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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