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Equilibrium of Deformable Body
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Review Static Equilibrium
external forces, the Newtons Second Law provides us
six scalar equations of equilibrium
0; 0; 0x y
F F Fz 0; 0; 0
x y zM M M
The sum of all the forces in x, y, and z coordinate
directions are zero
The sum of moments of all the forces about axes x,y,
and z are zero
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Considering Static Equilibrium of the element, we have
* Normal and shear stress components acting on oppositesides of an element must be equal in magnitude and
opposite in direction
* Shear stress components satisfy moment equilibrium
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PRINCIPAL STRESSESState of stress at a point in a material is
completely defined by the stresses acting on
Often it is im ortant to determine the state of stress on a lane
edges are parallel to coordinate directions.
at some angle to the coordinate axes. Mostly, the state of stress
on a plane on which no shear stresses act is important for design
.
A plane, where no shear stresses act is called a principal
plane and the normal stress that acts on such a plane is
called principal stress.
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Mohrs Circle - Its Use and Limitations
coordinate plane, there are no shear stresses and only normal
stress (may be zero or not) is present. That normal stress is one
o e pr nc pa s ress as s own n gure.
You can see that on an plane perpendicular to z-axis only sz ( may
be tensile, compressive or zero) is present. Then remaining two
rinci al stresses can be determined usin biaxial stress field and
Mohr Circle.
Biaxial stress field in xy-plane can be shown as above.
Using stress transformation equation, the normal stress alongx
x
.
be written as.
1 1
cos s n2 2
1sin 2 cos2
2
x x y x y xy
x y x y xy
(A)
will become principal stress if = 0. Using this condition,
the remaining principal stresses can be determined as.x
x y
2
21 1
2 2p x y x y xy
and the orientation of the principal planes with respect to x-
axis is given by
tan2 xy
x y
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The same results can be expressed graphically using Mohrs
Circle. Rearranging the equations (A), we obtain
1 1
cos2 sin 22 2
1sin 2 cos2
x x y x y xy
Squaring the above equations, then adding, gives
2
2 2
2 2
2 2x x y x y x y xy
This is the equation of a circle whose center is at 1
,02
x y
an w ose ra us s g ven y
2
21
2 x y xy
Every point on the circle defines the stress state acting on
planes at any angle q from the original x or y axis.
For the correct construction of Mohrs circle, certain rules are followed and a consistent
handling of positive and negative stress is essential, only if proper orientation of planes isdesired. No such concern is re uired if onl the ma nitudes of the rinci al stresses are
sought.
Although various conventions are in use, we follow the convention given in Hibbler Book.
1.Normal stresses are plotted to scale along the abscissa (horizontal axis) with tensile
stresses considered positive and compressive stresses negative.
2.Shear stresses are plotted along the ordinate (vertical axis) with positive direction
downward to the same scale as used for normal stresses.
A shear stress that would tend to cause counter-clock wise rotation of the stress element in
rotation.
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3.Angle between lines of direction on the Mohr plot are twice the
indicated angle on the physical plane.
The angle 2' on the Mohr circle is measured in the same directionas the angle for the orientation of the plane in physical plane.
According definition, the values of corresponding to the
points D and B (where = 0) are the principal stresses. Theradius of the Mohrs circle gives the maximum in-plane shear
stress
Three -Dimensional Mohrs Plot
As mentioned previously, Mohrs circle can be drawn to determine principal stresses
only if one of the three principal stresses is known.
Since the known principal stress is also a normal stress, it can be plotted on s axis and
circles can be drawn between all the principal stresses as shown.
then the maximum absolute shear stress is equal to radius of largest Mohrs circle.
Absolute maximum shear stress is given as
max min1
2 pmax p
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Static Failure Theories
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2
4 3 3
*( / 2) *( / 2) 32 32*600018108 /
/ 64 *(1.5)x
M d M d Mlb in
I d d
2
4 3 3
( / 2) 16 16*800048288 /
/ 32 1.5zx
Tr T d T lb in
J d d
2
2 2
4 4 4*1000754.5 /3 3 / 4 3* * 1.5 / 4
v
V V
lb inA d
At Point A
2
2 2
1 2, 58184, 40076 /2 2
x z x zxz
lb in
2
max 49130 /lb in
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248288 754 49042 /lb in
At point B
21 22
max
,
49042 /lb in