Ao 3 KN VB C 6 kN VD E -1.26 m-1.26 m-1.26 m 1 kN /F ↑ 1.2 m G 2. Determine the force in each member of the half hip truss shown. State whether each member is in tension of compression. Reaction at G is a roller support. (Ans: FAB = 5.80 KN (C), FAC = 4.20 KN (T), FBC = 0 KN, FCE = 4.20 KN (T), FBE = 1.450 kN (T), FBD = 5.25 KN (C), FDE = 6 KN (C), FDF = 5.25 KN (C), FEF = 7.25 KN (T), FEG = 0 KN, FFG = 6 KN (C)) FOLLOW THIS FORMAT BELOW I WILL UPVOTE
Ao 3 KN VB C 6 kN VD E -1.26 m-1.26 m-1.26 m 1 kN /F ↑ 1.2 m G 2. Determine the force in each member of the half hip truss shown. State whether each member is in tension of compression. Reaction at G is a roller support. (Ans: FAB = 5.80 KN (C), FAC = 4.20 KN (T), FBC = 0 KN, FCE = 4.20 KN (T), FBE = 1.450 kN (T), FBD = 5.25 KN (C), FDE = 6 KN (C), FDF = 5.25 KN (C), FEF = 7.25 KN (T), FEG = 0 KN, FFG = 6 KN (C)) FOLLOW THIS FORMAT BELOW I WILL UPVOTE
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
100%
![Ao
2.25 m
E
pivot point
2.25 m
Ex E
3 KN
1.26 m-1.26 m-1.26 m
Ey
VB
C
E
1.8 m
FOLLOW THIS FORMAT BELOW I WILL UPVOTE
1.8 m
Fy
Joint E:
For FEF:
VD
Solve for Ey: (1)
For FAB:
6 kN
E
1.8 m
2.8814, 2.25
1.8
E₁, = 30 kN
For FAF:
FAB =
B
B
1.8 m 1.8 m
4
FAF =
FEF 30
1.8 2.25
FAB
38.4187
1.8 2.8814
VF
=
1 kN
1.8 m
FAB = 24 KN (T)
↑
1.2 m
FAF 38.4187
2.25
2.8814
G
1.8(38.4187)
2.8814
FAF = 30 KN (C)
с
H
C
2. Determine the force in each member of the half
hip truss shown. State whether each member is in
tension of compression. Reaction at G is a roller
support. (Ans: FAB = 5.80 KN (C), FAC = 4.20 KN (T),
FBC = 0 KN, FCE = 4.20 KN (T), FBE = 1.450 KN (T),
FBD = 5.25 KN (C), FDE = 6 KN (C), FDF = 5.25 KN (C),
FEF = 7.25 KN (T), FEG = 0 KN, FFG = 6 KN (C))
10 kN
2.25(38.4187)
2.8814
1.8 m
H
10 kN
1.8 m
2.8814,
E₁ +F₁ = 10
E, +40 = 10
1.8
D
D
FEF
FEA
Reactions:
→
FEF=
SHOW FBD
→
ΣΕ
= 0
ΣΕ = 0
2.25 E,= 30 kN
E, +F, -10=0
E, +F, = 10
ΣΜ = 0
F,(1.8)-10(7.2)=0
[1.8F, = 72]
F, = 40 KN
E = -30 KN
E, = 30 kN,↓↓
1.8(30)
2.25
FEF = 24 KN (C)
Joint A:
For FEA:
2.8814
1.8
FAE = 38.4187 kN
FAF
;
A
30
2.8814 2.25
2.8814(30)
2.25
F = 38.4187 N (T)
EA
2.25
2.25
F
FAR
E, = 0
→ F, = 40 kN, 1
1.8
(1)
2.8814
FAE38.4187 kN
B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fc70597-7de9-4d95-bb98-47fe79963ee6%2Ff4817f31-9b15-4509-a1a9-b9d50f5c0ff1%2Fzk97t8t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Ao
2.25 m
E
pivot point
2.25 m
Ex E
3 KN
1.26 m-1.26 m-1.26 m
Ey
VB
C
E
1.8 m
FOLLOW THIS FORMAT BELOW I WILL UPVOTE
1.8 m
Fy
Joint E:
For FEF:
VD
Solve for Ey: (1)
For FAB:
6 kN
E
1.8 m
2.8814, 2.25
1.8
E₁, = 30 kN
For FAF:
FAB =
B
B
1.8 m 1.8 m
4
FAF =
FEF 30
1.8 2.25
FAB
38.4187
1.8 2.8814
VF
=
1 kN
1.8 m
FAB = 24 KN (T)
↑
1.2 m
FAF 38.4187
2.25
2.8814
G
1.8(38.4187)
2.8814
FAF = 30 KN (C)
с
H
C
2. Determine the force in each member of the half
hip truss shown. State whether each member is in
tension of compression. Reaction at G is a roller
support. (Ans: FAB = 5.80 KN (C), FAC = 4.20 KN (T),
FBC = 0 KN, FCE = 4.20 KN (T), FBE = 1.450 KN (T),
FBD = 5.25 KN (C), FDE = 6 KN (C), FDF = 5.25 KN (C),
FEF = 7.25 KN (T), FEG = 0 KN, FFG = 6 KN (C))
10 kN
2.25(38.4187)
2.8814
1.8 m
H
10 kN
1.8 m
2.8814,
E₁ +F₁ = 10
E, +40 = 10
1.8
D
D
FEF
FEA
Reactions:
→
FEF=
SHOW FBD
→
ΣΕ
= 0
ΣΕ = 0
2.25 E,= 30 kN
E, +F, -10=0
E, +F, = 10
ΣΜ = 0
F,(1.8)-10(7.2)=0
[1.8F, = 72]
F, = 40 KN
E = -30 KN
E, = 30 kN,↓↓
1.8(30)
2.25
FEF = 24 KN (C)
Joint A:
For FEA:
2.8814
1.8
FAE = 38.4187 kN
FAF
;
A
30
2.8814 2.25
2.8814(30)
2.25
F = 38.4187 N (T)
EA
2.25
2.25
F
FAR
E, = 0
→ F, = 40 kN, 1
1.8
(1)
2.8814
FAE38.4187 kN
B
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