Introduction to Finite Element Analysis and Design Ch 01

8/9/2019 Introduction to Finite Element Analysis and Design Ch 01

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19

CHAP 1

STRESS

STRAIN ANALYSIS

1. A vertical force F is applied to a two-bar truss as shown in the figure. Let cross-sectional areas of the members 1 and 2 be A1 and A2, respectively. Determine thearea ratio A1/A2in order to have the same magnitude of stress in both members.

Solution:

From force equilibrium atB,

1 1sin 45 0 2yF f F f F

2 1 2

1cos 45 0 2

2xF f f f F F

Since truss is two-force member, 1 1 1f A and 2 2 2f A . Thus,

1 1 1 1

1 22 2 2 2

2

( )

f A AF

f F A A

,

1

2

2A

A

l

C B

A

45

F

45

f1

f2

F

B


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20 Finite Element Analysis and Design

2. The stress at a point P is given below. The direction cosines of the normal n to aplane that passes through P have the ratio nx:ny:nz = 3:4:12. Determine (a) the

traction vector T(n)

; (b) the magnitude Tof T(n)

; (c) the normal stress n; (d) the shear

stress n; and (e) the angle between T(n)

andn.

Hint: Use 2 2 2 1x y zn n n .

13 13 0

[ ] 13 26 13

0 13 39

Solution:

(a) First, we need unit normal vector n:

2 2 2

3 0.23081

4 0.30773 4 12

12 0.9231

n

Then, the traction vector on this plane becomes

( )13 13 0 0.2308 7

13 26 13 0.3077 1

0 13 39 0.9231 40

nT n

(b) Since T(n)is a vector, its magnitude can be obtained using the norm as

( ) ( )2 ( )2 ( )2 2 2 27 ( 1) ( 40) 40.6202n n n n

x y zT T T T

(c) ( )0.2308

7 1 40 0.3077 35.6154

0.9231n

nT n

(d)2

( ) 2 2 240.6202 ( 35.6154) 19.5331n n n

T

(e) ( ) ( ) cosn n n

T n T

1 0

( )cos 2.64 151.3n

nT


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CHAP 1 Stress-Strain Analysis 21

3. At a pointPin a body, Cartesian stress components are given by xx= 80 MPa, yy=40 MPa, zz= 40 MPa, and xy= yz= zx= 80 MPa. Determine the traction vector,

its normal component, and its shear component on a plane that is equally inclined toall three coordinate axes.

Hint: When a plane is equally inclined to all three coordinate axes, the directioncosines of the normal are equal to each other.

Solution:

The unit normal in this case is:

1 0.577

11 0.577

31 0.577

n

The traction vector in this direction becomes

( )

80 80 80 0.577 138.56

80 40 80 0.577 69.28 MPa

80 80 40 0.577 69.28

nT n

The normal component of the traction vector is

( )0.577

138.56 69.28 69.28 0.577 160MPa

0.577n

nT n

The shear component of the traction vector is:

2( ) 2 2 2169.71 160 56.58 MPan n

nT


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4. If xx= 90 MPa, yy= 45 MPa, xy= 30 MPa, and zz= xz= yz= 0, compute the

surface traction T(n)on the plane shown in the figure, which makes an angle of =

40 with the vertical axis. What are the normal and shear components of stress onthis plane?

Solution:

Unit normal vector: cos(40) sin(40) 0T n

Traction vector: ( )90 30 0 .776 88.23

[ ] 30 45 0 .643 5.94 MPa

0 0 0 0 0

nT n

Normal stress: ( ) 63.77MPan n

T n

Shear stress:2

( ) 2 61.27MPan n n

T

nxxxx

yy

yy

xy

xy

xy

xy


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5. Find the principal stresses and the corresponding principal stress directions for thefollowing cases of plane stress.

(a) xx= 40 MPa, yy= 0 MPa, xy= 80 MPa

(b) xx= 140 MPa, yy= 20 MPa, xy= 60 MPa

(c) xx= 120 MPa, yy= 50 MPa, xy= 100 MPa

Solution:

(a) The stress matrix becomes

40 80MPa

80 0xx xy

xy yy

To find the principal stresses, the standard eigen value problem can be written as

0 I n

The above problem will have non-trivial solution when the determinant of the coefficientmatrix becomes zero:

40 800

80 0xx xy

xy yy

The equation of the determinant becomes:

240 80 80 40 6400 0

The above quadratic equation yields two principal stresses, as

1 102.46MPa and 2 62.46MPa .

To determine the orientation of the first principal stresses, substitute 1 in the original

eigen value problem to obtain

40 102.46 80 0

80 0 102.46 0x

y

n

n

Since the determinant is zero, two equations are not independent

62.46 80x yn n and 80 102.46x yn n .

Thus, we can only get the relation between nxand ny. Then using the condition |n| = 1 we

obtain


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(1)0.788

0.615x

y

n

n

To determine the orientation of the second principal stress, substitute 2in the original

eigen value problem to obtain

40 62.46 80 0

80 0 62.46 0x

y

n

n

102.46 80x yn n and 80 62.46x yn n .

Using similar procedures as above, the eigen vector of 2can be obtained as

(2)0.615

0.788x

y

n

n

Note that if nis a principal direction, nis also a principal direction

(b) Repeat the procedure in (a) to obtain

1 164.85MPa and 2 4.85MPa .

(1)0.924

0.383x

y

n

n

and

(2)0.383

0.924x

y

n

n

(c) Repeat the procedure in (a) to obtain

1 96.24MPa and 2 166.24MPa .

(1)0.420

0.908x

y

n

n

and

(2)0.908

0.420x

y

n

n

Note that for the case of plane stress 3=0 is also a principal stress and the corresponding

principal stress direction is given by n(3)=(0,0,1)


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2 2

2 2

(105 7) 56 0

56 (21 7) 0

x y

x y

n n

n n

The solution of above equations is 2 { 0.4472, 0.8944} n , which is principal

direction corresponding to 2 . Two principal directions are plotted on the following

graph. Note that the two principal directions are perpendicular each other.

y

x

n2

n1

n1'

n2'

-26.57o

135.43o


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7. Determine the principal stresses and their associated directions, when the stressmatrix at a point is given by

1 1 1

[ ] 1 1 2 MPa

1 2 1

Solution:

Use the Eq. (0.46) of Chapter 0 with the coefficients ofI1=3,I2= 3, andI3= 1,

3 23 3 1 0

By solving the above cubic equation using the method described in Section 0.4,

1 2 33.73MPa, 0.268MPa, 1.00MPa

(a) Principal direction corresponding to 1:

1 1 1

1 1 1

1 1 1

(1 3.7321) 0

(1 3.7321) 2 0

2 (1 3.7321) 0

x y z

x y z

x y z

n n n

n n n

n n n

Solving the above equations with |n1| = 1 yields

{ 0.4597, 0.6280, 0.6280} 1n

(b) Principal direction corresponding to 2:

1 1 1

2 2 2

2 2 2

(1 0.2679) 0

(1 0.2679) 2 0

2 (1 0.2679) 0

x y z

x y z

x y z

n n n

n n n

n n n


2 { 0.8881, 0.3251, 0.3251} n

(c) Principal direction corresponding to 3:

3 3 3

3 3 3

3 3 3

(1 1) 0

(1 1) 2 0

2 (1 1) 0

x y z

x y z

x y z

n n n

n n n

n n n



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3 {0, 0.7071, 0.7071} n


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8. Let xyz coordinate system be defined using the three principal directions obtainedfrom Problem 7. Determine the transformed stress matrix []xyz in the new

coordinates system.

Solution:

The three principal directions in Problem 6 can be used for the coordinate transformation

matrix:

(1) (2) (3)

(1) (2) (3)

(1) (2) (3)

0.460 0.888 0

0.628 0.325 0.707

0.628 0.325 0.707

x x x

y y y

z z z

n n n

n n n

n n n

N

To determine the stress components in the new coordinates we use Eq. (1.30):

1 0 0

0 .268 0

0 0 3.732

T

x y z

N N

Note that the transformed stress matrix is a diagonal matrix with the original principalstresses on the diagonal.


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9. For the stress matrix below, the two principal stresses are given as 3= 3 and 1= 2,respectively. In addition, the two principal stress directions corresponding to the two

principal stresses are also given below.

1 0 2

[ ] 0 1 0

2 0 2

, 1

2

50

1

5

n and 3

1

50

2

5

n

(a) What is the normal and shear stress on a plane whose normal vector is parallel to(2, 1, 2)?

(b) Calculate the missing principal stress 2and the principal direction n2.

(c) Write stress matrix in the new coordinates system that is aligned with n1, n2, andn3.

Solution:

(a) Normal vector: 13{2 1 2}T n

Traction vector ( ) 13

2

0

nT n

The normal component of the stress vector on the plane can be calculated

( )

2( ) 2

1.4444

1.4229

n

n n

n

n

T n

T

(b) Using Eq. (0.46) of Chapter 0, the eigen values are governed by

3 21 2 3 0I I I

We can find the coefficients of the above cubic equation from Eq. (0.47) by I1 = 0,I2 =

7, andI3 = 6. Thus, we have

3 27 6 ( 1)( 6) 0

Thus, the missing principal stress is 2 1 .

Since three principal directions are mutually orthogonal, the third principal direction

can be calculated using the cross product. To establish a defined sign convention for the

principal axes, we require them to form a right-handed triad. If n1and n3are unit vectors

that define the directions of the first and third principal axes, then the unit vector n2for

the second principal axis is determined by the right-hand rule of the vector multiplication.

Thus we have


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2 3 1 {0 1 0}T n n n

(c) Coordinate transformation matrix can be obtained from three principal directions as

2 10

5 50 1 0

1 20

5 5

1 2 3N n n n

The stress matrix at the transformed coordinates becomes

2 1 2 10 01 0 2 2 0 0

5 5 5 50 1 0 0 1 0 0 1 0 0 1 0

1 2 2 0 2 1 2 0 0 30 0

5 5 5 5

T

N N


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10. With respect to the coordinate systemxyz, the state of stress at a pointPin a solid is

20 0 0

[ ] 0 50 0 MPa

0 0 50

(a) m1, m

2and m

3are three mutually perpendicular vectors such that m

1makes 45

with both x- and y-axes and m3is aligned with the z-axis. Compute the normal

stresses on planes normal to m1, m2, and m3.

(b) Compute two components of shear stress on the plane normal to m1 in thedirections m2and m3.

(c) Is the vector n= {0, 1, 1}Ta principal direction of stress? Explain. What is thenormal stress in the direction n?

(d) Draw an infinitesimal cube with faces normal to m1, m2and m3and display the

stresses on the positive faces of this cube.

(e) Express the state of stress at the point P with respect to the xyz coordinates

system that is aligned with the vectors m1, m2and m3?

(f) What are the principal stress and principal directions of stress at the pointPwithrespect to thexyz coordinates system? Explain.

(g) Compute the maximum shear stress at the point P. Which plane(s) does thismaximum shear stress act on?

Solution:

(a)

1 2 31 1(1,1,0) ( 1,1, 0) (0,0,1)2 2

T T T m m m

1 1

2 2

3 3

1 1

2 2

3 3

[ ] 15 MPa[ ] 15 MPa

[ ] 50 MPa

m m

m m

m m

m m

m m

m m

(b)

1( ) 1 1[ ] { 20 50 0}2

T mT m

x

y

z

m1

m2

m3

P


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1

1 2

1

1 3

( ) 2

( ) 3

35 MPa

0 MPa

m

m m

m

m m

T m

T m

(c) Yes,

1 (0,1,1)2

n

( )

50 11 1

[ ] 50 50 1 502 2

0 0

nT n n

Since T(n)// n, nis a principal direction with principal stress = 50 MPa.

(d)

(e)

0.707 0.707 0 20 0 0

-0.707 0.707 0 0 50 0

0 0 1 0 0 50

0.707 -0.707 0

0.707 0.707 0

0 0 1

Tx y z

N N

15 35 0

[ ] 35 15 0 MPa

0 0 50x y z

(f) Principal stresses = 50, 50, and 20 MPa

3 1 (1, 1, 0)2

n

n1and n2are any two perpendicular unit vectors that is on the plane perpendicular to n3.

15

15

3535

-20

m

m

m3


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(g) The maximum shear stress occurs on a plane whose normal is at 45 o from the

principal stress direction. Since 1= 2, all directions that are 45ofromx-axis (3axis)

will have the maximum shear stress whose value is

1 3max 35MPa2

The maximum shear stress planes are in the shape of a cone whose axis is parallel to x-

axis and has an angle of 45o.


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11. A solid shaft of diameter d= 5 cm, as shown in the figure, is subjected to tensile

forceP= 13,000 N and a torque T= 6,000 Ncm. At point A on the surface, what is

the state of stress (write in matrix form), the principal stresses, and the maximumshear stress? Show the coordinate system you are using.

Solution:

Let us establish a coordinate system as shown in the figure. The axial force will cause

normal stress xx, while the torque will cause shear stress xy. Their magnitudes are

6.62 MPaP

A

2.44 MPaT rJ

Then, the stress matrix becomes

6.62 2.44 0

[ ] 2.44 0 0 MPa

0 0 0A

By solving the eigen value problem, the principal stress can be obtained as

1 2 37.43, 0, 0.81 MPa

Maximum shear stress is

1 2max 4.11 MPa2

P

T

A

A

y

z


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12. If the displacement field is given by

2 2

2

2

2

2 ( )

2

x

y

z

u x y

u y x y z

u z xy

(a) Write down 33 strain matrix.

(b) What is the normal strain component in the direction of (1,1,1) at point (1,3,1)?

Solution:

(a) 33 symmetric strain matrix can be calculated from its definition as

2

2( ) 0

0 2

x y z y

y z x y

y z

In addition, the unit normal vector in the direction of (1, 1, 1) is

1{1 1 1}

3

T n

b) Thus, the normal component of strain is

1 2(2 2 2 2 )

3 3x y z y y z x y y z y n n

Thus, the normal component of strain reduces as they-coordinate of a point increases. At

point (1, 3, 1), y= 3

3 2

y n n .


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13. Consider the following displacement field in a plane solid:

( , ) 0.04 0.01 0.006

( , ) 0.06 0.009 0.012

u x y x y

v x y x y

(a) Compute the strain components xx, yy, and xy. Is this a state of uniform strain?

(b) Determine the principal strains and their corresponding directions. Express theprincipal strain directions in terms of angles the directions make with thex-axis.

(c) What is the normal strain at Point Oin a direction 45oto thex-axis?

Solution:

(a) Strain components:

0.01xxu

x

0.012yyv

y

0.009 0.006 0.015xyv u

x y

Yes, this is a state of uniform strain, because the strains are independent of positionx,y,z.

(b) Principal strains and principal directions.

10.0075

2xy xy

0.01 0.0075

0.0075 0.012xx xy

xy yy

Find the eigen values (principal strains) and eigen vectors (principal direction) by solving

the eigen value problem:

0.01 0.0075 0

0.0075 0.012 0x

y

n

n

The above equation yields two principal strains: 1 = 1 =

0.01231 and 2 = 2 =

0.01431. The principal direction corresponding to the first principal strain is

(1) 0.9556 0.2948 n ,

The angle the direction makes with the x-axis can be found from the relation

cos 0.9556, sin 0.2948 . Solving 163o

The principal direction corresponding to the second principal strain is


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(2) 0.2948 0.9556n ,

and the angle is found to be 73o

(c)

Strain at point O

0.01 0.0075

0.0075 0.012

,

direction vector

1

21

2

n

Thus the normal strain in the direction of nbecomes

45

1 1

0.01 0.00752 2 0.00851 0.0075 0.012 1

2 2

o

T

n n


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14. The displacement field in a solid is given by

2

22

( )

x

y

z

u kx

u kxy

u k x y z

where kis a constant.

(a) Write down the strain matrix.

(b) What is the normal strain in the direction of n= {1, 1, 1}T?

Solution:

(a) From the definition of strain

2

2 , 4 , ( )

1

2

1 1

2 2

1 1

2 2

yx zxx yy zz

yx

xy

y zyz

x zxz

uu ukx kxy k x y

x y z

uuky

y x

u ukz

z y

u ukz

z x

Thus, the strain matrix is

2 12

2 1

21 12 2

2

[ ] 4

( )

kx ky kz

ky kxy kz

kz kz k x y

(b) Unit normal vector

1{1 1 1}

3

T n

Thus, the normal strain in the direction of nis

21

[ ] 2 4 3 23 ky kxy kx ky kz n n


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15. Draw a 22-inch square OABC on the engineering paper. The coordinates of Oare(0, 0) andBare (2, 2). Using the displacement field in Problem 13, determine the u

and vdisplacements of the corners of the square. Let the deformed square be denotedas O'A'B'C'.

(a) Determine the change in lengths of OAand OC. Relate the changes to the straincomponents.

(b) Determine the change in AOC . Relate the change to the shear strain.(c) Determine the change in length in the diagonal OB. How is it related to the

strain(s)?

(d) Show that the relative change in the area of the square (change in area/original

area) is given by A/A= xx+ yy= 1+ 2.

Hint: You can use the old-fashioned method of using set-squares (triangles) and

protractor or use Excel to do the calculations. Place the origin somewhere in thebottom middle of the paper so that you have enough room to the left of the origin.

Solution:

(a) Let ', ', ', 'O O A A B B C C after deformation, suppose the coordinates

of each point are O(0, 0),A(0, 2),B(2, 2), C(2, 0). From the displacement field, we can

obtain the displacement of each point:

: (0,0) 0.04, (0,0) 0.06O u v '(0.04, 0.06)O

: (0,2) 0.052, (0,2) 0.084A u v '(0.052,2.084)A

: (2, 0) 0.02, (2, 0) 0.078C u v '(2.02, 0.078)C

2 2' ' (0.052 0.04) (2.084 0.06) 2 0.024OA O A OA ;

A'(0.052, 2.084)

C'(2.02, 0.078)

O'(0.04, 0.06)

B'(2.032, 2.102)

A B

COx

y


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2 2' ' (2.02 0.04) (0.078 0.06) 2 0.0199OC O C OC .

0.0240.012

2 yyOA

OA

;

0.01990.01

2 xxOC

OC

.

(b)

1 1

' ' '2

0.052 0.04 0.078 0.06sin ( ) sin ( )

2.024 1.980.005929 0.00909

0.015

xy

AOC A O C

(c)

B: (2,2) 0.032, (2, 0) 0.102u v '(2.032,2.102)C ;

2 2' ' (2.032 0.04) (2.102 0.06) 2 2 0.0243OB O B OB

45

0.02430.0086

2 2o

OB

OB

(d)

' ' ' ' sin ' ' ' 2 2

_ 2 22.024 1.98 sin1.5558 4 0.00707

0.001774 4

Area O A O C A O C

Orignal Area

Note that the change is area is close to the sum of two normal strains:

1 2 0.012 0.01 0.002xx yy


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16. Draw a 22-inch square OPQR such that OP makes +73o to the x-axis. Repeatquestions (a) through (d) in Problem 15 for OPQR. Give physical interpretations to

your results.

Note: The principal strains and the principal strain directions are given by

2 2

1,2 2 2 2

tan2

xx yy xx yy xy

xy

xx yy

Solution:

(a) Let ', ', ', 'O O P P Q Q R R after deformation. The coordinates of each

point are O(0, 0), P(0.585, 1.913), Q(2.497, 1.328), R(1.913, -0.585). From the

displacement field, we can obtain the displacement of each point:

: (0, 0) 0.04, (0,0) 0.06O u v '(0.04, 0.06)O

: (0.585,1.913) 0.0456, (0.585,1.913) 0.0882P u v '(0.6306,2.0012)P

: (1.913, 0.585) 0.0174, (1.913, 0.585) 0.0702R u v '(1.9304, 0.5148)R

2 2' ' (0.6303 0.04) (2.0012 0.06) 2 0.0291OP O P OP

P'(0.631,2.001)

R'(1.930, -0.515)

O'(0.04, 0.06)

Q'(2.520, 1.426)P(0.585, 1.913)

Q(2.497, 1.328)

R(1.913, -0.585)

Ox

y

73o


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2 2' ' (1.9304 0.04) ( 0.5148 0.06) 2 0.0241OR O R OR

2

0.02910.0146

2

OP

OP

; 1

0.02410.0121

2

OR

OR

The change in length of OP and OR equal to the principal strains since 73o is the

principal direction.

(b)

1 10.6303 0.04 0.5148 0.06sin ( ) sin ( )2 2.0291 1.9759

0.2952 0.29522 2

P O R

0POR .

On principal direction, there is no shear deformation.(c) The Point Qis moved to:

(2.4973,1.3279) 0.023, (2.4973,1.3279) 0.09841u v '(2.520,1.4264)Q

2 2' ' (2.520 0.04) (1.4264 0.06) 2 2 0.003OQ O Q OQ

0.0030.0011

2 2

OQ

OQ

2 228 cos (28 ) sin (28 ) sin(28 )cos(28 ) 0.0011o xx yy xy

Thus, the meaning of the length of diagonal change is the same as in (c) in Problem 15.

(d)

' ' ' ' sin ' ' ' 2 2

_ 2 2

2.0291 1.9759 sin( ) 42 0.0023

4

Area O P O R P O R

Orignal Area

1 2 0.0025


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17. For steel, the following material data are applicable: Youngs modulusE= 207 GPaand shear modulus G = 80 GPa. For the strain matrix at a point shown below,

determine the symmetric 33 stress matrix.

0.003 0 0.006

[ ] 0 0.001 0.003

0.006 0.003 0.0015

Solution:

From Eq. (1.58) the elasticity matrix becomes

12

12

12

1 0 0 0

1 0 0 0

1 0 0 0[ ]

0 0 0 0 0(1 )(1 2 )

0 0 0 0 0

0 0 0 0 0

E

C

From the relation / 2(1 )G E , we calculate ( / 2 ) 1 0.294E G .

0.003 0.879

0.001 0.239

0.0015 0.639[ ]

0.003 0.240

0.006 0.480

0 0

xx

yy

zz

yz

xz

xy

C GPa

In the matrix notation

0.879 0 0.480

0 0.239 0.240 GPa

0.480 0.240 0.639


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18. Strain at a point is such that xx= yy= 0, zz= 0.001, xy= 0.006, and xz= yz= 0.Note: You need not solve the eigen value problem for this question.

(a) Show that n1= i+jand n2= i+jare principal directions of strain at this point.

(b) What is the third principal direction?

(c) Compute the three principal strains.

Solution:

(a) The strain matrix is

3

0 6 0

[ ] 6 0 0 10

0 0 1

In order to show a direction nis a principal direction, it is enough to show that[ ] n n .

After normalizing n1and n2,

3 31 1

0 6 0 1 610 10

[ ] 6 0 0 1 62 2

0 0 1 0 0

n n

3 32 2

0 6 0 1 610 10

[ ] 6 0 0 1 62 2

0 0 1 0 0

n n

Thus, n1and n2are principal directions.

(b) From the orthogonal property of principal directions, the third principal direction canbe found using the cross product as

3 1 2 {0 0 1}T n n n

Note that n3in the above equation is normalized.

(c) Since the third principal direction is parallel to the z-axis, zz is the third principal

strain; i.e., 3= zz= 0.001. From Part (a), the principal strain 1and 2can be obtained

because [ ] n n . Thus, the three principal strains are

1 2 30.006, 0.001, 0.006

Note that the three principal strains are reordered.


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19. Derive the stressstrain relationship in Eq. (1.60) from Eq. (1.55) and the plane stressconditions.

Solution:

Three-dimensional stress-strain relation is given in Eq. (1.57). From the third equation of

Eq. (1.57),

(1 ) 0(1 )(1 2 )

( )1

zz xx yy zz

zz xx yy

E

Then, from the first equation of Eq. (1.57),

2

2

(1 )(1 )(1 2 )

(1 ) ( )

(1 )(1 2 ) 1

1

xx xx yy zz

xx yy xx yy

xx yy

E

E

E

In a similar way,

21yy xx yy

E

Thus, if we combine these equations, we can obtain Eq. (1.60):

212

1 01 0

10 0 (1 )

xx xx

yy yy

xy xy

E


29/64


20. A thin plate of width b, thickness t, and length L is placed between two frictionlessrigid walls a distance b apart and is acted on by an axial force P. The material

properties are Youngs modulusEand Poissons ratio .

(a) Find the stress and strain components in thexyzcoordinate system.

(b) Find the displacement field.

Solution:

(a) From the given force conditions, we can calculate the stress components, as

xx

P

bt 0yy 0zz 0xy yz zx (1)

We dont know 0 yet, but it is clear that there must be a compressive stress in the y-

direction because of the effect of Poissons ratio. Since all shear stresses are zero, all

shear strain s are also zero:

0xy yz zx

From the geometry, we can calculate the following strain components:

0yy xxL (2)

We dont know yet.

Lets calculate unknown parameter 0and using the stress-strain relation.

0

xx

L

L

P P

y

x

y

zbt


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1( )

1( ) 0

( )

xx xx yy

yy yy xx

zz xx yy

E L

E

E

By substituting the relations in Eq. (1) in the above second equation, we obtain

0

10

P

E bt

0

P

bt

And from the first relation, the unknown parameter can be calculated as

22(1 ) (1 )

PL PL

E bt Ebt

Thus, the stress components are

xx

P

bt yy

P

bt

And the normal strain in thez-direction is

2

(1 ) (1 ) (1 )

1(1 )zz

P P

E bt Ebt LL

(b) Displacement components can be calculated through integration as

0

1

x

y

z

u xL

u

u zL


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21. A solid with Youngs modulusE = 70 GPa and Poissons ratio = 0.3 is in a state ofplane strain parallel to the xy-plane. The in-plane strain components are measured

as follows: xx= 0.007, yy= 0.008, and xy= 0.02.

(a) Compute the principal strains and corresponding principal strain directions.

(b) Compute the stresses including zz, corresponding to the above strains.

(c) Determine the principal stresses and corresponding principal stress directions.Are the principal stress and principal strain directions the same?

(d) Show that the principal stresses could have been obtained from the principalstrains using the stress-strain relations.

(e) Compute the strain energy density using the stress and strain components in xy-

coordinate system.

(f) Compute the strain energy density using the principal stresses and principalstrains.

Solution:

(a) The eigen value problem for the strain matrix is

0.007 0.01 0{ }

0.01 0.008 0x

y

n

n

I n

The eigen values can be calculated by making the determinant of the coefficient matrix

zero, as

0.007 0.010 0.013, 0.012

0.01 0.008

Thus, the principal strains are: 1 30.012, 0.013 (notice: 2 0 in z-direction)To find principal directions, substitute the principal strains into the characteristic

equation and solve for {n} with 2 2 1x yn n .

1 0.8944

0.4472

n for 1 0.012

3 -0.4472

0.8944

n for 3 0.013

(Notice: {0, 0, 1}T

is the principal direction corresponding to 2= 0)

(b) From the constitutive relation for a plane strain solid in Eq. (1.62),

0.3365

0.4712 GPa

0.5385

xx

yy

xy

C


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where { } { }T xx yy xy and

12

1 0

[ ] 1 0(1 )(1 2 )

0 0

E

C

zz component can be calculated from the condition the zero strain condition:

0 0.0404GPazzzz xx yy zz v

E E

Note that 0xz xz G and 0yz yz G

(c) From Part (b),

0.3365 0.5385 0

0.5385 0.4712 0 GPa

0 0 0.0404

Solving the eigen-value problem, we obtain the following principal stresses:

1,2,3 0.6059, 0.0404, 0.7404 GPa

And the following principal directions

1

0.8944

0.4472

0

n ,

2

0

0

1

n ,

3

0.4472

0.8944

0

n

Thus, the principal strain directions are the same as that of the principal stresses.

(d) If stress-strain relation for plane stress in Eq. (1.57) is applied to the principal strains,

1 1

2 2

3 3

1 0.6059

1 0.0404 GPa1 1 2

1 0.7404

E

Note that all shear stresses are zero because it is in the principal directions. Note also thatthree-dimensional constitutive relation is used rather than two-dimensional. However,

the same results will be expected if the plane strain relation is used.

(e) 0

6 3

1

28.45 10 J/m

x x y y z z xy xy xz xz yz yz U


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(f) 6 30 1 1 2 2 3 31

8.45 10 J/m2

U


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22. Assume that the solid in Problem 21 is under a state of plane stress. Repeat (b)through (f).

Solution:

(b)

0.3538

0.4538 GPa

0.5385

xx

yy

xy

C

where2

12

1 0

[ ] 1 01

0 0 (1 )

E

C

Note that for plane stress, 0zz xz yz .

(c)

0.3538 0.5385 0

0.5385 0.4538 0

0 0 0

Solving the eigen-value problem, we obtain:

1,2,3 0.6231, 0, 0.7231GPa

The principal stress directions are

10.8944

0.4472

0

n for 1 0.6231

20

0

1

n for 2 0

30.4472

0.8944

0

n for 3 0.7231

(d) Substitute principal stresses into equation (1.60) in the textbook to obtain principal

strains. Notice that 2 0.0004 0zz


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(e) 0

6 3

1

28.4 10 J/m

x x y y z z xy xy xz xz yz yz U

(f) 6 30 1 1 2 2 3 31

8.4 10 J/m2

U


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23. A strain rosette consisting of three strain gages was used to measure the strains at a

point in a thin-walled plate. The measured strains in the three gages are: A= 0.001,

B= 0.0006, and C= 0.0007. Not that Gage C is at 45owith respect to thex-axis.

(a) Determine the complete state of strains and stresses (all six components) at that

point. AssumeE = 70 GPa, and = 0.3.

(b) What are the principal strains and their directions?

(c) What are the principal stresses and their directions?

(d) Show that the principal strains and stresses satisfy the stress-strain relations.

Solution:

(a) From figure it is obvious xx= A= 0.001 and yy= B= 0.0006. Shear strain can be

found using the transformation relation in Eq. (1.50). The 2-D version of Eq. (1.50)

becomes

2 2nn xx x yy y xy x y n n n n

where nx= cos(45o) and ny= sin(45

o). Thus,

2 2(45 ) cos 45 sin 45 sin 45 cos 45 0.0007C nn xx yy xy

By solving the above equation, we obtain xy = 0.003. Since the strain rosette only

measures plane stress state, zz is unknown. But, there is no shear strain in the z-

direction, xz = yz = 0. In order to calculate the unknown stress zz , we use the

constitutive relation for plane stress. Since the plate is in a state of plane stress, zz= xz

= yz = 0. Other stresses can be obtained from stress-strain relations for plane stress

conditions as shown below:

2

1 63.1MPa

1 23.11

26.9MPa

x x

y y

xy xy

E

G

For plane stress condition the through-the-thickness strain is obtained from Eq. (1.59), as

0.000171zz xx yy E

x

y

A

B C


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Also, all shear strains and stresses are zero because they are in the principal directions.

Thus, the stress-strain relation satisfies in the principal stresses and strains.


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B= 0.004, and C= 0.016. Determine the complete state of strains and stresses (all

six components) at that point. AssumeE = 100 GPa and = 0.3.

Solution:

(a) The angle and direction cosines of each rosette are listed in the table below.

nx ny

A 90o 0 1

B 210o 3 / 2 1 / 2

C -30o

3 / 2 1 / 2

Then, we can use the following transformation equation to related Cartesian components

to the strains in the rosettes

2 2

2 2

( ) cos sin sin cosnn x y xy

xx x yy y xy x y n n n n

The three rosette equations become

3

16 10y A

33 1 3 4 104 4 4B x y xy

33 1 3 16 104 4 4C x y xy

The last two equations can be solved for the shear strain as

3 3 3

3 2412 10 10 13.86 102 3

xy xy

Then, from the second equation, we have

6 3 3 33 4 10 6 10 4 10 8 104x x

y

A

B C

x

120o120o


40/64


Since it is the plane stress condition, 0z yz zx . From the stress-strain

relation for the plane stress problem, we have

39

32

6

6

1 1 0.3 8 10100 10

1 0.3 11 0.09 16 101

12.8 1407 1.407100 1010 GP18.4 2022 2.0221 0.09

x x

y y

E

a

9

3100 10 13.86 10 0.533GPa2(1 ) 2.6xy xy xy

EG

.

0, 0yz zx

yz zx G G

9

9

0.3( ) (1.407 2.022) 10 0.01

100 10

z x y

E

.


41/64




B= 0.002, and C= 0.008. Determine the complete state of strains and stresses (all

six components) at that point. AssumeE = 100 GPa and = 0.3.

Solution:

(a) The angle and direction cosines of each rosette are listed in the table below.

nx ny

A 0o 1 0

B 120o 1 / 2 3 / 2 C 240o 1 / 2 3 / 2

Then, we can use the following transformation equation to relate the strains measured by

the strain gages to the strain components:

2 2

2 2

( ) cos sin sin cosnn x y xy

xx x yy y xy x y n n n n

The three rosette equations become

38 10x A

31 3 3( ) 2 104 4 4B x y xy

31 3 3 8 104 4 4C x y xy

The last two equations can be solved for the shear strain as

3 3 33 126 10 10 6.93 102 3

xy xy

Then, from the second equation, we have

6 38 310 4 103 3 3

xy xy

yA

B

C

x

120o

120o


42/64


Since it is the plane stress condition, 0z yz zx . From the stress-strain

relation for the plane stress problem, we have

39

32

6

8

1 1 0.3 8 10100 10

1 0.3 11 0.09 16 101

9.2 9.2 1.011100 101.099 106.4 6.4 0.7031 0.09

x x

y y

E

GPa

0.267GPa2(1 )xy xy xy

EG

.

0, 0yz zx

yz zx G G

3( ) 5.142 10z x y

E

.


43/64


26. The figure below illustrates a thin plate of thickness t. An approximate displacementfield, which accounts for displacements due to the weight of the plate, is given by

2 2( , ) (2 )2

( , ) ( )

x

y

u x y bx x y E

u x y y b x E

(a) Determine the corresponding plane stress field.

(b) Qualitatively draw the deformed shape of the plate.

Solution:

(a) From the definition of strain

( )

( )

10

2

xxx

y

yy

yxxy

ub x

x Eu

b xy E

uu

y x

Also, from the stress-strain relation for the plane stress problem,

212

1 0 ( )

1 0 01

00 0 (1 )

xx xx

yy yy

xy xy

b xE

Thus, ( )xx b x is the only non-zero stress component.(b) The deformed geometry is sketched below

A B

y

xa a

b


44/64



45/64


27. The stress matrix at a particular point in a body is

7

2 1 3

[ ] 1 0 4 10 Pa

3 4 5

Determine the corresponding strain ifE= 20 1010Pa and = 0.3.

Solution:

7 4

11

1 1[ ( )] [ 2 0.3(0 5)] 10 1.75 10

2 10xx xx yy zz

E

7 5

11

1 1[ ( )] [0 0.3( 2 5)] 10 4.5 10

2 10yy yy xx zz

E

7 4

11

1 1

[ ( )] [5 0.3( 2 0)] 10 2.8 102 10zz zz xx yy E

7 4

11

2(1 ) 2(1 0.3)1 10 1.3 10

2 10xy xy

E

7 4

11

2(1 ) 2(1 0.3)4 10 5.2 10

2 10yz yz

E

7 4

11

2(1 ) 2(1 0.3)( 3) 10 3.9 10

2 10xz xz

E


46/64


28. For a plane stress problem, the strain components in the xyplane at a point Parecomputed as

2 2.125 10 , .25 10xx yy xy

(a) Compute the state of stress at this point if Youngs modulus E= 21011

Pa and

Poissons ratio = 0.3.(b) What is the normal strain in thezdirection?

(c) Compute the normal strain in the direction of n= {1, 1, 1}T.

Solution:

(a) Compute the state of stress at this point if Youngs modulus E = 21011 Pa and

Poissons ratio = 0.3

6

2

1 0 357

1 0 357 10 Pa1

0 0 (1 ) / 2 385

0 (Plane Stress)

xx xx

yy yy

xy xy

zz xz yz

E

(b) What is the normal strain in thez-direction?

2( ) 0.1071 10zz xx yy

E

(c) Compute the normal strain in the direction of n= {1, 1, 1}T

2

1{1, 1, 1}

3[ ] 0.2143 10

T

nn

n

n n


47/64


29. The state of stress at a point is given by

80 20 40

[ ] 20 60 10 MPa

40 10 20

(a) Determine the strains using Youngs modulus of 100 GPa and Poissons ratio of0.25.

(b) Compute the strain energy density using these stresses and strains.

(c) Calculate the principal stresses.

(d) Calculate the principal strains from the strains calculated in (a).

(e) Show that the principal stresses and principal strains satisfy the constitutiverelations.

(f) Calculate the strain energy density using the principal stresses and strains.

Solution:

(a) From Eq. (1.53),

6 3

11

3

3

3

1 .25 .25 80 0.61

.25 1 .25 60 10 10 0.3510

.25 .25 1 20 0.15

0.5 10

1.0 10

0.25 10

xx

yy

zz

xy

xy

yz

yz

zx

zx

G

G

G

(b) Strain energy density:

1

59.25kPa2 xx xx yy yy zz zz xy xy yz yz xz xz

U

(c) Principal stresses: 1 2 3110, 50, 0MPa

(d)

Strain matrix:3

0.6 0.25 0.5

[ ] 10 0.25 0.35 0.1250.5 0.125 0.15

Principal strains: 3 3 31 2 30.975 10 , 0.225 10 , 0.4 10

(e) From Eq. (1.55)


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16 3

2 11

3

1 .25 .25 110 0.9751

.25 1 .25 50 10 10 0.22510

.25 .25 1 0 0.4

Thus, the principal stresses and principal strains satisfy the constitutive relations.

(f) Strain energy density

1 1 2 2 3 31

59.25kPa2

U


49/64


30. Consider the state of stress in Problem 29 above. The yield strength of the material is100 MPa. Determine the safety factors according to the following: (a) maximum

principal stress criterion, (b) Tresca Criterion, and (c) von Mises criterion.

Solution:

(a) Maximum principal stress criterion

1

0.91YFS

(b) Tresca criterion

maxmax

110 050, 55, 0.91

2 2Y Y

Y FS

(c) Von Mises criterion

2 2 21 2 3 1 2 2 3 1 3 95.39

1.048

VM

YF

VM

S


50/64


31. A thin-walled tube is subject to a torque T. The only non-zero stress component is

the shear stress xy, which is given by xy = 10,000 T(Pa), where T is the torque in

N.m. When the yield strength Y = 300 MPa and the safety factorN= 2, calculate themaximum torque that can be applied using

(a) Maximum principal stress criterion (Rankine)

(b) Maximum shear stress criterion (Tresca)

(c) Distortion energy criterion (Von Mises)

Solution:

Since it is a pure shear stress state, the three principal stresses are

1 2 3, 0,xy xy

(a) Maximum stress criterion,

1

15,000N m10, 000 10,000

Yxy

xy Y

N

TN

(b) Maximum shear stress criterion

10, 0002 2

7,500N m20,000

Y Yxy

Y

TN

TN

(c) Von Mises criterion

23

10,0003

8,660N m

YVM xy

Yxy

N

TN

T


51/64


32. A thin-walled cylindrical pressure vessel with closed ends is subjected to an internalpressurep = 100 psi and also a torque T around its axis of symmetry. Determine T

that will cause yielding according to von Mises yield criterion. The design requires asafety factor of 2. The nominal diameter D of the pressure vessel = 20 inches, wall

thickness t = 0.1 inch, and yield strength of the material = 30 ksi. (1 ksi = 1000 psi).

Stresses in a thin walled cylinder are: longitudinal stress l, hoop stress h,and shear

stress due to torsion. They are given by

4lpD

t ,

2hpD

t ,

2

2T

D t

Solution:

2

5,000psi4

10, 000psi2

2

xx l

yy h

xy

pD

tpD

tT

D t

2 2 23 15, 0002

Y YVM xx yy xx yy xy

N

37.071 10 psixy

2 31 444 10 lb-in2 xy

T D t


52/64


53/64


2 232

Y Yvm x xy N

Substituting stress components in the above expression, we can solve for the unknown

diameter,D= 46.02 mm.

(b) Maximum shear stress criterion: In order to calculate the maximum shear stress, the

principal stresses are calculated first

2

21,2 2 2

x xxy

Then, the maximum shear stress becomes

max minmax

/ 2 / 2

2 2Y Y

N

By substituting stress components in the above expression, we can solve for the unknowndiameter, D = 47.33 mm.


54/64


34. For the stress matrix below, the two principal stresses are given as 1= 2 and 3=3,respectively. In addition, two principal directions corresponding to the two principal

stresses are also given below. The yield stress of the structure is given as Y= 4.5.

1 0 2

[ ] 0 1 0

2 0 2

, 1

2

50

1

5

n and 3

1

50

2

5

n

(a) Calculate the safety factor based on the maximum shear stress theory anddetermine whether the structure is safe or not.

(b) Calculate the safety factor based on the distortion energy theory and determinewhether the structure is safe or not.

Solution:Continuation from Problem 9.

From Problem 9, 1 2 32, 1, 3 . Thus, the von Mises stress becomes2 2 32 1 ( 3) (2 1 1 3 2 3) 21

VM

. Also, the maximum shear

stress becomes max 1 3( ) / 2 2.5 .

(1)max

2.250.9

2.5YN

. Thus, the structure is not safe.

(2)max

4.50.982

21

YVMN

. Thus, the structure is not safe.


55/64


35. The figure below shows a shaft of 1.5 in. diameter loaded by a bending momentMz=5,000 lbin, a torque T= 8,000 lbin, and an axial tensile force N= 6,000 lb. If the

material is ductile with the yielding stress Y= 40,000 psi, determine the safety factorusing: (a) the maximum shear stress theory and (b) the maximum distortion energy

theory.

Solution:

From the given loading conditions, the magnitude of shear will be the same for all outer

surfaces, whereas the bottom surface will have the maximum tensile stress due to bending

and tension. Thus, if the material fails, it will fail at the bottom surface first. Lets take an

infinitesimal rectangle at the bottom surface. Then, the non-zero stress component will be

andxx xz .

Each component of stress can be calculated from the mechanics of materials by

3 2

3

32 5000 4 600018, 486 psi

(1.5) (1.5)16 8000

12,072 psi(1.5)

xx

xz

M r N

I A

T r

J

Principal stresses

2 2

0

0 0 ( ) 0

0

xx xz

xx xz

xz

2 2

1 2 3

424,447, 0, 5,961 psi

2

xx xx xz

(a) The maximum shear stress theory

1 3max

max

15,2042

20, 0001.315

15,204Y

psi

N

xz

xxxx

xz

xzxz

xz

x

y

NN MzMz

T

T


56/64


(b) Maximum distortion energy theory

2 224447 ( 5961) 24447( 5961) 27,909 psi

40, 0001.4332

27,909

VM

VMN


57/64


36. A 20-mm. diameter rod made of a ductile material with a yield strength of 350 MPais subject to a torque of T= 100 Nm and a bending moment ofM = 150 Nm. An

axial tensile force P is then gradually applied. What is the value of the axial forcewhen yielding of the rod occurs? Solve the problem two ways using (a) the

maximum shear stress theory and (b) the maximum distortional energy theory.

Solution:

(a) The yielding occurs at the bottom surface in which both M and Pproduce tensile

stress. At this bottom surface, the stress components are

6

3 2

63

32 4191 10 3183

1663.662 10

xx

xz

M PP

d dT

d

And all other components are zero. Now, the maximum shear stress is expressed in term

of stress components:

2 21 3max

14 175

2 2 xx xy

In the above equation, the following relations are used:

2 2 2 2

1 34 4,

2 2xx xx xz xx xx xz

Then, 2 2 2350 4xx xz

The above equation can be solved for axial forceP= 41,413 N.

(b) The von Mises stress can be written in terms of stress components, as

2 2 2 2 61 3 1 3

6 6

3 350 10

332 10 191 10 3183

VM xx xz

xx N

After solving for the axial force, we have P= 44,353 N. The distortion energy theory

allows a larger axial force.

x

y

PP MzMz

T

T


58/64


37. A circular shaft of radius rin the figure has a moment of inertia Iand polar momentof inertiaJ. The shaft is under torsion Tzin the positive z-axis and bending moment

Mx in the positive x-axis. The material is mild steel with yield strength of 2.8 MPa.Use only the given coordinate system for your calculations.

(a) If TzandMxare gradually increased, which point (or points) will fail first amongfour points (A,B, C, andD)? Identify all.

(b) Construct stress matrix []A at point A in xyz-coordinates in terms of givenparameters (i.e., Tz,Mx,I,J, and r).

(c) Calculate three principal stresses at point B in terms of given parameters.

(d) When the principal stresses at point C are 1 = 1, 2 = 0, and 3 = 2 MPa,calculate safety factors (1) from maximum shear stress theory and (2) fromdistortion energy theory.

Solution:

(a) The bending moment will produce maximum stress at points A and C. Thus, A and

C will fail first.

(b) At point A, non-zero stress components are

,x zxx xz

M r T r

I J

Thus, the stress matrix becomes

/ 0 /

[ ] 0 0 0

/ 0 0

x z

z

M r I T r J

T r J

(c) At point B, only non-zero stress component is

zyz

T rJ

Thus, the three principal stresses are

1 2 3, 0, ,z zT r T r

J J

(d) For maximum shear stress criterion,

x

y

z

A

B

C

D

x

yMx

Tz


59/64


1 3max

max

1.5, 1.42 2

0.933

YY

YN

For von Mises criterion,

2 2 21 2 3 1 2 2 3 1 3 7 2.645

1.06

VM

Y

VM

N


60/64


38. A rectangular plastic specimen of size 10010010 mm3is placed in a rectangularmetal cavity. The dimensions of the cavity are 1011019 mm

3. The plastic is

compressed by a rigid punch until it is completely inside the cavity. Due to Poissoneffect, the plastic also expands in thexandydirections and fills the cavity. Calculate

all stress and strain components and the force exerted by the punch. Assume there isno friction between all contacting surfaces. The metal cavity is rigid. Elastic

constants of the plastic areE = 10 GPa, = 0.3.

Solution:

The strains in the specimen are calculated as the ratio of change in length to original

length.

9 10 101 1000.1, 0.01

10 100zz xx yy

We have assumed that the plastic expands laterally and fill the cavity completely. If it

does not, then we will get positive values for xxand/or yy, which will indicate that our

assumption is wrong. Then we can assume xxand/or yy= 0, and redo the problem and

obtain corresponding strains xxand/or yywhich will be less than that calculated above.

Since there is no friction between contacting surfaces, all shear stresses and hence all

shear strains will be identically equal to zero.

The normal stresses can be obtained from three-dimensional stress strain elations:

1

11 1 2

1

x x

y y

z z

E

Substituting for the strains and elastic constantsEand we obtain the stresses as

{ } { 385 385 1,231} MPaxx yy xx

Since xxand yyare negative (compressive), our initial assumption about the strains is

correct. The punch force is obtained from zand the area of cross section:

0.1 0.1 1,231 12.31MNz

F A

Rigid punch

Plastic

Rigid die

Rigid punch

Plastic

Rigid die

F


61/64



62/64


39. Repeat Problem 38 with elastic constants of the plastic asE= 10 GPa and = 0.485.

Solution:

The strains in the plastic specimen are calculated as the ratio of change in length to

original length.

9 10 101 100

0.1, 0.0110 100z x y

We have assumed that the plastic expands laterally and fill the cavity completely. If it

does not, then we will get positive values for xxand/or yy, which will indicate that our

assumption is wrong. Then we can assume xxand/or yy= 0, and reiterate the problem

and obtain corresponding strains xx and/or yy which will be less than that calculated

above.

Since there is no friction between contacting surfaces, all shear stresses and hence all

shear strains will be identically equal to zero.

The normal stresses can be obtained from three-dimensional stress strain relations:

1

11 1 2

1

x x

y y

z z

E

Substituting for the strains and elastic constantsEand we obtain the stresses as

{ } { 8,642 8,642 9, 383} MPaxx yy xx

Since xxand yyare negative (compressive), our initial assumption about the strains is

correct. The punch force is obtained from zzand the area of cross section:

0.1 0.1 9,383 93.83MNzF A

Note: Punch force for this problem is almost 8 times that for Problem 38. The increase is

due to Poissons ratio. As the material compressibility decreases, Poissons ratio

increases. For example, as 0.5 the material becomes incompressible, i.e., its

volume cannot be changed, and the stresses become unbounded. Note the term 1 2

in the denominator of the above constitutive relation.


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40. Repeat Problem 38 with the specimen of size 10010010 mm3and the dimensions

of the cavity 1041049 mm3. Elastic constants of the plastic are E = 10 GPa, =

0.3.

Solution:

The strain in thez-direction remains the same as (9 10)/ 10 0.1z . As before,

if we assume that the specimen fills the cavity completely, the strains will be

104 1000.04,

100x y

The stresses are calculated using

1

11 1 2

1

x x

y y

z z

E

We obtain { } { 192 192 885} MPaxx yy xx .The above stresses are not physically possible as the cavity walls cannot exert tensile

stresses on the specimen. We will repeat the calculations with 0x y . This is

actually uniaxial state of stress, and the strains are obtained as 0.03x y z .

The extension of the plate in the x and y-directions is given by

104 0.03 3.12mmx x . Note that the expansion of the specimen is less

than the 4 mm-clearance.


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Introduction to Finite Element Analysis and Design Ch 01

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