307 tutorial2 2014.pdf

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METU – ME 307 Machine Elements I – Fall 2014

1

30 MPa

40 MPa

25 MPa

35 MPa

20 MPa

50 MPa

Prepared by: Ufuk Can KADIOĞLURoom: C-206 Phone: 5271E-mail: [email protected] Date: 13/10/2014

ME 307 – MACHINE ELEMENTS I

TUTORIAL 2

“3D Stress State, Thick-Walled Cylinders, Thermal Stresses”

Problem 1:

Consider the cubic stress element taken from a shaft. For the element, with the stresscomponents as shown:

a.

Write the stress matrix.b. Determine the principal stresses for the given stress state.c.

Draw Mohr’s circle diagram of the principal stresses.d.

Find the maximum shear stress.

Solution:

a. The stress matrix will be;

40 25 2025 30 35

20 35 50

x xy xz

yx y yz

zx zy z

Note that each stress component which is in the plane with i-surface normal and

which is in the j-direction is represented byij

. That is xy

represents the stress in

the plane whose normal is parallel to x-axis, with y-direction in the , , x y z coordinatesystem.

b. For a stress element, there exists a particular orientation in space for which all shearcomponents are zero. When an element has this particular orientation, the normals to thefaces correspond to principal directions and the normal stresses associated with thesefaces are the principal stresses.

Suppose that, this particular orientation of the element is as shown and can be expressed

in another coordinate system; ', ', ' x y z .

According to definition, since the shear stress components will be zero for thisorientation, the stress matrix will be diagonal. This corresponds to an eigenvalue problemwhere the magnitudes of the principal stresses will be the eigenvalues and their directions(directional cosines) will be the eigenvectors. That is:

pn n

which can also be written as:

0 p I n

z

x

y

z

x

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2

Here, I is the identity matrix, p is the eigenvalue and n is the eigenvectorcorresponding to each eigenvalue.

To avoid the trivial solution 0n the determinant of p I should bezero.

0 p I

40 25 2025 30 35

20 35 50

p xx p xy xz

p yx yy p yz

p zx zy zz p

3 2 2 2 2

2 2 2

( ) ( )

( 2 ) 0

x y z p x y x z y z xy yz zx p

x y z xy yz zx x yz y zx z xy

This is a cubic equation in the unknown p

where the three solutions are the principal

stresses 1 , 2 , 3 .Note that, the solutions to p are independent of the coordinate

system chosen. In other words; principle stresses occur only in one particular

orientation and no matter the coordinate system of the stress element chosen, thesolutions to must be the same. Therefore, the coefficients of

p in the cubic

equation are constant:

z y x I 1 222

2 zx yz xy z y z x y x I 222

3 2 xy z zx y yz x zx yz xy z y x I

I1, I2 and I3 are called as stress invariants.

Calculating the coefficients yields;

1 40 I , 2 3950 I , 3 19250 I

Now the roots of the cubic equation can be found from the formula:

3cos 1

I r ii , where r and i can be found from:

31 1 2 3

3

2 21 2

2 9 27cos3

2( 3 )

I I I I

I I

Now calculate r and , by inserting the numerical values,

77.316r

cos 3 0.33024 Here cosine equation has infinitely many solutions,

cos 0.33024 2

3

a k

, where 0,1,2..k

Take 3 of the possible solutions as we have 3 principal stresses.

1

cos 0.33024

0.41143

a

rad

2

cos 0.33024 22.5058

3

a

rad

3

cos 0.33024 21.6830

3

a

rad

Then the principal stresses will be

11 1cos 84.198 MPa3

I r

12 2cos 48.876 MPa

3

I r

13 3cos 4.678 MPa3

I r

Arrange σ1, σ2, σ3 such that σ1>σ2>σ3.

1 2 384.198 MPa , 4.678 MPa , 48.876 MPa

2

1

22

1 )3(3

2 I I r

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3Of course this solution will be much shorter if software was used. For example, theMATLAB command given below gives the eigenvectors and eigenvalues of the stressmatrix.

The diagonal elements of eigval matrix are the eigenvalues which are the principal

stresses, and the columns of eigvec are the corresponding eigenvectors which are thedirection cosines of the principal stresses.

The eigenvalues and eigenvectors can also be obtained using the MATHCADcommands “eigenvals” and “eigenvecs” respectively:

Eigenvectors can also be found from the following equation:

0 p I n

For each value of p

, there exists a corresponding eigenvector. For example for the first

eigenvalue 1 84.198 ,

0 p I n

40 84.168 25 20 0

25 30 84.168 35 020 35 50 84.168 0

x

y

z

n

nn

124.168 25 20 0 x y z

n n n (1)

25 54.168 35 0 x y z

n n n (2)

20 35 24.168 0 x y z

n n n (3)

Note that this matrix equation has infinitely many solutions, as det 0 p I ,these 3 equations are not linearly independent.

From Eq(1) & Eq(2)

4845.88 1958.36 0 2.4745 x y y x

n n n n

From Eq(2) & Eq(3)

1958.36 1304.2 0 1.5016 z z y

n n n n

40

25

20

25

30

35

20

35

50

eigenvals ( )

84.198

48.876

4.678

eigenvecs ( )

0.244

0.603

0.76

0.96

0.259

0.103

0.135

0.755

0.642

>> %calculating the eigenvectors and eigenvalues of the>> %stress matrix using "eig" command :>> stress=[-40 25 20; 25 30 35; 20 35 50]

stress =

-40 25 2025 30 3520 35 50

>> [eigvec,eigval]=eig(stress)

eigvec =

0.9604 0.1348 0.2437-0.2589 0.7546 0.6030-0.1026 -0.6422 0.7596

eigval =

-48.8757 0 00 4.6777 00 0 84.1979

>>

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4As there are infinitely many solutions say,

xn a then 2.4745 yn a and

3.7171 z n a

Now let’s normalize n and find the directional cosines,

We know that 2 2 2 x y z

n n n should be equal to 1, to be the directional cosines.

Then

2 2 20.2815 cos 0.2815 73.65

2.4745 3.7171

o x

an aa a a

2 2 2

2.47450.5408 cos 0.5408 57.26

2.4745 3.7171

o

x

an a

a a a

2 2 2

3.71710.8123 cos 0.8123 35.68

2.4745 3.7171

o

x

an a

a a a

The directional cosines of the first principal stresses are shown in figure below.

c. Mohr’s circle for the calculated principal stresses will be

d. Maximum in-plane principal shear stresses will be,

1 21/2 39.762

MPa

2 32/3 26.772

MPa

1 31/3 66.54

2

MPa

The maximum shear stress is the radius of the bigger circle,

max 1/3 66.54 MPa

z

x

y

2

3

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5Problem 2

A composite bar made of aluminum and steel is held between the supports as shownin the figure. The bars are stress free at a temperature of 37o C. What will be the stressin the two bars when the temperature is 20 oC. If;

a)

The supports are unyielding b)

The supports yield and come nearer each other by 0.10 mm?

It can be assumed that the change of temperature is uniform all along the length of the bar. TakeEs= 210 GPa; Ea=74 GPa; αs=11.7x10-6 /

oC and αc=23.4x10-6 /oC

Solution:

Given T1= 37o C; T2= 20o CEs= 210x109 N/mm2 ; Ea= 74x109 N/mm2Let us assume that the right support at B is removed and the bar is allowed to contractfreely due to fall in temperature. We know that the fall in temperature.T=T1-T2= 37-20=17o CContraction in steel bar= αs.ls.T=11.7x10-6x600x17=0.12 mmContraction in aluminium bar= αc.lc.T=23.4x10-6x300x17=0.12 mmTotal contraction= 0.12+0.12=0.24 mm= 0.24x10-3 mIt may be noted that even after this contraction (i.e. 0.24 mm) in length, the bar stillstress free as the right hand end was assumed free.

Let an axial force P be applied to the right end till this is brought in contactwith the right hand support at B, as shown in the figure below.

We know that cross-sectional area of the steel bar,

A =π

4(d)

=π

4(0.05) = 1.964x10 m

And cross sectional area of the aluminum bar,

A =π

4(d)

=π

4(0.025) = 0.491x10 m

We know that elongation of the steel bar;

δl =Pxl

AxE=

Px0.6

1.964x10x210x10 = 1.455x10xP m

And elongation of the aluminum bar;

δl =Pxl

AxE=

Px0.3

0.491x10x74x10 = 8.257x10xP m

Total elongation: 1.455+8.257= 9.712x10xP m Let σs= stress in the steel barσa= stress in the aluminum bar

a) When the supports are unyieldingThen the total contraction is equated to the total elongation.i.e. 0.24x10-3 =9.712x10-9 P or P= 24712 NStress in the steel bar;

σ

=P

A=

24712

1.964x10 = 12582x10

N

m = 12.582 MPa

And stress in the aluminum bar;

σ =P

A=

24712

0.491x10 = 50328x10

N

m = 50.328 MPa

b)

When the supports yield by 0.1 mmSupports come nearer to each other by 0.10 mm the net contraction in length=

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7

,which gives:

21

2

x y

(2)

Substituting eqn. (2) into eqn. (1) gives:

21 12

x x p

(3)

Summation of elongations of cubes in y direction is also equal to zero due to rigidwall. The total elongation in y direction expression can be written in terms of strainssince the side length of cubes are equal to each other.

Since the material properties and the stresses applied by cube 1 are the same for cubes2 and 3, it can be written as:

1 2 1 1 1 1 2 21 2

1 12 0, 2 0 y y y x y x p

E E

(4)

Solving eqn. (2), (3) and (4) simultaneously for unknowns 1 x , 1 y and 2 x gives:

Problem 4

A gun barrel is assembled by shrinking an outer barrel over an inner barrel so that themaximum principal stress equals 70 percent of the yield strength of the material. Both

members are steel and have as their properties a yield strength Sy=540MPa, E=207GPaand ν=0.292. The nominal radii of the barrels are 5, 10 and 15 mm.

a.

Plot the stress distribution for both parts of the barrels. b.

What value of interference should be used in assembly?c.

When the gun is fired, an internal pressure of 280MPa is created. Plot theresulting stress distributions. What are the maximum radial and tangentialstresses?

d.

If the gun barrel were made of a single barrel with the inner and outer radii of 5

and 15 mm, respectively, when the gun is fired, what would be the maximumradial and tangential stresses?

Solution:The material properties are;

GPa E 207 and MPaS y 540

The maximum principal stress on the gun barrel equals 70 percent of the yield strength ofthe steel. Hence;

yS 7.0max

a.

The tangential and radial stress formulas for thick-walled cylinders are given

below:

r t

io

iooiooii

for and for r r

r

P P r r P r P r

22

2

2222

(1)

For the inner barrel r i=5mm and r o=10mm. Also it is obvious that there i s onlyoutside pressure on the inner barrel. Then;

mmr mmr oi 10,5

unknown P P P oi *,0

Substituting these values into equation (1) and solve the equation for r=5mm andr=10mm, it can be found as;

*67.25 P mmt

*67.110 P mmt

05 mmr *10 P mmr

The same procedure can be followed for the outer barrel. This time there is no outside pressure on the outer barrel. There will be only inside pressure and this pressure will be equal to the outside pressure on the inner barrel.

2x1

E22x 2 2y 2z

1

E22x 2 1y 0

1y 2y 3y 0

1x 1 pE2 2 E1 2 E1 2

2 E2 1

E2 E2 12

2 E1 2 E1 22

1y E2 1 p1 1

E2 E2 12

2 E1 2 E1 22

2x E2 1 p1 1

E2 E2 12 2 E1 2 E1 22

2

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8

mmr mmr oi 15,10 *,0 P P P

io

Again substituting the values into the equation (1), one can get;

*6.210 P mmt

*6.115 P mmt

*10 P mmr 015 mmr It is clearly seen that the maximum principal stress occurs on the tangential stress

of inner surface on the inner barrel. Pressure can be found as: max

*67.25 P mmt

yS P 7.067.2*

MPa P 6.141* Minus sign indicates that there is compression on the inner barrel.Since pressure is found, stress distributions can be drawn as:

5 8.333 11.667 15400

200

0

200

400

Stress Distributions across the Barrel

mm

M P a t r ( )

r r ( )

r

b. Interference pressure is found as;

MPa P 6.141* Then the interference can be found as;

222

2222

2 io

io

r r r

r r r r

b

E P

(2)

where mmr and mmr mmr oi 1510,5

2222

2232

io

io

r r r r E

r r r P

After substituting given values, interference can be found as:mm029.0

c. When the gun is fired, an internal pressure of 280MPa is created. Since the outside pressure is 0.1MPa, it can be neglected. Additionally, since the members are thesame. The barrels can be considered as a single body. Hence, again using equation(1);

r t

io

iooiooii

for and for r r

r

P P r r P r P r

22

2

2222

where MPa P and P mmr mmr iooi 2800,15,5

After substituting the values into the equation, t and r can be drawn as a function

of r from r=5mm to r=15mm as:

6 8 10 12 14400

200

0

200

400

Stress Distributions due to Firing

mm

M P a t.f

r ( )

r.f r ( )

r

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9 Note that these stress distributions are due to firing. These stresses should besuperimposed on the stress distributions due to interference fit. Hence, the resultingstress distribution can be drawn as:

5 8.333 11.667 15400

200

0

200

400

600

Resulting Stress Distributions across the Barrel

mm

M P a

t r ( ) t.f r ( )

r r ( ) r.f r ( )

r

As it is seen clearly from the graph, the maximum tangential stress occurs on theinterference fit and has a value of 482MPa. Additionally, the maximum radial stress

occurs on the inner surface on the barrel inside and has a value of -280MPa. Hence; MPa

t 482max, MPar 280max,

d. If the gun barrel were made of a single barrel with the inner and outer diameters of5 and 15 mm, there will be only stresses due to firing. Again from the graph, one cansee that maximum tangential and radial stresses occur on the inner surface of the barrel and have values 350MPa and -280MPa, respectively.

MPat

350max, MPar 280max,

Note that, when the pressure values which are found above are compared with theatmospheric pressure, one can see the outside pressure is negligible. Hence, the

assumption is correct.

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