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2. 2
Section IV
Failures Due to Static
Loading
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3. 3
Talking Points
 Static Loading?
 Failure Theories:
 Maximum-Normal Stress Theory
 Maximum-Shear Stress Theory
 The Distortion-Energy Theory
 Factor of Safety for each Failure
Theory
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4. 4
?Static Loading
 Static Load is a stationary force or moment
acting on a member.
 Stationary means that the force or moment
does not change in magnitude.
 Sometime the load is assumed to be static
when it is known that some variation to be
expected. Such assumptions are made to
simplify the design computations when
variations in load are few or minor in nature.
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5. 5
Failure Theories
i. Maximum-Normal Stress Theory
 This theory states that failure occurs whenever
the largest principal stress equals the strength.
 Suppose the principal stresses are ordered as:
 Then if yielding is the criteria of failure, then failure
occurs whenever:
where:
tensile yield stress
=
yt
S
 If ultimate tensile strength is used (in case of brittle
materials), failure occur whenever:
where:
Principal
Element
1 2 3 s >s >s
yt yc ³ S £ -S 1 3 s or s
compressive yield stress
=
yc
S
ut uc ³ S £ -S 1 3 s or s
tensile ultimate stress
=
compressive ultimate stress
=
ut
S
uc
S
Pure tension
Pure torsion
Note: For pure torsion s1 = t = s3 and s2 = 0. This
means that the part fails in torsion when t = Sy. But
experiments show that a part in torsion will deform
at about 60% of the yield strength. This is one of
the reasons the maximum-normal stress theory is
not recommended.
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6. Failure Theories
ii. Maximum-Shear Stress Theory
 This theory is used only in the ductile materials.
 It states that yielding begins whenever the maximum shear stress becomes equal to
the maximum shear stress in a tension test specimen of the same material when that
specimen begins to yield.
6
2 (for simple tension)
t =
s
max 1
t s s
( ) 2 (for pure torsion)
= -
max 1 3
 The maximum-shear stress theory predicts that failure will occur whenever:
t s s s s
max or
 Shear yield strength is given by:
y
y S
S
= - 1 2
³ - ³ 1 3
2 2
sy y S = 0.5S school.edhole.com

7. a) both volume change
and angular distortion
 For figure (a):
 For unit cube, the work done in any of the principal
(1 2 )[ 2 ( )] (2) 1 2 2 3 1 3
7
Failure Theories
iii. The Distortion-Energy Theory
 This theory is called the shear-energy theory.
 It is used also for ductile materials.
 Like the maximum-shear stress theory, this
theory is employed to define the beginning of
yield. It predicts that yielding will occur
whenever the distortion-energy in a unit volume
equals the distortion energy in the same volume
when uniaxially stressed to the yield strength.
 The distortion energy is obtained by subtracting
equation (5) from equation (2):
= 1
+ ¢
3
n s
¢ = - + - + -
E
 So yielding is predicted to occur when:
y school.edsh¢ ³oS le.com
 Shear yield strength is given by:
b) volume change c) distortion without
volume change
directions is: , where n= 1,2,3
 The total strain energy is: or
U = E s 2
+s +s 2
- n s s +s s +s s
1 3
 The average value savg:
 The remaining stress, (s1-savg), (s2-savg), (s3-savg) will
produce angular distortion, figure (c). Substituting (savg
for s1, s2, s3) in equation (2) gives the amount of
strain energy producing only volume change.
s 2 = s +s +s 3 avg
 Substituting in equation (4)
gives:
1 2 3 s >s >s
2 (1) n n n U =s e
1 2 3 U =U +U +U
[ 2( )] (5)
= 1 - 2
n s +s +s + s s +s s +s s
v 1 6
1 2 2 3 1 3
2
3
2
2
2
E
U
( ) 3 (3) 1 2 3 s = s +s +s avg
(3 2 2 )(1 2 ) (4)
v U = s E - n avg
[( ) ]2
1 2 3
2
2
where, [( ) 2
( ) 2
( ) 2
] 2
1 2
2 3
3 1
2
d
s s s s s s s
U
sy y S = 0.577S

8. 8
Factor of Safety for each
Failure Theory
 For maximum-normal stress theory
n = S n = -
S
= = -
s s
or (for ductile materials)
s y s y
n S n S
1 3
s s
or (for brittle materials)
s u s u
1 3
 For maximum-shear stress theory
( ) 1 3 = s -s s y n S
 For the distortion-energy theory:
= ¢ s y n S
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s
¢ = - + - + -
where, s [( s s ) 2
( s s ) 2
( s s
)2 ] 2
1 2 2 3
3 1

9. 9
Example
A material has a yield strength of 600 MPa. Compute the fator of safety for each of the
failure theories for ductile materials. Use the following stress states:
a)- s1 = 420 MPa: s2 = 420 MPa, s3 = 0
b)- s1 = 420 MPa, s2 = 180 MPa, s3 = 0
c)- s1 = 420 MPa, s2 = 0 MPa, s3 = -180 MPa
d)- s1 = 0 MPa, s2 = -180 MPa, s3 = -420 MPa.
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10. 9
Example
A material has a yield strength of 600 MPa. Compute the fator of safety for each of the
failure theories for ductile materials. Use the following stress states:
a)- s1 = 420 MPa: s2 = 420 MPa, s3 = 0
b)- s1 = 420 MPa, s2 = 180 MPa, s3 = 0
c)- s1 = 420 MPa, s2 = 0 MPa, s3 = -180 MPa
d)- s1 = 0 MPa, s2 = -180 MPa, s3 = -420 MPa.
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