A copper tube with wall thickness of 8 mm must carry an axial tensile force of 175 kN. The allowable tensile stress is 90 MPa. The minimum required outer diameter is approximately: $\text { MPa. }$

Question

Quizplus · Accepted Answer

The stress in the tube can be calculated as:
Stress = Force/Area
Allowable Stress = Applied Force/((Outer Diameter - Wall Thickness)/2 x Wall Thickness)
Rearranging and substituting the given values in the above equation, we get:
Outer Diameter = (2 x Applied Force/Allowable Stress x Wall Thickness) + Wall Thickness
= (2 x 175 x 10^3)/(90 x 10^6 x 8 x 10^-3) + 8 x 10^-3
= 0.085 m or 85 mm

Therefore, the minimum required outer diameter is approximately 85 mm, which is option C.

A Copper Tube with Wall Thickness of 8 Mm Must MPa. \text { MPa. } MPa.

A Copper Tube with Wall Thickness of 8 Mm Must $\text { MPa. }$