pe2113-chapter 10 - moment of inertia_draft

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Chapter 10 –

Moments ofInertia



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Objectives

• To develop a method of determining themoment of inertia for an area

• To introduce the mass moment of inertia



APPLICATIONS

Why do they usually not have solid rectangular, square, orcircular cross sectional areas?

What primary property of these members influences design

decisions?

Many structural members like beams and columns have cross

sectional shapes like an I, H, C, etc..



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10.1. Definition of Moments of Inertia for Area

• Centroid is determine in Chapter 9 by considering the first

moment of the area about an axis. That is, we evaluatethe integral xd A

• Moment of inertia (

) is the integral of the second

moment of an area such as x2d A

– The moment of inertia for an area is a quantity that relates the

normal stress σ or force per unit area, acting on the transverse

cross section of an elastic beam, to the applied external

moment M, which causes bending of the beam.

– The theory of mechanics of materials shows that the stress

within the beam varies linearly with its distance from an axis

passing through the centroid C of the beam’s cross-sectional

area.



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Moment of Inertia

• Considering the areawhich lies in the x-y

plane, the moments of

inertia are determined

by integration asshown in the

equations.



MoI FOR AN AREA BY INTEGRATION

The step-by-step procedure is:

1. Choose the element dA: There are two choices: a vertical strip or a

horizontal strip. Some considerations about this choice are:

a) The element parallel to the axis about which the MoI is to be

determined usually results in an easier solution. For example,

we typically choose a horizontal strip for determining Ix and a

vertical strip for determining Iy.

For simplicity, the area element used has a

differential size in only one direction(dx or dy). This results in a single integration

and is usually simpler than doing a double

integration with two differentials, i.e., dx·dy.



2. Integrate to find the MoI. For example, given the element shown in

the figure above:

Iy = x2 dA = x2 y dx and

Ix = d Ix = (1 / 3) y3 dx (using the information for Ix for a

rectangle about its base from the inside back cover of the textbook).

Since in this case the differential element is dx, y needs to be

expressed in terms of x and the integral limit must also be in terms of

x. As you can see, choosing the element and integrating can be

challenging. It may require a trial and error approach plus

experience.

MoI FOR AN AREA BY INTEGRATION

b) If y is easily expressed in terms of x (e.g.,

y = x2 + 1), then choosing a vertical strip

with a differential element dx wide may

be advantageous.



EXAMPLE

Solution

Ix = y2 dA

dA = (2 – x) dy = (2 – y2/2) dy

Ix = O y2 (2 – y2/2) dy

= [ (2/3) y3 – (1/10) y5 ]0 = 2.13 m4

2

2

Given: The shaded area shown in the

figure.

Find: The MoI of the area about the

x- and y-axes.

Plan: Follow the steps given earlier.



EXAMPLE (continued)

In the above example, it would be difficult to determine Iy

using a horizontal strip. However, Ix in this example can bedetermined using a vertical strip. So,

Ix = (1/3) y3 dx = (1/3) (2x)3 dx .

Iy = x2 dA = x2 y dx

= x2 (2x) dx

= √2 0 x 2.5 dx

= [ (√2/3.5) x 3.5 ]0

= 4.57 m 4

2

2



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Polar Moment of Inertia• The second moment of d A about “pole” O or z-axis from

Figure 10.2 is referred to as the polar moment of inertia.

Polar Moment of Inertia is used to determine the torsionalstress in a shaft.

• It is defined as dJ o = r 2d A, where r is the perpendiculardistance from the pole (z-axis) to the element A.

• For the entire area the polar moment of inertia is givenbelow. The relationship between J o and I x, I y, is possiblesince r 2 = x2 + y2



10. 2. PARALLEL-AXIS THEOREM FOR AN

AREA

Consider an area with centroid C. The x' and y' axes pass through

C. The MoI about the x-axis, which is parallel to, and distance dy

from the x ' axis, is found by using the parallel-axis theorem.

This theorem relates the moment ofinertia (MoI) of an area about an axis

passing through the area’s centroid to

the MoI of the area about a

corresponding parallel axis. Thistheorem has many practical

applications, especially when working

with composite areas.



PARALLEL-AXIS THEOREM (Cont.)

Using the definition of the centroid:

y' = (A y' dA) / (A dA) . Now

since C is at the origin of the x' – y' axes,

y' = 0 , and hence A y' dA = 0 .

Thus IX = IX' + A dy 2

Similarly, IY = IY' + A dX2 and

JO = JC + A d 2

IX = A y 2 dA = A (y' + dy)2 dA

= A y' 2 dA + 2 dy A y' dA + dy 2 A dA



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10.3. Radius of Gyration of an Area

• The radius of g yrat ion of a planar area has

uni ts of leng th and is a quantity that is oftenused for the design of columns in structural

mechanics.

•Provided the areas and moment of inertia areknown, the radius of gyration are determined

from the following formulas:

(10.6)



Exercise: Prob. 10-14



10.4. MOMENT OF INERTIA FOR A COMPOSITE

AREA

The MoI about their centroidal axes of these “simpler” shaped

areas are found in most engineering handbooks as well as theinside back cover of the textbook.

Using these data and the parallel-axis theorem, the MoI for a

composite area can easily be calculated.

A composite area is made by adding or

subtracting a series of “simple” shapedareas like rectangles, triangles, and

circles.

For example, the area on the left can bemade from a rectangle minus a triangle

and circle.



1. Divide the given area into its

simpler shaped parts.

STEPS FOR ANALYSIS

4. The MoI of the entire area about the reference axis is

determined by performing an algebraic summation of the

individual MoIs obtained in Step 3. (Please note that MoI

of a hole is subtracted ).

3. Determine the MoI of each “simpler” shaped part about thedesired reference axis using the parallel-axis theorem

( IX = IX’ + A ( dy )2 ) .

2. Locate the centroid of each part

and indicate the perpendicular

distance from each centroid to

the desired reference axis.



Exercise Prob. 10-48



Moments of Inertia of Composite Areas

9 - 21

• The moment of inertia of a composite area A about a given axis is

obtained by adding the moments of inertia of the component areas

A1, A2, A3, ... , with respect to the same axis.



10.8. MASS MOMENT OF INERTIA

T and are related by the equation T = I . In this equation, I

is the mass moment of inertia (MMI) about the z axis.

The MMI of a body is a property that measures the resistance

of the body to angular acceleration. This is similar to the role

of mass in the equation F = m a. The MMI is often used

when analyzing rotational motion (done in dynamics).

Consider a rigid body with a center of

mass at G. It is free to rotate about the zaxis, which passes through G. Now, if we

apply a torque T about the z axis to the

body, the body begins to rotate with an

angular acceleration .



DEFINITION OF THE MMI

The MMI is always a positive quantity

and has a unit of kg ·m2 or slug · ft2.

Consider a rigid body and the arbitrary axis p shown in the figure. The MMI about the

p axis is defined as I = m r 2 dm, where r,

the “moment arm,” is the perpendicular

distance from the axis to the arbitraryelement dm.



RELATED CONCEPTS

Finally, the MMI can be obtained by integration or by the

method for composite bodies. The latter method is easier

for many practical shapes.

m

Parallel-Axis Theorem

Just as with the MoI for an area, the parallel-axis theorem can be used to

find the MMI about a parallel axis z

that is a distance d from the z’ axisthrough the body’s center of mass G.The formula is Iz = IG

+ (m) (d)2

(where m is the mass of the body).

The radius of gyration is similarly defined as

k = (I / m)



EXAMPLE

Given: The pendulum consists of a

slender rod with a mass 10 kg and

sphere with a mass of 15 kg.

Find: The pendulum’s MMI about anaxis perpendicular to the screen

and passing through point O.

Plan: Follow steps similar to finding the

MoI for a composite area.

Solution:

1. The wheel can be divided into a slender rod (r)and sphere (s).



EXAMPLE (continued)

IO = IG + (m) (d) 2

IOr = (1/12) (10)(0.45)2 +10 (0.225)2 = 0.675 kg·m2

IOs = (2/5) (15) (0.1)2 + 15 (0.55)2 = 4.598 kg·m2

4. Now add the three MMIs about point O.

IO = IOr + IOs = 5.27 kg·m2

3. The MMI data for a slender rod and sphere are

given on the inside back cover of the textbook.

Using those data and the parallel-axis theorem,calculate the following.

O

2. The center of mass for rod is at point Gr , 0.225 m

from Point O. The center of mass for sphere is atGs, 0.55 m from point O.

Gr

Gs



Exercise Prob. 10-86



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Practical Application of

Moments of Inertia



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Bottom Hole Assembly (BHA)

Buckling

● BHA Design contributing factors;1. Resistance of the collars to buckling (OD, ID, material).

Resistance is higher with a larger OD or a smaller ID

2. Length between supports (stabilizers). The greater the distance,

the less the resistance to buckling

● Downhole conditions contributing factors;

1. Torque – high torque will make buckling more likely as more

stress is imposed on the collars

2. Hole angle – high angle hole gives support to the DC and

decreases the likelihood of buckling

3. In gauge or over gauge hole – If hole is over gauged, then lateral

support is less effective and buckling is more likely



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35

Helical bucklingSinusoidal buckled



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Critical buckling load of a tubular

2

2Critical buckling load,

c

EI P L

Where E = Young's Modulus, 30 x 106

psi for steelI = 2nd moment of area, in4

L = Length between stabilizers, inches.

Be careful of units !

4 4

OD ID

64 I



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Critical Buckling - Example

• A 90-ft build assembly is to be run; bit - NB stab – 90-ft of 8” OD x 3” ID DC’s - FG stab. What is the

maximum load on the column before buckling of

these drill collars occurs?



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Critical buckling – Example 2

• A 90-ft build assembly is to be run; bit - NBstab – 90-ft of x 9” x 3” ID DC’s - FG stab.

What is the maximum load on the column

before buckling of these drill collars occurs?

4 4

49 3

318 in64

I

2

280,700 lbs

c

EI P

L



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Effect of Length Between Stabilizers

• 8” x 3” drill collars. I = 197 in4.

• With 90-ft between stabilizers, Pc = 50,000 lbs.

• What is the effect of reducing length to 60-ft?

• Recalculate the Pc for L = 60-ft,the reducedlength

2 2 6

22

30 10 197

112,518 lbs60 12c

EI P

L

More resistance to buckling !



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BHA Buckling - Conclusions

1. Drilling BHA’s can and do buckle under high weight on bit(WOB).

2. OD has a major influence on the critical load to buckle, Pc.

3. Length between stabilizers has a major influence on Pc.4. If the BHA buckles dynamically , there are potential problems;

a. Fatigue damage on the connections.

b. Physical damage to collars and hole at contact points.

c. High torques required to turn the string.

pe2113-chapter 10 - moment of inertia_draft

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