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Module 1Energy Methods inStructural Analysis Version 2 CE IIT, Kharagpur
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Lesson 3Castigliano’s Theorems Version 2 CE IIT, Kharagpur
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Instructional ObjectivesAfter reading this lesson, the reader will be able to;1. State and prove first theorem of Castigliano.2. Calculate deflections along the direction of applied load of a statically determinate structure at the point of application of load.3. Calculate deflections of a statically determinate structure in any direction at a point where the load is not acting by fictious (imaginary) load method.4. State and prove Castigliano’s second theorem.3.1 IntroductionIn the previous chapter concepts of strain energy and complementary strainenergy were discussed. Castigliano’s first theorem is being used in structuralanalysis for finding deflection of an elastic structure based on strain energy of thestructure. The Castigliano’s theorem can be applied when the supports of thestructure are unyielding and the temperature of the structure is constant.3.2 Castigliano’s First TheoremFor linearly elastic structure, where external forces only cause deformations, thecomplementary energy is equal to the strain energy. For such structures, theCastigliano’s first theorem may be stated as the first partial derivative of thestrain energy of the structure with respect to any particular force gives thedisplacement of the point of application of that force in the direction of its line ofaction. Version 2 CE IIT, Kharagpur
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Let P1 , P2 ,...., Pn be the forces acting at x1 , x 2 ,......, x n from the left end on a simplysupported beam of span L . Let u1 , u 2 ,..., u n be the displacements at the loadingpoints P1 , P2 ,...., Pn respectively as shown in Fig. 3.1. Now, assume that thematerial obeys Hooke’s law and invoking the principle of superposition, the workdone by the external forces is given by (vide eqn. 1.8 of lesson 1) 1 1 1 W = P1u1 + P2 u 2 + .......... + Pn u n (3.1) 2 2 2 Version 2 CE IIT, Kharagpur
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Work done by the external forces is stored in the structure as strain energy in aconservative system. Hence, the strain energy of the structure is, 1 1 1 U= P1u1 + P2 u 2 + .......... + Pn u n (3.2) 2 2 2Displacement u1 below point P1 is due to the action of P1 , P2 ,...., Pn acting atdistances x1 , x 2 ,......, x n respectively from left support. Hence, u1 may be expressedas, u1 = a11 P1 + a12 P2 + .......... + a1n Pn (3.3)In general, u i = ai1 P1 + ai 2 P2 + .......... + ain Pn i = 1,2,...n (3.4)where a ij is the flexibility coefficient at i due to unit force applied at j .Substituting the values of u1 , u 2 ,..., u n in equation (3.2) from equation (3.4), weget, 1 1 1U= P1 [ a11 P1 + a12 P2 + ...] + P2 [ a 21 P1 + a 22 P2 + ...] + ....... + Pn [ a n1 P1 + a n 2 P2 + ...] (3.5) 2 2 2We know from Maxwell-Betti’s reciprocal theorem a ij = a ji . Hence, equation (3.5)may be simplified as, 1 U= ⎡ a11 P 2 + a22 P22 + .... + ann Pn2 ⎤ + [ a12 P P2 + a13 P P3 + .... + a1n P Pn ] + ... (3.6) 2⎣ ⎦ 1 1 1 1Now, differentiating the strain energy with any force P1 gives, ∂U = a11 P1 + a12 P2 + .......... + a1n Pn (3.7) ∂P1It may be observed that equation (3.7) is nothing but displacement u1 at theloading point.In general, ∂U = un (3.8) ∂PnHence, for determinate structure within linear elastic range the partial derivativeof the total strain energy with respect to any external load is equal to the Version 2 CE IIT, Kharagpur
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displacement of the point of application of load in the direction of the appliedload, provided the supports are unyielding and temperature is maintainedconstant. This theorem is advantageously used for calculating deflections inelastic structure. The procedure for calculating the deflection is illustrated withfew examples.Example 3.1Find the displacement and slope at the tip of a cantilever beam loaded as in Fig.3.2. Assume the flexural rigidity of the beam EI to be constant for the beam.Moment at any section at a distance x away from the free end is given by M = − Px (1) L M2Strain energy stored in the beam due to bending is U = ∫ dx (2) 0 2 EISubstituting the expression for bending moment M in equation (3.10), we get, L ( Px) 2 P 2 L3 U =∫ dx = (3) 0 2 EI 6 EI Version 2 CE IIT, Kharagpur
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Now, according to Castigliano’s theorem, the first partial derivative of strainenergy with respect to external force P gives the deflection u A at A in thedirection of applied force. Thus, ∂U PL3 = uA = (4) ∂P 3EITo find the slope at the free end, we need to differentiate strain energy withrespect to externally applied moment M at A . As there is no moment at A , applya fictitious moment M 0 at A . Now moment at any section at a distance x awayfrom the free end is given by M = − Px − M 0Now, strain energy stored in the beam may be calculated as, ( Px + M 0 ) 2 L 2 P 2 L3 M 0 PL2 M 0 L U =∫ dx = + + (5) 0 2 EI 6 EI 2 EI 2 EITaking partial derivative of strain energy with respect to M 0 , we get slope at A . ∂U PL2 M 0 L = θA = + (6) ∂M 0 2 EI EIBut actually there is no moment applied at A . Hence substitute M 0 = 0 inequation (3.14) we get the slope at A. PL2 θA = (7) 2 EIExample 3.2A cantilever beam which is curved in the shape of a quadrant of a circle is loadedas shown in Fig. 3.3. The radius of curvature of curved beam is R , Young’smodulus of the material is E and second moment of the area is I about an axisperpendicular to the plane of the paper through the centroid of the cross section.Find the vertical displacement of point A on the curved beam. Version 2 CE IIT, Kharagpur
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The bending moment at any section θ of the curved beam (see Fig. 3.3) is givenby M = PR sinθ (1)Strain energy U stored in the curved beam due to bending is, π /2 P 2 R 2 (sin 2 θ ) Rdθ P 2 R 3 π π P 2 R 3 s M2 U =∫ ds = ∫ = = (2) 0 2 EI 0 2 EI 2 EI 4 8 EIDifferentiating strain energy with respect to externally applied load, P we get ∂U b π PR 3 uA = = (3) ∂P 4 EIExample 3.3Find horizontal displacement at D of the frame shown in Fig. 3.4. Assume theflexural rigidity of the beam EI to be constant through out the member. Neglectstrain energy due to axial deformations. Version 2 CE IIT, Kharagpur
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The deflection D may be obtained via. Castigliano’s theorem. The beamsegments BA and DC are subjected to bending moment Px ( 0 < x < L ) and thebeam element BC is subjected to a constant bending moment of magnitude PL .Total strain energy stored in the frame due to bending L L ( Px) 2 ( PL) 2 U = 2∫ dx + ∫ dx (1) 0 2 EI 0 2 EIAfter simplifications, P 2 L3 P 2 L3 5P 2 L3 U= + = (2) 3EI 2 EI 6 EIDifferentiating strain energy with respect to P we get, ∂U 5 P L3 5 P L3 = uD = 2 = ∂P 6 EI 3EI Version 2 CE IIT, Kharagpur
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Example 3.4Find the vertical deflection at A of the structure shown Fig. 3.5. Assume theflexural rigidity EI and torsional rigidity GJ to be constant for the structure.The beam segment BC is subjected to bending moment Px ( 0 < x < a ; x ismeasured from C )and the beam element AB is subjected to torsional moment ofmagnitude Pa and a bending moment of Px ( 0 ≤ x ≤ b ; x is measured from B) . Thestrain energy stored in the beam ABC is, a b M2 ( Pa) 2 b ( Px) 2 U =∫ dx + ∫ dx + ∫ dx (1) 0 2 EI 0 2GJ 0 2 EIAfter simplifications, P 2 a 3 P 2 a 2b P 2b3 U= + + (2) 6 EI 2GJ 6 EIVertical deflection u A at A is, ∂U Pa 3 Pa 2 b Pb 3 = uA = + + (3) ∂P 3EI GJ 3EI Version 2 CE IIT, Kharagpur
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Example 3.5Find vertical deflection at C of the beam shown in Fig. 3.6. Assume the flexuralrigidity EI to be constant for the structure.The beam segment CB is subjected to bending moment Px ( 0 < x < a ) andbeam element AB is subjected to moment of magnitude Pa .To find the vertical deflection at C , introduce a imaginary vertical force Q at C .Now, the strain energy stored in the structure is, ( Pa + Qy) 2 a b ( Px) 2 U =∫ dx + ∫ dy (1) 0 2 EI 0 2 EIDifferentiating strain energy with respect to Q , vertical deflection at C is obtained. ∂U 2( Pa + Qy ) y b = uC = ∫ dy (2) ∂Q 0 2 EI b 1 EI ∫ uC = Pay + Qy 2 dy (3) 0 Version 2 CE IIT, Kharagpur
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1 ⎡ Pab 2 Qb 3 ⎤ uC = ⎢ + ⎥ (4) EI ⎣ 2 3 ⎦ But the force Q is fictitious force and hence equal to zero. Hence, verticaldeflection is, Pab 2 uC = (5) 2 EI3.3 Castigliano’s Second TheoremIn any elastic structure having n independent displacements u1 , u 2 ,..., u ncorresponding to external forces P1 , P2 ,...., Pn along their lines of action, if strainenergy is expressed in terms of displacements then n equilibrium equations maybe written as follows. ∂U = Pj , j = 1, 2,..., n (3.9) ∂u jThis may be proved as follows. The strain energy of an elastic body may bewritten as 1 1 1 U= P1u1 + P2 u 2 + .......... + Pn u n (3.10) 2 2 2We know from Lesson 1 (equation 1.5) that Pi = ki1u1 + ki 2u2 + ..... + kinun , i = 1, 2,.., n (3.11)where kij is the stiffness coefficient and is defined as the force at i due to unitdisplacement applied at j . Hence, strain energy may be written as, 1 1 1U= u1[k11u1 + k12u2 + ...] + u2 [ k21u1 + k22u2 + ...] + ....... + un [k n1u1 + kn 2u2 + ...] (3.12) 2 2 2We know from reciprocal theorem kij = k ji . Hence, equation (3.12) may besimplified as, 1 U= ⎡ k11u12 + k22u2 + .... + knn un ⎤ + [ k12u1u2 + k13u1u3 + .... + k1n u1un ] + ... ⎣ 2 2 ⎦ (3.13) 2 Version 2 CE IIT, Kharagpur
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Now, differentiating the strain energy with respect to any displacement u1 givesthe applied force P at that point, Hence, 1 ∂U = k11u1 + k12 u2 + ........ + k1n un (3.14) ∂u1Or, ∂U = Pj , j = 1, 2,..., n (3.15) ∂u jSummaryIn this lesson, Castigliano’s first theorem has been stated and proved for linearlyelastic structure with unyielding supports. The procedure to calculate deflectionsof a statically determinate structure at the point of application of load is illustratedwith examples. Also, the procedure to calculate deflections in a staticallydeterminate structure at a point where load is applied is illustrated with examples.The Castigliano’s second theorem is stated for elastic structure and proved insection 3.4. Version 2 CE IIT, Kharagpur
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