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________________________________________________________________________
PROBLEMS (10.125 through *10.128) An overhanging beam is loaded as shown in Fig.
P10.125 through P10.128. Determine:
(a) The slope at B.
(b) The deflection at point C.
SOLUTION (10.125)
00 0
12
1(2 )
2
M
Ma Ma
Aa A
EI EI EI
== =
2
00 0
/513
() ()
326
CA
M
aMa Ma
aa
tEI EI EI
=+=
2
00 /0
/12
(2) ,
33 23
BA
BA A
M
aMa tMa
ta
EI EI a EI
θ
= = =− =−
(a) 0
1
BA
M
a
AEI
θθ
−==
00 0
4
33
B
M
aMa Ma
EI EI EI
θ
=+=
(b) We have 2
/0
3
'"2BA
CC t Ma EI== . Thus,
2
0
/'" (136)
6
CCA Ma
vt CC EI
=− = − 2
0
7
6Ma
EI
=
↑
Continued on next slide
x
x
B
C’
’
A
C
M/EI
A
2
M0
M0/2a
M0/EI
a
2a
A
θ
y
C
v
t
B
/A
C’
A
tC/A
M0/2a
a/2
C
A
1
5a/3
C’C’’
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SOLUTION (10.126)
Beam is symmetric about midspan D, 0
D
θ
=
.
23
1()
44
wa wa
Aa
EI EI
=− =−
23
21()
32 6
wa wa
Aa
EI EI
=− =−
/1BBD
A
θ
θ
=
=
(a) 3
4
Bwa
EI
θ
=
(b) 4
/1
()
28
BD awa
tA EI
==− 4
/1 2
3
()()
242
CD aawa
tAa A EI
=++ =−
44
// 28
CCDBD wa wa
vt t EI EI
=−=− + 4
3
8wa
EI
=
↓
Continued on next slide
w
w
x
B
tC/D
A
D
B
M/EI
-wa2/2EI
2E
I
EI
CC
v
wa
-wa2/4EI
tB/D
D
a/4
wa
3a/4
Spandrel
parabola
A
EI
a
/
2
a a a a
A
1
A
2
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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SOLUTION (10.127)
22
12
11
()
224 2 2
PL L PL PL PL
AAL
EI EI EI EI
= = =− =−
23
3/3
1()
22 2 8 6 48
BA
PL L PL L PL
AtA
EI EI EI
=− =− = =−
2
48
BA
AtPL
LEI
θ
=− =
AD is a straight line.
(a) /3BA B A A
θ
θθ
=−=−
22
348 8
BA
PL PL
AEI EI
θθ
=− + = − 2
7
48 PL
EI
=
(b) /1 2
() ()
63
CA LL
tA A
=+ 33
(1 4)
24 8
PL PL
EI EI
=−=−
//
3
2
CCA BA
vt t=− 33
832
PL PL
EI EI
=− + 3
3
32 PL
EI
=
↓
Continued on next slide
x
x
B
D
A
C
M/EI
B
B
A
θ
y
A
tA/B
C
A
2
- PL/EI
D
C
v
P
P
tC/A
RB=2P
RA=0
A
P
L/E
I
B’
M/EI
D
L/
6
A
3
A
D
-PL
/
2E
I
L/
2
L/
2
L/
2
L/
6
L/
3
A
1
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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SOLUTION (*10.128)
23 2 3
12
117 51 3
()3 (3)
22 4
wa wa wa wa
AaAa
EI EI EI EI
= = =− =−
23
313 9
(3 )
22
wa wa
Aa
EI EI
=− =− 23
419 9
(3 )
32 2
wa wa
Aa
EI EI
=− =−
23
51()
22
wa wa
Aa
EI EI
=− =−
/1 2 3 4
39
(2 ) ( ) (2 ) ( )
24
BA aa
tAaA AaA
=+ ++
444 4 4
51 9 9 81 15
22 8 8
wa wa wa wa wa
EI EI EI EI EI
= −−− =
3
/
15
38
ABA
wa
t
aEI
θ
=− =
(a) 3
/1234
3
4
BA B A wa
AAAA EI
θθθ
=−=+++=
33 3
351
488
Bwa wa wa
EI EI EI
θ
=−=
Continued on next slide
B
x
B
A
C
M/EI
B
B
A
θ
A
tB/A
C
7wa/6
C
v
wa
A
-4wa2/E
I
C’
C
M/EI
a
17wa2
/
2E
I
-wa2/E
I
x
17wa/6
-9wa2/2
w
3a
Spandrel
parabola
tan
g
ent at
A
C’
C’
’
A
5
A
1
A
2
A
3
A
4
3a/2
2a
9a
/
4
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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(b) /1 2 3 4 5
5132
(3 ) ( ) (3 ) ( ) ( )
243
CA aaa
tAaA AaA A
=+ ++ +
4153 15 27 117 1
()
42283
wa
EI
=−−−− 4
55
24 wa
EI
=
Then,
///
4
'" 3
CCABACA
vCCt tt=− + =− +
44
415 55
()
38 24
wa wa
EI EI
=− + 4
5
24 wa
EI
=
↓
1
/
5
100%