In this Article we will find the young Modulus of Elasticity for the material.
Example no 2 : A tie is 7.5cm wide, 15cm deep and 200cm long. It is subjects to an axial force of 4200kn. The stretch of the material is found to be 2.67cm.find young modulus of elasticity for the material
Breadth =b= 7.5cm
Depth =d= 15cm
Length = L= 200 cm
Stretch = Δ L= 2.67cm
Axial force = p= 4200kN
Young modulus of elasticity =E= ?
Young modulus = stress/strain
Stress = force/area, strain = stretch/original length
we find stress and strain for young modulus.
X-sectional area =A= b×d = 7.5×15 = 112.5cm2
stress = force/area = f= p/A= f= 4200/112.5 = 37.33 KN /cm2
Strain = stretch /original length = Δ L/L = 2.67/200 = 0.01335
Young modulus of elasticity =E= stress/strain = E=f/e = 37.33/0.01335 = 2796.25KN/cm2