Hibbeler R.C. Structural Analysis
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640 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F4–6. Reactions
Segment CB
Ans
Ans
AnsM
C
= 5370 lb
#
ft = 5.37 k
#
ft1345162- 150162132- M
C
= 0d+©M
C
= 0;
V
C
=-445 lbV
C
+ 1345 - 150162= 0+
c
©F
y
= 0;
N
C
= 0
:
+
©F
x
= 0;
B
y
= 1345 lbB
y
1152- 150192110.52- 600162- 800132= 0d +©M
A
= 0;
F4–7. Left segment
Ans
AnsM = 5-x
3
- 6x6 kN
#
mM +
1
2
a
18
3
xb1x2a
x
3
b + 6x = 0d+©M
O
= 0;
V = 5-3x
2
- 66 kN-6 -
1
2
a
18
3
xb1x2- V = 0+
c
©F
y
= 0;
F4–8. Reaction
Left segment
Ans
AnsM = e12.0x -
1
3
x
3
f kN
#
mM +
1
2
a
12
6
xb1x2a
x
3
b - 12.0x = 0d+©M
O
= 0;
V = 512.0 - x
2
6 kN12.0 -
1
2
a
12
6
xb1x2- V = 0+
c
©F
y
= 0;
A
y
= 12.0 kN
1
2
1122162122- A
y
162= 0d+©M
B
= 0;
F4–9. Reactions
left segment
Ans
Ans
right segment
Ans
AnsM = 5-4x
2
+ 40x - 646 kN
#
m24.018 - x2- 818 - x2a
8 - x
2
b - M = 0d+©M
O
= 0;
V = 540 - 8x6 kNV + 24.0 - 818 - x2= 0+
c
©F
y
= 0;
4 m 6 x 6 8 m
M = 58x6 kN
#
mM - 8.00x = 0d +©M
O
= 0;
V = 586 kN8.00 - V = 0+
c
©F
y
= 0;
0 … x 6 4 m
A
y
= 8.00 kN8142122 - A
y
182= 0d+©M
B
= 0;
B
y
= 24.0 kNB
y
182- 8142162= 0d +©M
A
= 0;
F4–10.
Ans
Ans
Ans
AnsM = e-
5
2
x
2
+ 10x - 45 f kN
#
mM + 51x - 22a
x - 2
2
b + 15 + 20 = 0d +©M
O
= 0;
V = 510 - 5x6 kN-51x - 22- V = 0+
c
©F
y
= 0;
2 m 6 x … 4 m
M =-20 kN
#
mM + 20 = 0d +©M
O
= 0;
V = 0+
c
©F
y
= 0;
0 … x 6 2 m
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 641
V (kN)
0
11
24
3
x (m)
F4–11. Reactions
left segment
Ans
Ans
right segment
Ans
Ans
F4–12. Support reactions
left segment
Ans
Ans
right segment
Ans
Ans
F4–13.
M = 5-15x + 3606 k
#
ft15.0124 - x2- M = 0d +©M
O
= 0;
V = 5-15 k6V + 15.0 = 0+
c
©F
y
= 0;
12 ft 6 x … 24 ft
M = 5- x
2
+ 27x6 k
#
ftM + 2xa
x
2
b- 27.0x = 0d +©M
O
= 0;
V = 527 - 2x6 k27.0 - 2x - V = 0+
c
©F
y
= 0;
0 … x 6 12 ft
A
y
= 27.0 k181122+ 211221182 - A
y
1242= 0d +©M
B
= 0;
B
y
= 15.0 kB
y
1242- 21122162- 181122= 0d +©M
A
= 0;
M = 515x - 606 kN
#
m-M - 1514 - x2= 0d+©M
O
= 0;
V = 15 kNV - 15 = 0+
c
©F
y
= 0;
2 m 6 x … 4 m
M = e-
5
2
x
2
+ 25x - 70 f kN
#
mM + 5xa
x
2
b + 70.0 - 25.0x = 0d+©M
O
= 0;
V = 525 - 5x6 kN25.0 - 5x - V = 0+
c
©F
y
= 0;
0 … x 6 2 m
M
A
= 70.0 kN
#
mM
A
- 5122112- 15142= 0d+©M
A
= 0;
A
y
= 25.0 kNA
y
- 5122- 15 = 0+
c
©F
y
= 0;
M (kN⭈m)
0
⫺28
24
⫺6
x (m)
V (kN)
0
4
8.5
0.5
68
x (m)
⫺6
F4–14.
M (kN⭈m)
0
468
⫺24
⫺7
⫺6
x (m)
642 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
V (k)
0
10.5 12 24
⫺9
⫺27
x (ft)
63
M (k⭈ ft)
0
10.5 12
330.75
324
24
x (ft)
10
M (k⭈ ft)
⫺70
30
x (ft)
30
V (k)
10
⫺20
x (ft)
⫺4.5
V (kN)
0
4.5
4.5 9
x (m)
M (kN⭈m)
0
6.75
4.5 9
x (m)
F4–19.
⫺4
⫺3
V (kN)
0
4
3
2.5
3.51.5 5
x (m)
3.51.5
M (kN⭈m)
⫺1.5 ⫺1.5
0
2.5
0.5
5
x (m)
M (kN⭈m)
0
246
x (m)
⫺12
⫺12
V (kN)
0
2
⫺12
4
12
6
x (m)
F4–18.
F4–17.
F4–16.
F4–15.
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 643
F4–20.
⫺6
V (k)
0
6
6
12
x (ft)
0
6
24
12
x (ft)
M (k⭈ft)
F6–1.
F6–2.
F6–3.
F6–4.
A
B
x
A
y
A
B
C
x
V
C
A
B
C
x
M
C
A
C
B
x
A
y
A
C
D
B
x
V
D
A
x
M
B
CB
A
C
B
x
A
y
A
C
DB DB
x
V
D
A
x
M
D
C
A
C
x
A
y
A
C
B
B
x
V
B
A
x
M
B
C
644 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F6–5.
D
E
B
A
x
A
y
D
E
C
BA
x
D
E
CBA
x
V
C
M
C
F6–6.
F6–7.
Ans
Ans= 6.50 kN
+ c
1
2
12210.52d122+ c
1
2
14 - 221-0.52d122+ c
1
2
16 - 4210.52d122
1V
C
2
max 1+2
= 810.52+ c
1
2
12210.52d11.52+ c
1
2
16 - 4210.52d11.52
= 13.0 kN
#
m
1M
C
2
max 1+2
= 8112+ c
1
2
16 - 22112d11.52+ c
1
2
1221-12d122 + c
1
2
16 - 22112d122
A
y
A
D
ECB
x
A
D
EC
B
x
A
D
EC
B
x
V
E
⫺
M
A
6
⫺1
1
24
x (m)
V
C
0
6
⫺0.5
0.5
0.5
24
x (m)
0
M
C
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 645
F6–8.
a) Ans
b)
Ans =-54 kN
#
m
1M
A
2
max 1-2
= 61-32+ c
1
2
16 - 021-32d122+ c
1
2
16 - 021-32d142+ c
1
2
19 - 62132d142
1C
y
2
max 1+2
= 6122+ c
1
2
19 - 32122d122+ c
1
2
19 - 32122d142= 48 kN
9
⫺3
3
36
x (m)
C
y
0
9
2
36
x (m)
0
M
A
F8–1.
.
646 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F8–2. F8–3.
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 647
F8–4. For
(1)
(2)
For
(3)
(4)
at From Eq (2),
at From Eq (1),
at From Eq (3),
at From Eq (4),
Ans
Ans
F8–5.
(1)
(2)
at . From Eq (1),
at . From Eq (2),
Ansv =
Px
6EI
(x
2
- 3Lx)
C
2
= 0x = 0v = 0
C
1
= 0x = 0
dv
dx
= 0
EI v =
P
6
x
3
-
PL
2
x
2
+ C
1
x + C
2
EI
dv
dx
=
P
2
x
2
- PLx + C
1
EI
d
2
v
dx
2
= Px - PL
M = Px - PL
v
2
=
P
48EI
1-4x
3
2
+ 12Lx
2
2
- 9L
2
x
2
+ L
3
2
v
1
=
Px
1
48EI
14x
2
1
- 3L
2
2
C
4
=
PL
3
48
x
2
= L.v
2
= 0
C
3
=-
3PL
2
16
x
2
=
L
2
.
dv
2
dx
2
= 0
C
1
=-
PL
2
16
x
1
=
L
2
.
dv
1
dx
1
= 0
C
2
= 0x
1
= 0.v
1
= 0
EI v
2
=
PL
4
x
2
2
-
P
12
x
3
2
+ C
3
x
2
+ C
4
EI
dv
2
dx
2
=
PL
2
x
2
-
P
4
x
2
2
+ C
3
EI
d
2
v
2
dx
2
2
=
PL
2
-
P
2
x
2
M
2
=
P
2
1L - x
2
2=
PL
2
-
P
2
x
2
L
2
6 x
2
… L
EI v
1
=
P
12
x
3
1
+ C
1
x
1
+ C
2
EI
dv
1
dx
1
=
P
4
x
2
1
+ C
1
EI
d
2
v
1
dx
2
1
=
P
2
x
1
M
1
=
P
2
x
1
0 … x
1
6
L
2
648 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F8–6.
(1)
at From Eq (1),
at From Eq (1),
Ans
F8–7. For
(1)
(2)
For
(3)
(4)
at . From Eq (2),
at . From Eq (4), (5)
at . From Eqs (1) and (3), (6)
at . From Eqs (2) and (4), (7)C
1
L - C
3
L - 2C
4
=
M
0
L
2
4
x
1
= x
2
=
L
2
v
1
= v
2
C
1
- C
3
=
M
0
L
2
x
1
= x
2
=
L
2
dv
1
dx
1
=
dv
2
dx
2
0 = C
3
L + C
4
+
M
0
L
2
3
x
2
= Lv
2
= 0
C
2
= 0x
1
= 0v
1
= 0
EIv
2
=
M
0
2
x
2
2
-
M
0
6L
x
3
2
+ C
3
x
2
+ C
4
EI
dv
2
dx
2
= M
0
x
2
-
M
0
2L
x
2
2
+ C
3
EI
d
2
v
2
dy
2
2
= M
0
-
M
0
L
x
2
M = M
0
-
M
0
L
x
2
L
2
6 x
2
… L
EIv
1
=-
M
0
6L
x
3
1
+ C
1
x
1
+ C
2
EI
dv
1
dx
1
=-
M
0
2L
x
2
1
+ C
1
EI
d
2
v
1
dx
2
1
=-
M
0
L
x
1
M =-
M
0
L
x
1
0 … x
1
6
L
2
v =
M
0
6EIL
1-x
3
+ 3Lx
2
- 2L
2
x2
C
1
=-
M
0
L
3
x = L.v = 0
C
2
= 0x = 0.v = 0
EI v =
M
0
2
x
2
-
M
0
6L
x
3
+ C
1
x + C
2
EI
dv
dx
= M
0
x -
M
0
2L
x
2
+ C
1
EI
d
2
v
dx
2
= M
0
-
M
0
L
x
M = M
0
-
M
0
L
x
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 649
Solving Eqs (5), (6) and (7)
Ans
Ans
F8–8.
. From Eq (1),
. From Eq (2),
Ans
F8–9.
. From Eq (1),
. From Eq (2),
Ansv =
w
0
120EIL
1-x
5
+ 5L
4
x - 4L
5
2
C
2
=-
w
0
L
4
30
x = Latv = 0
C
1
=
w
0
L
3
24
x = Lat
dv
dx
= 0
EI v =-
w
0
120L
x
5
+ C
1
x + C
2
122
EI
dv
dx
=-
w
0
24L
x
4
+ C
1
112
EI
d
2
v
dx
2
=-
w
0
6L
x
3
M =-
w
0
6L
x
3
v =
w
24EI
1-x
4
+ 4Lx
3
- 6L
2
x
2
2
C
2
= 0x = 0atv = 0
C
1
= 0x = 0at
dv
dx
= 0
EI v =-
w
24
x
4
+
wL
6
x
3
-
wL
2
4
x
2
+ C
1
x + C
2
122
EI
dv
dx
=-
w
6
x
3
+
wL
2
x
2
-
wL
2
2
x + C
1
(1)
EI
d
2
v
dx
2
=-
w
2
x
2
+ wLx -
wL
2
2
M =-
w
2
x
2
+ wLx -
wL
2
2
v
2
=
M
0
24EIL
1-4x
3
2
+ 12Lx
2
2
- 11L
2
x
2
+ 3L
3
2
v
1
=
M
0
24EIL
1-4x
3
1
+ L
2
x
1
2
C
1
=
M
0
L
24
C
3
=-
11M
0
L
24
C
4
=
M
0
L
2
8