Hibbeler R.C. Structural Analysis
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630 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F2–8.
A
x
B
x
B
x
C
x
C
y
C
y
C
x
B
y
B
y
2 m
3 m
6 kN
4 kN
6 kN
3 m
2 m 2 m
6 m
A
y
D
y
M
D
D
x
Member AB
Ans
Ans
Member BC
Ans
Ans
Ans
Member AB
Ans
Member CD
Ans
Ans
AnsM
D
= 12.0 kN
#
mM
D
- 2.00162= 0d+©M
D
= 0;
D
y
= 6.00 kND
y
- 6.00 = 0+
c
©F
y
= 0;
D
x
= 2.00 kN2.00 - D
x
= 0
:
+
©F
x
= 0;
A
y
= 6.00 kNA
y
- 6.00 = 0+
c
©F
y
= 0;
C
y
= 6.00 kNC
y
162- 6122- 6142= 0d+©M
B
= 0;
B
y
= 6.00 kN6122+ 6142- B
y
162= 0d+©M
C
= 0;
C
x
= 2.00 kN2.00 - C
x
= 0
:
+
©F
x
= 0;
A
x
= 2.00 kN4132- A
x
162= 0d+©M
B
= 0;
B
x
= 2.00 kNB
x
162- 4132= 0d+©M
A
= 0;
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 631
A
x
B
x
B
x
C
x
C
y
C
y
C
x
B
y
B
y
4 ft
3 ft
2(8) k
0.5(6) k
3 ft
4 ft
4 ft
A
y
D
y
M
D
D
x
F2–9.
Member AB
Ans
Ans
Member BC
Ans
Ans
Ans
Member AB
Ans
Member CD
Ans
Ans
AnsM
D
= 6.00 k
#
ftM
D
- 1.50142= 0d+©M
D
= 0;
D
y
= 8.00 kD
y
- 8.00 = 0+
c
©F
y
= 0;
D
x
= 1.50 k1.50 - D
x
= 0
:
+
©F
x
= 0;
A
y
= 8.00 kA
y
- 8.00 = 0+
c
©F
y
= 0;
C
x
= 1.50 k1.50 - C
x
= 0
:
+
©F
x
= 0;
C
y
= 8.00 kC
y
182- 2182142= 0d +©M
B
= 0;
B
y
= 8.00 k2182142- B
y
182= 0d+©M
C
= 0;
A
x
= 1.50 k0.5162132- A
x
162= 0d+©M
B
= 0;
B
x
= 1.50 kB
x
162- 0.5162132= 0d +©M
A
= 0;
632 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F2–10.
2 m 2 m 2 m
6 kN
8 kN 8 kN
6 kN
A
x
B
x
B
x
C
x
C
y
C
y
C
x
B
y
B
y
3 m
3 m
1.5 (6) kN
A
y
D
y
M
D
D
x
Member BC
Ans
Ans
Member AB
Ans
Ans
Ans
Member BC
Ans
Member CD
Ans
Ans
AnsM
D
= 27.0 kN
#
m1.5162132- M
D
= 0d+©M
D
= 0;
D
y
= 14.0 kND
y
- 14.0 = 0+
c
©F
y
= 0;
D
x
= 9.00 kND
x
- 1.5162= 0
:
+
©F
x
= 0;
C
x
= 0
:
+
©F
x
= 0;
A
y
= 14.0 kNA
y
- 14.0 = 0+
c
©F
y
= 0;
A
x
= 0
:
+
©F
x
= 0;
B
x
= 0d+©M
A
= 0;
B
y
= 14.0 kN8122+ 8142+ 6162- B
y
162= 0d+©M
C
= 0;
C
y
= 14.0 kNC
y
162- 8122- 8142- 6162= 0d+©M
B
= 0;
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 633
F3–1. Joint C
Ans
AnsF
CA
= 30.0 kN 1T250.01
3
5
2- F
CA
= 0+
c
©F
y
= 0;
F
CB
= 50.0 kN 1C240 - F
CB
1
4
5
2= 0
:
+
©F
x
= 0;
F3–2. Joint B
Ans
AnsF
BA
= 6.00 kN 1C2F
BA
- 8.485 cos 45° = 0
:
+
©F
x
= 0;
F
BC
= 8.485 kN 1T2= 8.49 kN 1T2F
BC
sin 45° - 6 = 0+
c
©F
y
= 0;
F
CA
F
CB
40
C
3
5
4
N
B
F
BA
B
3
5
4
F
CB
⫽ 50.0 kN
Joint B
Ans
N
B
= 30.0 kNN
B
- 50.01
3
5
2= 0+
c
©F
y
= 0;
F
BA
= 40.0 kN 1T250.01
4
5
2- F
BA
= 0
:
+
©F
x
= 0;
F
BA
F
BC
B
6 kN
45⬚
Joint C
Ans
AnsF
CA
= 6.00 kN 1C2F
CA
- 8.485 sin 45° = 0+
c
©F
y
= 0;
F
CD
= 6.00 kN 1T28.485 cos 45° - F
CD
= 0
:
+
©F
x
= 0;
F
CD
F
CA
C
8.485 kN
45⬚
F3–3. Joint C
Ans
AnsF
CB
= 10.0 kN 1C2F
CB
- 14.14 sin 45° = 0+
c
©F
y
= 0;
F
CD
= 14.14 kN 1T2= 14.1 kN 1T210 - F
CD
cos 45° = 0
:
+
©F
x
= 0;
F
CD
F
CB
C
45⬚
10 kN
F
DA
F
DB
D
14.14 kN
Joint D
Ans
AnsF
DB
= 0a
+
©F
y¿
= 0;
F
DA
= 14.14 kN 1T2= 14.1 kN 1T214.14 - F
DA
= 0
+
Q©F
x¿
= 0;
634 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F3–4. Joint D
Ans
AnsF
DA
= 0+
c
©F
y
= 0;
F
DC
= 2.00 k 1T2F
DC
- 2 = 0
:
+
©F
x
= 0;
10 kN
F
BA
N
B
B
2 k
F
DA
F
DC
D
F
CA
F
CB
2
C
3
5
4
F
AB
N
A
3.333 k
3
5
4
Joint B
Ans
N
D
= 10.0 kNN
B
- 10.0 = 0+
c
©F
y
= 0;
F
BA
= 0
:
+
©F
x
= 0;
Joint C
Ans
AnsF
CB
= 2.667 k 1T2= 2.67 k 1T23.3331
4
5
2- F
CB
= 0+
c
©F
y
= 0;
F
CA
= 3.333 k 1C2= 3.33 k 1C2F
CA
1
3
5
2- 2 = 0
:
+
©F
x
= 0;
Joint A
Ans
N
A
= 2.667 kN
A
- 3.3331
4
5
2= 0+
c
©F
y
= 0;
F
AB
= 2.00 k 1T2F
AB
- 3.3331
3
5
2= 0
:
+
©F
x
= 0;
F3–5. Joint D
Ans
AnsF
DA
= 0+
c
©F
y
= 0;
F
DC
= 0
:
+
©F
x
= 0;
F
DA
F
DC
F
CA
F
CB
C
8 kN
45⬚ 60⬚
Joint C
Ans
AnsF
CB
= 10.93 kN 1C2= 10.9 kN 1C2F
CB
- 5.657 sin 45° - 8 sin 60° = 0+
c
©F
y
= 0;
F
CA
= 5.657 kN 1T2= 5.66 kN 1T28 cos 60° - F
CA
cos 45° = 0
:
+
©F
x
= 0;
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 635
F
AB
N
B
10.93 kN
Joint B
Ans
N
B
= 10.93 kN+
c
©F
y
= 0;
F
AB
= 0
:
+
©F
x
= 0;
Joint F
Ans
AnsF
FD
= 1000 N 1T2= 1.00 kN 1T21414.21 sin 45° - F
FD
= 0+
c
©F
y
= 0;
F
FG
= 1000 N 1C2= 1.00 kN 1C2F
FG
- 1414.21 cos 45° = 0
:
+
©F
x
= 0;
F
ED
F
EF
1000 N
45⬚
F
F
FD
F
FG
1414.21 N
45⬚
F
DC
F
DG
600 N
1000 N
1000 N
45⬚
F
CG
800 N
F3–6. Entire truss
Joint E
Ans
AnsF
ED
= 1000 N 1T2= 1.00 kN 1T21414.21 cos 45° - F
ED
= 0
:
+
©F
x
= 0;
F
EF
= 1414.21 N 1C2= 1.41 kN 1C21000 - F
EF
sin 45° = 0+
c
©F
y
= 0;
E
y
= 1000 NE
y
182- 600122- 800142- 600162= 0d+©M
A
= 0;
Joint D
Ans
AnsF
DC
= 1400 N 1T2= 1.40 kN 1T21000 + 565.69 cos 45° - F
DC
= 0
:
+
©F
x
= 0;
F
DG
= 565.69 N 1C2= 566 N 1C21000 - 600 - F
DG
sin 45° = 0+
c
©F
y
= 0;
Joint C
Ans
Due to symmetry,
Ans
AnsF
AB
= F
ED
= 1.00 kN 1T2F
AH
= F
EF
= 1.41 kN 1C2F
HB
= F
FD
= 1.00 kN 1T2
F
HG
= F
FG
= 1.00 kN 1C2F
BG
= F
DG
= 566 N 1C2F
BC
= F
DC
= 1.40 kN 1T2
F
CG
= 800 N 1T2F
CG
- 800 = 0+
c
©F
y
= 0;
636 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F3–7. For the entire truss
For the left segment
Ans
Ans
AnsF
BC
= 4.00 k 1T2F
BC
152+ 2152- 3.001102= 0d +©M
G
= 0;
F
HG
= 3.00 k 1C2F
HG
152- 3152= 0d+©M
B
= 0;
F
BG
= 1.41 k 1C23.00 - 2 - F
BG
sin 45° = 0+
c
©F
y
= 0;
A
x
= 0
:
+
©F
x
= 0;
A
y
= 3.00 k2152+ 21102 + 21152- A
y
1202= 0d +©M
E
= 0;
F3–8. For the entire truss
For the left segment
AnsF
BC
= 1200 lb 1T2F
BC
132+ 600142- 1500142= 0d+©M
I
= 0;
F
HI
= 1600 lb 1C2F
HI
132+ 600142+ 600182- 1500182= 0d+©M
C
= 0;
A
x
= 0
:
+
©F
x
= 0;
A
y
= 1500 lb6001162 + 6001122+ 600182 + 600142- A
y
1162= 0d+©M
E
= 0;
5 ft
3.0 k
2.0 k
5 ft
5 ft
45⬚
F
BG
F
BC
G
B
F
HG
3 ft
4 ft 4 ft
1500 lb
600 lb
600 lb
3
5
4
F
BC
F
CI
F
HI
I
C
Joint H
Ans
AnsF
HC
= 600 lb 1C2F
HC
- 600 = 0+
c
©F
y
= 0;
F
HG
= 1600 lb 1C21600 - F
HG
= 0
:
+
©F
x
= 0;
1600 lb
600 lb
F
HC
F
HG
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 637
F3–9. For the entire truss
Consider the right segment
Ans
Ans
AnsF
BC
= 00 - F
BC
122= 0d +©M
D
= 0;
F
ED
= 1.00 kN 1C27.00122- 6122- F
ED
122= 0d +©M
B
= 0;
F
BD
= 9.899 kN 1T2= 9.90 kN 1T27.00 - F
BD
sin 45° = 0+
c
©F
y
= 0;
N
C
= 7.00 kNN
C
142- 8122- 6122= 0d +©M
A
= 0;
F3–10. For the entire truss
Consider the right segment
Ans
Ans
AnsF
CD
= 800 lb 1T21000182- 400182- F
CD
162= 0d+©M
F
= 0;
F
GF
= 666.67 lb 1C2= 667 lb (C)10001162- 4001162- 400182- F
GF
A
3
5
B
1162= 0d+©M
C
= 0;
F
CF
= 333.33 lb (C) = 333 lb 1C2400182- F
CF
A
3
5
B
1162= 0d+©M
E
= 0;
N
E
= 1000 lbN
E
1322- 400182- 4001162- 4001242- 4001322= 0d +©M
A
= 0;
2 m
7.00 kN
6.00 kN
2 m
45⬚
F
ED
F
BD
F
BC
B
D
6 ft
8 ft 8 ft
C
F
E
1000 lb
400 lb
400 lb
3
3
5
5
4
4
F
CD
F
CF
F
GF
638 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F3–11. For the entire truss
Consider the right segment
Ans
Ans
AnsF
FE
= 2.00 kN 1C24.0011.52- 211.52- F
FE
11.52= 0d+©M
C
= 0;
F
BC
= 4.00 kN 1T24.00132- 2132- F
BC
11.52= 0d+©M
F
= 0;
F
FC
= 2.828 kN 1C2= 2.83 kN 1C24.00 - 2 - F
FC
sin 45° = 0+
c
©F
y
= 0;
N
D
= 4.00 kNN
D
162- 2162- 4132= 0d+©M
A
= 0;
F3–12. For the entire truss
Consider the right segment
Ans
Ans
AnsF
CF
= 0F
CF
A
3
5
B
(16) + 5001122- 750182= 0d+©M
O
= 0;
F
GF
= 1030.78 lb = 1.03 k 1C2750182- 500142- F
GF
a
1
217
b1162= 0d+©M
C
= 0;
F
CD
= 1000 lb 1T2750142- F
CD
132= 0d+©M
F
= 0;
N
E
= 750 lbN
E
1162- 500142- 500182- 5001122= 0d+©M
A
= 0;
3 m
2 kN
F
4 kN
1.5 m
1.5 m
45⬚
F
FC
F
FE
F
BC
C
3 ft
4 ft 4 ft
8 ft
C
F
O
750 lb500 lb
1
3
17
5
4
4
F
CD
F
CF
F
GF
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 639
F4–1.
Segment CB
Ans
Ans
AnsM
C
=-20 kN
#
m-M
C
+ 10112- 10132= 0d +©M
C
= 0;
V
C
= 0V
C
+ 10 - 10 = 0+
c
©F
y
= 0;
N
C
= 0
:
+
©F
x
= 0;
B
y
= 10.0 kNB
y
122+ 20 - 10142= 0d+©M
A
= 0;
F4–2.
Segment CB
Ans
Ans
AnsM
C
= 6.75 kN
#
m10.511.52- 811.5210.752- M
C
= 0d+©M
C
= 0;
V
C
= 1.50 kNV
C
+ 10.5 - 811.52= 0+
c
©F
y
= 0;
N
C
= 0
:
+
©F
x
= 0;
B
y
= 10.5 kNB
y
132- 411.5210.752- 811.5212.252= 0d+©M
A
= 0;
F4–3.
Segment AC
Ans
Ans
AnsM
C
= 12.4 kN
#
mM
C
+
1
2
13211.5210.52- 9.0011.52= 0d +©M
C
= 0;
V
C
= 6.75 kN9.00 -
1
2
13211.52- V
C
= 0+
c
©F
y
= 0;
N
C
= 0
:
+
©F
x
= 0;
A
x
= 0
:
+
©F
x
= 0;
A
y
= 9.00 kN
1
2
162162132- A
y
162= 0d+©M
B
= 0;
F4–4.
Segment AC
Ans
Ans
AnsM
C
= 112.5 lbM
C
+ 30011.5210.752- 30011.52= 0d©M
C
= 0;
V
C
=-150 lb300 - 30011.52- V
C
= 0+
c
©F
y
= 0;
N
C
= 0
:
+
©F
x
= 0;
A
x
= 0
:
+
©F
x
= 0;
A
y
= 300 lb30013211.52-
1
2
(3002132112- A
y
132= 0d +©M
B
= 0;
F4–5. Reactions
Segment AC
Ans
Ans
AnsM
C
=-5.625 kN
#
mM
C
+ 511.5210.752= 0d+©M
C
= 0;
V
C
=-7.50 kN-511.52- V
C
= 0+
c
©F
y
= 0;
N
C
= 30.0 kNN
C
- 30.0 = 0
:
+
©F
x
= 0;
A
y
= 042.43 sin 45° - 5162- A
y
= 0+
c
©F
y
= 0;
A
x
= 30.0 kN42.43 cos 45° - A
x
= 0
:
+
©F
x
= 0;
F
B
= 42.43 kNF
B
sin 45°132- 5162132= 0d +©M
A
= 0;