ME16A: CHAPTER FOUR

ANALYSIS OF STRESSES IN TWO DIMENSIONS

4.1 DERIVATION OF GENERAL EQUATIONS

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

= 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

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 )

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

Special Cases of Plane Stress Contd.

Maximum Shear Stress

Example

Solution

Principal Stresses and Maximum Shear Stresses

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

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

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

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

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.

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

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

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.

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.

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:

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

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

Fig. 4.5 Mohr's stress circle.

 

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

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.

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).

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 .

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.

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.