Question

L0=2 M=3

-Lot Wo=2 KM ow - L=4m

a)Free-body Diagram for the beam, and TWO useful equilibrium equations

b) breakdown for the beam, into a minimum of 3 cases , with
clear indication of either deflections, and slopes (angles) to be used in following compatibility
equation

c) Write compatibility equation, and values of deflections or angles as they are obtained from textbook-
formulas

d)Solution for extra-unknown-reaction, including required algebra in compatibility equation:

e)Solutions for the rest of unknown-reactions, from part d above-solution and equilibrium equations of part a

Answer 1

2m zm Wo=21N Mo=3kNm " To Calculate a Free - body Diagram for the beam and two useful equilibrium equations MA MA 7 wo - m to Lo B е Firot equation ER =0 RA+RcZ wolo

MA O A+ Wolot +Mo-Rex 4 = 0 MA= 4Rc-Walo²_Mo KN mm D Breakdown for the, Beam into a minimum d. 3 cases with clear indication of either deflections, slopes to be used in compatibility equation. A I am B am. c Rc VMO

(x=L, Y OEM:0wxvM 0 M=- wx? [x= Lino] EI : ME-fwx? El - (Wx + c Ex [x=, 0) 0 = -5WL + C, C, - ŚwL3 EI -Ćwxá + WLS EIy - wx* + 6 w L’x + C [x=b, yoo! 0:- wľ + & WK+CZ O C2-(8-)W = WL" (a) Elastic curve. y -- EI x' – 44**-+ 3{"> Bol yerso: Select the best you were to get (c) Ex=0: selon enter a

In mod end +) My = 0: Mo-M=O MEMO - L x=0, y=0] [x-u sto] EI - Me EI = Max + C. Ely = Max* + C, % + C, [x=0, 0] 0:0+C, C; = 0 KL-X - 0 Ely: $MX+C [2=0, y =o) (Elastic curve. Ce=0 0:0+ Ca y: 201 M70 Max EI @ ZL. Yo = Me Le ML? ZEI (b) ye nal. con la exL. Dames

DEM0: M P(L-x) = 0 M : - P (L-x) [x=0; y=o] [x=0, 0] a. L-X- EI - P(t-x), - PL + Px EI - PLx + $ PX+C [x=0, o] 0:0010-C. C0 Ely: - PLx* * $ PX + C.x + Co 0:-0+0+0+C, C=0 yo-pa (31-) [x= 0, y:o] (a) Elastic curve. (b) y @ xol. Yo = - (3L-L): - Play (c) mexsli on (22-) -

c from the derived formulas • Dueto Moment Mo A = MOLE REI • Due to Single point ļoad Rc 3EI I • Due to UDL at point a wo AB= Wolf 8EI B = WOL GEI I WOLT प

d) To Cakulate the reaction at pocntc. - Adnet zo is extra equation a so -Ac + I WOLAB = RBX LAC TEX SEX Mol. 2 MOLAC + 24 Subsituting Mo = 3 kNm LAC = 4m wo = 2 kN/m 3842 + 22x2x16 = Rexhep 24 + 28 = RBX G4 RB - 29 KN = 1.5625 KN- Answer