Question

PLEASE FAST I HAVE 15 MINS LEFT !!

Question 2: [50 Marks] For the beam shown, use the Conjugate Beam Method to determine a) The deflection at F, b) The slopes on both sides of the hinge at C. 90 KN 160 kN.m E B C D F 2 m 2 m 2 m 2 m 2 m EI = Constant

Answer 1

gokm T60 KN-m MA ܠܓ B BD A A E f Pove VE Solm reactions. finding go VA + VE EM2=0, VEX 4 = 160 40 KN VA 50 KN Z MA TO VEX 8 - 160 - 90x2 +A=0 40x8 - 16or 180 + MA =0 MA 20 ka-m. drawing M/EI diagram 80 ym ME A IN (2m libm B 2 min am c - 20 VE conjugale -80 Beam

now finding reactions in conjugate Beam Яге - б, — — X20 хоча 76 + очх2 а 5 +) XRUX 2 x ST33 - Vox 4 +1 x 90 x 1, 6Х6: 2 у XRUX2 X 2. - Сб +) x86 x 2-х ч а 3 VLхи- — To6. Ч3 Vc= - 16.4) 2) со Vc tVe ) X20 X:Чt X8 Х3 и а + 1 x RX 2 —) XBox 2 D स - 17% C EVE ч + Tчч - 0 VE = 36. LI FI Ле -Me + 36. ) х = 0 EI DR - ME +3, 22 - m3.

CLASSMALL Date Page Slope at both sides of 들 o (left). Vc (left) at left side to - 200o. 4 t l x 8083.6 2 2 Oc= KN-m2 2 BI at Right sede Oc (aight) = Ve (Right + Oc light) -20x0.4 t1X8083.6-176.611 O (right) = -36.61 KN-m² EI done thy in horry so please forgive any f done mis lake