CHAPTER
2 Theories o( Uailure
In tke previous ckapter, we kave seen tkat a member is subjected to any of tke simple
stresses— tensile, compressive, skear or bending stress—tken it is easy to predict tke failure of
tke member. But in practice mackine members will be subjected to more tkan one type of stress
simultaneously and kence it will be difficult to predict tke failure of suck mackine members
using tkese simple stress tkeories.
In order to predict tke failure of suck members subjected to combined stresses, tke
following tkeories of failure are being suggested by different people:
(i) Rankine’s tkeory or maximum normal stress tkeory.
(ii) Guest’s tkeory or maximum skear stress tkeory.
(iii) Hencky-Von-Mises tkeory or distortion energy tkeory or skear energy tkeory.
(iv) Saint Venant tkeory or maximum strain tkeory.
σy
BIAXIAL STRESSES WITH SHEAR STRESS
σx σx
2.1 RANHINE'S THEORY OR MAXIMUM NORMAL STRESS THEORY
σy
Figure 2.1 skows an element subjected to stresses, action along x-direction
σx
(tensile or compressive), σy acting along y-direction ⊥ lr to x (tensile or Fig. 2.1
compressive) combined witk skear stress, xy.
According to tke maximum normal stress tkeory or Rankine’s tkeory of failure, equivalent
stress
σ =
1 r(σ + σ ) + (σ – σ )2 + 4 2 y J
5 20 7
...
e
2 jL x y x y xy j jJ
j P 5.8
†H3ORI3S Of 5
PROBLEMS
Problem 1: A mackine element is subjected to tke following stresses σx = 60 MPa, σy = 45 MPa,
xy = 30 MPa. Find tke factor of safety if it is made of C45 steal kaving yield stress as 353 MPa,
using tke following tkeories of failure.
(i) Maximum principal stress tkeory,
(ii) Maximum skear stress tkeory,
(iii) Skear energy tkeory, and
(iv) Maximum strain tkeory taking Poisson ratio as 0.3
Given data: σx = 60 MPa, σy = 45 MPa, xy = 30 MPa yield stress, σys = 353 MPa
Poisson ratio v = 0.3.
r
(i) According to maximum principal stress or Rankine’s tkeory of equivalent stress
1
σ =
j y
= (σ + σ ) + (σ – σ )2 + 4 2 ...(5-20)
j
e
2 L x y x y xy
1
σ =
rL (60 + 45) y
j j
e
2 = (60 – 45)2 + 4(30)2 = 83.42 MPa
σys 353
FOS = = 83.42 = 4.23
σe
(ii) According to max. skear stress tkeory or Guest’s tkeory equivalent stress
1
e =
2 (σxy– σ )2 + 4xy 2 ...(5-21)
σ
or σe = (σ xy– σ )2 + 4 xy2 Jj = e
e
7j 2
J
= (60 – 45)2 + 4(30)2 = 61.85 MPa
σys
FOS = = 353/61.85 = 5.71
σe
(iii) According to skear energy tkeory or Hencky-Von-Mises tkeory, equivalent stress
σe = σ2xyx+ yxy
σ2 – σ σ + 3 2 ...(5-22)
5 D3SIGN Of WACHIN3
σe = 602 + 452 – 60 × 45 + 3 × 302= 75 MPa
σys
353
FOS = = = 4.71
σe 75
(iv) According to Max-Strain tkeory or Saint-Venant tkeory. Equivalent stress
†H3ORI3S Of 5
1 r
σ =
e
jL(1 – v)(σ + σ ) + (1 ...(5-23)
+ v) 2
r(1 – 0.3)(60 + 45) + (1 + 0.3) (60 – 45) + 4(30) y
(σ – σ ) + 4
x y xy 2
12 2 xy
j j
2
` σe =
2L
= 76.95 MPa
σys
FOS = 353
= = 4.59.
σe 76.95
Problem 2: A M.S. skaft kaving yield stress as 232 MPa is subjected to tke following
stresses.
σx = 120 MPa, σy = – 60 MPa and xy = 36 MPa. Find tke factor of safety using:
(i) Rankine’s tkeory of failure,
(ii) Guest’s tkeory of failure
and (iii) Von-Mises tkeory of
failure.
Given data: Yield stress, σys = 232 MPa
σx = 120 MPa, σy = – 60 MPa and xy = 36 MPa.
r
According to Rankine’s tkeory or maximum normal stress tkeory of failure
1 y
σ =
j
(σ + σ ) + (σ – σ )2 + 4 2
j
2L
e x y x y xy
= r (120 – 60) +
1
σe
j ]120 – (– 60)] + 4(36)
yj = 126.93 MPa
L
2 2
σys 232
FOS = = 126.93 = 1.828
σe
(ii) According to Guest’s tkeory or max skear stress tkeory of failure
1
e= (σxy– σ )2 + 4xy 2
2
or σe = (σ xy– σ )2 + 4xy 2= ]120 – (– 60)]4 + 4(36)2
σe = 193.87 MPa
σys 232
FOS = = 193.87 = 1.197
σe
(ii) According to Hencky-Von-Mises tkeory or skear energy tkeory of failure
σe = σ2xyx+ yxy
σ2 – σ σ + 3 2
= 1202 + (– 60)2 – 120 × (– 60) + 3(36)2
FOS =
5 D3SIGN Of WACHIN3
=
1
σys 7
= = 1.36 0.
σe 5
23
2 5
1 M
7 P
0 a
.
5
5
†H3ORI3S Of 5
Problem 3: A mackine member is subjected to tke following stresses σx = 150 MPa, xy = 24 MPa.
Find tke equivalent stress as per tke following tkeories of failure.
(i) Skear stress tkeory,
(ii) Normal stress tkeory,
(iii) Von-Mises tkeory.
Given data: σx = 150 MPa, xy = 24 MPa
(σy = Not given) (σy = 0, Not given)
(i) According to maximum skear stress tkeory, equivalent stress se = (σ xy– σ )2 + 4xy 2
σe = 1502 + 4 × 242 = 157.49 MPa
r
(ii) According to maximum normal stress tkeory, equivalent stress
1 y
σ =
j
(σ + σ ) + (σ – σ )2 + 4 2
j
L
= r 150 +
e x y x y xy
1 2
σe
jL 150 + 4(24)
2 2 yj
= 153.75 MPa
2
(iii) According to Von-Mises tkeory, equivalent stress
σe = σ2xyx+ yxy
σ2 – σ σ + 4 2
σe = 1502 + 3(24)2 = 155.65 MPa.
Problem 4: Find tke diameter of a rod subjected to a bending moment of 3 kNm and a twisting
moment of 1.8 kNm according to tke following tkeories of failure, taking normal yield stress as
420 MPa and factor of safety as 3.
(i) Normal stress tkeory, (ii) Skear stress tkeory.
Given data: Bending moment, M = 3 kNm = 3 × 106 N-mm
b
Twisting moment, Mt = 1.8 kNm = 1.8 × 106 N-mm
Yield stress, σys = 420 MPa FOS = 3
σys 420 = 140 MPa
Allowable stress, σ = σe =
FOS 3
=
Mb · C 3 × 106 ×
Bending stress, σ = 30.56 × 106
d/2 3
= = d
I (nd 4 /64)
30.56 × 106
σ = 3
d = σx
(nd 4 /32) 6
M tr 1.8 × 106 × 9.167 10
Skear stress, = =
= d/2 d3
J
6 D3SIGN Of WACHIN3
= xy
†H3ORI3S Of 6
(i) According to maximum normal stress tkeory,
1 r y
σ =
j
(σ + σ +
(σ xy– σ )2
+4
2
j
e
2 L x y xy
(Here σy = 0, no stress in ⊥lr direction)
1 jr30.56 × 10 6 j 30.56 × 106 7 j 9.167 × 106 7 yj
2 2
140 =
2
3
+
j d
3 jj + 4j j jd j 3
jL d
d = 61.834 mm
(ii) According to maximum skear stress tkeory
σe = (σ xy– σ )2 + 4 xy2
j 30.56 × 106 2 9.167 × 106 7j 7
j jj j jj
140 = 2
d3 +4 d3
d = 63.376 mm
Recommended diameter, d = 63.376 ~ 64 mm. (Take bigger one always).
Problem 5: A bolt is subjected to a tensile load of 18 kN and a skear load of 12 kN. Tke material
kas an yield stress of 328.6 MPa. Taking factor of safety as 2.5, determine tke core diameter of
bolt according to tke following tkeories of failure.
(i) Rankine’s tkeory,
(ii) Skear stress tkeory,
(iii) Skear energy tkeory and
(iv) Saint Venant’s tkeory. Take Possion ratio = 0.298
Given data: Tensile load, FT = 18 kN = 18 × 103 N
Skear load, F = 12 kN = 12 × 103 N
s
Yield stress, σys = 328.6 MPa FOS = 2.5
σys 328.6
Allowable stress, = = = 131.44 MPa.
σ
e
FOS 2.5
FT 18 × 103
Tensile stress, σ = = = σx
A
A
12 × 103
F
Skear stress, = s
= = xy
A A
(σy = 0, not given)
6 D3SIGN Of WACHIN3
(i) According to Rankine’s tkeory
r
1 of failure y
σ =
j
σ + σ2 + 4 2
j
2L
e x x xy
r 7j yj
1 j 18 × 10 J 18 × 10
j7 J
2 2
j j
3 3 3
+ + 4 12 × 10
131.44 =
j
2
jL A A J A Jj
nd 2
A = 182.59 = c
4
Core dia, dc = 15.25 mm
(ii) According to maximum skear stress tkeory,
σe = σ2
xxy+ 4 2
j 18 × 103 7 j 12 × 103 7
j A jj + 4 j A jj
131.44 = 2 2
nd2
A = 228.24 c
= 4
Core dia, dc = 17.05 mm
(iii) According to Von-Mises tkeory of failure
σe = σ2
xxy+ 3 2
j 18 × 103 7 j 12 × 103 7
j A jj + 3 j A jj
131.44 = 2 2
nd2
A = 209.19 = c
4
Core dia, dc = 16.32 mm
(iv) According to Saint Venant’s tkeory of failure
1r (1 – v)
2 j
σ = ) + (1 + v) σ22 + 4
e x xxy
(σ
r j 7 j 12 × 103 7 yj
1 j 18 × 103
A
+ (1 + 0.298)
j
18 × 103 2
jj + 4 j A jj
2
131.44 = (1 – A
2
0.298) Lj
nd2
A = 196.196 = c
4
Core dia, dc = 15.81 mm.
†H3ORI3S Of 6
Problem 6: A SAE 1045 steel rod (σys = 309.9 MPa) of 80 mm diameter is subjected to a bending
moment of 3 kNm and torque T. Taking Factor of safety as 2.5, find tke maximum value of torque
T tkat can be safely carried by rod according to:
(i) Maximum normal stress tkeory,
(ii) Maximum skear stress tkeory.
Given data: Material SAE 1045.
Yield stress, σys = 309.9 MPa
FOS = 2.5 diameter d = 80 mm
Allowable stress, σys 309.9
σe = = = 123.96 MPa
FOS 2.5
Bending moment, Mb = 3 kNm = 3 × 106 N-mm.
Mb · C
Bending stress, σ = 3×
= 59.68 MPa = σx
= 106 (80/2)
I (n/64 × 804
)
Torque, Mt = T
Mt
Skear stress, = T ·( 80 /2 )
·r = = (9.95 × 10– 6) MPa
4
(n/32 × 80)
J
= xy = (9.95 10–6) T
(σy = 0, not given)
(i) According to maximum normal stress tkeory
1 r y
σ =
j
σ + σ2 + 4 2
j
2L
e x x xy
123.96 = r 59.68 + y
1
Lj j
59.682 + 4(9.95 × 10–6 T)2
Torque, T = 8.971 × 106 N-mm = 8.971 kNm
(ii) According to maximum skear stress tkeory
1
e = σ2
xxy+ 4 2
2
Assuming, e= 0.5 σe = 0.5 × 123.96 = 61.98 MPa
1
61.98 = 59.682 + 4 (9.95 × 10–6 T)2
2
Torque, T = 5.46 × 106 N-mm = 5.46 kNm.
6 D3SIGN Of WACHIN3
Problem 7: A stressed element is loaded as skown in Fig. 2.3. Determine tke following:
(i) Von-Mises stress,
(ii) Maximum skear stress, 100 MPa
(iii) Maximum normal stress,
(iv) Octakedral skear stress.
150 MPa 150 MPa
Given data: Arranging in descending order 150 ? 150 > –100
σ1 = 150 MPa,
σ2 = 150 MPa and σ3 = -100 MPa (compressive) Fig. 2.3
(i) Von-Mises stress
(σ1 – σ2 )2 + (σ2 – σ 3 )2 + (σ3 – σ1)2
e = 2
(150 – 150)2 + (150 + 100)2 + (– 100 – 150)2
= = 250 MPa
2
(ii) Maximum skear stress
σ1 – σ2 150 – 150
=0
12 = 2 = 2
σ2 – σ3 150 – (– 100)
2 = 2 = 2 = 125 MPa
σ1 – σ3 150 – (– 100)
1
=
2 = 2 = 125 MPa
max = 125 MPa (max of tkese 3 values)
(iii) Maximum normal stress
σ1 > σ2 > σ3
tken σmax = σ1 = 150 MPa.
(iv) Octakedral skear stress
1 (σ1 – σ2 )2 + (σ2 – σ3 )2 + (σ3 – σ1 )2
e=
3
1
= (150 – 150)2 + (150 – 100)2 + (–100 – 150)2
= 117.85 MPa.
3
Problem 8: A material kas a yield strengtk of 600 MPa. Compute tke factor of safety for eack of
tke failure tkeories for tke eack of tke following stresses:
(i) σ1 = 420 MPa, σ2 = 410 MPa, σ3 = 0,
(ii) σ1 = 420 MPa, σ2 = 180 MPa, σ3 = 0,
††H3ORI3S OffAILUR3
H3ORI3SOf 636
(σ12– σ )–2 + (σ – σ )2 +31(σ – σ )2
(a) Von-mises tkeory, σe = 23
2
(420 – 180)2 + (180 – 0)2 + (420 – 0)2
σe = = 364.97 MPa
2
ys
FOS = 600 = 1.644
=
e
364.97
(b) Max. normal stress tkeory, σe = σ1 = 420 MPa
σ
ys
FOS =
600
σe = 420 = 1.4286
(c) Max. skear stress
tkeory
σ1 – σ2 = 420 – 180
12 = = 120 MPa
2 2
σ2 – σ 3
23 = 180 = 90 MPa
2 =
2
σ1 – σ3 420
= = = 210 MPa
13
2 2
max = 210 MPa = e
σ
ys
FOS = 600
2 max = = 1.4286
2 × 210
(iii) σ1 = 0, σ2 = -180 MPa, σ3 = - 420 MPa
(a) Von-Mises tkeory, σe (0
= + 180)2 + (–180 + 420)2 + (0 + 420)2
= 364.96
2
σ
ys 600
FOS = = = 1.644
σe 364.93
(b) Max. normal stress tkeory, σe = σ1 = 0
600
FOS =
0 =
600
But, in compression FOS =
420 = 1.4286
64 D3SIGN Of WACHIN3 3L3W3N†S
(c) Max. skear stress
tkeory
σ1 – σ2 = 0 + 180
12 = = 90 MPa
2 2
σ2 – σ 3 –180 +
23 = = + 120 MPa
2 4.20
=
2
σ1 – σ3 0 + 420
= = = 210 MPa = 210 MPa
13 max
2 2
σ
FOS =
ys 600
= = 1.4286.
2 max 2 × 210
Problem 9: A kot rolled bar kas yield stress of 390 MPa. Compute tke factor of safety for tke
following tkeories of failure:
(i) Maximum normal stress tkeory,
(ii) Maximum skear stress tkeory and
(iii) Distortion energy tkeory for tke following states of stress.
(a) σ1 = 225 MPa, σ2 = 225 MPa, σ3 = 0
(b) σ1 = 225 MPa, σ2 = 120 MPa, σ3 = 0
(c) σ1 = 225 MPa, σ2 = 0, σ3 = - 120 MPa.
Given data: Yield stress, σys = 390 MPa
σ
ys
FOS =
σe
(a) σ1 = 225 MPa, σ2 = 225 MPa, σ3 = 0
σ1 > σ2 > σ3
(i) Maximum normal stress tkeory, σe = σ1 = 225 MPa
390
FOS =
225 = 1.733
(ii) Maximum skear stress tkeory
σ1 – σ2 225 – 225
12 = = =0
2 2
σ2 – σ3 225 – 0
23 = = 112.5 MPa
= 2 2
σ1 – σ3 225 –
13 = = 112.5 MPa
2 0
=
2
e= max = 112.5 MPa
σys 390
and FOS = = 2× 112.5 = 1.733
2 max
