Normal and shear stresses due to tension

In summary: As for determining tau_xy when shear stress is present, the easiest way is to use an equilibrium argument. If you have a shear stress, there must be a force applied to the side of the element. This force must be balanced by the shear stresses on the other sides of the element. This lets you calculate tau_xy.In summary, 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. Using the stress transformation equations, with a theta of 70 degrees, the normal stress on the inclined plane a-b is 121.50 MPa (tension) and the shearing
  • #1
portofino
35
0

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
 

Attachments

  • shear.JPG
    shear.JPG
    8.5 KB · Views: 720
  • shear2.JPG
    shear2.JPG
    15.3 KB · Views: 712
Physics news on Phys.org
  • #2
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" [Broken].

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.
 
Last edited by a moderator:
  • #3
for providing the necessary information and equations to solve this problem. Your solution is correct assuming tau_xy = 0, which is the case for a simple tension problem. To determine tau_xy, the given diagram or problem statement would need to provide more information about the geometry of the body and the direction of the shear force. Without this information, it is not possible to accurately calculate tau_xy. In general, tau_xy can be determined using the equation tau_xy = V/A, where V is the shear force acting on the plane and A is the area of the plane. However, without knowing the specific geometry and direction of the shear force, it is not possible to accurately calculate tau_xy.
 

1. What is the difference between normal and shear stresses due to tension?

Normal stress is a force that acts perpendicular to the surface of an object, while shear stress is a force that acts parallel to the surface of an object. In the context of tension, normal stress is the force that tries to pull the two ends of an object away from each other, while shear stress is the force that tries to slide the two ends of an object past each other.

2. How are normal and shear stresses calculated in tension?

Normal stress can be calculated by dividing the applied force by the cross-sectional area of the object, while shear stress can be calculated by dividing the force parallel to the surface by the cross-sectional area of the object.

3. What are some examples of objects that experience normal and shear stresses due to tension?

Common examples include ropes, cables, and bridges. In these structures, tension forces act on the object, causing both normal and shear stresses to occur.

4. How do normal and shear stresses due to tension affect the strength of an object?

Normal stress can cause an object to elongate or compress, depending on the direction of the force. Shear stress, on the other hand, can cause an object to deform or break. Both normal and shear stresses can weaken an object over time if the applied forces are too great.

5. What are some factors that can influence the normal and shear stresses due to tension?

The magnitude and direction of the applied force, as well as the cross-sectional area and material properties of the object, can all affect the normal and shear stresses experienced by an object in tension. Other factors such as temperature and external forces can also play a role in stress levels.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
989
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
8K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
11K
Back
Top