Question

Shown in the figure below is a solid cylinder. The x-axis is along the longitudinal direction. The angle is measured with respect to the x-axis, positive if counterclockwise, i.e. towards the y-axis as shown below; and negative if clockwise. 01 and 02, with 01202 are the principal stresses, and Imax is maximum shear stress. х 8 Fracture surface Normal to Fracture surface Suppose that the cylinder is subjected to a uniform tensile stress of 5000 psi in the longitudinal direction. The numerical value for Jy is: 10000 psi O 5000 psi O 2500 psi O Opsi

Answer 1

Drawing the State of stress we can solve this problem. I have shown below-
Sol: The data provided in the question are as below Uniform applied on the tensile stress 5000 psi longitudinas direction we have to find out Stress in the direction. so a Long Longitudinal direction Ort nac of → Transverse direction: The State of stress for the given condition will be as below-

Tx = 5000 PSN be х Therefore, there is stress in the in the y direction. so, numerical Value for 60 is o Psi Ans option @ is correct.