Chapter 4 Deflection-f (1)

7/25/2019 Chapter 4 Deflection-f (1)

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Mechanical Design I (MCE 321) L. Romdhane, SS 2016, 11:33 AM -- 1--

Summer 2016

Chapter 4Deflection and Stiffness

Mechanical Design 1(MCE 321)

Dr. Lotfi

Romdhane

[email protected]


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Chapter Outline

4-1 Spring Rates

4-2 Tension, Compression, and Torsion

4-3 Deflection Due to Bending

4-4 Beam Deflection Methods

4-5 Beam Deflections by Superposition

4-6 Beam Deflections by Singularity Functions

4-7 Strain Energy

4-8 Castiglianos Theorem

4-9 Deflection of Curved Members

4-10 Statically Indeterminate Problems

4-11 Compression MembersGeneral

4-12 Long Columns with Central Loading

4-13 Intermediate-Length Columns with Central Loading

4-14 Columns with Eccentric Loading

4-15 Struts or Short Compression Members

4-16 Elastic Stability

4-17 Shock and Impact


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Spring Rates

A spring is a mechanical element that exerts a forcewhen deformed.

If we designate the general relationship between forceand deflection by the equation

then spring rate is defined as

where y must be measured in the direction of F and atthe point of application of F.

For linear force-deflection problems, k is a constant,also called the spring constant


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Tension, Compression, and Torsion

The total extension or contraction of a uniform bar inpure tension or compression, is given by

The spring constant of an axially loaded bar is then

The angular deflection of a uniform round barsubjected to a twisting moment Tis

where is in radians

The torsional spring rate is


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Deflection Due to Bending

The curvature of a beam subjected to a bendingmoment M is given by

where is the radius of curvature

The slope of the beam at any pointxis

Therefore


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Deflection Due to Bending

(410)

(411)

(412)

(413)

(414)


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Beam Deflection Methods

There are many techniques employed to solve

the integration problem for beam deflection.

Some of the popular methods include :

Superposition

Singularity functions

Energy


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Example 1: Superposition

Beam Deflections bySuperposition :Superpositionresolves the effect ofcombined loading on a structure bydetermining the effects of each load

separately and adding the resultsalgebraically.

+


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Example 42 (continued)


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+


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Tables


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For tension and compression

The strain energy for torsion isgiven by

The strain energy due to directshear

Strain Energy

The external work done on an elastic

member in deforming it, is transformed

into strain, or potentialenergy.

The energy is equal to the product of the

average force and the deflection, or


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Strain Energy due to

Bending and shear loading

The strain energy storedin a section of the elasticcurve of length ds is

=

2 =

2

For small deflections,

= and

=

. Then,for the entire beam

Summarized to includeboth the integral andnon integral form, thestrain energy for bending is

The strain energy dueto shear loading of abeam can beapproximated as


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Example 1: Strain Energy

Determine the

strain energy for the

simply supported

beam


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Example 2: Strain Energy

A

B

P

R

Determine the BendingStrain Energy for the

curved beam


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Castiglianos Theorem

Castiglianos theorem states that

when forces act on elastic systems subject to small displacements, the

displacement corresponding to any force, in the direction of the force, is

equal to the partial derivative of the total strain energy with respect to

that force.

where i is the displacement of the point of application of the force Fiin the direction of Fi


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Castiglianos Theorem (applications)

Castiglianostheorem can be used to find the deflection at a

point even though no force or moment acts there.

Set up the equation for the total strain energy U

Find an expression for the desired deflection

Since Q is a fictitious force, solve the expression by setting Q equal to zero.

Tension/ Compression

2

2

F L FL

F AE AE

2

2T L TL

T GJ GJ

Torsion


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Castiglianos Theorem (applications)

2 3 3

8 4A

P R PRy

P EI EI

P

A

B

R

Bending (Shear Contributions)


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Example 410


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Example 3: Castiglianos Theorem


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Long Columns with Central Loading

If the axial force P shown acts along thecentroidal axis of the column, simplecompression of the member occurs for lowvalues of the force.

Under certain conditions, when P reaches a

specific value, the column becomesunstable and bending develops rapidly.

The critical force for the pin-endedcolumn is given by

which is called the Euler column formula

Euler Column formula can be extended to applyto other end-conditions by writing

where the constant C depends on the end conditions


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Long Columns with Central Loading

Critical Buckling Load

2

2cr

C EIP

l

2I Ak2

2cr

P C E

A l

k


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Euler Column Formula : General

Using the relation = 2, whereis the area and the radius of

gyration. Euler Column Equation can be rearranged as

where

is called the slenderness ratio

The quantity

is the critical unit load. It is the load per unit areanecessary to place the column in a condition of unstable equilibrium.

The factor C is called the

end-condition constant,


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Euler Column Formula : Application

In practical engineering applications where defects such as initialcrookedness or load eccentricities exist, the Euler equation can only be

used for slenderness ratio greater than

Most designers select point T such that

=

with corresponding

value of


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Euler Column Formula : Example


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Columns with Eccentric Loading

The magnitude of the maximum compressive

stress at mid span is found by superposing theaxial component and the bending component.

By imposing the compressiveyield strength as the

maximum value of

The term

is called the

eccentricity ratio.


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Struts or Short Compression Members

A strut is a short compression member such thatthe maximum compressive stress in thex directionat point B in an intermediate section is the sum of asimple component P/A and a flexural componentMc/I

where =

is the radius of gyration, is the

coordinate of point B, and is the eccentricity ofloading.

How long is a short member?

If we decide that the limiting percentage is to be 1percent of , then, the limiting slenderness ratioturns out to be


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Mechanical Design I (MCE 321) L Romdhane SS 2016 11:33 AM -- 34--

Summary

=

1 +

Struts

Short columnsIntermediate

length columnsLong columns

Chapter 4 Deflection-f (1)

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