Correct the error and write one to two sentences explaining the error
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Correct the error and write one to two sentences explaining the error
Transcribed Image Text:Ax
Ext eq:
Joint A
FAF
TAY
+3
Ys Initials.
2. (5) Determine the force in each member of the truss and state if the members are in tension or
compression. Use Method of Joints.
30 FAB
Mc=0 Ay (4) - 600N (2m) = 0 AX_
Ay
= 300N
FAF
Joint F:
Σ Fx = 0:
SEF
ENGR&214
Spring 2017
AX
+↑ ZFY = 0. AY + FAF SIM 30º = 0
AY
sin 306
Ax +
FAB +
FAB
FAB
FAB
Ext eq:
(EMB=0: Cy (2) + Ay (2) = 0
|CY = 300 N/
FAF
FAF 600N/CNT?
=
FAF Cos 30º = 0
FAF COS 30°
519.6 N
-
2m
5 19.6 N
+EFx=0 - F AF COS 30° + FE# sim30°
FEF
FAF Cos30°
.FEF
600 N
sim 30°
1039 N (c)
FEF 1039 N(C)
+3.50
+3+miot
D
2 m
Acy
Page 3 of 5
Transcribed Image Text:Joint C
FCD
FCB
FEF
FABE
Joint B.
FBE
个
Joint E.
FEB
CY
L
600 N
1+ 2FY=0: Cy + FCD sim 30° = 0
FCD
600 N
FED
FBC
1SFy:0
EFx=0:- FcB
Fc B
Fcb =
1 E Ex=0:1
⇒>
- 600 N +
062
2X61347
+↑ {FY=0;
=
T
MODE
FAB - FB C = 0
FAB =
FAB
FCD cos30° = 0(2)
FCD cos 30°
519.6 N (T) Pra
H30
FBE
(0) 100 E
FED
FBE O
-
FBC
519.6 N (T)
Fx = 0 : FED cos30 - FEF cos 30° = 6
647
FED =
FED =
1751.
600 N T
FEF Cos 30°
cos 30°
FED = FEF
FED= 600 N (C)
FEB - FEFSIM 30° - FEDsim30°: 0
FEB + FEFSIM30°
sim 30°
1800N (1))
report 2.f.1
Fmiot
M008 - 12.
m
tmist
3. (5) Determine the force inc
compression. Use Mo
LLH
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