Formulation of the Equilibrium Equations of Transversely Loaded Elements Taking Beam-Column Effect into Consideration

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1 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3 (): 0-07 Scholarlink Research Institute Journals, 0 (ISSN: 4-70) jeteas.scholarlinkresearch.org Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) Formulation of the Euilibrium Euations of Transversely oae Elements Taking Beam-Column Effect into Consieration Okonkwo V. O an Onyeyili I. O Department of Civil Engineering Nnami Azikiwe University, Awka Corresponing Author: Okonkwo V. O Abstract In this work a mathematical moel for the consieration of beam-column effect in structural analysis was formulate. The stiffness matrix for a prismatic element taking beam-column effect into account was evelope. The force/loa vectors for various cases of transversely loae elements taking beam-column effect into consieration were also formulate an these were presente in tables synonymous to the tables of en forces ue to unit en isplacements of prismatic elements an fixe en moments on transversely loae elements foun in many structural analysis textbooks. In the analysis of structures by the stiffness metho there is nee to obtain the fixe en moments of the transversely loae elements an also the en stiffness (forces ue to unit en isplacement of the elements) of these elements. These are evaluate using formulas obtaine from textbooks. Unfortunately in the erivation of such formulas beam-column effect was ignore. In this work similar tables that incorporate beam-column effects were evelope enabling an easy implementation of the effects of axial forces on the en stiffnesses of elements in structural analysis. Keywors: stiffness, beam-column effect, egrees of freeom, prismatic members, moulus of elasticity INTRODUCTION When an axial force is present in a member, two effects are noticeable. One is a change in geometry arising from a change in length of members an the secon is a change in stiffness of the member arising from bening by the axial force. The first effect is known as axial eformation. It is the only eformation use for the generation of the compatitibiliy euations use in the analysis of ineterminate trusses (Hibbeler, 00). The secon effect also known as the beam column effect (cguire et al, 000) increases or reuces the forces reuire to cause a unit rotation or translation at the en of a member. The force ecrease (the stiffness of the member ecrease) if the member is subjecte to an axial compressive force an conversely increase (the stiffness of the member increase) if the axial force is tensile (Ghali an Neville, 99). ODE The analysis of structures by the stiffness metho involves the writing of euilibrium euations for the egrees of freeom (coorinates) of the structure (Jenkins, 990). The euilibrium euation for the analysis of an element is given by [k]{} = {} () Where [k] is the element stiffness matrix, it is a x matrix for a space element (elements that can eform in all three coorinate axes) an a x for a plane element (elements that eform in only one plane). {} is the vector of external forces applie at any of the noes an which coincie with one of the egrees of freeom for which the euilibrium 9 euations were written. {} is the vector of isplacements at the coorinates or egrees of freeom of the element. When there is no external force on any of the egrees of freeom or coorinates euation () is rewritten as [k]{} = 0 () But for transversely loae elements euation () is written as { } + [k]{} = {} (3) (eet an Unang, 00) where { o } is the vector of reactive en forces on the transversely loae element when isplacements at its coorinates (egrees of freeom) are restraine. En Forces on Transversely oae Elements Here it will be illustrative to represent each coorinate (egree of freeom) with a number. This is shown in Figure (a) below. Figure (a) & (b) shows the egrees of freeom of a space element an their representation with numbers. If the plane element is lying in the xy plane (see Figure ) an all the external loas on it act in the same plane then = = = = = = 0 If the eformation in these coorinates is zero then the forces (or moments for rotations) in these coorinates will also be zero. F = F = F = F = F = F = 0

2 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) F F = F F F F [k] = k k k k k k k k k k k k k k k k k k (4) k k k k k k k k k (5 ) k k k k k k k k k The Euler s ifferential euation governing the eflection of a transversely loae prismatic member subjecte to a compressive force is y () y () = u A sin ux u A sin ux + But () = EI(Chajes, 974; Nash, 998) () = EIu A sin ux + EIu A cos ux (9) () = EIu A cos ux EIu A sin ux (0) Q () = () () Euation () is encountere in the erivation of the Euler s euation; see Timoshenko an Gere (9). By substituting euation (8) an (0) into euation () we obtain Q () = EIu A EIu () et the initial values of these parameters (ie values at x = 0) be y o, o, o an Q o respectively. By applying these initial conitions into these general solution of the euation of euilibrium of the axially loae uniform column (euations 7, 8, 9 an ) the arbitrary constants A A 4 are obtaine as A = y, A =, A = + an A = +. These are substitute into euations (7), (8),( 9) an () an rearrange to obtain y () = y Q + cos ux + () = cos ux + Q (4) + () = EIu sin ux + cos ux + Q + EI (5) Q () = Q () The above euations (euations 3 -) were use to obtain the fixe en forces of axially loae elements with a uniformly istribute loa. For a case of a uniformly varying loa such that = kx where k is a constant, euation () is moifie to + u = kx (7) The solution of the ifferential euation is y () = A + A x + A sin ux + A cos ux + The first erivative of euation (8) gives the slope + u = () y () = () = A + ua cos ux + Where u =, is the compressive force an the ua sin ux + (9) uniformly istribute loa. But () = EIy () The solution of the ifferential euation is () = y () = A + A x + A sin ux + A cos ux + (7)... (7) EIu A sin ux + EIu A cos ux (0) The first erivative of euation (7) gives the slope = () = A + ua cos ux ua sin ux + (8) () = EIu A cos ux EIu A sin ux () (8) Substituting euation (9) an () into euation (4) will give Q () = EIu A + () et the initial values of these parameters (ie values at x = 0) be y o, o, o an Q o respectively. By applying these initial conitions into these general solution of the euation of euilibrium of the axially loae uniform column (euations 8, 9, 0 an ) the arbitrary constants A A 4 are obtaine as A = y,a =, A = + + an A =. These are substitute into euations (8), (9), (0) an () an rearrange to obtain y () = y Q + sin ux ux + () = cos ux cos ux + + Q (3) + (4) () = EIu sin ux + cos ux + Q + (sin ux ux) (5) Q () = Q () Euations (3) () were use to obtain the fixe en forces of axially loae elements with a.... (3) uniformly varying loa. The above expressions were erive for members in compression. For prismatic members in tension the 9

3 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) same can be written by replacing with so that u is replace with iu. where i =. if β = u then sin iβ = i sin hβ cos iβ = cos β (7) (iβ) = β (iβ) = iβ (Strou, 995; Duffy, 998) From the calculations above the table of fixe en forces of transversely loae beam columns was obtaine an presente as table below (See Appenix). Note that in the table above clockwise moments were taken as positive while anticlockwise moments were negative. In the erivation of the en moments an forces moments that keep the bottom fibres in tension were taken as positive while those that keep the top fibres in tension were negative. ikewise shearing forces that cause the system to move upwars were treate as positive in the table. In the erivation, shearing forces that cause the system to rotate clockwisely were positive while the opposite were negative En Forces ue to En Displacements of Elements The Euler s ifferential euation governing the eflection of a prismatic member (without transverse loas) subjecte to a compressive force is + u = 0 (8) Where u = an is the compressive force. The solution of the ifferential euation is y () = A + A x + A cos ux + A sin ux (9) The first erivative of euation (39) gives the slope y () = () = A ua sin ux + ua cos ux (30) = u A cos ux u A sin ux y () But () = EIy () () = EIu A cos ux + EIu A sin ux (3) By substituting euation () an (30) into euation () we obtain Q () = EIu A (3) et the initial values of these parameters (ie values at x = 0) be y o, o, o an Q o respectively. By applying these initial conitions into these general solution of the euation of euilibrium of the axially loae uniform column, euations (9, 30, 3 an 3) the arbitrary constants A A 4 are obtaine as A = y, A =, A = an A = + are substitute into euations (9) (3) an rearrange to obtain y () = y + Q + + (33) () = cos ux + Q (34) () = EIu sin ux + cos ux + Q (35) Q () = Q (3) Euations (33) (3) were use to obtain the stiffness coefficients of axially loae uniform elements ue to unit en isplacements an is presente in table. The stiffness of a prismatic member in tension were erive by replacing with so that u is replace with iu, β with iβ where i =. Note that in the table above clockwise moments were taken as positive while anticlockwise moments were negative. In the erivation of the en moments an forces moments that keep the bottom fibres in tension were taken as positive while those that keep the top fibres in tension were negative. ikewise shearing forces that cause the system to move upwars were treate as positive in the table. In the erivation, shearing forces that cause the system to rotate clockwisely were positive while the opposite were negative. SUARY AND CONCUSION Beam column effect is often ignore in structural analysis. It is implemente mostly in non-linear analysis by commercial software. The evelopment of the fixe en moments on transversely loae prismatic elements an The en forces ue to unit en isplacements of prismatic elements an which are presente in tables an woul facilitate an easy implementation of beam column effect in structural analysis of frames. It is important to note that the use of tables an reuires the knowlege of the axial force in the element. Hence the frame has to be analyse first to etermine the forces in the members before the effects of beam column can be consiere. When beam column effect is ignore i.e the axial force in the element is taken to be zero in the calculation the element stiffness an loa vectors, tables an will give the same values as obtaine from similar tables in structural engineering textbooks like Reynols an Steeman (00) an Davison an Owens (007). REFERENCES Chajes A., (974), rinciples of Structural Stability Theory, rentice-hall Inc New Jersey Davison B., Owens G. W.,(007). Steel Designer s aual, th Eition, Blackwell ublishing t, UK Duffy, D. G., (998) Avance Engineering athematics, CRC ress New York 93

4 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) Ghali A, Neville A.. (99) Structural Analysis: A Unifie Classical an atrix Approach. 3 r Eition. Chapman & Hall onon Hibbeler, R. C. (00). Structural Analysis. Sixth Eition, earson rentice Hall, New Jersey Jenkins, W.., (990). Structural Analysis using computers. First eition, ongman Group imite, Hong Kong eet, K.., Uang, C.,(00). Funamental of Structural Analysis. cgraw-hill New York cguire, W., Gallagher R. H., Ziemian, R. D.(000). atrix Structural Analysis. Secon Eition, John Wiley & Sons, Inc. New York Nash, W.,(998). Schaum s Outline of Theory an roblems of Strength of aterials. Fourth Eition, cgraw-hill Companies, New York Reynols, C. E.,Steeman J. C. (00). Reinforce Concrete Designer s Hanbook, 0 th Eition) E&FN Spon, Taylor & Francis Group, onon Strou K. A., (995). Engineering athematics. Fourth Eition, acmillan ress t, onon Timoshenko, S.., Gere, J..,(9). Theory of Elastic Stability. Secon Eition, cgraw-hill Kogakusha t, Tokyo AENDIX Table : En forces on transversely loae prismatic members uner axial loas S/N Beam Force = β β = β β + cos β + cos β = = 3 = k = k = β ( cos β + β ) ( β)(cos β ) (cos β ) cos β ( β) = β cos β +β cos β (cos β ) cos β ( β) = β cos β +β cos β (cos β ) cos β ( β) β = (cos β )( β + β3 ) 3( β)( cos β + β ) = k3 cos β (cos β )( β + β3 ) 3( β)( cos β + β ) k3 (β β3 ) 3(cos β )( cos β + β ) EIk( β) β = β (β β3 ) 3(cos β )( cos β + β ) = β (β β3 ) 3(cos β )( cos β + β ) 3β 4 94

5 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) = β 3 β + β3 β = cos β + β β β3 = cos β + β β β3 3β 5 = k = β 3 β 3 + β = β + β3 cos β 3β = β + β3 cos β 3β 3β 7 = β β + cosh β sinh β = + cosh β β sinh β = = = β (sinh β β)(cosh β ) sinh β ( cosh β + β ) sinh β(cosh β ) cosh β (sinh β β) = β +β cosh β cosh β sinh β(cosh β ) cosh β (sinh β β) = β +β cosh β cosh β sinh β(cosh β ) cosh β (sinh β β) β 8 = k = 3(sinh β β)( cosh β β ) (cosh β )( sinh β β β 3 ) cosh β + β sinh β = 3(sinh β β)( cosh β β ) (cosh β )( sinh β β β 3 ) cosh β + β sinh β + β3 sinh β (β sinh β + β3 ) 3(cosh β )( cosh β + β ) cosh β + β sinh β EIk(sinh β β) + β = β sinh β (β sinh β + β3 ) 3(cosh β )( cosh β + β ) cosh β + β sinh β = β sinh β (β β3 ) 3(cosh β )( cosh β + β ) cosh β + β sinh β 3β 95

6 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) 9 = k = β 3 β + β3 + β β cosh β sinh β = cosh β sinh β + β β β3 β cos hβ sinh β = cosh β sin hβ + β β β3 3β β cosh β sinh β 0 = k = sinh β β 3 β 3 β cosh β sinh β + β sinh β = β + β3 cosh β 3β sinh β β cosh β sinh β = β + β3 cosh β 3β sinh β 3β β cosh β sinh β Where β = u an u = Table : En forces cause by en isplacement of prismatic members putting beam column effects into consieration. S/No Beam Force F = F = EIβ = cos β β = = EIβ cos β cos β β 3 = = F = F = EIβ cos β cos β β = EIβ β cos β β = EIβ β cos β cos β β F = F = EIβ cos β β cos β = EIβ β cos β 4 = F = F = EIβ β cos β = EIβ β cos β 5 F = F = EIβ sin hβ cos hβ + βsin hβ F F 9

7 Journal of Emerging Trens in Engineering an Applie Sciences (JETEAS) 3():0-07(ISSN: 4-70) = = EIβ cos hβ cos hβ + βsin hβ = F = F = EIβ cos hβ cos hβ + βsin hβ = EIβ sin hβ β cos hβ + βsin hβ = EIβ β cos hβ sin hβ cos β + β 7 = F = F = EIβ cos hβ β cos hβ sin hβ = EIβ sin hβ β cos hβ sin hβ 8 F = F = EIβ sin hβ = β cos hβ sin hβ = EIβ sin hβ β cos hβ Where β = u an u = y x z (a) A space element showing the twelve coorinates or egrees of freeom all labelle appropriately. is for translation an is for rotation y x z (b) A space element showing the twelve coorinates or egrees of freeom represente with numbers N S 8 7 Figure : A D representation of the six egrees of freeom (coorinates) of a plane element in the xy plane 97