Question

1) A mile-long bar (like an oil well rilling stem) hangs freely from its upper end. Model the pip with 5 equal length elements and simulate the load of gravity by putting half the weight of each element on its ends. Presuming a cross- sectional area A = 1 in2, an elastic modulus E = 30.0 x 10 psi, a weight density of y 0.3 lb/in', and length L= in, solve for the nodal displacements using the linear finite element formulation. Then, make graphs of 63,360 a) The displacement at all points of the pipe as a function Element of x b) The internal force p as a function of x Hanging pipe Model

Hey brother. first of all I want to thank you for solving this problem for me. I really don't understand how you get k1,k2, k3, k4, k5, and k6. I don't understand neither how you get F1 to F6... overall I do understand. can you explain me how u get these following values. please. u can also use matlab coding to safe time if u want to. I just don't have a clue idea how u get them. I just want u to write down the step of how you get k's and D's. thank you

Answer 1

-Ans: equation for the edongation Destving ot long bar the internal farce that A, E Constant w.rt L but we P is 5 0 took 'de strip length Over mc. (A.x).3.8 n-weight densi = A.2.n dre S strip 8 Tatal E Px.ch Sstrip A.E nA.da AE 2 Y 2 E E Tota 2 E

NovDetermining displaement /elongaton ysidawing individual elements Considering element only T.tl length 2 63360 1262 in 2 E x262 0. 3 = 0.802 in 2 30 x1o en sidering element FE Tstal dength is denidad into eual num ber Ports a s z2x 2 672 2 25344 in 0.3 x25344) S2 2 E 6 2x 30xlo 3.21 inch

for element 321 - E 3 8ol 6 in = o. 3 x 38016) 2 E 7.23 in 2 3Dx l0 for clement =ムズ」ニ 50 688 in (50688) S4 O.3X 12.85 2- E 2メ30 Xl06 Aoelement 5)- 5x, 6336 0 ih 0. 3 x 63360)2 2o.o 7 in 2 30 X 1D fanction as Displacement at all point 22 8 1 6 12 4 2 Dspbcemant

Load weight of element one x V W = W 0.3x 1 x 13672 3801.6 l b Half weight o the element 900. 8 b for element O adding halt weights at the end of element ) of element W7603.2 lb W1900.8 - 7 C03.2 b elencat elemrentl weight 1900.8), (19 w1100.8) + W 5702.4 Lb - Pe

Por elenent ( wa +( 100.)s 900.8 190 380 6 b element ( Aor 1900.8 b for clement Ps fanction The internal for P as gooolo Fo0ols SDool 3000s 2000