Mechanical Engineering Exam > Mechanical Engineering Questions > The state of plane stress in a plate of 100 m...
Start Learning for Free
The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:
- a)Zero
- b)90 N/mm2
- c)100 N/mm2
- d)200 N/mm2
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
The state of plane stress in a plate of 100 mm thickness is given as &...
**Given Data**
- Thickness of the plate (h) = 100 mm
- Stress in the x-direction (σxx) = 100 N/mm^2
- Stress in the y-direction (σyy) = 200 N/mm^2
- Young's modulus (E) = 300 N/mm^2
- Poisson's ratio (ν) = 0.3
**Understanding Plane Stress**
Plane stress refers to a state of stress in a thin plate where the stresses are acting only in the x and y directions, and there is no stress in the z-direction (thickness direction). In this state, the stresses in the z-direction are assumed to be zero.
**Calculating Stress in the Thickness Direction**
To determine the stress developed in the direction of thickness (z-direction), we can use Hooke's Law for plane stress:
σz = -ν(σxx + σyy)
where σz is the stress in the z-direction and ν is the Poisson's ratio.
Substituting the given values:
σz = -0.3(100 + 200) = -0.3(300) = -90 N/mm^2
Since stress is a measure of force per unit area, it cannot be negative. Therefore, the stress developed in the direction of thickness is zero (0 N/mm^2).
- Thickness of the plate (h) = 100 mm
- Stress in the x-direction (σxx) = 100 N/mm^2
- Stress in the y-direction (σyy) = 200 N/mm^2
- Young's modulus (E) = 300 N/mm^2
- Poisson's ratio (ν) = 0.3
**Understanding Plane Stress**
Plane stress refers to a state of stress in a thin plate where the stresses are acting only in the x and y directions, and there is no stress in the z-direction (thickness direction). In this state, the stresses in the z-direction are assumed to be zero.
**Calculating Stress in the Thickness Direction**
To determine the stress developed in the direction of thickness (z-direction), we can use Hooke's Law for plane stress:
σz = -ν(σxx + σyy)
where σz is the stress in the z-direction and ν is the Poisson's ratio.
Substituting the given values:
σz = -0.3(100 + 200) = -0.3(300) = -90 N/mm^2
Since stress is a measure of force per unit area, it cannot be negative. Therefore, the stress developed in the direction of thickness is zero (0 N/mm^2).
Free Test
FREE
| Start Free Test |
Community Answer
The state of plane stress in a plate of 100 mm thickness is given as &...
Zero
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
|
|
Similar Mechanical Engineering Doubts
Top Courses for Mechanical EngineeringView all
The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer?
Question Description
The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct 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 state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct 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 state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer?.
The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct 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 state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct 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 state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer?.
Solutions for The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct 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 state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer?, a detailed solution for The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The state of plane stress in a plate of 100 mm thickness is given as σxx = 100 N/mm2, σyy = 200 N/mm2, Young's modulus = 300 N/mm2, Poisson's ratio = 0.3. The stress developed in the direction of thickness is:a)Zerob)90 N/mm2c)100 N/mm2d)200 N/mm2Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
|
|
Explore Courses for Mechanical Engineering exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup