Dr.Hani Aziz Ameen
Principal Stresses
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
 xz  0 ,  zx  0 ,  yz  0 ,  z  0
 zx  0 ,  xz  0 ,  zy  0
and
 x  0 ,  y  0 ,  xy  0 ,  yx  0
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 .
٣١٠
Dr.Hani Aziz Ameen
11-2
Principal Stresses
Analysis of Plane Stress
From the plane Fig(11-2) shown .
Fig(11-2)
Taking moment about “O” , yields
 Mo  0
–  xy (dydz)dx   yx (dxdz)dy  0
 xy   yx
Similarly  yz   zy &  xz   zx
This is called Boltzmon’s principle which states 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
the stresses acting on any inclined planes “ t ” defined by the angle  and
then to determine the normal (  n ) & shear ( n ) stresses . as shown in
Fig(11-3) .
Fig(11-3)
٣١١
Dr.Hani Aziz Ameen
Principal Stresses
Applying the equilibrium equation to the incline plane ( t ) as shown in
Fig (11-4) yields
 FN  N  0
Fig(11-4)
 n .dA   x cos .dA cos    y sin .dA sin 
+  xy sin .dA cos    yx cos .dA sin   0
 n   x cos2    y sin 2   2 xy sin  cos 
 n   x cos 2    y sin 2    xy sin 2 ……………………(11-1)
using the trigonometric identities
1  cos 2
1  cos 2
sin2  
& cos2  
2
2
:. Eq.(11- 1) will be
x  y x  y
n 

cos 2   xy sin 2 ………………..(11- 2)
2
2
Similarly , summation of all forces along the direction of “ t “ leads to
equation
x  y
n 
sin 2   xy cos 2
...................... (11-3)
2
In order to ascertain the orientation of XnYn corresponding to the max ,
or min . (  n ) , the necessary condition
d n
0
is applied to Eq(11-2) yielding :
d
٣١٢
Dr.Hani Aziz Ameen
Principal Stresses
d n
 2 x cos  sin   2 y sin  cos   2 xy cos 2  0
d
  x sin 2   y sin 2  2 xy cos 2  0
( –  x   y ) sin 2  2 xy cos 2  0
tan2  p 
 2 xy
.............. (11-4)
x  y
where the subscript ( p ) denotes the principal stress .
Similarly to ascertain the orientation of XnYn corresponding to max . of
min  n , the necessary condition
d n
 0 is applied to eq.(11- 3) yielding :
d
tan 2 s 
x  y
.................. (11-5)
2 xy
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)  0 
tan2  
 2 xy
x  y
2
sin 2   xy cos 2
x  y
which is equivalent to Eq(11- 4) , hence substituting Eq(11-4) into Eq(11-2)
yields the  max &  min . and substituting Eq(11- 5) into eq(11-3) yields  max .
٣١٣
Dr.Hani Aziz Ameen
 max 
min
Principal Stresses
x  y
2
2
 x  y 
   xy 2
 
2


........ (11- 6)
&
2
 max
 x  y 
   xy 2
  
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 Mohr’s Circle For 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
Mohr’s circle .
The procedure for obtaining Mohr’s circle is as follows :1 – Establish a rectangular coordinate system , indicating + . and +  .
Both stress scales must be identical .
2 – Locate the center “ C “ of the circle on the horizontal axis , a distance
1
(  x   y ) from the origin .
2
3 – Locate point “A” by coordinates  x ,  xy , or A (  x ,  xy )
4 – Draw a circle with center at “ C “ and of radius of AC .
From the above raduis can be deduced
٣١٤
Dr.Hani Aziz Ameen
Principal Stresses
2
 x  y 
   xy 2
R = 
2


5 – Draw a line AB through point “ C “
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
x ,  y & z
Fig(11-5 a)
٣١٥
Dr.Hani Aziz Ameen
Principal Stresses
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 =
Fig(11-5 c)
Lateral strain in the x-direction due to  y only =  
x
E
y
Fig(11-5 d)
E
z
E
Thus ,the total strain in the x- direction due to  x ,  y &  z is
Lateral strain in the x-direction due to  z only =  
x 
y
x


  z  ( x   y   z ) / E
E
E
E
Similarly
 y  (  y   x   z ) / E
 z  (  z   x   y ) / E
where  y &  z are the total strain in y & z direction
٣١٦
Dr.Hani Aziz Ameen
Principal Stresses
11-5 Principal Stresses in Terms of Principal Strains
Have from previous sections
 x  ( x   y   z ) / E
 y  (  y   x   z ) / E
In the case of two dimensional stress system , and for an element which is
subjected to  x &  y only , the stress in the z- direction = 0 , i.e.  z  0 .
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 &  y are
the max. and min. principal strains and may be written as 1 &  2
respectively .
 x  1 ,  y   2 ,  z  0
Putting
 x  1 ,  y   2
, z  0
Hence
1
1  ( 1   2 )
E
1
 2  (  2  1 )
E
when solving the above two equations simultaneously , we obtain
1  E( 1   2 ) /(1   2 )
and  2  E(  2  1 ) /(1   2 )
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 .
1
1
1,2   x   y  
*  x   y 2   2
2
2
Since  x and  y each equals to zero ,hence
1,2  0  0   2   
1   &  2   
٣١٧
Dr.Hani Aziz Ameen
Principal Stresses
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.
 n  0  0   sin 2 * 45  
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 ,
1
E

The strain in the direction of 1 due to  2   2
E
The strain in the direction of 1due to 1 
 the total strain in the direction of 1
٣١٨
Dr.Hani Aziz Ameen
Principal Stresses

1
 2
E
E
,  2  
1  



 1 =
E
E

1 =
………………… ( 11-8 )
( 1–  )
E

Similarly  2 =
( 1 –  ) ……………….… ( 11-9 )
E
The total Strain 1 &  2 can be derived in other way :
E 1=
In the direction of  1
oa   oa oa 
1 =
=
–1
oa
oa
.................. (11-10)
Eq(11-8 ) & Eq(11-10) yields:

oa 
( 1–  ) =
–1
E
oa

oa   oa[1  (1  )]
…………….. (11-11)
E
in the direct of  2
2 = – (
ob  ob
) (negative)
ob
٣١٩
Dr.Hani Aziz Ameen
ob  ob
ob 
) = –1+
ob
ob
Eq(11-9) & Eq(11-12) gives
i.e.  2 = – (
 

ob  ob 1  (1   )
 E

ac
tan(45  (  / 2)) 
oa 
ac  ob

Principal Stresses
..................... (11-12)
.......................... (11-13)
ob
tan 45  tan(  / 2) 1  tan(  / 2))
 tan(45  (  / 2)) 

oa 
1  tan 45 tan(  / 2)) 1  tan(  / 2))
for small angle tan(  /2)) =  /2
therefore ,
ob 1  (  / 2)

oa  1  (  / 2)
................ (11-14)
where  is the shearing strain
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)

E

 2(1  )
G
E
2(1  )
٣٢٠
Dr.Hani Aziz Ameen
Principal Stresses
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
PD
= (6*1) / (2*20*103) = 150 MPa
2t
 x  0.5  y =75 MPa
y 
(a)
(b)
n 
x  y
x  y
cos 2    xy sin2 
2
2
75  150 75  150
n 

cos(2*120 ) = 131.25 MPa
2
2
x  y
n 
sin 2   xy cos 2
2
75  150
n 
sin ( 2 * 120 ) = 32.48 MPa
2
+
٣٢١
Dr.Hani Aziz Ameen
Principal Stresses
Example (11-2)
Fig(11-8) shows an element . Find 1 ,  2 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
 x   y  20  40

 10MPa.
the center C =
2
2
the radius R =
 xy
2
 x  y 

 
2


2
2
=
1  OC + R = 10 + 30 = 40 MPa
 2  OC– R = 10 – 30 – 20 MPa
 2 xy
0

0
tan2  p 
 x   y  20  40
2 p  0
p  0
٣٢٢
  20  40 

  30 MPa.
2


Dr.Hani Aziz Ameen
Principal Stresses
( b ) Numerically
x  y 1

1 

 x   y 2  4 2 xy
2
2
 20  40 1
 20  402 =40 MPa

2
2
x  y 1
 20  40 1

2 

 x   y 2  4 2 xy 

2
2
2
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
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 =
Pd
2t
&
 x =
Pd
4t
where d ..... is the internal diameter
The longitudinal stress due to the axial load is given by :=
F
Dt
,
where
D ......... is the mean diameter
٣٢٣
Dr.Hani Aziz Ameen
Principal Stresses
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.
x  y
 max 
80 = (140 - y) / 2
y = -20 N/mm2
2
For max. shear  = 45 o
x  y
x  y
 n 45 o 
cos 2
+
2
2
n )45 = (140–20)/2 + ((140+20)/2)* cos(2*45) =60 N/mm2
٣٢٤
Dr.Hani Aziz Ameen
Principal Stresses
when  =30
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/mm2
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
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°
 45  OQ  60MPa
 30  OD  100MPa
30  C 2 D  69.3 MPa
٣٢٥
Dr.Hani Aziz Ameen
Principal Stresses
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
x  y x  y
n 

cos 2
2
2
x  y
n 
sin 2
2
x  y
x  y
m
let
and n 
2
2
then
 n  m  n cos 2   95=m+n cos 2  …(i)
 n  n sin 2
 65=n sin 2  ……..(ii)
55=m+n cos 2 *45=m
………(iii)
sub.Eq(iii) into Eq(i) yields  45  55  n cos 2 
40 = n cos 2  ……..(iv)
Eq(ii) & Eq(iv) are
 tan 2 
65
 2  58.4
40
n  40 2  65 2  76.3
From the triangle
m = 55
& n=76.3
x  y
x  y
55 
& 76.3 
.
2
2
Solving this two equations give
 x  131.3 N/mm2  y  2103 N/mm2.
The position of xx in relation to  x is shown in Fig(11-10)
٣٢٦
Dr.Hani Aziz Ameen
Principal Stresses
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
Pd 5.5 * 10 6 * 0.075
x 

 20.6MN / m 2
4t
4 * 0.005
Pd 5.5 * 10 6 * 0.075
y 

 41MN / m 2
2t
2 * 0.005
T=F.r
F
T
torque
 xy  

A r.A mean raduis * cross - sectional area
٣٢٧
Dr.Hani Aziz Ameen
 xy
Principal Stresses
1.6 *103

 31.8 MPa
0.04 *  * 0.08 * 0.005



1
( x   y )   x  y ) 2  4 2 xy
2
1
1, 2  (120.6  41.2)  (20.6  41.2) 2  4 * (31.8) 2
2
1  64.3MPa.
 2  2.5MPa.
1, 2 
tan 2 
2 xy
x  y


2 * 31.8
20.6  41.2
2   180  72.4    5356
 max
x  y
20.6  41.2 2
)  (31.8) 2
2
2
 33.4 MPa acting on planes at 45  to the principal planes.
 max  (
) 2   xy 2  (
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)
٣٢٨
Dr.Hani Aziz Ameen
Principal Stresses
Solution
At the bearing M = 15*103*9.81*2= 294.3 kN.m
power 15 * 10 6 * 60
T=

 1.433MN.m
2n / 60
2 * 100
Pv
Engine power = where P is the propulsive force

P
15 * 10 6 * 0.85 * 3600
32 * 103
 10435N
Direct stress due to bending =
M
d 3
32
Direct stress due to end thrust = 
4

P
d
2
294.3 * 103 * 32
 * 0.45

3
 32.9MN / m 2
1.435 * 10 6 * 4
 * 0.45
2
 9.02MN / m 2
The total direct stress  x  32.9  9.02  41.92MPa
Shear stress due to torque
T
1.433 * 10 6 * 16
  3 
 80MN / m 2
3
d
 * 0.45
16
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
1
1, 2  { x   x 2  4 2 xy } 2
2
1
1, 2  {41.42  (41.92) 2  4 * 80 2 }
2
1  103.7 MPa
 2  – 61.8 MPa
2
 max =
2
 x 
 41.92 
2
2

   xy  
  80  80.75MPa
 2 
 2 
٣٢٩
Dr.Hani Aziz Ameen
Principal Stresses
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)
 n  m  n cos2 
 n  n sin2 
x  y
x  y
and n =
where
m=
2
2
90 = m+ n cos2  ……………….. ( i )
30 = n sin 2  …………………… ( ii )
60 = m + n cos2(  + 60 )
1
3
sin 2 )
60 = m – n ( cos 2 
2
2
i.e.
120 = 2m – n cos2   3n sin2 
………… ( iii )
solving Eq.( i ) , Eq( ii ) & Eq( iii ) , yields
m = 87.32 N / mm2
n=30.12
N/mm 2

  42 27
Fig(11-13)
from which
 x  117 .44 N/mm 2
 y  57.14 N/mm2
on the plane of the 60 N/mm2 normal stress
  30.12 sin2 (2427  60 )  -12.68 N/mm2
٣٣٠
Dr.Hani Aziz Ameen
Principal Stresses
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
A
B
C
Inclination to plane A
0°
45°
90°
Stress (N/mm2)
97
133
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 )
 n  m  n * cos 2
x  y
x  y
m
&n 
2
2
97 = m + n cos2 
…… ……. ( i )
133 = m + n cos2 (   45 ) = m – n sin 2  ………….. ( ii )
27 = m + n cos2(   90 ) = m – n cos2  ……………( iii )
Adding Eq( i ) and Eq( iii ) ,
2m = 124  = 62
n* sin 2  = – 17
n* cos 2  = 35
 n  712  352  79.2
  x  62 + 79.2 = 141.2 N / mm2
 y = 62 – 79.2 = – 17.2 N / mm2
٣٣١
Dr.Hani Aziz Ameen
Principal Stresses
 71
= –2.028
35
2  =360 – 63 46  29614
  148.7
Since n has been assumed position , sin2  negative & cos2  positive ,
hence 2  lies in the 4th quadrant .
  n sin 2
when   1847    79.2 sin 246.14  –71 N/mm2
when   139.7    79.2 sin 386.14 = 35 N/mm2
when
  238.7    79.2 sin 476.14  71 N/mm2
 max = 79.2 N / mm2
when
 n = 0  62 + 79.2 cos2   0
from which
  90  19.4
The relative positions of the various planes are as shown in Fig(11-14) .
tan2  =
Fig(11-14)
Example(11-11)
Fig(11-15) shows a point in the structural member , the stresses
( in MPa ) are represented as in Fig. Employ Mohr’s Circle to find
graphically:aThe magnitude and orientation of the principal stresses
bThe magnitude and orientation of the maximum shearing stresses
and associated normal stresses.
٣٣٢
Dr.Hani Aziz Ameen
Principal Stresses
In each case show the results on a properly oriented element.
Fig(11-15)
Solution
Scale 1 cm = 10 MPa
The center of the circle “ C “ is at ((40+80)/2) = 60 MPa on the  - 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
x y
tan2  p 
  x   y 
 2 
2 p  tan 1
30
 56 30
20
 p  2815
٣٣٣
Dr.Hani Aziz Ameen
Principal Stresses
( b ) The max. shearing stresses are given by points D and E , thus
 max =  36.05 MPa .
 x   y 80  40
40  20



tan2  s 
 2 xy
 2 * 30  60 30
s  28.15  45  73.15
s = 163  15
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
 x   y  14  28

Center “ C “ =
2
2
C = 27 MPa
Locate point A ( 28 , 0 )
 R = CR =21 MPa
٣٣٤
Dr.Hani Aziz Ameen
Principal Stresses
 A   7  21 cos 60  17.5
 B  3.5MPa
 A . B  21sin 60
=  18.186MPa
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  n &  n the plane incline 30 on
the direction of loading and  max . numerically & graphically .
Fig(11-7)
٣٣٥
Dr.Hani Aziz Ameen
Principal Stresses
Solution.
( a ) Numerically
P 60 * 103
x  
 70.6MPa
A
850
1
 n   x (1  cos 2)
2
1
 n  ( 70.6 ) ( 1– cos 60 ) = 17.65 MPa
2
1
 n   x sin 2
2
1
 n  ( 70.6 ) sin 60 = 30.6 MPa .
2
the , max Value of  n at   45 °
1
 n ) max  (70.6) sin 90  35.3MPa
2
(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 =  n  17.65 MPa.
The value of Kd =  n  30.6 MPa
٣٣٦
Dr.Hani Aziz Ameen
Principal Stresses
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
٣٣٧
Dr.Hani Aziz Ameen
Principal Stresses
(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
٣٣٨
Dr.Hani Aziz Ameen
Principal Stresses
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.
٣٣٩
Dr.Hani Aziz Ameen
Principal Stresses
Answer
11.1) 159 Mpa , -21.7 Mpa
11.2) –20 N/mm2 , 100 N/mm2 , 69.3 N/mm2
11.3) 57.5 MN/m2 , - 13 MN/m2
11.4) 1  71.8 MN/m 2 ,  2  36.1 MN/m 2
11.5) 80.5 rev/min , 53.63 MPa
11.6) 9.62 , - 21.2 Mpa , 25o 6 & 115 o 6
11.7) 18.1 MPa , 10.78 MPa , 7.9 Mpa , 7.9 Mpa
11.8)  max  0.760 MPa ,  min  4.360 MPa
11.9) ( a ) 113.2 N/mm2 ,   40 o15
( b ) 21.9 N/mm2 ,   130 o15
( c ) 113.2 N/mm2 ,
( d ) 45.6 N/mm2 ,   85o15
( e ) 81.4 N/mm2 ,   106 o 30
11.10)
0.3
٣٤٠