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7/27/2019 Che 159 - Stress Equilibrium
1/22
ChE 159 Particle
Technology
MSU-IIT DChE&T
MECHANICAL PROPERTIES
OF BULK SOLIDS
7/27/2019 Che 159 - Stress Equilibrium
2/22
ChE 159 Particle
Technology
MSU-IIT DChE&T
STRESS EQUILIBRIUM
IN BULK SOLIDS
Reporter: Ruel B. Cedeno
7/27/2019 Che 159 - Stress Equilibrium
3/22
ChE 159 Particle
Technology
MSU-IIT DChE&T
Learning Objectives
After completing this module, you should be able to
do the following:
Define and differentiate between normal and shear
stress
Visualize the physical context of stress equilibrium
in bulk solids.
Draw and Interpret Mohr Circles
Calculate principal stress, principal planes and
maximum & minimum shear stresses.
Apply Mohr Circles in Failure Analysis
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ChE 159 Particle
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Fundamentals of Forces & Stresses
The state of load on a bulk solid is described using
the similar methods youve learned in your
Mechanics of Deformable Bodies (ES 64).
You do not consider the forces at the individual
particles of the bulk solid, but the forces on the
boundary areas of individual volume elements as a
whole.
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ChE 159 Particle
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Visualizing Stress
the normal stress =FN/A : stress acting
perpendicularly(normally) on areaA;
the shear stress =FS/A : stress parallel to areaA. Within the bulk solid the horizontal stress, h, is a
result of the vertical stress, v, where the resulting
horizontal stress is less than the vertical stress exerted
on the bulk solid from the top. The ratio of horizontalstress, h, to vertical stress, v, is the lateral stress
ratio,
Typical values of K are between 0.3 and 0.6
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ChE 159 Particle
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Pop Quiz #1 (1.5 min)
In Figure b, the shearing stress was found to be 4 kPa while thenormal stress was determined to be 3 kPa. The contact area is 1 m2.
What is the magnitude and direction of force F?
a. 5kN, 30.960
b. 7 kN, 59.04o
c. 5 kN, 59.04o
d. 7 kN, 30.96o
7/27/2019 Che 159 - Stress Equilibrium
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ChE 159 Particle
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MSU-IIT DChE&T
Pop Quiz #1 (1.5 min)
In Figure b, the shearing stress was found to be 4 kPa while thenormal stress was determined to be 3 kPa. The contact area is 1 m2.
What is the magnitude and direction of force F?
F=[(4x1) 2+ (3x1)2]1/2= 5 kN ; = tan-1(3/5) = 30.960 Ans: A
I b lk lid t h l h t l d t f i ti l ff t
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ChE 159 Particle
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Visualizing Stress
In bulk solids technology, shear stresses always emerge due to frictional effects
a) when not inclined, shear stress is zero
b) if inclined at an angle, shear stress acts to prevent the bulk solid
from sliding
c) if the angle becomes steeper, shear stress wont be enough
causing them to slide
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Pop Quiz #2 ( 1.5 minutes)
A bulk solid with total mass of 2 kg is on an
inclined plane with contact area of 1 m2.
The coefficient of static friction between the
bulk solid and the plane was determined to
be 0.2 and the coefficient of kinetic friction
was 0.17. At what angle with the horizontal
will the solid start to fall?
a. 10.31o b. 11.31 o c. 9.650 d. none of these
7/27/2019 Che 159 - Stress Equilibrium
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ChE 159 Particle
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Fundamentals of Stresses
Pop Quiz #1 ( 1.5 minutes) A bulk solid with total mass of 2 kg is on an inclined plane
with contact area of 1 m2. The coefficient of static friction
between the bulk solid and the plane was determined to be
0.2 and the coefficient of kinetic friction was 0.17. At what
angle with the horizontal will the solid start to fall?
Solution:
Wsin- Wcos (us) = 0
=tan -10.2 = 11.31o
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The Mohr Circle The Mohrs circle represents the possible
combinations of normal and shear stresses acting
on any plane in a body (or powder) under stress.
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The Mohr Circle How to Draw Mohr Circle
Set a Cartesian Plane with normal stress in +x-axis upward
and shear stress on +y-axis downward.
Determine x, y , xyfrom the plane stress state. Rememberthe sign convention ; (+) for tension, (-) for compression;
(+) for shear stress towards both +y and +x, otherwise, (-)
Locate the center of the circle by calculating ave. C(x+y
2, 0)
Plot a point X(x, xy) and connect to the center. The distance|XC| is the radius of Mohr circle.
Since you have the center and the radius, you can easily draw
the Mohr Circle!
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The Mohr Circle
How to Interpret Mohr Circle Principal Stressesthe maximum and minimum normal
stress in a plane which corresponds to the point of the circle
touching the x-axis. Principal Planesassociated with the angle from A to the x-
axis. Angle of Rotation in the Mohr Circle is twice the actual
angle. Point A corresponds to 0oin the actual plane.
Maximum & Minimum Shear Stresstop and bottom ofMohr circle.
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The Mohr Circle Example 4.3.1
The bulk solid is under the state of stress shown below,
determine using Mohr circle
Principal stresses Principal planes
Maximum & minimum
shearing stress
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ChE 159 Particle
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The Mohr Circle Example 4.3.1
The bulk solid is under the state of stress shown below,
determine using Mohr circle
Principal stresses Principal planes
Maximum & minimum
shearing stress
Given:
x= +50 Mpa
y = -10 Mpa
xy
= 40 MPa
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ChE 159 Particle
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The Mohr Circle
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The Mohr Circle Answers:
a. Principal Stresses
max = 70 Mpa
min = -30 Mpa b. Principal Planes
max = 26.6o counterclockwise
min = 63.4ocounterclockwise
c. Maximum & Minimum Shear Stress
max = 50 MPa
min = -50 MPa
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Application of Mohrs Circle to
Failure Analysis Each point on a yield locus represents that point on a
particular Mohrs circle for which failure or yield of
the powder/bulk solid occurs. A yield locus is thentangent to all the Mohrs circles representing stress
systems under which the powder will fail (flow).
A yield locus is usually a slightly convex upward
curve. The curvature increases towards smallernormal stresses. With free-flowing, cohesionless bulk
solids one usually obtains a nearly straight-lined
yield locus that goes through the origin.
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Application of Mohrs Circle to
Failure Analysis Example 4.3.2.Determine which Mohr Circle will
cause the bulk solids to flow.
MSU IIT DChE&T
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ChE 159 Particle
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Application of Mohrs Circle to
Failure Analysis Solution:
Mohrs circles (a) and (b) represent stress systems under which
the powder would fail. In circle (c) the stresses are insufficient to
cause flow. Circle (d) is not relevant since the system underconsideration cannot support stress combinations above the yield
locus. It is therefore Mohrs circles which are tangential to yield
loci that are important to our analysis.
ChE 1 9 P i l MSU IIT DChE&T
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Application of Mohrs Circle to Failure Analysis
Example 4.3.3.A bulk solid shown below was subjected to stress
and its Mohr circle was drawn. The yield locus touches the Mohr
circle at =1 ksi. If the maximum shear stress is 2 ksi,
what is the angle of failure of the bulk powder at the right?
Ans: 60o
A
B
C
ChE 159 P ti l MSU IIT DChE&T
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Pop Quiz #3 (2 minutes)
Example 4.3.3.A bulk solid shown below was subjected to stress
and its Mohr circle was drawn. The yield locus touches the Mohr
circle at =1 ksi. If the maximum shear stress is 2 ksi,
What is the maximum normal stress? (Self Test)
What is the shear stress at the yield point? (Self Test)
A
B
C