Question1 The failure force for an average size mans humerus

Question1:

The failure force for an average size man’s humerus in compression is about 39 kN. At what force will a 10-year old child’s arm break? The average humerus bone length for a child is 80% of that for an adult. The modulus of elasticity for the bone of a child is 90% of that for an adult.

Solution

>> As, Compressive Force is related to elasticity and length in the following ways :

1). Fc is directly proportional to E³ , i.e.cube of modulus of elasticity

2). Fc is inversely proportional to L² , i.e. Square of Length

=> Fc = k*E³/L²

So, for, Man, 39 =  k*E³/L²

and, for child, Fc = k*(0.9*E)³/(0.8*L)²

=> Fc = 39*0.9³/0.8²

=> Fc = 44.42 kN