Che 159 - Stress Equilibrium

7/27/2019 Che 159 - Stress Equilibrium

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MECHANICAL PROPERTIES

OF BULK SOLIDS


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STRESS EQUILIBRIUM

IN BULK SOLIDS

Reporter: Ruel B. Cedeno


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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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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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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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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


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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?

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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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


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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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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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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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Failure Analysis Example 4.3.2.Determine which Mohr Circle will

cause the bulk solids to flow.

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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.

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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

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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

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Che 159 - Stress Equilibrium