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________________________________________________________________________
PROBLEMS (10.116 through 10.119) A simple beam is loaded as shown in Figs. P10.116
through P10.119. Determine:
(a) The slope at point C.
(b) The deflection at point C.
SOLUTION (10.116)
2
11
12 2 1 7
239 27 3 33 9
LPL PL L L
AxL
EI EI
== =+=
2
22
12 2 2 22 4
239 27 33 9
LPL PL L
AxL
EI EI
== ==
Thus,
33
11 2 2 5
(7 8)
243 81
BA PL PL
tAxAx EI EI
=+ = +=
2
5
81
BA
A
tPL
LEI
θ
=− =−
(a) 22
/1 / 5
,27 81
CA C CA A PL PL
AEI EI
θθθθ
==+=− 2
2
81 PL
EI
=
Continued on next slide
y
x
A
θ
P
tangent at A
M
EI
x
C
2L
/
3
C
v
B
2
9
P
L
EI
L/
3
2
x
1
x
RB=P/3
RA=2P/3
t
B
/A
C’
A
C’
’
tC/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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(b) 3
/1
1
'()
3 3 243
CA LPL
tCCA EI
== =
and
///
1
''' 3
CCA CA BA
vt CCt t=− =−
33
5
243 243
PL PL
EI EI
=− 3
4
243 PL
EI
=↓
SOLUTION (10.117)
00
11
125
23 2 12 2 32 6
MML
LLLL
Ax
EI EI
== =+=
00
22
12
26 2 24 2 6 3
MML
LLLL
Ax
EI EI
== =+=
00
33
12
23 2 12 32 3
MML
LLL
Ax
EI EI
== ==
Continued on next slide
C
M0
x
A
θ
M
tan
g
ent at
A
y
C C
v
B
2
x
M0
t
B
/A
C’
A
C’
’
tC/A
A
1
A
2
A
M0/L
M0/2
3E
I
L/
2
L/
2
x
M
0/L
EI
A
3
y
M/EI
x
B
M
0/6EI
3
x
M
0/3EI 1
x
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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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We have
/112233BA
tAxAxAx=+ +
22
00
1
(5 2 2)
72 8
M
LML
EI EI
=++=
0
1
8
BA
A
M
L
tLEI
θ
=− =−
(a) 00
/12 1
12 24 8 o
CA
M
LML ML
AA EI EI EI
θ
=+= + =
00
/110
88
CCAC ML ML
EI EI
θθ θ
=+= − =
(b) /1 2
21
() ()
32 32
CA LL
tA A
=+
22
00
5
(4 1)
144 144
M
LML
EI EI
=+=
22
00
//
15 1
2 144 16
CCA BA
M
LML
vt t EI EI
=− = − 2
0
1
36 ML
EI
=
↓
SOLUTION (10.118)
Continued on next slide
Parabolic
spandrel
x
B
θ
tan
g
ent at B
x
C
a
C
v
B
2
4
wa
EI
2
x
1
x
t
A
/B C’
A
C’
’
tC/B
A
1
a wa/4
A
C
B
3wa/4
w
B
y
M/EI
-wa2/2EI
wa2/2EI
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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23
11
11 2
(2 )
22 2 3
wa wa a
Aa x
EI EI
== =
2
22
1
332 4
bh wa a
Aax
EI
==− =
/1122AB
tAxAx=+
4
40
7
(8 1)
24 24
M
L
wL
EI EI
=−=
3
/
17
248
BBA
wa
t
aEI
θ
=− =
(a) 23
/11
()
24 8
CB wa wa
a
EI EI
θ
==
/CCBB
θ
θθ
=− +
33
17
848
wa wa
EI EI
=− + 3
1
48 wa
EI
=
(b) 34
/1()
8324
CB wa a wa
tEI EI
==
44
//
17
22448
CCB BAwa wL
vt t EI EI
=− = − 4
5
48 wL
EI
=
↓
Continued on next slide
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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 (10.119)
00
11
39
32 1 3
43 2 24 4 32
M
ML
LL L
xA
EI EI
= = =− =−
00
22
31 5 1
434 6 24432
M
ML
LLL L
xA
EI EI
=+ = = =
/1122BA
tAxAx=+
22
00
11
(5 27)
192 96
M
LML
EI EI
=−=−
0
11
96
AB
B
M
L
tLEI
θ
==−
(a) 0
/2
32
CB
M
L
AEI
θ
==
00
211
96 32
CB
M
LML
AEI EI
θθ
=−=− − 0
7
48
M
L
EI
=
(b) 2
0
/2
1
()
3 4 384
CA
M
L
L
tA EI
==
22
00
//
111
4 384 384
CCB AB
M
LML
vt t EI EI
=− = + 2
0
1
32 ML
EI
=
↑
x
B
θ
tan
g
ent at B
x
C
C
v
B
2
x
tC/B
C’
A
C’
’
A
1
A
C
B
M/EI
A
2
M0
M0/L M0/L
L/
4
3L/4
1
x
M
0/4EI
-3M0/4EI
tA/B
1
/
5
100%