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A body is subjected to two normal stresses 20 kN/m2 (tensile) and 10 kN/m2 (compressive) acting mutually perpendicular to each other. The maximum shear stress is
- a)30 kN/m2
- b)5 kN/m2
- c)15 kN/m2
- d)10 kN/m2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A body is subjected to two normal stresses 20 kN/m2 (tensile) and 10 ...
Τmax = 
= 15 kN/m2
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= 15 kN/m2Most Upvoted Answer
A body is subjected to two normal stresses 20 kN/m2 (tensile) and 10 ...
Given:
Normal stress in x-direction (tensile) = 20 kN/m2
Normal stress in y-direction (compressive) = 10 kN/m2
To find: Maximum shear stress
Solution:
Shear stress is given by the formula:
τ = σx - σy / 2
where, τ is the shear stress, σx is the normal stress in x-direction, and σy is the normal stress in y-direction.
Substituting the given values, we get:
τ = 20 - (-10) / 2
τ = 15 kN/m2
Therefore, the maximum shear stress is 15 kN/m2.
Answer: Option (c)
Normal stress in x-direction (tensile) = 20 kN/m2
Normal stress in y-direction (compressive) = 10 kN/m2
To find: Maximum shear stress
Solution:
Shear stress is given by the formula:
τ = σx - σy / 2
where, τ is the shear stress, σx is the normal stress in x-direction, and σy is the normal stress in y-direction.
Substituting the given values, we get:
τ = 20 - (-10) / 2
τ = 15 kN/m2
Therefore, the maximum shear stress is 15 kN/m2.
Answer: Option (c)
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