Given figure forces on each cross-section are shown. The surface area of cross-sections is like the area of AB=Area of CD=2mm2, Area of EF= Area of GH=1mm2. Then find the stress on the surface AB, GH, EF and CD.
Solution:
- If we start to analyse from the left side then, Net force on the left side of AB=50-10=40N
Hence total load on either side of AB=40N (positive sign means tensile force )
Surface area of AB= 2mm2
Hence stress on AB= 402N/mm2= 20MPa
- Similarly, the net force on the right side of GH= 60N
As the rod is in equilibrium, then we can say that the Total force on either side of the surface GH= 60N
Surface area GH= 1mm2
Hence stress on GH= 601N/mm2=60 MPa
- Force on right side of CD= 60N
Surface area of CD= 2mm2
Hence stress on CD= 602N/mm2= 30MPa
- For surface EF similar manner, we can calculate the force that is 50N
Surface area EF= 1mm2
Hence stress on EF= 501N/mm2=50 MPa
Problem 2
A balloon is filled with air having an internal pressure of 3Pa. The diameter of the balloon is 4cm. Then find the volumetric stress.
Solution:
As the balloon is in equilibrium, its volume is constant and the total force on either side of the surface is the same. Hence pressure is equal to the volumetric stress here.
Hence volumetric stress = 3Pa
Conclusion
There are various studies and applications of stress in material engineering and science. By the information of stress on a body, we can calculate the elongation produced on the body. When the stress on a material increases, various changes that are described by various limits can be explained based on stress-strain diagrams. By using Hooke’s law, we can derive Elastic constants from stress information.
In this study material, we discussed stress, types of stress and solved some stress-related questions.