PRINCIPAL STRESSES-INTRODUCTION CLASS NOTES

PRINCIPAL STRESSES

INTRODUCTION CLASS NOTES

Every object in nature is under complex

loading. Hence it is under complex

stresses. But failure occurs in a simple

mode. It may be in simple tension, simple

compression or simple shear. Hence

convert complex stresses into

simple stresses. Simple stresses are

principal stresses. To achieve this,

use analytical or graphical method.

PRINCIPAL STRESSES-INTRODUCTION

Number of forces are acting simultaneously almost on each body. Thus, it is complex  loading condition. When these forces increase, a body fails by a simple principal stress or by maximum shear stress.

COMPLEX STRESS SYSTEM

When normal and shear stresses act simultaneously  on an area, it is a complex stress system. Complex stress is there when an inclined force acts on an area. Resolve this inclined force into normal and tangential components. Normal component causes normal stress and tangential component causes the shear stress. Hence complex stress is there.

Practical Applications of Complex stresses

(I) Beams

(ii) Shafts

(iii) Springs

(iv) Chimneys etc.

FOR A PRINCIPAL STRESS

Only normal stress (with zero shear stress on an area) is  principal stress.

There are at the most three principal stresses.

These act at right angles to each other.

 σ1 , σ2 , and  σare principal stresses.

The units are N/mm2.

Principal stresses  act on principal planes.

PLANE

It is two dimensional.

Area is 2-Dimensional.

 Rectangle, square, circle, triangle, ellipse or T-section are planes.

PRINCIPAL PLANE

The area on which there is only normal stress with zero shear stress. P

θpp represents a principal plane.

DOUBLE SUBSCRIPT NOTATION ON STRESSES

First suffix represents direction of axis normal to the area considered.

Second suffix represents direction of axis in the direction of the stress considered.

The normal stress on vertical plane BC will be σxx. The shear stress on the vertical plane BC will be zxy.

 There is a double suffix notation for 3-D stress system.

Use single suffix for sigma stress

Use no suffix for shear stresses in 2-D stress system

Firstly Write  σxx  as σx

 Secondly write σyy  as σy

 Thirdly  write zyx and zxy  as z

since zyx. = zxy (numerically equal)

These are at right angles to each other.

These are complementary of each other.

Thus one finds the followings from the given complex stress system of σx, σand z   

(i) Principal planes

(ii)  Angle between principal planes

(iii) Principal stresses

(iv) Planes of maximum shear stress

(v)  Angle between planes of maximum shear

(vi) Maximum shear stress

(vii) Angle between a principal plane and a plane of maximum shear

Find Principal stresses. When a body under number of forces fail in a simple manner.  The failure is because of any one reason given below:

(a) by tension alone

or

(b) by compression alone

or

( c) by shear alone

(d) By maximum principal strain

Or

(e) maximum principal strain energy

OR

(f) maximum shear strain energy

PRINCIPAL STRESSES-ANALYTICAL METHOD

In this, it is required  to determine

(i) principal planes

(ii) angle between principal planes

(iii) principal stresses

(iv) planes of maximum shear

(v) angle between planes of maximum shear

(vi) maximum shear stress

(vii) angle between a principal plane and a plane of maximum shear.

There are two methods to determine all the above quantities.

(i) Analytical Method

(ii) Graphical Method or Mohr’s Stress Circle Method

ANALYTICAL METHOD

Fig. Complex, Principal stresses, Principal Planes,

        maximum shear stress & Planes of Maximum Shear Stress

(i) Given a vertical plane AB.

(ii) Given complex stresses on this vertical plane

(a)  Given a tensile stress (+ ve) σx in the x direction on the given vertical plane. (assumed)

(b) Given a counterclockwise (+ve) shear stress τ on the given vertical plane. (assumed)

(iii) Given a horizontal plane BC.

Given stresses on this horizontal plane.

(a) Given a tensile stress (+ve) σin the y direction on the given horizontal plane. (assumed)

(b) Given a  clockwise shear stress (-ve) τ on the given horizontal plane. (assumed)

    USE ACTUAL NATURE OF STRESSES IN THE NUMERICAL PROBLEMS. 

STRESSES ON ANY INCLINED PLANE WITHIN THE BODY UNDER COMPLEX STRESSES

 Inclined plane AD is at an angle ‘θ’ with the given vertical plane AB. Stresses on this inclined plane (IP) will also be complex.

σIp = σθ = (σx + σy)/2  + ((σx — σy)/2) (Cos2θ) + τ Sin2θ

τIP =τθ = (( σx — σy)/2) (Sin2θ) — τ Cos2θ

(i) To find a plane means to find θ with reference plane AB.

(ii) To find principal planes means to find θpp , put τθ  = 0

We get

tan2 θpp = 2 τ/( σx — σy)

Put values of σx, σy and τ.

Two values of tan 2θ will give two values of 2θpp. These will be 2θpp1 and 2θpp2

such that  2θpp2 =  2θpp1 +1800

Hence θpp2 =  θpp1 +900

 PRINCIPAL PLANES ARE AT RIGHT ANGLES TO EACH OTHER.

 (iii) To find principal stresses σ and σ, draw the triangle for tan 2θpp and put the values of cos2θpp and sin2θpp  in the σθ equation. We get

σ1 = (σx + σy)/2  + (1/2)  ((σx — σy)2 +4τ2 ) 0.5

σ2 = (σx + σy)/2  –  (1/2)   ((σx — σy)2 +4τ2 ) 0.5

(iv) To find planes of maximum shear and the maximum shear stress ,

put dτθ/dθ = 0

We get

tan2θms = –( σx — σy)/2τ

Put values of σx, σy and τ.

Two values of tan2θms will give  2θms1 and 2θms2

such that  2θms2 =  2θms1 +1800

Thus θms2 =  θms1 +900

 PLANES OF MAXIMUM SHEAR ARE AT RIGHT ANGLES TO EACH OTHER.

 To find τmax, draw the triangle for tan2θms, put the values of sin2θms and cos2θms in the τθ  equation.

We get

 τmax =  (1/2)   (σx — σy)2 + 4 τ2

(v) To find angle between principal plane and plane of maximum shear

(between θpp and θms)

Find the product of slopes tan2θpp tan2θms

We get tan2θpp tan2θms = –1

Therefore angle between 2θpp and 2θms will be 900.

Therefore angle between θpp and θms will be 450.

 THE ANGLE BETWEEN A PRINCIPAL PLANE AND A PLANE OF MAXIMUM SHEAR IS 450.

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