me16a: chapter four analysis of stresses in two dimensions
TRANSCRIPT
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ME16A: CHAPTER FOUR
ANALYSIS OF STRESSES IN TWO DIMENSIONS
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4.1 DERIVATION OF GENERAL EQUATIONS
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Resolving perpendicular to EC: x 1 x EC = x x BC x 1 x cos
+ y x EB x 1 x sin
+ xy x 1 x EB x cos
+ xy x 1 x BC x sin
Note that EB = EC sin and BC = EC cos
x EC = x x EC cos2 + y x EC sin 2
+ xy x EC x sin cos
+ xy x EC sin cos
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= x cos2 + y sin 2 + 2 xy sin cos
Recall that : cos2 = (1 + cos 2)/2, sin 2 = (1 – cos 2)/2 and
sin 2 = 2 sin cos
= x/2 (1 + cos 2) + y/2 (1 - cos 2) + xy sin 2
x y x yxy2 2
2 2cos sin ………………… (4.1)
Resolving parallel to EC:
x 1 x EC = x x BC x 1 x sin + y x EB x 1 x cos
+ xy x 1 x EB x sin + xy x 1 x BC x cos
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Derivation of General Equation Concluded
x E C = x x E C s i n c o s - y x E C s i n c o s +
x y x E C x s i n 2 - x y x E C c o s 2
= x s i n c o s - y s i n c o s + x y s i n 2 - x y c o s 2
R e c a l l t h a t s i n 2 = 2 s i n c o s a n d c o s 2 = c o s 2 - s i n 2
x yx y2
2 2s i n c o s … … … … … … … . ( 4 . 2 )
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SPECIAL CASES OF PLANE STRESS
T h e g e n e r a l c a s e o f p l a n e s t r e s s r e d u c e s t o s i m p l e r s t a t e s o f s t r e s s u n d e r s p e c i a l c o n d i t i o n s : 4 . 1 . 1 U n i a x i a l S t r e s s : T h i s i s t h e s i t u a t i o n w h e r e a l l t h e s t r e s s e s a c t i n g o n t h e x y
e l e m e n t a r e z e r o e x c e p t f o r t h e n o r m a l s t r e s s x , t h e n t h e e l e m e n t i s i n u n i a x i a l
s t r e s s . T h e c o r r e s p o n d i n g t r a n s f o r m a t i o n e q u a t i o n s , o b t a i n e d b y s e t t i n g y a n d
x y e q u a l t o z e r o i n t h e E q u a t i o n s 4 . 1 a n d 4 . 2 a b o v e :
x x
21 2
22( c o s ) , s i n
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Special Cases of Plane Stress Contd.
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Maximum Shear Stress
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Example
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Solution
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Principal Stresses and Maximum Shear Stresses
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Principal Stresses and Maximum Shear Stresses Contd.
T h e s o l u t i o n o f e q u a t i o n 4 . 4 y i e l d s t w o v a l u e s o f 2 s e p a r a t e d b y 1 8 0 o , i . e . t w o v a l u e s
o f s e p a r a t e d b y 9 0 o . T h u s t h e t w o p r i n c i p a l s t r e s s e s o c c u r o n m u t u a l l y p e r p e n d i c u l a r
p l a n e s t e r m e d p r i n c i p a l p l a n e s ,
S u b s t i t u t i n g i n e q u a t i o n 4 . 1 :
x y x y
2 2( )
( )
x y
x y x y
2 24 + x y
2
42 2
x y
x y x y( )
x y
2( )
( )
x y
x y x y
2
2 22 4 +
2
4
2
2 2
x y
x y x y( )
x y
212
4
4
2 2
2 2
( )
( )
x y x y
x y x y
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Shear Stresses at Principal Planes are Zero
1 o r 2 =
x yx y x y
212
42 2) … … . . ( 4 . 5 )
T h e s e a r e t e r m e d t h e p r i n c ip a l s t r e s s e s o f t h e s y s t e m . B y s u b s t i t u t i o n f o r
f r o m e q u a t i o n 4 . 4 , i n t o t h e s h e a r s t r e s s e x p r e s s i o n ( e q u a t i o n 4 . 2 ) :
x yx y2
2 2s i n c o s … … … … … … … . ( 4 . 2 )
x y
22
42 2
x y
x y x y( ) - x y
( )
( )
x y
x y x y
2 24
x y x y
x y x y
( )
( )
2 24 -
x y x y
x y x y
( )
( )
2 24 = 0
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Principal Planes and Stresses Contd.
Thus at principal planes, = 0. Shear stresses do not occur at the principal planes.
The complex stress system of Figure 4.1 can now be reduced to the equivalent system
of principal stresses shown in Figure 4.2 below.
Figure 4.3: Principal planes and stresses
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Equation For Maximum Shear Stress
From equation 4.3, the maximum shear stress present in the system is given by:
max ( )12x y
=12
42 2 x y xy )
and this occurs on planes at 45o to the principal planes. Note: This result could have been obtained using a similar procedure to that used for determining the principal stresses, i.e. by differentiating expression 4.2, equating to
zero and substituting the resulting expression for
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4.4 PRINCIPAL PLANE INCLINATION IN TERMS OF THE ASSOCIATED PRINCIPAL
STRESS It has been stated in the previous section that expression (4.4), namely
tan( )
22 xy
x y
yields two values of , i.e. the inclination of the two principal planes on which the
principal stresses 1 or 2. It is uncertain, however, which stress acts on which
plane unless eqn. (4.1 ) is used, substituting one value of obtained from eqn. (4.4)
and observing which one of the two principal stresses is obtained. The following
alternative solution is therefore to be preferred.
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PRINCIPAL PLANE INCLINATION CONTD.
Consider once again the equilibrium of a triangular block of material of unit depth (Fig. 4.3); this time EC is a principal plane on which a principal stress acts, and the shear stress is zero (from the property of principal planes).
p
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PRINCIPAL PLANE INCLINATION CONTD.
Resolving forces horizontally,
(,x x BC x 1) + ( xy x EB x 1) = ( p x EC x l) cos
x EC cos + xy x EC sin = p x EC cos
x + xy tan = p
tan
p x
xy
… (4.7)
E
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PRINCIPAL PLANE INCLINATION CONTD.
Thus we have an equation for the inclination of the principal planes in terms of the principal stress. If, therefore, the principal stresses are determined and substituted in the above equation, each will give the corresponding angle of the plane on which it acts and there can then be no confusion.
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PRINCIPAL PLANE INCLINATION CONTD.
The above formula has been derived with two tensile direct stresses and a shear stress system, as shown in the figure; should any of these be reversed in action, then the appropriate minus sign must be inserted in the equation.
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Graphical Solution Using the Mohr’s Stress Circle
4.5. GRAPHICAL SOLUTION-MOHR'S STRESS CIRCLE Consider the complex stress system of Figure below. As stated previously this represents a complete stress system for any condition of applied load in two dimensions. In order to find graphically the direct stress p and shear stress on any plane inclined at to the plane on which x acts, proceed as follows: (1) Label the block ABCD.
(2) Set up axes for direct stress (as abscissa) and shear stress (as ordinate)
(3) Plot the stresses acting on two adjacent faces, e.g. AB and BC, using the following
sign conventions:
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Mohr’s Circle Contd.
Direct stresses: tensile, positive; compressive, negative;
Shear stresses: tending to turn block clockwise, positive; tending to turn block
counterclockwise, negative. This gives two points on the graph which
may then be labeled AB and BC respectively to denote stresses on these planes
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Mohr’s Circle Contd.
Fig. 4.5 Mohr's stress circle.
(4) Join AB and BC.
(5) The point P where this line cuts the a axis is then the centre of Mohr's circle, and
the
line is the diameter; therefore the circle can now be drawn. Every point on the
circumference of the circle then represents a state of stress on some plane
through C.
y
xy
xy
x
A B
CD
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Fig. 4.5 Mohr's stress circle.
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Proof
C o n s i d e r a n y p o i n t Q o n t h e c i r c u m f e r e n c e o f t h e c i r c l e , s u c h t h a t P Q m a k e s a n a n g l e 2 w i t h B C , a n d d r o p a p e r p e n d i c u l a r f r o m Q t o m e e t t h e a a x i s a t N .
C o o r d i n a t e s o f Q :
O N O P P N Rx y 12
2( ) c o s ( )
12
2 2( ) c o s c o s s i n s i n x y R R
R a n d Rx y x yc o s ( ) s i n 12
O N x y x y x y 12
12
2 2( ) ( ) c o s s i n
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Proof Contd.
On inspection this is seen to be eqn. (4.1) for the direct stress on the plane inclined at to BC in the figure for the two-dimensional complex system.
Similarly,
QN sin ( 2 - )
= R sin 2 cos - R cos 2 sin
12
2 2( )sin cos x y xy
Again, on inspection this is seen to be eqn. (4.2) for the shear stress on the plane inclined at to BC.
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Note
Thus the coordinates of Q are the normal and shear stresses on a plane
inclined at to BC in the original stress system.
N.B. - Single angle BCPQ is 2 on Mohr's circle and not , it is evident that angles are doubled on Mohr's circle. This is the only difference, however, as they are
measured in the same direction and from the same plane in both figures (in this case
counterclockwise from
~BC).
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Further Notes on Mohr’s Circle
F u r th e r p o in t s t o n o te a r e :
( 1 ) T h e d ir e c t s t r e s s is a m a x im u m w h e n Q is a t M , i . e . O M is t h e le n g th r e p re s e n t in g
t h e m a x im u m p r in c ip a l s t r e s s 1 a n d 2 1 g iv e s t h e a n g le o f th e p la n e 1 f r o m
B C . S im i la r ly , O L is t h e o th e r p r in c ip a l s t r e s s .
( 2 ) T h e m a x im u m s h e a r s t r e s s is g iv e n b y t h e h ig h e s t p o in t o n th e c ir c le a n d is
r e p r e s e n te d b y th e r a d iu s o f t h e c ir c le . T h is f o l lo w s s in c e s h e a r s t r e s s e s a n d
c o m p le m e n ta r y s h e a r s t r e s s e s h a v e t h e s a m e v a lu e ; t h e r e fo r e t h e c e n t r e o f t h e
c ir c le w il l a lw a y s l ie o n t h e 1 a x is m id w a y b e tw e e n x ya n d .
( 3 ) F ro m th e a b o v e p o in t t h e d ir e c t s t r e s s o n t h e p la n e o f m a x im u m s h e a r m u s t b e
m id w a y b e tw e e n x ya n d .
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Further Notes on Mohr Circle Contd.
(4) The shear stress on the principal planes is zero.
(5) Since the resultant of two stresses at 90° can be found from the parallelogram of
vectors as the diagonal, as shown in Figure below, the resultant stress on the
plane at to BC is given by OQ on Mohr's circle.
Resultant stress r on any plane.
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Preference of Mohr Circle
The graphical method of solution of complex stress problems using Mohr's circle is a very powerful technique since all the information relating to any plane within the stressed element is contained in the single construction.
It thus provides a convenient and rapid means of solution which is less prone to arithmetical errors and is highly recommended.