Question

P12.026

The stresses shown act at a point on the free surface of a machine component. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are Sx = 17.9 ksi, Sy = 2.3 ksi, and Sxy = 7.4 ksi. Assume β=30∘. Determine the normal stresses σx and σy and the shear stress τx⁢yat the point.

Sxy S Oy Txy σα B Answers: Ox = ksi. Oy ksi. Try ksi.

I tired those number and gave me wrong answers

x = 16.21 and 20.40

y = -10.81 and -0.2

xy = -20.91 and -3.054

0 0
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Answer #1

From Mohr's circle
we can transform stress tensors from S frame to xy frame
given beta = 30 deg
so, let area of the side of the cube be A
then

From mohr's circle
sigma-x = 0.5(Sx + Sy) + 0.5(Sx - Sy)*cos(2*beta) + Txy*sin(2*beta)
and
Tn = -0.5(s-x - s-y)*sin(2*beta) + Txy*cos(2*beta)

given values
Sx = 17.9 ksi, Sy = 2.3 ksi, Sxy = 7.4 ksi, beta = 30 deg

Sigma-x = 0.5(17.9 + 2.3) + 0.5(17.9 - 2.3)*cos(60) + 7.4*sin(60)
sigma-x = 20.408587988004 ksi
and
Tn = -0.5(17.9 - 2.3)*sin(60) + 7.4*cos(60)
Tn = -3.0549981495 ksi

hence, sigma-x = 20.408587988004 ksi
sigma-y = 0 ksi
Txy = -3.0549981495 ksi

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