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At a point in a stressed material, normal stresses are 120 N/mm2 in x-direction and 80 N/mm2 in y-direction. The normal stres on a plane at 30° to the plane of 120 N/mm2 is-
- a)90 N/mm2
- b)100 N/mm2
- c)110 N/mm2
- d)120 N/mm2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
At a point in a stressed material, normal stresses are 120 N/mm2in x-d...
The normal stress on a plane at an angle of 30degree to the plane of 120 N/mm2 can be found using the following formula:
Normal stress = (120 N/mm2 * cos 30
degree
) + (80 N/mm2 * sin 30degree
)Plugging in the values and simplifying gives:
Normal stress = (120 N/mm2 * 0.866) + (80 N/mm2 * 0.5)
= 105.92 N/mm2 + 40 N/mm2
= 145.92 N/mm2
Rounding to the nearest whole number gives a normal stress of 146 N/mm2.
Therefore, the normal stress on a plane at 30
degree
to the plane of 120 N/mm2 is approximately 146 N/mm2. The closest answer choice to this value is (c), 110 N/mm2.Most Upvoted Answer
At a point in a stressed material, normal stresses are 120 N/mm2in x-d...
° to the x-direction can be calculated using the formula:
σn = (σx + σy) / 2 + (σx - σy) / 2 cos2θ + τxy sin2θ
where σn is the normal stress on the plane, σx and σy are the normal stresses in x and y directions, respectively, τxy is the shear stress in the xy plane, and θ is the angle between the plane and x-axis.
In this case, σx = 120 N/mm2, σy = 80 N/mm2, and θ = 30°. We assume that there is no shear stress, i.e., τxy = 0.
σn = (120 + 80) / 2 + (120 - 80) / 2 cos(2 × 30°) + 0 sin(2 × 30°)
= 100 + 20 cos(60°)
= 100 + 20 × 0.5
= 110 N/mm2
Therefore, the normal stress on a plane at 30° to the x-direction is 110 N/mm2.
σn = (σx + σy) / 2 + (σx - σy) / 2 cos2θ + τxy sin2θ
where σn is the normal stress on the plane, σx and σy are the normal stresses in x and y directions, respectively, τxy is the shear stress in the xy plane, and θ is the angle between the plane and x-axis.
In this case, σx = 120 N/mm2, σy = 80 N/mm2, and θ = 30°. We assume that there is no shear stress, i.e., τxy = 0.
σn = (120 + 80) / 2 + (120 - 80) / 2 cos(2 × 30°) + 0 sin(2 × 30°)
= 100 + 20 cos(60°)
= 100 + 20 × 0.5
= 110 N/mm2
Therefore, the normal stress on a plane at 30° to the x-direction is 110 N/mm2.
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