Question

The yield stress of titanium is 830 MPa (830,000,000 Pa). A titanium rod used for an femur implant needs to withstand 29 kN (29,000 N). What should the diameter of the rod be to not deform? a. 14.62 mm b. 20.00 mm c. 10.00 mm d. 25.50 mm

Answer 1

To prevent deformation, the titanium rod used as a femur implant must have a minimum diameter of 20.00 mm to withstand a force of 29 kN without exceeding its yield stress of 830 MPa.

Step-by-step explanation:

The question involves calculating the minimum diameter of a titanium rod designed to be used as a femur implant, which must not deform under a given load. To ensure that the rod does not deform, the stress applied to the rod must be less than the yield stress of titanium, which is 830 MPa.

To find the required diameter, we use the formula for stress, σ = F/A, where F is the force and A is the cross-sectional area of the rod. The cross-sectional area A for a circular rod is π(d/2)^2, where d is the diameter of the rod. Rearranging the formula to solve for the diameter, we get d = 2 * √(F/(π*σ)).

Substituting the given values (F = 29,000 N and σ = 830,000,000 Pa), we calculate the minimum diameter of the rod required to prevent deformation.

After calculations, we find that the correct answer is: b. 20.00 mm