Normal and shear stresses due to tension

[portofino]

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"

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

 

[Mapes]

Correct, area doesn't enter into it. You're dealing with the stress state in an infinitesimal element inside the material, trying to determine if the normal or shear stress on any plane exceeds the material's capabilities.

It was also correct to assume no shear stress in the original configuration. The presence of shear stress is indicated graphically by four arrows (always four!) next to the sides of the element, as shown http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/images/MaxShearStress.gif" .

The normal force arrows will always intersect at the center of the square. Technically, the square has no physical extent; it has a side length of dx = dy (that is, as small as you can make it). All normal forces must go through the center.