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Question 6 0.5 points Save Answer • 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 principal stresses, and Tmax is maximum shear stress. 012021 are the Fractute 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 Txy is: 5000 psi Oopsi 10000 psi 2500 psi

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hear cylinder is subjected to longitudinal direction re Je = 50oopsi Jy = Tx = 5000 Txy -o (No sheas applied) 7 09 - Numerical valye of Tay Txy 38 for A 36 steel ( duetile material under tension Fail at 45 angle (under shear) It will 459 Foilur 8.9 For longitudinal diration Jy=0 as mention above Q. 10 Max shear under longitudinal tension max (304)+4440 to Tmax old None of the above 2

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