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8/20/2019 Chapter 9 Solution
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0, 0] [ , 0] x y x L y= = = =
PROBLEM 9.12
For the beam and loading shown, (a) express the magnitude and location
of the maximum deflection in terms of w0, L, E , and I . (b) Calculate thevalue of the maximum deflection, assuming that beam AB is a W460 74× rolled shape and that 0 60 kN/m,w = 6 m, L = and 200 GPa. E =
SOLUTION
Using entire beam as a free body,
0
0
10: 0
2 3
16
B A
A
L M R L w L
R w L
Σ = − + =
=
Using AJ as a free body,
20
0
300
1 10: 0
6 2 3
1 1
6 6
J
w x x M w Lx M
L
w M w Lx x
L
Σ = − + + =
= −
230
02
2 400 1
3 500 1 2
1 1
6 6
1 112 24
1 1
36 120
d y w EI w Lx x
Ldx
dy w EI w Lx x C dx L
w EIy w Lx x C x C
L
= −
= − +
= − + +
2 2
34 4 0
0 0 1 1
[ 0, 0]: 0 0 0 0 0
1 1 7[ , 0]: 0 0
36 120 360
x y C C
w L x L y w L w L C L C
= = = − + + ∴ =
= = = − + + ∴ = −
Elastic curve:5
3 30
42 30
1 1 7
36 120 360
1 1 712 24 360
w x y Lx L x
EI L
dy w x Lx Ldx EI L
= − −
= − −
USE DISCONTINUITY FUNCTIONS UP TO PROBLEMS 9.64
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PROBLEM 9.12 (Continued)
To find location of maximum deflection, set 0.dy
dx =
2 4 44 2 2 4 2
2 2 2
30 900 42015 30 7 0
30
81 0.2697
15
m m m
m
L L L x L x L x
x L L
− −− + = =
= − =
0.5193m x L=
53 30 1 1 (0.5193 ) 7
(0.5193 ) (0.5193 )36 120 360
m
w L y L L L L
EI L
= − −
4 40 00.00652 or 0.00652
w L w L
EI EI = −
Data: 30 60 kN/m 60 10 N/m 6 mw L= = × =
For 6 4 6 4W460 74, 333 10 mm 333 10 m I −× = × = ×
3 43
9 6
(0.00652)(60 10 )(6)7.61 10 m
(200 10 )(333 10 )m y
−
−
×= = ×
× × 7.61 mmm y =
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PROBLEM 9.15
For the beam and loading shown, determine the deflection at
point C. Use 200 GPa. E =
SOLUTION
Reactions: 0 0 / , / A B R M L R M L= ↑ = ↓
0 : x a< <
0
0
0:
0
=
− + =
=
J M
M
x M L
M M x
L
[ 0, 0] [ , 0]
[ , ]
,
x y x L y
x a y y
dy dy x a
dx dx
= = = =
= =
= =
:a x L< <
00
0
0:
0
( )
=
− + + =
= −
K M
M x M M
L
M M x L
L
0 x a< <
20
2
201
1
2
d y M EI x
Ldx
dy M EI x C
dx L
=
= +
(1)
301 2
1
6
M EIy x C x C
L
= + +
(2)
a x L< <
20
2
203
( )
1
2
d y M EI x L
Ldx
dy M EI x Lx C
dx L
= −
= − +
(3)
3 203 4
1 1
6 2
M EIy x Lx C x C
L
= − + +
(4)
[ ] 2 20, 0 Eq. (2): 0 0 0 0 x y C C = = = + + =
2 20 01 3
3 1 0
1 1, Eqs. (1) and (3):
2 2
dy dy M M x a a C a La C
dx dx L L
C C M a
= = + = − +
= +
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PROBLEM 9.15 (Continued)
[ ] 30 1, Eqs. (2) and (4):6
M x a y y a L= = 1C a
+
30 1
6 M a
L= 2 11 (2
La C
− +
0 4
24 0
)
1
2
M a a C
C M a
+ +
= −
[ ] 3 3 201 0 0
2 201
1 1 1, 0 Eq. (4): ( ) 0
6 2 2
1 1
3 2
M x L y L L C M a L M a
L
M C L a aL
L
= = − + + − =
= + −
Elastic curve for 0 : x a< < 3 2 20 1 1 1
6 3 2
M y x L a aL x
EIL
= + + −
Make . x a= 3 2 3 2 3 2 20 01 1 1 2 1
6 3 2 3 3C
M M y a L a a a L a L a La
EIL EIL
= + + − = + −
Data: 9200 10 Pa, E = × 6 4 6 434.4 10 mm 34.4 10 m , I −= × = × 30 60 10 N m M = × ⋅
1.2 m, 4.8 ma L= =
3 3 2 2
3
9 6
(60 10 ) (2)(1.2) / 3 (4.8) (1.2) / 3 (4.8)(1.2)6.28 10 m
(200 10 )(34.4 10 )(4.8)C y −
−
× + − = = ×× ×
6.28 mmC y = ↑
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PROBLEM 9.45
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at point C . Use 200 GPa. E =
SOLUTION
Units: Forces in kN, lengths in m
0:
1.6 (9.6)(0.8) (20)(0.4) 0
9.8 kN
D
A
A
M
R
R
=
− + + =
=
0 0
0 0
( ) 12 0.4 12 1.2 kN/m
( ) 12 0.4 12 1.2 kN/m
w x x x
dV w x x x
dx
= − − −
= − = − − + −
1 1 0
22 2 1
2
2 3 3 2 2
1
3 4 4 3 31 2
9.8 12 0.4 12 1.2 20 1.2 kN
9.8 6 0.4 6 1.2 20 1.2 kN m
4.9 2 0.4 2 1.2 10 1.2 kN m
1 1 101.63333 0.4 1.2 1.2 kN m
2 2 3
dM V x x x
dx
d y EI M x x x x
dx
dy
EI x x x x C dx
EIy x x x x C x C
= = − − + − − −
= = − − + − − − ⋅
= −
−
+
−
−
−
+ ⋅
= − − + − − − + + ⋅
2 2[ 0, 0] : 0 0 0 0 0 0 0 x y C C = = − + − + + = =
3 4 4 31
1 1 10[ 1.6, 0] : (1.63333)(1.6) (1.2) (0.4) (0.4) (1.6) 0 0
2 2 3 x y C = = − + − + + =
21 3.4080 kN mC = − ⋅
Data: 9 6 4 6 4
4 6 6 2 2
200 10 Pa, 6.83 10 mm 6.83 10 mm
(200 10 )(6.83 10 ) 1.366 10 N m 1366 kN m
E I
EI
−
−
= × = × = ×
= × × = × ⋅ = ⋅
(a) Slope at A: at 0dy
xdx
=
20 0 0 0 3.4080 kN mdy
EI dx
= − + − − ⋅
33.40802.49 10 rad
1366 Aθ
−= − = − × 32.49 10 rad Aθ −= ×
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PROBLEM 9.45 (Continued)
(b) Deflection at C : ( at 1.2 m) y x =
3 4
3
1(1.63333)(1.2) (0.8) 0 0 (3.4080)(1.2) 0
2
1.4720 kN m
C EIy = − + − − +
= − ⋅
31.47201.078 10 m
1366C y −= − = − × 1.078 mmC y = ↓
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SOLUTION
Units: Forces in kN; lengths in
0: 4.8 (
( B A
M R= − +
+
6 4
9
2
212 10 mm
(200 10 )(212
42400 kN m
I
EI
= ×
= ×
= ⋅
( ) 30 30 2.4w x x= − −
30 30dV
w
dx
= − = − +
79 30dM
V xdx
= = − +
2
79 15d y
EI M x xdx
= = −
2 3795 5
2
dy EI x x
dx= − +
3 479 5 5
6 4 4 EIy x x= − +
[ 0, 0] 0
7[ 4.8, 0]
x y
x y
= =
= =
PROBLEM 9.64
The rigid bar DEF is welded at point D to
For the loading shown, determine (a) thedeflection at midpoint C of the beam. Use
meters.
30)(2.4)(3.6)
0)(2.4) 0
79 kN A R
=
= ↑
6 4
6 6 2
212 10 m
10 ) 42.4 10 N m
−
−
×
× = × ⋅
0
02.4 x −
1 030 2.4 50 3.6 x x − − −
2 1 015 2.4 50 3.6 60 3.6 x x x+ − − − − −
3 2 112.4 25 3.6 60 3.6 x x x C − − − − − +
4 3 21
252.4 3.6 30 3.6
3 x x x C x − − − − − + +
( )
2 2
43 4
3 21
0 0 0 0 0 0 0
9 5 5(4.8) 4.8 (2.4)
4 4
25(1.2) (30)(1.2) 4.8 0
3
C C
C
+ − − + + = =
− +
− − + =
1 161.7C = −
the rolled-steel beam AB.
slope at point A, (b) the200 GPa. E =
kN/m
kN
kN m⋅
2kN m⋅
2 3kN m⋅
26 kN m⋅
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PROBLEM 9.64 (Continued)
(a) Slope at point A: at 0dy
xdx
=
2
3
0 0 0 0 0 161.76
161.76 kN m
161.763.82 10
42400
A
A
dy EI
dx
dy
dx
−
= − + − − −
= − ⋅
− = = − ×
33.82 10 rad. Aθ
−= ×
(b) Deflection at midpoint C : ( at 2.4) y x =
3 4
3
3
79 5
(2.4) (2.4) 0 0 0 (161.76)(2.4) 06 4
247.68 kN m
247.685.84 10 m
42400
C
C
EIy
y −
= − + − − − +
= − ⋅
−= = − × 5.84 mmC y = ↓
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PROBLEM 9.77
For the beam and loading shown, determine (a) the slope at
end A, (b) the deflection at point C . Use 200 GPa. E =
SOLUTION
Units: Forces in kN; lengths in m.
Loading I: Moment at B.
Case 7 of Appendix D: 80 kN m, 5.0 m, 2.5 m M L x= ⋅ = =
3 2 3 2
(80)(5.0) 66.667
6 6
80 125( ) [2.5 (5.0) (2.5)]
6 6 (5.0)
A
C
ML
EI EI EI
M y x L x
EIL EI EI
θ = = =
= − − = − − =
Loading II: Moment at A: (Case 7 of Appendix D.)
80 kN m, 5.0 m, 2.5 m
(80)(5.0) 133.333
3 3 A
M L x
ML
EI EI EI θ
= ⋅ = =
= = =
125C y
EI = (Same as loading I.)
Loading III: 140 kN concentrated load at C : 140 kNP =
2 2
3 3
(140)(5.0) 218.75
16 16
(140)(5.0) 364.583
48 48
A
C
PL
EI EI EI
PL y
EI EI EI
θ = − = − = −
= − = − = −
Data: 9 6 4 6 4200 10 Pa, 156 10 mm 156 10 m E I −= × = × = ×
9 6 6 2 2(200 10 )(156 10 ) 31.2 10 N m 31200 kN m EI −= × × = × ⋅ = ⋅
(a) Slope at A: 367.667 133.333 218.750.601 10 rad
31200 Aθ
−+ −= = − ×
30.601 10 rad Aθ −= ×
(b) Deflection at C : 3125 125 364.5833.67 10 m
31200C y
−+ −= = − × 3.67 mmC y = ↓
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PROBLEM 9.86
The two beams shown have the same cross section and
are joined by a hinge at C . For the loading shown,determine (a) the slope at point A, (b) the deflection at
point B. Use 200 GPa. E =
SOLUTION
Using free body ABC ,
3 3 4
6 12 2
0: 0.45 (0.3)(3.5) 0 2.33 kN
200 GPa
1 1
(30)(30) 67500 mm12 12
(200 10 )(67500 10 ) 13.5 kNm
A C C M R R
E
I bh
EI −
Σ = − = =
=
= = =
= × × =
Using cantilever beam CD with load ,C R
Case 1 of Appendix D:
3 3(2.33)(0.3)0.001555 mm
3 (3)(13.5)
C CDC
R L y
EI = − = − =
Calculation of Aθ ′ and B y′ assuming that point C does not move.
Case 5 of Appendix D:
2 2 2 2
2 2 2 2
3.5 kN, 0.45 m, 0.3 m, 0.15 m
( ) (3.5)(0.15)(0.45 0.15 )0.00259 rad
6 (6)(13.5)(0.45)
(3.5)(0.15) (0.3)0.000389 m
3 (3)(13.5)(0.45)
A
B
P L a b
Pb L b
EIL
Pb a y
EIL
θ
= = = =− −
′ = − = − = −
′ = − = − = −
Additional slope and deflection due to movement of point C .
0.0015550.003456 rad
0.45
(0.3)(0.001555)0.0010367 mm
0.45
C A
AC
B C
y
L
a y y
L
θ ′′ = = − = −
′′ = = − = −
(a) Slope at A: 0.00259 0.003456 A A Aθ θ θ ′ ′′= + = − −
0.006046 rad 0.00605 rad− =
(b) Deflection at B: 0.000389 0.0010367 B B B y y y′ ′′= + = − −
31.4257 10 m 1.43 mm−= − × =
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SOLUTION
Using portion AB and applying
3
1
2
1
1 1
3
(13( )
3(1
( )2
( ) ( )
1.56 10
B
B
C B
PL y
EI PL
EI
y y L
θ
−
= =
= =
= +
= ×
Due to load P at point C : (C
( y
Total deflection at point C :
PROBLEM 9.87
Beam DE rests on the cantilever beam
that a square rod of side 10 mm is usedthe deflection at end C if the 25-N m⋅ c
E of beam DE , (b) to end C of beam AC .
9
3 4
2
200 10 Pa
1(10)(10) 833.33 mm
12
166.667 N m
E
I
EI
= ×
= = =
= ⋅
(a) Couple applied to beam DE .
Free body DE . 0: 0.180 25 0 M PΣ = − =
Loads on cantilever beam ABC are P at point
as shown.
Due to P at point B.
ase 1 of Appendix D,
33
23
3 31
8.889)(0.120)0.480 10 m
(3)(166.667)8.889)(0.120)
6.00 10(2)(166.667)
( ) 0.480 10 (0.180)(6.00 10 )
m
C Bθ
−
−
− −
= ×
= ×
= × + ×
se 1 of App. D applied to ABC .)
3 3
2
(138.889)(0.120 0.180)) 7.50 1
3 (3)(166.667)
PL
EI
+= − = − = − ×
31 2( ) ( ) 5.94 10 mC C C y y y −
= + = − ×
(b) Couple applied to beam AC :
Case 3 of Appendix D.
2 23(25)(0.300)
6.75 10 m2 (2)(166.667)
C
ML y
EI
−= − = − = − ×
AC as shown. Knowing
for each beam, determineuple is applied (a) to endUse 200 GPa. E =
12 433.33 10 m−×
138.889 NP =
B and P at point C
30 m−
5.94 mmC y =
6.75 mmC y =
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PROBLEM 9.91
Before the load P was applied, a gap, 0 0.5mm,δ = existed
between the cantilever beam AC and the support at B.Knowing that 200 GPa, E = determine the magnitude of P for which the deflection at C is 1 mm.
SOLUTION
Let length 0.5 m AB L= =
length 0.2 m BC a= =
Consider portion AB of beam ABC .
The loading becomes forces P and B R at B plus the couple Pa.The deflection at B is 0.δ Using Cases 1 and 3 of Appendix D ,
3 2
0
3 2 3
0
( )
3 2
3 2 3
B
B
P R L PaL
EI EI
L L a LP R EI
δ
δ
−= +
+ − =
(1)
The deflection at C depends on the deformation of beam ABC
subjected to loads P and . B R For loading I, using Case 1 of
Appendix D,
3
1
( )( )
3C
P L a
EI δ +=
For loading II, using Case 1 of Appendix D,
3 2
3 2
B B B B
R L R L y
EI EI θ = =
Portion BC remains straight.
3 2
3 2
BC B B
L L a R y y a
EI θ
= + = +
By superposition, the downward deflection at C is3 3 2( )
3 3 2
BC
P L a L L a R
EI EI δ
+= − +
3 3 2( )
3 3 2 B C
L a L L aP R EI δ
+− + =
(2)
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PROBLEM 9.91 (Continued)
Data:9
200 10 Pa E = × 3 6 4 6 4
3 2
3 30
1(60)(60) 1.08 10 mm 1.08 10 m12
216 10 N m
0.5 10 m 1.0 10 mC
I
EI
δ δ
−
− −
= = × = ×
= × ⋅
= × = ×
Using the data, eqs (1) and (2) become
0.06667 0.04167 108 BP R− = (1)′
0.11433 0.06667 216 BP R− = (2)′
Solving simultaneously,
3
5.63 10 NP = ×
5.63 kNP = ↓
36.42 10 N B R = ×
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PROBLEM 9.93
A 22-mm-diameter rod BC is attached to the lever AB and
to the fixed support at C . Lever AB has a uniform crosssection 10 mm thick and 25 mm deep. For the loadingshown, determine the deflection of point A. Use
200 GPa E = and 77 GPaG =
SOLUTION
Deformation of rod BC : (Torsion)
4 4
9 12
1 1(22) 11 mm
2 2
22998 mm2
(350)(0.25) 87.5 N m
0.5 m
(87.5)(0.5)
(77 10 )(22998 10 )
0.0247 rad
B
C d
J C
T Pa
L
TL
GJ
π
ϕ −
= = =
= =
= = = ⋅
=
= =× ×
=
Deflection of point A assuming lever AB to be rigid:
1( ) (0.25)(0.0247) B B y aϕ = =
0.006175 m=
Additional deflection due to bending of lever AB.
Refer to Case 1 of Appendix D.
3 4
3 3
9 12
1(10)(25) 13021 mm
12
(350)(0.25)( )
3 (3)(200 10 )(13021 10 ) A
I
PL y
EI −
= =
= =× ×
32.1 10 m−= ×
Total deflection at point A:
1 2( ) ( ) 8.28 mm A A A y y y= + =
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PROBLEM 9.94
A 16-mm-diameter rod has been bent into the shape shown.
Determine the deflection of end C after the 200-N force is applied.Use 200 GPa E = and 80 GPa.G =
SOLUTION
Let 200 N .P=
Consider torsion of rod AB.
2
3
( ) B
C B
TL PL L PL
JG JG JG
PL y L
JG
φ
φ
= = =
′ = − = −
Consider bending of AB. (Case 1 of Appendix D.)
3
3C B
PL y y
EI ′′ = = −
Consider bending of BC . (Case 1 of Appendix D.)
3
3C
PL y EI
′′′ = −
Superposition:
3 3 3 3 2
3 3 3
C C C C y y y y
PL PL PL PL EI
JG EI EI EI JG
′ ′′ ′′′= + +
= − − − = − +
Data:
9 4 9 4
9 9 4
2 2
33
180(10 ) Pa (0.008) 6.4340(10 ) m
2
1200(10 ) Pa 3.2170(10 ) m2
643.40 N m 514.72 N m
(200)(0.25) 643.40 29.3093(10 ) m
643.40 514.72 3C
G J
E I J
EI JG
y
π −
−
−
= = =
= = =
= ⋅ = ⋅
= − + = −
9.31 mmC y = ↓