Question

Spiral fracture of bone: Spiral fracture of bone occurs due to twisting of the limb, and is a very common skiing accident. The fracture plane is helical, and is very difficult to heal. Mechanically, it occurs due to an applied torsion load on the bone. Recall that a state of pure shear occurs within the material when torsion is applied on a cylinder, and the bone can be idealized as a cylinder. Let a femur bone be subjected to a torque of T 50 N-m. Assume body weight of the person to be W- 80 Kgs, while each leg is subjected to half of that weight. Given radius of the bone r 10 mm. Compute the principal stresses and shear stresses, as well as orientation of planes on which these stresses are realized.

Answer 1

Answer:

principal stresses :б1 = 32.62mPa  б2 = 31.38mPa

Max Shear stress : 16.31 mPa

Orientation of max principle plane = 44.43°

Orientation of minimum principal plane = 134.43°

Explanation:

Given data:

Torque = 50 N-m

weight = 80 kgs

half of weight is subjected to each leg

radius of bone = 10 mm = 0.010 m

a) Determine the principal stresses and shear stress

first calculate the max shear stress ( this will occur in the outermost element

= 16T / π*d^3   where : T = 50 , d = 0.020 m

hence max shear stress = 32 mPa

next determine compressive stress

= ( 40*g)  / π/4*d^2 . where : d = 0.020 m , g = 9.81

hence compressive stress = 1.24 mPa

draw and calculate the radius of Mohr's circle

radius of Mohr's circle = 32.0060

Hence principal stresses = 32.0060 ± 0.62

б1 = 32.62mPa  

б2 = 31.38mPa

attached below is the remaining part of the solution