shaft design and theory of failure for UG

shaft design

1. Failure Theories
Sunil Kumar P

2. THEORIES OF FAILURE
 Failure in a ductile material is specified by the
initiation of yielding.
 Failure in a brittle material is specified by fracture.
THEORIES OF FAILURE FOR DUCTILE MATERIALS
Maximum Shear Stress Theory (Tresca Yield Criterion)
The most common cause of yielding of a ductile material such
as steel is slipping, which occurs along the contact planes of
randomly ordered crystals that make up the material.
 This slipping is due to shear stress :
2
minmaxabs
max





3. THEORIES OF FAILURE
The maximum shear stress theory for plane stress can be
expressed for any two in-plane principal stresses as s
1 and s
2 by
the following criteria:





Y2
Y1


s
1 and s
2 have same sign
Y21   s
1 and s
2 have opposite sign

4. THEORIES OF FAILURE
Maximum Distortion Energy Theory (von Mises Yield Criterion)
s
1
s
3
s
2
von Mises Stress for 3-Dimension
   
2
13
2
32
2
21o
 
s
2
s
1
von Mises Stress for 2-Dimension
 
2
221
2
1o  
Yield stress, s
y , occurs when s
o = s
y

5. THEORIES OF FAILURE
Allowable Stress, s
allow & Safety Factor, SF
In designing a component or structure, it is introduced what so-
called Allowable Stress, which is defined as
oallow
SF
Y
O

where

6. THEORIES OF FAILURE
THEORIES OF FAILURE FOR BRITTLE MATERIALS
Failure of a brittle material
in tension
45
o
Failure of a brittle material
in torsion
If the material is subjected to plane stress, we require that
ultimate1
ultimate2

7. Stress in Shafts Due to Axial Load and Torsion
An axial force of 900 N and a torque of
2.50 N.m are applied to the shaft as shown
in the figure. If the shaft has a diameter of
40 mm and the safety factor is 5,
determine the minimum yield stress of the
material used.
Internal Loadings
(a)(a)
(b)
PROBLEM-1
The internal loadings consist of the torque
and the axial load is shown in Fig.(b)

8. (a)
(b)
Maximum stress Components
 Due to torsional load

4
(0.02)
.50)(0.02)(
2
2
J
cT

 198.9 kPa
 Due to axial load
kPa716.2
2(0.02)
900
2

A
F

PROBLEM-1

9. The state of stress at point P is defined by
these two stress components
Principal Stresses:
2
xy
2
yy
2,1
22


 









We get s
1
= 767.8 kPa
s
2 = – 51.6 kPa
The orientation of the principal plane:













2
tan
y
xy1
p


2 = – 29
o
PROBLEM-1

10. von Mises equivalent stress for 2-D
 
2
221
2
1o  
22
o
51.6)(51.6)(767.8)((767.8)  = 794.8 kPa
PROBLEM-1
Yield Stress of the shaft material can be found from:
SF
Y
o


oy
SF )( = (5)(794.8) kPa
= 3974 kPa
= 3.974 MPa

11. Solid shaft has a radius of 0.5 cm and made of steel having yield
stress of 
Y = 360 MPa. Determine if the loadings cause the shaft to
fail according to the maximum-shear-stress theory and the maximum-
distortion-energy theory.
PROBLEM-2

12. State of stress in shaft caused by axial force and torque. Since
maximum shear stress caused by torque occurs in material at outer
surface, we have
 
 
 
MPa165.5kN/cm16.55
cm0.52π
cm0.5cmkN3.25
MPa191kN/cm19.10
cm0.5π
kN15
2
4
2
2




J
Tc
A
P
σ
xy
x

PROBLEM-2

13. Stress components acting on an element of material at pt A. Rather
than use Mohr’s circle, principal stresses are obtained using stress-
transformation eqns 9-5:

MPa6.286
MPa6.95
1.1915.95
5.165
2
0191
2
0191
22
2
1
2
2
2
2
2,1








 















σ
σ
σ
xy
yxyx


PROBLEM-2

14. Maximum-shear-stress theory
Since principal stresses have opposite signs, absolute maximum
shear stress occur in the plane, apply Eqn 10-27,
Thus, shear failure occurs by maximum-shear-stress theory.
 
Fail!,
σσσ
Y21
360382.2
?360286.695.6Is



PROBLEM-2

15. Stress in Shaft due to Bending Load and Torsion
T
x
y
z
A shaft has a diameter of 4 cm. The
cutting section shows in the figure is
subjected to a bending moment of 2
kNm and a torque of 2.5 kNm.
Determine:
1.The critical point of the section
2.The stress state of the critical point.
3.The principal stresses and its orientation.
4.Select the material if SF = 6
PROBLEM-3

16. T
x
y
z
Analysis to identify the critical point
Maximum shear stresses occur at the
peripheral of the section.
 Due to the torque T
 Due to the bending moment M
Maximum tensile stress occurs at the
bottom point (A) of the section.
Conclusion: the bottom point (A) is the critical point
A
PROBLEM-3

17. T
x
y
z
Stress components
 Due to the torque T
 Due to the bending moment M
A
198.9 kPa
)(0.02
.50)(0.02)(
4
2
2
J
cT


318.3 kPa
4
4
(0.02)
2)(2.00)(0.0


z
I
cM
PROBLEM-3

18. 318.3 kPa
198.9 kPa
Stress state at critical point A
s
x = 318.3 kPa
t
xy
= 198.9 kPa
Principal stresses
2
2
22
xy
xx
1,2 

 






We get s
1
= 413.9 kPa
s
2 = – 95.6 kPa
The orientation of the principal plane:








2
tan
x
xy1



p
2 = 51.33
o
25.65
o
s
1
s
2
PROBLEM-3

19. 25.65
o
s
1
s
2
von Mises equivalent stress for 2-D
 
2
221
2
1o  
22
o 95.6)(95.6)(413.9)((413.9) 
= 469 kPa
PROBLEM-3
Yield Stress of the shaft material can be found from:
SF
Y
o


oy
SF )( = (6)(469) kPa
= 2814 kPa
= 2.814MPa

20. PROBLEM-4
Steel pipe has inner diameter of 60 mm and outer diameter of 80 mm. If
it is subjected to a torsional moment of 8 kN·m and a bending moment of
3.5 kN·m, determine if these loadings cause failure as defined by the
maximum-distortion-energy theory. Yield stress for the steel found from
a tension test is 
Y
= 250 MPa.

21. Investigate a pt on pipe that is
subjected to a state of maximum
critical stress.
Torsional and bending moments are
uniform throughout the pipe’s length.
At arbitrary section a-a, loadings
produce the stress distributions
shown.
PROBLEM-4

22. Stress in Shafts Due to Axial Load, Bending and Torsion
A shaft has a diameter of 4 cm. The cutting section shows in the figure
is subjected to a compressive force of 2500 N, a bending moment of 800
Nm and a torque of 1500 Nm.
Determine: 1. The stress state of point A.
2. The principal stresses and its orientation
3. Determine the required yield stress if SF = 4.
PROBLEM-5

23. Analysis of the stress components at point A
 Due to comprsv load:
 Due to torsional load:
J
cT
A
A
F
A'

 Due to bending load:
z
'A'
I
cM

(compressive stress)
Stress state at point A
Shear stress: t = t
A
Normal stress: s = s
A’ + s
A”
t
s
PROBLEM-5