Question
Download Solution PDFA cylindrical specimen of length L and cross - sectional area A extends by δ under a gradually applied tensile load P. The strain energy of the specimen is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The strain energy of a cylindrical specimen subjected to a gradually applied tensile load can be calculated using the formula for strain energy in elastic materials. Strain energy (
For a cylindrical specimen of length L, cross-sectional area
U=1/2×P×δ
Additional InformationThe elastic strain energy stored in a member of length (it may be curved or straight) due to axial force, bending moment, shear force and torsion are summarized below:
|
Axial Force, P |
|
|
Bending, M |
|
|
Shear Force, V |
|
|
Torsion, T |
|
Last updated on Jun 4, 2025
-> BPSC AE 2025 exam date has been revised. The exam will be conducted on July 5, 6 & 7 now.
-> Candidates who were facing technical issues while filling form can now fill the BPSC AE application form 2025 without any issue.
->BPSC AE age limit 2025 has been revised.
->BPSC AE application form 2025 was released on April 30. The last date to fill BPSC AE form 2025 was May 28.
->BPSC AE interview call letters released for Advt. 32/2024.
->BPSC AE notification 2025 has been released.
->A total of 1024 vacancies are announced through BPSC AE recruitment 2025
->The BPSC Exam Calendar 2025 has been released for the Assistant Engineer Recruitment.
-> The selection will be based on a written exam and evaluation of work experience.
-> Candidates with a graduation in the concerned engineering stream are eligible for this post.
-> To prepare for the exam solve BPSC AE Previous Year Papers. Also, attempt the BPSC AE Civil Mock Tests.