Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > In a 2D stress system, the two principal stre...
Start Learning for Free
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?
- a)240 N/mm2
- b)180 N/mm2
- c)195 N/mm2
- d)200 N/mm2
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (te...
Yielding may occur either in tension or compression.
For yielding in tension,
According to normal strain theory,
For yielding in compression,
For yielding in tension,
According to normal strain theory,
For yielding in compression,
Most Upvoted Answer
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (te...
's ratio is 0.3. Determine the value of p2 and the maximum shear stress.
To determine the value of p2, we can use the formula for the sum and product of the principal stresses:
p1 + p2 = σx + σy
p1p2 = σxσy - τxy^2
where σx and σy are the normal stresses in the x and y directions, and τxy is the shear stress.
Substituting the given values, we have:
p1 = 180 N/mm2 (tensile)
σx = p1 = 180 N/mm2
σy = p2 (unknown)
σx + σy = p1 + p2 = 180 + σy
p1p2 = σxσy - τxy^2
p1p2 = 180σy - τxy^2
We know that the yield stress in compression for the material is 240 N/mm2, so the maximum compressive stress that the material can withstand is -240 N/mm2. Therefore, we can write:
-240 ≤ σy ≤ 0
Substituting the lower bound, we have:
p1p2 = 180σy - τxy^2
180p2 = 240τxy^2 - p1p2
180p2 + p1p2 = 240τxy^2
p2(180 + p1) = 240τxy^2
p2 = (240τxy^2) / (180 + p1)
To find the maximum shear stress, we can use the formula:
τmax = (σ1 - σ2) / 2
where σ1 and σ2 are the principal stresses.
Substituting the given values, we have:
τmax = (p1 - p2) / 2
τmax = (180 - p2) / 2
τmax = (180 - (240τxy^2) / (180 + p1)) / 2
To maximize τmax, we can differentiate it with respect to τxy and set the derivative equal to zero:
dτmax/dτxy = (240(180 + p1) - 480τxy^2) / (2(180 + p1)^2) = 0
240(180 + p1) - 480τxy^2 = 0
τxy^2 = (240(180 + p1)) / 480
τxy = sqrt[(240(180 + p1)) / 480]
Substituting this value back into the equation for τmax, we have:
τmax = (180 - p2) / 2
τmax = (180 - (240(sqrt[(240(180 + p1)) / 480])) / (180 + p1)) / 2
Simplifying this expression, we get:
τmax = 60 / sqrt(2(180 + p1))
Substituting the given values, we have:
τmax = 60 / sqrt(2(180 + 180))
τmax = 15.81 N/mm2
Therefore, the value of p2 is:
p2 = (240τxy^2) / (180 + p1)
p2 = (240(sqrt[(240(180 + 180)) /
To determine the value of p2, we can use the formula for the sum and product of the principal stresses:
p1 + p2 = σx + σy
p1p2 = σxσy - τxy^2
where σx and σy are the normal stresses in the x and y directions, and τxy is the shear stress.
Substituting the given values, we have:
p1 = 180 N/mm2 (tensile)
σx = p1 = 180 N/mm2
σy = p2 (unknown)
σx + σy = p1 + p2 = 180 + σy
p1p2 = σxσy - τxy^2
p1p2 = 180σy - τxy^2
We know that the yield stress in compression for the material is 240 N/mm2, so the maximum compressive stress that the material can withstand is -240 N/mm2. Therefore, we can write:
-240 ≤ σy ≤ 0
Substituting the lower bound, we have:
p1p2 = 180σy - τxy^2
180p2 = 240τxy^2 - p1p2
180p2 + p1p2 = 240τxy^2
p2(180 + p1) = 240τxy^2
p2 = (240τxy^2) / (180 + p1)
To find the maximum shear stress, we can use the formula:
τmax = (σ1 - σ2) / 2
where σ1 and σ2 are the principal stresses.
Substituting the given values, we have:
τmax = (p1 - p2) / 2
τmax = (180 - p2) / 2
τmax = (180 - (240τxy^2) / (180 + p1)) / 2
To maximize τmax, we can differentiate it with respect to τxy and set the derivative equal to zero:
dτmax/dτxy = (240(180 + p1) - 480τxy^2) / (2(180 + p1)^2) = 0
240(180 + p1) - 480τxy^2 = 0
τxy^2 = (240(180 + p1)) / 480
τxy = sqrt[(240(180 + p1)) / 480]
Substituting this value back into the equation for τmax, we have:
τmax = (180 - p2) / 2
τmax = (180 - (240(sqrt[(240(180 + p1)) / 480])) / (180 + p1)) / 2
Simplifying this expression, we get:
τmax = 60 / sqrt(2(180 + p1))
Substituting the given values, we have:
τmax = 60 / sqrt(2(180 + 180))
τmax = 15.81 N/mm2
Therefore, the value of p2 is:
p2 = (240τxy^2) / (180 + p1)
p2 = (240(sqrt[(240(180 + 180)) /
Attention Civil Engineering (CE) Students!
To make sure you are not studying endlessly, EduRev has designed Civil Engineering (CE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Civil Engineering (CE).
|
Explore Courses for Civil Engineering (CE) exam
|
|
Similar Civil Engineering (CE) Doubts
Top Courses for Civil Engineering (CE)View all
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer?
Question Description
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer?.
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer?.
Solutions for In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.
Here you can find the meaning of In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer?, a detailed solution for In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In a 2D stress system, the two principal stress are p1 = 180 N/mm2 (tensile) and p2 (compressive). For the materials, yield stress in simple tension and compression is 240 N/mm2 and Poisson’s ratio is 0.25. According to maximum normal strain theory for what value of p2 shall yielding commence?a)240 N/mm2b)180 N/mm2c)195 N/mm2d)200 N/mm2Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.
|
|
Explore Courses for Civil Engineering (CE) 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
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup