Question

n=9

The equation for the deflection along a particular uniform beam under a given load is given by da y cos(x) H(x – 2n7) = with 0 < x < 4nt dr4 y'" (0) = 0, y" (0) = 0, y (4n7) = 0, y(4nn) = 0 dy 1. Integrate once and write down your expression for and then apply the boundary d.p3 condition y" (0) = 0. Write down your value for the integration constant. day 2. Integrate again and write down your expression for and then apply the boundary dx2 condition y" (0) = 0. Write down your value for the new integration constant. dy 3. Integrate again and write down your expression for and then apply the boundary dx condition y' (4n7) = 0. Write down your value for the new integration constant. 4. Integrate again to obtain an expression for the deflection y(x) and then apply the boundary condition y(4nt) = 0. Write down your value for the final integration constant.

Answer 1

• 
• 
  EH1n-2nn) (-uosn) + Hsinn - this - antiH dn2 Integration again white +fnisine-fonde - dy - (H/n-an) (-won) dut dn fannede dy - H(n-an ) (-snm) - H (- sinn Dehu + +fnsin-/moda-fantasia da Q dy = Hn-an) 1-sinn) 2H Con - Hur - 200 Hnt Cz dn put of y' (Ant) = O = H (4087-2007) (-sin 407 )-2M. Cosimo - M (403)2 3.0 2 - 201TH (495) + (3 동 - 2H + 648H 12
  
  4 dy = Hn-2nn) (-sinn) -24cosh - H1² - 200 H x + (2+ 6+81) dn ď Integration y (41o =åt») ksion)-fantes da -Impedan - Jannten din fis.de y = H(-2) con - files fansados - you - HN - H² + Conta 6 y = 41-215) csn H/n-2017) On - 34 Sinx - H3 H x² + (nt ca 6 at n = 400 0 0= (ann-205) cos (and) - 34 SinAnn-H (Ah) HH (47)2 6 + z(en) & le a = 120835. 147 H
  
• 
  final y = H (u-2017) issn - 34 sinn - Hus - noun ² + (2+ 6487²) Hn 4 120835,147H
  ​​​​​​​