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A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa?
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A body is subjected to a tensile stress of 1200 MPa on one plane and a...
Solution:
Given,
Tensile stress on one plane, σx = 1200 MPa
Tensile stress on a plane at right angles to the former, σy = 600 MPa
Shear stress on the same planes, τxy = 400 MPa
We need to find the maximum shear stress, τmax.
Steps to solve the problem:
1. Calculate the normal stress on the plane inclined at 45 degrees to the x-axis and y-axis.
2. Calculate the principal stresses using the formula:
σ1,2 = (σx + σy)/2 ± [(σx - σy)/2]^2 + τxy^2)^1/2
3. Calculate the maximum shear stress using the formula:
τmax = (σ1 - σ2)/2
Step 1: Calculate the normal stress on the plane inclined at 45 degrees to the x-axis and y-axis.
The normal stress on the plane inclined at 45 degrees to the x-axis and y-axis is given by:
σn = (σx + σy)/2 + (σx - σy)/2 cos 2θ + τxy sin 2θ
Where θ = 45 degrees
Substituting the given values, we get:
σn = (1200 + 600)/2 + (1200 - 600)/2 cos 90 + 400 sin 90
σn = 900 MPa
Step 2: Calculate the principal stresses using the formula:
σ1,2 = (σx + σy)/2 ± [(σx - σy)/2]^2 + τxy^2)^1/2
Substituting the given values, we get:
σ1,2 = (1200 + 600)/2 ± [(1200 - 600)/2]^2 + 400^2)^1/2
σ1 = 1000 MPa
σ2 = -200 MPa (negative sign indicates that it is a compressive stress)
Step 3: Calculate the maximum shear stress using the formula:
τmax = (σ1 - σ2)/2
Substituting the values, we get:
τmax = (1000 - (-200))/2
τmax = 600 MPa
Therefore, the maximum shear stress is 600 MPa, which is option (B).
Given,
Tensile stress on one plane, σx = 1200 MPa
Tensile stress on a plane at right angles to the former, σy = 600 MPa
Shear stress on the same planes, τxy = 400 MPa
We need to find the maximum shear stress, τmax.
Steps to solve the problem:
1. Calculate the normal stress on the plane inclined at 45 degrees to the x-axis and y-axis.
2. Calculate the principal stresses using the formula:
σ1,2 = (σx + σy)/2 ± [(σx - σy)/2]^2 + τxy^2)^1/2
3. Calculate the maximum shear stress using the formula:
τmax = (σ1 - σ2)/2
Step 1: Calculate the normal stress on the plane inclined at 45 degrees to the x-axis and y-axis.
The normal stress on the plane inclined at 45 degrees to the x-axis and y-axis is given by:
σn = (σx + σy)/2 + (σx - σy)/2 cos 2θ + τxy sin 2θ
Where θ = 45 degrees
Substituting the given values, we get:
σn = (1200 + 600)/2 + (1200 - 600)/2 cos 90 + 400 sin 90
σn = 900 MPa
Step 2: Calculate the principal stresses using the formula:
σ1,2 = (σx + σy)/2 ± [(σx - σy)/2]^2 + τxy^2)^1/2
Substituting the given values, we get:
σ1,2 = (1200 + 600)/2 ± [(1200 - 600)/2]^2 + 400^2)^1/2
σ1 = 1000 MPa
σ2 = -200 MPa (negative sign indicates that it is a compressive stress)
Step 3: Calculate the maximum shear stress using the formula:
τmax = (σ1 - σ2)/2
Substituting the values, we get:
τmax = (1000 - (-200))/2
τmax = 600 MPa
Therefore, the maximum shear stress is 600 MPa, which is option (B).
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A body is subjected to a tensile stress of 1200 MPa on one plane and a...
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A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa?
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A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa? 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 A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa?.
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa? 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 A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum shear stress will be (A) 400 MPa (B) 500 MPa (C) 900 MPa (D) 1400 MPa?.
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