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Contribution of Axial Deformation in the Analysis of Rigidly Fixed Portal Frames
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Contribution of Axial Deformation in the Analysis of Rigidly Fixed Portal Frames

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http://www.ijmer.com/papers/Vol2_Issue6/BY2642444255.pdf

http://www.ijmer.com/papers/Vol2_Issue6/BY2642444255.pdf

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  • 1. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 Contribution of Axial Deformation in the Analysis of Rigidly Fixed Portal FramesOkonkwo V. O.1, Aginam C. H.2, and Chidolue C. A.3123 Department of Civil Engineering Nnamdi Azikiwe University, Awka Anambra State-NigeriaABSTRACT: In this work the stiffness equations forevaluating the internal stress of rigidly fixed portal framesby the displacement method were generated. But obtainingthe equations for the internal stresses required non-scalaror parametric inversion of the structure stiffness matrix. Tocircumvent this problem, the flexibility method was usedtaking advantage of the symmetrical nature of the portalframe and the method of virtual work. These were used toobtain the internal stresses on rigidly fixed portal framesfor different cases of external loads when axial deformationis considered. A dimensionless constant s was used tocapture the effect of axial deformation in the equations.When it is set to zero, the effect of axial deformation isignored and the equations become the same as what can beobtained in any structural engineering textbook. Theseequations were used to investigate the contribution of axialdeformation to the calculated internal stresses and howthey vary with the ratio of the flexural rigidity of the beamand columns and the height to length ratio of the loadedportal frames. Figure 1: A simple portal frame showing its dimensions andKeywords: Axial deformation, flexural rigidity, flexibility the 6 degrees of freedommethod, Portal frames, stiffness matrix, The analysis of portal frames by the stiffness method I. INTRODUCTION requires the determination of the structure’s degrees of Structural frames are primarily responsible for freedom and the development of the structure’s stiffnessstrength and rigidity of buildings. For simpler single storey matrix. For the structure shown in Figure 1a, the degrees ofstructures like warehouses, garages etc portal frames are freedom are as shown in Figure 1b. I1 and A1 areusually adequate. It is estimated that about 50% of the hot- respectively the second moment of inertia and cross-rolled constructional steel used in the UK is fabricated into sectional area of the columns while I2 and A2 are the secondsingle-storey buildings (Graham and Alan, 2007). This moment of inertia and cross-sectional area of the beamshows the increasing importance of this fundamental respectively. The stiffness coefficients for the variousstructural assemblage. The analysis of portal frames are degrees of freedom considering shear deformation can beusually done with predetermined equations obtained from obtained from equations developed in Ghali et al (1985)structural engineering textbooks or design manuals. The and Okonkwo (2012) and are presented belowequations in these texts were derived with an underlyingassumption that deformation of structures due to axial The structure’s stiffness matrix can be written as:forces is negligible giving rise to the need to undertake thisstudy. The twenty first century has seen an astronomical use 𝑘11 𝑘12 𝑘13 𝑘14 𝑘15 𝑘16of computers in the analysis of structures (Samuelsson and 𝑘21 𝑘22 𝑘23 𝑘24 𝑘25 𝑘26Zienkiewi, 2006) but this has not completely eliminated the 𝑘 𝑘32 𝑘33 𝑘34 𝑘35 𝑘36use of manual calculations form simple structures and for 𝐾 = 31 . (1) 𝑘41 𝑘42 𝑘43 𝑘44 𝑘45 𝑘46easy cross-checking of computer output (Hibbeler, 2006). 𝑘51 𝑘52 𝑘53 𝑘54 𝑘55 𝑘56Hence the need for the development of equations that 𝑘61 𝑘62 𝑘63 𝑘64 𝑘65 𝑘66capture the contribution of axial deformation in portalframes for different loading conditions. Where kij is the force in coordinate (degree of freedom) i when there is a unit displacement in coordinate (degree of II. DEVELOPMENT OF THE MODEL freedom) j. They are as follows: 𝐸𝐴1 𝑘11 = ℎ 𝑘12 = 0 www.ijmer.com 4244 | Page
  • 2. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 6𝐸𝐼2 𝑑1𝑘13 = 𝑙2 𝑑2 12𝐸𝐼1 𝑑3𝑘14 = − ℎ3 {𝐷} = . . . . . (5) 𝑑4 𝑑5𝑘15 = 0 𝑑6 6𝐸𝐼2𝑘16 = 𝑙2 Where di is the displacement in coordinate i. 12𝐸𝐼1 𝐸𝐴2 𝐹1𝑘22 = ℎ3 + 𝑙 𝐹2 6𝐸𝐼1 𝐹3𝑘23 = {𝐹} = . . . . . . (6) ℎ2 𝐹4 𝐹5𝑘24 = 0 𝐹6 𝐸𝐴2𝑘25 = − 𝑙 . . (2) Where Fi is the external load with a direction coinciding with the coordinate i (McGuire et al, 2000).𝑘26 = 0 By making {D} the subject of the formula in equation (3) 4𝐸𝐼2 4𝐸𝐼1𝑘33 = + 𝑙 ℎ −1 𝐷 = 𝐾 {𝐹 − 𝐹𝑜 } . . . . (7) 6𝐸𝐼2𝑘34 = − 𝑙2 Once {D} is obtained the internal stresses in the frame can be easily obtained by writing the structure’s compatibility𝑘35 = 0 equations given as 2𝐸𝐼2𝑘36 = 𝑙 𝑀 = 𝑀 𝑟 + 𝑀1 𝑑1 + 𝑀2 𝑑2 + 𝑀3 𝑑3 . . (8) 12𝐸𝐼2 𝐸𝐴1 Where M is the internal stress (bending moment) at any𝑘44 = + 𝑙3 ℎ point on the frame, Mr is the internal stress at the point under consideration in the restrained structure while Mi is𝑘45 = 0 the internal stress at the point when there is a unit 6𝐸𝐼2 displacement in coordinate i.𝑘46 = − To solve equation (8) there is need to obtain {D}. {D} can 𝑙2 be obtained from the inversion of [K] in equation (7). 12𝐸𝐼1 𝐸𝐴2𝑘55 = + Finding the inverse of K parametrically (i.e. without ℎ3 𝑙 substituting the numerical values of E, h, l etc) is a difficultFrom Maxwell’s Reciprocal theorem and Betti’s Law k ij = task. This problem is circumvented by using the flexibilitykji ( Leet and Uang, 2002). method to solve the same problem, taking advantage of the symmetrical nature of the structure and the principle ofWhen there are external loads on the structure on the virtual work.structure there is need to calculate the forces in therestrained structure Fo as a result of the external load. III. APPLICATION OF THE FLEXIBILITY MODELThe structure’s equilibrium equations are then written as The basic system or primary structure for the structure in Figure 1a is given in Figure 2. The removed redundant{𝐹} = {𝐹𝑜 } + 𝐾 {𝐷} . . . . (3) force is depicted with X1. 𝑘10 𝑘20 𝑘30{𝐹𝑜 } = . . . . . (4) 𝑘40 𝑘50 𝑘60Where kio is the force due to external load in coordinate iwhen the other degrees of freedom are restrained. Figure 2: The Basic System showing the removed redundant forces www.ijmer.com 4245 | Page
  • 3. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645The flexibility matrix of the structure can be determined 1 12𝐸𝐼1 𝐼2 𝐴1 = . . . (15a) 𝑑 11 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24𝐼1 𝐼2 ℎusing the principle of virtual work.By applying the unit load theorem the deflection in beamsor frames can be determined for the combined action of theinternal stresses, bending moment and axial forces with 𝑑 33 3𝐸𝐼1 𝐴2 (𝑙𝐼1 +2ℎ𝐼2 ) 𝑀 𝑀 𝑁 𝑁 2 𝑑 22 𝑑 33 −𝑑 23 = 2𝐴 3 2 2 4 . (15b) 𝐷= 𝑑𝑠 + 𝑑𝑠 . . . (9) 2 ℎ 𝑙𝐼1 +3𝐼1 𝑙 +𝐴2 ℎ 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝐸𝐼 𝐸𝐴 𝑑 32 −3𝐸𝐼1 𝐼2 𝐺𝐴2 ℎ 2Where 𝑀 and 𝑁 are the virtual internal stresses while M 2 𝑑 22 𝑑 33 −𝑑 23 = 2𝐴 3 2 2 4 (15c) 2 ℎ 𝑙𝐼1 +3𝐼1 𝑙 +𝐴2 ℎ 𝐼2 +6ℎ𝑙𝐼1 𝐼2and N are the real/actual internal stresses. 𝑑 22 𝐸𝐼1 𝐼2 2𝐴2 ℎ 3 +3𝐼1 𝑙E is the modulus of elasticity of the structural material 2 = 3 𝑙𝐼 +3𝐼 2 𝑙 2 +𝐴 ℎ 4 𝐼 +6ℎ𝑙𝐼 𝐼 . (15d) 𝑑 22 𝑑 33 −𝑑 23 2𝐴2 ℎ 1 1 2 2 1 2A is the cross-sectional area of the element (McGuire et al, Equation (12) is evaluated to get the redundant forces and2000; Nash, 1998) these are substituted into the structures force equilibrium (superposition) equations to obtain the internal stresses atIf dij is the deformation in the direction of i due to a unit any point.load at j then by evaluating equation (9) the following areobtained: 𝑀 = 𝑀 𝑜 + 𝑀1 𝑋1 + 𝑀2 𝑋2 + 𝑀3 𝑋3 . . (16) 𝐼1 𝐺𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +12𝐼1 𝐼2 ℎ Where M is the required stress at a point, Mo is the stress at𝑑11 = . . (10a) 12𝐸𝐼1 𝐼2 𝐴1 that point for the reduced structure, Mi is stress at that point when the only the redundant force Xi =1 acts on the reduced𝑑12 = 0 . . . . . (10b) structure. For the loaded portal frame of Figure 3, the deformations of𝑑13 = 0 . . . . . . (10c) the reduced structure due to external loads are 2𝐴2 ℎ 3 +3𝐼1 𝑙𝑑22 = 3𝐸𝐼1 𝐴2 . . . (10d) 𝑑10 = 0 . . . . . . (17a) −ℎ 2𝑑23 = 𝐸𝐼1 . . . (10e) 𝑤 ℎ 2 𝑙2 𝑑20 = . . . . . 8𝐸𝐼1 𝑙𝐼1 +2ℎ 𝐼2 . (17b)𝑑33 = . . . . (10f) 𝐸𝐼1 𝐼2 −𝑤 𝑙 2 (𝑙𝐼1 +6ℎ𝐼2 )From Maxwell’s Reciprocal theorem and Betti’s Law dij = 𝑑30 = . . (17c) 24𝐸𝐼1 𝐼2dji. By substituting the values of equations (17a) – (17c) intoThe structure’s compatibility equations can be written thus equations (12) 𝑑11 𝑑11 𝑑11 𝑋1 𝑑10 𝑋1 = 0 . . . . . (18a) 𝑑21 𝑑21 𝑑21 𝑋2 + 𝑑20 = 0 . (11a) 𝑑31 𝑑31 𝑑31 𝑋3 𝑑30 −𝐴2 𝐼1 𝑤ℎ 2 𝑙 3 𝑋2 = 4 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . (18b)i.e. 𝑓𝐹 + 𝑑 𝑜 = 0 . . . (11b) 2 𝑤 𝑙 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +3𝐴2 ℎ 4 𝐼2 +18ℎ𝑙𝐼1 𝐼2 𝑋3 = 2 24 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . (18c)Where F is the vector of redundant forces X1, X2, X3 and dois the vector deformation d10, d20, d30 due to external load Evaluating equation (16) for different points on theon the basic system (Jenkins, 1990). structure using the force factors obtained in equations (18a)𝐹 = 𝑓 −1 (−𝑑 𝑜 ) . . . . (12) – (18c) 2𝑤𝑙 3 𝐼1 ℎ 3 𝐴2 −3𝑙𝐼1 𝑀 𝐴 = 12 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . (19) 𝐴𝑑𝑗 𝑓𝑓 −1 = 𝑑𝑒𝑡 𝑓 . . (13) −2𝑤𝑙 3 𝐼1 2ℎ 3 𝐴2 +3𝑙𝐼1(Stroud, 1995) 𝑀 𝐵 = 24 2 . . (20) 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 1 𝑑 11 0 0 𝑑 33 −𝑑 32𝑓 −1 = 0 2 𝑑 22 𝑑 33 −𝑑 23 2 𝑑 22 𝑑 33 −𝑑 23 . (14) −2𝑤 𝑙 3 𝐼1 2ℎ 3 𝐴2 +3𝑙𝐼1 𝑀𝐶 = 2 . (21) −𝑑 23 𝑑 22 24 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 0 2 2 𝑑 22 𝑑 33 −𝑑 23 𝑑 22 𝑑 33 −𝑑 23 −𝑤 𝑙 3 𝐼1 8𝐼2 ℎ 3 +𝑠2 𝑙 3 𝐼1 𝑀𝐶 =From equations 10a – 10f 12 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 www.ijmer.com 4246 | Page
  • 4. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 2𝑤𝑙 3 𝐼1 ℎ 3 𝐴2 −3𝑙𝐼1 −𝑤 ℎ 3 𝑀𝐷 = 2 . . (22) 𝑑30 = . . . . . (30c) 12 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 6𝐸𝐼1 By substituting the values of equations (30a) – (30c) into equations (12)For the loaded portal frame of Figure 4, the deformations ofthe reduced structure due to external loads are 𝑤 ℎ 3 𝑙𝐼2 𝐴1 𝑋1 = 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 . . . (31a) −𝑤𝑙 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ 𝐼1 𝐼2 𝑑10 = . (23a) −𝑤ℎ 4 𝐴2 3𝑙𝐼1 +2ℎ 𝐼2 128 𝐸𝐼1 𝐼2 𝐴1 𝑋2 = 8 3 𝑙𝐼 +3𝐼 2 𝑙 2 +𝐴 ℎ 4 𝐼 +6ℎ𝑙𝐼 𝐼 2𝐴2 ℎ 1 . (31b) 1 2 2 1 2 𝑤 ℎ 2 𝑙2 𝑑20 = 16𝐸𝐼1 . . . (23b) 𝑤ℎ 3 𝐼2 −ℎ 3 𝐴2 +12𝑙𝐼1 𝑋3 = 24 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . (31c) −𝑤 𝑙 2 (𝑙𝐼1 +6ℎ𝐼2 ) 𝑑30 = . . (23c) Evaluating equation (16) for different points on the 48𝐸𝐼1 𝐼2 structure using the force factors obtained in equations (31a)By substituting the values of equations (23a) – (23c) into – (31c)equations (12) 𝑀𝐴 = 3𝑤𝑙 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ𝐼1 𝐼2 𝑤 ℎ 2 𝐼1 𝐴1 𝑙 3 +5𝐴1 𝑙 2 ℎ 𝐼2 +24ℎ𝐸𝐼1 𝐼2 𝑋1 = . . (24a) − + 32 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑤 ℎ 3 9ℎ 2 𝑙𝐼1 𝐴2 +5𝐴2 ℎ 3 𝐼2 +12𝑙𝐼1 𝐼2 −2𝑤 ℎ 2 𝑙 3 𝐴2 𝐼1 2 . . . . (32) 𝑋2 = 16 . (24b) 24 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑤 ℎ 3 𝑙 2 𝐴2 3ℎ 𝐼2 +2𝑙𝐼1 +3𝑤𝑙 3 𝐼1 𝑙𝐼1 +6ℎ 𝐼2 𝑋3 = 2 . (24c) 𝑀𝐵 = 48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑤 ℎ 3 𝑙 2 𝐼2 𝐴1 𝑤ℎ 3 𝐼2 −ℎ 3 𝐴2 +12𝑙𝐼1 3 +6𝐼 𝐴 𝑙 2 ℎ +24ℎ 𝐼 𝐼 + 24 3 𝑙𝐼 +3𝐼 2 𝑙 2 +𝐴 ℎ 4 𝐼 +6ℎ𝑙𝐼 𝐼 . 𝑠2 2 𝐼1 𝐴1 𝑙 2 1 1 2 2𝐴2 ℎ 1 1 2 2 1 2Let 𝛽 = . . . . . . (25) . . (33) 𝑠1Evaluating equation (16) for different points on the 𝑀𝐶 =structure using the force factors obtained in equations (24a) 𝑤 ℎ 3 𝑙 2 𝐼2 𝐴1 𝑤ℎ 3 𝐼2 −ℎ 3 𝐴2 +12𝑙𝐼1 −2 + 24– (24c) 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . . . (34) 𝑀𝐴 = 𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +8𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2 𝑤 𝑙 4 𝐴1 5𝑙𝐼1 +24ℎ 𝐼2 𝑀𝐷 = 2 − 64 𝑤 ℎ 3 𝑙 2 𝐼2 𝐴1 𝑤ℎ 3 9ℎ 2 𝑙𝐼1 𝐴2 +5𝐴2 ℎ 3 𝐼2 +12𝑙𝐼1 𝐼248 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2 −2 3 +6𝐼 𝐴 𝑙 2 ℎ +24ℎ𝐼 𝐼 + 24 2. . . . (26) 𝐼1 𝐴1 𝑙 2 1 1 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . . . (35) 𝑀𝐵 = 𝑤 𝑙 4 𝐴1 5𝑙𝐼1 +24ℎ 𝐼2 For the loaded portal frame of Figure 6, the deformations of− 64 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 + the reduced structure due to external loads are 𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +2𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2 2 . . . (27)48 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑃𝑎 2 𝑙 𝑑10 = − . . . (36a) 2𝐸𝐼1 𝑀𝐶 = 3𝑤 𝑙 2 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ𝐼1 𝐼2 𝑑20 = 0 . . . . (36b)− + 64 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +2𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2 𝑑30 = 0 . . . (36c) 248 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . . (28) By substituting the values of equations (36a) – (36c) into equations (12) 𝑀𝐷 = 6𝑃𝑎 2 𝑙𝐼2 𝐴1 3𝑤 𝑙 2 𝑙 3 𝐼1 𝐴1 +8ℎ 𝑙 2 𝐼2 𝐴1 +64ℎ𝐼1 𝐼2 𝑋1 = 𝐴1 𝐼1 𝑙 3 +6𝐴1 ℎ 𝑙 2 𝐼2 +24ℎ 𝐼1 𝐼2 . . . (37a)− 64 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 +𝑤 𝑙 2 ℎ 3 𝐴2 3ℎ 𝐼2 +8𝑙𝐼1 +3𝑙𝐼1 𝑙𝐼1 +6ℎ𝐼2 𝑋2 = 0 . . . . (37b) 248 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . . (29) 𝑋3 = 0 . . . . . . (37c)For the loaded portal frame of Figure 5, the deformations ofthe reduced structure due to external loads are Evaluating equation (16) for different points on the structure using the force factors obtained in equations (37a) 𝑤𝑙 ℎ 3 𝑑10 = − . . . . (30a) – (37c) 12𝐸𝐼1 𝑤ℎ4 𝑑20 = . . . . (30b) 8𝐸𝐼1 www.ijmer.com 4247 | Page
  • 5. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 3𝑎𝑙 2 𝐼2 𝐴1 For the loaded portal frame of Figure 8, the deformations of 𝑀 𝐴 = −𝑃𝑎 1 − 𝐴1 𝐼1 𝑙 3 +6𝐴1 ℎ 𝑙 2 𝐼2 +24ℎ 𝐼1 𝐼2 . (38) the reduced structure due to external loads are 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 𝑎 2 𝐼1 𝐴1 3𝑙 −2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1 𝑀𝐵 = 𝐴1 𝐼1 𝑙 3 +6𝐴 ℎ 𝑙 2 𝐼 +24ℎ 𝐼 𝐼 . . (39) 𝑑10 = −𝑃 . (48a) 1 2 1 2 12𝐸𝐼1 𝐼2 𝐴1 𝑃𝑎ℎ 2 𝑑20 = 2𝐸𝐼1 . . . (48b) 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 𝑀𝐶 = − 𝐴1 𝐼1 𝑙 3 +6𝐴1 ℎ 𝑙 2 𝐼2 +24ℎ𝐼1 𝐼2 . . (40) 𝑃𝑎 𝑎𝐼1 +2ℎ 𝐼2 𝑑30 = − . . . (48c) 2𝐸𝐼1 𝐼2 3𝑎𝑙 2 𝐼2 𝐴1 𝑀 𝐷 = 𝑃𝑎 1 − . . (41) 𝐴1 𝐼1 𝑙 3 +6𝐴1 ℎ 𝑙 2 𝐼2 +24ℎ 𝐼1 𝐼2 By substituting the values of equations (48a) – (48c) into equations (12) 𝑃 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ𝐼2 𝑎𝑙 𝐴1 +2𝐼1For the loaded portal frame of Figure 7, the deformations of 𝑋1 = . (49a) 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2the reduced structure due to external loads are −3𝑃𝑎 ℎ 2 𝐼1 𝐴2 𝑙−𝑎 𝑃𝑎 2 𝑙 𝑋2 = 2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 (49b) 𝑑10 = − 4𝐸𝐼 . . . (42a) 1 𝑃𝑎 ℎ 3 𝐴2 2𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 𝑃𝑎 2 3ℎ −𝑎 𝑋3 = 2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 (49c) 𝑑20 = . . (42b) 6𝐸𝐼1 𝑃𝑎 2 𝑑30 = − 2𝐸𝐼 . . . . (42c) 1 Evaluating equation (16) for different points on theBy substituting the values of equations (42a) – (42c) into structure using the force factors obtained in equations (49a)equations (12) – (49c) 3𝑃𝑎 2 𝑙𝐼2 𝐴1 𝑃𝑙 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1 𝑋1 = . . . (43a) 𝑀 𝐴 = −𝑃𝑎 + + 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑃𝑎 𝐴2 ℎ 3 −𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 . . (50) −𝑃𝑎 2 𝐴2 𝑙𝐼1 3ℎ−𝑎 +ℎ 𝐼2 3ℎ −2𝑎 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑋2 = 2 2 . (43b) 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑃 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1 −𝑃𝑎 2 𝐼2 −ℎ 3 𝐴2 +𝑎ℎ 2 𝐴2 +3𝑙𝐼1 𝑀 𝐵 = −𝑃𝑎 + + 2(𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 ) 𝑋3 = 2 . (43c) 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑃𝑎 ℎ 3 𝐴2 2𝑎𝐼1 +ℎ𝐼2 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 . . . (51) 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑀𝐶 =Evaluating equation (16) for different points on the 𝑃𝑙 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1structure using the force factors obtained in equations (43a) − 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 +– (43c) 𝑃𝑎 ℎ 3 𝐴2 2𝑎𝐼1 +ℎ𝐼2 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 . . . . (52) 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑀𝐴 = 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 𝑀𝐷 =−𝑃𝑎 + + 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 𝑃𝑙 𝑎 2 𝐼1 𝐴1 3𝑙−2𝑎 +6ℎ 𝐼2 𝑎𝑙 𝐴1 +2𝐼1𝑃𝑎 2 𝐴2 𝐼1 𝑙 3ℎ 2 −𝑎ℎ +𝐼2 2𝐴2 ℎ 3 −𝑎𝐴2 ℎ 2 +3𝐼1 𝑙 − + 2 . . (44) 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 𝑃𝑎 𝐴2 ℎ 3 −𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 +3𝑙𝐼1 𝑎𝐼1 +2ℎ 𝐼2 2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . (53) 𝑀𝐵 = 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 𝑃𝑎 2 𝐼2 −ℎ 3 𝐴2 +𝑎ℎ 2 𝐴2 +3𝑙𝐼12 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ 𝐼1 𝐼2 +2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 .. . (45) IV. DISCUSSION OF RESULTS The internal stress on the loaded frames is summarized in 𝑀𝐶 = 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 𝑃𝑎 2 𝐼2 −ℎ 3 𝐴2 +𝑎ℎ 2 𝐴2 +3𝑙𝐼1 table 1. The effect of axial deformation is captured by the−2 𝐼1 𝐴1 𝑙 3 +6𝐼 𝐴 𝑙 2 ℎ +24ℎ𝐼 𝐼 +2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 dimensionless constant s taken as the ratio of the end 2 1 1 2. . . (46) translational stiffness to the shear stiffness of a member. 12𝐸𝐼1 ℎ 12𝐼 𝑀𝐷 = 𝑠1 = ℎ3 ∙ 𝐸𝐴1 = ℎ 2 𝐴1 . . . (54) 1 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1− + 2 𝐼1 𝐴1 𝑙 3 +6𝐼2 𝐴1 𝑙 2 ℎ +24ℎ𝐼1 𝐼2 12𝐸𝐼2 𝑙 12𝐼2𝑃𝑎 2 𝐴2 𝐼1 𝑙 3ℎ 2 −𝑎ℎ +𝐼2 2𝐴2 ℎ 3 −𝑎𝐴2 ℎ 2 +3𝐼1 𝑙 𝑠2 = 𝑙3 ∙ 𝐸𝐴2 = 𝑙 2 𝐴2 . . (55) 2 2 2𝐴2 ℎ 3 𝑙𝐼1 +3𝐼1 𝑙 2 +𝐴2 ℎ 4 𝐼2 +6ℎ𝑙𝐼1 𝐼2 . . (47) When the axial deformation in the columns is ignored 𝑠1 = 0 and likewise when axial deformation in the beam is www.ijmer.com 4248 | Page
  • 6. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645ignored 𝑠2 = 0 . If axial deformation is ignored in the (about 40% drop in calculated bending moment value)whole structure, 𝑠1 = 𝑠2 = 0. while at ℎ 𝐿 > 0, ∆𝑀 𝐴 dropped in magnitude exponentiallyThe internal stress equations enable an easy investigation to values below 1.into the contribution of axial deformation to the internalstresses of statically loaded frames for different kinds of V. CONCLUSIONexternal loads. The flexibility method was used to simplify the analysisFor frame 1 (figure 3), the moment at A, MA is given by and a summary of the results are presented in table 1. Theequation (19). The contribution of axial deformation in the equations in table 1 would enable an easy evaluation of thecolumn ∆𝑀 𝐴 is given by internal stresses in loaded rigidly fixed portal frames considering the effect of axial deformation.∆𝑀 𝐴 = 𝑀 𝐴(𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 19) − 𝑀 𝐴(𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 19 𝑤ℎ𝑒𝑛 𝑠2 =0) From a detailed analysis of frame 1 (Figure 3), it was observed that the contribution of axial deformation is 𝑤 𝑙 3 𝐼1 4ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 1 generally very small and can be neglected for reasonable = 3 𝐼 2𝑙𝐼 +ℎ 𝐼 +𝑠 𝑙 3 𝐼 𝑙𝐼 +2ℎ 𝐼 − . 12 4ℎ 2 1 1 2 1 1 2 2𝑙𝐼1 +ℎ 𝐼2 values of 𝐼1 𝐼2 . However, its contribution skyrockets at. . (56) very low values of ℎ 𝑙 i.e. as ℎ 𝑙 => 0. This depicts the case of an encased single span beam and a completeEquation (56) gives the contribution of axial deformation to departure from portal frames under study. This analysis canMA as a function of h, l, I1, I2 and s be extended to the other loaded frames (Figures 4 – 8) usingBy considering the case of a portal frame of length l = 5m, the equations in Table 1. These would enable theh = 4m. Equation 56 was evaluated to show how the determination of safe conditions for ignoring axialcontribution of axial deformation varied with 𝐼1 𝐼2 . The deformation under different kinds of loading.result is shown in Table 2. When plotted on a uniform scale(Figure 9) the relationship between ∆𝑀 𝐵 and 𝐼1 𝐼2 is seen REFERENCESto be linear. This was further justified by a linear regressionanalysis of the results in Table 2 which was fitted into the [1] Ghali A, Neville A. M. (1996) Structural Analysis: A 𝐼 Unified Classical and Matrix Approach (3rd Edition)model ∆𝑀 𝐴 = 𝑃1 𝐼1 + 𝑃2 to obtain 𝑃1 = −0.004815 and Chapman & Hall London 2 𝑃2 = 0.0001109 and the fitness parameters sum square of [2] Graham R, Alan P.(2007). Single Storey Buildings:errors (SSE), coefficient of multiple determination (R2 ) and Steel Designer’s Manual Sixth Edition, Blackwellthe root mean squared error (RMSE) gave 1.099 x 10 -7, Science Ltd, United Kingdom0.999952 and 0.0001105 respectively. From Table 2 when [3] Hibbeler, R. C.(2006). Structural Analysis. Sixth 𝐼1 𝐼2 = 0 , ∆𝑀 𝐴 ≅ 0 and when 𝐼1 𝐼2 = 10; ∆𝑀 𝐴 = Edition, Pearson Prentice Hall, New Jersey−0.0479 which represent only a 5% reduction in the [4] Leet, K. M., Uang, C.,(2002). Fundamental ofcalculated bending moment. Structural Analysis. McGraw-Hill ,New YorkIn like manner by evaluating the axial contribution in thebeam, ∆𝑀 𝐵 for varying 𝐼1 𝐼2 of the portal frame, Table 3 [5] McGuire, W., Gallagher R. H., Ziemian, R.was produced. This was plotted on a uniform scale in D.(2000). Matrix Structural Analysis,SecondFigure 10. From Figure 10 it would be observed that there Edition, John Wiley & Sons, Inc. New Yorkis also a linear relationship between ∆𝑀 𝐵 and 𝐼1 𝐼2 . When [6] Nash, W.,(1998). Schaum’s Outline of Theory and 𝐼 Problems of Strength of Materials. Fourth Edition,fitted into the model ∆𝑀 𝐵 = 𝑃3 𝐼1 + 𝑃4 it gave 𝑃3 = 2 McGraw-Hill Companies, New York−0.006502 and 𝑃4 = −0.0003969 for the fitness [7] Okonkwo V. O (2012). Computer-aided Analysis ofparameters sum square of errors (SSE), coefficient of Multi-storey Steel Frames. M. Eng. Thesis,multiple determination (R2 ) and the root mean squared Nnamdi Azikiwe University, Awka, Nigeria.error (RMSE) of 6.51 x 10-7, 0.9999 and 0.000269 [8] Reynolds, C. E.,and Steedman J. C. (2001).respectively. Reinforced Concrete Designer’s Handbook, 10thBy pegging 𝐼1 𝐼2 to a constant value of 0.296 and the Edition) E&FN Spon, Taylor & Francis Group,variation of ∆𝑀 𝐴 with respect to ℎ 𝐿 investigated, Table 4 Londonwas generated. A detailed plot of Table 4 was presented inFigure 11. From Table 4 when ℎ 𝐿 = 0; ∆𝑀 𝐴 = −3.125 [9] Samuelsson A., and Zienkiewicz O. C.(2006), Review: History of the Stiffness Method.(about 30% drop in calculated bending moment value) International Journal for Numerical Methods inwhile at ℎ 𝐿 > 0, ∆𝑀 𝐴 dropped in magnitude exponentially Engineering. Vol. 67: 149 – 157to values below 1. In like manner, when the variation of∆𝑀 𝐵 with respect to ℎ 𝐿 was investigated, Table 5 was [10] Stroud K. A.,(1995). Engineering Mathematics.generated. A detailed plot of Table 5 was presented in Fourth Edition, Macmillan Press Ltd, London.Figure 12. From Table 5 when ℎ 𝐿 = 0; ∆𝑀 𝐴 = −4.1667 www.ijmer.com 4249 | Page
  • 7. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645Table 1: Internal stresses for a loaded rigid frameS/No LOADED FRAME REMARKS www.ijmer.com 4250 | Page
  • 8. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝑤 𝑙 3 𝐼1 4ℎ 3 𝐼2 −𝑠2 𝑙 3 𝐼1 𝑀 𝐴 = 𝑀 𝐷 = 12 4ℎ 3 𝐼 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 −𝑤𝑙 3 𝐼1 8𝐼2 ℎ 3 +𝑠2 𝑙 3 𝐼1 𝑀 𝐵 = 𝑀 𝐶 = 12 4ℎ 3 𝐼 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2 𝑤𝑙 𝑤 ℎ 2 𝑙 3 𝐼1 𝐼2𝑉𝐴 = 𝑉 𝐷 = 2 𝐻 𝐴 = 𝐻 𝐷 = 4ℎ 3 𝐼 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2See equations (19) – (22) 𝑤𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +8𝑙𝐼1 +𝑙 3 𝐼1 𝑠1 𝑙𝐼1 +6𝑠2 ℎ 𝐼2 𝑤 𝑙 4 5ℎ 𝐼1 +24ℎ 𝐼2 𝑀𝐴 = − 48 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝑙𝐼1 +2ℎ𝐼2 64 𝑙 2 𝐼1 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 𝑤𝑙 4 5ℎ𝐼1 +24ℎ𝐼2 𝑤 𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +2𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +6ℎ 𝐼2 𝑀 𝐵 = − 64 𝑙 2 𝐼1 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 48 4ℎ 3 𝐼2 2𝑙ℎ 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2 𝑤𝑙 3𝐼1 𝑙 3 +24𝐼2 𝑙 2 ℎ +16𝑠1 ℎ 3 𝐼2 𝑤 𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +2𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +6ℎ𝐼2 𝑀𝐶 = − + 64 𝐼1 𝑙 3 +6𝐼2 𝑙 2 ℎ+2𝑠1 ℎ 3 𝐼2 48 4ℎ 3 𝐼2 2𝑙ℎ 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑤𝑙 2 4ℎ 3 𝐼2 3ℎ 𝐼2 +8𝑙𝐼1 +𝑙 3 𝐼1 𝑠1 𝑙𝐼1 +6𝑠2 ℎ 𝐼2 𝑤𝑙 3𝐼1 𝑙 3 +24𝐼2 𝑙 2 ℎ +16𝑠1 ℎ 3 𝐼2 𝑀𝐷 = 48 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝑙𝐼1 +2ℎ𝐼2 − 64 𝐼1 𝑙 3 +6𝐼2 𝑙 2 ℎ +2𝑠1 ℎ 3 𝐼2 𝑤𝑙 3𝑙 3 𝐼1 +24ℎ𝑙 2 𝐼2 +16𝑠1 ℎ 3 𝐼2 𝑤𝑙𝑉𝐷 = 𝑉𝐴 = − 𝑉𝐷 32 𝑙 3 𝐼1 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 2 𝑤 ℎ 2 𝑙 3 𝐼1 𝐼2𝐻 𝐴 = 𝐻 𝐷 = 2 4ℎ 3 𝐼 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝑙𝐼1 +2ℎ𝐼2See equations (26) – (29) 𝑤ℎ 2 𝐼1 𝑙 3 +5ℎ 𝑙 2 𝐼2 +2𝑠2 ℎ 3 𝐼2 𝑤 ℎ 3 𝐼2 9ℎ 2 𝑙𝐼1 +5ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐴 = − 2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 6 4ℎ 3 𝐼 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑤ℎ 3 𝑙 2 𝐼2 𝑤 ℎ 3 𝐼2 −4ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐵 = 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 6 4ℎ 3 𝐼 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑤 ℎ 3 𝑙 2 𝐼2 𝑤 ℎ 3 𝐼2 −4ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐶 = −2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 6 4ℎ 3 𝐼 3 2 2𝑙 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2 𝑤ℎ 3 𝑙 2 𝐼2 𝑤 ℎ 3 𝐼2 9ℎ 2 𝑙𝐼1 +5ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐷 = − + 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 6 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑤 ℎ 3 𝑙𝐼2𝑉𝐴 = 𝑉 𝐵 = 𝐼1 𝑙 3 +6𝐼2 𝑙 2 ℎ +2𝑠1 𝐼2 ℎ 3 www.ijmer.com 4251 | Page
  • 9. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝑤 ℎ 4 𝐼2 3𝑙𝐼1 +2ℎ𝐼2𝐻𝐷 = 𝐻 𝐴 = 𝑤ℎ − 𝐻 𝐷 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2See equations (32) – (35) 3𝑎 𝑙 2 𝐼2 3𝑃𝑎 2 𝑙 2 𝐼2 𝑀 𝐴 = −𝑃𝑎 1 − 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 𝑀𝐵 = 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 HD 3𝑃𝑎 2 𝑙 2 𝐼2 𝐴1 3𝑎𝑙 2 𝐼2 𝑀𝐶 = − 𝐴1 𝐼1 𝑙 3 +6𝐴1 ℎ 𝑙 2 𝐼2 +24ℎ𝐼1 𝐼2 𝑀 𝐷 = 𝑃𝑎 1 − 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 6𝑃𝑎 2 𝑙𝐼2𝑉𝐴 = 𝑉 𝐵 = 3 𝐼 +6𝑙 2 ℎ 𝐼 +2𝑠 ℎ 3 𝐼 𝑙 1 𝐻𝐴 = 𝐻𝐵 = 𝑃 2 1 2See equations (38) – (41) 3𝑃𝑎 2 𝑙 2 𝐼2 𝑃𝑎 2 4𝐼1 𝐼2 ℎ𝑙 3ℎ −𝑎 +𝐼2 8ℎ 3 𝐼2 −4𝑎ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀 𝐴 = −𝑃𝑎 + 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2 3𝑃𝑎 2 𝑙 2 𝐼2 𝑃𝑎 2 𝐼2 −4ℎ 3 𝐼2 +4𝑎ℎ 2 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐵 = 2 + 2 4ℎ 3 𝐼 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 3 2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 3𝑃𝑎 2 𝑙 2 𝐼2 𝑃𝑎 2 𝐼2 −4ℎ 3 𝐼2 +4𝑎ℎ 2 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐶 = −2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 2 4ℎ 3 𝐼 3 2 2𝑙 𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 𝐼1 𝑙𝐼1 +2ℎ𝐼2 3𝑃𝑎 2 𝑙 2 𝐼2 𝑃𝑎 2 4𝐼1 𝐼2 ℎ𝑙 3ℎ −𝑎 +𝐼2 8ℎ 3 𝐼2 −4𝑎ℎ 3 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑀𝐷 = − + 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 3𝑃𝑎 2 𝑙𝐼2𝑉𝐴 = 𝑉 𝐵 = 𝑙 3 𝐼1 +6𝑙 2 ℎ 𝐼2 +2𝑠1 ℎ 3 𝐼2 2𝑃𝑎 2 𝐼2 𝑙𝐼1 3ℎ −𝑎 +ℎ𝐼2 3ℎ −2𝑎𝐻𝐷 = 4ℎ 3 𝐼2 2𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2 𝐻𝐴 = 𝑃 − 𝐻 𝐷See equations (44) – (47) www.ijmer.com 4252 | Page
  • 10. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ𝐼2 6𝑎𝑙 +𝑠1 ℎ 2 𝑃𝑎 4ℎ 3 𝐼2 −𝑎𝐼1 +ℎ𝐼2 +3𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ𝐼2 𝑀 𝐴 = −𝑃𝑎 + 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ𝐼2 𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ 𝐼2 6𝑎𝑙 +𝑠1 ℎ 2 𝑃𝑎 4ℎ 3 𝐼2 2𝑎𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ 𝐼2 𝑀 𝐵 = −𝑃𝑎 + 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ 𝐼2 6𝑎𝑙 +𝑠1 ℎ 2 𝑃𝑎 4ℎ 3 𝐼2 2𝑎𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ 𝐼2 𝑀𝐶 = − + 2 𝐼1 𝑙 3 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑃𝑙 𝑎 2 𝐼1 3𝑙−2𝑎 +ℎ 𝐼2 6𝑎𝑙 +𝑠1 ℎ 2 𝑃𝑎 4ℎ 3 𝐼2 −𝑎𝐼1 +ℎ 𝐼2 +3𝑙𝐼1 +𝑠2 𝑙 3 𝐼1 𝑎𝐼1 +2ℎ𝐼2 𝑀𝐷 = − 2 𝐼1 𝑙 3 +6ℎ 𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 + 2 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2 𝑃 𝑎 2 𝐼1 3𝑙−2𝑎 +6𝑎ℎ𝑙𝐼2 +𝑠1 ℎ 3 𝐼2𝑉𝐷 = 𝑉𝐴 = 𝑃 − 𝑉 𝐷 𝑙 3 𝐼1 +6ℎ𝑙 2 𝐼2 +2𝑠1 ℎ 3 𝐼2 6𝑃𝑎 ℎ 2 𝐼1 𝐼2 𝑙−𝑎𝐻𝐴 = 𝐻𝐵 = 4ℎ 3 𝐼2 2𝑙𝐼1 +ℎ 𝐼2 +𝑠2 𝑙 3 𝐼1 𝑙𝐼1 +2ℎ 𝐼2See equations (50) – (53) 𝑰𝟏Table 2: Axial deformation contribution ∆𝑴 𝑨 for different values of 𝑰𝟐w =1kN/m L = 5m H = 0.4m h = 4m 𝑰𝟏 𝑰𝟐 0 1 2 3 4 5 ∆𝑴 𝑨 0 -0.0045 -0.0095 -0.0144 -0.0192 -0.0241 𝑰𝟏 𝑰𝟐 6 7 8 9 10 ∆𝑴 𝑨 -0.0289 -0.0337 -0.0384 -0.0432 -0.0479 𝑰𝟏Table 3: Axial deformation contribution ∆𝑴 𝑩 for different values of 𝑰𝟐w =1kN/m L = 5m H = 0.4m h = 4m 𝑰𝟏 𝑰𝟐 0 1 2 3 4 5 ∆𝑴 𝑩 0 -0.0066 -0.0135 -0.0201 -0.0267 -0.0332 𝑰𝟏 𝑰𝟐 6 7 8 9 10 ∆𝑴 𝑩 -0.0396 -0.0460 -0.0524 -0.0587 -0.0651Table 4: Axial deformation contribution ∆𝑴 𝑨 for different values of 𝒉 𝒍 𝐼1w =1kN/m L = 5m 𝐼2 = 0.296 𝒉 0 0.1 0.2 0.3 0.4 0.5 𝒍 ∆𝑴 𝑨 -3.1250 -0.5409 -0.0823 -0.0239 -0.0096 -0.0047 𝒉 0.6 0.7 0.8 0.9 1.0 𝒍 ∆𝑴 𝑨 -0.0026 -0.0015 -0.0010 -0.0006 -0.0004Table 5: Axial deformation contribution ∆𝑴 𝑩 for different values of 𝒉 𝒍 www.ijmer.com 4253 | Page
  • 11. International Journal of Modern Engineering Research (IJMER) www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 𝐼1 w =1kN/m L = 5m 𝐼2 = 0.296𝒉 0 0.1 0.2 0.3 0.4 0.5 𝒍∆𝑴 𝑨 -4.1667 -0.7668 -0.1208 -0.0359 -0.0146 -0.0072𝒉 0.6 0.7 0.8 0.9 1.0 𝒍∆𝑴 𝑨 -0.0040 -0.0024 -0.0015 -0.0010 -0.0007 𝑰𝟏 Figure 9: A graph of ∆𝑴 𝑨 against 𝑰𝟐 𝑰𝟏 Figure 10: A graph of ∆𝑴 𝑩 against 𝑰𝟐 www.ijmer.com 4254 | Page
  • 12. International Journal of Modern Engineering Research (IJMER)www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645 Figure 11: A graph of ∆𝑴 𝑨 against 𝒉 𝑳 Figure 12: A graph of ∆𝑴 𝑩 against 𝒉 𝑳 www.ijmer.com 4255 | Page