Question

9. For the beam loaded and supported as shown in Figure (see Week 4), use the integration method to determine (a) The equation of the elastic curve using the x1 and x2 coordinates (b) The slope at A. (c) The deflection at C. Take E = 200 GPa and I= 4 x 108 mm 30 KN 4 m *20 kNm tem2m* -

Answer 1

301N B cl 220 wine 120 kN-me - RA mmm Calculate support Reactions. # E MA = 0 30X6 +20 - RB X 4 = 0 i RB = 50KN.T . . Efy=0 » RA TRB = 30 - RA +50=30 . » Ra=-20knd #using Equation M = EI dy o dxz i calculate moment at a distance x, ... from A.. Mai = RAX 1 =-2001 From ® [0321= 4m] 21 Y = -2031 .: Integrating worot xí El d y = - 20 x1 ta => €147 = -loxi +61 . Again Integaating eqh ® ely = -10% sej! +9X+6 - 3 # Apply boundary conditions. : (ů At x,=0 7 y=0 (deflection, pinned) A substitute in equation (3) To=C2) ax2

(ii) At x1 = 4 m > y=0 (Roller support) from equation (3) 03-10 (4)?+49 +0 → 19 = 1607 substitute value of a and. Ca in egn (3) - (4) Deflection equation Esy = - 10 x + 160 x # Slope equation From equation (2) EI = -10%;"+160 — 5) slope at A %,= 0, EI 2009 pa = 200x100gr/m² put in eqh (5) and I = 4 _00x1062 4x10*1712 OA = .0+ 160 On - The Ox105* X( | QA = 0.000667 radi?

Now for BG partion : - Macz = -20-30x2 [ORX<2m] using Erde z. Mar EIdy = -20-3022 - Integrating, EI OY -.-20%2 - 30 m2 + B;. Again Integaating, EIY = - 20x22 - 15 x 3 m + Bast Bz . Ely =-108% - 5%,+B1&₂7 B2 # Apply Boundary conditions. ů At x = 2m y=0 0= = 10x 22- 5x2 + 2B1 + B2 . 2 B1 + B2 = 80 - (fo) (11) At support B, equation of denkeition . would be same therefore Ly=2mand) Equation (5) - Equation (6)

2. - 10x²+ 160 = -200₂-15 %, +B1 - 1x 4+ 160 = -20x2 -15x2? +B1 | B = - 6.62 | B2 - 93.24) substitute in eah a Ety = - 10x, _ 5 x _ 6.6200277 93.24) put X₂ = 0 notex B. 24 E1 Jo = 0 -0 -0+ 93.24 Yo = 1324 - 93.24x10N) (200x107x 4 x 100 x 10 12 = 0.001165m (46 = 1.165 mm EI (200x109)x 441008