Example- axial and shear deformation.

  • A steel rod 20mm in diameter and 300mm long is subjected to an axial pull of 40 KN. If the young modulus is 2 x10⁵ N/mm². Calculate the elongation of bars.

 

==> Given Data

  • Dia. of rod =20 mm

  • Length of rod = (L) =300mm

  • Axial pull (P=F) = 40 KN

    • P =40 x10³ N

  • young modulus is(E) = 2 x10⁵ N/mm²

  • Stress (σ) = P/A

= {40 x10³ /[(π/4) x 20²]}

  • σ= 127.324N/mm²


  • Young Modulus = (E) = (σ / e)


    2 x10⁵ = ( 127.324 /  e)


    e = 6.3662 x10㆒⁴ (i.e. e= δL / L)


    6.3662 x10㆒⁴  = ( δL/ 300)


    δL = 0.190986 mm---Ans


    Check (δl) = {(P L) /(A E)}

    δl = [(40 x10³ x300) / ((π/4) x 20² x2 x10⁵)]


    δL= 0.190986 mm---Ans

  • ---x---

    • A cast column has an external diameter of 300mm and 20 mm thick. Find the safe compressive load on the column with a Factor of safety of 5. If the crushing strength is 550 N/mm²

     ==> Given data

    Area = (π/4) x (300² - 260²)

    = 70371.674 mm²

    Safe load=P=?      (F. Of  S. =5)

    Crushing strength = 550 N/mm²  

    • Safe stress = (Crushing strength / F. Of  S.)

    = (550 / 5)

    = 110 N/mm²


    • Safe stress =( Safe load /Area)

    110 = ( Safe load / 70371.674 )

    Safe load =1.53610㆒³ N --- Ans.


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