Question

Consider the frame shown in (Figure 1). Assume A and Care pins and the joint at B is fixed connected. El is constant. Take w = 27 kN/m. Part A Determine the 2 component of reaction at C Express your answer to three significant figures and include the appropriate units. View Available Hint(s) THÅ ? Value Units Submit Part B Complete previous part(s) Part C Complete previous part(s) Part D Complete previous part(s) Provide Feedback Figure < 1 of 1 > 12 kN/m

Please find X and Y components at C and A. Thank you!

Answer 1

Answer 2

Egiven data, W = 27KN/m f J 12 KNM vly A God es constant Since frame the is in determinate given frame adopt moment distribution method Distribution factors: Calccllation of Joint Member Relative o stiffness Total 2 relative Stiffness Distribution o factor BA 315 31 40 37 31 4x4 16 -31 B SI BC 16 3 31 41 486 소 2/5 Moment distribution table fined end momentos -wp² MEAB 27x42 wp² twe² २ २ - - 36 KNM MEBA + we = +27x42 = 36 KNN 12 12 x 62 MA BC - we 20 = -21-6 KNM 20 М{ср - - + w p2 ما wee 20 + 12x6² 30 14.6 KNM 30 30

B 315 215 -36 36 -21-6 14.4 13612 +36 - 7.2 14tely 12 I 18 14.4 - 28.8 -1512 -10.08 38.88 -38.88 38-88 ta Horizontal reaction Taking moments at B EMB=0 GM CX XG + Cyxo - 38.88 Hiq x3 x6 )a&*6).-0 С 6 CX = 33:12 Cx = 5.52 KN Acy

Apply equillibaium equations Vertical forces equilliboium Iv=o Ayt. cy=27x4 Aytcy Horizontal forces equilibrium TH=0 Ax+ (x = ½ x 128G 108 3+ 2 Ax + x = 36 Apply moment equillibrium at A. consider Anticlockwise clocrwise tve EMA=0 - CyX4 – Cyx6 Háx12x673x6)+(2+84)x2 = 0 4Cy F6Cx=288 2Cy+3CX = 144 Đl 3

Ax +5.52=36 Ax = 30.48 KN 2xy + 3x5.52=14 4 Cy= 63.72 KN Ay+63-72 =108 Ay = 44.28 KN 0

W = 2#kN/m mm 12 kN/m 30-H9 KNS A 4m Ay A = 44.28KN 6 m 5.52 kN & cy - 63-72 KN