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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