Extensiometria

II']

6.

-7­

.']

I-

.

112

-]

THEORY CONNECTED WITH COMPLEX STRESS SYSTEMS

The following diagrams represent (a) the stresses and (b) the forces acting upon an element of material under the action of a two dimensional stress system •

ii]

III


II]
II]

IIJ


_---r-.. . . .

1:


(a) Stress Diagram

_J
"

_J,
'1

I:'

_I

'~I

;I

j


(b) Force Diagram Assuming (b) to be a 'wedge' of material of unit depth and the side AS
to be of unit length:
Resolving along 0e will give:
0e = (Oy case) case + (ox sine) sine
+

t

I

1­

(TCOSe) sine

+

(T sire) case


-8­

0 = 0y (cos 20 + 0 2
oo

1)

+

Ox

(l -

cos 20) + L sin 20
2

• • • • 1

=
Le
will give:

Resolvingalong to

=

(0y cos0) sin0- (0 x
sinG) cos0+ (tsinG) sin0- (Tcos0) cos0
Ox

T0 = 0y sin 20 2

sin 20 + Tsin 20 2


T cos 2 0


• • • • 2


From equation 2 it can be seen that there are values for 0 for which 10
is zero, and the planes on which the shear component is zero are called
'Principal Planes'.
For equation 2:

o =• t

(0 Y -

°x ) sin 2 0 -

Tcos 20T cos 2 e = t (0Y T

°x ) sin 2 0


=t

(0Y - 0 x ) tan 20


This will give two values of 20 differing by 180° and, therefore, values of 0 differing by 90°. This shows that Principal Planes are two planes at right angles to each other.

B
21

A'-----'----- C

(c)

Diagrammatic Representation of Equation 3


-9 ­

From the diagram:

sin

26

=+

2T -",t;::;:(0==-=0:::::;::);:2=+=4=T::;:2

• • • • 4

y

x

and cos 20=

+

0y - Ox j (Oy - 0 x) 2 + 4 T2

• • • • 5

The stresses on the principal planes are normal to these planes and are called principal stresses. From equation 1 and substituting the above values:
• • • • 6

Principal stresses are the maximum and minimum values of normal stress in the system. The sign will denote thetype of stress. i.e. Negative sign - Compressive stress Positive sign - Tensile stress

I I I I
1

B
----I ...... 01

sin e

A

1
02

cos e

j

(d) Force Diagram for an Element Assuming BC and AC are Principal Planes, i. e. the' principal stresses T0 = t (0'2- O'J.) sin 20 Now maximum shear stress when a = 45 0 •
T

i I I I

= 0,

and q and

O2

are
• 7

·.•fa will be seen to occur when sin 20 = 1, i.e.

Therefore the maximum shear stress occurs on planes at 45 0 to the prin ­

-10­

and ~e
1\

=i

(0

2

-

0)
0

• • • • 8
) 2 + 4T2

or (using eqn 6)

Te =

J

(O

X

Y

• • • • 9

Two Dimensional stress System

1

1
I I I
(e) Diagram of Principal Stresses on an Element
~

Strain in direction of

E

E\10 1

Strain in direction of O 2 : c 2

o =-.: _

E
£1

E

and £2 are the values of the principal strains A negative quantity denotes compressive strain. A positive quantity denotes tensile strain.

j

j

These strains can be used to construct a 'Mohr Strain Circle' same way as stresses. In the usual manner, referring to Fig (f): OR is the maximum principal strain.

I I

.... OP is the minimum principal strain at right angles to maximum value.

Q is the centre of the strain circle.

-11­

...
Esfuerzos directos

£

(f)

Representacioo ( f) :

ae. :Ssfuer.zO

en-.U11 'Cireulo Mohr

lli 1 eli a~rama

em

= E • (E I 2- .,) 2

• ( ., 2 -

"Jeos 26
• • • • 12


• • • • 13


•

IIIIi

1

I

-12­

I
I

7.

THEORY ASAPPLIED TO THE TECQUIPMENT THIN CYLINDER

I I
I I I I I I I I

Because this is a thin cylinder, i.e. the ratio of wall thickness to in­ ternal diameter is less than about 1/20, the value of 0H and 0L may be assumed reasonably constant over the area, i.e. throughout the wall thickness, and in all subsequent theory the radial stress, which is small, will be ignored. By Symmetry the two principal...