stress strain analysis

7/29/2019 stress strain analysis

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Stress Transformation

9.1-9.3

Plane Stress

Stress Transformation in Plane Stress

Principal Stresses & Maximum Shear Stress



Introduction

We have learned Axially

In Torsion

In bending

These stresses act on cross sections of

the members. Larger stresses can occur on inclined

sections.



Introduction

We will look at stress elements toanalyze the state of stress produce by asingle type of load or by a combinationof loads.

From the stress element, we will derive

the Transformation Equations

Give the stresses acting on the sides of anelement oriented in a different direction.



Introduction

Stress elements: only one intrinsic state of stress exists at a point in a stressed body,

regardless of the orientation of the elementfor that state of stress.

Two elements with different orientations at

the same point in a body, the stress acting onthe faces of the two elements are different,but represent the same state of stress

The stress at the point under consideration.



Introduction

Remember, stresses are not vectors.

Are represented like a vector withmagnitude and direction

Do not combine with vector algebra

Stresses are much more complex

quantities than vectors Are called Tensors (like strain and I)



Plane Stress

Plane Stress – The state of stresswhen we analyzed bars in tensionand compression, shafts in torsion,and beams in bending.

Consider a 3 dimensional stresselement

Material is in plane stress in the xy

plane Only the x and y faces of the element

are subjected to stresses

All stresses act parallel to the x and yaxis



Plane Stress

Normal stress –

subscript identifies the face on which thestress acts

Sign Convention

Tension positive

compression negative

x



Plane Stress

Shear Stress - Two subscripts

First denotes the face on which the stress acts Second gives the direction on that face

Sign convention Positive when acts on a positive face of an

element in the positive direction of an axis (++)or (--)

Negative when acts on a positive face of anelement in the negative direction of an axis (+-)

or (-+)

xy



Plane Stress

A 2-dimensionalview can depict the

relevant stressinformation, fig. 9.1c

Special cases

Uniaxial Stress Pure shear

Biaxial stress



Stresses on Inclined Planes

First we know x, y, andxy,

Consider a new stresselement Located at the same point in

the material as the original

element, but is rotatedabout the z axis

x’ and y’ axis rotatedthrough an angle




The normal and shear stresses actingon they new element are:

Using the same subscript designationsand sign conventions described.

Remembering equilibrium, we knowthat:

'''' ,, y x y x

'''' x y y x




The stresses in the x’y’ plane can be expressed in

terms of the stresses onthe xy element by usingequilibrium.

Consider a wedge shapedelement

Inclined face same as the x’ face of inclined element.




Construct a FBD showing all theforces acting on the faces

The sectioned face is A. Then the normal and shear

forces can be represented onthe FBD.

Summing forces in the x and ydirections and rememberingtrig identities, we get:

2cos2sin2

2sin2cos22

xy

y x

y x

xy

y x y x

x




These are called the transformation equationsfor plane stress.

They transfer the stress component form one setof axes to another.

The state of stress remains the same.

Based only on equilibrium, do not depend onmaterial properties or geometry

There are Strain Transformation equations thatare based solely on the geometry of deformation.




Special case simplifications

Uniaxial stress- y

& Txy

= 0

Pure Shear - x & y = 0

Biaxial stress - Txy = 0

Transformation equations are simplifiedaccordingly.



Principal & Maximum Shear

Stresses Since a structural member can fail due

to excessive normal or shear stress, weneed to know what the maximumnormal and stresses are at a point.

We will determine the maximum and

minimum stress planes for whichmaximum and minimum normal andshear stresses act.




Stresses Principal stresses – maximum and minimum

normal stresses.

Occurs on planes where:

Applying to eq 9.1 we get:

p=the orientation of the principal planes

The planes on which the principal stresses act.

0'

d

d x

y x

xy

p

2

2tan




Stresses Two values of the angle 2p are obtained

from the equation.

One value 0-180, other 180-360 Therefore p has two values 0-90 & 90-180

Values are called Principal Angles.

For one angle x is maximum, the other x isminimum.

Therefore: Principal stresses occur onmutually perpendicular planes.




Stresses We could find the principal stress by

substituting this angle into thetransformation equation and solving

Or we could derive general formulas forthe principal stresses.



Principal Stresses

Consider the right triangle

Using the trig from the triangle

and substituting into thetransformation equation fornormal stress, we get

Formula for principal stresses.

2

2

2,122

xy

y x y x



Shear Stresses on the

Principal Planes If we set the shear stress x’y’ equal to

zero in the transformation equation and

solve for 2, we get equation 9-4. The angles to the planes of zero shear

stress are the same as the angles to the

principal planesTherefore:The shear stresses are zero on

the principal planes



The Third Principal Stress

We looked only at the xy plane rotating aboutthe z-axis.

Equations derived are in-plane principalstresses

BUT, stress element is 3D and has 3 principal

stresses. By Eigenvalue analysis it can be shown thatz=0 when oriented on the principal plane.



Maximum In-Plane Shear

Stress Consider the maximum shear stress and

the plane on which they act.

The shear stresses are given by thetransformation equations.

Taking the derivative of x’y’

withrespect to and setting it equal to zerowe can derive equation 9-7



Maximum Shear Stress

The maximum negative shear stress min has the same magnitude but opposite

sign.

The planes of maximum shear stressoccur at 45 to the principal planes



Maximum Shear Stress

If we use equation 9-5, subtract 2 from 1, and compare with equation 9-

7, we see that:

Maximum shear stress is equal to ½ thedifference of the principal shear stress.

2

21

max



Average Normal Stress

The planes of maximum shear stressalso contain normal stresses.

Normal stresses acting on the planes of maximum positive shear stress can bedetermined by substituting the

expressions for the angle

s into theequations for x’ .

Result is Equation 9-8.



Important Points

The principal stresses are the max and min normalstress at a point

When the state of stress is represented by theprincipal stresses, no shear stress acts on theelement

The state of stress at the point can also berepresented in terms of max in-plane shear stress .

In this case an average normal stress also acts onthe element

The element in max in-plane shear stress is oriented45° from the element in principal stresses.

























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