2-Lectures LEC 11

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    Chapter 5

    Deflection and Stiffness

    A. Aziz Bazoune

    10/1/2007 Chapter 5 1

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    Outline

    1. Spring Rates2. Deflection in Tension, Compression & Torsion

    3. Deflection due to Bending

    4. Strain Energy5. Castiglianos Theorem

    6. Statically Indeterminate Problems

    7. Compression MembersLong Columns with Central Loading

    Intermediate Length Columns with Central Loading

    10/1/2007 Chapter 5 2

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    Strain Energy Casti lianos Theorem

    Lec. 11

    10/1/2007 Chapter 5 3

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    5-7 Strain Energy The strain energy is a potential energy stored in a deformed

    component.

    If a member is deformed a distance y, and if the force deflection-relationship is linear, this energy is equal to the product of the average

    10/1/2007 Chapter 5 4

    forceF and the deflectiony, or,

    2

    2 2 2

    F F F F U y

    k k

    = = =

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    Strain Energy

    10/1/2007 Chapter 5 5

    (5-15) (5-16)

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    (5-17)

    10/1/2007 Chapter 5 6

    Figure 5-8(a) Pure shear element

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    (5-18)

    10/1/2007 Chapter 5 7

    Figure 5-8(b) Beam bending element

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    10/1/2007 Chapter 5 8

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    Sometimes the strain energy stored in a unit volume u is a useful

    quantity. By dividing Eqs. (5-15) to (5-17) by the total volume , andsetting for tension or compression, for directshear, and for torsion , we obtain

    l AF A = F A =

    ( ) max2Td J =

    10/1/2007 Chapter 5 9

    (5-19)

    Eq. (5-19) shows that thedevelopment of high stressin a material with a lowmodulus of elasticity, or

    rigidity, will result in greatamount of energy storage

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    Example-5-8 (Textbook)Find the strain energy due to shear in a rectangular cross-section beam,simply supported, and having a uniformly distributed load

    Solution

    Using Table A-9-7, we find the shear force to be

    wLV wx=

    10/1/2007 Chapter 5 10

    Using Table 5-1, C=1.2 for a rectangular cross-section. Substituting into Eq.

    (5-20) gives

    22 3

    0

    1.2

    2 2 20

    lwL w l

    U wx dxG AG

    = =

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    Example-5-9 (Textbook)A cantilever has a concentrated load F at the end, as shown in Fig. 5-9. Find thestrain energy in the beam by neglecting shear

    At any point x along the beam,the moment is

    M Fx=

    10/1/2007 Chapter 5 11

    Substituting this value of M intoEq. (5-18) gives

    Figure 5-9

    2 2 2 3

    0 2 6

    lF x F x

    U dxEI EI

    = =

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    5-8 Castiglianos Theorem When a body is elastically deflected by any combination of

    loads, the deflection at any point in any direction is equal to the

    partial derivative of the strain energy, computed with all theloads, with respect to the load acting in that direction.

    10/1/2007 Chapter 5 12

    (5-21) (5-22)

    is the displacement of the pointof application of the force in thedirection of .

    i

    iF

    is the rotational displacement,in radians, of the moment inthe direction of .

    i

    iF

    iM

    iM

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    5-8 Castiglianos Theorem In performing any integration, it is generally better to take the partial

    derivative with respect to the loadFi first. This is true especially if the

    force is a fictitious force Qi since it can be set to zero as soon as thederivative is taken.

    10/1/2007 Chapter 5 13

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    10/1/2007 Chapter 5 14

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    Example

    Find the total deflection atL due to bending only using Castiglianos method.

    Solution

    ( ) ( )andM x F x V x F= =

    0

    1l

    i

    l

    MM

    EI F

    =

    10/1/2007 Chapter 5 15

    ( )0

    3

    2

    0 0

    3

    1

    3

    3

    ll

    F x x dxEI

    F xF x dx

    EI EI

    F l

    EI

    =

    = =

    =

    This is in agreement with case A-9-1 Page 969 ofyour Textbook.

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    Example

    Find the deflection at the free end of the beam due to bending only usingCastiglianos method.

    Solution

    ( ) ( )2( 2) 2M x Q x wx x

    MxQ

    Q x w x= =

    =

    ( ) ( )( )21 1

    2

    l l

    Q

    MM xQ x w x dx

    EI Q EI

    = =

    10/1/2007 Chapter 5 16

    This is in agreement with case A-9-3 Page 970 of your Textbook. Notice that the

    displacement is in the same direction (downward) as the assumed Q.

    At this stage (After differentiating the strain energy expressionw.r.t Q, The fictitious fore Q is SET EQUAL TO ZERO

    1Q

    IQ

    E = ( ) ( )( )

    ( ) ( )( ) ( )0

    0

    43

    0

    0

    2

    212 2

    8

    2

    l

    l l

    Q

    x dx

    x dx dxEI EI EI

    x w x

    w wlw x x

    =

    = = =

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    QUESTIONS ?

    10/1/2007 Chapter 5 17