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When a set of co-planer external forces and moments act on a body,
the stress developed at any point ‘P’ inside the body
can be completely defined by the two dimensional state of stress:
sx = normal stress in X direction,
sy = normal stress in Y direction, and
txy = shear stress which would be equal but opposite in
X (cw) and Y (ccw) directions, respectively.
The 2D stress at point P is described by a box drawn with its faces perpendicular to X
& Y directions, and showing all normal and shear stress vectors (both magnitude and
direction) on each face of the box. This is called the stress element of point P.
Two dimensional state of stress, and the stress element
X
Y
F1 F2
F3 F4
Fn
MM
P
Stress
Element
sxsx X
txycw
Y
sy
sy
txy
txy
ccw
txy
txy t x y
sx
t x y
sy
sy
P sx
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The stress formulae that we have learnt thus far, can determine the 2D stresses
developed inside a part, ONLY ALONG A RECTANGULAR AXIS SYSTEM X -Y, that
is defined by the shape of the part.
For example, X axis for a cantilever beam is parallel to its length,
and Y axis is perpendicular to X.
For a combined bending and axial
loading (F1, F2 etc.) of this cantilever
beam:
the normal and shear stress at a point
P, can be determined using the
formulae, such as,
sx= Mv/I+P/A,
txy
=VQ/(Ib).
Note that, these formulae can only determine stresses parallel to X and Y
axis, and the stress element is aligned with X-Y axis.
The question is, what would be the values of normal and shear stresses at
the same point P, if the stresses are measured along another rectangular
axis system U-V, rotated at an angle f with the X-Y axis system ?
X
Y
F1
F2P
txy
txy
t x y
sx
t x
y
sy
sy
P sx
X
Y f
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f
txy X
Y
txy
t x y
sx
t x y
sy
sy
sx
Xf
F
Knowing the 2D stresses at point P along XY coordinate system,
we want to determine the 2D stresses for the same point P, when measured
along a new coordinate system UV,
which is rotated by an angle f with respect to the XY system.
The Problem is: given sx, sy, txy and f,
can we determine su
, sv
, tuv
?
X
F1 Y
F2PX
Y
f
F2P
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1. We cut the stress
element by an arbitraryplane at an angle f. This
plane is normal to u-axis
txy
sx
t x y
sy
f
sx
txy X
Y
txy
t x y
sx
t x y
sy
sy
f
f
Lsinf
L c o s f
txy(LBsinf)
sx(LBcosf)
t x y (
L B c o s f )
sy(LBsinf)
f
2. To maintain static equilibrium,
let the internal normal and shear
stresses su & tuv, respectively
are developed on the cut plane
3. Let, L be the length
of the cut side. Then
the other two sides are
Lsinf & Lcosf
4. If the thickness of the
element is B, then the force
acting on each face of the
element will be equal to the
stress multiplied by the area
of the face.
THIS IS HOW WE CAN ACHIEVE THAT
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txy(LBsinf)
sx(LBcosf)
t x y
( L B c o
s f )
sy(LBsinf)
f
f
f
f
f
f
Equating forces in u-direction:
suLB = sxLBcos2f + syLBsin
2f + 2txyLBsinfcosf
Or, su = sxcos2f + sysin
2f + 2txysinfcosf ………..(1)
Equating forces in v-direction:
tuvLB = txyLBcos2f - txyLBsin
2f - sxLBsinfcosf+ syLBsinfcosf
Or,t
uv
=
t
xy
cos
2
f
- sin
2
f
) –
s
x
-
s
y
)
sin
f
cos
f
……. 2)
5. Forces
acting on the
faces = force xarea
6. Resolving
each force in u
& v directions
CONTINUING
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Replacing the square terms of trigonometric
functions by double angle terms and rearranging :
Equations 3, 4 & 5 gives us the 2D stress
values, if measured along U-V axis which isat an anglef
from X-Y axis.
Since both sets of stresses refer to the
stress of the same point, the two sets of
stresses are also equivalent.
)5.......(2sin2cos22
,
,
)4(....................2sin2
2cos
cossincossin)sin(cos
)3(..........2sin2cos22
2sin)2cos1(2
)2cos1(2
cossin2sincos
v
yx
yx
22
u
22
u
f t f s s s s
s
f s s
f t
f f s f f s f f t t
f t f
s s s s
f t f s
f s
s
f f t f s f s s
xy
y x y x
xy
xyuv
xy
y x y x
xy
y x
xy y x
that shownbecanit thenaxisvtheto
lar perpendicu planeabyelement stressthecut weif Also
--
-+
-
--
+--
+
-
+
+
+-++
++
txy X
Y
txy
t x y
sx
t x y
sy
sy
f
X
f
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f t f s s s s
s
f s s
f t t
f t f s s s s
s
2sin2cos22
2sin2
2cos
2sin2cos22
v
yx
u
xy
y x y x
xyuv
xy
y x y x
---+
--
+-
++
Mohr’s circle
implements these three equations
by a graphical aid, which simplifies
computation and visualization of thechanges in stress values (su, sv & tuv)
with the rotation angle f of the
measurement axis.
Mohr circle is plotted on a rectangular coordinate system
in which the positive horizontal axis represents positive
(tensile) normal stress (s), and the positive vertical axis
represents the positive (clockwise) shear stress (t).
Thus the plane of the Mohr circle is denoted as s-t
plane.
In this s-t plane, the stresses acting on two faces of the
stress element are plotted.
txy
Y
sx
sy
sx
sytxy
X
txy
sxsx Xcw
txy
Y
sy
sytxy
ccw
s-s
t
-t
For a stress element
Y faces have stress:
(sy,-txy)
x faces have stress:(sx & txy)
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X
1. Start by drawing the original stress element
with its sides parallel to XY axis, and show the
normal and the shear stress vectors on the
element.
2. Draw the s-t rectangular axis and label them.3. On the s-t plane, plot X with normal and
shear stress values of sx and txy, and Y with
values sy and –txy.4. Join X and Y points by a straight line, which
intersects the horizontals
axis at C. C
denotes the average normal stress
savg=(sx+sy)/2 .5. The line CX denotes X axis, and line CY
denotes Y axis in Mohr circle. Name them.
6. Draw the Mohr circle using C as the center,
and XY line as the diameter.
7. To find stress along the new UV axis system,
draw a line UV rotated at an angle 2 from
the XY line. CU line denotes U axis, and CV
denotes V axis.
8. The normal and shear stress values of the
points U and V on the s-t plane denote the
stresses in U and V directions, respectively.
9. This way we can find stresses for an element
rotated at any desired angle f .
Y
sx
txy
sy
sx
sytxy
X
Normal stress
axis (s)
S h e a r s t r
e s s
a x i s ( t )
-s
t x y
Y(sy,-txy)
sxsy
savg(sx+sy)/2
t x y
C
su
t u v
sv
t u v
s
t
-t
DRAWING MOHR CIRCLE
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Y
sxtxy
sy
sx
sy
txy
X
f t f s s s s
s
f s s
f t t
f t f s s s s s
2sin2cos
22
2sin2
2cos
2sin2cos22
v
yx
u
xy
y x y x
xyuv
xy y x y x
--
-+
--
+-
++
X
f t f s s
f t
f s s
f f
f
2sin2cos2
)2sin2cos2
(
)2sin2sin2cos2(cos
)22cos(
xy
y x
xy y x
aaa
a
a
+-
+-
+
-
-s
t x y
S h e a r s t r e s s a x i s ( t )
Y(sy,txy)
sxsy
Normal Stress axis (s)
savg(sx+sy)/2
f s s
f t
f s s
f t
f f
f
2sin2
2cos
)2sin2
2cos(
)2sin2cos2cos2(sin
)22sin(
y x
xy
y x xy
aaa
a
a
--
--
-
-PROOF
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Y(sy,-txy)
sxsy
savg-s
-t
o
txy
txy
t
s
X (sx,txy)
Similarly, if the XY axis line is
rotated by an angle 2 ‘ to make
it vertical, then the shear stress
maximizes and the element will
have normal stress = savg andMaximum shear stress = tmax
In the Mohr circle, for a rotation
of 2 angle, the XY axis line
becomes horizontal. In therotated axis s1-s2, the shear
stress vanishes.
The element will have only
normal stresses s1 & s2, and s1
being the maximum normal
stress. s1 & s2 are called thePrincipal normal stresses. tmax
Principal Normal Stressess
1
s
2
,
and Max Shear Stresstmax
2
x
Y
’
s2 s1
2’
(savg,tmax)
(savg,-tmax)
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2
x y
avg
s s s
+
2
2
2
x y
xy Rs s
t -
+
1 2
2 tan xy
x y
t
s s
-
-
1
2
max
avg
avg
R
R
R
s s
s s
t
+
-
x
Y
f
Y
sxtxy
sy
sx
sy
txy X
Formulea for Principal Normal Stresses & Max Shear Stress
X (sx,txy)
Y(sy,-txy)
sxs2 s1
sy
savg
tmax
2 -s
-t
o
2’ txy
txy
(savg,tmax)
(savg,-tmax)
s
t
Maximum shear
stress element
Principal normal
stress element
2902 -
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Determining su, sv & tuv
Given sx, sy, txy & f
X
Y
sx
txy
sy
sx
sytxy
X
-s
t x y
Y(sy,-txy)
sxsy
savg(sx+sy)/2
t
x y
C
su
t u v
sv
t u v
s
t
-t
2: y xavg C
s s s
+
2
2
2 xy
y x R Radius t
s s +
-
-
-
y x
xy
s s
t
2tan2 1
)22sin( f s s -+ Ravg u
)22( f s s -- Sin Ravg v
)22( f t - Cos Ruv
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Y
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Y
X
5,000 psi
20,000 psi20,000 psi
4,000 psi
4,000 psi
X (20k,5k)
Y(-4k,-5k)
20k
s2 -5k s121k-4k 8k
tmax
22.6 s -s
t
-t
o 67.4 5k
5k
(8k,13k)
(8k,-13k)
For a stress element with
1. Draw the stress element along XY axis.
2. Draw the s-t axes for mohr circle
3. Plot point X for sx=20K, txy=5k
4. Plot point Y for sy= -4K, txy=-5k
5. Draw line XY and show X & Y axes.
6. Draw the circle with XY as the diameter
20 48
2 2
x y
avg
k k k psi
s s s
+ -
o
y x
xy6.22)417.0(tan
420
52tan
2tan2 111
+
- ---
s s
t
Kpsi R 13max t
sx=20,000 psi,
sy= -4000 psi, and
txy= 5000 psi.
Draw the Mohr Circle and, draw two stress elements
properly oriented for (i) the principal normal stresses,
and (ii) max shear stresses element.
oo
o
4.676.2290
2902
-
- kpsi Ravg 211381 ++ s s
kpsi Ravg 51382 --- s s
kpsi R xy y x
135
2
)4(20
2
2
2
2
2
+
--+
- t
s s
This completes the Mohr circle. Next, the stress elements
Y
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Y
X
5,000 psi
20,000 psi20,000 psi
4,000 psi
4,000 psi
X (20k,5k)
Y(-4k,-5k)
20k
s2 -5k s121k-4k 8k
tmax
22.6 s -s
t
-t
o 67.4 5k
5k
(8k,13k)
(8k,-13k)
11.3
x
Y
33.7
The principal normal stress axis
will be rotated CW
Draw the principal stress axis
11.3o
CW from XY axis.Show the principal stresses.
o3.112
6.22
The tmax axis will be
rotated CCW
Draw the tmax stress axis33.7o CCW from XY axis.
Show the the stresses.
o7.332
4.67
PRINCIPAL NORMAL STRESS ELEMENT
STRESS ELEMENT FOR tMAX
That completes the drawing of
the two stress elements
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This ends the
presentationand thanks for watching it
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