1- Spiral fractures of the tibia are common in ski accidents and are caused by torsional load on bone. Ski binding are designed to release the boot when a desire torque is applied to the leg. The bone has a shear modulus of G=3.5 GPa. The bone can be modelled as a hollow cylinder with inner diameter of 1.5 cm and outer diameter of 5cm. If the bone will fracture (failure) at the shear stress of 17MPA and the design of binding must not allow the bone to reach to the maximum shear stress of 60% of the failure shear stress. a) Determine the value for the maximum torque that the release mechanism should act to prevent bone fracture b) Determine the angle of twist when torque reach to the value obtain from part a 45cm (Answer: a) 248.31 N.m; b) 0 = 0.052 rad)

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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1- Spiral fractures of the tibia are common in ski accidents and are caused by torsional load on bone.
Ski binding are designed to release the boot when a desire torque is applied to the leg. The bone
has a shear modulus of G-3.5 GPa. The bone can be modelled as a hollow cylinder with inner
diameter of 1.5 cm and outer diameter of 5cm. If the bone will fracture (failure) at the shear stress
of 17MPA and the design of binding must not allow the bone to reach to the maximum shear stress
of 60% of the failure shear stress.
a) Determine the value for the maximum torque that the release mechanism should act to
prevent bone fracture
b) Determine the angle of twist when torque reach to the value obtain from part a
45cm
(Answer: a) 248.31 N.m; b) o = 0.052 rad )
%3D
A R
Transcribed Image Text:1- Spiral fractures of the tibia are common in ski accidents and are caused by torsional load on bone. Ski binding are designed to release the boot when a desire torque is applied to the leg. The bone has a shear modulus of G-3.5 GPa. The bone can be modelled as a hollow cylinder with inner diameter of 1.5 cm and outer diameter of 5cm. If the bone will fracture (failure) at the shear stress of 17MPA and the design of binding must not allow the bone to reach to the maximum shear stress of 60% of the failure shear stress. a) Determine the value for the maximum torque that the release mechanism should act to prevent bone fracture b) Determine the angle of twist when torque reach to the value obtain from part a 45cm (Answer: a) 248.31 N.m; b) o = 0.052 rad ) %3D A R
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