BIAXIAL (PLANE) STRESS SITUATION
APPLYING MAXIMUM SHEAR STRESS THEORY
THE CASE OF TRI-AXIAL STRESS WHEN YIELDING OCCURS
The MAXIMUM SHEAR STRESS theory of failure states:
When Yielding occurs in any material, the maximum shear stress at the point of failure equals or exceeds the maximum shear stress when yielding occurs in the tension test specimen .
MAXIMUM SHEAR STRESS THEORY
BI-AXIAL (PLANE) STRESS SITUATION
Plane stress situation
The plane stress situation is the stress situation in which the stress elements are
 x
,
 y
, and
 xy
, and the stresses on the z-axis are zero,
Principal stresses in biaxial stress situation
In such a plane stress situation, the principal stresses are expressed in terms of the stress elements by the expression

1
=
 x

2
 y
+


 x

2
 y


2
  xy
2
(1)

2
=
 x

2
 y
-


 x
  y
2


2
  xy
2
(2)

3

0 The bi-axial (plane) stress situation (3)
The three principal stresses above describe the bi-axial stress situation
PRINCIPAL STRESSES AND PLANE STRESS ELEMENTS
THE BI-AXIAL STRESS SITUATION
Maximum shear stress at a location of the element
The extreme values of shear stresses

12
,

13
,

23
, in each of the three principal planes are then given by the expressions:

University of Nairobi Engineering design II

12


1
 
2
,

13


1
 
3
,

23


2
 
3
2 2 2
Expressing the principal stresses in the order of magnitude and sign

1


2


3
Then the maximum shear stress is given by

13


1
 
3
2
MAXIMUM SHEAR STRESS IN TERMS OF PLANE STRESS ELEMENTS
This is given by the expression
 max



 x
  y
2


2
  xy
2
PLANE STRESS SITUATION WITH
 y

0
Substituting for
 y

0 into the equation for maximum shear stress yields
 max

  x
2

0
2
  xy
2
=
2 x
2
  xy
2
=

4 x
2
  xy
2
 x
2
 max

4
  xy
2
1
=
2
 x
2 
4
 xy
2
SOLID CIRCULAR SHAFT SUBJECT TO BENDING AND TORSION
COMBINED TORSION AND BENDING OF SHAFT
In the case of combined torsion and bending, the stress elements in the plane stress situation are:
 x
 y
 normal

0 stress due to bending

32 M
 d
3
 xy
 shear stress due to torque

16 T
 d
3
Nyangasi Page 2 of 4 30 January 2005

University of Nairobi Engineering design II
APPLYING MAXIMUM SHEAR STRESS THEORY OF FAILURE
PLANE STRESS SITUATION WITH
 y

0
In this situation, the equation maximum shear stress becomes
 max

 x
2
  xy
2
4
Substituting for
 x
,
 y
 y

0 ,
,
1
=
2
 x
2 
4
 xy
2 and
 xy
in equation for equivalent stress with
Yields
 x
 xy

Normal

Shear stress stress due due to to bending

32 M
 d
3 torque

16 T
 d
3
 max

1
2
 x
2 
4
 xy
2 
1
2
32 M
 d
3
2


4

16 T
 d
3
2
(1)
 max

16
 d
3
M
2 
T
2
(2)
DESIGN OF A TRANSMISSION SHAFT
SELECTION OF MATERIAL FOR PART
Material selected is medium carbon steel, to British Standard specification BS
970:080M040(H&T), whose mechanical properties are:
(a) Tensile yield strength S y

385 Mpa
(b) Ultimate tensile strength S ut

625

775 Mpa
(c) Elongation % =16%
(d) Hardness number=179-229 HB
The material is a ductile material, having an elongation % of 16 %>5%.
Nyangasi Page 3 of 4 30 January 2005

University of Nairobi Engineering design II
EXTERNAL LOAD CARRIED BY TRANSMISSION SHAFT
CONSISTS OF TORQUE LOAD T

32 .
9 N
 m

32900 N
BENDING MOMENT M max

16 .
5 N
 m

16500 N
 mm
 mm
From (2), Maximum shear stress
 max

16
 d
3
M
2 
T
2
But the same maximum shear stress theory predicts that S sy

S
2 y
Where
S sy

Shear strength of the material
S y

Yield strength of the material in tension
Design equation then becomes
 max

16
 d
3
M
2 
T
2
=
S sy f .
s .

2 *
S y f .
s .
OR d
3 
16

M
2 
T
2
*
2 *
S y f .
s .
Substituting for working (design, allowable) stress
 w

S y f .
s .
d
3 
16

M
2 
T
2
*
2 *
S y f .
s .
; OR d
3 
32
 w
M
2 
T
2
Where
Substituting the TORQUE and BENDING MOMENT loads into design equation d
3 
32


16500
 
32900

2
* f .
s .
S y
Substituting for yield strength of chosen material and the factor of safety
Factor of safety =2.5 and Tensile yield strength S y

385 Mpa d
3 
32
 d

13 .
45

16500
 
32900

2 mm .
2 .
5
*
385
Nyangasi Page 4 of 4 30 January 2005