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
0 votes
242 views
in General by (115k points)
closed by
The flow stress (in MPa) of a material is given by \(\sigma = 500{\ \epsilon^{0.1}}\) , Where ϵ is true strain. The Young’s modulus of elasticity of the material is 200 GPa. A block of thickness 100 mm made of this material is compressed to 95 mm thickness and then the load is removed. The final dimension of the block (in mm) is _________

1 Answer

0 votes
by (152k points)
selected by
 
Best answer

Concept:

True strain \({\epsilon _T}\; = \;ln\left( {\frac{{original\;area}}{{instantaneous\;area}}} \right) = \;ln\left( {\frac{{original\;thickness}}{{instantaneous\;thickness}}} \right)\), when length is not changing.

Flow stress σf = KϵTn, where K is yield strength in MPa and n is strain hardening component.

Calculation:

True strain ϵT \(= ln\frac{{100}}{{95}} = 0.05129\)

σ = 500 × (0.05129)0.1 ⇒ 371.5147523 = 371.5147523 MPa

Upto elastic limits using hooks law

\(\sigma = E\epsilon = E.\frac{{{\rm{\Delta }}l}}{l}\)

\(371.514 \times {10^6} = 200 \times {10^9} \times \left( {\frac{{{\rm{\Delta }}l}}{{100}}} \right)\)

⇒ Δl = 0.18575 mm = considering this for elastic recovery

∴ This will be added to 95 mm

Final dimension = 95.18575 mm

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

...