PRINCIPAL STRESSES ACTING ON MATERIALS

In 2D and 3D

The inclined plane has an area of A/cosq; the stress normal to the plane and shear stress along theplane (in the direction of maximum inclination)are;

SIMPLE AXIAL STRESS – 2D

N = F cosq

T = F sinq

snq = F/A

2coscosn

N F

A Aq

qs q

cossin 2

2

T F

A Aq

q q

The maximum normal stress (s) is F/A which acts on radial planes. The magnitude and direction ofmaximum shear stress is extracted from the differentiation;

The maximum value of shear stress is obtained by putting dq/dq=0;

Note: Maximum shear stress (qmax) acts on a plane with q=45˚ andmaximum normal stress (snqmax) acts on a plane with q=0 ˚.

cos2d F

d A

q qq

max

cos 2 0

45 ( 135 )

2

or

F

Aq

q

q

o o

Problem

A cylindrical rock sample is subjected to an axial compressive force of 5kN. The diameter of the sample is 50 mm. Please determine;a. Normal stress and shear stress on an inclined plane of 30˚. b. Maximum shear stressc. Inclination of planes on which the shear stress is half of maximum shear stress.

Solution

a. Unit area; A=πr2=1.96x10-3 m2

Normal stress; snq = (5 kN/1.96x10-3)cos230 = 1913 kPaShear stress; q= (5 kN/2x1.96x10-3)sin60 = 1105 kPa

b. Maximum Shear stress; qmax= (F/2A) =(5 kN/2x1.96x10-3) = 1275 kPa

c. Maximum Shear stress; 1/2qmax= qmax sin2q; q=15˚ or 75˚

SIMPLE BIAXIAL STRESS – 2D

Consider a rectangular plate (a) of unit thickness with normal principal stresses s1 and s2. The shearstresses along the edges are assumed to be zero. A square element of the plate is shown in 2D (b). Thenormal and shear stresses acting on a plane inclined at an angle direction of the plane on which s1 actsare found by considering forces acting on the triangular element (c).

Unit length along CD = l, normal stress for a plate of unit thickness

Fl=s1l

1 1

1 1

2 2

2 2

2 2

cos

sin

tan

tan sin

tan cos

N l

T l

F l

N l

T l

s q

s q

s q

s q q

s q q

1 2secn l N Nqs q

2 2

1 2cos sinnqs s q s q

Forces in normal stress direction

Forces in shear stress direction;

1 2 / secT T q q

1 2

1( )sin 2

2q s s q

max 1 2

1( )

2q s s

1max 1 2( )

2ifq

s s s

Maximum shear stress on 45˚ plane;

Problem

Solution

1

1 2

0.8

4

MPa

and

s

s s

max 1 2

1( )sin 2

2q s s q

Maximum shear stress is on 45˚ plane;

1

2

1.6

0.4

MPa

and

MPa

s

s

Shear stress is on 60˚ plane;

1 2

1( )sin 2

2q s s q

0.866MPaq

1

2

1.48

0.37

MPa

and

MPa

s

s

2

2

1 cos 2cos

2

1 cos 2sin

2

qq

qq

1 1( ) ( )cos2 sin 2

2 2n z y z y zyqs s s s s q q

2 2

2 21 1( ) ( )

2 2n z y z y zyq qs s s s s

2 2sin 2 cos 2 1q q

2 2 2

1( )

2

1( )

2

z y

z y zy

s

and

r

s s

s s

( 2 2 2

n s rq qs

Which is the equation of a circle with radius “r” and with a center on “-s” plot

MOHR STRESS CIRCLE

The graphical stress relations was discovered by Culmann (1866) and developed by Mohr (1882) based on the equations given below

Biaxial Compression-2DBiaxial stresses are represented by a circle which plots in “+s” space, passing through s1 and s2 on“=0” axis. Centre of circle is on “=0” axis at point “1/2(s1+s2)”.Radius of circle has the magnitudeof “1/2(s1-s2)” which represents “max”

Biaxial Tension-2DThe stress circle extends into both positive and negative “s” space. Center of circle is on “=0” axis at point “1/2(s1+s2)”.Radius is “1/2(s1-s2)=max” which occurs at 45˚ to s1 direction. Normal stress is zero in directions “±q” to the direction of s1;

1 2

1 2

cos2s s

qs s

Biaxial Shear-2D

The stress circle has a radius of “zy” which is opposite to “yz”. Center of the circle is at “s=0; =0”. Principal normal stresses “s1 and s2” are equal but opposite in sign which have magnitudes equal to “zy”. The directions of principal normal stresses are at 45˚ in directions of “zy” and “yz”

General Considerations on Principal Stress Relations

Problem

A plane element is subjected to the stresses given below. Determine the principal stresses and directions by Mohr’s circle.

The principal stresses are represented by points G and H. Since the coordinate of “C” is 40; CD= (402+302)0.5 = 50Minimum principal stress issmin=OG=OG-CG=40-50=-10 Mpa

Maximum principal stress issmax=OH=OC+CH=40+50=90 Mpa

The angle 2qp;tan 2qp =30/40; qp =18.43

STRESS in 3D

In the body of a stressed material, 3D stresses at any point can be represented as if acting on a smallcubical element. The nine stresses in three Cartesian space are in form of a matrix “STRESS TENSOR”