Castigliano Method for Frame Analysis

1
Castigliano Method for Frame Analysis

• Castiglianos theorem for a linear elastic system
  subjected to
• External forces Pi
• External moments Mi

• where
• xi is the deflection in the direction of Pi
• qi is the rotation in the direction of Mi
• U is the total strain energy of the system given
  by
• U0 is the strain energy density per unit volume

2
Castigliano Method for Frame Analysis

• Method applied to linear elastic systems
  consisting of slender, straight or curvilinear
  members (rods, beam, shafts etc) with uniform
  cross-section.
• General member configuration in 3D
• Internal forces/moments
• Axial force N(s)
• Bending moment M(s)
• Shear force V(s)
• Torque T(s)
• Stresses
• N and M generate axial stress s
• V and T generate shear stress t

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Castigliano Method for Frame Analysis

• Strain energy density

• Total strain energy

• Contributions to the strain energy U
• Axial force N

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Castigliano Method for Frame Analysis

• Contributions of M, V, and T to the strain energy
  U obtained by a similar, albeit more complicated
  process

• Final expressions
• Axial force N

• Bending moment M

• Shear force V

• Torque T

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Example 5.7 Shaft-beam mechanism

• Beams CD, FH rectangular section
• Shaft AB circular section

• Determine the rotation of end section at B.
• Castigliano equation (neglecting shear force
  contributions to U)

• Reactions must be determined SFy 0 ? RC RH
  0
• SMH 0 ? 1.0?RC - T0 0 ? RC - RH T0

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Example 5.7 Shaft-beam mechanism

• Expressions for V, M, T Sections - Free body
  diagrams (FBD)
• Beam CD
• V RC T0
• M xRC xT0 ?
• T 0

• Shaft AB
• V M 0
• T T0 ?

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Example 5.10

• Problem data
• d 20 mm
• R 200 mm
• P Q 150 N
• E 200 GPa
• G 77.5 GPa
• Determine deflection at C
• Sections FBDs to determine expressions for N,
  V, M

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Example 5.10

• Part BC 0 ? f ? p/2
• N Pcosf ?N/ ?P cosf
• V Psinf ?V/ ?P sinf
• M PR(1-cosf) ?M/ ?P R(1-cosf)

• Part AB 0 ? q ? p/2
• N - (PQ)sinq , ?N/ ?P - sinq
• V (PQ)cosq , ?V/ ?P cosq
• M PR(1sinq) QRsinq, ?M/ ?P R(1sinq)

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Example 5.10

• Deflection at C (considering only contributions
  from M)

• Determine, evaluate and compare contributions
  from N, V and M. (take shear constant k 10/9)

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Deflection/rotation in any direction

• (not necessarily in the direction of applied
  force/moment)
• A fictitious force/moment X is assumed acting at
  a point P in any direction q. Then the
  displacement/rotation of/about that point in the
  chosen direction is given according to
  Castigliano by

The bending moment in a member can be considered
as the combination of M(z) due to actual
applied forces m(z) due to a fictitious unit
force acting at P in the direction of q
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Deflection/rotation in any direction

• The principle of superposition applies to linear
  elastic systems. Therefore, the total bending
  moment
• M(X, z) M(z) Xm(z)
• Hence
• M(X 0, z) M(z)
• ?M/?X m(z)
• General expression for the deflection/rotation

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Example 5.13

• Determine deflection of free end B
• Sections and FBDs
• Part OA
• Actual forces
• M Rsinq ?P
• T R(1 cosq ) ?P
• Fictitious unit force
• m Rcosq ?1
• t R(1 sinq ) ?1

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Example 5.13

• Part AB
• Actual forces Fictitious unit force
• M 0 m Rsinf ?1
• T 0 t R(1 cosf ) ?1
• Substituting into the Castigliano equation

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Statically indeterminate systems

• If the number of unknown forces/moments is
• gt 6 in 3D
• gt 3 in 2D
• the equations of equilibrium are not sufficient
  for a complete analysis of the frame, i.e. for
  finding stresses and deflections.
• If the redundant unknowns are X1, X2, , Xn,
  Castiglanos theorem provides n additional
  equations

• The validity of the above is obvious when Xi is a
  reaction but it can also be proved in the case of
  internal redundant forces/moments

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Example 5.16

• 2D frame, 5 reactions, therefore two redundant
  reactions
• Reactions H, Q at C are selected for applying
  Castigliano equations

• Axial, shear force contributions to U are
  neglected
• Sections and FBDs

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Example 5.16

• (i) Part BC
• 0 ? q ? p
• M -R(1-cosq)H (Rsinq)Q
• ?M/ ?H -R(1-cosq), ?M/ ?Q Rsinq

• (ii) Part AB
• 0 ? z ? 2R
• M z(P-Q) - 2RH
• ?M/ ?H -2R, ?M/ ?Q -z

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Example 5.16

• Castigliano equations

Integrating (3p/2 8)H 2Q 4P 2H (p/2
8/3)Q 8P/3 Solution Q 0.5193P H 0.2329P
Vertical deflection at P since
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Example 5.17 Inverted king post

• Determine max P with SF 2.0
• Statically indeterminate structure with one
  redundant internal unknown axial force NAD NDC
  
• Castigliano is applied neglecting N, V
  contributions in beam AC, considering only N
  contributions in AD, DC and DB
• Sections and FBDs, accounting for symmetry

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Example 5.17 Inverted king post

• Force equilibrium at joint D
• NBD 2NAD sina
• where
• tana 0.5/2 0.25

• Bending moment in AB
• M z(P/2 - NAD sina)
• Hence
• ?M/ ?NAD - (z sina)