ce2155 - stress and strain transformation (part 1)

CE2155 Structural Mechanics and Materials

byAssoc Professor T. H. Wee

Department of Civil EngineeringEmail: [email protected]

Stress and Strain Transformation(Part 1)


The knowledge of stress and strain transformation will help to: Establish the state of plane stress (where the stresses in the out-

of-plane axis is zero) and the state of plane strain (where the strains in the out-of-plane axis is zero) for various orientations of reference axes

Determine the principal stress and principal strain; and establish the principal planes for plane stress and plane strain conditions

Evaluate the maximum shearing stress and maximum shearing strain for both in-plane and 3-D cases and

Establish the state of plane strain using the strain rosettes

The following syllabus will be covered in this topic. Plane stress Transformation equations for plane stress Principal stresses and maximum shear stress Mohrs circle of stress Plane strain Mohrs circle of strain Strain measurement

Introduction


Introduction


Introduction


Elasticity

Most engineering structures are designed to undergo relatively small deformations, involving only linear portion of the stress-strain relationship. Within this linear portion, the stress, is directly proportional to the strain , given by

= Ewhere E is the modulus of elasticity of the material, also known as Youngs modulus and the relationship is known as Hookes Law. The largest value of the stress for which Hookes Law can be used for a given material is known as the proportional limit of that material.

If the strains induced in a test specimen by the application of a given load disappear when the load is removed, the material is said to behave elastically. The largest value of the stress for which the material behaves elastically is called the elastic limit.


Stress State at a Point

The state of stress at a point can be represented most generally by six independent normal and shear stress components which act on the faces of an element of material located at the point. These stresses are referred to the corresponding coordinate axes.

If the coordinate axes are rotated, the same state of stress will be represented by a different set of component stresses.


Generally, in failure criteria, the respective component stresses are compared against the critical stress or strain (e.g. yield stress, tensile strain capacity) of the materials.

Since it is prudent to compare the maximum of the respective component stress or strain with the critical stress or strain, transformation of stresses and strains would be necessary to identify this maximum component stress or strain as well as the plane in which this stress or strain acts by rotating the coordinate axes.



Consider the case of a structural element subjected to a generalized stress state. The subscript of normal stress, denotes the direction along the axis which the stress is directed. The first subscript of shear stress, denotes the plane on which the stress acts (plane designation x, y or z corresponds to the plane in which the axis x, y and z acts perpendicularly to) and the second subscript denotes the direction of stress (along the axis). The direct and shear strains, and , associated with the normal and shear stresses respectively are also accordingly denoted.

x, y, z = normal stressesxy, xz, yz = shear stressesx, y, z = direct strainsxy, xz, yz = shear strainsxy = yx; zx = xz; yz = zyxy = yx; zx = xz; yz = zy

xyplane (perpendicular to the x-axis) on which the stress acts

direction of stress




Sign convention:For brevity, the front side of the element is defined as one where the normal to the face is pointing in the positive direction. Accordingly, the rear side of the element is one where the normal to the face is pointing in the negative direction. Stresses on the front side of the element are positive if they act in the positive direction of the axes. Stresses on the rear side of the element are positive if they act in the negative direction of axes. In other words,

Stress(+ve) if surface (+ve) & direction (+ve) (tension)if surface (-ve) & direction (-ve) (tension)

Stress (-ve) if surface (+ve) & direction (-ve) (compression)if surface (-ve) & direction (+ve) (compression)



When an axial load, P is applied to a homogeneous, slender bar of cross-sectional area A along its axis x, the resulting stress and strain would satisfy Hookes law. The axial stress and strain can be expressed as

However, it also causes lateral transverse strains along y and z axes. The ratio of lateral strain over axial strain is called Poissons ratio and is denoted by . Hence,

Ex

x =

Ex

zy ==

strainaxialstrainlateral=

AP

x =

Uni-axial

Poissons ratio


Now extending Hookes law to the case of multi-axial loading, the resulting strain conditions would be:

EEEzyx

x+=

EEEzxy

y+=

EEExyz

z+=

Ex

x=

Ey

y

=

Ez

z=

Ey

zx

==E

xzy

==

Ez

yx==

The relations are referred to as generalized Hookes law for multi-axial loading.

Generalized Hookes law for multi-axial loading


A concrete block of dimension 1500mm by 1200mm by 800mm and cubecompressive strength 5 MPa is dropped into the sea of depth 1000m. When the concrete block come to rest at the seabed, what would be the change in volume of the concrete block? Would the concrete block crush? (Given for the concrete, E = 4000 MPa, = 0.2 and assume for seawater, = 1000 kg/m3).Solution:

At 1000m depth, x = y = z = p = gh (hydrostatic)= 1000 x 9.81 x 1000= 9.81 MPa

From generalized Hookes law,

Substituting, we obtain x = y = z = ( 9.81/4000) x (1 2 x 0.2)= 0.0015

)21( ===Ep

zyx

Example Problem


Reduction in volume = l x b x h l (1 + x ) x b (1 + y ) x h (1 + z )= (1500 x 1200 x 800) 1500(1 0.0015) x

1200(1 0.0015) x 800(1 0.0015)

= 6470 cm3

Cube compressive strength = 5 MPa

Applied hydrostatic pressure = 9.81 MPa

Will the concrete block crush? No

Failure can only occur when surface dislocate either in shear ortension. Compression cannot cause dislocation directly but can cause dislocation in other planes in shear or tension.

Example Problem (Contd)


Example of stresses acting on stressed element

1. A link plate subjected to axial load will induce normal and shear stresses. In the plane perpendicular to the direction of axial load, a normal stress which is also the maximum stress, will be present. By transformation it can be found that the maximum shear stress acts in a plane inclined 45 to the direction of axial load. The axial load may be tensile or compressive.


Link plate

pin

Link plates with pin connections used as tension members.



2. An axial compressive load would produce a maximum shear stress along a plane inclined at 45o to the plane of the applied load. A material, such as wood, which is weaker in shear than tension or compression, would fail in a plane 45 to the plane of the applied load.



However, by transformation it can be shown that the maximum normal stress exist in a plane orientated 45 to pure shear stress ( = 0) plane and also equal to . For brittle material which fails in tension, failure would result in a plane perpendicular to the direction of maximum normal stress.

3. When a bar is subjected to only torsion (T), the element abcd would be in a state of pure shear.



4. When a beam is subjected to bending, the following normal and shear stresses are induced.

The two equations provide only the normal and shear stress in the longitudinal direction of the beam. Sometimes, when the beam is non-homogeneous or non-isotropic, such as timber beams, the stresses obtained from the two equations would have to be transformed to find the critical stresses that acts on the weaker planes in the beam.

MyI

= VQIb =compression

tension



At point A, for the element orientated parallel to the beam, only normal compressive stress, which is also the maximum, would exist. By transformation, the maximum shear stress is found in the direction 45 to the beam.At point B, for the element orientated parallel to the beam, both normal compressive and shear stress, would exist. By transformation, it is found that the maximum normal compressive and tensile stresses are orientated in a direction less than 45 to the beam. The maximum shear stress is found in a direction less than 45 to the beam.

compression

tension

VQIb

=MyI

=


compression

tension

VQIb

=MyI

=

At point C (which lies on the neutral axis), for the element orientated parallel to the beam, only shear stress, would exist.

By transformation, it can be found that the maximum normal compressive and tensile stresses are orientated in the direction 45 to the beam.

The maximum shear stress is found in the direction parallel and perpendicular to the beam.


compression

tension

VQIb

=MyI

=

At point D, for the element orientated parallel to the beam, both normal tensile and shear stress, would exist. The maximum normal compressive and tensile stresses are found by transformation to be orientated in a direction less than 45 to the beam. The maximum shear stress is found in a direction less than 45 to the beam.

At point E, for the element orientated parallel to the beam, only normal tensile stress which is also the maximum, would exist. By transformation, the maximum shear stress is found in a direction 45 to the beam.


Earlier, the maximum and minimum normal stress and its direction were demonstrated at five points along a cross section of a beam. If this were extended to a larger number of sections and a larger number of points in each section, it would be possible to draw two orthogonal systems of curves on the side of the beam.

As shown above, the two systems of orthogonal curves (Stress Trajectories) represents the directions of maximum normal compressive and tensile stresses. Solid lines show the tensile stresses, and dashed lines show the compressive stresses.

Maximum Stresses in Beams

Stress Trajectories for cantilever and simply supported rectangular beams

stress at point C

C


Stress Trajectories for cantilever and simply supported rectangular beams

stress at point C

C

The curves for maximum tensile and compressive stresses always intersect at right angles, and every trajectory crosses the longitudinal (centroidal) axis at 45o (example see point C).

At top and bottom surfaces of the beam, where the shear stress is zero, the trajectories are either horizontal or vertical. Location where the trajectories are predominantly concentrated and is in the same direction indicate susceptibility to failure.

Maximum Stresses in Beams


Plane Stress Problems

In a plane (2D) problem, two conditions can be imposed.1) The out-of-plane components of the stresses, that is the stresses

acting in the direction perpendicular to the plane in consideration, is zero. Problems subjected to this condition is known as PLANE STRESS PROBLEMS.i.e. for a xy-plane stress problem, z = xz = yz = 0Examples of plane stress problems include thin plate loaded by forces parallel to plane of plate only, pressure vessels, thin shell structures, membrane structures.


Membrane

Compression strut

Tension tie

Example of membrane structures which can be analyzed as a plane stress problem.

Plane Stress Problems


Plane Strain Problems

2) The out-of-plane components of the strains, that is the strains in the direction perpendicular to the plane in consideration, is zero. Problems subjected to this condition is known as PLANE STRAIN PROBLEMS.i.e. for a xy-plane strain problem, z = xz = yz = 0Examples of plane strain problems includes dams, tunnels and retaining walls. As conditions can be assumed to be the same at all cross sections for these structures, it is only required to consider a slice between two sections, a unit distance apart with two fixed supports at the ends. Note that strain between the fixed support would be zero.


Hookes Law for Plane Stress

Consider an element of material in plane stress (z = xz = yz = 0) subjected to biaxial stress in the x and y direction.

1 ( )x x yE

= 1 ( )y y xE

= ( )z x yE = +

Due to the effect of Poissons ratio, the strain will be present in all the three directions. The strains can be obtained by substituting z = 0 into the generalized Hookes law equations to obtain:


Now extending Hookes law to the case of multi-axial loading, the resulting strain conditions would be:

EEEzyx

x+=

EEEzxy

y+=

EEExyz

z+=

Ex

x=

Ey

y

=

Ez

z=

Ey

zx

==E

xzy

==

Ez

yx==

The relations are referred to as generalized Hookes law for multi-axial loading.

Generalized Hookes law for multi-axial loading


2 ( )(1 )x x y

E = + 2 ( )(1 )y y xE = +

Now, by rearranging the strain equations obtained earlier, 1 ( )x x yE

= 1 ( )y y xE

= The stresses in a plane stress problem can be expressed in terms of strains as:

Similarly, by introducing the conditions of uniaxial loading

xx

E = xy z

E = =

x xE =0y z = =in the generalized Hookes law equations, the strains in a uniaxialloading condition can be obtained as

Note that only two strain components (x and y) are sufficient to express the stresses in a plane stress problem. Knowledge of the out-of-plane strain (z) is not required although it is not zero.



0x y z = = =0x y z = = =

xyxy

G =

Where G is the shear modulus. The stress and strain in a pure shear loading condition is given by

xy xyG =

2(1 )EG = +

Beside direct strains () induced by normal stress, elements can also be subjected to shear strains, induced by shear stresses. The deformation due to shear strain is illustrated below. Applying Hookes law, the stress and strain relationship for shear can be expressed as

Note that, so far, three material parameters, the Youngs modulus, E, the Poissons ratio, and shear modulus, G have been introduced. However, only two of these parameters, E and are independent as G can be deduced from them:



Consider the point O being subjected to a state of plane stress. The stress components acting at the point O with respect to the xy-coordinate axes can be represented by the set of component stresses acting on anelement as shown in the figure. The normal stresses are defined by the stress components, x and y; and the shear stress by xy and yx. To satisfy rotational equilibrium, xy = yxPositive normal stress indicate tension and negative normal stress indicate compression.

Transformation of Plane Stress


However, the same state of plane stress at point O can be represented by different set of component stresses. To illustrate, let us rotate the coordinate axes counter-clockwise through an angle and the new coordinate axes named as x1, y1 and z1, with z1 axis coinciding with z axis. The same state of plane stress at point O can now be represented by the stress components, x1 and y1, and the shear stress by x1y1 and y1x1.Next we look at how the stress components x1, y1 and x1y1 associated with the element after it has been rotated through an angle , can be expressed in terms of x, y and xy.



In order to determine the normal stress x1 and the shearing stress x1y1exerted on the face perpendicular to the x1 axis, we consider a prismatic element with faces respectively perpendicular to x, y and x1 axes. This will allow the horizontal and vertical components of the stresses, x ,yand xy to be expressed as a function of the normal stress x1 and the shearing stress x1y1.



If the area of the vertical face is denoted by A0, the areas of the horizontal and oblique face are respectively equal to A0tan and A0sec. The forces exerted on the three face can be given by the stress multiplied by the respective area as shown in the figure.

STRESS DIAGRAM FORCE DIAGRAM



1 1 0 0 0

0 0

0 : sec cos sintan sin tan cos 0

x x x xy

y yx

F A A AA A

= =

1 1 1 0 0 0

0 0

0 : sec sin costan cos tan sin 0

y x y x xy

y yx

F A A AA A

= + + =

Taking the equilibrium of forces along x1and y1 axes, the following equilibrium equations can be obtained. Note here that the forces on the horizontal and vertical faces of the prismatic element would have to be resolved into component forces in the x1 and y1 axes first.



Note that xy= yx, and after simplifying the equilibrium equations, we obtain

( ) ( )2 2

12 2

1 1

cos sin 2 sin cos= - sin cos cos sin

x x y xy

x y x y xy

= + + +

For case when =0,1 1 1andx x x y xy = =

For case when =90o,1 1 1andx y x y xy yx = = =

Shear stress yx acts to the right, while positive stress x1y1, after rotating 900, acts to the left

1 1 0 0 0

0 0

0 : sec cos sintan sin tan cos 0

x x x xy

y yx

F A A AA A

= =

1 1 1 0 0 0

0 0

0 : sec sin costan cos tan sin 0

y x y x xy

y yx

F A A AA A

= + + =



Now, substituting the trigonometric identities

( ) ( )2 21sin cos sin 22

1 1cos 1 cos 2 sin 1 cos 22 2

== + =

into the equations

1

1 1

cos2 sin22 2

= sin2 cos22

x y x yx xy

x yx y xy

+ = + + +

The Transformation Equations for Plane Stress can be obtained as follows:

( ) ( )2 2

12 2

1 1

cos sin 2 sin cos= - sin cos cos sin

x x y xy

x y x y xy

= + + +

Since the transformation equations were derived solely from equilibrium of an element, they are applicable to stresses in any kind of material, whether linear or nonlinear, elastic or inelastic.




The expression for the normal stress y1 is obtained by replacing in the equation

1 cos2 sin22 2x y x y

y xy

+ = Summing expressions for x1 and y1 , we obtain

1 1x y x y + = +This equation shows that sum of normal stresses acting on perpendicular faces of plane-stress elements is constant and independent of angle .


x xy

+ = + +by the angle ( + 90o). This is possible because, the rotation of the y-axis anti-clockwise by an angle would coincide with the rotation of the x-axis by an angle ( + 90o). Since cos(2 + 180o) = cos2 and sin(2 + 180o) = sin2, normal stress y1 is given by



Stresses vary continuously as the orientation of the element is changed. The graph shows the variation of the stress components x1 and x1y1 with respect to the axes orientation(). At certain angles, normal stress reaches a maximum or minimum value. At other angles, it becomes zero. Similarly for shear stress. Note that the normal stress x1 is maximum or minimum when x1y1 is zero.



In the case of biaxial stress, there is no shear stress. Substituting xy= 0 into the transformation equations,

1. Biaxial Stress

Biaxial stress occurs in many kinds of structures, including thin-walled pressure vessels.

Transformation Equations for Special Cases of Plane Stress

1

1 1

cos2 sin22 2

= sin 2 cos22

x y x yx xy

x yx y xy

+ = + + +

1

1 1

cos 22 2

= sin 22

x y x yx

x yx y

+ = +

The transformation equations for the case of biaxial stress can be obtained as follows:


2. Uniaxial Stress

3. Pure Shear

In the case of uniaxial stress, only the normal stress component, x is not zero. By setting y and xyequal to zero in the transformation equations, we obtain

In the case of pure shear, substituting x= 0 and y= 0 into the transformation equation would give

( )11 1

1 cos22

sin 22

xx

xx y

= +=

1

1 1

sin2cos2

x xy

x y xy

==

Pure shear occurs in many kinds of structures, including cylinders subject to pure torsion.

Uniaxial stress occurs in many kinds of structures, including members in a truss structure.

Transformation Equations for Special Cases of Plane Stress


SUMMARY


y xy

+ =

Where is the rotation in counter-clockwise direction. Summing expressions for x1 and y1 , we obtain

1 1x y x y + = +This equation shows that sum of normal stresses acting on perpendicular faces of any plane-stress elements is constant and independent of angle .(EXAMPLE)


x xy

+ = + +The Transformation Equations for Plane Stress are as follows:

1 1= sin2 cos22x y

x y xy

+

Transformation Equations for Plane Stress


Principal Stresses

The transformation equations for plane stress show that normal stresses x1 and shear stresses x1y1 vary continuously as the axes are rotated through angle .

The maximum normal stress is known as the principal stress, 1 and the plane on which it acts is the principal plane. The stress orthogonal with the principal stress, 1 is the minimum normal stress, also known as principal stress, 2 and the shear stress acting on all the four principal planes is zero.

Most failures of structures are often associated with the maximum tensile or compressive stresses, and thus their magnitudes and orientations should be determined.

x

x1

yy1

2 = p

x = max = 1


x

x1

yy 1

2 = p

x = max = 1

By taking the derivative of x1 of the transformation equation with respect to and setting it to zero, we obtain an equation for which we can find the values of x1 at which it is a maximum or minimum. The equation for the derivative is

( )1 sin2 2 cos2x x y xydd = +and setting this equation to zero, we obtain

2tan2 xyp

x y

= The orientation of the principal stresses can therefore be obtain from the above equation. Subscript p indicates that the angle p defines the orientation of principal stresses. The angle p is known as the principal angle. Substituting this angle into the transformation equations, the principal stresses can be obtained.

Principal Stresses


General Formulae for Principal Stresses

Now, consider the right-angled triangle, which is constructed from the equation

22

2x y

xyR = +

cos2 sin22x y xy

p pR R

= =From the triangle, we obtain

2tan2 xyp

x y

=

By Pythagorean theorem, the hypotenuse R is given by

&


22

2x y

xyR = + cos2 sin22x y xy

p pR R

= =Next, substituting the relationships,

22

1 2 2x y x y

xy + = + +

into the transformation equations


x xy

+ = + +


y xy

+ = The following equations for the principal stresses can be obtained, in which 1 > 2.

22

2 2 2x y x y

xy + = +

The equations for the principal stresses, 1 and 2 can be combined into one as:2

21,2 2 2

x y x yxy

+ = +

and



The principal angle, which is the angle p1 corresponding to the principal stress 1 can be obtained from the equations

1 1cos2 sin22x y xy

p pR R

= =

Only one angle exists between 0 and 360o that satisfies both of these equations. For example, if both cos2p1 and sin2p1 are positive, angle 2p1 can only be between 0 to 90o. Otherwise, if both cos2p1 and sin2p1are negative, angle 2p1 can only be between 180 to 270o.

x

x1

yy1

2 = p

x = max = 1

Thus, value of p1 can be determined uniquely. Angle p2 corresponding to 2, defines a plane that is perpendicular to the plane defined by p1. Thus, p2 can be taken as 90o larger or 90osmaller than p1.



Uniaxial stress Biaxial stress

Pure shear

The principal planes for elements in uniaxial stress and biaxial stress are the x and y planes themselves, because xy = 0 and hence tan2p= 0, and the two values of p are 0 and 90o. In another word, when shear stress is zero, the normal stresses are the principal stresses.

For element in pure shear, the principal planes are orientated at 45o to the x axis, obtained from the condition x = y = 0 and hence cos2p = 0 and sin2p = 1.

Principal Stresses for Special Cases of Plane Stress


Maximum Shear Stresses

The shear stresses x1y1 acting on inclined planes are given by the transformation equation,

( )1 1 cos2 2 sin2x y x y xydd =

tan22x y

sxy

= and setting it to zero, we obtain

1tan2 cot 2tan2s pp

= = This results in the relationship o45s p =

The planes of maximum shear stress occur at 45o to the principal planes.

Taking derivative of the above equation with respect to 1 1= sin2 cos22

x yx y xy

+

2tan2 xyp

x y

= determines the maximum principle stress, we deduce that

where s is the angle of the plane

of maximum shear stress, and comparing with the equation which

we obtain:

x1y 1 = max

y 1 y

x

x1 = s


Now, from the two transformation equations

Next, adding the left and right hand sides of the two equations respectively, the following can be obtained:

the following equations can be obtained by rearranging and squaring the transformation equations.

1

1 1

cos2 sin22 2

= sin 2 cos22

x y x yx xy

x yx y xy

+ = + + +

2

xyyx2

x1y1

2

xyyx

2yx

x1

cos2sin22

sin2cos22

2

+=

+=

+

2xy

2yx2

x1y1

2yx

x1 2

2

+

=+

+



Note that the equation

Since at points A and B, x1y1 = 0, then

2xy

2yx2

x1y1

2yx

x1 2

2

+

=+

+

+= 0,

2 yxC

2xy

2yx

2 +

=R

and the radius at

2

2

21minmaxave+=+=

2min1x ==1max1x == at A,

at B and

x1y1

x1

represent the equation of a circle plotted in a rectangular coordinate system with abscissa x1 and ordinate x1y1 with the centre point at



At points D and E in the figure, observe that

Note also the normal stress corresponding to the condition of maximum shear stress is

and

2

yxx1+=

Since at points D and E, x1y1 is the positive and negative of the maximum shear respectively, therefore the maximum shear stress would be given by

2xy

2yx

x1y1 2

+

== R

2xy

2yx

max 2

+

=

2

yxx1+== ave

x1y1

x1



Recall that the maximum shear stress would be given by

and the corresponding normal stress acting in the plane of the maximum shear stress would be given by

2xy

2yx

max 2

+

=

2

yxx1+== ave

If an element is subjected to principal stresses 1 and 2 whereby the shear stress xy is zero, the maximum shear stress would then be given by

2

21max=

and the corresponding normal stress acting in the plane of the maximum shear stress would be given by

2

21x1+== ave


SUMMARY

The direction of the plane in which the principal stresses and the maximum shear stress acts could be obtained, respectively, from:

2xy

2yx

max 2

+

=

2

yxavey1x1+===

The principal stresses are given by:

2

21max=

The maximum shear stress and the corresponding normal stress acting in the plane of the maximum shear stress would be given by

2

21avey1x1+===

tan22x y

sxy

=

22

1,2 2 2x y x y

xy + = +

2tan2 xyp

x y

=

For an element subjected to principal stresses, the maximum shear stress and the corresponding normal stress acting in the plane of the maximum shear stress would be given by


Example 1


Example 1 (contd)


Example 1 (contd)


Example 2


Example 2 (Contd)


Writing this equation in a simpler form by using

would result inThis is the equation of a circle in standard algebraic form. The coordinates are x1and x1y1, the radius is R, and the center of the circle has coordinates x1= aver and x1y1=0.

2xy

2yx2

x1y1

2yx

x1 2

2

+

=+

+

2xy

2yx

2 +

=R

2 yx

ave

+= and( ) 221y1x2ave1x R=+

Recall earlier that from the two transformation equations, the following equation was obtained:

The transformation equations for plane stress can be represented in graphical form by a plot known as Mohrs Circle. This graphical representation is extremely useful for visualising the relationships between normal and shear stresses acting on various inclined planes at a point in a stressed body. It also provides a means for calculating principal stresses, maximum shear stresses, and stresses on inclined planes. Mohrs Circle is valid not only for stresses but also for other quantities of a similar mathematical nature, including strains and moments of inertia.

Mohrs Circle for Plane Stress


Construction of Mohrs Circle

Mohrs Circle can be constructed in a variety of ways, depending upon which stresses are known and which are unknown. Assume that we know the stresses x, y and xy acting on the x and y planes of an element in plane stress and we wish to know the stresses x1, y1 and x1y1 acting on the x1 and y1 planes.

Known stress state

Stress state to be determined for rotated element

For the Mohrs Circle, adopt the convention to plot clockwise shear stress aspositive, anticlockwise shear stress as negative, tension stress as positive, compression stress as negative and counter-clockwise angle () as positive.


Introduction



Sign convention:For brevity, the front side of the element is defined as one where the normal to the face is pointing in the positive direction. Accordingly, the rear side of the element is one where the normal to the face is pointing in the negative direction. Stresses on the front side of the element are positive if they act in the positive direction of the axes. Stresses on the rear side of the element are positive if they act in the negative direction of axes. In other words,

Stress(+ve) if surface (+ve) & direction (+ve) (tension)if surface (-ve) & direction (-ve) (tension)

Stress (-ve) if surface (+ve) & direction (-ve) (compression)if surface (-ve) & direction (+ve) (compression)


Note that after transformation in Mohrs Circle, the stresses would have to be converted to this convention


+

--

1. Draw a set of coordinate axes with x1 as abscissa (+ve to the right), and x1y1 as ordinate (+ve upward). Note that x1 and x1y1 are the variables.

2. Locate center C of circle at point having coordinates x1 = aveand x1y1 = 0

2 yx

ave

+=

Procedure for Construction of Mohrs Circle


+

-- 3. Locate point A,

representing stress condition on x face of element labelled A by plotting its coordinates x1= x and x1y1 = -xy.



+

-- 4. Locate point B,

representing stress condition on y face of element labelled Bby plotting its coordinates x1= y and x1y1 = + xy.



+

--

5. Draw line from point A to point B. This line is the diameter of the Mohrs circle and passes through center C.

6. Using point C as center, draw the Mohrs Circle passing through points A and B.



Stresses on Inclined Element (with angle from x-axis )

--

+

The stresses on faces D and Dare represented by the stresses at the point D and D on the Mohrs circle, respectively. These points are located by rotating the line AB about C through an angle 2 in the same direction as the element is rotated, which in the anticlockwise direction.


Point P1 and P2 represents the principal stresses as the corresponding shear stress, xyat the principal plane is zero.

Principal angle p1 between x-axis and the axis of 1 (the algebraically larger principal stress) is one-half the angle 2p1 (which is the angle between radii CA and CP1)

+

--

RCPOC yx +==222

RCPOC yx ++=+=211

x

x1

yy1

2 = p

x = max = 1

Principal Stresses (with angle p from x-axis )



Points S1 and S2, representing the maximum negative and maximum positive shear stresses, respectively, are located at the bottom and top of the Mohrs Circle. These points are at angles 2 =90ofrom points P1 and P2. That is the maximum and minimum shear stress acts on the plane inclined 45o from the principle plane.

--

+



The shear stresses x1y1 acting on inclined planes are given by the transformation equation,

( )1 1 cos2 2 sin2x y x y xydd =

tan22x y

sxy

= and setting it to zero, we obtain

1tan2 cot 2tan2s pp

= = This results in the relationship o45s p =

The planes of maximum shear stress occur at 45o to the principal planes.

Taking derivative of the above equation with respect to 1 1= sin2 cos22

x yx y xy

+

2tan2 xyp

x y

= determines the maximum principle stress, we deduce that

where s is the angle of the plane

of maximum shear stress, and comparing with the equation which

we obtain:

x1y 1 = max

y 1 y

x

x1 = s


Example

Example


Example contd

Example


Example contd


Example

Example


Example-contd

Example contd


Example-contd

Example contd


Element in Triaxial Stress

Because the transformation equations of plane stress are based on force equilibrium in the xy plane, they are independent of the normal stress z..Therefore, we can use these equations for determining the stresses and . Note that when considering the stresses in the xy-plane, the stress z is known as the out-of-plane stress while the stresses x and y are known as the in-plane stresses.


In order to determine the normal stress x1 and the shearing stress x1y1 exerted on the face perpendicular to the x1 axis, we consider a prismatic element with faces respectively perpendicular to x, y and x1 axes. This will allow the horizontal and vertical components of the stresses, x ,y and xy to be expressed as a function of the normal stress x1 and the shearing stress x1y1.

Element in Triaxial Stress


If the area of the vertical face is denoted by A0, the areas of the horizontal and oblique face are respectively equal to A0tan and A0sec. The forces exerted on the three face can be given by the stress multiplied by the respective area as shown in the figure.

STRESS DIAGRAM FORCE DIAGRAM


1 1 0 0 0

0 0

0: sec cos sintan sin tan cos 0

x x x xy

y yx

F A A AA A

= =

1 1 1 0 0 0

0 0

0: sec sin costan cos tan sin 0

y x y x xy

y yx

F A A AA A

= + + =

Taking the equilibrium of forces along x1 and y1 axes, the following equilibrium equations can be obtained. Note here that the forces on the horizontal and vertical faces of the prismatic element would have to be resolved into component forces in the x1 and y1 axes first.


General State of Stress

Recall that the generalized state of stress at a point Q can be represented by six components of stress, namely three normal stresses, x, y and z and three shear stresses xy, xz and yz. When the coordinate axes, x, y and z are rotated, the stress state can be represented by another set of six stresses components x, y, z, xy, xz and yz. However, for every stress state, there exist one orientation of the coordinate axes where the shear stress on all faces of the cubic element would vanish. Only normal stresses would remain and these normal stresses are, therefore, the principal stresses 1, 2 and 3 for the stress state at point Q. The principal stresses may also be denoted as a, b and c corresponding to coordinate axes a, b and c as shown.


Application of Mohrs Circle to Three Dimensional Analysis of StressAs mentioned earlier the transformation equations of plane stress in the xy-plane are independent of the out-of-plane stress, z. Assume the point Q is subjected to a generalized (3D) stress state and the coordinate axes a, b and c are the principal axes of stress. Therefore, when the element is rotated about one of the principal axes, e.g. c-axis, the corresponding transformation of stress may be analyzed by means of Mohrs circle as if it was a transformation of plane stress. We may therefore use the circle of diameter AB to determine the normal and shearing stresses exerted on the faces of the element as it is rotated about the c-axis.


Similarly, circles of diameter BC and CA may be used to determine the stresses on the element as it is rotated about the a and b axes, respectively. It can be shown that any other transformation of axes would lead to stresses represented by a point located within the shaded area. Thus, the radius of the largest of the three circles yields the maximum value of the shearing stress at point Q. Therefore,

where max and min are the maximum and minimum value of the three principal stress and which also represent the algebraic values of the maximum and minimum stresses at point Q.

minmax21

max =

Application of Mohrs Circle to Three Dimensional Analysis of Stress


Returning to the case of plane stress, we observe z = zx = zy = 0, and therefore the z-axis is one of the principal axes since the shear stress, zx = zy = 0 in the x-yplane. Hence, in the Mohrs circle, this axis would corresponds to the origin O where = = 0. We also recall that the other two principal stresses corresponds to another two points, A and B in Mohrs circle. If A and B are located on opposite sides of the origin O, the corresponding principal stresses represent the maximum and minimum normal stresses at point Q, and the maximum shearing stress is equal to the maximum in-plane shearing stress. The in-plane refers to a plane which is perpendicular to the plane in consideration. The planes of maximum shearing stress correspond to points D and E of Mohrs circle and are 45o to the principal planes corresponding to points A and B, shown as shaded diagonal planes in figures (a) and (b).

(a) (b)



If, on the other hand, A and B are on the same side of the origin O, that is, if a and b have the same sign, then the circle defining max, min and max is not the circle corresponding to a transformation of stress within the xy plane. If a > b > 0, as assumed, we have max = a, min = 0, and max is equal to the radius of the circle defined by points O and A, that is

We also note that the normals Qd and Qe to the planes of maximum shearing stress in figures (a) and (b) respectively, are obtained by rotating the axis Qathrough 45o within the za plane. In other words, the maximum shear stress is in-plane to the za-plane and therefore out-of-plane to the ab-plane that is being considered as the plane of plane stress. Thus, the planes of maximum shearing stress, are the shaded diagonal planes shown.

max21

max =

(a)

(b)


ce2155 - stress and strain transformation (part 1)

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