Lecture 2

Stress analysis:
Principal Stresses & Mohr’s Circle
Bending & Torsion
Thermal Stresses & Strains
 1. Stress Analysis
1.1 Stress Conditions
Fig. 2-1a shows general 3D stress element, while
Fig. 2-1b illustrates a state of biaxial / plane stress.
Suppose the element of Fig. 2-1b is cut by an
oblique plane at an angle  to the x-axis as shown
in Fig. 2-2. Now we are concerned with the stresses
that act upon the oblique plane, σ and τ.

 ø
 1.2 Derivation of Normal and Shear
Stress
Consider now the stress element of Fig. 2-2
subjected to normal stresses σx and σy and
complementary shear stresses τxy= τyx. (2-1)
By summing forces caused by all the stress
components along the normal and tangential
directions, the stresses σ and τ are found to be

 x  y  x  y
  cos 2   xy sin 2 (2 - 2)
2 2

 x  y
 sin 2   xy cos 2 (2 - 3)
2
 1.3 Principal Stresses
Differentiating Eq.(2-2) with respect to  and
setting the result equal to zero gives:

2 xy
tan 2  (2 - 4)
 x  y

Eq.(2-2) defined two values for the angle 2, one

of which defines the max. normal stress σ1 and
the other the min. normal stress σ2.
These stresses are called the principal stresses
and their directions - principal directions. The
angle between the principal directions is 90o.
 Principal Stresses
Similarly we differentiate Eq. (2-3), set the result to
zero, and obtain the expression

 x  y
tan 2   (2 - 5)
2 xy

Eq. (2-5) defines the two values of 2 at which

the shear stress τ reaches an extreme value.

The equations for the principal stresses can be

obtained by substituting the values of angle 2
from Eq. (2-4) into Eq. (2-2).
 Principal Stresses
The two principal stresses are found to be

 x  y  x  y 
2

 1 , 2       xy2 (2 - 8)
2  2 

In a similar manner by substituting the angle 2

from Eq. (2-5) into Eq. (2-3), the two extreme-
value shear stresses are found to be

 x  y 
2

 1 , 2       xy
2
(2 - 9)
 2 
 2. Mohr’s Circle
A graphical method for expressing the relations
between σ1, σ2, and τmax known as Mohr’s circle
diagram, as practiced.

90o
 3. Triaxial Stress State
Fig. 2-1a shows a 3D stress state, or triaxial stress.

To determine the 3 principal stresses & directions

the following cubic equation has to be solved
 3  ( x   y   z ) 2  ( x y   x z   y z   xy2   yz2   zx2 )
 ( x y z  2 xy yz zx   x yz2   y zx2 )   z xy2 )  0 (2 - 10)
 4. Uniformly Distributed Stresses
The assumption of a uniform distribution of stresses
is frequently made in design. These require that:

 The bar is straight and of homogeneous material

 The line of action of the force coincide with the
centroid of the cross section
 The section be taken remote from the ends and from
any discontinuity on the bar

The following equations are used for uniform

tension (compression) and direct shear stress.

F F
 (2 - 12)  (2 - 13)
A A
 5. Elastic Strain
A bar subjected to tensile load becomes longer and
the amount of stretch (elongation), is called strain.
The elongation per unit length of a bar is called
unit strain and is defined as 
 (2 - 14)

The Hooke’s law for the normal stress is given by
  E (2 - 15)
Where E is the modulus of elasticity.
The shear strain  is the change in right angle of a
stress element subjected to pure shear. Here G is
the modulus of rigidity   G (2 - 16)
When a bar is placed in tension there exist not
only axial strain but also a lateral strain.
Poisson found that these strains were proportional
to each other within the range of Hooke’s law
Lateral strain
  (2 - 17)
Axial strain
where  is known as Poisson’s ratio.

The three elastic constant are related to each

other as follows:
E  2G (1   ) (2 - 18)
 6. Stress Strain Relations
There are many experimental techniques which
can be used to measure the strains.
If the relationship between stress and strain is
known, the stress state at a point can be obtained
after the state of strain has been measured.
We define the principal strains as the strains in
the direction of the principal stresses.
From Eq. (2-17) the three principal strains
corresponding to a state of uniaxial stress are
1
1  ,  2  -1 ,  3  -1 (2 - 19)
E
 6.1 Biaxial Stresses
Let us assume that for the case of biaxial stress
state σ1 & σ2 are the non-zero principal stresses &
σ3=0.
Then the principal strains in terms of principal
stresses are given by
 1  2  2 1 1  2
1   , 2   , 3  -  (2 - 20)
E E E E E E

Solving Eqs. 2-20 for σ1 and σ2 gives

E ( 1   2 ) E ( 2  1 )
1  (2 - 21) 2  , (2 - 22)
1  2 1  2
7. Shear & Moments Conventions
8. Straight Members in Flexure
The following idealization is made based on:
• The beam is subjected to pure bending
• The material is isotropic and homogeneous
• The beam is initially straight and the material obeys Hooke’s law
• The beam has an axis of symmetry in the plane of bending

+
 Normal Stresses in Bending

• Fig. 2-12a shows a portion of the beam subjected

to a positive bending moment M, where:

• The x axis is the axis of symmetry of the beam,

• The x axis coincides with the neutral axis,

• The xz plane, which contains the neutral axis is

called the neutral plane,

• The points of the beam coincident with xz plane

have no strain.
 Normal Stresses in Bending
Eq. (2-29) gives the second moment of area,
I   y 2 dA (2 - 29)
while Eq. (2-30) gives the expression for the radius
of the beam curvature after being bended
1 M
 (2 - 30)
 EI
Eq. (2-31) states that the bending stress is directly
proportional to the distance y from the neutral
axis and the bending moment M
My
  (2 - 31)
I
 Normal Stresses in Bending
• Fig. 2-13 shows the stress distribution across
the beam section;
• Compression on top
• Tension at the bottom
• Also it shows the neutral axis

M M
 Normal Stresses in Bending
It is common to designate c  y max , to omit the
negative sign, and to write the equation for the
maximum normal stress in the form
Mc
 (2 - 32)
I
Further substituting I/c=Z called section modulus
in Eq. (2-32), gives another equation for the
maximum normal stress
M
 (2 - 33)
Z
 9. Torsion
Any moment vector that is collinear with an axis of
a mechanical element is called a torque vector.
This vector causes the element to twist about that
axis and hence it is said to be in torsion.

The assumptions used

in the analysis are:
• The bar is acted upon
by a pure torque,
 max
T • The material obeys
Hooke’s law,
r • Any radial line in the
section remains straight
 max
 Torsion
The angle of twist for round bar is given by
T
 (2 - 45)
GJ
Where: T = torque; ℓ = length of the bar; G =
modulus of rigidity; J = polar moment of area.
For a solid round bar, the shear stress is zero at
the centre and maximum on the surface. The
distribution is proportional to the radius ρ
T
 (2 - 48)
J
Designating ρ=r as the radius to the outer surface
gives Tr
 max  (2 - 49)
J
 Torsion
In using Eq. (2-49) it is often necessary to obtain
the torque from the power input or output and
the speed of the rotating shaft (round bar).

The relations are:  nT

H  T  (2 - 50)
30

H
T  9.55 (2 - 51)
n

Where: H = power, watt; T = torque, N-m;

ω = angular velocity, rad/s;
n = angular speed, rev/min.
 Torsion
Fig. 2-22 shows the relation between the torque and the
corresponding shear stresses in a round bar
x
y
 xy

 yx
Gaborone

(cw)

(acw)

(acw)
(cw)
 9.1 Torsion of Noncircular Members
Determination of the torsional stresses in
noncircular members is difficult problem. However,
the following approximate formula is useful for
obtaining the maximum torsional stress in a
rectangular section. T  t 
 max  . 3  1.8  (2 - 52)
2 
t = the shortest dim. .t  
 19. Thermal Stresses & Strains
When the temperature of an unstrained bar is
uniformly increased, the body expands, and the
normal strain is
 x   y   z   T  (2 - 67)

where α is the coefficient of thermal expansion,

ΔT is the temperature change. (Table 2-5)
In this case the body experiences a simple volume
increase with all shear strains being zero.
If a bar is restrained at the ends so as to prevent
the lengthwise expansion, and then is subjected
to a uniform increase in temperature, a
compressive stress will develop in the bar.
 Thermal Stresses & Strains
The stress due to the ends constrain is given by
   E   T E (2 - 68)
If a uniform plate is restrained at the edges the
compressive stress developed due to increase in toC
is  T E
 (2 - 69)
1 
The stresses represented by Eqs. (2-68) and (2-69)
although due to temperature, are not thermal
stresses as they result from the restrained edges.
A thermal stress is one which arises because of the
existence of a temperature gradient in a body.
 Thermal Stresses & Strains
Fig. 2-27 shows the internal stresses within a slab
of infinite dimensions during heating and cooling.
 During cooling, the
max. stress is a surface
tension at the same time
a compressive stress is
set at the centre.
 During heating, the
external surfaces are hot
and tend to expand but
are restrained from the
cooler centre.
SOLVE THE FOLLOWING PROBLEMS

1. Determine the diameter of a solid round shaft, 450 mm long, which

is supported by self-aligning bearings at the ends. Mounted upon the
shaft is a V-belt sheave, carrying a radial load of 1800 N, and a gear,
creating a radial load of 700 N. Both loads are in the same plane and
have the same direction. The bending stress should not exceed 75
MPa (Problem 3-5, Shigley 2003).

(b)

(a)

(c)
2. The countershaft shown in the figure supports two V-pulleys. The
shaft runs at 1100 rpm and the belt tension on the loose side of pulley
A is 15% of the tension on the tight side. The shaft has a uniform
diameter of 25 mm.
a) What torques is transmitted between the pulleys?
b) Calculate the angle of twist of the shaft between the pulleys.
c) Find the maximum torsion stress in the shaft (Problem 2.7, Shigley
1986)
3. A V-belt drive operates on two sheaves having pitch diameters of 250 mm and 300 mm, groove angle of 36 o, and a
center distance of 804 mm (Fig. Q3). The permissible belt load is 850 N and the mass of the belt is 0.525 kg/m. The
driver rotates at 1150 rev/min and 28 kW is to be transmitted to the larger sheave. If the coefficient of friction between
the belt and the driver is 0.2 and between the belt and the driven sheave is 0.13, calculate:

• The angles of contact and load carrying capacities of the sheaves. By using a proper criterion state which sheave
governs the design and explain why? Sketch the belt drive showing the contact angles, the preferred direction of
rotation of the driver, the belt velocity, and the tension forces in the tight and slack side of the belt.

• The force in the slack side of the belt. By making another sketch of the belt drive state, which direction of rotation
of the driver is preferred and explain why?

• The number of the V belts required for this application.

B A 1150
rpm
28 kW
Ø 250 mm ?
Ø 300
mm

804 mm