sm10 73 76

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________________________________________________________________________
PROBLEMS (10.73 through 10.76) A propped cantilever beam is loaded as shown in Figs.
P10.73 through P10.76. Using a direct-integration method, obtain:
(a) All of the reactions.
(b) The midspan deflection for the given data:
w = w0 /2 =20 kN/m, P = 30 kN, M0 = 8 kN
⋅
m, L = 5 m, EI = 3 MN⋅m2
as required.
SOLUTION (10.73)
(a)
Segment AC
2
111
1
'' ' 2
AA
EIv R x EIv R x C==+
3
112
1
6A
EIv R x C x C=++ (1)
Segment CB
2'' ( )
2
AL
EIv R x P x=−−
22
23
1
'()
222
APL
EIv R x x C=−−+
33
234
1()
662
APL
EIv R x x C x C=−−++
Boundary and continuity conditions:
121213
(0) 0: 0 '( ) '( ):
22
LL
vCvvCC== = =
22
23
11
'( ) 0: 0
28
A
vL RL PLC=−+= (2)
12 4
() (): 0
22
LL
vv C==
33
23
11
() 0: 0
648
A
vL RL PL CL=−+= (3)
From Eqs.(2) and (3):
5
16
A
R
P=↑
and
Continued on next slide
M
B
L/
2
L/
2
R
A
A
C
B
R
B
y
x
P

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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.
2
13 1
32
CC PL==−
Then
11
0: 16
yB
FRP==↑
∑
3
0: 16
AB
M
MPL==
∑
(b) Substitute constants into Eq.(1):
3
1[5( ) 3( )]
96
PL x x
vEI L L =−
Make x=L/2 in the above
3
7
768
CPL
vEI =↓
33
6
7(30 10 )(5)
768(3 10 )
×
=×11.39 mm
=
↓
SOLUTION (10.74)
(a)
00
'' ( )
BBB
EIv R L x M R L R x M=−+=−+
201
1
'2
BB
EIv R Lx R x M x C=− ++
23 2
012
111
262
BB
EIv R Lx R x M x C x C=−+++ (1)
Continued on next slide
L/
2
L/
2
R
A
A
C
B
R
B
y
x
M
0
M
A

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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 conditions:
21
(0) 0: 0 '(0) 0: 0vCv C== ==
33 2
0
111
() 0: 0
262
BB
vL RL RL ML=−+=, 0
3
2
BM
RL
=
↓
Then, 0
3
0: 2
yA
M
FR
L
==↑
∑
0
1
0: 2
yA
M
MM==
∑
(b) Introduce the constants into Eq.(1):
23
0()
4M
vLxx
EIL
=−+
Let x=L/2,
2
0
32
CML
vEI =↓
32
6
810(5) 2.08
32(3 10 ) mm
×
==↓
×
SOLUTION (10.75)
(a)
Segment AC
2
11
'' 2
AA
EIv R x M wx=−−
23
11
11
'26
AA
EIv R x M x wx C=−−+
324
112
11 1
62 24
AA
EIv R x M x wx C x C=− −++ (1)
Segment BC
2'' ( )
24
AA
wL L
EIv R x M x=−− −
22
23
1
'()
244
AA
wL L
EIv R x M x x C=−−−+
Continued on next slide
L/
2
R
A
A
C
B
L/
2
M
A
R
B
w

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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.
32 3
234
11 ()
62 124
AA
wL L
EIv R x M x x C x C=− −−++
Boundary & continuity conditions:
111 2
'(0) 0: 0 (0) 0: 0vCv C== = =
33 3
12 33
'( ) '( ) : ,
2 2 48 64 192
L L wL wL wL
vv CC=−=−+ =−
444 4
12 44
() (): ,
2 2 384 768 384 768
L L wL wL wL wL
vv CC=−=−−+ =
32444
2915
() 0: 0, 3
6 2 256 192 763 64
AA A
A
RL ML M
wL wL wL
vL R wL
L
=− − − + + = = + (b)
Statics:
2
9
0: 128
BA
M
MwL==
∑
57
0: 128
yB
FRwL== ↑
∑
(b) Equation (1) becomes
3224
1(57 27 32 )
768
w
vLxLxx
EI
==−−
Make x=L/2 in the above
13
6144
CwL
vEI =↓
34
6
13(20 10 )(5)
6144(3 10 )
×
=×8.82 mm
=
↓
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
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.
SOLUTION (10.76)
(a) See Table B.6.
4
02
'' 12
Aw
EIv M R x x
L
== −
25
01
2
1
'260
Aw
EIv R x x C
L
=− +
36
012
2
1
6360
Aw
EIv R x x C x C
L
=− ++ (1)
Boundary conditions:
2
(0) 0: 0vC==
23
01
1
'( ) 0 : 0
260
Aw
vL RL L C=−+= (2)
34
01
1
() 0: 0
6 360
Aw
vL RL L CL=−+= (3)
From Eqs.(2) and (3): 3
0
1240
wL
C=−
and
0
1
24
A
R
wL=↑
Statics:
0
1
0: 3
yAB
FRRwL=+=
∑, 0
7
24
B
R
wL
=
↑
2
00
11
0: ( ) 0
24 3 4
BB
L
MwLMwL=+−=
∑, 0
1
24
B
M
wL=
Continued on next slide
2
0()
x
ww
L
=
A
L
0
w
y
M
B
R
A
R
B
M
A
x
2
0()
x
wL
2
0()
3
wx
x
L
x/
4
R
AV
O