Using the method of joints, determine the forces in members AB, AC and CD (b) Using the method of sections, determine the force in member DF. Indicate if tension or compression. Reaction exerted by the roller at C is 1400 lb and the x and y components of the rection by the pin at G are 600 lb and 800 lb respectively.
Using the method of joints, determine the forces in members AB, AC and CD (b) Using the method of sections, determine the force in member DF. Indicate if tension or compression. Reaction exerted by the roller at C is 1400 lb and the x and y components of the rection by the pin at G are 600 lb and 800 lb respectively.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Hello 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 (b) Using the method of sections, determine the force in member DF. Indicate if tension or compression. Reaction exerted by the roller at C is 1400 lb and the x and y components of the rection by the pin at G are 600 lb and 800 lb respectively.
![& 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a222d5e-4945-4514-b048-8d0ce2657a63%2F12eebad1-7639-4e75-827e-3628ea5e2a0a%2Ft6c4ze7_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a222d5e-4945-4514-b048-8d0ce2657a63%2F12eebad1-7639-4e75-827e-3628ea5e2a0a%2Fyy44npk_processed.png&w=3840&q=75)
Transcribed Image Text:800 lb
600 lb
20
A
20
20
20-
400 lb
1000 lb
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