Compressional stress & shear stress

[petep]

A force of 500 N is applied at an angle of 37 degrees to the surface of the end of a square bar. that surface is 4.0 cm on a side. what are the compressional & shear stress on the bar? material processing class and i don't know the equation or where to start someone help please

 

[Astronuc]

You need to specify more information on this problem. What are the boundary (fixed point) conditions, i.e. is it a cantilever beam (one end fixed, one end free)? I assume that is the geometry for this problem.

Also I presume the horizontal component of the force vector is pointing toward the fixed end, in order to put the entire beam in compression.

So assuming it this is a cantilevered beam, what are the equations for a cantilever beam in deflection? Think of the horizontal and vertical components. You may ignore torsion.

Are you familiar with superposition?

 

[wais]

Compressional stress is a type of stress that occurs when a material is being pushed or compressed from opposite ends, causing it to become shorter and thicker. This type of stress is typically seen in structures such as columns or pillars. On the other hand, shear stress is a type of stress that occurs when a material is being subjected to forces that are parallel to its surface, causing it to slide or deform.

To calculate the compressional and shear stress on the bar in this scenario, we can use the following equations:

Compressional stress = Force/Area
Shear stress = Force/Area x sin(angle)

First, we need to convert the given surface area of the square bar from centimeters to meters, as the SI unit for area is square meters. This gives us a surface area of 0.04 square meters.

Next, we can plug in the values into the equations. For compressional stress, we have a force of 500 N and an area of 0.04 square meters. This gives us a compressional stress of 12,500 Pa.

For shear stress, we need to first find the component of the force that is parallel to the surface of the bar. This can be done by using the sine function to find the opposite side of the triangle, which represents the component of the force. We have a force of 500 N and an angle of 37 degrees, so the component of the force is 500 x sin(37) = 300.8 N.

Plugging this value into the equation for shear stress, we have a force of 300.8 N and an area of 0.04 square meters. This gives us a shear stress of 7,520 Pa.

In conclusion, the compressional stress on the bar is 12,500 Pa and the shear stress is 7,520 Pa. It is important to note that these values are within the elastic limit of most materials, meaning that the bar will return to its original shape once the force is removed. If the stress exceeds the elastic limit, the material will undergo permanent deformation.