Theories of Failure Under Static Load

University of Al-Qadisiya
Collage of Engineering
Mechanical Engineering
By
Mr. HUSAM K. MOHSIN
18/10/2014
1

•
Introduction
Strength is a property or characteristic of a mechanical element.
• This
property results from the material identity, the treatment
and processing incidental to creating its geometry, and the
loading, and it is at the controlling or critical location.
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•
Static load– a stationary load that is gradually applied
having an unchanging magnitude and direction.
•
•
A static load can produce axial tension or compression, a shear
load, a bending load, a torsional load, or any combination of
these.
Failure– a part is permanently distorted and will not
function properly. Thus it has had reliability downgraded.
•
A part has been separated into two or more pieces.
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•
Under any load combination, there is always a combination of
normal and shearing stresses in the material.
•
Generally, mechanical components fail because the applied
stresses exceeds the material’s strength.

Ductile and Brittle Materials
•
A ductile material deforms significantly before fracturing.
•
Ductility is measured by % elongation at the fracture point.
•
Materials with 5% or more elongation are considered ductile.
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•
The limiting strength of ductile materials is the stress at yield
point.
•
A brittle material yields very little before fracturing, the yield
strength is approximately the same as the ultimate strength in
tension.
•
The ultimate strength in compression is much larger than the
ultimate strength in tension.
•
The limiting strength of brittle materials is the ultimate stress.
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Tensile test diagrams of ductile and brittle materials
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F.S.= failure stress/allowable stress
or F.S.= failure load/working load
The allowable stress is the stress value which is used in design
to determine the dimensions of the component. It is considered
as a stress which will not reach or exceed the limiting stress
under normal operating conditions.
For ductile materials
Allowable stress = yield stress/F.S.
For brittle materials
Allowable stress = ultimate stress/F.S.
 F.S ≥ 1
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
•
Static failure theories
Predicting failure in members subjected to uni-axial stress is
both simple and straight-forward. But, predicating the failure
stresses for members subjected to bi-axial or tri-axial stresses is
much more complicated.
•
A large numbers of different theories have been formulated.
The principal theories of failure for a member subjected to biaxial stress are as follows:
I.
Maximum principal (or normal) stress theory (Rankine’s th.)
II. Maximum shear stress theory (Guest’s theory)
III. Maximum principal (or normal) strain theory (Saint’s theory)
IV. Maximum strain energy theory (Haigh’s theory)
V. Maximum distortion energy theory (Hencky&Von Mises th.)
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
Maximum shear stress theory (Guest’s theory)
According to this theory, the failure or yielding of mechanical
component subjected to bi-axial or tri-axial stress occurs when
the maximum shear stress at any point in the component becomes
equal to the shear stress at yield point in a simple tension test.
Mathematically,

Where,



Max

yt
F .S
: Maximum shear stress in an axial stress system.
yt: Shear stress at yield point from simple tension test.
F.S: Factor of safety.
Max
Since,  yt = 0.5 
yt
→

Max

yt
2 F .S
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

For a general state of stress, three principal stresses can
be determined and ordered such that  1   2   3 .



.
2
The maximum shear stress is then  Max
Thus, for a general state of stress, the maximum-shearstress theory predicts yielding when,
1



≥  yt
   
3



Assuming that  A   B ,there are three cases to consider
when using equation above for plane stress:
1
Max
2
3
or
1
3
y
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Case 1:  A   B  0 . For this case,  1   A and  3  0 . it
reduces to a yield condition of,
 
A
y
Case 2:  A  0   B . Here,  1   A and  3   B , let to
   
A
B
y
Case 3: 0   A   B . For this case,  1  0 and
and it gives,

B
 
3
B
,
  y
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
Note ,  y is same S y .
The maximum-shear-stress (MSS) theory yield envelope for plane stress, where
σA and σB are the two nonzero principal stresses.
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Example 1: The load on a bolt consists of an axial pull of 10 kN
together with a transverse shear force of 5 kN. Find the diameter
of bolt required according to Maximum shear stress theory
(MSS). Take permissible tensile stress at elastic limit = 100 MPa.
Solution. Given : Pt1 = 10 kN ; PS = 5 kN ; σt(el) = 100 MPa = 100
N/mm2.
Let
d = Diameter of the bolt in mm.
∴ Cross-sectional area of the bolt,
A

d
2
4
 0.7854 d
2
2
mm
We know that axial tensile stress,


1
P
t1
A

10
0.7854 d
2

12.73
d
2
kN
2
mm
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and transverse shear stress,


P
s
A

5
0.7854 d
2

6.365
d
2
kN
2
mm
We know that maximum shear stress,


Max
1
2

( 1 3)


Max
1
2

( 1)

1
 
2



2
2

4
2
12.73
(
)
2
d
1 6.365
*
2
2
d

2

4
2



, 2  0



2
6.365
 4(
)
2
d
44






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


Max
9
d
kN
2
9000
d
2
N
2
mm
2
mm
According to maximum shear stress theory,

Max


y ( el )
2

9000
2
100

 50
2
d
9000
d  50  180 mm
2
2
 d  13.42mm
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
The distortion-energy theory or (Von Mises theory)
predicts that yielding occurs when the distortion strain
energy per unit volume reaches or exceeds the
distortion strain energy per unit volume for yield in
simple tension or compression of the same material.

The total strain energy per unit volume (u) is the
distortive strain energy + the volumetric strain energy
U = 1/2[σ1ε1 + σ2ε2 + σ3ε3]
U = 1/2E [σ1^2+ σ2^2+ σ3^2− 2ν(σ1σ2+ σ2σ3+ σ3σ1)]
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
u
: the distortive strain energy (pure angular distortion
of element, that is no volume change).
d


u
1d


2d


3d
0
: the volumetric strain energy (pure volume change of
element, that is no angular distortion).
v
 

 
3
1
2
3
av
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Since, 

av
  
1
2
3
3
∴


Note that the distortion energy is zero if  1   2   3 .
For the simple tensile test at yield,  1  S y & 2   3  0 ,
the distortion energy is
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So for the general state of stress is given, yield is predicted if,

If we had a simple case of tension σ , then yield would occur
when   s y . Thus, the left of above equation can be thought
of as a single, equivalent, or effective stress for the entire
general state of stress given by 1 , 2 and 3 .

This effective stress is usually called the Von Mises stress (σˊ ).
for yield, can be written as
,
→
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For plane stress, the von Mises stress can be represented
by the principal stresses  A , B , and zero.
∴
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Using xyz components of three-dimensional stress, the
von Mises stress can be written as
and for plane stress,
Considering the factor of safety (F.S), we get
'
Sy
′


F .S
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
Example 2: A hot-rolled steel has a yield strength
s yt  s yc  100 kpsi and a true strain at fracture of ε = 0.55.
Estimate the factor of safety (F.S) for the following
principal stress states:
(a)  x  70Kpsi, y  70Kpsi, xy  0kpsi
(b)  x  60Kpsi, y  40Kpsi, xy  15kpsi
(c)  x  0Kpsi, y  40Kpsi, xy  45kpsi
(d) x  40Kpsi, y  60Kpsi, xy  15kpsi
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
Solution:
Since ε > 0.05 and Syt and Syc are equal, the material is
ductile and both the distortion-energy (DE) theory and
maximum-shear-stress (MSS) theory apply. Both will
be used for comparison. Note that cases a to d are plane
stress states.
(a) Since there is no shear stress on this stress element,
the normal stresses are equal to the principal stresses.
The ordered principal stresses are σA = σ1 = 70, σB = σ2
= 70, σ3 = 0 kpsi.
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DE
σ′ = [702 − 70(70) + 702]1/2
= 70 kpsi
F.S = Sy/σ′
= 100/70= 1.43
MSS
Noting that the two nonzero principal stresses are equal,
τmax will be from the largest Mohr’s circle, which will
incorporate the third principal stress at zero.
τmax = (σ1 − σ3)/2
= (70 − 0)/2
= 35 kpsi
F.S = (Sy/2τmax)
= 100/70= 1.43
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(b) the nonzero principal stresses are
σA, σB = (60 + 40)/2 ± [((60 − 40)/2)2+ (−15)2 ]0.5
= 68.0, 32.0 kpsi
DE
σ′ = (682 − 68(32) + 322)1/2
= 59.0 kpsi
F.S = Sy/σ′
= 100/59.0
= 1.70
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MSS
Noting that the two nonzero principal stresses are both
positive, τmax will be from the largest Mohr’s circle
which will incorporate the third principle stress at zero.
τmax = (σ1 − σ3)/2
= (68.0 − 0)/2
= 34.0 kpsi
F.S = Sy/2τmax
= 100/68
= 1.47
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(c) This time, we will obtain the factors of safety directly
from the xy components of stress.
DE
σ′= [σx2− σxσy + σy2+ 3τxy2]1/2
= [(40)2 + 3(45)2]1/2
= 87.6 kpsi
F.S = Sy/σ′
= 100/87.6
= 1.14
NOTE. For comparison purposes later in this problem, the nonzero
principal stresses can be obtained to be 70.0 kpsi and −30 kpsi.
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MSS
Taking care to note from a quick sketch of Mohr’s
circle that one nonzero principal stress will be positive
while the other one will be negative, τmax can be
obtained from the extreme-value shear stress without
finding the principal stresses.
τmax=[(σx − σy/2)2+ τxy2]0.5
=[(0 − 40/2)2+ 452]0.5
= 49.2 kpsi
F.s = Sy/2τmax
= 100/98.4
= 1.02
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(d) the nonzero principal stresses are
σA, σB = (-40 + -60)/2 ± [((-40 +60)/2)2+ (15)2 ]0.5
= −32.0,−68.0 kpsi
The ordered principal stresses are σ1 = 0, σA = σ2 =−32.0,
σB = σ3 = −68.0 kpsi.
DE
σ′= [(−32)2 − (−32)(−68) + (−68)2]1/2
= 59.0 kpsi
n = Sy/σ′
= 10059.0
= 1.70
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MSS
τmax = (σ1 − σ3)/2
= (0 +68.0)/2
= 34.0 kpsi
F.S = Sy/2τmax
= 100/68.0
= 1.47
A tabular summary of the factors of safety is included for comparisons.
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
Since the MSS theory is on or within the boundary of
the DE theory, it will always predict a factor of safety
equal to or less than the DE theory, as can be seen in
the table.

For each case, the coordinates and load lines in the σA,
σB plane are shown in figure below.

Note that the load line for case (a) is the only plane
stress case given in which the two theories agree, thus
giving the same factor of safety.
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
Maximum principal (or normal) stress theory

The maximum-normal-stress (MNS) theory states that failure
occurs whenever one of the three principal stresses equals or
exceeds the strength.

Again we arrange the principal stresses for a general stress
state in the ordered form  1   2   3 . This theory then
predicts that failure occurs whenever
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


where S ut and S uc are the ultimate tensile and compressive
strengths, respectively, given as positive quantities.
For plane stress, with  A   B , the equations can be written as
which is plotted in below figure.
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
As before, the failure criteria equations can be converted
to design equations. We can consider two sets of equations
where  A   B as
A
S
ut
F .S
or
S uc


 B F.S
Example 3:The load on a bolt consists of an axial pull of 10
kN together with a transverse shear force of 5 kN. Find the
diameter of bolt required according to Maximum principal
stress (MPS) theory. Take permissible tensile stress at
elastic limit equals 100 MPa and poisson’s ratio equals
0.3.
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Solution:
d = Diameter of the bolt in mm.
∴ Cross-sectional area of the bolt,
A

d
4
2
 0.7854 d
2
2
mm
We know that axial tensile stress,
10
12.73
P



kN mm
 A 0.7854
d
d
and transverse shear stress,
5
6.365
P



kN mm
 A 0.7854
d
d
We know that maximum principal stress,
2
t1
2
1
2
2
s
2
2
σt1 = (σx + σy )/2 + [((σx - σy )/2)2+ (τ)2 ]0.5
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= (12.73/d2+ 0)/2 + [((12.73/d2 - 0)/2)2+ (6.365/d2 )2 ]0.5
= (6.365/d2)[1+0.5(4+4)0.5]
σt1 = (15.365/d2) kN/mm2
According to maximum principal stress theory,
σt1 = σt(el) = (15.365/d2) = 100
→ d2 = 153.65 mm2
∴ d= 12.4 mm
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Mathematical analysis and
experimental measurement
show that in a loaded structural
member, near changes in the
section, distributions of stress
occur in which the peak stress
reaches much larger magnitudes
than does the average stress
over the section. This increase
in peak stress near holes, grooves,
notches, sharp corners, cracks,
and other changes in section is called stress concentration.
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
A theoretical, or geometric, stress-concentration factor Kt or
Kts is used to relate the actual maximum stress at the
discontinuity to the nominal stress. The factors are defined by
the equations

Max

Kt




Max

K ts


where Kt is used for normal stresses and Kts for shear stresses.
The nominal stress σ0 or τ0 is the stress calculated by using
the elementary stress equations and the net area, or net cross
section.
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
The stress-concentration factor depends for its value only on the
geometry of the part.

Most stress-concentration factors are found by using experimental
techniques.

In static loading, stress-concentration factors are applied as
follows:
In ductile materials (εf ≥ 0.05), the stress-concentration factor is
not usually applied to predict the critical stress.
In brittle materials (εf < 0.05), the geometric stress-concentration
factor is applied to the nominal stress before comparing it with
strength.


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
Chart of Thin plate in tension or simple compression with a
transverse central hole. The net tensile force is F = σwt.
Where,
t is the thickness of the plate.
The nominal stress is given
by
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Example 4: The 2-mm-thick bar shown in figure below
is loaded axially with a constant force of 10 kN. The
bar material has been heat treated and quenched to raise
its strength, but as a consequence it has lost most of its
ductility. It is desired to drill a hole through the center
of the 40-mm face of the plate to allow a cable to pass
through it. A 4-mm hole is sufficient for the cable to fit,
but an 8-mm drill is readily available. Will a crack be
more likely to initiate at the larger hole or the smaller
hole?
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Solution:
A. For a 4-mm hole,


F
F
10000
 

 139Mpa
A ( w  d )t (40  4)2
The theoretical stress concentration factor, from
previous chart, with d/w= 4/40= 0.1, is Kt= 2.7. The
maximum stress is

Max
 K t    2.7(139)  380Mpa
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A. Similarly, for an 8-mm hole,


F
F
10000
 

 156Mpa
A ( w  d )t (40  8)2
The theoretical stress concentration factor, from
previous chart, with d/w= 8/40= 0.2, is Kt= 2.5. The
maximum stress is

Max
 K t    2.5(156)  390Mpa
∴ The crack will most likely occur with the 8-mm hole
and next likely would be the 4-mm hole.
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