It is known that the principal stresses acting in the vertical plane are 50 kPa and 100 kPa in the horizontal plane. What is the magnitude of the normal and shear stresses acting on a plane with a slope of 45 degrees?

A. 50 kPa and 50 kPa
B. 75 kPa and 25 kPa
C. 75 kPa and 50 kPa
D. 50 kPa and 25 kPa

Answer :

The normal and shear stresses acting on a plane with a slope of 45 degrees, given that the principal stresses acting in the vertical plane are 50 kPa and 100 kPa in the horizontal plane can be calculated as follows: Normal and shear stresses acting on a plane with a slope of 45 degrees can be calculated as:

Normal stress = 1/2[(σ1−σ2)+ (σ1+σ2) cos2θ]Shear stress = 1/2[(σ1−σ2) sin2θ]Whereσ1 and σ2 are the principal stresses acting in the vertical and horizontal planesθ is the angle of inclination of the plane The inclination angle of the plane, θ = 45° σ1 = 100 kPaσ2 = 50 kPa.

Normal stress = 1/2[(σ1−σ2)+ (σ1+σ2) cos2θ]=1/2[(100−50)+ (100+50) cos2×45°]= 75 kPa Shear stress = 1/2[(σ1−σ2) sin2θ]=1/2[(100−50) sin2×45°]= 25 kPa Therefore, the magnitude of the normal and shear stresses acting on a plane with a slope of 45 degrees is 75 kPa and 25 kPa, respectively. Hence, option (B) is the correct choice.

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