιS = 2.2 × 108 N/m

ιS = 220 GPa

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...