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Average stress and factor of safety
Two steel tube are joined at B by four pins (dp = 11 mm), as shown in the cross section a-a in the figure. The outer diameters of the tubes are dAB = 40 mm and dBC = 28 mm. The wall thicknesses are tAB = 6 mm and tBC = 7 mm. The yield stress in tension for the steel is sY = 200 MPa and the ultimate stress in tension is sU = 340 MPa. The corresponding yield and ultimate values in shear for the pin are 80 MPa and 140 MPa respectively. Finally, the yield and ultimate values in bearing between the pins and the tubes are 260 MPa and 450 MPa, respectively. Assume that the factors of safety with respect to yield stress and ultimate stress are 4 and 5 respectively.
a. Calculate the allowable tensile force Pallow considering tension in the tubes.
b. Recompute Pallow for shear in the pins.
c. Finally, recompute Pallow for bearing between the pins and the tubes.
d. Which is the limiting value of P?
Summary
The question belongs to Mechanical Engineering and the question is about average stress and factor of safety where the tensile strength of a steel tubing is tested.
Total Word Count 105
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