2005 Pearson Education South Asia Pte Ltd 9. STRESS TRANSFORMATION 1 CHAPTER OBJECTIVES To show how...

2005 Pearson Education South Asia Pte Ltd

9. STRESS TRANSFORMATION

1

CHAPTER OBJECTIVES

To show how to transform the stress components that are associated with a particular coordinate system into components associated with a coordinate system having a different orientation.



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General State of stressPlane stress

(a simplification)

Plane stress

(two dimensional view)

9-1 PLANE–STRESS TRANSFORMATION

General state of stress at a point is characterized by six independent

normal and shear stress components; x , y , z , xy , yz , and zx

General plane stress at a point is represented by x , y and xy , which act on four faces of the element



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x

y

x

y

xy

x’

y’ x’

y’

x’y’

If we rotate the element of the point in different orientation, we will have different values of the stresses

The stresses are now x’ , y’ and x’y’

It is said that the stress components can be transformed from one orientation of an element to the element having a different orientation



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Failure of a brittle material

in tension

45o

Failure of a brittle material

in torsion

Failure of a brittle material will occur when the maximum normal stress in the material reaches a limiting value that is equal to the ultimate normal stress



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y

x

xy

x

y

9-2 GENERAL EQUATION OF PLANE–STRESS TRANSFORMATION

Sign ConventionA normal or shear stress component is positive provided it acts in the positive coordinate direction on the positive face of the element,

or it acts in the negative coordinate direction on the negative face of the element



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x

y

x

y

x’y’

The orientation of the inclined plane, on which the normal and shear stress components are to be determined, will be defined using the angle

The angle is measured from the positive x to the positive x’



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The element in Fig.(a) is sectioned along the inclined plane and the segment shown in Fig.(b) is isolated.

Assuming the sectioned area is A, then the horizontal and vertical faces of the segment have an area of A sin and A cos, respectively

(a)(b)



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Resulting free-body diagram

The unknown normal and shear stress

components in the inclined plane, x’

and x’y’, can be determined from the

equations of force equilibrium

Fx’ = 0 Fy’ = 0

2222 xy

yxyx'x sincos

We get



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y’ x’y’ x’


Three stress components, x’ , y’ and x’y’ , oriented along the x’, y’ axes

2222 xy

yxyx'x sincos

2222 xy

yxyx'y sincos

222 xy

yx'y'x cossin



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9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES

To determine the maximum and minimum normal stress, we must differentiate equation of x’ with respect to and set the result equal to zero. This gives

02222d

dxy

yx'x

cos)(2sin

Solving this equation, we obtain the orientation = p of the planes of maximum and minimum normal stress

2yx

xyp

2tan



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2yx

xyp

2tan

The solution has two roots; p1 and p2

2p2 = 2p1 +

180o

Based on the equation of tan2p above, we can construct two shaded triangles as shown in the figure



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2xy

2yx

xyp1 2

2sin

2xy

2yxyx

p1 22

2cos

sin 2p2 = – sin 2p1

cos 2p2 = – cos 2p1

The equation of the maximum/minimum normal stresses can be found by substituting the above equations into x’



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In-plane principal stresses

Principal stresses = Maximum and minimum normal stress

2xy

2yxyx

2,1 22



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Depending upon the sign chosen, this result gives the maximum or minimum in-plane normal stress acting at a point, where 1 2.

2xy

2yxyx

2,1 22

This particular set of values, 1 and 2, are called the in-plane principal stresses, and the corresponding planes on which they are act are called the principal planes of stress

Furthermore, if the trigonometric relations for p1 and p2 are substituted into equation of x’y’, it can be seen that x’y’ = 0; that is, no shear stress acts on the principal planes



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Maximum In-Plane Shear Stress

Similarly, to get the maximum shear stress, we must differentiate equation of x’y’ with respect to and set the result equal to zero. This gives

xy

yxs

22tan

The solution has two roots; s1 and s2

2s2 = 2s1 + 180o



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The maximum shear stress can be found by taking the trigonometric values of sin 2s and cos 2s from the figure and substituting them into equation of x’y’ . The result is

2xy

2yxmax

planein 2

Substituting the values for sin 2s and cos 2s into equation of x’, we see that there is also a normal stress on the planes of maximum in-plane shear stress. We get

2yx

avg

(9-7)



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9-4 MOHR’S CIRCLE PLANE STRESS

We rewrite the stress component x’ and x’y’ as follows

2222 xy

yxyx'x sincos

222 xy

yx'y'x cossin

Squaring each equation and adding the equation together can eliminate the parameter . The result is

22'y'x

2yx

'x R2

2

xy

2yx

2R



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22'y'x

2yx

'x R2

Since x , y and xy are known constants, the above equation can be written in a more compact form as

22'y'x

2avg'x R

2yx

avg



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2

2

2 xyyxR

2yx

avg

x

xy

This circle is called MOHR’S CIRCLE

22'y'x

2avg'x R

This equation represents a circle having a radius R and center on axis at point C(avg, 0) as shown in the Figure



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x

y

x

y

xy

9-4 MOHR’S CIRCLE PLANE STRESS Procedure how to draw and use Mohr’s circle

A stress state of a point which all stresses x , y and xy are positive (just for example)

CONSTRUCTION OF MOHR’S CIRCLE

• Establish a coordinate system; axis

• Plot the center of the Mohr’s circle C(avg, 0) on s axis avg = (x + y)/2

• Plot the reference point A(x, xy). This represents = 0

C

avg

Ax

xy

• Connect point A with the center C, and CA becomes the radius of the circle

• Sketch the circle



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ANALYSIS OF MOHR’S CIRCLE

C

avg

Ax

xy

Principal Stresses 1 and 2

B

1

• Point B: 1

• Point D: 2

D

2

Orientation of principal plane, p1

p1

Maximum In-Plane Shear Stress: max = CE = CF

E

F

max

max

Orientation of maximum in-plane shear stress, s1

s1



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9-5 STRESS IN SHAFT DUE TO COMBINED LOADINGS

Occasionally, circular shafts are subjected to the combined effects of torsion and axial load, or torsion and bending, or in fact the combined effects of torsion, axial load, and bending load.

Provided the material remains linear elastic, and is only subjected to small deformation, and then we can use the principle of superposition to obtain the resultant stress in the shaft due to the combined loadings.

The principal stress can then be determined using either the stress transformation equations or Mohr’s circle



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EXAMPLE 9-1

Stress in Shafts Due to Axial Load and Torsion

An axial force of 900 N and a torque of 2.50 N.m are applied to the shaft as shown in the figure. If the shaft has a diameter of

40 mm, determine the principal stresses at a point P on its surface.

Internal Loadings

The internal loadings consist of the torque and the axial load is shown in Fig.(b)

(a)(a)(b)



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EXAMPLE 9-1

(a)(b)

Stress Components

Due to axial load

Due to torsional load

4(0.02)

.50)(0.02)(

2

2

J

cT

198.9 kPa

kPa716.22(0.02)

9002

AF



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The state of stress at point P is defined by these two stress components

EXAMPLE 9-1

Principal Stresses:

2xy

2yy

2,1 22

We get 1 = 767.8 kPa

2 = – 51.6 kPa

The orientation of the principal plane:

2tan

y

xy1p

2 = – 29o

p = 14.5O



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EXAMPLE 9-2

Stress in Shaft due to Bending Load and Torsion

T

xy

z

A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a bending moment of 2 kNm and a torque of 2.5 kNm.

Determine:

1. The critical point of the section

2. The stress state of the critical point.

3. The principal stresses and its orientation



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EXAMPLE 9-2

T

xy

z

Analysis to identify the critical point

Maximum shear stresses occur at the peripheral of the section.

Due to the torque T

Due to the bending moment M

Maximum tensile stress occurs at the bottom point (A) of the section.

Conclusion: the bottom point (A) is the critical point

A



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EXAMPLE 9-2

T

xy

z

Stress components at point A

Due to the torque T

Due to the bending moment M

A

198.9 kPa4(0.02)

.50)(0.02)(

2

2

J

cT

318.3 kPa 44 (0.02)

2)(2.00)(0.0

zIcM

318.3 kPa

198.9 kPa

Stress state at critical point A

x = 318.3 kPa xy = 198.9 kPa



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318.3 kPa

198.9 kPa Principal stresses

22

22 xyxx

1,2

We get 1 = 413.9 kPa

2 = – 95.6 kPa

The orientation of the principal plane:

2tan

x

xy1

p2 = 51.33o

25.65o

1

2

EXAMPLE 9-2

p = 25.65O



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EXAMPLE 9-3

Stress in Shafts Due to Axial Load, Bending Load and Torsion

A shaft has a diameter of 4 cm. The cutting section shows in the figure is subjected to a compressive force of 2500 N, a bending moment of 800 Nm and a torque of 1500 Nm.

Determine: 1. The stress state of point A.

2. The principal stresses and its orientation



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EXAMPLE 9-3

Analysis of the stress components at point A

Due to comprsv load:

Due to torsional load: JcT

A

AF

A'

Due to bending load:z

'A' IcM

(compressive stress)

Stress state at point A

Shear stress: = A

Normal stress: = A’ + A”

SOLVE THIS PROBLEM AT HOME !!!



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2005 Pearson Education South Asia Pte Ltd 9. STRESS TRANSFORMATION 1 CHAPTER OBJECTIVES To show how...

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