Question

URGENT, please be as neat as possible and make sure to use/explain concepts and equations being used for correct solution. Thank you again and do not copy solutions from Coursehero/Chheegg please. I really need the correct! solution for this

Problem 4 (20 Points) A cantilever beam of length L with constant EI is subjected to n point loads (each of magnitude P) that are evenly spaced. The beam is fixed at x = L (see Figure below). (a) Use the free-body diagram of the beam to determine the reactions at = L. (8 Points) (b) Use discontinuity functions to determine the deflection, va, at point A, which is the left end of the beam (x = 0). Your answer for va should be in terms of n, P, L, E, and I. There should be no summations present. (12 Points) You can use the following identities: m m(m +1) Đi=12+3+...+ m = 2 i=1 m m m (m – 1)2 = 12 + +...+ m2 (m + 1)2m + 1), IM:IM=IM 6 i=0 i=0 m m (m – i)* = 12 + 23 +33 + ... + m3 Гm (mt 1)], i=0 i=0 where m is a natural number. Р Р Р Р Р P А L/n L/n L/n L/n L/n

Answer 1

P B vm Yr Yn R (9) Ray Reachta force Mr > reaction moment Efyzo R-(P+P tp 4 upto on teens).=o R-nP = 0 2 RanP (EM) P. n. I + P (0-1) 4 / + pln-2) / + t-pot- an - Mp=o MRE P {n+(n-1) + (0-2) +---+ +11 MR = PL {it It 2+ + (n-1) tn n} MRE ma MR = PL Di Klnti) 2 me=(n+1)PL W

(b) EI day = M(x) dx² d²y da? → EI -P (x-4134%) =-Px-P(2-4)-P(2-2%). + day EI da? + EI (1) ا с. + -P[2+ X + (2-ń)+(- 2) +(2-6m):)] ETS --- - f[2*+(2-) P+(2.photo (2. tai yw @xe, lie slope: o) ? [2+(9:14.2)2 + ( 2 L.)+. (L)] +4)*7*] [met + i] [1 +2° +32 + +indlet me) pen 2 * C= Р 2 L² + L + + ( 92 932 → G PL² n²+ (0-1)²+(2-2)2 + en2 » G= PL² 202 * Ga i=1 G = Pl? 2n2 X na In+i) (2n+1) . G (+1)(2n+1) PL² 12n

Again, from (i). EI dy - G + + dx P 2 X2+)+(x-25 23 (*-10-1) 533] * {2+{z-+ (x-74) + - +(2-1-)3) L * Ely = C2 + ax سا (ii) @x=4, Y=0 + - - Pl? is 6n3 tie deflection =o) 0 = Cz+GL- & 2 + ( en raha na ) + +(640] 0 = (2+(L [mP+ 13+(n-1)3 + (0-2) + et de [1+28+3*+--+ (muget n) G =-612+ er fase .{plants + C2 = -6,Lt P² * 6 =-GL + 623 (n+1) PL² * C = -(L + 24 m (n+1)(2n+1) P2? + 12 m+112 24N PL² PL? (2= (n+112 (n+1)(20+1) 120 2 mp+20+1 -22n2+3n+1) & Q = PB [***, plun'azon) [ un 3n?+47 + 24M +1) PL m

from (ii) Put x=0 Deflechon at A 1 FIYA = C to- & EIYA = ₂ +1 YA 3n2+4071 PL 249