Modulus of Elasticity =120GPa Yield Strength = 758 MPa A football player require

Question

Modulus of Elasticity =120GPa

Yield Strength = 758 MPa

A football player requires a total hip joint replacement surgery. The artificial joint is shown in Figure 1. Find the smallest radius of the hip neck, when the angle of inclination (Figure 2) is 90 degree, 135 degree and 160 degree respectively. Given: The hip neck has to be hollow with at least 5-mm inner diameter. Body weight of the patient is 200 lb Loading on the hip head is limited in coronal plane and in the perpendicular direction. Size of femur head is pre-determined and identical for all cases. No friction between the head of femur and hip acetabulum. Length of the hip neck is 40 mm. Material of the artificial joint neck is Ti alloy, Ti-6Al-4V, which has excellent bio compatibility The material is isotropic, homogeneous, linear elastic.

Explanation / Answer

solution:

1)here for out of given condition we will solve for maximum angle that is 160 degree first and it will gives eccentric loading on hip head perpendicular to horizontal plane

here W=200 lbf or 889.64 N

here bending moment is

M=W*l=889.64sin(90-70)*40=12170.99 N mm

here bending stresses are

Sb=M*y/I=12170.99*(do/2)/[(pi/64)*(do^4-di^4)]

where compressive stress isSc=889.64cos(90-70)/(pi/4)*(do^2-di^2)

where as maximum compressive stress is sumof both of them

Syt=Sc+Sb

on putting value we get equation as

-758do^6+0do^5+20014.41do^4+123972.7do^3+473750do^2-3099317.5do-12509006.25=0

on solving we get

do=6.4972 mm

hence outer diameter of neck is 6.4972 mm for 160 degree angle of inclination

2) here for out of given condition we will solve for maximum angle that is 135 degree first and it will gives eccentric loading on hip head perpendicular to horizontal plane,135-90=45 degree

here W=200 lbf or 889.64 N

here bending moment is

M=W*l=889.64sin(90-45)*40=25162.81 N mm

here bending stresses are

Sb=M*y/I=25162.8*(do/2)/[(pi/64)*(do^4-di^4)]

where compressive stress isSc=889.64cos(90-45)/(pi/4)*(do^2-di^2)

where as maximum compressive stress is sumof both of them

Syt=Sc+Sb

on putting value we get equation as

758do^6+0do^5-19750.95do^4-256306.37do^3-473750do^2+6407659.25do+12344343.75=0

on solving we get

do=7.5611 mm

hence outer diameter of neck is 7.5611 mm for 135 degree angle of inclination

3)in the similar way we can solve for 90 degree angle as above

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