A bar of solid circular crosssection is loaded by a tensile

A bar of solid circular cross-section is loaded by a tensile axial force F as shown in the accompanying sketch. The diameter of the bar is 9 mm. In this case, the magnitude of the applied force is 0.6 kN and the plane a-a is inclined at an angle of -39 degrees to the vertical. Here, a positive angle implies that the plane a-a is inclined counterclockwise to the vertical as shown, while a negative value implies a clockwise angle. Calculate the normal stress on plane a-a. Give your answer in MPa to 1 decimal place.

Solution

ANS - Normal stress acting in the bar N= (F/A)

= (600/(3.14*.0045*.0045))

= 9.43 MPa

Normal force in Plane a-a = N*sin(x)^2

=9.43*sin(39)^2

=3.74 MPa