Hibbeler R.C. Structural Analysis
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650 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
Ans
Ans¢
B
=
ƒ
t
B>A
ƒ
= ca
8 kN
#
m
EI
b14 m2dc
1
2
14 m2d=
64 kN
#
m
3
EI
c
u
B
=
ƒ
u
B>A
ƒ
= a
8 kN
#
m
EI
b14 m2 =
32 kN
#
m
2
EI
(3 m)
(3 m)
( )
2
—
3
1
—
2
18 kN⭈m
—————
EI
18 kN⭈m
—————
EI
M¿
A
V¿
A
Ans
Ans
F8–11.
¢
A
=
ƒ
t
A>B
ƒ
=
`
c
1
2
a
-18 kN
#
m
EI
b13 m2dc
2
3
13 m2d
`
=
54 kN
#
m
2
EI
T
u
A
=
ƒ
u
A>B
ƒ
=
`
1
2
a
-18 kN
#
m
EI
b13 m2
`
=
27 kN
#
m
2
EI
F8–10.
tan A
0
3
x (m)
M
—
EI
⫺18 kN⭈m
—————
EI
u
A/B
u
A
t
A/B
⌬
A
Ans
Ans
F8–12.
=
54 kN
#
m
3
EI
TM¿
A
=¢
A
=-
54 kN
#
m
3
EI
-M¿
A
- c
1
2
a
18 kN
#
m
EI
b13 m2dc
2
3
(3 m2d= 0d+©M
A
= 0;
u
A
=
27 kN
#
m
2
EI
V¿
A
-
1
2
a
18 kN
#
m
EI
b13 m2 = 0+
c
©F
y
= 0;
tan A
tan B
4
x (m)
M
—
EI
8 kN⭈ m
————
EI
u
B/A
u
B
⌬
B
⫽ t
B/A
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 651
Ans
Ans
F8–14.
M¿
B
=¢
B
=
64 kN
#
m
3
EI
c
M¿
B
- ca
8 kN
#
m
EI
b14 m2d12 m2= 0d +©M
B
= 0;
u
B
=
32 kN
#
m
2
EI
a
8 kN
#
m
EI
b14 m2 - V¿
B
= 0+
c
©F
y
= 0;
F8–13.
2 m
(4 m)
( )
8 kN⭈ m
————
EI
M¿
B
V¿
B
Ans
Ans¢
C
=¢¿-t
C>A
=
7.5 kN
#
m
3
EI
-
4.6875 kN
#
m
3
EI
=
2.81 kN
#
m
3
EI
T
u
A
=
ƒ
t
B>A
ƒ
L
AB
=
15 kN
#
m
3
>EI
3 m
=
5 kN
#
m
2
EI
¢¿ =
1
2
t
B>A
=
1
2
a
15 kN
#
m
3
EI
b =
7.5 kN
#
m
3
EI
=
4.6875 kN
#
m
3
EI
t
C>A
= c
1
2
a
2.5 kN
#
m
EI
b11.5 m2dc
2
3
11.5 m2d+ ca
2.5 kN
#
m
EI
b11.5 m2dc
1
2
11.5 m2d
t
B>A
= c
1
2
a
5 kN
#
m
EI
b13 m2dc
2
3
13 m2d=
15 kN
#
m
3
EI
tan B
tan C
31.5
x (m)
M
—
EI
5 kN⭈ m
————
EI
2.5 kN⭈ m
————
EI
u
A
t
C/A
⌬
C
t
B/A
tan A
⌬
¿
652 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
Ans
Ans
F8–16.
¢
C
= M¿
C
=-
2.8125 kN
#
m
3
EI
=
2.81 kN
#
m
3
EI
T
c
1
2
a
2.5 kN
#
m
EI
b11.5 m2d10.5 m2- a
2.5 kN
#
m
2
EI
b11.5 m2- M¿
C
= 0d+©M
C
= 0;
u
A
= V¿
A
=-
5 kN
#
m
2
EI
=
5 kN
#
m
2
EI
-V¿
A
-
5 kN
#
m
2
EI
= 0+
c
©F
y
= 0;
F8–15.
(3 m)
( )
1
—
2
5 kN⭈ m
2
————
EI
2.5 kN⭈m
2
—————
EI
5 kN⭈ m
————
EI
(1.5 m)
( )
1
—
2
2.5 kN⭈m
————
EI
2.5 kN⭈m
2
—————
EI
0.5 m
1.5 m
1 m 2 m
M¿
C
V¿
C
u
A
tan C
tan A
u
C/A
⌬
C
⌬¿
t
C/A
630
x (m)
M
—
EI
12 kN⭈m
————
EI
18 kN⭈m
2
————
EI
18 kN⭈m
2
————
EI
18 kN⭈m
2
————
EI
(3 m)
( )
1
—
2
12 kN⭈m
————
EI
(6 m)
( )
1
—
2
12 kN⭈m
————
EI
1 m
3 m
M¿
C
V¿
C
Ans
Ans
F8–17.
¢
C
=¢¿-t
C>A
=
54 kN
#
m
3
EI
-
18 kN
#
m
3
EI
=
36 kN
#
m
3
EI
T
¢¿ = u
A
L
AC
= a
18 kN
#
m
2
EI
b13 m2 =
54 kN
#
m
3
EI
t
C>A
= c
1
2
a
12 kN
#
m
EI
b13 m2dc
1
3
13 m2d=
18 kN
#
m
3
EI
u
A
= u
C>A
=
1
2
a
12 kN
#
m
EI
b13 m2 =
18 kN
#
m
2
EI
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 653
Ans
Ans
F8–18.
M¿
C
=¢
C
=-
36 kN
#
m
3
EI
=
36 kN
#
m
3
EI
T
M¿
C
+ a
18 kN
#
m
2
EI
b13 m2 - c
1
2
a
12 kN
#
m
EI
b13 m2d11 m2= 0d+©M
C
= 0;
-V¿
A
-
18 kN
#
m
2
EI
= 0
V¿
A
= u
A
=-
18 kN
#
m
2
EI
=
18 kN
#
m
2
EI
+
c
©F
y
= 0;
u
A
tan C
tan A
u
C/A
⌬
C
⌬¿
t
C/A
84260
x (m)
M
—
EI
8 kN⭈ m
————
EI
8 kN⭈ m
————
EI
24 kN⭈ m
2
————
EI
24 kN⭈ m
2
————
EI
24 kN⭈ m
2
————
EI
(4 m) ⫹
1
—
2
48 kN⭈ m
2
————
EI
(4 m) ⫽
( )
8 kN⭈ m
———
EI
( )
8 kN⭈ m
———
EI
(2 m)
( )
8 kN⭈ m
———
EI
(2 m)
( )
8 kN⭈ m
———
EI
1 m
2.667 m
4 m
4 m 4 m
M¿
C
V¿
C
1
—
2
Ans
Ans
F8–19.
¢
C
=¢¿-t
C>A
=
96 kN
#
m
2
EI
-
37.33 kN
#
m
3
EI
=
58.7 kN
#
m
3
EI
T
¢¿ = u
A
L
AC
= a
24 kN
#
m
2
EI
b14 m2 =
96 kN
#
m
3
EI
t
C>A
= c
1
2
a
8 kN
#
m
EI
b12 m2dc2 m +
1
3
(2 m2d+ ca
8 kN
#
m
EI
b12 m2d11 m2=
37.33 kN
#
m
3
EI
u
A
= u
C>A
=
1
2
a
8 kN
#
m
EI
b12 m2 + a
8 kN
#
m
EI
b12 m2 =
24 kN
#
m
2
EI
Ans
Ans¢
C
= M¿
C
=
58.7 kN
#
m
3
EI
T
M¿
C
+ a
24 kN
#
m
2
EI
b14 m2- c
1
2
a
8 kN
#
m
EI
b12 m2d12.667 m2- a
8 kN
#
m
EI
b12 m211 m2= 0d+©M
C
= 0;
-V¿
A
-
24 kN
#
m
2
EI
= 0 u
A
= V¿
A
=
24 kN
#
m
2
EI
+
c
©F
y
= 0;
654 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
F8–20.
Ans
F8–21.
¢
B
=
ƒ
t
B>A
ƒ
=
`
c
1
2
a-
18 kN
#
m
EI
b12 m2dc2 m +
2
3
12 m2d
`
=
60 kN
#
m
EI
T
u
B
=
ƒ
u
B>A
ƒ
=
`
1
2
a-
18 kN
#
m
EI
b12 m2
`
=
18 kN
#
m
EI
Ans
AnsM¿
B
=¢
B
=-
60 kN
#
m
3
EI
=
60 kN
#
m
3
EI
T
M¿
B
+ c
1
2
a
18 kN
#
m
EI
b12 m2dc
2
3
12 m2+ 2 m d= 0d+©M
B
= 0;
u
B
=-
18 kN
#
m
2
EI
=
18 kN
#
m
2
EI
-V¿
B
-
1
2
a
18 kN
#
m
EI
b12 m2= 0+
c
©F
y
= 0;
tan A
tan B
42
x (m)
M
—
EI
18 kN⭈m
————
EI
u
B/A
u
B
t
B/A
(2 m)
(2 m)
( )
2
—
3
2 m
1
—
2
18 kN⭈m
————
EI
M¿
B
V¿
B
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 655
F9–1.
Member
n
(lb)
N
(lb)
L
(ft)
AB –1.667 –250 10 4166.67
AC 1 150 6 900.00
BC 1.333 200 8 2133.33
©
7200
nNL 1lb
2
#
ft2
Member
NL
(ft)
AB –1.667P –1.667 –250 10 4166.67
AC P 1 150 6 900.00
BC 1.333P 1.333 200 8 2133.33
7200©
N a
dN
dP
bL 1lb
#
ft2N 1P = 150 lb2
dN
dP
Member
n
(kN)
N
(kN)
L
(m)
AB 1 – 4.041 2 –8.0829
AC 0 8.0829 2 0
BC 0 –8.0829 2 0
CD 0 8.0829 1 0
– 8.0829©
nNL 1k
N
2
#
m2
Member
N
(kN)
L
(m)
AB P – 4.041 1 –4.041 2 –8.083
AC 8.083 0 8.083 2 0
BC –8.083 0 –8.083 2 0
CD 8.083 0 8.083 1 0
– 8.083©
N a
dN
dP
bL 1k N
#
m2N 1P = 02 1k N2
dN
dP
Thus,
Ans
F9–2.
¢
B
v
=
7200 lb
#
ft
AE
T
1 lb
#
¢
B
v
=
a
nNL
AE
=
7200 lb
2
#
ft
AE
Ans
F9–3.
¢
B
v
=
a
Na
dN
dP
b
L
AE
=
7200 lb
#
ft
AE
T
Thus,
Ans
F9–4.
¢
A
h
=-
8.08 kN
#
m
AE
=
8.08 kN
#
m
AE
:
1 kN
#
¢
A
h
=
a
nNL
AE
=-
8.0829 kN
2
#
m
AE
Ans¢
A
h
=
a
Na
dN
dP
b
L
AE
=-
8.083 kN
#
m
AE
=
8.08 kN
#
m
AE
:
656 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
Ans ¢
D
v
=
237.5 kN
#
m
AE
T
1 kN
#
¢
D
v
=
a
nNL
AE
=
237.5 kN
2
#
m
AE
Member
n
(kN)
N
(kN)
L
AB 003 0
AC 1.414 8.485 50.91
BC –1 – 6 3 18.00
AD 0–6 3 0
CD –1 0 3 0
68.91©
322
nNL 1kN
2
#
m2
F9–5.
Ans
F9–6.
¢
D
h
=
68.9 kN
#
m
AE
:
1 kN
#
¢
D
h
=
a
nNL
AE
=
68.91 kN
2
#
m
AE
Ans
F9–7.
: =
68.9 kN
#
m
AE
¢
D
h
=©Na
dN
dP
b
L
AE
Member
N
(kN)
L
(m)
AB 0003 0
AC 50.91
BC –1 – 6 3 18.00
AD –60–63 0
CD – P –1 0 3 0
68.91©
-1P + 62
322
62222221P + 62
N a
dN
dP
bL 1kN
#
m2N 1P = 02 (kN)
dN
dP
Member
n
(kN)
N
(kN)
L
(m)
AB 0.375 18.75 3 21.09
BC 0.375 18.75 3 21.09
AD – 0.625 – 31.25 5 97.66
CD – 0.625 – 31.25 5 97.66
BD 0504 0
237.5©
nNL 1kN
2
#
m2
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 657
Ans¢
B
v
=
a
Na
dN
dp
b
L
AE
=
44.75 kN
#
m
AE
T
Member
N
(kN)
L
(m)
AB 0.375 18.75 3 21.09
BC 0.375 18.75 3 21.09
AD – 0.625 – 31.25 5 97.66
CD – 0.625 – 31.25 5 97.66
BD 50 0 50 4 0
237.5©
-1
5
8
P + 31.252
-1
5
8
P + 31.252
3
8
P + 18.75
3
8
P + 18.75
N a
dN
dP
bL 1kN
#
m2N 1P = 02 (kN)
dN
dP
F9–8.
Ans
F9–9.
¢
D
v
=
a
Na
dN
dP
b
L
AE
=
237.5 kN
#
m
AE
T
Ans
F9–10.
¢
B
v
=
44.75 kN
#
m
AE
T
1 kN
#
¢
B
v
=
a
nNL
AE
=
44.75 kN
2
#
m
AE
Member
n
(kN)
N
(kN)
L
(m)
AB 0 – 6 1.5 0
BC 0 – 6 1.5 0
BD 1020
CD 0 10 2.5 0
AD – 1.25 – 10 2.5 31.25
DE 0.75 12 1.5 13.5
44.75©
nNL 1k N
2
#
m2
Member
N
(kN)
L
(m)
AB – 6 0 – 6 1.5 0
BC – 6 0 – 6 1.5 0
BD P 1020
CD 10 0 10 2.5 0
AD – – 1.25 – 10 2.5 31.25
DE 0.75 12 1.5 13.5
44.75©
0.75P + 12
11.25P + 102
N a
dN
dP
bL 1k N
#
m2N 1P = 02 1k N2
dN
dP
658 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND A NSWERS
Member
n
(kN)
N
(kN)
L
(m)
AB 0.5 50 2 50.00
DE 0.5 50 2 50.00
BC 0.5 50 2 50.00
CD 0.5 50 2 50.00
AH – 0.7071 – 70.71 141.42
EF – 0.7071 – 70.71 141.42
BH 0302 0
DF 0302 0
CH 0.7071 28.28 56.57
CF 0.7071 28.28 56.57
CG 002 0
GH – 1 – 70 2 140.00
FG – 1 – 70 2 140.00
878.98©
222
222
222
222
nNL 1k N
2
#
m2
Member
N
(kN)
L
(m)
AB 0.5 50 2 50.00
DE 0.5 50 2 50.00
BC 0.5 50 2 50.00
CD 0.5 50 2 50.00
AH – 0.7071 – 70.71 141.42
EF – 0.7071 – 70.71 141.42
BH 30 0 30 2 0
DF 30 0 30 2 0
CH 0.7071P 0.7071 28.28 56.57
CF 0.7071P 0.7071 28.28 56.57
CG 00 02 0
GH – 1 – 70 2 140.00
FG – 1 –70 2 140.00
875.98©
-1P + 302
-1P + 302
222
222
222-10.7071P + 42.432
222
-10.7071P + 42.432
0.5P + 30
0.5P + 30
0.5P + 30
0.5P + 30
N a
dN
dP
bL (kN
#
m)N 1P = 40 k N)
dN
dP
F9–11.
Ans
F9–12.
¢
C
v
=
876 kN
#
m
AE
T
1 kN
#
¢
C
v
=
a
nNL
AE
=
875.98 kN
2
#
m
AE
Ans¢
C
v
=
876 kN
#
m
AE
T
¢
C
v
=
a
Na
dN
dP
b
L
AE
=
875.98 kN
#
m
AE
FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 659
F9–13. For the slope,
Ans
For the displacement,
Ans
F9–14. For the slope, Then, Set Then, .
Ans
For the displacement, Then
Set Then
F9–15. For the slope, and .
Ans
For the displacement, and
Ans
F9–16. For the slope, Then
Set Then
Ans
For the displacement, Then
Set Then
Ans¢
A
v
=
L
L
0
Ma
0M
0P
b
dx
EI
=
L
3
m
0
41x2dx
EI
=
18 kN
#
m
3
EI
c
M = 4 kN
#
m.P = 0.
0M
0P
= x.M = 1Px + 42 kN
#
m.
u
A
=
L
L
0
Ma
0M
0M¿
b
dx
EI
=
L
3
m
0
4112dx
EI
=
12 kN
#
m
2
EI
M = 4 kN
#
m.M¿=4 kN
#
m.
0M
0M
= 1.M = M¿.
¢
A
v
=
18 kN
#
m
3
EI
c
1 kN
#
¢
A
v
=
L
L
0
mM
EI
dx =
L
3
m
0
x142dx
EI
=
18 kN
2
#
m
3
EI
M = 4 kN
#
m.m = x kN
#
m
u
A
=
12 kN
#
m
2
EI
1 kN
#
m
#
u
A
=
L
L
0
m
u
M
EI
dx =
L
3
m
0
112142dx
EI
=
12 kN
2
#
m
3
EI
M = 4 kN
#
mm
u
= 1 kN
#
m
¢
A
v
=
L
L
0
Ma
0M
0P
b
dx
EI
=
L
3
m
0
(-30x)(-x)dx
EI
=
270 kN
#
m
3
EI
T
M = 1-30x2 kN
#
m.P = 30 kN .
0M
0P
=-x.M =-Px.
u
A
=
L
L
0
Ma
0M
0M¿
b
dx
EI
=
L
3
0
m
1-30x21-12dx
EI
=
135 kN
#
m
2
EI
M = 1-30x2 kN
#
mM¿=0.
0M
0M¿
=-1.M =-30x - M¿.
¢
A
v
=
270 kN
#
m
3
EI
T
1 kN
#
¢
A
v
=
L
L
0
mM
EI
dx =
L
3
m
0
1-x21-30x2
EI
dx =
270 kN
2
#
m
3
EI
u
A
=
135 kN
#
m
2
EI
1 kN
#
m
#
u
A
=
L
2
0
m
u
M
EI
dx =
L
3
m
0
1-121-30x2
EI
dx =
135 kN
2
#
m
3
EI