METHOD OF LEAST WORK

METHOD OF EAST WORK 91 METHOD OF EAST WORK CHAPTER TWO The method of least work is used for the analysis of statically indeterminate beams, frames and trusses. Indirect use of the Castigliano s nd theorem is made and the following steps are taken. (1) The structure is considered under the action of applied loads and the redundants. The redundants can be decided by choosing a particular basic determinate structure and the choice of redundants may vary within a problem. () Moment expressions for the entire structure are established in terms of the applied loads and the redundants, which are assumed to act simultaneously for beams and frames. (3) Strain energy stored due to direct forces and in bending etc. is calculated and is partially differentiated with respect to the redundants. (4) A set of linear equations is obtained, the number of which is equal to that of the redundants.solution of these equations evaluates the redundants. NOTE: Special care must be exercised while partially differentiating the strain energy expressions and compatibility requirements of the chosen basic determinate structure should also be kept in mind. For the convenience of readers, Castigliano s theorem are given below:.1. CASTIGIANO S FIRST THEOREM: The partial derivative of the total strain energy stored with respect to a particular deformation gives the corresponding force acting at that point. Mathematically this theorem is stated as below: U = P and U θ = M It suggests that displacements correspond to loads while rotations correspond to moments... CASTIGIANO S SECOND THEOREM : The partial derivative of the total strain energy stored with respect to a particular force gives the corresponding deformation at that point. Mathematically, and U P = U M = θ

9 THEORY OF INDETERMINATE STRUCTURES.3. STATEMENT OF THEOREM OF EAST WORK. In a statically indeterminate structure, the redundants are such that the internal strain energy stored is minimum. This minima is achieved by partially differentiating strain energy and setting it to zero or to a known value. This forms the basis of structural stability and of Finite Element Method..4. Example No.1: 1st Degree Indeterminacy of Beams. Analyze the following loaded beam by the method of least work. Ma A wkn/m x B Ra Number of reactions = 3 n Number of equations = Rb The beam is redundant to first degree. In case of cantilever, always take free end as the origin for establishing moment expressions. Choosing cantilever with support at A and Rb as redundant. Apply loads and redundant simultaneously to BDS. Ma A WwKN/m x B Ra Rb Taking B as origin (for variation of X) M X = RbX wx 0 < X < 1 U = M EI dx. o A generalized strain energy expression due to moments. Therefore, partially differentiating the strain energy stored w.r.t. redundant, the generalized form is: U R = 1 M M dx Where R is a typical redundant. EI R o Putting moment expression alongwith its limits of validity in strain energy expression. 1 U = RbX wx dx EI o Partially differentiate strain energy U w.r.t. redundant Rb, and set equal to zero. U So Rb = b = 0 = 1 RbX wx (X) dx, because at B, there should be no deflection. EI o

METHOD OF EAST WORK 93 0 = 1 EI o RbX wx3 RbX 3 3 0 = 1 EI wx4 8 o dx Or and Rb 3 3 = w4 8 Rb = +3 8 w The (+ve) sign with Rb indicates that the assumed direction of redundant Rb is correct. Now calculate Ra. Ra + Rb = w Ra = w Rb = w 3 8 w = 8 w 3 w 8 Ra = 5 8 w Put X = and Rb = 3 8 w in moment expression for M X already established before to get Ma. Ma = 3 8 w. w = 3 8 w w = 3 w 4 w 8 Ma = w 8 The ( ve) sign with Ma indicates that this reactive moment should be applied such that it gives us tension at the top at point A.

94 THEORY OF INDETERMINATE STRUCTURES Example No.: Solve the following propped cantilever loaded at its centre as shown by method of least work. Ma x P x A B C Ra Rb Ma A B.D.S. is a cantiever supported at A. I Rb is a redundant as shown. x P x B C Ra Rb BDS under loads and redundant. Taking point B as origin. / Mbc = RbX 0 < X < and Mac = RbX P U = U Rb 1 EI / o = b = 0 = 1 EI 0 = 1 EI / o x (RbX) dx + 1 EI / o RbX dx + 1 EI 0 = 1 EI Rb. X3 3 / o [RbX] [X] dx + 1 EI < X <. Now write strain energy expression. / RbX P X / Rbx P X RbX PX + P X dx / RbX 3 3 + 1 EI PX3 3 + P 4 X / dx. Partially differentiate. Put limits 0 = 1 EI Rb 3 4 0 + 1 EI Rb 3 P3 3 3 + P3 4 Rb3 4 + P3 4 P3 16 Rb 3 4 + Rb3 3 0 = 1 EI Rb3 Rb 3 4 P3 3 + P3 4 + P3 4 P3 16 16P 3 + 1P 3 + P 3 3P 3 0 = 1 EI + 3 48 w.r.t redundant Rb. [X] dx

METHOD OF EAST WORK 95 Or 0 = Rb3 3 Rb 3 3 5P3 48 = 5P3 48 Rb = +5P 16 The (+ve) sign with Rb indicates that the assumed direction of redundant Rb is correct. Now Ra can be calculated. Ra + Rb = P Ra = P Rb Ra = P 5P 16P 5P = 16 16 Put X = and Rb = 5P 16 Ma = 5P 16 P 5 P 8 P = 16 Ra = 11P 16 in expression for Mac to get Ma. Ma = 3 P 16 The ( ve) sign with Ma indicates that this reactive moment should be acting such that it gives us tension at the top..5. ND DEGREE INDETERMINACY: EXAMPE NO. 3: Analyze the following fixed ended beam loaded by Udl by least work method. Ma A WwKN/m B Mb Ra Rb B.D.S. is chosen as a cantilever supported at A. Rb and Mb are chosen as redundants. Ma A WwKN/m x Mb B Ra Rb BDS UNDER OADS AND REDUNDANTS

96 THEORY OF INDETERMINATE STRUCTURES Mx = RbX wx Mb 0 < X < Choosing B as origin. Write strain energy expression. 1 U = EI o RbX wx Mb dx Differentiate strain energy partially w.r.t. redundant Rb and use castigations theorem alongwith boundary condition. U = b = 0 = 1 RbX wx Mb [X] dx Rb EI 0 = 1 EI o o RbX wx Mb dx 0 = 1 EI Rb X3 3 wx4 8 MbX 0 = 1 EI Rb 3 Mb 3 w4 8 o 0 = Rb 3 3 w4 8 Mb (1) As there are two redundants, so we require two equations. Now differentiate strain energy expression w.r.t. another redundants Mb. Use castigations theorem and boundary condition. U Mb = θb = 0 = 1 EI o RbX wx Mb ( 1) dx 0 = 1 EI o RbX + wx + Mb dx 0 = 1 EI RbX + wx3 + MbX 6 o 0 = Rb + w3 6 + Mb. Rb w3 6 = Mb So Mb = Rb w 6 () Put Mb in equation 1, we get 0 = Rb3 3 w4 8 Rb w 6

METHOD OF EAST WORK 97 0 = Rb3 3 w4 8 Rb3 4 + w4 1 0 = Rb3 1 w4 4 Put Rb value in equation, we have Mb = w w 6 Rb = w Mb = +w 1 The (+ve) value with Rb and Mb indicates that the assumed directions of these two redundants are correct. Now find other reactions Ra and Mb by using equations of static equilibrium. Ra + Rb = w Ra = w Rb = w w Ra = w Put X =, Rb = w & Mb = w 1 in M X expression to get Ma Ma = w. w w 1 Ma = w 1 The ( ve) sign with Ma indicates that this moment should be applied in such direction that it gives us tension at the top. Example No. 4: Solve the same previous fixed ended beam by taking a simple beam as B.D.S.: Choosing Ma and Mb as redundants. A Ma WwKN/m x B Mb Ra BDS UNDER OADS AND REDUNDANTS B.D.S. is a simply supported beam, So Ma and Mb are redundants. Rb

98 THEORY OF INDETERMINATE STRUCTURES Ma = 0 Rb + Ma = Mb + w Rb = (Mb Ma ) + w Rb = Mb Ma + w So taking B as origin. Write M X expression. M X = RbX Mb wx 0 < X < Put Rb value M X = Mb Ma + w X wx Mb 0 < X <. Set up strain energy expression. 1 U = EI o Mb Ma + w X wx Mb dx. Differentiate w.r.t. Ma first. Use castigations theorem and boundary conditions. U Ma = θa = 0 = 1 EI 0 = 1 EI 0 = 1 EI o o MbX o MaX Mb Ma + w X wx + w X wx Mb X MbX + MaX wx + wx3 + MbX Mb X dx dx. In general R.H.S. is 1 EI N.m.dX. dx. Integrate it. 0 = 1 EI Mb X 3 + Ma X3 3 wx3 6 + wx4 8 + MbX o. Simplify it. 0 = Mb + Ma 6 3 w3 (1) 4 Now differentiate U Partially w.r.t. Mb. Use castiglianos theorem and boundary conditions. U Mb = θb = 0 = 1 EI o Mb Ma + w X wx X 1 dx Mb 0 = 1 EI 0 = o o MbX MaX MbX MaX + wx + w X wx X 1 dx Mb wx3 MbX MbX + MaX wx + wx + Mb dx

METHOD OF EAST WORK 99 0 = MbX 3 3 MaX3 3 + wx3 6 wx4 8 MbX Put limits now. MbX + MaX wx 4 + wx3 6 + MbX o 0 = Mb 3 3 Ma3 3 + w3 6 w4 8 Mb Mb + Ma w 4 + w3 6 + Mb Simplifying we get. or 0 = Mb 3 Mb 3 + Ma 6 w3 4 = Ma 6 + w3 4 so Mb = w 8 Ma (), Put Mb in equation (1) we get. 0 = w 8 Ma 6 + Ma 3 w3 4 Simplify to get Ma. 0 = w3 48 Ma 1 + Ma 3 w3 4 Put Ma in equation (), we have Mb = w 8 w 1 1 Ma = w 1 or Mb = w Ma + Mb 1 ; Now Rb = + w Putting Ma and Mb we have. w Rb = 1 w 1 Ra + Rb = w Ra = w Rb Ra = w w + w Rb = w Put value of Rb. Ra = w, Calculate Ra now. We get same results even with a different BDS. The beam is now statically determinate. SFD and BMD can be drawn. Deflections at can be found by routine methods.

100 THEORY OF INDETERMINATE STRUCTURES.6. ND DEGREE INDETERMINACY OF BEAMS: Exmaple No. 5: Solve the following loaded beam by the method of least work. W W A B / / C EI=Constant B.D.S. is a cantilever supported at A. Rb & Rc are chosen as redundants. Ma W x W A C B Ra Rb Rc x / / BDS UNDER OADS AND REDUNDANTS Choosing C as origin, Set-up moment expressions in different parts of this beam. Mbc = Rc.X wx 0 < X < Mab = Rc.X + Rb U = 1 EI / X o Rc.X wx wx dx + 1 EI < X <. Write strain energy expression for entire structure. / Rc.X + Rb X wx Partially differentiate it w.r.t. redundant Rc first. Use castiglianos theorem and boundary conditions. U Rc = c = 0 = 1 / EI 0 = 1 EI / o Rc.X wx o Rc.X wx3 0 = 1 EI Rc. X3 3 wx 4 8 / o dx + 1 EI [X]dX + 1 EI / Rc.X + Rb X dx wx / wx3 Rc.X + Rb.X Rb.X + 1 EI Rc. X3 X3 + Rb. 3 3 RbX. wx 4 4 8 / [X] dx dx. Integrate it.. Insert limits and simplify. 0 = Rc.3 3 + 5Rb.3 48 w4 8 (1)

METHOD OF EAST WORK 101 Now partially differentiate strain energy w.r.t. Rb. Use Castiglianos theorem and boundary conditions. U Rb = b = 0 = 1 EI 0 = 0 + 1 EI / o Rc.X wx (0) dx + 1 EI / Rc.X + Rb X wx X / wx3 + Rb. + w.x 4 4 Rc.X 3 3 Rc.X + RbX RbX Rc..X Rb.X 0 = 1 EI + Rb.X3 Rb..X wx4 Rc..X Rb.X + w.x3 3 4 8 4 4 + Rb..X 4 1 dx. dx Integrate. /. Put limits 0 = Rc.3 3 + Rb.3 3 Rb.3 4 w4 8 Rc.3 4 Rb.3 4 + Rb.3 4 + w4 1 Rc.3 4 Rb.3 4 Simplify to get + Rb.3 16 + w4 18 + Rc.3 16 + Rb.3 16 Rb.3 w4 8 96 Rc. = 17 Rb. + w () Put this value of Rc in equation ( 1), to get Rb 5 40 0 = 5 Rb. + 17 40 w 3 3 + 5 48 Rb.3 w4 8 (1) Simplify to get 0 = 15 Rb.3 + 17 10 w4 + 5 48 Rb.3 w4 8 Rb. = 1 1 w Put value of Rb in equation () and evaluate Rc, Rc = 5 1 17 w + 1 40 w Rc = 11 56 w The (+ve) signs with Rb & Rc indicate that the assumed directions of these two redundants are correct. Now calculate Ra. Ra + Rb + Rc = w or Ra = w Rb Rc. Put values of Rb and Rc from above and simplify.

10 THEORY OF INDETERMINATE STRUCTURES Ra = = w 1 11 w w 1 56 373 1176 w Ra = 91 39 w Putting the values of these reactions in Mx expression for span AB and set X =, we have Ma = Rc. + Rb. w. Put values of Rb and Rc from above and simplify. = 11 w 1. + 56 1 w w Ma = 1 1176 w Ma = 7 39 w The ( ve) sign with Ma indicates that this reactive moment should be applied in such a direction that gives us tension at the top. Now the beam has been analyzed and it is statically determinate now..7. INTERNA INDETERMINACY OF STRUCTURES BY FORCE METHOD : The question of internal indeterminacy relates to the skeletal structures like trusses which have discrete line members connected at the ends. The structures which fall in this category may include trusses and skeletal frames. For fixed ended portal frames, the question of internal indeterminacy is of theoretical interest only. 1 Relative displacement of horizontal number = Consider he truss shown in the above diagram. If this truss is to be treated as internally indeterminate, more than one members can be considered as redundants. However, the following points should be considered for deciding the redundant members. (1) The member which is chosen the redundant member is usually assumed to be removed or cut. The selection of redundant should be such that it should not effect the stability of the remaining structure.

METHOD OF EAST WORK 103 () The skeletal redundant members will have unequal elongations at the two ends and in the direction in which the member is located. For example, if a horizontal member is chosen as redundant, then we will be concerned with the relative displacement of that member in the horizontal direction only. (3) Unequal nodal deflection ( 1 ) of a typical member shown above which is often termed as relative displacement is responsible for the self elongation of the member and hence the internal force in that member..7.1. FIRST APPROACH: WHEN THE MEMBER IS REMOVED : With reference to the above diagram, we assume that the redundant member (sloping up to left) in the actual structure is in tension due to the combined effect of the applied loads and the redundant itself. Then the member is removed and now the structure will be under the action of applied loads only. B Apart Together B 1 B C A D Due to the applied loads, the distance between the points B and D will increase. et us assume that point B is displaced to its position B. This displacement is termed as apart. Now the same structure is considered under the action of redundant force only and let us assume that point B comes to its position B 1 (some of the deflections have been recovered). This displacement is termed as together. The difference of these two displacements ( apart together) is infact the self lengthening of the member BD and the compatibility equation is apart together = self elongation..7.. ND APPROACH We assume that the member is infact cut and the distance between the cut ends has to vanish away when the structure is under the action of applied loads and the redundant. In other words, we can say that the deformation produced by the applied loads plus the deformation produced by the redundant should be equal to zero. B C B C 1 1 A D A D F-Diagram U-Diagram

104 THEORY OF INDETERMINATE STRUCTURES Total Deflection produced by redundants R = n i = 1 U i i X AiEi n FiUii Total Deflection produced by loads = i = 1 AiEi If deflection is (+ve), there is elongation. If deflection is ( ve), there is shortening. Now U = P Elastic strain energy stored due to axial forces AE P AE PROOF: Work done = 1/ P. = shaded area of P diagram. Now f α (Hooke s aw) P or P A α P A = E = P AE (For direct stresses) where E is Yung s Modulus of elasticity. Therefore work done = P = 1 P P. AE ( Shaded area under P line By putting value of ) Work done = P AE Work done = P AE (for single member) (for several members) We know that Work done is always equal to strain energy stored.

METHOD OF EAST WORK 105 EXAMPE NO 6: Analyze the truss shown below by Method of east work. Take (1) Member U 1 as redundant. () Member U 1 U as redundant. Number in brackets ( ) are areas 10 3 m. E = 00 10 6 KN/m 0U1 = 7.5m Cos = 0.8 Sin = 0.6 (3.0) (1.) U1 (4) (1.8) (1.8) U (3.0) 6m 0 (.4) 1 (.4) (.4) 3 3 @ 4,5m 48KN Note: In case of internally redundant trusses, Unit load method (a special case of strain energy method) is preferred over direct strain energy computations followed by their partial differentiation. SOUTION: Case 1 Member U 1 as redundant 0 U1=7.5m U 1 U Cos = 0.8 Sin = 0.6 3.0 1. 1.8 1..0 0.4 1.4 3@4.5m 48KN (1) U 1 is redundant.: STEPS F-Diagram 3.0 3 6m 1 Remove this member. (See diagram) Assume that tensile forces would be induced in this member. 3 Analyze the structure without U 1 (B.D.S.) or F' diagram. 4 Displacement of members due to redundant + that due to loads should be equal to zero. OR + R = 0 5 Analyze the truss with unit tensile force representing U 1 or U diagram.

106 THEORY OF INDETERMINATE STRUCTURES Condition: apart = 8 1 F U AE together= 8 1 U AE P U 1 U 1.4 U 1.8 3.0 1. 1. 3.0 6m 0 1.4 3 16 0 + 3@4.5m (BDS under loads) F - diagram / SFD 3 144 7 + 0 0 B.M.D. + We shall determine member forces for F / - diagram by method of moments and shears as explained earlier. These are shown in table given in pages to follow. Member forces in U-diagram are determined by the method of joints. U1 0.60 48 +1.0 U 1 1 Cos o 3 O 1 o 0 JOINT ( ) Sin (BDS under) U-diagram redundant unit force. 1 U 1 F X = 0 1 Sinθ + 1 = 0 1 = Sinθ = 0.60 U + 1 Cosθ = 0 U = Cosθ = 0.80

METHOD OF EAST WORK 107 Joint (1) U 1 1 1 U 0.6 1 F X =0 1 U Sinθ 0.6 = 0 1 U = 0.6 0.6 = + 1 1 U 0.80 + U 1 = 0 U 1 1 = 0.80 Now Book F / forces induced in members as determined by moments and shears method and U forces as determined by method of joints in a tabular form. Member A 10-3 (m ) (m) Fi (KN) Ui F U AE 10-3 (m) U AE 10-3 (m) Fi=Fi +UiX (KN) U 1 U.4 4.5 1 0.6 +0.0675 3.375 10-3 5.15 o 1.4 4.5 +1 0 0 0 +1 1.4 4.5 +4 0.6 0.135 3.375 10-3 +10.84 3.4 4.5 +4 0 0 0 +4 ou 1 3.0 7.5 0 0 0 0 0 1 U 4.8 7.5 0 +1.0 0.416 0.83 10-3 + 1.93 U 3 3.0 7.5 40 0 0 0 40 U 1 1 1. 6.0 +16 0.8 0.3 16 10-3 1.54 U 1. 6.0 +48 0.8 0.96 16 10-3 +30.456 U 1 1.8 7.5 0 +1.0 0 0.83 10-3 +1.96 Compatibility equation is + R = 0 F U = AE n 1 1.7635 10 3 80.91 10 6 R = n 1 U AE. X Putting values from above table in compatibility equation. Where R = X = force in redundant Member U 1

108 THEORY OF INDETERMINATE STRUCTURES 1.7635 10-3 + 80.41 10-6. X = 0 or 1.7635 10-3 + 0.08041 10-3. X = 0 1.7635 + 0.08041 X = 0 0.08041 X = 1.7635 X = 1.7635 0.08041 X = + 1.93 KN (Force in members U 1 ) Now final member forces will be obtained by formula Fi = Fi' + Ui X. These are also given in above table. Apply check on calculated forces. Check on forces Joint o 0 1 Note: 16 Tensile forces in above table carry positive sign and are represented as acting away from joint. Compressive forces carry negative sign and are represented in diagram as acting towards the joint. Fx = 0 1 0 Sin θ = 0 1 0 0.6 = 0 0 = 0 16 0 Cos θ = 0 16 0 0.8 = 0 0 = 0 Checks have been satisfied showing correctness of solution. EXMAPE NO. 7: CASE : Analyze previous loaded Truss by taking U 1 U as Redundant F / =Diagram 0 U 1 U 40 3 0 64 Cos = 0.8 Sin = 0.6 40 0 3 1 36 4 16 48 3

METHOD OF EAST WORK 109 In this case member forces in BDS (F / diagram) have been computed by method of joints due to obvious reasons.) Joint o:- ou 1 16 o 1 Joint U 1 16 + ou 1 Cosθ = 0 ou 1 = 16 0.8 = 0 F X = 0 o 1 + ou 1 Sinθ = 0 o 1 + ou 1 0.6 = 0 o 1 0 0.6 = 0 o 1 = + 1 0 U 1 1 Joint 1 : 0 Sinθ+ U 1 Sinθ = 0 0 0.6 + U 1 0.6 = 0 U 1 = 0 0 0.8 U 1 1 U 1 0.8 = 0 0 0.8 U 1 1 + 0 0.8 = 0 U 1 1 = 3 1 1 U Cosθ + 3 = 0 3 U1 1 U 1

110 THEORY OF INDETERMINATE STRUCTURES 1 U = 3 0.8 1 U = 40 1 + 1 U Sinθ 1 = 0 1 40 0.6 1 = 0 1 = 36 Joint U 40 U U 3 40 Sinθ + U 3 Sinθ = 0 40 0.6 + U 3 0.6 = 0 U 3 = 40 40 Cosθ U 3 Cosθ U = 0 40 0.8 ( 40) 0.8 U = 0 U = 64 Joint 0 64 36 3 3 + 0 Sinθ 36 = 0 3 + 0 0.6 36 = 0 3 4 = 0 3 = 4 48

METHOD OF EAST WORK 111 Joint 3 (Checks) 40 4 3 40 Sinθ 4 = 0 40 0.6 4 = 0 0 = 0 3 40 Cosθ = 0 3 40 0.8 = 0 0 = 0 Checks are satisfied. Results are OK and are given in table at page to follow: Now determine member forces in U diagram. U 1 1 1 U 0 0 1.38 1.66 1.38 1.66 0 1 1 0 U-Diagram (BDS under unit redundant force) 0 3 Joint U 1 1 U 1 1 U 1 1 + U 1 Sinθ = 0 1 + U 1 0.6 = 0

11 THEORY OF INDETERMINATE STRUCTURES U 1 = 1.66 U 1 1 +U 1 Cosθ = 0 U 1 1 + ( 1.66) 0.8 = 0 U 1 1 = 1.38 Joint 1 :- 1.38 1 U 1 1.38 + 1 U 0.8 = 0 1 U = 1.38 0.8 = 1.66 1 + 1 0.6 = 0 1 1.66 0.6 = 0 1 = +1 Entering results of member forces pertaining to F / diagram and U diagram alongwith member properties in a tabular form. Member A 10-3 (m) (m) Fi (KN) U 1 F U AE 10-3 (m) U AE 10-3 (m) Fi=Fi+UiX (KN) U 1 U.4 4.5 0 +1 0 9.375 10-3 5.34 o 1.4 4.5 +1 0 0 0 + 1 1.4 4.5 + 36 + 1 +0.3375 9.375 10-3 +10.66 3.4 4.5 +4 0 0 0 + 4 ou 1 3.0 7.5 0 0 0 0 0 1 U 1.8 7.5 40 1.66 +1.383 57.4 10-3 +.06 U 3 3.0 7.5 40 0 0 0 40 U 1 1 1. 6.0 + 3 1.38 1.064 44.09 10-3 + 65.65 U 1. 6.0 + 64 1.38.148 44.09 10-3 + 97.65 U 1 1.8 7.5 0 1.66 0.691 57.4 10-3 6.06 5.6 10-3 1.73 10-6

METHOD OF EAST WORK 113 Compatibility equation is + R = 0 Putting values of and R due to redundant from above table. 56 10-3 + 1.73 10-6 X = 0, where X is force in redundant member U 1 U. or 5.6 10-3 + 0.173 10-3 X = 0 X = 5.6 10-3 0.173 10-3 X = 5.34 KN. FU 1 U = 0 + Ui.x = 0 5.34 1 = 5.34 Fo 1 = 1 5.34 0 = + 1 F 1 = 36 5.34 1 = + 10.66 F 3 = 4 0 = + 4 FoU 1 = 0 0 5.34 = 0 F 1 U = 40 + 1.66 5.34 = +.06 FU 3 = 40 + 0 5.34 = 40 FU 1 1 = + 3 + 1.38 5.34 = + 65.65 FU = + 64 + 1.38 5.34 = + 97.65 Therefore forces in truss finally are as follows. (by using formula (Fi = Fi' + UiX and are given in the last column of above table) FU 1 = 0 1.66 5.34 = 6.06. Now based on these values final check can be applied. Joint o. 0 1 1 0 Sinθ = 0 1 0 0.6 = 0 0 = 0 16 0 Cosθ = 0 16 0 0.8 = 0 16 16 = 0 0 = 0 Results are OK. 16

114 THEORY OF INDETERMINATE STRUCTURES.8. STEPS FOR TRUSS SOUTION BY METHOD OF EAST WORK. Now instead of Unit load method, we shall solve the previous truss by direct use of method of least work. (1) Consider the given truss under the action of applied loads and redundant force X in member U 1 () The forces in the relevant rectangle will be a function of applied load and redundant force X. (As was seen in previous unit load method solution) (3) Formulate the total strain energy expression due to direct forces for all the members in the truss. (4) Partially differentiate the above expressions with respect to X. (5) Sum up these expressions and set equal to zero. Solve for X. (6) With this value of X, find the member forces due to applied loads and redundant acting simultaneously (by applying the principle of super positions). EXAMPE NO. 8 :- Analyze the loaded truss shown below by least work by treating member U 1 as redundant. Numbers in ( ) are areas 10-3 m. E = 00 10 6 KN/m. SOUTION:- 48 x 4.5 48 = 16KN 3 b = 10 r = 3 j = 6 b + r = j 10 + 3 = 6 13 = 1 D = 13 1 = 1

METHOD OF EAST WORK 115 Stable Indeterminate to 1st degree. X X 48 16 3 F Diagram (Truss under loads and redundant) NOTE: Only the rectangle of members containing redundant X contains forces in terms of X as has been seen earlier. Now analyze the Truss by method of joints to get Fi forces. JOINT 0 16KN 0 U 1 0 1 Joint U 1 ou 1 Cosθ + 16 = 0 ou 1 = 16 Cosθ = 16 0.8 ou 1 = 0 KN o 1 + ou 1 Sinθ = 0 o 1 + ( 0) 0.6 = 0 o 1 1 = 0 o 1 = 1 KN 0 U 1 U U 1 U + X Sinθ + 0 Sinθ = 0 U 1 1 X

116 THEORY OF INDETERMINATE STRUCTURES U 1 U + X 0.6 + 0 0.6 = 0 U 1 U = (0.6 X +1) U 1 1 X Cosθ + 0 Cosθ = 0 U 1 1 X 0.8 + 0 0.8 = 0 U 1 1 = 0.8 X + 16 U 1 1 = (0.8 X 16) Joint 1 :- 0.8X - 16 1 U 1 1 (0.8X 16) + 1 U Cosθ = 0 1 U 0.8 = 0.8 X 16 1 U = (X 0) 1 + 1U Sinθ 1 = 0 Put value of 1 U. 1 + (X 0 ) 0.6 1 = 0 1 + 0.6 X 1 1 = 0 1 = (0.6X 4) Joint U (0.6X+1) (X-0) U U 3 (0.6 X + 1) + U 3 Sinθ (X 0) Sinθ = 0 0.6 X + 1 + U 3 0.6 (X 0) 0.6 = 0

METHOD OF EAST WORK 117 0.6 X + 1 + 0.6U 3 0.6 X + 1 = 0 U 3 = 4 0.6 U 3 = 40 KN U (X 0) Cosθ U 3 Cosθ = 0 U (X 0) 0.8 ( 40) 0.8 = 0 U 0.8 X + 16 + 3 = 0 0.8 X + 48 = U Joint :- U = (0.8X 48) X 0.6 X-4 3 + 0.6 X 4 X Sinθ = 0 3 = 0.6 X + 4 + 0.6 X 0.8 X- 48 48 3 3 = 4 KN (0.8X 48) 48 + X Cosθ = 0 0.8X + 48 48 + 0.8X = 0 0 = 0 (Check) Joint 3 :- At this joint, all forces have already been calculated. Apply checks for corretness. 40 4 3

118 THEORY OF INDETERMINATE STRUCTURES 40 Sinθ 4 = 0 40 0.6 4 = 0 4 4 = 0 0 = 0 O.K. 40 Cosθ + 3 = 0 40 0.8 + 3 = 0 3 + 3 = 0 O.K. Checks have been satisfied. 0 = 0 This means forces have been calculated correctly. We know that strain energy stored in entire Truss is U = Fi AE Fi Fi U So, X = = 0 = X. i AE Fi Fi X. i = 0 = 80.41 10 AE 6 X 1764.17 10 6 Values of Fi and i for various members have been picked up from table annexed. 0 = 80.41 X 1764.17 or 80.41 X = 1764.17 X = 1764.17 80.41 X = 1.94 KN Now putting this value of X in column S of annexed table will give us member forces. Now apply equilibrium check on member forces calculated. You may select any Joint say 1. Joint 1 :- 1 15.5 1.74 10.84, 10.84 1 + 1.94 Sinθ = 0 or 10.84 1 + 1.94 0.6 = 0, or 0 = 0 (Check) It means that solution is correct.

METHOD OF EAST WORK 119 Insert here Page No. 138 A

10 THEORY OF INDETERMINATE STRUCTURES EXAMPE NO. 9:- By the force method analyze the truss shown in fig. below. By using the forces in members 1 U and U 3 as the redundants. Check the solution by using two different members as the redundants. E = 00 10 6 KN/m SOUTION:- 0U1 = 7.5m U1 (1.8) U (1.8) U Cos = 0.8 3 (1.) Sin = 0.6 (.4) 0 0 (1.) 6m (0.90) (1.) (1.) (0.60) (0.90) 48+96+7-114 = 10KN 0 (1.5) (1.5) (1.5) (1.5) 4 1 3 48KN 96KN 7KN 4@4.5m F - Diagram 48x4.5 18 + 96x9 18 + 7x13.5 = 114KN 18 10KN 10KN 0 (.4) (1.) (1.) (0.90) (0.60) (1.) (1.)(0.90) (.4) loads only. 6m Or F-Diagram (1.5) (1)(1.5) ()(1.5) (1.5) 4 0 1 3 4 48KN 96KN 7KN 114KN 4@4.5m + 54KN 4KN - S.F.D. 0 114KN 459 KN-m 70KN-m 513KN-m + 0 0 0.6 0 B.M.D. 0 0 0 0.8 0.8 0 0 1 0 3 0 0.6 0 4 6m U 1 -Diagram for redundant X 1 0 0 0 0 0.8 0.8 6m U -diagram for redundant X 0 0 1 0 0.6 3 0 4 Compatibility equations are: X 1 + X 1 R 1 + X 1 R = 0 (1) Change in length in member 1 due to loads and two redundants should be zero. X + X R 1 + X R = 0 () Change in length in member due to loads and two redundants should be zero. Here R 1 = X 1 R = X

METHOD OF EAST WORK 11 Where X 1 =.F U 1 AE = Deflection produced in member (1) due to applied loads. X 1 R 1 = Deflection produced in member (1) due to redundant R 1 = x 1 R = Deflection produced in member (1) due to redundant R = x = Deflection produced in member () due to loads = F U AE x R 1 = Deflection produced in member () due to redundant R 1 = x R = Deflection produced in member () due to redundant R = U 1 AE. X 1 U 1 U. X AE U 1 U. X AE 1 U AE. X From table attached, the above evaluated summations are picked up and final member forces can be seen in the same table. All member forces due to applied loads (Fi' diagram) have been determined by the method of moments and shears and by method of joints for U 1 and U diagrams. Evaluation of member forces in verticals of F Diagram :- Forces in verticals are determined from mothod of joints for different trusses shown above. (Joint 1 ) U 1 1 76.5 76.5 U 1 1 48 = 0 48 U 1 1 = 48 (Joint U ) 117 85.5 U + 5.5 Cosθ = 0 U + 5.5 0.8 = 0 U = 5.5 0.8 U 5.5 U = + 4

1 THEORY OF INDETERMINATE STRUCTURES (Joint U 3 ) 85.5 U 3 3 + 14.5 Cosθ = 0 U 3 3 = 14.5 0.8 U 3 3 14.5 U 3 3 = + 114 Evaluation of forces in verticals of U 1 Diagram:- (Joint 1 ) U 1 1 1 1 1 + 1 Sin θ = 0 1 = 0.6 U 1 1 + 1 Cos θ = 0 U 1 1 = 0.8 (Joint U 1 ) U 1 U U 1 U + U 1 Sinθ = 0 0.8 U 1

METHOD OF EAST WORK 13 + 0.8 U 1 Cos θ = 0 0.8 = U 1 0.8 U 1 = 1 so U 1 U + 1 0.6 = 0 Putting value of U 1 in FX. U 1 U = 0.6 Now from the table, the following values are taken. X 1 = 0.671 10-3 X 1 R 1 = 15.7 10 6 X1 = 0.157 10-3 X1 X 1 R = 3 10-6 X = 0.03 10-3 X X = 6.77 10-3 X R 1 = 0.03 10-3 X1 X R = 15.6 10-6 X = 0.156 10 3 X Putting these in compatibility equations, we have. 0.671 10 3 +0.157 10 3 X1+0.03 10 3 X = 0 (1) 6.77 10 3 +0.03 10 3 X1+0.156 10 3 X = 0 () dividing by 10 3 0.671+0.157X1 + 0.03X = 0 (1) 6.77 + 0.03X1 + 0.156X = 0 () From (1), X 1 = 0.671 0.03X 0.157 (3) Put X 1 in () & solve for X 6.77 + 0.03 0.671 0.03X + 0.156X 0.157 = 0 6.77 + 0.171 8.146 10-3X + 0.156X = 0 6.599 + 0.1174X = 0 0.1174X = 6.599 X = 56.19 KN 0.671 0.03 56.19 From (3) X1 = 0.157 X1 = 8.96 KN After redundants have been evaluated, final member forces can be calculated by using the formula shown in last column of table. Apply checks on these member forces.

14 THEORY OF INDETERMINATE STRUCTURES CHECKS:- (Joint o) 17.5 76.5 10 76.5 17.5 Sinθ = 0 76.5 17.5 0.6 = 0 0 = 0 10 17.5 Cosθ = 0 10 17.5 0.8 = 0 0 = 0 The results are O.K. Follow same procedure if some other two members are considered redundant. See example No. 1.

METHOD OF EAST WORK 15 Insert Page No. 143 A

16 THEORY OF INDETERMINATE STRUCTURES.9. SIMUTANEIOUS INTERNA AND EXTERNA TRUSS REDUNDANCY EXAMPE NO. 10: Determine all reactions and member forces of the following truss by using castiglianos theorem or method of least work. Consider it as: (i) internally redundant; (ii) internally and externally redundant. SOUTION: Nos. in ( ) are areas in 10-3 m. E = 00 10 6 KN/m 3KN (3) 6KN B 0KN C (4) (5) DEGREE OF INDETERMINACY :- A (3) () (4) E () (5) 6m () F D = (m + r ) j = (10 + 4 ) 6 = 0KN D Therefore, the truss is internally statically indeterminate to the nd degree. There can be two approaches, viz, considering two suitable members as redundants and secondly taking one member and one reaction as redundants for which the basic determinate structure can be obtained by cutting the diagonal CE and replacing it by a pair of forces X 1 X 1 and replacing the hinge at F by a roller support with a horizontal redundant reaction HF = X. Applying the first approach and treating inclineds of both storeys sloping down to right as redundants. (I) WHEN THE TRUSS IS CONSIDERED AS INTERNAY REDUNDANT :- 3KN 0KN C (4) 0KN D 8m 8m (3) 6KN B (5) A X 1 (3) () (4) X1 E () (5) X 6m () F 8m 8m Applying method of joints for calculating member forces.

METHOD OF EAST WORK 17 Consider Joint (C) and all unknown forces are assumed to be in tension to begin with, acting away from the joint. ength AE= 10 m, cos θ = 0.6, sin θ = 0.8 Joint (C) 0KN 3KN SCD Scd + 3 + X 1 Cos θ = 0 Scd = (3 + 0.6 X 1 ) Sbc X 1 Sin θ 0 = 0 Sbc = ( 0 + 0.8 X 1 ) Joint (D) SBC X1 0KN (3+0.6X1) S BD SDE 3 + 0.6X 1 S BD 0.6 = 0 S BD = ( 5 + X 1 ) S DE 0 S BD Sinθ = 0 S DE 0 ( 5 + X 1 ) 0.80 = 0 S DE = ( 4 + 0.8X 1 ) Joint (B) (0+0.8X1) (5+X1) 6KN S BE X S AB S BE + (5+X 1 ) 0.6 + X 0.6 + 6 = 0 S BE = ( 9 + 0.6 X 1 + 0.6 X )

18 THEORY OF INDETERMINATE STRUCTURES S AB X Sinθ (0 + 0.8 X 1 ) + (5+X 1 ) Sinθ = 0 S AB 0.8 X 0 0.8 X 1 + 4 + 0.8 X 1 = 0 S AB = (16 + 0.8 X ) Joint (E) (4 + 0.8 x 1) X 1 9+0.6X + 0.6X 1 S AE S EF 9 + 0.6 X 1 + 0.6 X X 1 x 0.6 S AE 0.6 = 0 9 + 0.6 X = S AE 0.6 S AE = ( 15 + X ) S EF 4 0.8 X 1 + X 1 0.8 (15 + X ) 0.8 = 0 S EF = 4 0.8 X 1 + 0.8 X 1 1 0.8 X = 0 S EF = 36 0.8 X S EF = (36 + 0.8 X ) Enter Forces in table. Now applying Catiglianos theorem and taking values from table attached. S. S. X 1 AE = 0 = 485.6 + 65.64X 1 +.7X = 0 (1) and S. S X. AE = 0 = 748.3 +.7X 1 + 6.94 X = 0 () or 485.6 + 65.64 X 1 +.7 X = 0 (1) 748.3 +.7 X 1 + 6.94 X = 0 () From (1) X = 485.6 + 65.64 X 1 putting in ().7 748.3 +.7 X 1 6.94 485.6 + 65.64 X 1 = 0 ().7 748.3+.7X 1 11319.875 1530.141X 1 10571.575 157.441 X 1 = 0 (3) From (3) X = X 1 = 6.91 KN 485.6 65.64 6.91.7 X = 11.59 KN Now put values of X 1 and X in 5 th column of S to get final number forces SF as given in last column of table. Apply equilibrium check to verify correctness of solution.

METHOD OF EAST WORK 19 Insert Page No. 148 A

130 THEORY OF INDETERMINATE STRUCTURES EQUIIBRIUM CHECKS :- Joint (A) 6.76KN HA 3.408KN 4KN 3.408 Cosθ H A 0 H A =.045 KN 6.76 + 4 + 3.408 Sinθ = 0 0 = 0 Check is OK. Joint (F) 11.59KN 6.76KN HF 36KN H F + 11.59 Cosθ = 0 H F = + 6.955 KN 36 7.76 11.59 Sinθ = 0 0 = 0 (check) It means solution is correct. Now calculate vertical reactions and show forces in diagram.

METHOD OF EAST WORK 131 3KN 6KN C 14.463 B 6.76 6.76 11.59 HA=.045Kn 3.408 A VA=4KN 0KN 1.153 6.91 1.91 4.46 6m 0KN D 18.463 E HF=6.955KN F VF=+36KN 8m 8m ANAYZED TRUSS M A = 0 V F 6 0 6 3 16 6 8 = 0 V F = + 36 KN V A + V F = 40 KN V A = + 4 KN EXAMPE NO. 11: CASE II : When the Truss is considered as both externally & internally redundant. Taking S CE & H F as redundants. Now Truss is determinate and calculate vertical reactions. Cos =0.6 8m Sin =0.8 8m 3kn C 6KN B 0KN X X 0KN D E Fy = 0 VA + VF = 40 MA = 0 VFx6-3x16-0x6-6x8=0 VF = 36KN and VA = 4KN (9-H F ) A F H F 4KN 36Kn 6m Fig..51

13 THEORY OF INDETERMINATE STRUCTURES Compatibility Equations are: S. S HF. = 0 AE (1) Partial differentiation of strain energy w.r.t. HF = H = 0. (Pin support) S. S X. = 0 AE () Partial differentiation of strain energy w.r.t. X = elongation of member CE due to X = 0. As before determine member forces Si in members by method of joints. Joint (A) (9-HF) S AB SAE FX = 0 S AE Cosθ (9 H F ) = 0 S AE 0.6 (9 H F ) = 0 S AE = 9 H F 0.6 4 S AE = 15 1.67 H F 4 + S AB + S AE Sinθ = 0 4 + S AB + (15 1.670 H F ) 0.8 = 0 4 + S AB + 1 1.33 H F = 0 S AB = 16 + 1.33 H F S AB = (16 1.33 H F ) Joint (F) S BF SEF H F 36

METHOD OF EAST WORK 133 H F S BF Cosθ = 0 H F 0.6 S BF = 0 H F = 0.6 S BF S BF = 1.67 H F 36 + S EF + S BF Sinθ = 0 36 + S EF 1.67 H F 0.8 = 0 S EF = (36 1.33 H F ) Joint (E) X S DE SBE (36-1.33H F) (15-1.67HF) S BE X Cosθ (15 1.67 H F ) Cosθ = 0 S BE 0.6X ( 15 1.67 H F ) 0.6 = 0 S BE 0.6X 9 + H F = 0 H F 0.6X 9 = S BE S BE = (H F 0.6 X 9) S DE +36 1.33 H F + X Sinθ (15 1.67H F ) Sinθ = 0 by putting Sinθ = 0.08 S DE + 36 1.33 H F + 0.8X 1 + 1.33 H F = 0 S DE = 0.8X 4 S DE = ( 4 + 0.8X) Joint (C) 0KN 3KN SCD SBC X

134 THEORY OF INDETERMINATE STRUCTURES S CD + 3 + X Cosθ = 0 S CD = ( 3 + 0.6X) 0 S BC X Sin θ = 0 0 S BC 0.8X = 0 S BC = ( 0 + 0.8 X ) Joint (D) 0KN (3+0.6X) SBD (4+ 0.8X) FX = 0 3 + 0.6X S BD Cosθ = 0 3 + 0.6X 0.6 S BD = 0 S BD = ( 5 + X) 0 + 4 + 0.8X S BD Sinθ = 0 0 + 4 + 0.8X ( 5 + X ) 0.8 = 0 0 + 4 + 0.8X 4 0.8X = 0 0 = 0 (check) Calculation of H F & X : From the attached table, picking up the values of summations, we have.. S. S. H F AE = 0 = ( 147.03 + 175.4 H F 4.5 X) 10 6

METHOD OF EAST WORK 135 and. S. S X. AE = 0 = (460.6 4.5 H F + 65.64X) 10-6 147.03 + 175.4 H F 4.5X = 0 (1) + 460.6 4.5 H F + 65.64X = 0 () From (1) X = 147.03 + 175.4 H F 4.5 (3) Put in () to get H F 460.6 4.5 H F + 65.64 147.03 + 175.4 H F 4.5 460.6 4.5 H F 18190.01 + 556.17 H F = 0 1779.41 + 551.67 H F = 0 = 0 H F = 6.948 KN Put this value in 3 to get X. X = 147.03 + 175.4 6.948 4.5 (3) or X = 6.541 KN Now calculate number Forces by putting the values of X and H F in S expressions given in column 5 of the attached table. These final forces appear in last column for S F. 3KN 6KN 0KN 0.95 C 14.76 B 1.541 6.641 1.873 18.767 3.39 6.759 6.759 8m 11.603.05Kn A F 6.948KN D E 0KN 8m 4kn 6m 36KN Fig.5 ANAYZED TRUSS

136 THEORY OF INDETERMINATE STRUCTURES Insert Page No. 153 A

METHOD OF EAST WORK 137 Equilibrium checks for the accuracy of calculated member Forces:- Joint (A) 6.759 3.397.05 3.397 Cosθ.05 = 0 0 = 0 Check 6.759 + 4 + 3.397 0.8 = 0 0 = 0 Check Joint (F) 11.603 6.759 4 6.948 Joint (C) 6.948 + 11.603 0.6 = 0 0 0 Check 36 6.759 11.603 0.8 = 0 0 0 Check 36 0 3 0.95 14.767 6.541 0.95 6.541 0.6 + 3 = 0 0 = 0 Check 14.767 0 + 6.541 0.8 = 0 0 = 0 Check. This verifies correctness of solution.

138 THEORY OF INDETERMINATE STRUCTURES EXAMPE NO. 1:- By the unit load method analyze the internally indeterminate truss shown below. Take the forces in members 1 U and U 3 as the redundants. Note: The same truss has already been solved in Example No. 9, by taking 1 U and U 3 as redundants. E = 00 10 6 KN/m SOUTION:- U 1 (1.8) U (1.8) U 3 ou1 = 7.5 m Cos = 0.8 (.4) (.4) Sin = 0.6 (1.) (1.) (1.) 0.90 (0.90) (1.) 0 (1.5) (1.5) (1.5) (1.5) 1 3 10KN 0 10KN 10 + 0 0 459 48KN 96KN 4@4.5m U 1 U U 3 7KN 1 3 48KN 96KN 7KN 54 4 70 513 U 0.6 1 U 0 U 3 1 0.8 0 0 0 0.8 0.6 0 1 0 3 0 U 1 0 U 0.6 U 3 + 4 114KN 114KN 6m F-Diagram B.D.S. Under applied load only. / Or F -Diagram 4 S.F.D. 114 4 B.M.D. U 1 -Diagram 0 0 0 0.8 0.8 0 U -Diagam 0 0 1 0 0.6 3 0 4

METHOD OF EAST WORK 139 Compatibility equations are : X 1 + X 1 R 1 + X 1 R = 0 (1) Here X 1 = R 1 X = R Deflection created by applied loads and redundants shall be zero. X + X R 1 + X R = 0 () X 1 =. F U 1 AE (Change in length of first redundant member by applied loads) X 1 R 1 = U 1 AE X 1 (Change in length in first redundant member due to first redundant force) X 1 R = U 1 U. X AE (Change in length in first redundant member due to second redundant force) X = F U AE (Change in second redundant member due to applied load.) X R 1 = X R = U 1 U. X AE 1 (Change in length of second redundant member due to first redundant force.) U AE. X (Change in length of second redundant member due to redundant force in it.) Picking up the above deformations from the table (158 A) and calculate final member forces by following formula. F = F' + U 1 X 1 + U X Forces in chord members and inclineds are determined by the method of moments and shears as explained already, while for verticals method of joints has been used. Evaluation of force in verticals of F Diagram (Joint ) 67.5 U 5.5 76.5 85.5 96 85.5 76.5 + 5.5 Sinθ 67.5 Sinθ = 0 85.5 76.5 + 5.5 0.6 67.5 0.6 = 0 0 = 0 (Check) Fy + 0 U + 5.5 Cosθ + 67.5 Cos θ 96 = 0 U = 5.5 0.8 67.5 0.8 + 96 = 0 U = 0

140 THEORY OF INDETERMINATE STRUCTURES Insert Page No. 158 A

METHOD OF EAST WORK 141 Picking the following values from attached table (Table for example No.1) X 1 = + 1.009 10 3 X 1 R 1 = + 15.7 10 6 X 1 = + 0.157 10 3 X 1 X 1 R = + 3 10 6 X = + 0.03 10 3 X X = 0.171 10 3 X R 1 = + 3 10 6 X 1 = + 0.03 10 3 X 1 X R = + 15.7 10 6 X = + 0.157 10 3 X Putting these in compatibility equals. 1.009 10 3 +0.157 10 3 X 1 +0.03 10 3 X = 0 (1) 0.171 10 3 +0.03 10 3 X1+0.157 10 3 X = 0 () Simplify 1.009 + 0.157 X 1 + 0.03 X = 0 (1) 0.171 + 0.03 X 1 + 0.157X = 0 () From (1) X 1 = 1.009 0.03 X 0.157 (3) Put in () & solve for X 0.171 + 0.03 1.009 0.03 X + 0.157 X 0.157 = 0 0.171 0.57 8.146 10 3 X + 0.157X = 0 0.48 + 0.1176 X = 0 X = 0.48 0.1176 X = 3.64 KN Put this in equation (3) to get X 1 (3) X 1 = 1.009 0.03 3.64 0.157 X 1 = 8.95 KN So final forces in members are calculated by the following given formula. F = F + U 1 X 1 + U X Fo 1 = 76.5 + 0 + 0 = + 76.5 KN F 1 = 76.5 + ( 0.6) ( 8.95) + 0 = + 81.87 KN F 3 = 85.5 + 0 + 3.64) ( 0.6) = + 83.3 KN F 3 4 = 85.5 + 0 + 0 = + 85.5 KN FU 1 U = 117 + ( 0.6) ( 8.95) + 0 = 111.63 KN FU U 3 = 117 + o +( 0.6) (3.64) = 119.18 KN FU 1 1 = + 48 + ( 0.8) ( 8.95) + 0 = + 55.16 KN FU = 0 + ( 0.8) ( 8.95) + ( 0.8) (3.64) = + 4.5 KN

14 THEORY OF INDETERMINATE STRUCTURES FU 3 3 = + 7 + 0 + ( 0.8) (3.64) = + 69.09 KN Flo U 1 = 17.5 + 0 + 0 = 17.5 KN FU 1 = + 67.5 + (1) ( 8.95) + 0 = 58.55 KN F 1 U = 0 + (1) ( 8.95) + 0 = 8.95 KN FU 3 = 0 + 0 + (1) (3.64) = + 3.64 KN F U 3 = 5.5 + 0 + (1) (3.64) = + 56.14 KN FU 3 4 = 14.5 + 0 + 0 = 14.5 KN CHECK ON FORCE VAUES We may apply check at random at any joint. If solution is correct, equilibrium checks will be satisfied at all joint. Joint o. 17.5 76.5 76.5 17.5 Sinθ = 0 76.5 17.5 0.6 = 0 0 = 0 10 10 17.5 0.8 = 0 0 = 0 OK. Results seem to be correct. The credit for developing method of least work goes to Alberto Castiglianos who worked as an engineer in Italian Railways. This method was presented in a thesis in partial fulfillment of the requirement for the award of diploma engineering of associate engineer. He published a paper for finding deflections which is called Castiglianos first theorem and in consequence thereof, method of least work which is also known as Castiglianos second theorem. Method of least work also mentioned earlier in a paper by an Italian General Menabrea who was not able to give a satisfactory proof. eonard Euler had also used the method about 50 years ago for derivation of equations for buckling of columns wherein, Daniel Bernolli gave valuable suggestion to him. Method of least work or Castiglianos second theorem is a very versatile method for the analysis of indeterminate structures and specially to trussed type structures. The method does not however, accounts for erection stresses, temperature stresses or differential support sinking. The reader is advised to use some other method for the analysis of such indeterminate structures like frames and continuos beams. It must be appreciated in general, for horizontal and vertical indeterminate structural systems, carrying various types of loads, there are generally more than one structural actions present at the same time including direct forces, shear forces, bending moments and twisting moments. In order to have a precise analysis all redundant structural actions and hence strain energies must be considered which would make the method laborious and cumbersome. Therefore, most of engineers think it sufficient to consider only the significant strain energy. The reader should know that most of structural analysis approaches whether classical or matrix methods consider equilibrium of forces and displacement/strain compatibility of members of a system.

METHOD OF EAST WORK 143 The basis of the method of consistent deformation and method of least work are essentially the same. In consistent deformation method, the deformation produced by the applied loads are equated to these produced by the redundants. This process usually results in the evolution of redundants. However, in the method of least work, total strain energy expression of a structural system in terms of that due to known applied loads and due to redundants is established. Then the total strain energy is partially differentiated with respect to redundant which ultimately result in the evolution of the redundant. It must be appreciated that, for indeterminate structural system like trusses, the unknown redundants maybe external supports reaction or the internal forces or both. And it may not be very clear which type of redundants should be considered as the amount of work involved in terms of requisite calculation may vary. Therefore, a clever choice of redundants (or a basic determinate structure as was the case with consistent deformation method) may often greatly reduce the amount of work involved. There is often a debate going on these days regarding the utility or justification of classical structural analysis in comparison to the computer method of structural analysis. It is commented that in case of classical methods of structural analysis the student comes across basic and finer points of structural engineering after which a computer analysis of a complex structure maybe undertaken. In the absence of basic knowledge of classical structural analysis, the engineer maybe in a difficult position to justify to computer results which are again to be checked against equilibrium and deformation compatibility only. EXAMPE NO. 13: The procedure for analysis has already been given. Utilizing that procedure, analyze the following truss by the method of least work. Areas in ( ) carry the units of 10 3 m while the value of E can be taken as 00 10 6 KN/m. E 4 F A 4 B C 15 kn 3@4.5m 4 D 4.5m where i = total degree of indeterminacy b = number of bars. r = total number of reactive components which the support can provide. b + r = j 10 + 3 > 6 13 > 1 so i= 1. First degree internal indeterminancy. F U = Strain energy due to direct forces induced due to applied loads in a BDS Truss. AE U X = F. F X. AE = 0

144 THEORY OF INDETERMINATE STRUCTURES Note: We select the redundants in such a way that the stability of the structure is not effected. Selecting member EC as redundant. A 5KN B E x x C F 15KN D 4.5m 10KN F-diagram B.D.S. under the action of applied loads & redundant. 5 + S.F.D. due to applied load only..5 + 45 10 load only. Method of moments and shears has been used to find forces in BDS due to applied loads. A table has been made. Forces vertical in members in terms of redundant X may be determined by the method of joints as before. From table. F. F x. AE = 0 = 331. 10 6 + 51.49 10 6 X or 331. + 51.49X = 0 X = + 6.433 KN The final member forces are obtained as below by putting value of X in column 5 of the table. Member Force (KN) AB + 5 BC +5.45 CD + 10 EF 9.55 BE + 0.45 CF + 10.45 CE + 6.43 BF 0.64 AE 7.07 DF 14.14

METHOD OF EAST WORK 145 Insert page No. 164 A

146 THEORY OF INDETERMINATE STRUCTURES CHECK. Joint A. 5 5 5 7.07 Cosθ = 0 5 7.07 0.707 = 0 0 = 0 7.07 0.707 + 5 = 0 0 = 0 Check is OK. EXAMPE NO. 14: Analyze the following symmetrically loaded second degree internally indeterminate truss by the method of least work. Areas in ( ) are 10 3 m. The value of E can be taken as 00 10 6 KN/m D 4 E 4 F 3 3 A 4 4 B 15KN @3m 3 C 3m Selecting member BD and Before as redundants. D E F A X 1 7.5KN X 1 B X 15KN X C 3m 7.5KN BDS under loads and redundants. @3m

METHOD OF EAST WORK 147 SOUTION: Note : By virtue of symmetry, we can expect to have same values for X 1 and X. It is known before hand. 7.5 + S.F.D. 7.5.5 + B.M.D. SFD and BMD in BDS due to applied loads are shown above. As in previous case determine member forces in BDS due to applied loads by the method of moments shears while method of joints may be used to determine member forces due to redundants acting separately. Apply super position principal. Then these are entered in a table given. Summation of relavant columns due to X1 and X gives two equations from which these can be calculated. Putting values from table and solving for X1 and X. [.65 10 3 (7.5 0.707X 1 ).65 10 3 ( 0.707X 1 ) 3.53 10 3 ( 0.707X 1 ) 3.53 10 3 (15 0.707X 1 0.707X ) +10.6 10 3 ( 10.6+X 1 ) + 10.6 10 3 (X ) ]10 3 = 0 19.875 + 1.874X 1 + 1.874 X 1 +.450 X 1 5.45 +.50 X 1 +.5 X 11.36 + 10.6 X 1 + 10.6 X 1 = 0 9.898 X 1 +.50 X 185.185 = 0 (1) ( col 8 ).65 10 3 (7.5 0.707 X ).65 10 3 ( 0.707 X ) 3.53 10 3 (15 0.707 X 1 0.707 X ) 3.53 10 3 ( 0.707 X ) + 10.6 10 3 ( 10.6+X ) + 10.6 10 3 X = 0 19.875+1.874 X +1.874 X 5.95+.50 X 1 +.50X +.450 X 11.36+10.6X +10.6 X = 0.50 X 1 + 9.898 X 185.185 = 0 () ( col 9 ) From (1), X 1 = 185.185.50 X 9.898 (3) Put in above ().50 185.185.50 X + 9.898X 9.898 185.185 = 0 15.465 0.1 X + 9.898 X 185.185 = 0 9.689 X 169.7 = 0 X = + 5.716 KN Put X in equation 3 to get X 1. The final member forces are given in last column. These are obtained by putting values of X 1 and X, whichever is applicable, in column 5 of the table.

148 THEORY OF INDETERMINATE STRUCTURES Insert Page No. 166 A

METHOD OF EAST WORK 149 Then X 1 = 185.185.50 5.716 9.898 Equilibrium Check. X 1 = + 5.716 KN 4.04 4.884 3.459 3.459 4.884 Cosθ = 0 3.459 4.884 0.707 = 0 0 = 0 7.5 7.5 4.04 4.884 0.707 = 0 0 = 0 Checks are satisfied. Results are OK. EXAMPE NO. 15: Analyze the following internally indeterminate truss by the method of least work. Areas in ( ) are 10 3 m. The value of E can be taken as 00 10 6 KN/m. SOUTION: b = 13, r = 3, j = 7 so degree of indeterminacy I =( b + r ) j = Choosing members EB and BG as redundants, forces due to loads have been determined by the method of moments and shears for the BDS and are entered in a table. While forces due to redundants X 1 and X. E F G X 1 X1 X X 3m A 10KN B 15KN 3@3m C D 5KN 10 S.F.D + 0 0 5 30 15 B.M.D + 0 0

150 THEORY OF INDETERMINATE STRUCTURES A 10KN E X 1 X1 B F X X 15KN 3@3m G C D 3m 5KN 10 S.F.D + 0 0 5 30 15 B.M.D + 0 0 Member Forces Due to Redundants Only. Please number that due to separate action of redundants X 1 and X member forces will be induced only in the square whose inclineds are X 1 and X. There will be no reaction at supports. Joint D: DG DG Sinθ 0 CD DG = 0 DG Cos θ + CD = 0 Joint G : CD = 0 FG FG X Cos θ = 0 X CG FG = 0.707 X CG X Sin θ = 0 CG = 0.707 X

METHOD OF EAST WORK 151 Joint C : CF 0.707X CF Sin θ 0.707 X = 0 CF = 0.707 X 0.707 BC CF = + X BC CF Cos θ = 0 BC = 0.707 X Joint B. X 1 BF X AB 0.707X 0.707 X AB + X Cos θ X 1 Cos θ = 0 AB = 0.707 X 1 X 1 Sin θ + X Sin θ + BF = 0 BF = 0.707X 1 0.707X Joint A. AE AF 0.707X 1 0.707 X 1 + AF Cos θ = 0 AF = X 1

15 THEORY OF INDETERMINATE STRUCTURES AE + AF Sin θ = 0 Joint E. AE = 0.707X 1 EF X1 0.707 X 1 EF + X 1 Cos θ = 0 EF = 0.707 X 1 0.707 X 1 0.707 X 1 = 0 0 = 0 (Check) Entering the values of summations from attached table, we have. F. F X 1. F. F X. Simplifying From (1) AE = 0 = 9.443 10 6 +9.848 10 6 X 1 +.45 10 6 X AE = 0 = 168.9 10 6 +.45 10 6 X 1 +9.848 10 6 X X 1 = Put in () & solve for X 168.9 +.45 9.443 + 9.848 X 1 +.45 X = 0 (1) 168.9 +.45 X 1 + 9.848 X = 0 ().45 X + 9.443 9.848.45 X + 9.443 9.848 168.9 0.01 X + 18.833 + 9.848 X = 0 150.067 + 9.647 X = 0 X = 150.067 9.647 X = + 5.06 KN + 9.848 X = 0 (3)

METHOD OF EAST WORK 153 Insert page No. 170 A

154 THEORY OF INDETERMINATE STRUCTURES.45 5.06 + 9.443 So X 1 = 9.848 by putting value of X in (3) X 1 = + 7.7 KN EQUIIBRIUM CHECKS : E 5.141 F 8.579 G 6.87.000 7.07 5.141 6.8 5.66 1.41 7.7 A 4.859 B 6.41 C 5 10KN 15KN D 5KN Joint B: 7.7 6.8 5.06 4.859 6.41 15 Joint C: 6.41 + 5.06 Cosθ 7.7 Cosθ 4.859 = 0 0 = 0 6.8 15 + 5.06 Sinθ + 7.7 Sinθ = 0 0 = 0 The results are OK..008 1.41 6.41 5 5 +.008 Cosθ 6.41 = 0 0 = 0 1.41.008 Sinθ = 0 0 = 0 Results are OK.