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).
A body is subjected to a tensile stress of 1200 MPa on one plane and a...
Answer pls
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).