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________________________________________________________________________
PROBLEMS (10.68 through 10.71) A fixed-end beam constructed of a W150x30 wide-flange
section (see Table B.8) is loaded as shown in Figs. P10.68 through P10.71. Using a direct-integration
method, determine:
(a) All of the reactions.
(b) The mid span deflection for the given values:
w = w0/2 = 15 kN/m, P = 25 kN, M0 = 10 kN
⋅
m, L = 4 m, E = 200 GPa, I = 17.2x106 mm4
SOLUTION (10.68)
(a)
Due to symmetry:
1
2
AB
R
RwL== ↑ AB
M
M
=
We have
2
11
'' 22
A
EIv wLx M wx=−−
23
1
11
'46
A
EIv wLx M x wx C=−−+
324
12
11 1
12 2 24
A
EIv wLx M x wx C x C=−−++
Boundary conditions:
1
'(0) 0 : 0vC==, 2
(0) 0: 0vC
=
=
2
1
() 0: 12
A
vL M wL==
(b) Then Eq.(a) becomes
22
22 2
(2 ) ( )
24 24
wx wx
vLxLx Lx
EI EI
=−−=−−
Make x=L/2:
4
384
CwL
vEI
=↓
Substitute the given data
34
96
15 10 (4)
384(200 10 )(17.2 10 )
C
v−
×
=××
3
2.91 10 2.91mmm
−
=
×= ↓
Continued on next slide
A
R
A
y
2
1
2
AA
M
Rx wx M=− −
x
w
M
A
V
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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 (10.69)
(a)
Segment AC
2
111
1
'' , ' 2
AA A A
EIv R x M EIv R x M x C=− = − +
32
112
11
62
AA
EIv R x M x C x C=− ++ (1)
Segment BC
2
102 03
'' , ' 2
A
AA A
Rx
EIv R x M M EIv M x M x C=−+ = − + +
322
2034
11 1
62 2
AA
EIv R x M x M x C x C=− + ++ (2)
Boundary & continuity conditions:
1112
'(0) 0: 0 (0) 0: 0vCvC== ==
0
1123
6
() 0: '() '():
2222
A
A
M
L
M
LLL
vR vvC
L
== = =−
2
200
11
'( ) 0 : 0
22
AA
v L RL ML ML ML=−+−=
Solving,
0
1
4
A
M
M= 0
3
2
AM
RL
=
↓
Statics:
00
31
24
BB
M
R
MM
L
=↑ =
We have, from 2
40
() 0: 8vL C ML== .
(b) Equation (1) for x=L/2:
1() 0
2
CL
vv==
Continued on next slide
M
0
L/
2
L/
2
R
A
A
C
B
R
B
y
x
M
B
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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SOLUTION (10.70)
(a)
3
0
() ()
6
BB
w
M
RLx M Lx
L
=−−− −
32 23
0
'' ( 3 3 )
6
BB B
w
EIv R L R x M L L x Lx x
L
=−−− −+ −
4
23223
01
13
'()
2624
BBB
wx
EIv R Lx R x M x L x L x Lx C
L
=− −− − −−+
32 23 4 5
23 2
012
111 ()
262622420
BBB
wLx Lx Lx x
EIv R Lx R x M x C x C
L
=−−− −−−++ (1)
Boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCv C== = =
244
244
03
'()0: ( )0
2624
B
BB
w
RL LL
vL RL ML L L
L
=−−−−+−=
0
212
B
BwL
M
RL
=+ (2)
32 5555
0
() 0: ( ) 0
32 22420
BB
w
RL ML LLLL
vL L
=−−−+−=
0
3210
B
BwL
M
RL
=+ (3)
Equations (2) and (3) yield
2
0
30
BwL
M= 0
3
20
B
R
wL
=
↑
Statics:
0
7
20
A
R
wL=↑ 0
20
AwL
M=
Continued on next slide
M
B
2
0()
2
wLx
L−
O
0()
wLx
L
−
M
R
B
L
-
x
V
x
y
A
w0
B
L
R
A
M
B
R
B
MA
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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) Equation (1) becomes
32 23 4 5
0(3 7 5 )
120
w
vLxLxLxx
EIL
=−+−+
Making x=L/2:
4
0
768
MwL
vEI
=↓
34
96
30 10 (4)
768(200 10 )(17.2 10 )
−
×
=××
2.91 mm
=
↓
SOLUTION (10.71)
(a)
Due to symmetry:
'0
AB A B C
RR M M v===
Statics:
0
1
4
AB
R
RwL== ↑
Thus, only segment AC need be considered.
3
00
11
'' 43
A
EIv w Lx M w x
L
=−+
24
001
11
'812
A
EIv w Lx M x w x C
L
=−− +
32 5
0012
11 1
24 2 60
A
EIv w Lx M x w x C x C
L
=−−++ (1)
Continued on next slide
y
A
w0
B
0
2
x
w
L
R
A
M
B
R
B
L/
2
C
M
A
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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(b) Boundary conditions:
12
'(0) 0: 0 (0) 0: 0vCv C== = =
33
00
'( ) : 0
2322192
A
wL wL
ML
L
v−−=
2
0
5
96
A
M
wL= 2
0
5
96
B
M
wL=
Equation (1) becomes
322 5
03.2
(8 5 )
192
w
vLxLxx
EI L
=−−
Let x=L/2:
4
0
7
3840
CwL
vEI
=↓
34
96
7(30 10 )(4)
3840(200 10 )(17.2 10 )
C
v−
×
=××
4.07 mm
=
1
/
5
100%