# 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.

#### σ_{1 }, σ_{2} , and σ_{3 }are principal stresses.

#### The units are N/mm^{2}.

#### 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** ** **

_{x, }σ

_{y }

#### (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 p****lane means to find θ with reference plane AB.**

**(ii) ****To find principal planes means to find θ**_{pp} , put τ_{θ} = 0

_{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} +180^{0}

#### Hence θ_{pp2} = θ_{pp1} +90^{0}

** ****PRINCIPAL PLANES ARE AT RIGHT ANGLES TO EACH OTHER.**

** ****(iii) ****To find principal stresses σ**_{1 } and σ_{2 }, draw the triangle for tan 2θ_{pp} and put the values of cos2θ_{pp} and sin2θ_{pp} in the σ_{θ} equation. We get

_{1 }and σ

_{2 }, draw the triangle for tan 2θ

_{pp}and put the values of cos2θ

_{pp}and sin2θ

_{pp}in the σ

_{θ}equation. We get

#### σ_{1} = (σ_{x} + σ_{y})/2 + (1/2) ((σ_{x} — σ_{y})^{2} +4τ^{2} )^{ 0.5}

#### σ_{2} = (σ_{x} + σ_{y})/2 – (1/2) ((σ_{x} — σ_{y})^{2} +4τ^{2} )^{ 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} +180^{0}

#### Thus θ_{ms2} = θ_{ms1} +90^{0}

** ****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 τ^{2}

#### (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 90^{0}.

#### Therefore angle between θ_{pp} and θ_{ms} will be 45^{0}.

** ****THE ANGLE BETWEEN A PRINCIPAL PLANE AND A PLANE OF MAXIMUM SHEAR IS 45**^{0}.

^{0}.

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