Chap9_slides.pdf

M. Vable Mechanics of Materials: Chapter 9Pr

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2nd.

htmStrain Transformation

.

Ideas, definitions, and equations in strain transformation are very sim-ilar to those in stress transformation. But there are also several differ-ences.

Learning objective Learn the equations and procedures of relating strains at a point in dif-

ferent coordinate systems. August 2014 9-1


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2nd.

htmLine Method

Plane Strain

Global coordinate system is x,y, and z.Local coordinate system is n, t, and z.We assume xx, yy, and xy are known at a point.Objective is to find nn, tt, and nt.ProcedureStep 1 View the n and t directions as two separate lines and determine the

deformation and rotation of each line as described in steps below.

Step 2 Construct a rectangle with a diagonal in direction of the line.

Step 3 Relate the length of the diagonal to the lengths of the rectangles sides.

Step 4 Calculate deformation due to the given strain component and draw the deformed shape.

Step 5 Find the deformation and rotation of the diagonal using small strain approximations.

Step 6 Calculate normal strains by dividing the deformation by the length of the diagonal.

Step 7 Calculate the change of angle from the rotation of the lines in the n and t directions. August 2014 9-2


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2nd.

htm C9.1 At a point, the only non-zero strain component is . Determine the strain components in n and t coordi-

nate system shown.xx 400=

30ox

y

t

nAugust 2014 9-3


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2nd.

htmVisualizing Principal Strain Directions

Principal coordinates directions are the coordinate axes in which the shear strain is zero.

The angles the principal axes makes with the global coordinate system are called the principal angles.

Normal strains in principal directions are called principal strains. The greatest principal strain is called principal strain one (1).

Observations Principal strains are the maximum and the minimum normal strain at a

point. A circle in undeformed state will become an ellipse during deforma-

tion with major axis as principal axis 1 and minor axis as principal axis 2.

Visualizing ProcedureStep 1 Visualize or draw a square with a circle drawn inside it.

Step 2 Visualize or draw the deformed shape of the square due to just normal strains.

Step 3 Visualize or draw the deformed shape of the rectangle due to the shear strain.

Step 4 Using the eight 45o sectors shown report the orientation of principal direction 1. Also report principal direction 2 as two sectors counter-clockwise from the sector reported for principal direction 1.

1

23

4

56 7

8

x

yAugust 2014 9-4


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2nd.

htm C9.2 The state of strain at a point in plane strain is as given in each problem. Estimate the orientation of the principal directions and report your results using sectors shown.

Class Problem 9.1

Estimate the orientation of the principal directions and report your results using sectors shown.

xx 400 = yy 600 = xy 500 =

1

23

4

56 7

8

x

y

xx 400 = yy 600 = xy 500 =

xx 400 = yy 600 = xy 500 =August 2014 9-5


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2nd.

htmMethod of Equations

Calculations for xx acting alone.

Calculations for yy acting alone

Calculations for xy acting along

Total Strains:

ny

xo

P P1

1

Pnn1

ny

n112 n

t1t

xxxx n( ) cos=

A

nn1( ) xx 2cos= tt1( ) xx 90+( )

2cos xx 2sin= =

1 xx sin cos=2 xx 90+( )sin 90+( )cos xx sin cos= =

nt1( ) 1 2+( ) 2xx sin cos= =

nn2( ) yy 2sin= tt2( ) yy 90+( )

2sin yy 2cos= =

nt2( ) 1 2+( ) 2yy sin cos= =

nn3( ) xy sin cos= tt3( ) xy sin cos=

nt3( ) 1 2( ) xy 2cos 2sin( )= =

nn nn1( )= nn2( ) nn3( )+ +August 2014 9-6


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2nd.

htmStrain Transformation Equations

Stress Transformation equations

The coefficient of the shear strain term is half the coefficient of the shear stress term. This difference is due to the fact that we are using engineering strain instead of tensor strain.

nn xx 2cos yy

2sin xy sin cos+ +=tt xx

2sin yy 2cos xy sin cos+=

nt 2xx sin cos 2yy sin cos xy 2cos 2sin( )+ +=

nn xx 2cos yy

2sin 2xy cossin+ +=tt xx

2sin yy 2cos 2xy sincos+=

nt xx sincos yy cossin xy 2cos 2sin( )+ +=August 2014 9-7


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2nd.

htmPrincipal Strains

The angle of principal axis one from the x-axis is only reported in describing the principal coordinate system in two dimensional prob-lems.

Maximum Shear strain The maximum shear strain in coordinate systems that can be obtained

by rotating about the z-axis is called the in-plane maximum shear strain.

The maximum shear strain at a point is the absolute maximum shear strain that can be obtained in a coordinate system by considering rota-tion about all three axes.

Maximum shear strain in plane stress and plane strain will be differ-ent.

1 2,xx yy+( )

2--------------------------

xx yy2

--------------------- 2 xy

2-------

2+=

nn tt+ xx yy+ 1 2+= =

2ptanxy

xx yy( )--------------------------=

30

1 ------------

xx yy+( ) 1 ------------ 1 2+( )=

=Plane Strain

Plane Stress

p2-----

1 22

----------------=

max2

------------ max1 2

2----------------

2 32

----------------3 1

2----------------, , =August 2014 9-8


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2nd.

htm C9.3 At a point in plane strain, the strain components in the x-y coordinate system are as given in each problem. Using Method of Equa-tions determine

(a) the principal strains and principal angle one. (b) the maximum shear strain. (c) the strain components in the n-t coordinate system shown in each problem.

20o x

yt

n

xx 600 =yy 800 =xy 500 =August 2014 9-9


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2nd.

htmMohrs Circle for Strains

Each point on the Mohrs circle represents a unique direction passing through the point at which the strains are specified.

The coordinates of each point on the circle are the strains (nn , nt/2). On Mohrs circle, lines are separated by twice the actual angle

between the lines.

nnxx yy+( )

2--------------------------

xx yy( )2

-------------------------- 2cos xy2

------- 2sin+ +=nt2

------- xx yy( )

2-------------------------- 2sin xy2-------

2cos+=

nnxx yy+

2----------------------

2 nt2

------- 2

+xx yy

2---------------------

2 xy2

------- 2

+=August 2014 9-10


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2nd.

htmConstruction of the Mohrs Circle for strain.Step 1. Draw a square with deformed shape due to shear strain xy. Label the

intersection of the vertical plane and x-axis as V and the intersection of the horizontal plane and y-axis as H.

Step 2. Write the coordinates of point V and H as:

Step 3. Draw the horizontal axis to represent the normal strain, with extension to the right and contractions to the left. Draw the vertical axis to represent half the shear strain, with clockwise rotation of a line in the upper plane and counter-clockwise rotation of a line of rotation lower plane.

x

y

xy > 0H

Vx

yH

Vxy < 0

V xx xy 2,( ) and H yy xy 2,( ) for xy 0>

(E)(C)

/2

C

V

HCW

CCWxx yy+

2------------------------

xy2

---------

yx2

---------

xx yy2

------------------------

DE

R

RAugust 2014 9-11


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2nd.

htmStep 4. Locate points V and H and join the points by drawing a line. Label the point at which the line VH intersects the horizontal axis as C.

Step 5. With C as center and CV or CH as radius draw the Mohrs circle.

Principal Strains & Maximum In-Plane Shear Strain

The principal angle one 1 is the angle between line CV and CP1. Depending upon the Mohr circle 1 may be equal to p or equal to (p+ 90o).

Maximum Shear Strain

(E)(C)

/2

R

C

V

H

P1P2

2

1

3 2p2pP3

CW

CCW

xy2

---------

yx2

---------

xx yy2

------------------------

DE

R

xx yy+2

------------------------

p 2

p 2

S2

S1August 2014 9-12


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2nd.

htmStrains in a Specified Coordinate System

Sign of shear strain

(E)(C)

/2CW

CCW

p 2

p 2 max 2

max 2

P1P2P3

C

DE

V

H

N

nnT

tt2V

2H nt 2

/2CW

CCW

(E)x

y

n

t

V

HN

T

V(C)August 2014 9-13


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2nd.

htmThe coordinates of point N and T are as shown below:

Increase in angle results in negative shear strain and decrease in angle results in positive shear strains.

N nn nt 2,( ) and T tt nt 2,( )

n

t

N

T

n1

t1August 2014 9-14


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2nd.

htm C9.4 At a point in plane strain, the strain components in the x-y coordinate system are as given in each problem. Using Mohrs circle determine

(a) the principal strains and principal angle one. (b) the maximum shear strain. (c) the strain components in the n-t coordinate system shown in each problem.

20o x

yt

n

xx 600 =yy 800 =xy 500 =August 2014 9-15


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2nd.

htm Class Problem 9.2

The Mohrs circle corresponding to a given state of strain are shown. Identify the circle you would use to find the strains in the n, t coordinate system in each question.

250x

y n

t

V

H

H

V

250 x

y

n

t

50o

N

T

NT

V

H

H

V50o

H

V

50o

N

T

T

N

N

T

H

V

50o

T

N

Circle A Circle B

Circle C Circle D

1.2.August 2014 9-16


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2nd.

htmGeneralized Hookes Law in Principal Coordinates

Generalized Hookes Law is valid for any orthogonal coordinate sys-tem.

Principal coordinates for stresses and strains are orthogonal. For isotropic materials, the principal directions for strains are the same

as principal directions for stresses.

1 1 2 3+( )[ ] E=2 2 3 1+( )[ ] E=3 3 1 2+( )[ ] E=August 2014 9-17


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2nd.

htmStrain Gages

Strain gages measure only normal strains directly. Strain gages are bonded to a free surface, i.e., the strains are in a state

of plane stress and not plane strain. Strain gages measure average strain at a point.

Strain Rosette

The change in strain gage orientation by makes no difference to the strain values.

a xx 2

acos yy 2

asin xy asin acos+ +=

b xx 2

bcos yy 2

bsin xy bsin bcos+ +=c xx

2ccos yy

2csin xy csin ccos+ +=

180oAugust 2014 9-18


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2nd.

htm C9.5 At a point on a free surface of aluminum (E = 10,000 ksi and G =4,000 ksi) the strains recorded by the three strain gages shown in Fig. C9.5 are as given. Determine the stresses 1, 2, 1 and max.

Fig. C9.5

b

a

c

600450 x

ya 600 in in=b 500 in in=c 400 in in=August 2014 9-19


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2nd.

htm C9.6 An aluminum (E = 70 GPa, and =0.25) beam is loaded by a force P and moment M at the free end as shown in Figure 9.6. Two strain gages at 30o to the longitudinal axis recorded the strains given. Deter-mine the applied force P and applied moment M.

Fig. C9.6

10 mm 10mm

30 mm

30 mm

0.5 m 0.5m

a

P

z

y

M

b

a 386 m m=b 4092 m m=

zAugust 2014 9-20

Strain Transformation. Ideas, definitions, and equations in strain transformation are very similar to those in stress transformation. But there are also several differences.Learning objective. Learn the equations and procedures of relating strains at a point in different coordinate systems.

Line MethodProcedureStep 1 View the n and t directions as two separate lines and determine the deformation and rotation of each line as described in steps below.Step 2 Construct a rectangle with a diagonal in direction of the line.Step 3 Relate the length of the diagonal to the lengths of the rectangles sides.Step 4 Calculate deformation due to the given strain component and draw the deformed shape.Step 5 Find the deformation and rotation of the diagonal using small strain approximations.Step 6 Calculate normal strains by dividing the deformation by the length of the diagonal.Step 7 Calculate the change of angle from the rotation of the lines in the n and t directions.

C9.1 At a point, the only non-zero strain component is . Determine the strain components in n and t coordinate system shown.

Visualizing Principal Strain Directions. Principal coordinates directions are the coordinate axes in which the shear strain is zero.. The angles the principal axes makes with the global coordinate system are called the principal angles.. Normal strains in principal directions are called principal strains.. The greatest principal strain is called principal strain one (e1).Observations. Principal strains are the maximum and the minimum normal strain at a point.. A circle in undeformed state will become an ellipse during deformation with major axis as principal axis 1 and minor axis as principal axis 2.

Visualizing ProcedureStep 1 Visualize or draw a square with a circle drawn inside it.Step 2 Visualize or draw the deformed shape of the square due to just normal strains.Step 3 Visualize or draw the deformed shape of the rectangle due to the shear strain.Step 4 Using the eight 45o sectors shown report the orientation of principal direction 1. Also report principal direction 2 as two sectors counter- clockwise from the sector reported for principal direction 1.

C9.2 The state of strain at a point in plane strain is as given in each problem. Estimate the orientation of the principal directions and report your results using sectors shown.Class Problem 9.1

Method of Equations. The coefficient of the shear strain term is half the coefficient of the shear stress term. This difference is due to the fact that we are using engineering strain instead of tensor strain.Principal Strains. The angle of principal axis one from the x-axis is only reported in describing the principal coordinate system in two dimensional problems.

Maximum Shear strain. The maximum shear strain in coordinate systems that can be obtained by rotating about the z-axis is called the in-plane maximum shear strain.. The maximum shear strain at a point is the absolute maximum shear strain that can be obtained in a coordinate system by considering rotation about all three axes.. Maximum shear strain in plane stress and plane strain will be different.

C9.3 At a point in plane strain, the strain components in the x-y coordinate system are as given in each problem. Using Method of Equations determine(a) the principal strains and principal angle one.(b) the maximum shear strain.(c) the strain components in the n-t coordinate system shown in each problem.

Mohrs Circle for Strains. Each point on the Mohrs circle represents a unique direction passing through the point at which the strains are specified.. The coordinates of each point on the circle are the strains (enn , gnt/2).. On Mohrs circle, lines are separated by twice the actual angle between the lines.Construction of the Mohrs Circle for strain.Step 1. Draw a square with deformed shape due to shear strain gxy. Label the intersection of the vertical plane and x-axis as V and the intersection of the horizontal plane and y-axis as H.Step 2. Write the coordinates of point V and H as:Step 3. Draw the horizontal axis to represent the normal strain, with extension to the right and contractions to the left. Draw ...Step 4. Locate points V and H and join the points by drawing a line. Label the point at which the line VH intersects the horizontal axis as C.Step 5. With C as center and CV or CH as radius draw the Mohrs circle.

Principal Strains & Maximum In-Plane Shear Strain. The principal angle one q1 is the angle between line CV and CP1. Depending upon the Mohr circle q1 may be equal to qp or equal to (qp+ 90o).

Strains in a Specified Coordinate System. Increase in angle results in negative shear strain and decrease in angle results in positive shear strains.

C9.4 At a point in plane strain, the strain components in the x-y coordinate system are as given in each problem. Using Mohrs circle determine(a) the principal strains and principal angle one.(b) the maximum shear strain.(c) the strain components in the n-t coordinate system shown in each problem.

Class Problem 9.2

Generalized Hookes Law in Principal Coordinates. Generalized Hookes Law is valid for any orthogonal coordinate system.. Principal coordinates for stresses and strains are orthogonal.. For isotropic materials, the principal directions for strains are the same as principal directions for stresses.

Strain Gages. Strain gages measure only normal strains directly.. Strain gages are bonded to a free surface, i.e., the strains are in a state of plane stress and not plane strain.. Strain gages measure average strain at a point.. The change in strain gage orientation by makes no difference to the strain values.C9.5 At a point on a free surface of aluminum (E = 10,000 ksi and G =4,000 ksi) the strains recorded by the three strain gages shown in Fig. C9.5 are as given. Determine the stresses s1, s2, q1 and tmax.Fig. C9.5

C9.6 An aluminum (E = 70 GPa, and n=0.25) beam is loaded by a force P and moment M at the free end as shown in Figure 9.6. Two strain gages at 30o to the longitudinal axis recorded the strains given. Determine the applied force P and applied moment M.Fig. C9.6

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Chap9_slides.pdf