# Normal and shear stresses due to tension

## Homework Statement

at a point in a stressed body, there are normal stresses of 95 MPa (tension) on a vertical plane and 125 MPa (tension) on a horizontal plane. determine the normal and shear stresses at this point on the inclined plane a-b (please see attachment)

the solution should be stress, sigma_ab = 121.5 MPa (tension) and sheering stress, tau_ab = -9.64 MPa

## Homework Equations

stress, sigma = F/A where F is force, A is area

sheering stress, tau = V/A where V is shear force, A is area

stress transformation equations:

stress:
sigma_n = sigma_x (cos^2(theta)) + sigma_y (sin^2(theta)) + 2tau_xy(sin(theta)cos(theta))

shear stress:
tau_nt = -(sigma_x - sigma_y)sin(theta)cos(theta) + tau_xy(cos^2(theta) - sin^2(theta))

where theta = 90 - 20 = 70 degrees, tau_nt is along the plane a-b and sigma_n is perpendicular to tau_nt = plane a-b

## The Attempt at a Solution

just by looking at the stress and shear stress formulas it seems i need to determine area, the diagram given in the problem has not dimensions, area dimensions irrelevant?

the point i am looking at is where the two tension forces intersect, this point appears to be at the center of the "square"

using the stress transformation equations:
sigma_x = 95 MPa, sigma_y = 125 MPa, theta = 90 - 20 = 70 degrees

i am not sure about tau_xy though, it may be = 0, correct?

so assuming tau_xy = 0, and using the stress transformation equations:

sigma_n = sigma_x (cos^2(theta)) + sigma_y (sin^2(theta)) + 2tau_xy(sin(theta)cos(theta))

sigma_n = 95(cos^2(70)) + 125 (sin^2(70)) + 2(0)(sin(70)cos(70))

sigma_n = 121.49 = 121.50 as solutions suggest

tau_nt = -(sigma_x - sigma_y)sin(theta)cos(theta) + tau_xy(cos^2(theta) - sin^2(theta))

tau_nt = -(95 - 125)sin(70)cos(70) + 0(cos^2(70) - sin^2(70))

tau_nt = -9.64 as solutions suggest

i did end up getting the solutions i should get assuming tau_xy = 0, how do i determine tau_xy?

thanks

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