Lesson 25. SOLVING NUMERICAL

Module 5. Properties of material, failures and factor of safety


Lesson 25
SOLVING NUMERICAL

The load on a bolt consists of an axial pull of 20 KN together with a transverse Shear force of 10 KN. Find the diameter of bolt required according to

  • Maximum principal stress theory.
  • Maximum distortion energy theory.
  • Maximum shear stress theory.
  • Maximum strain energy theory; and
  • Maximum principal strain theory.
     

Take Permissible Tensile stress at elastics limit=100Mpa and Poisson’s ratio=0.3

Nomenclature

Load=P

бt1 = normal tensile stress in x direction

бt2 = normal tensile stress in y direction

б1, б2= Principal stresses

Shear stress= ז

бt(el)= Permissible Tensile stress at elastics limit

Sol.

Given load=Ptensile =20KN

Pshear=10KN

(stress(sigma symbol))tensile(elongation)=100Mpa=100N/MM2

µ=1/m=0.3

Let d=diameter of the bolt in mm

Therefore Cross sectional area of the bolt

A = (Π / 4) x d2=0.785*d2 mm2

We know that axial tensile stress

бt1 = (Pt1 / A) = (20 / 0.785d2) = (25.47 / d2) kN / mm2

And transverse shear stress

Shear stress(ז) = ( Ps/ A) = (10 / 0.785d2 ) = ( 12.73/ d2) kN/mm2

1). According to maximum principal stress theory.

We know that maximum principal stress,

25.1

According to maximum principal stress theory

бt1 = бt(el)

(30365 / d2) = 100

d2 = ( 30365/ 100) = 303.65

= 17.42mm


2) According to maximum shear stress theory

We know that maximum shear stress


25.2

According to maximum shear stress theory

25.3


25.4


25.7


25.8


Last modified: Thursday, 27 September 2012, 10:13 AM