Chapter 4 Mechanical Deflection-f

7/25/2019 Chapter 4 Mechanical Deflection-f

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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 Outline4-1 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 Castigliano s Theorem

4-9 Deflection of Curved Members

4-10 Statically Indeterminate Problems

4-11 Compression Members General

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 force

when deformed. If we designate the general relationship between force

and 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 in

pure 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 T is

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 point x is

Therefore


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Deflection Due to Bending

(4 10)

(4 11)

(4 12)

(4 13)

(4 14)


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Beam Deflection Methods

There are many techniques employed to solvethe integration problem for beam deflection.Some of the popular methods include : Superposition Singularity functions Energy


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Example 1: Superposition Beam Deflections by

Superposition :Superposition resolves the effect ofcombined loading on a structure bydetermining the effects of each loadseparately and adding the resultsalgebraically.

+


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Example 4 2 (continued)


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Example 2: Superposition

+


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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 elasticmember in deforming it, is transformedinto strain , or potential energy .

The energy is equal to the product of theaverage force and the deflection, or


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Strain Energy due toBending 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 thestrain energy for thesimply supportedbeam


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Example 2: Strain Energy

A

B

P

R

Determine the BendingStrain Energy for thecurved beam


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Castiglianos Theorem Castiglianos theorem states that

when forces act on elastic systems subject to small displacements, thedisplacement corresponding to any force, in the direction of the force, isequal 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 F iin the direction of F i


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Castiglianos Theorem (applications)

Castiglianos theorem can be used to find the deflection at apoint 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 TLT GJ GJ

Torsion


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Castiglianos Theorem (applications)

2 3 3

8 4 A P R PR

y P EI EI

P

A

B

R

Bending (Shear Contributions)


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Example 4 10


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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 bywhich is called the Euler column formula

Euler Column formula can be extended to applyto other end-conditions by writingwhere the constant C depends on the end conditions


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Long Columns with Central Loading

Critical Buckling Load

2

2cr C EI

P l

2 I Ak 2

2cr P C E A l

k


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Euler Column Formula : General Using the relation = 2, where is 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 theend-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 beused 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 themaximum value of

The term is called theeccentricity ratio .


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Struts or Short Compression Members

A strut is a short compression member such thatthe maximum compressive stress in the x 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 thecoordinate of point B, and is the eccentricity ofloading.

How long is a short member? If we decide that the limiting percentage is to be 1

percent 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 +

StrutsShort columns

Intermediatelength columns

Long columns

Chapter 4 Mechanical Deflection-f

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