transformation of stresses

7/28/2019 Transformation of Stresses

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by the transformation equations for plane stress.

the principal stresses, 1, and, 2, principal angles,p1, and, p2, and maximum and minimum shear stressmax, and, min, and shear stress angles, s1, and, s2.

(2)

the normal, x1, and, y1, and shear, x1y1, stressesacting on an element inclined at an angle from theoriginal element, and

(1)

acting on an element, this worksheet can be used to

calculate and graph

shear stress, xy,(2)

normal stresses, x, and, y, and(1)

Given the

PLANE STRESSTransformation Equations

ANALYSIS OF STRESS

AND STRAIN

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45 deg:=Inclination of element:

xy 40 psi:=Shear stress on x and y faces:

y 10 psi:=Normal stress on y face:

x 50 psi:=Normal stress on x face :

Positive

Inclination

Convention

Inclination, , is positive when it is in acounterclockwise angle from original orientation.

(4)

Shear stress, , is positive when directions associatedwith subscripts are plus-plus or minus-minus.

(3)

Normal stress, , tension is positive andcompression is negative.

(2)

Right-hand face of the element is the positive x face.The top face is the positive y face.

(1)

Sign conventions:

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in min min +, max..:=Plotting range:

max min

100:=Plotting range increment:

max 360 deg:=Maximum angle of inclination:

Range Variable

min 45 deg:=Minimum angle of inclination:

The angle of inclination in is defined as a Mathcadrange variable.

To better understand normal and shear stresses, the given

stresses and the stresses as a function of element inclination

are plotted below.

Graph of Given Element Stresses

xy ( )x y

2

sin 2 ( ) xy cos 2 ( )+:=

Shear stress on xy face:

y ( )x y+

2

x y

2cos 2 ( ) xy sin 2 ( ):=

Normal stress on y face:

x ( )x y+

2

x y

2cos 2 ( )+ xy sin 2 ( )+:=

Function

Normal stress on x face:

Let us write the transformation equations asfunctions ofinclination .

Transformation Equations for Plane Stress

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The stresses are graphed below on anx-y plot.

100 0 100 200 300 40080

60

40

20

0

20

40

60

Normal stress on inclined x faces

Normal stress on inclined y facesShear stress on inclined x and y faces

Given normal stress on x faceGiven normal stress on y-face

Given shear stress on x and y faces

ELEMENT NORMAL AND SHEAR STRESS

angle

stress

x-y Plot

Click on the graph to

see the arguments

being plotted.

Inclined Element Normal and Shear Stresses

The normal stresses, x1, y1, and shear stress, x1y1, on theinclined element are calculated using the transformation

equations for plane stress defined above.

Normal stress on inclined x face.

x1

x ( ):= x1

60 psi=

Normal stress on inclined y face.

y1 y ( ):= y1 20psi=

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Shear stress on inclined x and y face.

x1y1 xy ( ):= x1y1 30psi=

Graph the normal and shear stress on the inclined element.

100 0 100 200 300 40080

60

40

20

0

20

40

60


Normal stress on inclined y facesShear stress on inclined x and y faces

Normal stress on inclined x face

Normal stress on inclined y face

INCLINED ELEMENT NORMAL & SHEAR STRESS

angle

stress

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Shear stress on inclined x and y faces

Principal Stresses

The principal stresses, ,and2, are calculated from equatioderived from the transformation equations.

Largest principal stress:

1x y+

2

x y

2

2

xy2

++:=

1 30psi=

Smallest principal stress:

2x y+

2

x y

2

2xy

2+:=

2 70 psi=

The shear stresses, x1y1, on the principal planes is zero.

Principal Angles

The principal angle p1 associated with the largest principalstress, 1, is found by finding the angle that uniquely satisfietwo equations below. These equations are derived from the

transformation equations.

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p2 26.6 deg=

p2 if p1 90 deg p1 90 deg+, p1 90 deg,( ):=

Conditional If

The principal angle p2, associated with principal stress 2 is90o larger or 90o smaller than p1. The text usually reports apositive value forp2. To be consistent with the text, aconditional ifstatement will be used to add or subtract 90o

from p1 to insure that a positive number forp2 is found.

p1 116.6 deg=

p1 Find p1( ):=

sin 2 p1( ) xyR

=

cos 2 p1( )x y

2 R=

0 deg p1 180 degGiven

p1 89.99 deg:=Initial solution estimate:

Solve BlockR

x y

2

2

xy2

+:=First quantity R must

be defined:

The two equations are solved simultaneous using a Mathcad

solve block.

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Graph the principal stresses.

100 0 100 200 300 40080

60

40

20

0

20

40

60


Normal stress on inclined y faces


Larger principal stress

Smaller principal stress

PRINCIPAL STRESSES

angle

stress

2

psi

1

psi

Maximum and Minimum Shear Stress

Maximum, max, and minimum, min, shear stress can becalculated from the principal stresses, 1, and, 2.

Maximum shear stress:

max1 2

2:=

max 50psi=

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Minimum shear stress:

min max:=

min 50 psi=

The normal stress on the planes of maximum and minimum

stress equals the average normal stress, aver.

averx y+

2:=

aver 20 psi=

Maximum and Minimum Shear-Stress Angle

The planes of maximum and minimum, s1 ands2, are 90o

apart and occur 45o from the principal planes p1

, p2

.

Maximum shear-stress plane orientation:

s1 p1 45 deg:= s1 71.6 deg=

Minimum shear-stress plane orientation:

s2 s1 90 deg+:= s2 161.6 deg=

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p2 26.6 deg=p1 116.6 deg=

2 70 psi=1 30psi=

Principal stresses and angles:

x1y1

30psi=

y1 20psi=x1 60 psi=

Normal and shear stresses:

45deg=xy 40 psi=

y 10psi=x 50 psi=

System parameters:

100 0 100 200 300 40080

60

40

20

0

20

40

60


Normal stress on inclined y faces


Maximum shear stress

Minimum shear stress

MAXIMUM AND MINIMUM SHEAR STRESS

angle

stress

min

psi

max

psi

Graph the maximum and minimum shear stresses.

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Use this worksheet to solve Problems 7.2-1 through 7.2-

7.3-1 through 7.3-8, and 7.3-10 through 7.3-17.

(4)

Use this worksheet to work Examples 7-1 through 7-6(3)

Use x = 50 psi, y = 0 psi, xy = 0 psi, and = 0 deg anconfirm that the maximum shear stress occurs on a plane

45o from the x axis as discussed in Section 2.6 Stresse

on Inclined Sections

(2)

Explore normal and shear stress on an element by choosi

your own System Parameters.(1)

Here are a few suggested applications:

This worksheet can be used to calculate normal and shear

stresses on an inclined element, principal stresses and angle

maximum and minimum shear stresses and angles, and grap

results.

aver 20 psi=

s2 161.6 deg=s1 71.6 deg=

min 50 psi=max 50psi=

Maximum and minimum shear stresses, angles, and

average normal stress:

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transformation of stresses

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