1. CHAPTER OBJECTIVES
To show how to transform the stress components that are associated with a particular coordinate system into components associated with a coordinate system having a different orientation.

2. 9-1 PLANE–STRESS TRANSFORMATION
General State of stress
Plane stress
(a simplification)
(two dimensional view)
General state of stress at a point is characterized by six independent normal and shear stress components; sx , sy , sz , txy , tyz , and tzx
General plane stress at a point is represented by sx , sy and txy , which act on four faces of the element

3. 9-1 PLANE–STRESS TRANSFORMATIONsx sy x y
txy
If we rotate the element of the point in different orientation, we will have different values of the stresses
sx’
sy’ x’ y’ q
tx’y’
The stresses are now sx’ , sy’ and tx’y’
It is said that the stress components can be transformed from one orientation of an element to the element having a different orientation

4. 9-1 PLANE–STRESS TRANSFORMATION
Failure of a brittle material
in tension
45o
Failure of a brittle material
in torsion
Failure of a brittle material will occur when the maximum normal stress in the material reaches a limiting value that is equal to the ultimate normal stress

5. 9-2 GENERAL EQUATION OF PLANE–STRESS TRANSFORMATION
Sign Convention
+sy
+sx
+txy x y
A normal or shear stress component is positive provided it acts in the positive coordinate direction on the positive face of the element,
or it acts in the negative coordinate direction on the negative face of the element

6. 9-2 GENERAL EQUATION OF PLANE–STRESS TRANSFORMATIONx y x’ y’ +q
The orientation of the inclined plane, on which the normal and shear stress components are to be determined, will be defined using the angle q
The angle q is measured from the positive x to the positive x’

7. 9-2 GENERAL EQUATION OF PLANE–STRESS TRANSFORMATION
(b)
The element in Fig.(a) is sectioned along the inclined plane and the segment shown in Fig.(b) is isolated.
Assuming the sectioned area is DA, then the horizontal and vertical faces of the segment have an area of DA sinq and DA cosq, respectively

8. 9-2 GENERAL EQUATION OF PLANE–STRESS TRANSFORMATION
Resulting free-body diagram
The unknown normal and shear stress components in the inclined plane, sx’ and tx’y’, can be determined from the equations of force equilibrium
 Fx’ = 0
 Fy’ = 0
We get

9. 9-2 GENERAL EQUATION OF PLANE–STRESS TRANSFORMATION
sy’
tx’y’
sx’
Three stress components, sx’ , sy’ and tx’y’ , oriented along the x’, y’ axes

10. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
To determine the maximum and minimum normal stress, we must differentiate equation of sx’ with respect to q and set the result equal to zero. This gives
Solving this equation, we obtain the orientation q = qp of the planes of maximum and minimum normal stress

11. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
The solution has two roots; qp1 and qp2
2qp2 = 2qp o
Based on the equation of tan2qp above, we can construct two shaded triangles as shown in the figure

12. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
sin 2qp2 = – sin 2qp1
cos 2qp2 = – cos 2qp1
The equation of the maximum/minimum normal stresses can be found by substituting the above equations into sx’

13. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
In-plane principal stresses
Principal stresses = Maximum and minimum normal stress

14. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
Depending upon the sign chosen, this result gives the maximum or minimum in-plane normal stress acting at a point, where s1  s2.
This particular set of values, s1 and s2, are called the in-plane principal stresses, and the corresponding planes on which they are act are called the principal planes of stress
Furthermore, if the trigonometric relations for qp1 and qp2 are substituted into equation of tx’y’, it can be seen that  tx’y’ = 0; that is, no shear stress acts on the principal planes

15. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
Maximum In-Plane Shear Stress
Similarly, to get the maximum shear stress, we must differentiate equation of tx’y’ with respect to q and set the result equal to zero. This gives
The solution has two roots; qs1 and qs2
2qs2 = 2qs o

16. 9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
The maximum shear stress can be found by taking the trigonometric values of sin 2qs and cos 2qs from the figure and substituting them into equation of tx’y’ . The result is
(9-7)
Substituting the values for sin 2qs and cos 2qs into equation of sx’, we see that there is also a normal stress on the planes of maximum in-plane shear stress. We get

17. 9-4 MOHR’S CIRCLE  PLANE STRESS
We rewrite the stress component sx’ and tx’y’ as follows
Squaring each equation and adding the equation together can eliminate the parameter q. The result is

18. 9-4 MOHR’S CIRCLE  PLANE STRESS
Since sx , sy and txy are known constants, the above equation can be written in a more compact form as

19. 9-4 MOHR’S CIRCLE  PLANE STRESSsx
txy s t
This equation represents a circle having a radius R and center on s axis at point C(savg, 0) as shown in the Figure
This circle is called MOHR’S CIRCLE

20. 9-4 MOHR’S CIRCLE  PLANE STRESS
Procedure how to draw and use Mohr’s circle sx sy x y
txy
A stress state of a point which all stresses sx , sy and txy are positive (just for example)
CONSTRUCTION OF MOHR’S CIRCLE
Establish a coordinate system;
s-t axis s t C
savg
Plot the center of the Mohr’s circle
C(savg, 0) on s axis
savg = (sx + sy)/2
Plot the reference point A(sx, txy).
This represents q = 0 A sx
txy
Connect point A with the center C, and
CA becomes the radius of the circle
Sketch the circle

21. 9-4 MOHR’S CIRCLE  PLANE STRESS
ANALYSIS OF MOHR’S CIRCLE B s1 C
savg E F
tmax
Principal Stresses s1 and s2
qs1
Point B: s1 D s2
qp1 A sx
txy
Point D: s2
Orientation of principal plane, qp1
Maximum In-Plane Shear Stress:
tmax = CE = CF
Orientation of maximum in-plane shear stress, qs1

22. 9-5 STRESS IN SHAFT DUE TO COMBINED LOADINGS
Occasionally, circular shafts are subjected to the combined
effects of torsion and axial load, or torsion and bending, or in fact
the combined effects of torsion, axial load, and bending load.
Provided the material remains linear elastic, and is only subjected
to small deformation, and then we can use the principle of
superposition to obtain the resultant stress in the shaft due to the
combined loadings.
The principal stress can then be determined using either
the stress transformation equations or Mohr’s circle

23. EXAMPLE 9-1 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, determine the principal stresses at a point P on its surface.
(a)
(b)
(a)
Internal Loadings
The internal loadings consist of the torque and the axial load is shown in Fig.(b)

24. EXAMPLE 9-1 Stress Components Due to axial load Due to torsional load
(b)
Due to axial load
Due to torsional load
198.9 kPa

25. EXAMPLE 9-1
The state of stress at point P is defined by these two stress components
Principal Stresses:
We get s1 = kPa
s2 = – kPa
FG09_23c.TIF
Notes:
Example 9-12: solution
The orientation of the principal plane:
= – 29o
qp = 14.5O

26. EXAMPLE 9-2 Stress in Shaft due to Bending Load and Torsion
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. T x y z
Determine:
The critical point of the section
The stress state of the critical point.
The principal stresses and its orientation
FG09_23c.TIF
Notes:
Example 9-12: solution

27. EXAMPLE 9-2 Analysis to identify the critical point
Due to the torque T T x y z
Maximum shear stresses occur at the peripheral of the section.
Due to the bending moment M
Maximum tensile stress occurs at the bottom point (A) of the section.
FG09_23c.TIF
Notes:
Example 9-12: solution A
Conclusion: the bottom point (A) is the critical point

28. EXAMPLE 9-2 Stress components at point A T Due to the torque Ty z
Stress components at point A
Due to the torque T
198.9 kPa A
Due to the bending moment M
FG09_23c.TIF
Notes:
Example 9-12: solution
318.3 kPa
318.3 kPa
198.9 kPa
Stress state at critical point A
sx = kPa
txy = kPa

29. EXAMPLE 9-2 198.9 kPa Principal stresses 318.3 kPa
We get s1 = kPa
s2 = – kPa
25.65o s1 s2
The orientation of the principal plane:
FG09_14-04UNP08_09.TIF
Notes:
Problems 9-08/09
= o
qp = 25.65O

30. EXAMPLE 9-3
Stress in Shafts Due to Axial Load, Bending Load 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.
FG09_17c.TIF
Notes:
Procedure for Analysis
Determine: 1. The stress state of point A.
2. The principal stresses and its orientation

31. EXAMPLE 9-3 t s Analysis of the stress components at point A
Due to comprsv load:
Due to torsional load:
Due to bending load:
(compressive stress) t s
FG09_17c.TIF
Notes:
Procedure for Analysis
Stress state at point A
Shear stress: t = tA
Normal stress: s = sA’ + sA”
SOLVE THIS PROBLEM AT HOME !!!

32. FG09_32-09UNP95_96.TIF
Notes:
Problem 9-95/9-96