0 votes
in Physics by (49.2k points)
closed by

The bar shown in figure is subjected to an axial pull of 150 kN. Determine diameter of the middle portion if stress there is limited to 125N/mm2 . Proceed to determine the length of this middle portion if total extension of the bar is specified as 0.15 mm. Take modulus of elasticity of bar material E = 2 × 105N/mm2 .

1 Answer

+1 vote
by (44.2k points)
selected by
Best answer

Each of the segment of this composite bar is subjected to axial pull P =150 kN. 

Axial Stress in the middle portion \(\sigma\)2 = Axial pull/Area = 150 × 103 /[(\(\pi\)/4).(d22)] 

\(\therefore\) d2 = 39.1 mm

Since stress is limited to 125 N/mm2 , in the middle portion

125 = 150 × 103 /[(\(\pi\)/4).(d22 )]

Diameter of middle portion

d2 = 39.1mm

(ii) Stress in the end portions, \(\sigma\)1 = \(\sigma\)3

= 150 × 103 /[(\(\pi\)/4).(502 )] = 76.43 N/m2

Total change in length of the bar,

= change in length of end portions + change in length of mid portion

δL = δL1 + δL2 + δL3

= \(\sigma\)1L1/E + \(\sigma\)2L2/E + \(\sigma\)3L3/E; Since E is same for all portions 

= \(\sigma\)1(L1 + L3)/E + \(\sigma\)2L2/E 

L1 + L3 = 300 – L2

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.