sm10 16 19

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________________________________________________________________________
PROBLEMS (10.16 through *10.19) A simple beam is loaded as shown in Figs. P10.16
through P10.19. Using the double-integration method, determine:
(a) The equation of the elastic curve.
(b) The slope at the end A.
(c) The deflection at midspan.
SOLUTION (10.16)
(a)
3
0
0
1
0: () ()0
66
ow
M M wLLx Lx
L
=−+ −− −=
∑
or
32
000
623
wx wx wLx
ML =−+
Thus,
32
000
'' 623
wx wx wLx
EIv L =−+
43 2
000 1
'24 6 6
wx wx wLx
EIv C
L
=−+ + (1)
54 3
000 12
120 24 18
wx wx wLx
EIv C x C
L
=−+ ++ (2)
Boundary conditions:
2
(0) 0: 0vC==
444 3
000 0
11
() 0: ,
120 24 18 45
wL wL wL wL
vL CL C=−++ =−
Equations (1) and (2):
43224
0
' (15 60 60 8 )
360
w
vxLxLxL
EIL
=−+− (3)
54234
0(3 15 20 8 )
360
w
vxLxLxLx
EIL
=−+− (4)
Continued on next slide
x
y
A
wo
2
L
B
2
L
w0L/6
w0L/3
C
2
0()
2
Lx
wL
−
O
0Lx
wL
−
M
w0L/6
L
-
x
V

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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
Copyright Act without the permission of the copyright owner is unlawful.
(b) Make x=0 in Eq. (3):
33
00
45 45
AwL wL
EI EI
θ =− =
(c) Make x=L/2 in Eq. (4):
44
00
55
768 768
CwL wL
vEI EI =− = ↓
SOLUTION (10.17)
11(0 )
82
L
MwLx x=≤≤
2
211
()
822
L
MwLxwx=−−
22
511 ()
828 2
L
wLx wx wL x L=−− ≤≤
(a) Segment AC
11
'' 8 EIv wLx=
2
11
1
'16 EIv wLx C=+ (1)
3
112
1
48
EIv wLx C x C=++ (2)
Segment BC
22
2511
'' 828 EIv wLx wx wL=−−
232
23
511
'16 6 8 EIv wLx wx wL x C=−−+
3422
234
511
48 24 16
EIv wLx wx wL x C x C=−− ++ (3)
Continued on next slide
L
/2
R
A=wL/8
A
C
B
L
/2
R
B=3wL/8
y
x
w

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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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Boundary and continuity conditions:
12
(0) 0: 0vC==
33
12 1 3
'( ) '( ) :
2 2 64 192
LLwL wL
vv C C=+=−+, 3
31
48
wL
CC=+ (a)
44444
11
12 4
5
() ():
2 2 384 2 384 384 64 2 96
CL CL
L L wL wL wL wL wL
vv C=+=−−+++,
4
4384
wL
C=−
444 4
23
5
() 0: 0
48 24 16 384
wL wL wL wL
vL CL=−−++=, 3
3384
wL
C=
Equation (a) gives 3
17 384CwL=− .
Equations (1) through (3):
23
1'(247)
384
w
vLxL
EI
=− (4)
33
1(8 7 )
384
w
vLxLx
EI
=− (5)
342234
2(40 16 24 )
384
w
vLxxLxLxL
EI
=−−+−
(b) Make x=L/2 in Eq. (4):
33
77
384 384
AwL wL
EI EI
θ =− =
(c) Make x=0 in Eq. (5):
44
55
768 768
CwL wL
vEI 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 (10.18)
(a)
0
1(0 )
M
M
xxa
L
=− ≤ ≤
0
20
(0 )
M
M
xM xa
L
=− + ≤ ≤
Segment AC
0
1'' M
EIv x
L
=−
2
0
11
'2
M
EIv x C
L
=− + (1)
3
0
112
6
M
EIv x C x C
L
=− + + (2)
Segment BC
2
00
202 03
'' ' 2
MM
EIv x M EIv x M x C
LL
=− + =− + +
3
00
234
62
MM
EIv x xCxC
L
=− + + + (3)
Boundary conditions:
12
(0) 0: 0vC==
2
0
234
() 0: 0
3
ML
vL CLC=++= (a)
Continuity conditions:
12 103
'( ) '( ) :va va C MaC==+ (b)
2
12 1 034
1
() (): 2 va va Ca Ma Ca C==++ (c)
Solving Eqs.(a) through (c):
2
00
10
32
M
LMa
CMa
L
=− + −
2
00 0
34
32 2
M
LMa Ma
CC
L
=− − =−
Continued on next slide
b
a
R
A=M0/L
A
C
B
R
B= M0/L
y
x
M
0
L

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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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Then, Eqs.(1) through (3):
22 2
0
1'(3263)
6M
vxLaLa
EIL
=− + − + (4)
32 2
0
1(2 6 3)
6M
vxLxaLxax
EIL
=− + − + (5)
32222
0
2(3 2 3 3)
6M
vxLxLxaxLa
EIL
=− − + + − (6)
(b) Make x=0 in Eq. (4):
22
0(2 6 3 )
6
AMLLaa
EIL
θ =− − +
(c) Make x=L/2 in Eq. (5):
22
09
(63)
64
ML
vaLa
EI
=− − +
*SOLUTION (10.19)
(a)
11(0 )
2
M
Px x a=≤≤
211
()
22
M
Px P x a Px Pa=−−=−+ ( 2 )ax a
≤
≤
Segment AC
11
'' 2 EIv Px=
2
11
1
'4 EIv Px C=+ (1)
3
112
1
12
EIv Px C x C=++ (2)
Continued on next slide
a
a
A
C
B
P
/2
y
x
P
P
/2
k