Stress and Strain Numerical No # 1




 

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

Given data

Breadth =b= 7.5cm

Depth =d= 15cm

Length = L= 200 cm

Stretch = Δ L= 2.67cm

Axial force = p= 4200kN




Required

Young modulus of elasticity =E= ?

Solution

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

Also read : Find Safe Compressive Load for a Column Using Euler’s Formula

Engr Sami Ullah Author

Founder & Admin of civilengineeringinfo.com I am Civil Engineer working as a Site Engineer. With a good subject knowledge in civil Engineering. And I have started the blog to share my valuable information with civil engineer students. Civil Global is the Civil Engineering learning Home around all over the world.

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