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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 ab (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 ab and sigma_n is perpendicular to tau_nt = plane ab
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"
please see second attachment
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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