Stress Transformation-chapterproblem.ppt

STRESS TRANSFORMATION
1
General State of stress
Plane stress
(a simplification)
Plane stress
(two dimensional view)
1. PLANE–STRESS TRANSFORMATION
General state of stress at a point is characterized by six independent
normal and shear stress components; x , y , z , xy , yz , and zx
General plane stress at a point is represented by x , y and xy ,
which act on four faces of the element

2
x
y
x
y
xy
x’
y’
x’
y’

x’y’
If we rotate the element of the point in
different orientation, we will have different
values of the stresses
The stresses are now x’ , y’ and x’y’
It is said that the stress components can be
transformed from one orientation of an
element to the element having a different
orientation

3
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

4
y
x
xy
x
y
2. GENERAL EQUATION OF PLANE–STRESS TRANSFORMATION
Sign Convention
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

5
x
y
x
y
x’
y’

The orientation of the inclined plane,
on which the normal and shear stress
components are to be determined, will
be defined using the angle 
The angle  is measured from the
positive x to the positive x’

6
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 A, then the horizontal and vertical
faces of the segment have an area of A sin and A cos,
respectively
(a)
(b)

7
Resulting free-body diagram
The unknown normal and shear stress
components in the inclined plane, x’
and x’y’, can be determined from the
equations of force equilibrium
 Fx’ = 0  Fy’ = 0
' 2 2
2 2
x y x y
x xy
   
   
 
  
cos sin
We get

8
y’ x’y’
x’
Three stress components, x’ , y’ and
x’y’ , oriented along the x’, y’ axes







 2
2
2
2
xy
y
x
y
x
'
x sin
cos 











 2
2
2
2
xy
y
x
y
x
'
y sin
cos 









 2
2
2
xy
y
x
'
y
'
x cos
sin 




9
3. PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
To determine the maximum and minimum normal stress, we must
differentiate equation of x’ with respect to  and set the result equal to
zero. This gives
0
2
2
2
2
d
d
xy
y
x
'
x 



 






cos
)
(2sin
Solving this equation, we obtain the orientation  = p of the
planes of maximum and minimum normal stress
  2
y
x
xy
p






2
tan

10
9-3 PRINCIPAL STRESSES & MAX.IN-PLANE SHEAR STRESSES
  2
y
x
xy
p






2
tan
The solution has two roots; p1 and p2
2p2 = 2p1 +
180o
Based on the equation of tan2p
above, we can construct two shaded
triangles as shown in the figure

11
2
xy
2
y
x
xy
p1
2




 







 

2
sin
2
xy
2
y
x
y
x
p1
2
2





 







 







 

2
cos
sin 2p2 = – sin 2p1
cos 2p2 = – cos 2p1
The equation of the maximum/minimum normal stresses can be
found by substituting the above equations into x’

12
In-plane principal stresses
Principal stresses = Maximum and minimum normal stress
2
xy
2
y
x
y
x
2
,
1
2
2





 







 




13
Depending upon the sign chosen, this result gives the maximum or
minimum in-plane normal stress acting at a point, where 1  2.
2
xy
2
y
x
y
x
2
,
1
2
2





 







 



This particular set of values, 1 and 2, 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 p1 and p2 are substituted
into equation of x’y’, it can be seen that  x’y’ = 0; that is, no shear
stress acts on the principal planes

14
Maximum In-Plane Shear Stress
Similarly, to get the maximum shear stress, we must
differentiate equation of x’y’ with respect to  and set the result
equal to zero. This gives
 
xy
y
x
s




2
2
tan



The solution has two roots; s1 and s2

2s2 = 2s1 + 180o

15
The maximum shear stress can be found by taking the trigonometric
values of sin 2s and cos 2s from the figure and substituting them
into equation of x’y’ . The result is
2
xy
2
y
x
max
plane
in
2



 







 


Substituting the values for sin 2s and cos 2s into equation of x’, we
see that there is also a normal stress on the planes of maximum in-
plane shear stress. We get
2
y
x
avg





(9-7)

16
4. MOHR’S CIRCLE  PLANE STRESS
We rewrite the stress component x’ and x’y’ as follows







 2
2
2
2
xy
y
x
y
x
'
x sin
cos 









 2
2
2
xy
y
x
'
y
'
x cos
sin 



Squaring each equation and adding the equation
together can eliminate the parameter . The result is
2
2
'
y
'
x
2
y
x
'
x R
2

















 
 


 2
xy
2
y
x
2
R 










 


17
2
2
'
y
'
x
2
y
x
'
x R
2

















 
 



Since x , y and xy are known constants, the above equation
can be written in a more compact form as
  2
2
'
y
'
x
2
avg
'
x R


 


2
y
x
avg






18
2
2
2
xy
y
x
R 










 

2
y
x
avg





x
xy


This circle is called MOHR’S CIRCLE
  2
2
'
y
'
x
2
avg
'
x R


 


This equation represents a circle having a radius R and
center on  axis at point C(avg, 0) as shown in the Figure

19
x
y
x
y
xy
Procedure how to draw and use Mohr’s circle
A stress state of a point which all stresses x ,
y and xy are positive (just for example)
CONSTRUCTION OF MOHR’S CIRCLE
• Establish a coordinate system;
 axis


• Plot the center of the Mohr’s circle
C(avg, 0) on s axis
avg = (x + y)/2
• Plot the reference point A(x, xy).
This represents  = 0
C
avg
A
x
xy
• Connect point A with the center C, and
CA becomes the radius of the circle
• Sketch the circle

20
ANALYSIS OF MOHR’S CIRCLE


C
avg
A
x
xy
Principal Stresses 1 and 2
B
1
• Point B: 1
• Point D: 2
D
2
Orientation of principal plane, p1
p1
Maximum In-Plane Shear Stress: max = CE = CF
E
F
max
max
Orientation of maximum in-plane shear stress, s1
s1

21
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

22
EXAMPLE 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.
Internal Loadings
The internal loadings consist of the torque
and the axial load is shown in Fig.(b)
(a)
(a)
(b)

23
EXAMPLE 1
(a)
(b)
Stress Components
 Due to axial load
 Due to torsional load



4
(0.02)
.50)(0.02)
(
2
2
J
c
T

 198.9 kPa
kPa
716.2
2(0.02)
900
2 


A
F


24
The state of stress at point P is defined by
these two stress components
EXAMPLE 1
Principal Stresses:
2
xy
2
y
y
2
,
1
2
2



 







 


We get 1 = 767.8 kPa
2 = – 51.6 kPa
The orientation of the principal plane:
  










 
2
tan
y
xy
1
p



2 = – 29o
p = 14.5O

25
EXAMPLE 2
Stress in Shaft due to Bending Load and Torsion
T
x
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

26
EXAMPLE 2
T
x
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

27
EXAMPLE 2
T
x
z
Stress components at point A
 Due to the torque T
 Due to the bending moment M
A
198.9 kPa



4
(0.02)
.50)(0.02)
(
2
2
J
c
T


318.3 kPa


 4
4 (0.02)
2)
(2.00)(0.0


z
I
c
M
318.3 kPa
198.9 kPa
Stress state at critical point A
x = 318.3 kPa xy = 198.9 kPa

28
318.3 kPa
198.9 kPa Principal stresses
2
2
2
2 xy
x
x
1,2 


 








We get 1 = 413.9 kPa
2 = – 95.6 kPa
The orientation of the principal plane:
  





 
2
tan
x
xy
1


p
2 = 51.33o
25.65o
1
2
EXAMPLE 2
p = 25.65O

29
EXAMPLE 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.
Determine: 1. The stress state of point A.
2. The principal stresses and its orientation

30
EXAMPLE 3
Analysis of the stress components at point A
 Due to comprsv load:
 Due to torsional load: J
c
T
A 

A
F
A' 


 Due to bending load:
z
'
A'
I
c
M



(compressive stress)
Stress state at point A
Shear stress:  = A
Normal stress:  = A’ + A”


SOLVE THIS PROBLEM !!!

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