Mechanical Engineering Exam  >  Mechanical Engineering Questions  >   The strain energy in bar of 20 mm diameter,... Start Learning for Free
The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N is
  • a)
    0.0796 N-mm
  • b)
    0.796 kNm
  • c)
    0.080 Nm
  • d)
    0.080 kNm
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The strain energy in bar of 20 mm diameter, 1.0 m length, and young’...
Diameter, d = 20 mm
Length, L = 1000 mm
E = 200 × 103 MPa
P = 100N
U =
=
=
U = 0.0796 N­mm
View all questions of this test
Most Upvoted Answer
The strain energy in bar of 20 mm diameter, 1.0 m length, and young’...
Calculation of Strain Energy in Bar

Given parameters are:

- Diameter (d) = 20 mm
- Length (L) = 1.0 m
- Young's Modulus (E) = 200.0 GPa
- Axial Force (F) = 100.0 N

Formula for Strain Energy:

Strain Energy (U) = (1/2) x (F/A) x δ

where,

- F = Axial Force
- A = Cross-sectional Area of Bar
- δ = Elongation or Deformation in Bar

Calculation of Cross-sectional Area:

Cross-sectional Area (A) = π/4 x d²

= 3.14/4 x (20 mm)²

= 314.16 mm²

Calculation of Elongation or Deformation:

Elongation or Deformation (δ) = F x L / A x E

= 100.0 N x 1.0 m / 314.16 mm² x 200.0 GPa

= 0.000159

Calculation of Strain Energy:

Strain Energy (U) = (1/2) x (F/A) x δ

= (1/2) x (100.0 N / 314.16 mm²) x 0.000159

= 0.0796 N-mm

Therefore, the strain energy in the bar is 0.0796 N-mm (Option A).
Attention Mechanical Engineering Students!
To make sure you are not studying endlessly, EduRev has designed Mechanical Engineering study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Mechanical Engineering.
Explore Courses for Mechanical Engineering exam

Top Courses for Mechanical Engineering

The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer?
Question Description
The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer?.
Solutions for The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free.
Here you can find the meaning of The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The strain energy in bar of 20 mm diameter, 1.0 m length, and young’s modulus 200.0 GPa under an axial force of 100.0 N isa) 0.0796 N-mmb) 0.796 kNmc) 0.080 Nmd) 0.080 kNmCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
Explore Courses for Mechanical Engineering exam

Top Courses for Mechanical Engineering

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev