Question

Problem 2 (18 points): The structure shown below consists of the beam AC which supported by a pin at A and a link at B. 3 kips/ft 1.6 kips/ft А D a) Determine the reactions. (3 pts) b) Write the equations for shear and moment for the region BC (8<x<12). Set the origin at point A. (6 pts) c) At x=4 ft, make a cut section and calculate internal forces. (3 pts) d) Locate the points of zero shear between A and C. (3 pts) e) Using the equations developed in part (b), confirm the results you got in part (C). (3 pts)

Answer 1

Solutions Draue Free Bodej diagram of Ac 3 Kibelft (a 1.6 kipslift tam Los = 3 (Ax) 0336-86 (Ay) e FBD 4 A

EMA ze Føp sin(e)a8 z 1:661246 + 1.41:441268 FBP = 38.00 Kibs z Fyzo Ay + FBD sim(@) z 1:60/24 141.4612 vertical Reaction at Ay = 4.80525 vite Ans Ay = 41805 Ribas EFxzo Ax- FBD Cos(e) A Horizontal Reaction An= 30.403 Kips. A) at Ang Force in BD FBD = 38.00 Ribs Ang b X Whe 3 Kips ift 106 kips/ft A FBOCSO B Ay Х Frosince # 4 Considering left hand side of section section Bc (82 x L 12 ft) X-X For

Shear force, Vre Vez Ay + FBp since) – 1622 - xxles (where их, 1:4. Nacz 27.6-1.6*- 114 x2 24 ac 12 4422= 1:4 Answer EX 12 Bending moment Mac Moez Ay.X + Frosincexies) 11632-4 na licu x 2 2 3 Imee 4.812+2268 (4-8)-0.822-0.0194 x3 C Answer At x = 4ft from A jus 14 *4 = 0.46 7 kebsite 12 116 kipslit AM N Az To aff Are Normal force, N = An= 30-403 Kips Shear force VE-2.529 kibe Beneling Moment M = 4.8052544 - 1.64442-24446.4677 M3 5:175 Ribas-ft As

@ Zero shear In between AB Vr=o= Ag 1.62 HIN 114 x 22/ 2 12 4152-7368 24 x = from A 20997 Answer In between BC 24 Vz) =0= 2706 - 1.6.2 - 1.4.x² 24 at x 12 Shear will be zero Again from. A So between A&C Shear will be zero at 1244 = 2.7347 From '? Answer