NOVEL THINKING

Lateral deflection of portal frames is largely dependent on the stiffness of structural joints and also the flexural rigidity of vertical (column) members. E = 200 GPa, I = 60(106) mm4, A = 600 mm2 0000050055 00000 n 244 0 obj << /Linearized 1 /O 246 /H [ 768 1198 ] /L 406396 /E 50958 /N 72 /T 401397 >> endobj xref 244 18 0000000016 00000 n This is confirmed by evaluating the relative stiffnesses: By inspection we can see that the core will resist in excess of 99% of the external loading due to its much larger relative stiffness. The the next post we’ll consider how this analysis plays out for asymmetrically propped structures and confirm the results of our hand analysis with a simple FE model analysis. Structville is a media channel dedicated to civil engineering designs, tutorials, research, and general development. <> Portal frame structures are most often encountered as steel frame structures, e.g. They possess all the three basic stiffness characteristics, i.e., they have local bending stiffness, global bending stiffness and shear stiffness. By. In the context of building structures, there are several well established structural schemes for facilitating this load transfer. The disposition of the multi-storey frame is shown below; The properties of the building are as follows; Geometrical Properties of the frameSecond moment of area of beams IB = (bd3)/12 = (0.4 × 0.753)/12 = 0.0140625 m4Second moment of area of columns IC = (bd3)/12 = (0.4 × 0.43)/12 = 0.0021333 m4Flexural rigidity of beams EIb = 25 × 106 × 0.0140625 = 351562.5 kNm2Flexural rigidity of columns EIC = 25 × 106 × 0.0021333 = 53333.333 kNm2Shear stiffness of beams = KB = ∑(12EIb)/(Lih) = 3 × [(12 × 351562.5 )/(6 × 3)] = 703125 kN, The part of the shear stiffness associated with the columns is;KC = ∑(12EIC)/h2 = 4 × [(12 × 53333.333 )/32] = 284444.443 kN From the above, the reduction factor r can be defined as; r = (KC )/(KB + KC) = (284444.443 )/(703125 + 284444.443) = 0.2880The shear stiffness of the frame work can now be defined as;K = KB × r = 703125 × 0.2880 = 202500 kNFor the local bending stiffness (EI = EIC.r), the sum of the moment of areas of the columns should be produced and multiplied by the reduction factor r. As the bays of the frame are identical, the second moment of area of one column is simply multiplied by n and r.I = r∑IC = 4 ×0.2880 × 0.0021333 = 0.002457216 m4 The second global moment of area Ig is given by; Ig = ∑Ac,itiWhere Ac,i is the cross-sectional area of the ith column, and ti is the distance of the ith column from the centroid of the cross-sections.Ig = ∑Ac,iti = 0.4 × 0.4 × (92 +32 + 32 + 92) = 28.8 m4The total second moment of area for the bending stiffness (If) is given by;If = IC.r + Ig = 28.8 + 0.002457216 = 28.8024 m4, The parameters S, , and H are also needed for the calculation of the maximum deflection;S = 1 + (IC.r)/Ig = 1 + (0.002457216)/28.8 = 1.00008532 ≈ 1.0 = √bs = √(K/EI) = √[202500/(25 × 106 × 0.002457216)] = 1.8156Therefore H = 1.9485 × 60 = 108.936With the above auxiliary quantities, the maximum total deflection of the frame work can now be calculated;ymax = y(H) = (wH4/8EIf) + (wH2/2KS2) – wEI/(K2S3) × [((1 + HsinhH)/cosh⁡H) – 1]yb(H) = wH4/8EIf = (5.05 × 604)/(8 × 25 × 106 × 28.8024) = 0.01136 mys(H) = wH2/2KS2 = (5.05 × 602)/(2 × 202500 × 12) = 0.044889 myi(H) = wEI/(K2S3) × [((1 + HsinhH)/cosh⁡H) – 1] = (5.05 × 25 × 106 × 0.002457216) / (2025002 × 13) × [(1+ 108.936 sinh(108.936)) /cosh⁡(108.936) – 1] = 0.00082413 mymax = 0.01136 + 0.044889 – 0.00082413 = 0.055424 m = 55.424 mm. 3 0 obj Hence, the behaviour of frames in resisting lateral loads may be characterised by three types of stiffnesses and corresponding deflection types which are; (a) Shear deformation(b) Global bending(c) Local bending. The result of this is that portal frames generally experience larger lateral (sway) deflections than braced frames. 0000050726 00000 n

A rigid frame derives its lateral stiffness mainly from the bending rigidity of frame members interconnected by rigid joints. Analysis of Sub-Frames Using Stiffness Method: A solved Example. As you will see, this can be accomplished by utilising quite rudimentary analysis techniques. Can you identify the cause of failure of this building. /Creator <652D50444620436F6E76657274657220616E642043726561746F722076322E31202D204275696C643A204175672031352032303035> 0000002290 00000 n Such a building is said to be propped by the cantilevering cores. Our previous analysis suggests one shear wall resists approximately 17.3% of the wind load or . The deflection behaviour of the rigid frame was investigated using a manual method proposed by Zalka (2013), and computer-based method (finite element analysis). LATERAL STIFFNESS OF INFILLED FRAME WITH DOOR & WINDOW OPENINGS FOR VARYING MODULUS OF MASONRY Syed Farooquddin 1, Renuka Devi M.V 2, K Madhavi3 and Manjunath.S 4 Address for Correspondence 1M-Tech Student, 2Associate Professor, 3,4 Assistant Professor

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