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 Ba ghdad Dies and Tools Eng. Dept. E-mail:[email protected] www.mediafire.com/haniazizameen

Strength of materials Handout No. 1111- Principal StressesStresses- Dr. Hani Hani Aziz Ameen 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 the z direction are equal to zero zero , the stress stress condition reduces to bi-axial ( or two dimensional or plane plane stresses ) state of stress . i.e. in the x , y planes

and

xz

0 ,

zx

0

,

zx

0 ,

xz

0

,

y

0 ,

x

0 ,

yz zy

xy

0

,

z

0

0 0 ,

yx

0

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


11-2

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

Fig(11-2) Taking moment a Mo 0 xy (dydz )dx xy

Similarly

yx (dxdz )dy

0

yx yz

zy

&

xz

zx

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 ( n ) & shear ( n ) 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)

FN

0

N

n .dA

x cos

+

sin xy sin

n

x

.dA cos

.dA cos

cos2

n

x

yx

y sin

cos 2

sin y sin

y

2

sin 2

.dA sin sin

cos .dA sin sin 2

xy sin

y

x

cos

xy sin 2

using the trigonometric identities 1 cos 2 sin2 & cos2 2 :. Eq.(11- 1) will be x

0

11-1) 1 cos 2 2

y

cos 2 xy sin 2 2 2 Similarly , summation of all forces along the direc equation n

x

y

11- 2)

sin 2 ...................... (11-3) xy cos 2 2 In order to ascertain the the orientation orientation of XnYn corresponding to the max , or min . ( n ) , the necessary condition n

d d

n

0

is applied to Eq(11-2) yielding :


d

n

2

d x

(

x

cos sin

2

sin 2

2

sin 2

y

sin 2 y ) sin

x

tan2

2 2

y sin xy

cos 2

cos 2

xy

xy

2

xy

cos 2

0

0

0 .............. (11-4)

p x

cos

y

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

n

d tan 2

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

x s

y

2

where the subscript

.................. (11-5)

xy

s

denotes the shear

Eq.(11Eq.(11-4) 4) is inve inverse rse of of Eq(11-5 Eq(11-5)) and and the valu valuee of 2 diffe differs rs at 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 known as principal planes. planes. Thus principal stresses are normal stresses acting on t he 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 equal to zero zero ( n=0) °

°

°

°

From Eq(11- 3) 2

tan2 x

°

0

x

y

2

sin 2

xy

cos 2

xy y

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


2 x max min min

y

x

2

y

2 xy

2

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

& 2 x max

y

2

2 xy

................... (11-7)

Note that the algebraically larger stress given in eq(10- 6) is the max . principal stress , denoted 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 1 ( x y ) from the origin . 2 3 x, xy , or A ( x , xy ) 4

Draw a circle with center

From the above raduis can be deduced

ius of AC .


2

R= 5

x

y

2

2 xy

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 circle , it can be stated that the value value of stress at point E is 1 2 ( min . principal stress ) and the value of stress at point D is (max . principal stress ) and the shear stress at point F is max . ( max ) i.e. 1 = OC + R R 2 = OC

11- 4 Strain in Three Perpendicular Perpendicu lar Directions The rectangular bar shown in Fig(11-5 a) is subjected subjected to three perpendicular forces in the x, y, and z directions to induce the normal stresses x, y & z

Fig(11-5 a)


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

Fig(11-5b) Hence , the axial strain in the x-direction x- direction due to

x

only =

x

E

Fig(11-5 c) Lateral strain in the x-direction due to

y only

=

z only

=

y

E

Fig(11-5 d) Lateral strain in the x-direction due to

Thus ,the total strain in the xx - direction due to x

x

y

z

E

E

E

(

x

y

z

E x, y &

z ) / E

Similarly

where

y

(

y

x

z ) / E

z

(

z

x

y ) / E

y

&

z

are the total strain in y & z direction

z

is


11-5

Principal Stresses in Terms of Principal Strains

Have from previous sections ( x x y z ) / E y

(

y

z ) / E

x

In the case of two dimensional stress system , and for an element which is subjected to the stress in the z- direction = 0 , i.e. z 0 . x & y only , the Note that when the element is free of shearing stresses , the normal stresses x & y are regarded as the maximum & minimum principal stresses and may be written as

1&

2

respectively . The resulting strains

x

the max. and min. principal strains and may be written as respectively . , z 0 x 1 , y 2 Putting , , z 0 x 1 y 2

&

y 1

&

Hence 1 1

E 1

(

2)

1

( 2 1) E when solving the above two equations simultaneously , we obtain 2

1

and

E(

E(

2

2

2 ) /(1

1

1 ) /(1

2

2

)

)

11- 6 The Relation Between the 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 p ure shearing stresses may be found by applying . 1 1 2 2 * x 1, 2 x y y 2 2 Since x and y each equals to zero ,hence 1,2

0

1

&

0 2

2

are 2


to find the planes on which

1

&

2 act

tan2

0 hence , 2 = 90 and 1=45 , 2 = 90 + 45 = 135 substitute 1=45 in the general equation. 0 0 sin sin 2 * 45 n 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 The strain in the direction of

1due

to

1

1 due

to

the total strain in the direction of

1

2

1

E 2

E


E 1=

1

2

E

E

,

1 1

1

=

Similarly

=

E

E

(1

E 2

2

)

=

E The total Strain 1 &

11-8 )

(1

)

2

11-9 )

can be derived in other way :

In the direction of 1 oa oa oa = = 1 oa oa

1

.................. (11-10)

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

oa

(1

oa[1

)=

E

in the direct of 2 =

(

oa oa

(1

1

)]

11-11)

2

ob ob ob

) (negative)


i.e.

2=

(

ob

ob

) = 1+

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

ob 1

tan( 45 ac

ob oa

E

(1

ob

)

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

.......................... (1 1-13)

ac

( / 2))

oa

ob

tan( 45

( / 2))

tan 45 1

therefore ,

tan( / 2)

tan 45 tan( / 2))

for smal smalll angl anglee tan( tan( /2)) /2)) =

where

ob

1 ( / 2)

oa

1 ( / 2)

from Eq(11-13) and Eq(11-11) & Eq(11-14) ob

ob[1

(1

) / E]

1 ( / 2)

oa

oa[1

(1

) / E]

1 ( / 2)

G

) E

2(1

)

tan( / 2))

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

is the shearing strain

2(1

1

/2

ob

E

1 tan( / 2))

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

11-7 Examples The following examples explain explain the different ideas of the principal principal stresses problems . Example (11-1) Fig(11-7) shows a tank of diameter diameter 1 m and wall thickness thickness t=20mm is subjected to an internal pressure of 6 MPa .Find : (a) The state of stress stress in the rectangular rectangular element shown in Fig. (b) The normal and shear stress stress along the inclined inclined plan plan m-m

Fig(11-7) Solution

PD

(a)

y

2t x

(b)

3

= (6*1) / (2*20*10 ) = 150 MPa

0.5

y =75

x n

y

2 75 150

n

+

MPa x

2 75 150

2 x

y

2 y

n

2 75 150

n

2

sin 2

xy

cos 2

xy

sin2

cos(2*120 ) = 131.25 MPa cos 2

sin ( 2 * 120 ) = 32.48 32 .48 MPa


Example (11-2) Fig(11-8) shows an element . Find

1,

2 and

p

using two methods .

(i.e. Mathematical Mathematical method and Graphical Graphical (Mohr s circle) Fig(11-9) Solution ( a ) Graphical Method take a scale that : 1 cm = 10 MPa 20 40 x y the center C = 10MPa. 2 2 2 2

the radius R =

1 2

tan2

x

xy

y

2

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

xy

p x

2

0

p p

y

0

0 20 40

0

=

20 2

40

2

30 MPa.


( b ) Numerically x

y

1

2 20 40 1 2

x 2

1

2 x

y

20 40

2

2

2 1

y

2

2

2 x

2

xy

=40 MPa

4

y

4

2

20

xy

2

40

1 2

20

40

2

20MPa

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 2. internal pressure of 3.5 MN / m Find the normal and shear stresses on a plane making an angle of o f 30 with the axis of cylinder . °

Fig(11-9) Solution y

=

Pd 2t

&

x

=

Pd 4t

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

F Dt

,


D ......... is the mean diameter =(Pd =(Pd/4 /4t) t)+( +(F/ F/ DT) DT) 6 3 *0.297*0.0 0.003)] 03)]=12 =121.5 1.5 MPa y= [(3.5*10 *0.294)/(4*0.003)] + [(100*10 )/ ( *0.297* 6 y= [ ( 3.5*10 *0.294)/ ( 2* 0.003)] = 171.5MPa n=( x+ y) /2 + (( x y) /2 )* cos2 xy sin2 =(121.5+171.5)/2 .5)/2 + ( (121.5 (121.5 171.5)/2)* 171.5)/2)* cos(2*60) cos(2*60) =159 MPa MPa n=(121.5+171 n = (( x y) /2) * sin2 xy *cos2 = ((121.5 171.5)/2)*sin(60*2) = 21.7 MPa x

Example(11-4) At a point in the cross section of a loaded beam the major principal 2 2 stress is 140 N/mm tension and the max. shear stress is 80 N/mm . 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. o c) The state of stress on a plane making an angle of 30 with the plane of the major principle tensile stress . Solution. x

y

80 = (140 -

max

2 For max. shear x

= 45 y

/2

2

y

= -20 N/mm

o

x

y

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

+

y)


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

Graphical solution scale 1 cm = 20 MPa center C = ( x y ) / 2 (140

y ) / 2

the max. shear max =Radius of the circle =80 MPa 2 set off OA=140 N/mm =140MPa i.e QA=80 MPa Then minor principal stress , y = OB = 20 MPa .(-ve) QC1=2*45 =90 QC2=2*45 =60 OQ 60MPa 45 OD 100MPa 30 C2 D 69.3 MPa 30 ° °

°

°



Example(11-6) At a point in a stressed material , the normal ( tensile) and shear stresses 2 2 on a certain plane xx are 95 N/ mm of max. shear is 55 N/mm and 2 65 N/mm respectively respectively . The tensile stress on the plane of max. shear is 2 55 N/mm . Find (a) The principal principal stresses 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 sketch . Solution x n

y

2 x

n

let

x

2 y

2 m

y

cos 2

sin 2

x

y

2

x

and n

y

2

then n n

m n cos 2 n sin 2

95=m+n cos 2 65=n sin 2

i) ii) iii) 45 55 n cos 2 40 = n cos 2

sub.Eq(iii) into Eq(i) yields

iv)

Eq(ii) & Eq(iv) are

65

tan 2

2

40

From the triangle m = 55 & n=76.3 55

x

y

&

76.3

58.4

n x

402

652

76.3

y

. 2 2 Solving this two equations give 2 2 2103 N/mm . y x 131.3 N/mm 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 2 pressure of 5.5 MN/m . 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 x

Pd

5.5 * 10 6 * 0.075

4t

4 * 0.005

Pd

5.5 * 10 6 * 0.075

y

2t 2 * 0.005 T=F.r F T xy

A

r.A

20.6MN / m 2 41MN / m 2 torque

mean raduis * cross - sectional area


1.6 *103 xy

0.04 * * 0.08 * 0.005 1

1, 2

1, 2

1 2

2 1

(

y)

x

(120.6

2

tan 2 180

max

max

(

y

4

2

xy

41.2) 2

(20.6

20.6

72 .4 x

2

4 * (31.8) 2

2 * 31.8

xy

x

2

y)

x

41.2)

2 64.3MPa. 2.5MPa.

31.8 MPa

41.2

53 56

y 2

)

2 xy

(

20.6

41.2

)

2

(31.8)

2

2 2 acting on planes planes at 45 to the the princi principal pal planes. planes. 33.4 MPa acting

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 can be considered considered as a load concentrated at the end of a cantilever canti lever 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

T=

M = 15*103*9.81*2= 294.3 kN.m 15 * 10 6 * 60

power

1.433MN.m 2 * 100 Pv Engine power = where P is the propulsive force 2 n / 60

6

15 * 10 * 0.85 * 3600

P

32 * 10

10435N

3

294.3 * 103 * 32

M

Direct stress due to bending =

3

* 0.453

d 32

1.435 * 10 6 * 4

P

Direct stress due to end thrust = 4

32.9 MN / m 2

d2

* 0.45 2

9.02MN / m

2

The total direct stress x 32.9 9.02 41.92MPa Shear stress due to torque 6

T 16

d

1.433 * 10 * 16 3

* 0.45

3

80MN / m 2

The stresses on the element on the upper surface of the shaft at the bearing are there free as shown in in Fig(11-12 b) these being the greatest applied stresses in the shaft 1 1, 2

2 1

1, 2

1 2

{

2 x

4

x

{41.42

2 103.7MPa 61.8 MPa

max =

2

xy }

(41.92) 2

2 x

2

2 xy

2

4 * 80 2 }

41.92 2

2

80

2

80.75MPa


Example(11-9) At a point in a piece of stressed material the normal stress on a certain 2 plane is 90 N/mm tension and the shearing stress on this plane is 2 30N/mm . On a plane inclined at 60 to the first named plane , there is a 2 tensile stress of 60 N/mm . Find :( a ) The principal stresses at the point . 2 (b) The intensity of shearing stress on the plane having 60 N/mm normal stress relative to the given planes , and show the relative positions in a clear diagram . °

Solution As in example ( 11- 6) m n cos2 n n sin2 n

where

m=

x

y

and

n=

x

y

2 2 90 = m+ n cos2 30 = n sin 2 60 = m + n cos2( + 60 ) 1 3 60 = m n ( cos 2 sin 2 ) 2 2 3n sin2 i.e. 120 = 2m n cos2 solving Eq.( i ) , Eq( ii ) & Eq( iii ) , yields m = 87.32 N / mm2 2 n=30.12 N/mm 42 27

Fig(11-13) from which x

117.44 N/mm

y

57.14 N/mm 2

2

2

on the plane of the 60 N/mm normal stress

30.12 sin2 (24 27

60 )

-12.68 N/mm2


The positions of the various planes are shown in Fig(11-13) Example(11-10) At a point in a material material under two-dimensional two-dimensional stress, the normal normal stresses , all tensile, on three planes are as follows: 2 Inclination to plane A Stress (N/mm ) 0 97 B 45 133 90 27

Plane A C

°

° °

Find (a) The shearing shearing stresses on planes A.B and C (b) The principal stresses and the inclination to plane A of the planes on which which they act. (c) The max. shearing stress. (d)The inclination to plane A of the plane plane whose the normal normal stress is zero. Show by a sketch the relative positions of the various planes . Solution As in example (11- 6 ) m n * cos 2 n

m

x

y

2

&n

x

y

2

97 = m + n cos2 133 = m + n cos2 ( 45 ) = m n sin 2 90 ) = m n cos2 27 = m + n cos2( Adding Eq( i ) and Eq( iii ) , 2m = 124 n* sin 2 = 17 n* cos 2 = 35

n 712 352 79.2 2 62 + 79.2 = 141.2 N / mm x 2 17.2 N / mm y = 62 79.2 =

= 62


tan2 =

71

= 2.028

35 2 =360 63 46

296 14

148 .7 Sinc Sincee n has has been been assum assumed ed posi positi tion on , sin2 sin2 nega negati tive ve & cos2 cos2 th hen hence 2 lies in the the 4 quadrant . n sin sin 2 2 when 71 N/mm 184 7 79.2 sin 246 .14 2 when 139 .7 79.2 sin 386 .14 = 35 N/mm when

238 .7 79.2 sin 476 .14 2 max = 79.2 N / mm when 62 + 79.2 cos2 0 n =0 from which

71 N/mm

posi positiv tivee ,

2

90 19 .4 The relative positions positions of the various planes planes are as shown in Fig(11-14) Fig(11-14) .

Fig(11-14) Example(11-11) Fig(11-15) shows

a point in the structural member , the

stresses

graphically:aThe magnitude and orientation of the principal stresses bThe magnitude magnitude and orientation of the maximum shearing stresses and associated normal stresses.


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

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

- axis.

The radius R = CA1 1

OC

R

96.05 MPa

2

OC

R

23.45 MPa

Locate point A(80-30) Draw line through C to B The plane on which the principle stress acts is given by tan2

xy p

x

y

2

2

p p

tan

1

28 15

30 20

56 30


( b ) The max. shearing shearing stresses are given given by points D and E , thus 36.05 MPa . max = 80 40 40 20 x y tan2 s 2 xy 2 * 30 60 30 s

28 .15

s

= 163 15

45

73.15

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

Fig(11-16) Solution . Scale 1 cm

Center

C

= 10 MPa 14 28 y

x

2

C = 27 MPa Locate point A ( 28 , 0 ) R = CR =21 MPa

2


A B A .B

7 21cos 60 17.5 3.5MPa 21sin 60 = 18.186MPa

Example(11-13) 2 Fig(11-17) shows a rod with 850 mm cross

sectional area . 60 kN is

applied axially to it at its ends , find the n & n the plane incline 30 on the direction of loading and max . numerically & graphically .

Fig(11-7)


Solution. ( a ) Numerically x

1 n

2 1

n

2 1

n

2 1

n

2

x (1

P

60 * 103

A

850

70.6MPa

cos 2 )

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

sin 2

( 70.6 ) sin 60 = 30.6 MPa .

the , max Value of

n

at

45

°

1

( 70.6) sin 90 35.3MPa 2 (b) Graphically Scale 1 cm = 10 MPa. n ) max

Point A = (70 , 0) Radius R=35 MPa Now the value value of 2 is measur measured ed anti-c anti-cloc lockwi kwise se from from OC Draw Cd & dK .: the value of OK = n 17.65 MPa. The value of o f Kd = n 30.6 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 the cross section of a loaded beam , the major major 2 principal stress is 140 N/mm tension and the max . shear stress is 80N/mm2 .Using either either graphical or or analytical methods, Find for this point, (a) The magnitude of the minor principal stress; str ess; (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 per pendicular direct stresses. A piece of steel plate is subjected to perpendicular stresses 2 of 80 and 50 MN/m , 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 2 not to exceed 60 MN /m . 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 2 produces a max . direct stress of 75 MN/m 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

Strength of materials Handout No. 1111- Principal StressesStresses- Dr. Hani Hani Aziz Ameen Ameen 2

75 MN/m . 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. 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 shows two two separate uni-axial states of stress. stress. Find (a) the state of stress, referred to as as an element 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/mm 2 acting along the planes AC and CB towards C, and normal tensile 2 stresses on AC and CB of 75 N/mm respectively. There is no stress on the plane perpendicular to planes AC and CB . Find the position of the 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 tangential component component along AB (e) The least inclination to AB Analytical or graphical methods may be used; in the case of a graphical g raphical


solution, indicate how how the diagrams are constructed. State State for each plane found its angular position relative to AC and the magnitude magnitude of the stress referred to. 11-10) Establish a relationship between the modulus of elasticity, modulus modulus of rigidity and and Possion s ratio for an elastic elastic material. material. A close-coil helicalspring of circular wire and and mean diameter 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. spring. Find the value value of Possion Possion s ratio ratio for the materia materiall of spring. °

Strength of Materials- Principal Stresses- Hani Aziz Ameen

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