Strength of Materials- Principal Stresses- Hani Aziz Ameen

Strength of Materials Handout No.11

Principal Stresses Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:[email protected]

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Strength of materials Handout No. 11- Principal Stresses- Dr. Hani Aziz Ameen

11-1 Introduction In the most general case normal stress ( ) and shear stress ( ) at a point in a body may be considered to act on three mutually perpendicular planes .

The most general state of stress is usually referred to as a tri-axial is shown in Fig(11-1)

Fig(11-1) If all stress components in the z direction are equal to zero , the stress condition reduces to bi-axial ( or two dimensional or plane stresses ) state of stress . i.e. in the x , y planes

0 , 0 , 0 , 0 zyzzxxz 0 , 0 , 0 zyxzzx

and 0 , 0 , 0 , 0 yxxyyx Many of the problems encountered in practice are such that they can be considered plane state of stress . e . g . thin shells , beams , plate etc .


11-2 Analysis of Plane Stress From the plane Fig(11-2) shown .

Fig(11-2) Taking moment a

0Mo 0dy)dxdz(dx)dydz( yxxy

yxxy Similarly zxxzzyyz & This s that the shear stress on any two mutually perpendicular planes through a point in a stressed body must be equal in magnitude and opposite in direction . It is desirable to be able to relate those stresses on the X and Y planes to

t defined by the angle and then to determine the normal ( )(shear&) nn stresses . as shown in Fig(11-3) .

Fig(11-3)


Applying the equilibrium equation to the incline plane ( t ) as shown in Fig (11-4) yields

Fig(11-4) 0F NN

sindA.sincosdA.cosdA. yxn

+ 0sindA.coscosdA.sin yxxy

cossin2sincos xy2

y2

xn

2sinsincos xy2

y2

xn 11-1) using the trigonometric identities

sin2

22cos1 & cos2

22cos1

:. Eq.(11- 1) will be

2sin2cos22 xy

yxyxn 11- 2)

Similarly , summation of all forces along the direcequation

2cos2sin2 xy

yxn ...................... (11-3)

In order to ascertain the orientation of XnYn corresponding to the max , or min . ( n ) , the necessary condition

0d

d n is applied to Eq(11-2) yielding :


02cos22sin2sin

02cos2cossin2sincos2d

d

xyyx

xyyxn

( 02cos22sin) xyyx

tan2yx

xyp

2 .............. (11-4)

where the subscript ( p ) denotes the principal stress . Similarly to ascertain the orientation of XnYn corresponding to max . of min n , the necessary condition

0dd n is applied to eq.(11- 3) yielding :

xy

yxs 2

2tan .................. (11-5)

where the subscript s denotes the shear Eq.(11-4) is inverse of Eq(11-5) and the value of 2 differs at 90° , so the plane of max. shear will be at =45° Eq(11-5) defines two values of 2 p differing by 180°or two values differing by 90°.(Similarly Eq(11-4) differing by 90°) As one of these two planes the normal stress n become max. i.e. 1 and on the normal stress the two plans are known as principal planes. Thus principal stresses are normal stresses acting on the principal planes

The principal planes are free of any shear stress and therefore another way of defining principal stresses is to say that they are normal stresses acting on planes or the shear stress is equal to zero ( n=0)

From Eq(11- 3) 2cos2sin2

0 xyyx

tan2yx

xy2

which is equivalent to Eq(11- 4) , hence substituting Eq(11-4) into Eq(11-2) yields the .minmax & and substituting Eq(11- 5) into eq(11-3) yields .max


2xy

2yxyx

min max 22

........ (11- 6)

&

2xy

2yx

max 2 ................... (11-7)

Note that the algebraically larger stress given in eq(10- 6) is the max . principal stress , denoted by &1 the min. represented by 2 . 11-3 Two Dimensional Stress .

A graphical technique , predicated from Eqs(11- 2) & (11-3) permits the rapid transformation of stress from one plane to another , and leads also to the determination of the max . normal and shear stresses . In this approach Eq(11-2)&Eq(11-3) are depicted by a stress circle , called

circle . -

1 Establish a rectangular coordinate system , indicating + . and + . Both stress scales must be identical .

2

yx(21 ) from the origin .

3 ,, xyx or A ( xyx , ) 4 Draw a circle with center ius of AC .

From the above raduis can be deduced


R = 2xy

2yx

2

5 Draw a line AB through point

At the points A & B the tensile stress is positive & the compressive stress is negative and also the shear stress is positive if the rotation is clockwise about the center . 6- From the circle , it can be stated that the value of stress at point E is

2 ( min . principal stress ) and the value of stress at point D is 1 (max . principal stress ) and the shear stress at point F is max . (

max ) i.e. 1 = OC + R 2 = OC R 11- 4 Strain in Three Perpendicular Directions The rectangular bar shown in Fig(11-5 a) is subjected to three perpendicular forces in the x, y, and z directions to induce the normal stresses

zyx &, Fig(11-5 a)


The strain in any direction indicated is due to simultaneous action of the normal stresses shown in Fig.(11-5 b ,c & d)

Fig(11-5b)

Hence , the axial strain in the x-direction due to x only =E

x

Fig(11-5 c)

Lateral strain in the x-direction due to y only =E

y

Fig(11-5 d)

Lateral strain in the x-direction due to z only =E

z

Thus ,the total strain in the x- direction due to zyx &, is

E/)(EEE zyx

zyxx

Similarly E/)( zxyy E/)( yxzz

where zy & are the total strain in y & z direction


11-5 Principal Stresses in Terms of Principal Strains Have from previous sections

E/)( zyxx E/)( zxyy

In the case of two dimensional stress system , and for an element which is subjected to yx & only , the stress in the z- direction = 0 , i.e. 0z . Note that when the element is free of shearing stresses , the normal stresses

yx & are regarded as the maximum & minimum principal stresses and may be written as 21 & respectively . The resulting strains yx & are the max. and min. principal strains and may be written as 21 & respectively .

Putting 0, ,

0 , ,

z2y1x

z2y 1x

Hence

)(E1

211

)(E1

122

when solving the above two equations simultaneously , we obtain

)1/()(E 2211

and )1/()(E 2

122 11- 6 The Relation Between the Modulus of Elasticity E

and the Modulus of Rigidity G The element of Fig(11- 6a) is subjected to pure shearing stresses .The max. & min. principal normal stresses due to pure shearing stresses may be found by applying .

22yxyx2,1 *

21

21

Since x and y each equals to zero ,hence 2

2,1 00

21 &


to find the planes on which 21 & act

tan20

hence , 2 = 90 and 1=45 , 2 = 90 + 45 = 135° substitute 1=45° in the general equation.

45*2sin00n therefore, the max .principal stress acts at an angle of 45° to the vertical and the minimum principal stress acts at an angle of 135° to the vertical.

-a- -b- Fig(11-6) The element in Fig(11-6 b) is equivalent to the two element shown in Fig(11- 6 c)

Fig(11-6 c) Thus ,

The strain in the direction of E

to due 111

The strain in the direction of 1due to E

22

the total strain in the direction of 1


E 1 = E

1 E

2

1 , 2

1 = EE

1 = E

( 1 ) 11-8 )

Similarly 2 = E

( 1 ) 11-9 )

The total Strain &1 2 can be derived in other way :

In the direction of 1

1 = oa

oaao = oaao 1 .................. (11-10)

Eq(11-8 ) & Eq(11-10) yields:

E( 1 ) =

oaao 1

)]1(E

1[oaao 11-11)

in the direct of 2

2 = (ob

boob ) (negative)


i.e. 2 = (

obboob ) = 1+

obbo ..................... (11-12)

Eq(11-9) & Eq(11-12) gives

)1(E

1obbo .......................... (11-13)

aoca))2/(45tan(

boca

))2/tan(1))2/tan(1

))2/tan(45tan1)2/tan(45tan))2/(45tan(

aobo

for small angle tan( /2)) = /2

therefore , )2/(1)2/(1

aobo ................ (11-14)

where is the shearing strain from Eq(11-13) and Eq(11-11) & Eq(11-14)

)2/(1)2/(1

]E/)1(1[oa]E/)1(1[ob

aobo

)1(2E

)1(2EG


11-7 Examples The following examples explain the different ideas of the principal stresses problems . Example (11-1) Fig(11-7) shows a tank of diameter 1 m and wall thickness t=20mm is subjected to an internal pressure of 6 MPa .Find :

(a) The state of stress in the rectangular element shown in Fig. (b) The normal and shear stress along the inclined plan m-m

Fig(11-7) Solution

(a) t2

PDy = (6*1) / (2*20*103) = 150 MPa

yx 5.0 =75 MPa

(b) 2

yxn +

2yx cos 2 xy sin2

215075

n 215075 cos(2*120 ) = 131.25 MPa

2cos2sin2 xy

yxn

215075

n sin ( 2 * 120 ) = 32.48 MPa


Example (11-2) Fig(11-8) shows an element . Find 21, and p using two methods .

(i.e. Mathematical method and Graphical (Mohr s circle) Fig(11-9) Solution ( a ) Graphical Method take a scale that : 1 cm = 10 MPa

the center C = .MPa102

40202

yx

the radius R = 2

yx2xy 2

= 302

4020 2 MPa.

1 OC + R = 10 + 30 = 40 MPa 2 OC R = 10 30 20 MPa

tan2 04020

02

yx

xyp

02 p 0p


( b ) Numerically

xy22

yxyx

1 421

2

2402021

24020 =40 MPa

MPa20402021

240204

21

22

xy22

yxyx

2

Example(11-3) Fig(11-9) shows a cylindrical vessel , 300 mm external diameter , wall thickness 3 mm , is subjected to an axial tensile force of 100 kN and an internal pressure of 3.5 MN / m2 . Find the normal and shear stresses on a plane making an angle of 30° with the axis of cylinder .

Fig(11-9) Solution

y = t2

Pd & x = t4

Pd

where d ..... is the internal diameter The longitudinal stress due to the axial load is given by :-

= DtF ,

where


D ......... is the mean diameter

x =(Pd/4t)+(F/ DT) y= [(3.5*106*0.294)/(4*0.003)] + [(100*103 )/ ( *0.297*0.003)]=121.5 MPa y= [ ( 3.5*106*0.294)/ ( 2* 0.003)] = 171.5MPa n=( x+ y) /2 + (( x y) /2 )* cos2 xysin2 n=(121.5+171.5)/2 + ( (121.5 171.5)/2)* cos(2*60) =159 MPa

n = (( x y) /2) * sin2 xy *cos2 = ((121.5 171.5)/2)*sin(60*2) = 21.7 MPa

Example(11-4) At a point in the cross section of a loaded beam the major principal stress is 140 N/mm2 tension and the max. shear stress is 80 N/mm2 . Using either graphical or analytical methods , find for this point :-

a) the magnitude of the minor principal stress. b) The magnitude of the direct stress on the plane of max. shear

stress. c) The state of stress on a plane making an angle of 30 o with the

plane of the major principle tensile stress . Solution.

2 yx

max 80 = (140 - y) / 2 y = -20 N/mm2

For max. shear = 45 o

2yx

45n o + 2cos2

yx

n )45 = (140 20)/2 + ((140+20)/2)* cos(2*45) =60 N/mm2


when =30

n)30=(140-20)/2 +((140+20))/2 cos(2*30)=100 N/mm 2 o30n ) = ((140 + 20 )/2 )*sin(2*30) = 69.3 N/mm2

Graphical solution scale 1 cm = 20 MPa center C = 2/)140(2/)( yyx the max. shear max =Radius of the circle =80 MPa set off OA=140 N/mm2 =140MPa i.e QA=80 MPa Then minor principal stress , y = OB = 20 MPa .(-ve) QC1=2*45° =90 ° QC2=2*45°=60°

MPa60OQ45 MPa100OD30

MPa 3.69DC230


Example(11-6) At a point in a stressed material , the normal ( tensile) and shear stresses on a certain plane xx are 95 N/ mm2 of max. shear is 55 N/mm 2 and 65 N/mm2 respectively . The tensile stress on the plane of max. shear is 55 N/mm2. Find (a) The principal stresses (b) The max .shear stress (c) The direction of the plane xx relation to the plane on which the major principal stress acts. Illustrate your answer to (c ) by a sketch . Solution

2cos22

yxyxn

2sin2

yxn

let 2

m yx and 2

n yx

then nmn cos 2 95=m+n cos 2 i)

2sinnn 65=n sin 2 ii) iii) sub.Eq(iii) into Eq(i) yields n5545 cos 2 40 = n cos 2 iv) Eq(ii) & Eq(iv) are

4.58240652tan

From the triangle 3.766540n 22 m = 55 & n=76.3

255 yx & 76.3

2yx .

Solving this two equations give 3.131x N/mm2 2103y N/mm2. The position of xx in relation to x is shown in Fig(11-10)


Fig(11-10) Example (11-7) Fig.(11-11) shows a thin cylindrical tube, 75 mm internal diameter and wall thickness 5mm, is closed at the ends subjected to an internal pressure of 5.5 MN/m2. A torque of 1.6 kN.m is also applied to the tube. Find the max. and min. principal stresses and also the max. shearing stress in the wall of the tube.

Fig(11-11) Solution

26

x m/MN6.20005.0*4

075.0*10*5.5t4

Pd

26

y m/MN41005.0*2

075.0*10*5.5t2

Pd

T = F . r

area sectional-cross*raduis meantorque

A.rT

AF

xy


MPa8.31

005.0*08.0**04.010*6.1 3

xy

xy22

yxyx2,1 4))(21

222,1 )8.31(*4)2.416.20()2.416.120(

21

.MPa3.641 .MPa5.22

2.416.208.31*22

2tanyx

xy

2 65534.72180

222xy

2yxmax )8.31()

22.416.20()

2(

4.33max MPa acting on planes at 45 to the principal planes. Example(11-8) Fig(11-12) shows a propeller shaft of a ship is 0.45 m diameter and it supports a propeller of mass 15t .The propeller can be considered as a load concentrated at the end of a cantilever of length 2m .The propeller is driven at 100 rev/min. When the speed of the ship is 32 km/h , if the engine develops 15 MW , find the principal stresses in the shaft and the max . shear stress. It may be assumed that the propulsive efficiency of the propeller is 85 percent.

- a - - b Fig(11-12)


Solution At the bearing M = 15*103*9.81*2= 294.3 kN.m

T= m.MN433.1100*2

60*10*1560/n2

power 6

Engine power = wherePv P is the propulsive force

N1043510*32

3600*85.0*10*15P 3

6

Direct stress due to bending = 23

3

32d

m/MN9.3245.0*

32*10*3.294M3

Direct stress due to end thrust = 22

6

24

m/MN02.945.0*

4*10*435.1dP

The total direct stress MPa92.4102.99.32x Shear stress due to torque

23

6

316

m/MN8045.0*

16*10*433.1d

T

The stresses on the element on the upper surface of the shaft at the bearing are there free as shown in Fig(11-12 b) these being the greatest applied stresses in the shaft

}4{21

xy22

xx2,12

}80*4)92.41(42.41{21 22

2,1

MPa7.1031 2 61.8 MPa

max = MPa75.8080292.41

22

22

xy

2x


Example(11-9) At a point in a piece of stressed material the normal stress on a certain plane is 90 N/mm2 tension and the shearing stress on this plane is 30N/mm2 . On a plane inclined at 60° to the first named plane , there is a tensile stress of 60 N/mm2 . Find :- ( a ) The principal stresses at the point . (b) The intensity of shearing stress on the plane having 60 N/mm2 normal stress relative to the given planes , and show the relative positions in a clear diagram . Solution As in example ( 11- 6) nmn cos2 nn sin2

where m = and2

yx n = 2

yx

90 = m+ n cos2 30 = n sin 2 60 = m + n cos2( + 60 )

60 = m n ( 2sin232cos

21 )

i.e. 120 = 2m n cos2 n3 sin2 solving Eq.( i ) , Eq( ii ) & Eq( iii ) , yields

m = 87.32 N / mm2 n=30.12 N/mm 2

7242

Fig(11-13) from which

N/mm 44.117x2

2y N/mm 14.57

on the plane of the 60 N/mm2 normal stress N/mm -12.68)6072(24 sin2 12.30 2


The positions of the various planes are shown in Fig(11-13) Example(11-10) At a point in a material under two-dimensional stress, the normal stresses , all tensile, on three planes are as follows:-

Plane Inclination to plane A Stress (N/mm2) A 0 ° 97 B 45° 133 C 90° 27

Find (a) The shearing stresses on planes A.B and C (b) The principal stresses and the inclination to plane A of the planes on which they act. (c) The max. shearing stress. (d)The inclination to plane A of the plane whose the normal stress is zero. Show by a sketch the relative positions of the various planes . Solution As in example (11- 6 )

2cos*nmn

2

n&2

m yxyx

97 = m + n cos2 133 = m + n cos2 )45( = m n sin 2 27 = m + n cos2( 90 ) = m n cos2

Adding Eq( i ) and Eq( iii ) , 2m = 124 = 62 n* sin 2 = 17 n* cos 2 = 35 2.793571n 22

x 62 + 79.2 = 141.2 N / mm2 y = 62 79.2 = 17.2 N / mm2


tan2 = 3571= 2.028

2 =360 63 4129664 7.148

Since n has been assumed position , sin2 negative & cos2 positive , hence 2 lies in the 4th quadrant .

2sinn when 41.246sin2.797184 71 N/mm2 when 41.386sin2.797.139 = 35 N/mm2 when 41.476sin2.797.238 71 N/mm2

max = 79.2 N / mm2 when n = 0 62 + 79.2 cos2 0 from which

4.1990 The relative positions of the various planes are as shown in Fig(11-14) .

Fig(11-14) Example(11-11) Fig(11-15) shows a point in the structural member , the stresses

graphically:- a- The magnitude and orientation of the principal stresses b- The magnitude and orientation of the maximum shearing stresses

and associated normal stresses.


In each case show the results on a properly oriented element.

Fig(11-15) Solution Scale 1 cm = 10 MPa The c - axis.

The radius R = CA1

MPa 23.45ROCMPa 96.05ROC

2

1

Locate point A(80-30) Draw line through C to B The plane on which the principle stress acts is given by

tan22

yxp

yx

5128

03562030tan2

p

1p


( b ) The max. shearing stresses are given by points D and E , thus max = 36.05 MPa .

tan23020

6040

30*24080

2 xy

yxs

15.734551.28s s = 163 51

Example(11-12) Fig(11-16) shows an element of a loaded body . The stresses ( in MPa ) act on an element. Apply Mohr s circle to find the normal and shear stresses acting on a plane defined by = 30

Fig(11-16) Solution . Scale 1 cm = 10 MPa

Center C 2

28142

yx

C = 27 MPa Locate point A ( 28 , 0 )

R = CR =21 MPa


5.1760cos217A

MPa5.3B 60sin21B.A

= MPa186.18

Example(11-13) Fig(11-17) shows a rod with 850 mm2 cross sectional area . 60 kN is applied axially to it at its ends , find the nn & the plane incline 30 on the direction of loading and .max numerically & graphically .

Fig(11-7)


Solution. ( a ) Numerically

MPa6.70850

10*60AP 3

x

)2cos1(21

xn

21

n ( 70.6 ) ( 1 cos 60 ) = 17.65 MPa

2sin21

xn

(21

n 70.6 ) sin 60 = 30.6 MPa .

the , max Value of 45 at n °

MPa3.3590sin)6.70(21

max)n

(b) Graphically Scale 1 cm = 10 MPa.

Point A = (70 , 0) Radius R=35 MPa Now the value of 2 is measured anti-clockwise from OC Draw Cd & dK .: the value of OK = 65.17n MPa. The value of Kd = 6.30n MPa


11-8 Problems 11-1) A cylindrical , 300mm external diameter, wall thickness 3mm, is

subjected to an axial tensile force of 100 kN and an internal pressure of 3.5 MN/m2 . Find the normal and shear stresses on a plane making an angle of 30° with the axis of the of cylinder?

11-2) At a point in the cross section of a loaded beam , the major

principal stress is 140 N/mm2 tension and the max . shear stress is 80N/mm2 .Using either graphical or analytical methods, Find for this point,

(a) The magnitude of the minor principal stress; (b)The magnitude of the direct stress on the plane of max . shear stress (c) The state of stress on a plane making an angle of 30° with the plane of the major principal tensile stress.

11-3) Derive formulae for the normal and tangential stresses on an

oblique plane within a material subjected to two perpendicular direct stresses. A piece of steel plate is subjected to perpendicular stresses of 80 and 50 MN/m2 , both tensile , find the normal and tangential stresses and the magnitude and direction of the resultant stress on the interface whose normal makes an angle of 30° with the axis of the second stress.

11-4) Show that the principal stresses are the extreme values of the

normal stress for any interface under conditions of complex stress. A 50mm diameter bar is subjected to a pull of 70 kN and a torque of 1.25 kN.m. Find stresses for a point on the surface of the bar and show by a diagram the relation between the principal planes and the axis of the bar.

11-5) A hollow propeller shaft , having 250 mm and 150 mm external

and internal diameters respectively transmits 1200 kW with a thrust of 400kN. Find the speed of the shaft if the max . principal stress is not to exceed 60 MN /m2 . what is the value of the max . shear stress at this speed ?

11-6) At a section of a rotating shaft there is a bending moment which

produces a max . direct stress of 75 MN/m2 and a torque which produces a max. shearing stress of 45 MN/m2. Consider a certain point on the surface of the shaft where the bending stress is initially


75 MN/m2. tension and find the principal stresses at the point in magnitude and direction

(a) When the point is at the initial position . (b) When the shaft has turned 45°. (c) When the shaft has turned through 90°. Make sketches to show the changes in the principal planes & stresses 11-7) A flywheel of mass 500 kg is mounted on shaft 80 mm in diameter

and midway between bearings 0.6 m apart in which the shaft may be assumed to be directionally free. If the shaft is transmitting 30 kW at 360 rev/min . Find principal stresses and the max. shearing stresses in the shaft at the ends of a vertical and horizontal diameter in a plane close to the flywheel.

11-8) Fig(11-18) shows two separate uni-axial states of stress. Find

(a) the state of stress, referred to as an element whose sides are parallel to the xy axes that results from a superposition of these two stress states and

(b) the magnitudes and directions of the principal normal stresses associated with the combined state.

Fig(11-18) 11-9) A right angle triangle ABC with the right-angle at C represents planes in an elastic material. There are sheaving stresses of 45 N/mm2

acting along the planes AC and CB towards C, and normal tensile stresses on AC and CB of 75 N/mm2 respectively. There is no stress on the plane perpendicular to planes AC and CB .

Find the position of the plane AB when the resultant stress on AB has (a) The greatest magnitude

(b) The least magnitude (c) The greatest component normal to AB (d) The greatest tangential component along AB (e) The least inclination to AB Analytical or graphical methods may be used; in the case of a graphical


solution, indicate how the diagrams are constructed. State for each plane found its angular position relative to AC and the magnitude of the stress referred to.

11-10) Establish a relationship between the modulus of elasticity,

modulus of rigidity and Possion s ratio for an elastic material. A close-coil helicalspring of circular wire and mean diameter 100 mm was found to extend 42.6mm under an axial load of 50 N . The same spring ,when firmly fixed at one end, was found to rotate through 90° under a torque of 6 N.m applied in a plane at right angles to the axis of the spring. Find the value of Possion s ratio for the material of spring.

Strength of Materials- Principal Stresses- Hani Aziz Ameen

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Transcript of Strength of Materials- Principal Stresses- Hani Aziz Ameen