7

Transformacin de Esfuerzos y
Deformaciones

1
Introduction
The most general state of stress at a point may
be represented by 6 components,
x , y , z normal stresses
xy , yz , zx shearing stresses
(Note : xy yx , yz zy , zx xz )

Same state of stress is represented by a

different set of components if axes are rotated.

The chapter is concerned with how the

components of stress are transformed under a
rotation of the coordinate axes.

7- 2
Introduction
Plane Stress - state of stress in which two faces of
the cubic element are free of stress. For the
illustrated example, the state of stress is defined by
x , y , xy and z zx zy 0.

State of plane stress occurs in a thin plate subjected

to forces acting in the midplane of the plate.

State of plane stress also occurs on the free

surface of a structural element or machine
component, i.e., at any point of the surface not
subjected to an external force.

7- 3
Transformation of Plane Stress
Consider the conditions for equilibrium of a
prismatic element with faces perpendicular to
the x, y, and x axes.
Fx 0 xA x A cos cos xy A cos sin
y A sin sin xy A sin cos
Fy 0 xyA x A cos sin xy A cos cos
y A sin cos xy A sin sin

The equations may be rewritten to yield

x y x y
x cos 2 xy sin 2
2 2
x y x y
y cos 2 xy sin 2
2 2
x y
xy sin 2 xy cos 2
2

7- 4
Principal Stresses
The previous equations are combined to
yield parametric equations for a circle,

x ave 2 x2y R 2
where
2
x y x y
ave R xy
2
2 2

Principal stresses occur on the principal

planes of stress with zero shearing stresses.
2
x y x y
max, min xy
2
2 2
2 xy
tan 2 p
x y
Note : defines two angles separated by 90o

7- 5
Maximum Shearing Stress
Maximum shearing stress occurs for x ave

2
x y
max R xy
2
2
x y
tan 2 s
2 xy

Note : defines two angles separated by 90o and

offset from p by 45o
x y
ave
2

7- 6
Example 7.01
SOLUTION:
Find the element orientation for the
principal stresses from
2 xy
tan 2 p
x y
Determine the principal stresses from
2
Fig. 7.13 x y x y
max, min xy
2
For the state of plane stress 2 2
shown, determine (a) the Calculate the maximum shearing stress
principal planes, (b) the with
principal stresses, (c) the 2
maximum shearing stress and x y
max xy
2
the corresponding normal stress. 2
x y

2

7- 7
Example 7.01
SOLUTION:
Find the element orientation for the
principal stresses from
2 xy 2 40
tan 2 p 1.333
x y 50 10
2 p 53.1, 233.1
Fig. 7.13
p 26.6, 116.6
x 50 MPa xy 40 MPa
x 10 MPa Determine the principal stresses from
2
x y x y
max, min xy
2
2 2
20 302 402
max 70 MPa
min 30 MPa
Fig. 7.14

7- 8
Example 7.01
Calculate the maximum shearing stress
with
2
x y
max xy
2
2
302 402
Fig. 7.13 max 50 MPa
x 50 MPa xy 40 MPa s p 45
x 10 MPa s 18.4, 71.6

The corresponding normal stress is

x y 50 10
ave
2 2
20 MPa

Fig. 7.16

7- 9
 Sample Problem 7.1
SOLUTION:
Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
Evaluate the normal and shearing
stresses at H.
Determine the principal planes and
calculate the principal stresses.
A single horizontal force P of 150 lb
magnitude is applied to end D of
lever ABD. Determine (a) the normal
and shearing stresses on an element
at point H having sides parallel to the
x and y axes, (b) the principal planes
and principal stresses at the point H.
7- 10
Sample Problem 7.1
SOLUTION:
Determine an equivalent force-couple
system at the center of the transverse
section passing through H.
P 150 lb
T 150 lb 18 in 2.7 kip in
M x 150 lb 10 in 1.5 kip in

Evaluate the normal and shearing

stresses at H.
y
Mc

1.5 kip in 0.6 in
I 1 0.6 in 4
4

xy
Tc

2.7 kip in 0.6 in
J 1 0.6 in 4
2

x 0 y 8.84 ksi y 7.96 ksi

7- 11
Sample Problem 7.1
Determine the principal planes and
calculate the principal stresses.
2 xy 27.96
tan 2 p 1.8
x y 0 8.84
2 p 61.0,119
p 30.5, 59.5

2
x y x y
max, min xy
2
2 2
2
0 8.84 0 8.84
7.96
2

2 2
max 13.52 ksi
min 4.68 ksi

7- 12
Mohrs Circle for Plane Stress
With the physical significance of Mohrs
circle for plane stress established, it may be
applied with simple geometric considerations.
Critical values are estimated graphically or
calculated.
For a known state of plane stress x , y , xy
plot the points X and Y and construct the
circle centered at C.
2
x y x y
ave R xy
2
2 2

The principal stresses are obtained at A and

B. max, min ave R
2 xy
tan 2 p
x y
The direction of rotation of Ox to Oa
is the same as CX to CA.
7- 13
Mohrs Circle for Plane Stress
With Mohrs circle uniquely defined, the
state of stress at other axes orientations may
be depicted.

For the state of stress at an angle with

respect to the xy axes, construct a new
diameter XY at an angle 2 with respect to
XY.

Normal and shear stresses are obtained

from the coordinates XY.

7- 14
Mohrs Circle for Plane Stress
Mohrs circle for centric axial loading:

P P
x , y xy 0 x y xy
A 2A

Mohrs circle for torsional loading:

Tc Tc
x y 0 xy x y xy 0
J J
7- 15
Example 7.02

Fig. 7.13

For the state of plane stress shown,

(a) construct Mohrs circle, determine
(b) the principal planes, (c) the SOLUTION:
principal stresses, (d) the maximum Construction of Mohrs circle
shearing stress and the corresponding
ave
x y

50 10 20 MPa
normal stress.
2 2
CF 50 20 30 MPa FX 40 MPa
R CX 302 402 50 MPa
7- 16
Example 7.02
Principal planes and stresses
max OA OC CA 20 50
max 70 MPa
min OB OC BC 20 50
min 30 MPa

FX 40
tan 2 p
CP 30
2 p 53.1
p 26.6

7- 17
Example 7.02

Maximum shear stress

s p 45 max R ave
s 71.6 max 50 MPa 20 MPa

7- 18
Sample Problem 7.2

For the state of stress shown,

determine (a) the principal planes
and the principal stresses, (b) the
stress components exerted on the
element obtained by rotating the SOLUTION:
given element counterclockwise Construct Mohrs circle
through 30 degrees. x y 100 60
ave 80 MPa
2 2
R CF 2 FX 2 202 482 52 MPa
7- 19
Sample Problem 7.2

Principal planes and stresses

XF 48 max OA OC CA max OA OC BC
tan 2 p 2.4
CF 20 80 52 80 52
2 p 67.4 max 132 MPa min 28 MPa
p 33.7 clockwise

7- 20
Sample Problem 7.2

180 60 67.4 52.6

Stress components after rotation by
x OK OC KC 80 52 cos 52.6
30o
y OL OC CL 80 52 cos 52.6
Points X and Y on Mohrs circle that
xy KX 52 sin 52.6
correspond to stress components on the
rotated element are obtained by rotating x 48.4 MPa
XY counterclockwise through 2 60 y 111.6 MPa
xy 41.3 MPa
7- 21