Question

QI. A vertical pile is used to transfer the vertical load from the soft ground surface to the rock surface. It is assumed that the stiffness of the rock is sufficient to prevent any vertical displacement so that the lower edge of the rod may be considered as fixed. The soft ground acts on the pile along its length with a force per unit length which is proportional to the displacement i.e. cu(t), where c is the elastic stiffness. For this problem the appropriate governing equation is du(e) EA +(I)-f(x) = 0. where E and A are the Young's modulus and the area of the rod respectively (both assumed constant), (I) is the displacement in the 3-direction, c is the stiffness of the elastic foundation and f(a) is the external force supplied per unit length of the rod. Prove that a stiffness matrix for the above problem with a two node finite element and linear interpolation of displacement u(x)may be expressed in the form (a) EA + C- 3 3 (b) EA 20 EA Symm. [13 marks] Using the above stiffness matrix and a three-element representation as indicated in Figure Q1, determine the vertical displacement distribution. Assume that the length of the pile is I = 18[m] and the cross-sectional area A = 0.003[mº]. Young's modulus is specified as E = 200,000[kN/m") while the traction, which acts along the length of the pile, is described by the elastic stiffness c = 10[kN/m]. The load is given in terms of the point load P = 100kN] on the upper end of the pile. [12 marks] P Geometry: 1 =18 A = 0,003 m) Material properties: E = 200,000 kN/m') e = 10[kN/m] Loading: P = 100kN) Figure Q1. Vertical displacement of a pile: Geometry, material characteristics and loading conditions TURN OVER

need to solve the mathematical model to prove
that we can get the equations i   Q1 a methematically

Answer 1

doe EA E A The Governing differential equation is (Au + Cu() - P(x) = 0 where, Young's is modulas of the rod Area of the rod U (1) - displacement in a-direction. stipeness of chetic foundation f(x) = External face Per mit length. (a) We are solving from Rayleigh - Ritz method. In first make weak form of gov. diff. equation. In this method choice of weight fincation is interpdation fincation C this Loe Same For an element of height it, 5 of 4 an dupan) it EH W ( 24. dulu) + Cult) – A(z))de (-5. dug:)) de +w.c. Ux) dr - Als de o are doing doing integration by parts. SEA En w dupe de +/c w ur) dic it EA W. de(x) -) +

W. flot) de o O منابع و مرا weight fincation. Now 2 rode limear interpolation fincation for displacement wie Age taking 5 and W = W; " Uj W wis Interpolation Firmesition and Wing is displaces a where Vis ment at node points linear element 오. 44 We = %, t it ta (19). 4L2 x 4]ox + + L v ]de 4. Holodse WER ( الله )1 - 0 En dulu & Posce at the node Points 'Od (6 4 4 + cxx)a) s-forove the way

dement from direction of noda at 1 nal M1 J . [u, colegios y me Wn. EA duſ) 4,4 CM-212 4; CA.dll) + W; En dulu) dit For linear element N=2 Here in l.2, and I=1,2 So the stiffness matrix the will be like, jerseys be + C4 - c. 48) de lén dudave) - eu valoda VEA.dll dy In Chat 1 p 47) LEA In integration, (11) (+)+(-3)* + EN: 64114843) EL + Cat (5) (+) t+ | 4 (1-3) I vett - FER |< (3)3 4 - Fan 0 -- ER it + EA + 27 3t

(1,3)+(994 + it 312 ER d cd + Co stiffness matrix is EA + ct 120 3 Kia 1 - + Ct + Ct -90 120 can Now assemble 3 elements, here it = 6m, foo 3 element & other data are given in question, ku 120 -90 90 120 Here Hla) = external applied load/ lengte (-3).x Plu).de it = Se lo k مه( - ) * odbe = 10 Li X- lox 30 x 10 30

One mildy 190 190 30+ 0 U, 14- 20-10 0 120+10 - U, --90 & 120 Elemenet Local Global 1-2 1-2 2 1-2 2-3 3-4 3 1204,- 900g = 100 +30 0 -904, + 240. Ug - 9045 - 60 2 -9049 + 2404 toll Из -0.0128 m, He = -0.70lm, 0.557 element (1) ам as his So displacement is given 14 = u, Wit ue 42 - (*), + similarly for the others, (7) .U2
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e-(02) 55. (8-1)-(414 (ѕ еvель" 9 *(14) + ?() po