MODULE 2
THEORIES OF FAILURE (STATIC LOADING)
v¥
Y
v
When loads are acting on the component, there will be some point in the material, which
is critically loaded and this critical condition will lead to failure.
Theory of failure gives the condition for the initiation of failure at a point in the material.
This theory helps us to find out the load or combination of load that will cause failure of
component, thereby helps in design process.
FAILURE ?
A machine component is considered to have failed when it no longer performs its mechanical
design function.
O
Yielding or Elastic failure (ductile material)
O
Fracture(brittle material )
In combined
the material
loadings,
behavior
failure theories are needed
based
on
yielding
and
for representing
fracture
Utility of theories of failure
Y
They help in finding out the load capacity of a component; the load capacity may be
single load or combination of loading.
v¥
When the component is subjected to combined loadings, failure theories are to be
utilised for determining the load capacity of machine component based on the
observations from uni axial tension test.
1. Briefly explain the theories of failure ?plot the region of safety for each
theory
** THEORIES
OF FAILURE
a. Maximum principal stress theory (Rankines Theory)
The failure of machine component subjected to combined action of normal and shear
stresses occurs whenever the maximum principal stress reaches the yield (elastic)
strength or ultimate strength of the material in uniaxial simple tension test.
The dimensions of the component are determined by using a factor of safety.
;
For design,
o
n
|
o, +6
:
r) 4
(
o,-6
Y)
2
+2?
Applicable only for brittle materials. This theory disregards the effects of other
principal stresses and the effects of shear stress on other planes and hence it is not used
for ductile materials.
b) Maximum shear stress theory (Tresca’s theory/Guest’s Theory)
The failure of a machine component subjected to combined action of normal and shear
stresses occurs when the maximum shear stress at any point in the component becomes
equal to the maximum shear stress in the standard specimen of tension test at yield point.
Since the shear stress at yield point in a simple tension test is equal to one-half the
yield stress in tension,
we have
sak
9 sie
Se
=
e2
«. (Eq. 2.25a) 2.8(c)/ Pg 22, DHB
"
Pow
av: -o,/ +4ti, =
or
vio, -o,)° +43,
0, -02 =6,
.»» (Eq. 2.25b) 2.8(c)/ Pg 22, DHB
..- (Eq. 2.25c) 2.8(c Pg 22, DHB
—
Go~,932
5
ioe
— Oy) + Athy
a,
oes
For design,
6, -9;
2 02
esses are o}, 02 and os, where 01
str
al
rm
l
no
pa
ci
in
pr
the
If
e:
Fora 3D cas
thenmre timay is the largest of:
;
a
"3
Oo, -—6
re
or
|
=
G,
=—
Sy
er ee
(Eq. 2.27b) yields...
.
o,
01-93 a
Applicable to ductile materials. The results are on the safer side.
c) Maximum principal strain theory or St Venants Theory
The failure of a machine component subjected to combined action of normal and shear
stresses occurs whenever the maximum principal strain of the component becomes equal
to the maximum strain of the material in uni axial tension.
or
Se = (0; — HO?)
If factor of safety is considered, then
--» (Eq. 2.20) 2.8(b)/ Pg 22, DHB
iji<_——_ i
(a, — p02)
or
=
= (0 — HO2)
where : = Poisson's ratio
This theory is not used in general.
d) Maximum strain energy density theory or Haighs theory
The failure of a machine component subjected to combined action of normal and shear
stresses occurs whenever the strain energy density(strain energy per unit volume) at the
most critically stressed point in the material becomes equal to strain energy per unit
volume of material at yield point.
e)Maximum shear energy theory or Distortion theory or Von mises
theory
The failure of a machine component subjected to combined action of normal and shear
stress occurs whenever the distortion energy density(shear strain energy per unit volume)
of material becomes equal to the distortion energy density (shear strain energy density) of
the material in uni axial simple tension test.
QN:2
A bolt is designed to take up direct tensile load of 30 KN and a shear load of
16 KN with a factor of safety of 4. The yield strength of the material used is
400 MPa. Calculate the size of the bolt based on various theories of failure. Take
p=0.3.
Solution:
F = 30 kN = 30 x 10° N, F, = 16 kN = 16 x 10° N, factor of safety n = 4,
yield of elastic stress, o, = 400 MPa,d =?
e Axialstress,
¢ Shear stress,
op =F
t= -
30210"
(A, = A)
3
_ 382010" wpa
... 1.1(a)/
Pg 2, DHB
.-. 1.1(c)/ Pg2, DHB
According to Maximum normal stress theory
According to Maximum shear stress theory
100
(2s
ll
dad =23.63m™m
According to Maximum principal strain theory
According to Distortion energy /Von mises Theory
... 2.8(d) / Pg 22, DHB
Se = Jor +05 - 0,0, + 3ty
38.20 x 10°
400
oo)
de
-(2se
3
2
+0-—
38.20 x 10
| (ee)
3
2
| |
xOl+
10 3 )
20.37 x e
2
(3.22
)
.. ad =22.80 mm
According to strain energy /Haighs Theory
e. Strain energy theory:
For design
Se
aa
= {c?
=|
Oo,
+03 —2p10,0,
5
+6
r)s
... 2.8(e) / Pg 22, DHB
o,-o
es
2
}
+0 xy
a[ 230s 10") “2s 10")
oy =
47 x 10°
oy =
_. 1.8(c)
(2)
—8820
& (d) / Pg 5, DHB
d = 22.44 mm
QN: 3
A mild steel shaft is subjected to 3500 N-m of bending moment at it critical point
and transmits a torque of 2500 N-m. The shaft is made of steel having yield
strength of 231 MPa. Estimate the size of the shaft based on various theories of
failure and specify the final size. Take FOS = 2 and p = 0.3.
VTU — Dec. 2013/ Jan. 2014 — 14 Marks
Solution: M = 3500 N-m = 3.5 x 10° N-mm, T = 2500 N-m = 2.5 x 10° N-mm, yield
stress, o, = 231 MPa, d=?,n=2,
* Bending stress, op =
M _32M __3.5x10°
Z
nd*
16T
* Shear stress,
= nd
35.65 10°
nd*/32
(16x(2.5x10°))
~
... 1.1(b)/ Pg 2, DHB
d°
12.73=10°
nd?
~
a
... L.d)/ Pg2, DHB
State of stress: Here
65te ,
[
a
;
Yn,
y
xy
P.
= 2275 * 10" spe
da
According to maximum normal stress theory
For design
#1 =| 2855) «
2
2
rae:Sr) 423,
2
231 ~ (868x108 +0),
2
Ww
_ (17.83 x =)
Z
115.5 -(
115.5
+
... 2.8(a)/
Pg 21, DHB
Bent =)
2d
[22 x =)
B
(1273z10°)
Ce
... Eq. (a)
39.73 x 10°
es)
d=70,06 mm
.. Standard shaft diameter = 71 mm
-+ Tb. 3.5(a)/ Pg 57, DHB
According to maximum shear stress theory
b. Maximum shear stress theory:
For design
= = \lo, -9,)° + 4x2,
231__|{ 35.65 «10°
2
da
115.5 -(
-0)°
... 2.8(¢)/ Pg 22, DHB
Tk
12.73
x 10°)*
43.80 x 10")
d°
d =72.38 mm
.. Standard shaft diameter = 80 mm
... Tb. 3.5(a)/ Pg 57, DHB
According to Maximum principal strain theory (St Venants Theory)
For design
Se:
=(1 -w) (2%)
2 |
(
“Ie
Las
02)
«|
[sean
2d°
‘ wah
(
d =21.52 mm
“. Standard shaft diameter = 71 mm
2
=*)
. a, |
_. 2.8(b)/ Pg 22, DHB
+f)
(2 .65 ee
x
-0
115.5 = me
“Y
Jo
o,-o
=
3
{ezxut } |
mai)
... Tb. 3.5(a)/ Pg 57, DHB
According to Distortion energy theory /Von mises Theory
d. Distortion energy theory or von Mises theory:
For design
Ge
=P
_
... 2.8(d)/ Pg 22, DHB
2
\(o2 +03 2 - 9,0, +3t3,)
231 _ (ssesx10
6\2
a
2
A
6
}x0].[3.(2220)
d
d
41.92
x 10°
115.5 =
=
d=71.33 mm
... Tb. 3.5(a)/ Pg 57, DHB
:. Standard shaft diameter = 80 mm
According to Maximum strain energy theory
For design
ts
eres
of
+0, - 26,0,
0,+6,\
ag +(e
a, +0
a°5)
?
+t,
P73 «10°
Te
mt
.. LBlel Pg 22, DHB
ae
.« LB(ehe(d)! Pg 5, DHE
155
———
\\
jx)
2
eae
+|
(407x108)
DP
P
{
10%~F
©. Standard shaft diameter
d = 80 ,
mm
A machine member is statically loaded and has a yield strength of 350 MPa. For
each of the stress state indicated below, find the factor of safety according to:
a. Maximum normal stress theory
shear stress theory
distortion energy theory.
MPa, a2 = 70 MPa
MPa, o, = 35 MPa
iii. o, = 70 MPa, o2 = —70 MPa
iv. o, = 70 MPa, oc, =0 MPa
VTU
Solution: o, = 350 MPa,
n =?
397310") | zs)
p
(=
a
a)/
vo Th. 3.54
Pg 57,
DHB
1 the maximum
i
i
isisd =7238 mm
shaft diameter
Based on above theories,
QN.4
b. Maximum
c. Maximum
i. a, = 70
ii. o; = 70
aso.
;
°, Standard shaft diameter = 80 mm
(407 x1
|
_| 41.1310"
nie
: 1783 10" see
Te
2
— Dec. 2011 — 12 Marks
Case i: ¢, = 70 MPa, 0; = 70 MPa
b. Maximum
shear stress theory:
a. Maximum
normal stress theory:
For design
> ~(S2%)
3 fe.-ojFoae
Fordesgn — %=0)-0;
=0;
2.81)
/ Pg 22, DHB
== 00-7)
-», 2.8(a)
/ Pg 21, DHB
n = (infinity)
Si
n=50
QN:5(KTU 2018)
The stresses acting at a critical point in a component are the following.
o,, = 60MPa, co, = 30MPa, o,, = 20MPa, c,, =40MPa, o,, = 25MPa and
The component
is made
of steel having
the following
material properties.
©,, = 2OMPa
Ultimate strength in tension
=600 MPa, Yield strength in tension =400 MPa, Yield strength in shear =200 MPa and the poisons ratio
=0.3.Determine the factor of safety using all the five static failure theories.
Given,
3D State of stress
* Material : Steel
* o,, = 60MPa =a,
* Oy = 600 Mpa
* o,, = 40MPa =<,
* oy = 400 Mpa
* 6, = 25MPa =a,
**,* 200MPa
* o, = 30MPa =t,
* o, = 20MPa =r,
* o,, = 20MPa = t,,
* Poisson's ratio, 1. = 0.3
* FOS?
MODULE
FATIGUE
2
LOADING
FATIGUE
O
When a material is subjected to repeated stresses (repeated
loads/alternating loads), it fails at a stress far below the yield
point stress. Such type of failure is referred to as fatigue.
Fatigue failure is always brittle and catastrophic in nature with
no visible warning prior to failure .
It is observed that about 80 % of failures of mechanical
components are due to fatigue failure resulting from fluctuating
stresses.
The decreased resistance of the materials
the main characteristics of fatigue failure .
Fatigue
failure
cyclic loading
is defined
as
the
time
Transmission
shafts
,connecting
Suspension
springs, ball bearings
failure.
to cyclic stresses
delayed
fracture
is
under
rods
=,gears _ _,vehicle
are subjected to fatigue
SExtruDesign.com
PN TTTTSSITTITRNT
TN TTTT SITTIN
FATIGUE FAILURE
ENGINEERING
IN VARIOUS
FIELD
Automobiles in Mechanical Engineering
Bridges in Civil Engineering
Aircrafts in Aeronautical Enginnering
Ship hull in Marine Engineering
Pressure vessels in Chemical Engineering
Tractors involving Agricultural Engineering
OF
FACTORS TO BE CONSIDERED
FATIGUE FAILURE
Variation in size of the component
as possible.
Holes ,
avoided.
Proper
notches
stress
and
other
should
stress
deconcentrators
TO AVOID
such
be as gradual
raisers
as
should
fillets
and
notches should be provided wherever necessary.
Components
Provide
should
smooth
be
be protected from corrosion.
finish
on
the
outer
component ,thereby increasing fatige life.
surface
of
the
Materials with high fatigue strength should
be selected.
Residual compressive stresses
increases its fatigue strength
parts
over
the
surface
Types of models for
cyclic stresses
> Fluctuating or alternating stresses
> Repeated stresses
> Reversed stresses
FLUCTUATING
STRESS
Fluctuating stress is the stress which varies from a maximum
minimum value of the same nature (tensile / compressive).
Stress
is
o
A
0
Time
(a) Fluctuating stresses
-1(6
max.
value to a
+0
min
)
REPEATED
STRESS
Repeated/released stress is the stress which varies from zero to a maximum
value (tensile /compressive).
Here
6,
oe=0,
o,i, =0,K=0,A=1
°
«a
=—
2
Omn= 0 R=0
Stress ratio
Stress
+h
A=1
Omax
,_
a,
R =—™
O max
Amplitud
mplitude ratioio
co,
A=—*=
cs.
I1-R
1+R
(b) Repeated stresses