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800 lb 600 b YD 20 20 20 20 400 lb 1000 lb

800 lb 600 b YD 20 20 20 20 400 lb 1000 lb

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Can you help me explain briefly how it come up with the solution (picture#2) about the given problem (picture#1) analysis of truss

Problem:

Refer to the given truss below, (a) Using the method of joints, determine the forces in members AB, AC and CD

-Please explain me why the solution is like that for roller pin(Rc)

-And also the method of joints part please explain to me the solution 

-please im having a hard time following how it happen

 

& MG=°] Rie la0) - fo0L60) - 800 (30)~1000(20)+ 6o0(20) = 0 Re (40) : 56000 t fy- o] RetR6= 400+8 004100 40 Re = SUODO 1900 Re: 1400 Ib Pin A 0: tan-' () : 63,43° AB fry> o] AB sin (6343) - 100 = 0 AB = 441. 23 1b/ T € Fx =01 Ac+ 147-23 cos (63.43)- 0 400 1b AC : - 200 16| C rin B € Fy: 0] - BC sin (6343) - 447. 23 sin l6a.ta) = D Çolb BD BU = -441.233 lb/c & fx - 07 60+ 600 H-+41-233)(03(6 3.43) - 447-233 cos (63149) - 0 441-13 Ib BC 8D= -20016 c Pin u - 447.23 1b t Fg -07 1400 + co sin C63-43) - 447-23 sin (63. 43) - 0 cD = - |18-08 16 C CE - 2001b 190016
Transcribed Image Text:& MG=°] Rie la0) - fo0L60) - 800 (30)~1000(20)+ 6o0(20) = 0 Re (40) : 56000 t fy- o] RetR6= 400+8 004100 40 Re = SUODO 1900 Re: 1400 Ib Pin A 0: tan-' () : 63,43° AB fry> o] AB sin (6343) - 100 = 0 AB = 441. 23 1b/ T € Fx =01 Ac+ 147-23 cos (63.43)- 0 400 1b AC : - 200 16| C rin B € Fy: 0] - BC sin (6343) - 447. 23 sin l6a.ta) = D Çolb BD BU = -441.233 lb/c & fx - 07 60+ 600 H-+41-233)(03(6 3.43) - 447-233 cos (63149) - 0 441-13 Ib BC 8D= -20016 c Pin u - 447.23 1b t Fg -07 1400 + co sin C63-43) - 447-23 sin (63. 43) - 0 cD = - |18-08 16 C CE - 2001b 190016
800 lb 600 lb 20 A 20 20 20- 400 lb 1000 lb
Transcribed Image Text:800 lb 600 lb 20 A 20 20 20- 400 lb 1000 lb
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