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  σ₃ are 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

θ_pp 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 σ_xx. The shear stress on the vertical plane BC will be z_xy.

 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  σ_xx  as σ_x

 Secondly write σ_yy  as σ_y

 Thirdly  write z_yx and z_xy  as z

since z_yx. = z_xy (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 σ_x, σ_y and 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) σ_x 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) σ_y in 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.

σ_Ip = σ_θ = (σ_x + σ_y)/2  + ((σ_x — σ_y)/2) (Cos2θ) + τ Sin2θ

τ_IP =τ_θ = (( σ_x — σ_y)/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 θ_pp , put τ_θ  = 0

We get

tan2 θ_pp = 2 τ/( σ_x — σ_y)

Put values of σ_x, σ_y and τ.

Two values of tan 2θ will give two values of 2θ_pp. These will be 2θ_pp1 and 2θ_pp2

such that  2θ_pp2 =  2θ_pp1 +180⁰

Hence θ_pp2 =  θ_pp1 +90⁰

 PRINCIPAL PLANES ARE AT RIGHT ANGLES TO EACH OTHER.

 (iii) To find principal stresses σ₁  and σ₂ , draw the triangle for tan 2θ_pp and put the values of cos2θ_pp and sin2θ_pp  in the σ_θ equation. We get

σ₁ = (σ_x + σ_y)/2  + (1/2)  ((σ_x — σ_y)² +4τ² ) ^0.5

σ₂ = (σ_x + σ_y)/2  –  (1/2)   ((σ_x — σ_y)² +4τ² ) ^0.5

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

put dτ_θ/dθ = 0

We get

tan2θ_ms = –( σ_x — σ_y)/2τ

Put values of σ_x, σ_y and τ.

Two values of tan2θ_ms will give  2θ_ms1 and 2θ_ms2

such that  2θ_ms2 =  2θ_ms1 +180⁰

Thus θ_ms2 =  θ_ms1 +90⁰

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

 To find τ_max, draw the triangle for tan2θ_ms, put the values of sin2θ_ms and cos2θ_ms in the τ_θ  equation.

We get

 τ_max =  (1/2)   (σ_x — σ_y)² + 4 τ²

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

(between θ_pp and θ_ms)

Find the product of slopes tan2θ_pp tan2θ_ms

We get tan2θ_pp tan2θ_ms = –1

Therefore angle between 2θ_pp and 2θ_ms will be 90⁰.

Therefore angle between θ_pp and θ_ms will be 45⁰.

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

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