(Answered): 8 kN 2 kN/m 2. Using superposition and the beam tables determine the displacement at point C and th ...

8 kN 2 kN/m 2. Using superposition and the beam tables determine the displacement at point C and the slope at point A. Assume EI constant. (Slope at A 56 kN-m'/EI in a cw direction) A Uc C 4 m 4 m Cantilevered Beam Slopes and Deflections Elastic Curve Deflection Slope Beam P -Px2 (3L x) 6EI -PL3 -PL2 max 8max Vmax = 3EI 2EI max - Px2 L-x) 0x L/2 6E max -5PL3 -PL2 0max Vmax = 8E 48EI - PL2 24E (3x L L/2x L - 78ma 2 wx (x2 24E -wL Umax SEI wL Umax -4Lx +6L). 0max= 6EI emax 0max Mor MoL MoL 0max= EI Umax 2EI 2E1 Mo max -wx x2 24E - 2Lx + Umax 0 x L/2 -7wL 384E wL3 Omax Vmax -wL 192ET 4x -L/2) 48EI 21= Omax L/2 xs L WoL War WOL3 Umax 10L2X+5LX2- r) (10L3 120EIL 22= max Umax= 30EI 24EI max L Simply Supported Beam Slopes and Deflections Deflection Elastic Curve Slope Beam -Px (3L2 48EI P 4x2) L L. -PL -PL2 2 2 0max Vmax 48EI 16EI 0x L/2 0max Umax P - Pab(L +b) -Pbx (L2 6?IL -B-) 02 6?IL - Pba (L2- b2-a) Pab(L+ a) 6?IL x-a 0xs a = 6?IL -MoL - MoL = L. max = - Max 6EI V243EI (L2-) 6EIL MoL 8 = 3EI at x = 0.5774L -5wL -wL3 2LX2L3) (x3 24EI Omax 21= Vmax = 24EI 384EI max max -5wL (16x3 - 24Lx2 9L3) -3wL3 V= 384E 768EI |x=L/2 128EI 0sxs L/2 7WL3 82 = 384EI wL -wL (8x - 24Lx Vmax = -0.006563 - EI L = 384EI 2 +17L2x- L3) L/2x < L at x = 0.4598L -7woL3 = WoL Wn Vmax -0.00652 EI 360EI -Wox (3x 360EIL 10L2X27L 2= WoL at x = 0.5193L 45EI