Capitolo 9
DEFORMAZIONI DELLE TRAVI
F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.1
SOLUTION
 M K = 0 : − M0 + M = 0
M = M0
d2y
= M = M0
dx
dy
EI
= M 0 x + C1
dx
EI
dy


 x = L, dx = 0  :


0 = M 0 L + C1
EIy =
[ x = L, y = 0]
(a)
0=
C1 = − M 0 L
1
M 0 x 2 + C1x + C2
2
1
M 0 L2 − M 0 L2 + C2
2
Elastic curve:
C2 =
1
M 0 L2
2
y =
M0 2
( x − 2 Lx + L2 ) 
2EI
y =
(b)
y at x = 0 :
(c)
dy
at x = 0 :
dx
yA =
M0
( L − 0) 2
2EI
M0
( L − x) 2 
2EI
yA =
M 0 L2
↑ 
2 EI
θA =
w0 L
EI
dy
M
M
M L
= − 0 ( L − x) = − 0 ( L − 0) = − 0
dx
EI
EI
EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.2
SOLUTION
ΣM J = 0: − M − P( L − x) = 0
M = − P ( L − x)
EI
d2y
= − P( L − x) = − PL + Px
dx 2
EI
dy
1
= − PLx + Px 2 + C1
dx
2
dy


 x = 0, dx = 0  :


0 = −0 + 0 + C1
C1 = 0
1
1
EIy = − PLx 2 + Px3 + C1x + C2
2
6
[ x = 0, y = 0] :
(a)
0 = −0 + 0 + 0 + C2
C2 = 0
y =−
Elastic curve.
Px 2
(3L − x) 
6 EI
dy
Px
=−
(2 L − x)
dx
2 EI
(b)
y at x = L.
(c)
dy
at x = L.
dx
yB = −
dy
dx
=−
B
PL2
PL3
(3L − L) = −
6EI
3EI
PL
PL2
(2 L − L) = −
2EI
2EI
yB =
θB =
PL3
↓ 
3EI
PL2
2EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.3
SOLUTION
ΣM J = 0: (wx)
x
+M =0
2
1
M = − wx 2
2
d2y
1
= M = − wx 2
2
dx 2
1
dy
EI
= − wx3 + C1
6
dx
EI
dy
1 3
1 3


 x = L, dx = 0  : 0 = − 6 wL + C1 C1 = 6 wL


EI
dy
1
1
= − wx3 + wL3
dx
6
6
EIy = −
1
1
wx 4 + wL3 x + C2
24
6
[ x = L, y = 0] 0 = −
1
1
wL4 + wL4 + C2 = 0
24
6
1
3
 1
C2 = 
−  wL4 = − wL4
24
 24 6 
(a)
Elastic curve.
(b)
y at x = 0.
(c)
dy
at x = 0.
dx
y =−
yA = −
dy
dx
=
A
3wL4
wL4
=−
24EI
8EI
wL3
6EI
w
( x 4 − 4 L3 x + 3L4 ) 
24EI
yA =
θA =
wL4
↓ 
8EI
wL3
6 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.4
SOLUTION
 Fy = 0:
1
wL = 0
2
1
RA = w0 L
2
RA =
 M A = 0:
2L wL
⋅
=0
3
2
1
= − w0 L2
3
− MA =
MA
ΣM J = 0:
1
1
w x2 x
w0 L2 − w0 Lx + 0 ⋅ + M = 0
3
2
2L 3
1
1
w x3
M = − w0 L2 + w0 Lx − 0
3
2
6L
EI
1
1
d2y
w0 x3
2
=
−
+
−
w
L
w
Lx
0
0
3
2
6L
dx 2
1
1
dy
w x4
= − w0 L2 x + w0 Lx 2 − 0 + C1
3
4
24L
dx
dy

= 0,
= 0 :
0 = −0 + 0 − 0 + C1
dx

EI

x

C1 = 0
1
1
w x5
EIy = − w0 L2 x 2 +
w0 Lx3 − 0 + C2
6
12
120L
[ x = 0, y = 0] :
0 = −0 + 0 − 0 + 0 + C2
(a)
Elastic curve:
y =−
C2 = 0
w0  1 3 2
1
1 5
Lx −
Lx 4 +
x 

EIL  6
12
120 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEM$ 9. (ContinuD)
(b)
y at x = L
yB = −
w0 L4  1
1
1 
11 w0 L4
+
 −
=−
EI  6 12 120 
120 EI
yB =
(c)
dy
at x = L
dx
dy
dx
=−
B
11 w0 L4
↓ 
120 EI
w0 L3  1 1
1 
1 w0 L4
−
+
=
−


EI  3 4 24 
8 EI
θB =
1 w0 L3
8 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.5
SOLUTION
FBD ABC:
Using ABC as a free body,
ΣFy = 0: RA + 2wa −
2
wa = 0
3
4
4
RA = − wa = wa ↓
3
3
2

ΣM A = 0: −M A + (2wa)(a) −  wa  (3a) = 0
3

MA = 0
Using AJ as a free body,
FBD AJ:
4

 x
ΣM J = 0: M +  wa  ( x) − (wx)   = 0
3

2
1
4
M = wx 2 − wax
2
3
d2y
1
4
= wx 2 − wax
2
2
3
dx
dy
1
2
EI
= wx3 − wax 2 + C1
dx 6
3
EI
dy


 x = 0, dx = 0  : 0 = 0 − 0 + C1 ∴ C1 = 0


EIy =
1
2
wx 4 − wax3 + C2
24
9
[ x = 0, y = 0]: 0 = 0 − 0 + C2 ∴ C2 = 0
(a)
Elastic curve over AB.


(b)
y at x = 2a.
(c)
dy
at x = 2a.
dx
y =
w
(3x 4 − 16ax3 ) 
72EI
dy
w 3
=
( x − 4ax 2 )
dx
6EI
yB = −
10wa 4
9 EI
4wa3
 dy 
=
−
 
3EI
 dx  B
yB =
θB =
10wa 4
↓ 
9 EI
4wa3
3EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.8
SOLUTION
Reactions:
1
1  1
 1 
ΣM B = 0: − RA L +  w0 L   L  −  w0 L  L  = 0
2
3  4
 6 
1
RA = w0 L
8
Boundary conditions: [ x = 0, y = 0] [ x = L, y = 0]
(0 ≤ x < L)
For portion AB only,
1
1w   x
w0 Lx +  0 x  ( x)   + M = 0
8
2 L  3
1
1 w0 3
M = w0 Lx −
x
8
6 L
ΣM J = 0: −
d2y 1
= w0 Lx −
8
dx 2
dy
1
EI
w0 Lx 2
=
dx 16
1
EIy =
w0 Lx3
48
EI
1 w0 3
x
6 L
1 w0 4
x + C1
−
24 L
1 w0 5
−
x + C1x + C2
120 L
[ x = 0, y = 0] : 0 = 0 − 0 + 0 + C2
[ x = L, y = 0] : 0 =
(a)
1
1
1
w0 L4 −
w0 L4 + C1L C1 = − w0 L3
48
120
80
y =
Elastic curve.
C2 = 0
w0  1 2 3
1 5
1 4 
Lx −
x −
L x 

120
80
EIL  48

dy
w  1
1 4
1 4
= 0  L2 x 2 −
x −
L
24
80 
dx
EIL  16
L
.
2
(b)
y at x =
(c)
dy
at x = L.
dx
yL 2 =
dy
dx
=
B
w0
EIL
 L5
L5
L5 
15w0 L4
−
−

 = −
3840EI
 384 3840 160 
w0  L4 L4 L4 
2w L3
−
−

 = + 0
EIL  16 24 80 
240 EI
yL 2 =
θB =
w0 L4
↓ 
256EI
w0 L3
120 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.11
SOLUTION
Using entire beam as a free body,
ΣM B = 0: M 0 − RA L = 0 RA =
M0
L
Using portion AJ ,
[ x = 0, y = 0]
[ x = L, y = 0]
ΣM J = 0: M 0 −
M =
M0
x+M =0
L
M0
( x − L)
L
d2y
M
= 0 ( x − L)
2
L
dx
dy
M 1

= 0  x 2 − Lx  + C1
EI
dx
L 2

EI
EIy =
[ x = 0, y = 0]
0 = 0 − 0 + 0 + C2
[ x = L, y = 0]
0=
y =
(a)
M0  1 3 1 2 
 x − Lx  + C1x + C2
2
L 6

M0  1 3 1 3 
 L − L  + C1L + 0
2 
L 6
M0  1 3 1 2 1 2 
 x − Lx + L x 
2
3
EIL  6

To find location maximum deflection, set
1 2
1
xm − Lxm + L2 = 0
2
3
C2 = 0
1
M 0L
3
1 
dy
M 1
= 0  x 2 − Lx + L2 
3 
dx
EIL  2
dy
= 0.
dx
xm = L −
 1  1  
L2 − (4)   L2  = 1 −
 2  3  
= 0.42265L
ym =
C1 =

M 0 L2  1 
1
1
3
2
  (0.42265) −   (0.42265) +   (0.42265) 
EI  6 
2
3

1
L
3 
xm = 0.423L 
ym = 0.06415
M 0 L2

EI
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEM$ 9.11 (ContinuD
Solving for M 0 ,
(b)
Data:
M0 =
EIym
0.06415L2
E = 200 × 109 Pa, I = 178 × 106 mm 4 = 178 × 10−6 m 4
L = 3.5m
M0 =
ym = 1 mm = 10−3 m
(200 × 109 )(178 × 10−6 )(10−3 )
= 45.3 × 103 N ⋅ m
2
(0.06415)(3.5)
M 0 = 45.3 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.14
SOLUTION
Using ACB as a free body and noting that L = 3a,
a
 M A = 0: RB L − (wa)   = 0
2
[ x = 0, y = 0]
[ x = L, y = 0]
RB = ( wa)
[ x = a, y = y]
a
1
= wa
2L 6
dy
dy 

 x = a, dx = dx 


 Fy = 0:
RA + RB − wa = 0
0≤ x≤a
MK = 0 :
− M + RB ( L − x) = 0
x
M − RA x + (wx)   = 0
2
M = RB ( L − x)
1 2
wx
2
2
2
1
d y
= RA x − wx 2
2
2
dx
1
dy 1
EI
= RA x 2 − wx3 + C1
6
dx 2
1
1
EIy = RA x3 −
wx 4 + C1 x + C2
6
24
EI
[ x = 0, y = 0]
5
wa
6
a≤x≤L
MJ = 0 :
M = RA x −
RA =
0 = 0 − 0 + 0 + C2
1
1
EIy = RA x3 −
wx 4 + C1x
6
24

1
dy 1
2
3
= RA x − wx + C1
EI
dx 2
6
C2 = 0
d y
= RB ( L − x)
dx 2
dy
1
EI
= − RB ( L − x) 2 + C3
dx
2
1
EIy = RB ( L − x)3 + C3 x + C4
6
EI
[ x = L, y = 0]
0 = 0 + C3 L + C4
C4 = − C3 L
1
RB ( L − x) 2 − C3 ( L − x) 
6
1
dy
EI
= − RB ( L − x) 2 + C3
2
dx
EIy =
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.14 (Continua)
dy dy 
1
1
1

RAa 2 − wa3 + C1 = − RB (2a) 2 + C3
 x = a, dx = dx 
2
6
2


1
1
1
7
C3 = C1 + RAa 2 − wa3 + RB (2a) 2 = C1 + wa3
2
6
2
12
1
1
1
7


R Aa 3 −
wa 4 + C1a = RB (2a)3 −  C1 +
wa3  (2a)
6
24
6
12


1
1
1
7
25
wa 4 + RB (2a)3 −
wa 2 (2a) = −
wa 4
3C1a = − RAa3 +
6
24
6
12
24
25 3
C1 = − wa
72
5
1
25
wax3 −
wx 4 −
wa3 x
For 0 ≤ x ≤ a, EIy =
36
24
72
dy
5
1
25
EI
wax 2 − wx3 −
wa3
=
dx 12
6
72
[ x = a,
Data:
y = y]
w = 50 × 103 N/m, a = 2 m, E = 200 × 109 Pa
I = 84.9 × 106 mm 4 = 84.9 × 10−6 m 4 , EI = 16.98 × 106 N ⋅ m 2
(a)
Slope at x = 0.
16.98 × 106
dy
dx
dy
dx
(b)
=0− 0−
A
25
(50 × 103 )(2)3
72
= θ A = −8.18 × 10−3
θ A = 8.18 × 10−3 rad

A
Deflection at x = 2 m.
5
1
25 4
1
wa 4 −
wa 4 −
wa = − wa 4
36
24
72
4
1
16.98 × 106 yC = − (50 × 103 )(2)4 yC = − 11.78 × 10−3 m
4
EIyC =
yC = 11.78 mm ↓ 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.15
SOLUTION
Reactions:
RA = M 0 /L ↑,
RB = M 0 /L ↓
 M J = 0:
0 < x < a:
M0
x+M =0
L
M
M= 0x
L
−
[ x = 0, y = 0]
[ x = L, y = 0]
[ x = a, y = y]
a < x < L:
M K = 0:
M0
x + M0 + M = 0
L
M
M = 0 ( x − L)
L
dy
dy 

 x = a, dx = dx 


−
0< x<a
a< x< L
d 2y M0
x
=
L
dx 2
dy M 0  1 2 
EI
=
 x  + C1
dx
L 2 
EI
EIy =
[ x = 0,
M0  1 3 
x + C1x + C2
L  6 
y = 0] Eq. (2):
d 2y M0
( x − L)
=
L
dx 2
dy M 0  1 2

EI
=
 x − Lx  + C3
dx
L 2

EI
(1)
(2)
0 = 0 + 0 + C2
EIy =
M0  1 3 1 2 
x − Lx  + C3 x + C4
2
L  6

(3)
(4)
C2 = 0
dy dy 
M0  1 2 
M0  1 2


 x = a, dx = dx  Eqs. (1) and (3): L  2 a  + C1 = L  2 a − La  + C3






C3 = C1 + M 0a
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEMA 9.15 (Continua)
[ x = a,
y = y ] Eqs. (2) and (4):
[ x = L,
y = 0] Eq. (4):
M0
L
C1 =
Elastic curve for 0 < x < a.
M0  1 3 
M0  1 3 1 2 
 a  + C1a =
 a − La  + ( C1 + M 0a)a + C4
L 6 
L 6
2

1
C4 = − M 0a 2
2
1
1 3 1 3
2
 L − L  + (C1 + M 0a) L − M 0a = 0
6
2
2


M0  1 2 1 2

 L + a − aL 
L 3
2

y=
M0 1 3  1 2 1 2
 
 x +  L + a − aL  x 
EIL  6
2
3
 
M0 1 3 1 2
1
M 2
1


a + L a + a3 − a 2 L  = 0  a3 + L2a − La 2 
EIL  6
3
2
EIL
3
3



Make x = a.
yC =
Data:
E = 200 × 109 Pa, I = 34.4 × 106 mm 4 = 34.4 × 10−6 m 4 , M 0 = 60 × 103 N ⋅ m
a = 1.2 m, L = 4.8 m
yC =
(60 × 103 ) (2)(1.2)3 / 3 + (4.8) 2 (1.2) / 3 − (4.8)(1.2)2 
(200 × 109 )(34.4 × 10−6 )(4.8)
= 6.28 × 10−3 m
yC = 6.28 mm ↑ 

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Esercizio 9.16
SOLUTION
Consider portion ABC only. Apply symmetry about C.
Reactions:
RA = RE = P
dy
dy  
L dy


=
, x = ,
= 0
Boundary conditions: [ x = 0, y = 0], [ x = a, y = y],  x = a,

dx
dx  
2 dx


0< x<a
a< x< L−a
2
EI
d y
= M = Px
dx 2
EI
dy
1
= Px 2 + C1
dx
2
EIy =
d2y
= M = Pa
dx 2
dy
EI
= Pax + C3
dx
1
EIy = Pax 2 + C3 x + C4
2
EI
(1)
1 3
Px + C1x + C2
6
(2)
L dy
1


 x = 2 , dx = 0  → C3 = − 2 PaL


[ x = 0, y = 0] → C2 = 0
L dy
dy 

 x = 2 , dx = dx 


1 2
1
Pa + C1 = Pa 2 − PaL
2
2
L


x = 2 , y = y



1 3 1 2
1
1
1
Pa +  Pa − PaL  a = Pa3 − Pa 2 L + C4
6
2
2
2
 2

C1 =
C4 =
(a)
1 2 1
Pa − PaL
2
2
1 3
Pa
6
Elastic curve for portion BD.
y =
1 1

Pax 2 + C3 x + C4 

EI  2

y =
P
EI
1 3
1 2 1
 2 ax − 2 aLx + 6 a  


PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEMA 9.16 (Continua)
For deflection at C,
x=
L
.
2
yC =
P
EI
set
=−
Data:
1 2 1 2 1 3
 aL − aL + a 
4
6 
8
Pa  1 2 1 2 
 L − a 
EI  8
6 
I = 23.9 × 106 mm 4 = 23.9 × 10−6 m 4 ,
E = 200 × 109 Pa
P = 17.5 × 103 N,
L = 2.5 m, a = 0.8 m
(b)
yC = −
 2.52 0.82 
(17.5 × 103 )(0.8)
−3
−

 = −1.976 × 10 m
6 
(200 × 109 )(23.9 × 106 )  8
yC = 1.976 mm ↓ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.17
SOLUTION
Boundary conditions are shown at right.
[ x = 0, y = 0]
[ x = L, V = 0]
dy
[ x = 0,
= 0]
dx
[ x = L, M = 0]
2

dV
x x 
= − w = − w0 1 − 4   + 3   
dx
 L   L  


2 x 2 x3 
+ 2  + CV
V = − w0  x −
L
L 

[ x = L, V = 0]: 0 = − w0 [ L − 2 L + L] + CV = 0
CV = 0

dM
2 x 2 x3 
= V = − w0  x −
+ 2
dx
L
L 

 x 2 2 x3
x4 
M = − w0  −
+ 2  + CM
3L 4 L 
2
2
1 
1
[ x = L, M = 0]: 0 = − w0  L2 − L2 + L2  + CM
3
4 
2
EI
CM =
1
w0 L2
12
1
d2y
2 x3 1 x 4 1 2 
= M = − w0  x 2 −
+
− L 
dx
3 L 4 L2 12 
2
EI
1
1 x 4 1 x5 1 2 
dy
= − w0  x3 −
+
− L x  + C1
6 L 20 L2 12
dx
6

[ x = 0,
dy
= 0]
dx
C1 = 0
1
1 x5
1 x6 1 2 2 
EIy = − w0  x 4 −
+
−
L x  + C2
30 L 120 L2 24
 24

[ x = 0, y = 0]
C2 = 0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.17 (Continua)
(a)
Elastic curve.
(b)
Deflection at x = L.
y=−
yB = −
w0  1 2 4 1
1 6 1 4 2
Lx −
Lx5 +
x −
Lx  
2 
24
30
120
24
EIL 

w0  1 6 1 6
w0 L4
1 6 1 6
−
+
−
=
−
L
L
L
L


30
120
24 
40 EI
EIL2  24
yB =
w0 L4
↓ 
40 EI
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Esercizio 9.18
SOLUTION
Boundary conditions at A and B are noted.
w0
(4 Lx − 4 x 2 )
L2
w
= −w = 20 (4 x 2 − 4Lx)
L
w 4

= V = 20  x3 − 2Lx 2  + C1
L 3

w=
dV
dx
dM
dx
M =
[ x = 0, M = 0]
0 = 0 + 0 + 0 + C2
[ x = L, M = 0]
0=
EI
w0  1 4 2 4 
 L − L  + C1L + 0
3 
L2  3
C1 =
1
w0 L
3
1
1
dy
w  1

= 20  x5 − Lx 4 + L3 x 2  + C3
6
6
dx
L  15

EIy =
1
1 3 3
w0  1 6
x −
Lx5 +
L x  + C3 x + C4
2 
30
18
L  90

[ x = 0, y = 0]
0 = 0 + 0 + 0 + 0 + C4
[ x = L, y = 0]
0=
Elastic curve.
C2 = 0
2
1
d2y
w 1

= M = 20  x 4 − Lx3 + L3 x 
2
3
3
dx
L 3

EI
(a)
w0  1 4 2 3 
 x − Lx  + C1x + C2
3
L2  3

C4 = 0
w0  1 6
1 6
1 6
L −
L +
L  + C3L + 0
2 
30
18 
L  90
y =
C3 = −
1
w0 L3
30
w0  1 6
1
1 3 3
1 5 
x −
Lx5 +
Lx −
L x 
2
30
18
30
EIL  90

dy
w0  1 5 1 4 1 3 2
1 5
=
 x − 6 Lx + 6 L x − 30 L 
dx
EIL2  15

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 9.18 (ContinuD
(b)
(c)
Set x = 0 in
Slope at end A.
Deflection at midpoint.
Set x =
dy
.
dx
dy
dx
=−
A
1 w0 L3
30 EI
θA =
1 w0 L3
30 EI
yC =
61 w0 L4
↓ 
5760 EI

L
in y.
2
6
5
3
w0 L4  1  1 
1 1
1  1  
 1  1 
yC =
   −    +   −
 
EI  90  2 
18  2 
30  2  
 30  2 
=
w0 L4  1
1
1
1
61 w0 L4
−
+
−

=−
EI  5760 960 144 60 
5760 EI
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Esercizio 9.19
SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown above.
Using free body AJ,
ΣM J = 0: M 0 − RA x + M = 0
M = RA x − M 0
d2y
= RA x − M 0
dx 2
1
dy
EI
= RA x 2 − M 0 x + C1
dx
2
EI
dy
1


2
 x = L, dx = 0  0 = 2 RA L − M 0 L + C1


C1 = M 0 L −
EIy =
1
RA L2
2
1
1
RA x3 − M 0 x3 + C1x + C2
6
2
[ x = 0, y = 0]
C2 = 0
[ x = L, y = 0] 0 =
1
1
1


RA L3 − M 0 L2 +  M 0 L − RA L2  L + 0
6
2
2


RA =
3 M0
↑ 
2 L
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.20
SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown above.
Using free body KB,
L − x
M K = 0: RB ( L − x) − w( L − x) 
−M =0
 2 
M = RB ( L − x) −
1
w( L − x) 2
2
d2y
1
= RB ( L − x) − w( L − x) 2
2
2
dx
dy
1
1
= − RB ( L − x) 2 + w( L − x)3 + C1
EI
dx
2
6
EI
dy
1
1 3


2
 x = 0, dx = 0  : 0 = − 2 RB L + 6 wL + C1


1
1
C1 = RB L2 − wL3
2
6
EI y =
1
1
RB ( L − x)3 −
w( L − x) 4 + C1x + C2
6
24
1
RB L3 −
6
1
C2 = − RB L3 +
6
[ x = 0, y = 0]: 0 =
1
wL3 + C2
24
1
wL4
24
[ x = L, y = 0]: 0 = 0 − 0 + C1L + C2
1
1
1
1
RB L3 − wL4 − RB L3 +
wL4 = 0
2
6
6
24
RB =
3
wL ↑ 
8
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.21
SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown above.
w0
( L − x)
L
w
= − w = − 0 ( L − x)
L
w 
1 
= V = − 0  Lx − x 2  + RA
L 
2 
w=
dV
dx
dM
dx
M =−
EI
w0  1 2 1 3 
 Lx − x  + RA x
L 2
6 
d2y
w 1
1 
= − 0  Lx 2 − x3  + RA x
2
L
2
6 
dx

EI
dy
w 1
1 4 1
x  + RA x 2 + C1
= − 0  Lx3 −
dx
L 6
24  2
EIy = −
w0  1
1 5 1
Lx 4 −
x  + RA x3 + C1x + C2
L  24
120  6
[ x = 0, y = 0]
dy


 x = L, dx = 0


0 = 0 + 0 + 0 + C2
−
w0  1 4
1 4 1
L −
L  + RA L2 + C1 = 0

L 6
24  2
C1 =
[ x = L, y = 0]
−
C2 = 0
1
1
w0 L3 − RA L2
8
2
w0  1 5
1 5 1
1
1

L −
L  + RA L3 +  w0 L3 − RA L2  L = 0

L  24
120  6
2
8

1
1 
1 1
1
 2 − 6  RA =  8 − 24 + 120  w0 L




1
11
RA =
w0 L
3
120
RA =
11
w0 L ↑ 
40
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.22
L
SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown above.
Using free body JB,
ΣM J = 0: −M + RB ( L − x) +
+
1
2
w0 ( L − x) ( L − x)
2
3
1 w0 x
1
( L − x) ( L − x ) = 0
2 L
3
w0
[2L( L − x) 2 + x( L − x) 2 ]
6L
w
= RB ( L − x) − 0 [2L3 − 4L2 x + 2Lx 2 + xL2 − 2 Lx 2 + x3 ]
6L
w
= RB ( L − x) − 0 ( x3 − 3L2 x + 2L3 )
6L
M = RB ( L − x) −
d2y
w
= RB ( L − x) − 0 ( x3 − 3L2 x + 2L3 )
2
6L
dx
1  w 1
3
dy


= RB  Lx − x 2  − 0  x 4 − L2 x 2 + 2L3 x  + C1
EI
2  6L  4
2
dx


EI
1  w  1
1
1

EIy = RB  Lx 2 − x3  − 0  x5 − L2 x3 + L3 x 2  + C1x + C2
6  6L  20
2
2

[ x = 0, y = 0]
→ C2 = 0
dy


 x = 0, dx = 0  → C1 = 0


1
 1 1  w L4  1

[ x = L, y = 0] 0 = RB L3  −  − 0 
− + 1
2
6
6
20
2




1
 1  11 
RB =    w0 L
3
 6  20 
RB =
11
w0 L ↑ 
40
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.23
SOLUTION
Reactions are statically indeterminate.
Boundary conditions are shown at left.
Using free body JB,
MJ = 0 :
[ x = 0, y = 0]
[ x = L, y = 0]
L
− M + x
dy
= 0]
[ x = 0,
dx
w0 2
ξ (ξ − x)d ξ + RB ( L − x) = 0
L2
w0 L 2
 ξ (ξ − x)dξ − RB (L − x)
L2 x
M =
L
EI
w0  1 4 1 3 
 ξ − xξ  − RB ( L − x)
3
L2  4
x
=
1 4
w0  1 4 1 3
L − Lx+
x  − RB ( L − x)
2 
3
12 
L 4
d2y
w 1
1
1 4
x  − RB ( L − x)
= 20  L4 − L3 x +
3
12 
dx 2
L 4
EI
dy
w 1
1
1 5
1 

x  − RB  Lx − x 2  + C1
= 20  L4 x − L3 x 2 +
dx
6
60 
2 
L 4

EIy =
dy


 x = 0, dx = 0 


[ x = 0, y = 0]
[ x = L, y = 0]
Data:
=
w0  1 4 2
1 3 3
1 6
1 
1
Lx +
x  − RB  Lx 2 − x3  + C1x + C2
 Lx −
18
360 
2
6 
L2  8

0 = 0 + 0 + C1
0 = 0 + 0 + 0 + C2
C1 = 0
C2 = 0
1 
1 1
1 1
4
3
+
 −
 w0 L −  −  RB L = 0
 8 18 360 
2 6
13
1
13
w0 L4 − RB L3 = 0
RB =
w0 L
180
3
60
L = 3m
w0 = 15 kN/m
13
RB =
(15)(3) = 9.75 kN
60
RB = 9.75 kN ↑ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.25
SOLUTION
Reactions are statically indeterminate.
ΣFy = 0:
RA + RB = 0
ΣM A = 0:
RA = − RB
− M A − M 0 + RB L = 0
0 <x <
M A = RB L − M 0
L
2
M = RB x + M A = −M 0 + RB L − RB x
d2y
= − M 0 + RB ( L − x)
dx 2
1 
dy

EI
= − M 0 x + RB  Lx − x 2  + C1
dx
2 

EI
1
1 
1
EIy = − M 0 x 2 + RB  Lx 2 − x3  + C1x + C2
2
6 
2
L
< x< L
2
M = RB ( L − x)
d2y
= RB ( L − x)
dx 2
1 
dy

= RB  Lx − x 2  + C3
EI
dx
2 

EI
1 
1
EIy = RB  Lx 2 − x3  + C3 x + C4
6 
2
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.25 (Continua)
dy


 x = 0, dx = 0  0 + 0 + C1 = 0


C1 = 0
[ x = 0, y = 0] 0 + 0 + 0 + C2 = 0
C2 = 0
L dy
dy 

 x = 2 , dx = dx 


−M 0
L
1 
1 
1
1
+ RB  L2 − L2  = RB  L2 − L2  + C3
2
6 
6 
2
2
C3 = −
M 0L
2
L


x = 2 , y = y


2
L
1
1 3
1 3
L
1
1
− M 0   + RB  L3 −
L  = RB  L3 −
L  + C3 + C4
2
2
48 
48 
2
8
8
1
1
C4 = − M 0 L2 − C3 L
8
2
1
 1 1
=  − +  M 0 L2 = M 0 L2
8
 8 4
[ x = L, y = 0]
1  M L
1
1
RB  L3 − L3  + 0 L + M 0 L2 = 0
6 
2
8
2
1 1
1 1
3
2
 2 − 6  RB L =  2 − 8  M 0 L




1
3 M0
RB =
3
8 L
RB =
MA =
9
1
M0 − M0 = M0
8
8
M C − = −M 0 +
9 M0
7
= − M0
8 L
16
L  9 M0  L 
9

M C + = RB  L −  =
M0
=


2  8 L  2  16

9 M0
↑ 
8 L
MA =
1
M0 
8
M C− = −
7
M0 
16
M C+ =
9
M0 
16
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.26
SOLUTION
Reactions are statically indeterminate.
ΣFy = 0 : RA + RB − P = 0 RA = P − RB
ΣM A = 0: − M A +
M A = RB L −
0< x<
1
PL − RB L = 0
2
1
PL
2
1
L:
2
M = M A + RA x
d2y
= M A + RA x
dx 2
dy
1
EI
= M A x + RA x 2 + C1
dx
2
1
1
EIy = M A x 2 + RA x3 + C1x + C2
2
6
EI
1
L< x<L:
2
1 

M = M A + RA x − P  x − L 
2 

EI
d2y
1 

= M = M A + RA x − P  x − L 
2 
dx 2

2
EI
dy
1
1 
1 
= M A x + RA x 2 − P  x − L  + C3
dx
2
2 
2 
3
L
1
1
1 
EIy = M A x 2 + RA x3 − P  x −  + C3 x + C4
2
6
6 
2
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.26 (Continua)
dy


 x = 0, dx = 0 


[ x = 0,
0 + 0 + C1 = 0
C1 = 0
y = 0 ] 0 + 0 + 0 + C2 = 0
C2 = 0
L dy
dy 

 x = 2 , dx = dx 


1
1
1
1
M A L + RA L2 + 0 = M A L + RA L2 − 0 + C3
2
8
2
8
C3 = 0
L


x = 2 , y = y


1
1
1
1
M A L2 +
RA L3 + 0 + 0 = M A L2 +
RAL3 − 0 + 0 + C4 C4 = 0
8
48
8
48
[ x = L, y = 0]
1
1
1
M A L2 + RA L3 −
PL3 + 0 + 0 = 0
2
6
48
1
1  3 1
1
3
PL3 = 0
 RB L − P  L + ( P − RB ) L −
2
2 
6
48
RB =
5
P↑ 
16
5
P
16
RA =
7
P↑
16
5
1
PL − PL
16
2
MA = −
3
PL 
16
MC =
5
PL 
32
RA = P −
MA =
 L   5  L 
M C = RB   =  P  
 2   16  2 
MB = 0 
Bending moment diagram
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.27
SOLUTION
Reactions are statically indeterminate.
0< x<
EI
L
2
d2y
= M = RA x
dx 2
EI
(1)
dy 1
= RA x 2 + C1
dx 2
EIy =
(2)
1
RA x3 + C1x + C2
6
(3)
L
< x< L
2
EI
d2y
1 
L
= M = RA x − w  x − 
2 
2
dx 2
2
(4)
3
EI
dy 1
1 
L
= RA x 2 − w  x −  + C3
6 
2
dx 2
(5)
4
EIy =
[ x = 0,
y = 0]
1
1 
L
R A x3 −
w x −  + C3 x + C4
6
24 
2
0 = 0 + 0 + C2
C2 = 0
2
2
L dy dy 

 x = 2 , dx = dx 


1
1
L
L
RA   + C1 = RA   + 0 + C3
2
2
2
2
L


x = 2 , y = y


L
L
1 L
1 L
RA   + C1 + C2 = RA   − 0 + C3 + C4
2
2
6 2
6 2
dy


 x = L, dx = 0 


1
1 L
Rx L2 − w   + C3 = 0
2
6 2
[ x = L,
1
1 L  1
1

RAL2 −
w
+
wL3 − RA L2  L + 0 = 0
6
24  2   48
2

y = 0]
(6)
3
C1 = C3
3
3
C3 =
C2 = C4 = 0
1
1
wL3 − RA L2
48
2
4
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.27 (Continua)
1  4
1 1
3  1
 2 − 6  RA L =  48 − 384  wL




From (1), with x =
L
,
2
1
7
RA =
wL
3
384
RA =
7
L
M C = RA   =
wL2
 2  256
7
wL ↑ 
128
M C = 0.02734 wL2 
2
From (4), with x = L,
MB
1 L
1
9
 7
= RA L − w   = 
−  wL −
wL2
2 2
128
 128 8 
M B = − 0.07031wL 
Location of maximum positive M:
L
< x< L
2
L

Vm = RA − w  xm −  = 0
2

xm =
From (4), with x = xm ,
xm −
L
R
7
= A =
L
2
w 128
L
7
71
+
L=
L
2 128
128
M m = RA xm −
1 
L
w xm − 
2 
2
2
 7
 71  1  7 
=
wL 
L  − w
L
128

 128  2  128 

2
M m = 0.02884 wL2 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.28
SOLUTION
Reactions are statically indeterminate.
ΣFy = 0 : RA + RB −
w0 L
=0
4
RB =
w0 L
− RA
4
 w L  2L 
ΣM B = 0 : − RA L +  0 
 + MB = 0
 4  3 
0≤ x ≤
w=
M B = RA L −
L
≤ x≤ L:
2
L
:
2
2w0
x
L
M = RA x −
w0 2
x
L
w
M = RA x − 0 x 3
3L
V = RA −
EI
d2y
w
= RA x − 0 x 3
2
3L
dx
1
1 w0 4
dy
EI
x + C1
= RA x 2 −
2
12 L
dx
1
1 w0 5
EIy = RA x3 −
x + C1x + C2
6
60 L
w0 L 
L
x − 
4 
3
d2y
1 
1
L
= RA x − w0 L  x −
2
12 
dx
4
EI
EI
[ x = 0, y = 0] : 0 = 0 − 0 + 0 + C2
w0 L2
6
dy
1
1
1

Lx  + C3
= RA x 2 − w0 L  x 2 −
dx
2
12 
8
EIy =
1
1
 1

RA x3 − w0 L  x3 −
Lx 2  + C3 x + C4
6
24
 24

∴ C2 = 0
L dy
dy  1
1
1
1
1

2
3
2
3
 x = 2 , dx = dx  : 8 RA L − 192 w0 L + C1 = 8 RA L + 96 w0 L + C3 ∴ C3 = C1 − 64 w0 L


L
1
1

 1
3
4
 x = 2 , y = y  : 48 RA L − 1920 w0 L + 2 C1L + 0


=
1
1 3
 1 3
RA L3 − w0 L 
L −
L
48
96 
 192
1

 L 
+  C1 −
w0 L3   + C4
64

 2 
∴ C4 =
1
w0 L4
480
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.28 (Continua)
dy
1 2
1
1

 1
1 2
2
2
3
 x = L, dx = 0 : 2 RA L − w0 L  8 L − 12 L  + C3 = 0 ∴ C3 = − 2 RA L + 24 w0 L




[ x = L, y = 0] :
Over
1
1 3  1
1
1
 1

RA L3 − w0 L  L3 −
L  +  − RA L2 +
w0 L3  ( L) +
w0 L4 = 0
6
24   2
24
480
 24

RB =
w0 L
21
−
w0 L
4
160
MB =
21
w L2
w0 L2 − 0
160
6
0< x<
21
w0 L ↑ 
160
RB =
19
w0 L ↑
160
17
w0 L2 = −0.0354w0 L2 
480
L
w
21
w
, V = RA − 0 x 2 =
w0 L − 0 x 2
2
L
160
L
V = 0 at
M =
MB = −
RA =
x = xm = 0.36228L
21
w
w0 Lx − 0 x3
160
3L
M A = M ( x = 0) = 0
L

M C = M  x =  = 0.0240w0 L2
2

M m = M ( xm = 0.36228L) = 0.0317 w0 L2 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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Esercizio 9.29
SOLUTION
Reactions are statically indeterminate.
ΣFy = 0: RA +
1
1
wL − wL + RB = 0 RA = − RB
2
2
1
L
ΣM A = 0: − M A −  wL  + RB L = 0
2
2
M A = RB L −
0< x≤
From A to C:
1 2
wL
4
L
2
d2y
1
= M = M A + RA x + wx 2
2
2
dx
1
1
dy
EI
= M A x + RA x 2 + wx3 + C1
2
6
dx
1
1
1
EIy = M A x 2 + RA x3 +
wx 4 + C1x + C2
2
6
24
EI
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEMA 9.29 (Continua)
From C to B:
L
≤ x< L
2
d2y
1
L 1 
L

EI 2 = M = M A + RA x + wL  x −  − w  x − 
2
4 2 
2
dx

2
2
3
dy
1
1
L
1 
L

EI
= M A x + RA x 2 + wL  x −  − w  x −  + C3
dx
2
4
4
6 
2

3
4
1
1
1
L
1 
L

EIy = M A x 2 + RA x3 +
wL  x −  −
w  x −  + C3 x + C4
2
6
12
4
24 
2

dy


 x = 0, dx = 0 


0 + 0 + 0 + C1 = 0
C1 = 0
[ x = 0, y = 0] 0 + 0 + 0 + 0 + C2 = 0
C2 = 0
L dy
dy 

 x = 2 , dx = dx 


2
MA
3
L
1 L
1 L
L
1 L
− RA   + w   = M A + RA  
2 2
6 2
2
2 2
2
2
2
+
1
L
wL   − 0 + C3
4
4
1  3
1
 1
3
C3 = 
−
 wL = 192 wL
48
64


L


x = 2 , y = y


2
3
1
1 L
1 L
L
M A   + RA   +
w 
2
6 2
24  2 
2
2
=
3
4
1
1 L
1
L
L
M A   + RA   +
wL  
2
6 2
12
2
4
−0+
3
1
L
wL3   + C4
192
2
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEMA 9.29 (Continua)
1
1  4
1
 1
4
C4 = 
−
−
 wL = − 768 wL
384
768
384


[ x = L, y = 0]
3
1
1
1
1 L
 3L 
M A L2 + RA L3 +
wL   −
w 
2
6
12
24  2 
 4 
+
4
1
1
wL3 ( L) −
wL4 = 0
192
768
1
1
1
1
1
1  4

 27
RB L − wL2  L2 + (− RB ) L3 + 
−
+
−

 wL = 0
2
4
6

 768 384 192 768 
7  4
1 1
1
3
 2 − 6  RB L =  8 − 192  wL




RA = − RB = −
M A = RB L −
Deflection at C.
1
17
RB =
wL
3
192
RB =
17
wL ↑ 
64
17
wL
64
1 2  17 1  2
1
wL = 
wL2
−  wL =
4
64
 64 4 
L

 y at x = 2 


2
EIyC =
3
1
1 L
1 L
L
M A   + RA   +
w 
2
6 2
24  2 
2
2
=
4
3
1 1
1  17
1 L
 L 
2  L 
w 
 wL   +  − wL   +
2  64
6  64
24  2 
 2 
 2 
17
1  4
1
 1
=
−
+
wL4
 wL = −
1024
 512 3072 384 
4
yC =
1 wL4
↓ 
1024 EI
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Esercizio 9.30
SOLUTION
Reactions are statically indeterminate.
0 < x<
L
2
d2y
1
= M = RA x − wx 2
2
2
dx
dy
1
1
EI
= RA x 2 − wx3 + C1
dx
2
6
1
1
EIy = RA x3 −
wx 4 + C1x + C2
6
24
EI
L
< x< L
2
(See free body diagram.)


 M K = 0: − RA x + wL  x −  + M = 0
2
4


1
EI
L
d2y
1
1 

= M = RA x − wL  x − L 
2
2
4 
dx

2
EI
dy
L
1
1

= RA x 2 − wL  x −  + C3
dx
2
4
4

3
EIy =
L
1
1

RA x 3 −
wL  x −  + C3 x + C4
6
12
4

[ x = 0, y = 0] :
0 − 0 + 0 + C2 = 0
L dy
dy 

 x = 2 , dx = dx  :


1 L
1 L
1 L
1
L
RA   − w   + C1 = RA   − wL   + C3
2 2
6 2
2 2
4
4
2
3
C1 = C3 +
3
C2 = 0
4
2
2
1
1
1
wL3 −
wL3 = C3 +
wL3
48
64
192
3
3
1 L
1
1 L
1
L
L

 1 L

L
3 L
 x = 2 , y = y  : 6 RA  2  − 24 w  2  +  C3 + 192 wL  2 = 6 RA  2  − 12 wL  4  + C3 2 + C4


 
 


 
 
C4 = −
1
1
1
1
wL4 +
wL4 +
wL4 =
wL4
384
384
768
768
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM$ 9.30 (ContinuD
dy


 x = L, dx = 0  :


2
1
1
 3L 
RAL2 − wL   + C3 = 0
2
4
 4 
C3 =
9
1
wL3 − RAL2
64
2
3
[ x = L, y = 0] :
1
1
1
1
 3L 
 9

RA L3 −
wL   +  wL3 − RA L2  L +
wL4 = 0
6
12
2
768
 4 
 64

27
1  4
1 1
 9
3
 2 − 6  RA L =  64 − 768 + 768  wL




C3 =
1
41
RA =
wL
3
384
41
wL ↑ 
128
9
1 41 3
5
wL3 −
wL = −
wL3
64
2 128
256
C1 = −
5
1
11
wL3 +
wL3 = −
wL3
256
192
768
L

 y at x = 2 


Deflection at C.
yC =
wL4  1 41
 ⋅
EI  6 128
3
1
1
⋅  −
24
2
4

11 1
1
⋅  −
⋅ + 0
768 2
2

1
11  wL4
19 wL4
 41
−
=
−
−
=

6144 EI
 6144 384 1536  EI
or
RA =
yC =
19 wL4
↓ 
6144 EI
3
3
wL4  1 41  1 
1 1
5 1
1 
⋅  +
⋅ +
yC =
 ⋅

  −
EI  6 128  2 
12  4 
256 2 768 
 41
1
5
1  wL4
19 wL4
=
−
−
+
=
−

768  EI
6144 EI
 6144 768 512
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.31
SOLUTION
Reactions are statically indeterminate.
Because of symmetry,
[ x = 0, y = 0]
[ x = L, y = 0]
dy
L
= 0 and V = 0 at x = .
dx
2
L

Use portion AC of beam.  0 < x ≤ 2 


dy
dy

 

 x = 0, dx = 0   x = L, dx = 0 

 

L


 x = 2 , V = 0


L dy


 x = 2 , dx = 0 


dV
w
= − w = −2 0 x
dx
L
EI
dM
w
= V = − 0 x 2 + RA
dx
L
(1)
d2y
1 w0 3
=M =−
x + RA x + M A
2
3 L
dx
(2)
EI
dy
1 w0 4 1
=−
x + RA x 2 + M A x + C1
dx
12 L
2
EIy = −
(3)
1 w0 2 1
1
x + RA x3 + M A x 2 + C1x + C2
60 L
6
2
dy


 x = 0, dx = 0  :


0 = 0 + 0 + 0 + C1
C1 = 0
[ x = 0, y = 0] :
0 = 0 + 0 + 0 + 0 + C2
C2 = 0
L


 x = 2 , V = 0 :


−
w0  L 
+ RA = 0
L  2 
L dy


 x = 2 , dx = 0  :


−
1 w0  L 
11
L
 L
+  w0 L    + M A + 0 = 0
12 L  2 
24
2
2
 
2
4
RA =
(4)
wL

4
2
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM$ 9.3 (ContinuD
1 
5
 1
2
2
M A = −2 
−
 w0 L = − 96 w0 L
32
192


From (2), with x =
M A = −0.05208w0 L2 
L
,
2
3
MC = −
1 w0  L 
1
 L  5
w0 L12
+  w0 L   −


3 L 2
4
 2  96
1
5 
1
 1
= −
+ −
w0 L2 =
w0 L2

32
 24 8 96 
M C = 0.03125w0 L2 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.40
SOLUTION
Loading I:
Case 6 in Appendix D.
yC = −
Loading II:
5wL4
384EI
θA = −
wL3
24EI
Case 7 in Appendix D.
3
M A  L 
1 M A L2
 L 
yC = −
  − L2    =
6EIL  2 
 2   16 EI
MA =
with
Loading III:
wL2
12
yC =
θA =
1 M B L3
16 EI
M L
θA = B
6EI

MB =
with
wL2
12
1 wL3
36 EI
(using Loading II result)
yC =
1 wL4
192 EI
θA =
1 wL3
72 EI
Deflection at C.
yC = −
5 wL4
1 wL4
1 wL4
1 wL4
+
+
=−
384 EI
192 EI
192 EI
384 EI
yC =
(b)
M AL
3EI
Case 7 in Appendix D.
yC =
(a)
1 wL4
192 EI
θA =
1 wL4
↓ 
384 EI
Slope at A.
θA = −
1 wL3
1 wL3
1 wL3
+
+
=0
24 EI
36 EI
72 EI
θ A = 0 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.41
SOLUTION
Units:
Forces in kN; lengths in m.
Loading I:
Concentrated load at B
Case 1 of Appendix D applied to portion AB.
θ B′ = −
PL2
(3)(0.75) 2
0.84375
=−
=−
2EI
2EI
EI
y′B = −
PL3
(3)(0.75)3
0.421875
=−
=−
3EI
3EI
EI
Portion BC remains straight.
θC′ = θ B′ = −
0.84375
EI
yC′ = y′B − (0.5)θ B′ = −
Loading II:
Concentrated load at C.
0.84375
EI
Case 1 of Appendix D.
θ ′′A = −
PL2
(3)(1.25) 2
2.34375
=−
=−
2EI
2 EI
EI
y′′A = −
PL3
(3)(1.25)3
1.953125
=−
=−
EI
3EI
3EI
3.1875
EI
2.796875
y A = y′A + y′′A = −
EI
By superposition,
θ A = θ ′A + θ A′′ = −
Data:
E = 200 × 109 Pa, I = 2.52 × 106 mm 4 = 2.52 × 10−6 m 4
EI = (200 × 104 )(2.52 × 10−6 ) = 504 × 103 N ⋅ m 2 = 504 kN ⋅ m 2
Slope at C.
θC = −
3.1875
= −6.32 × 10−3 rad
504
Deflection at C.
yC = −
2.796875
= −5.55 × 10−3 m
504
θC = 6.32 × 10−3 rad

yC = 5.55 mm ↓ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.42
SOLUTION
Units:
Forces in kN; lengths in m.
The slope and deflection at B depend only on the
deformation of portion AB.
Reduce the force at C to an equivalent force-couple system
at B and add the force already at B to obtain the loadings I
and II shown.
Loading I:
Loading II:
Case 1 of Appendix D.
θ B′ = −
PL2
(6)(0.75) 2
1.6875
=−
=−
2EI
2 EI
EI
y′B = −
PL3
(6)(0.75)3
0.84375
=−
=−
EI
3EI
3EI
Case 3 of Appendix D.
θ B′′ = −
ML
(1.5)(0.75)
1.125
=−
=−
EI
EI
EI
y′′B = −
(1.5)(0.75)2
0.421875
ML2
=−
=−
2EI
EI
EI
By superposition,
2.8125
EI
1.265625
yB = y′B + y′′B = −
EI
θ B = θ B′ + θ B′′ = −
Data:
E = 200 × 109 Pa, I = 2.52 × 106 mm 4 = 2.52 × 10−6 m 4
EI = (200 × 109 )(2.52 × 10−6 ) = 504 × 103 N ⋅ m 2 = 504 kN ⋅ m 2
Slope at B.
θB = −
2.8125
= −5.58 × 10−3 rad
504
Deflection at B.
yB = −
1.265625
= −2.51 × 10−3 m
504
θ B = 5.58 × 10−3 rad

yB = 2.51 mm ↓ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.45
SOLUTION
Units:
Forces in kN; lengths in m.
Loading I:
Moment at B.
M = 80 kN ⋅ m, L = 5.0 m, x = 2.5 m
Case 7 of Appendix D.
ML (80)(5.0) 66.667
=
=
6 EI
6 EI
EI
M
80
125
yC = −
( x3 − L2 x) = −
[2.53 − (5.0)2 (2.5)] =
EI
6 EIL
6 EI (5.0)
θA =
Loading II:
Moment at A.
Case 7 of Appendix D.
M = 80 kN ⋅ m, L = 5.0 m, x = 2.5 m
ML (80)(5.0) 133.333
θA =
=
=
3EI
3EI
EI
yC =
Loading III:
125
EI
140 kN concentrated load at C.
(Same as loading I.)
P = 140 kN
PL2
(140)(5.0)2
218.75
=−
=−
16 EI
16 EI
EI
PL3
(140)(5.0)3
364.583
=−
=−
yC = −
48 EI
48 EI
EI
θA = −
E = 200 × 109 Pa, I = 156 × 106 mm 4 = 156 × 10−6 m 4
Data:
EI = (200 × 109 )(156 × 10−6 ) = 31.2 × 106 N ⋅ m 2 = 31200 kN ⋅ m 2
(a)
Slope at A.
θA =
67.667 + 133.333 − 218.75
= −0.601 × 10−3 rad
31200
θ A = 0.601 × 10−3 rad
(b)
Deflection at C.
yC =
125 + 125 − 364.583
= −3.67 × 10−3 m
31200

yC = 3.67 mm ↓ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.46
SOLUTION
Units:
Forces in kN; lengths in m.
Loading I:
8 kN/m uniformly distributed.
Case 6:
w = 8 kN/m, L = 3.9 m, x = 1.3 m
WL3
(8)(3.9)3
19.773
=−
=−
24 EI
24EI
EI
w
8
yC = −
[ x 4 − 2Lx3 + L3 x] = −
[(1.3) 4 − (2)(3.9)(1.3)3 + (3.9)3 (1.3)]
24EI
24 EI
20.945
=−
EI
θA = −
Loading II:
35 kN concentrated load at C.
Case 5 of Appendix D.
P = 35 kN, L = 3.9 m, a = 1.3 m, b = 2.6 m, x = a = 1.3 m
Data:
θA = −
Pb( L2 − b 2 )
(35)(2.6)(3.92 − 2.6) 2
32.861
=−
=−
6EIL
6EI (3.9)
EI
yC = −
Pa 2b 2
(35)(1.3) 2 (2.6) 2
34.176
=−
=−
3EIL
3EI (3.9)
EI
E = 200 × 109 , I = 102 × 106 mm 4 = 102 × 10−6 m 4
EI = (200 × 109 )(102 × 10−6 ) = 20.4 × 106 N ⋅ m 2 = 20, 400 kN ⋅ m 2
(a)
Slope at A.
θA = −
θ A = 2.58 × 10−3 rad

(b)
19.773 + 32.861
= −2.58 × 10−3 rad
20, 400
Deflection at C.
yC = −

20.934 + 34.176
= −2.70 × 10−3 m
20, 400
yC = 2.70 mm ↓ 
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Esercizio 9.48
SOLUTION

Consider RC as redundant and replace loading system by I and II.

Loading I:
(Case 4 of Appendix D.)
Loading II:
(Case 7 of Appendix D.)
yC′ = −
RC L3
48 EI
yC′′ = −
=
3
M 0  L 
 L 
  − L2   
6 EIL  2 
 2  
M 0 L2
16 EI
Superposition and constraint:
yC = yC′ + yC′′ = 0
−
RC L3 M 0 L2
+
=0
48EI 16 EI
 3M 0  L 
ΣM B = 0: − RA L + 
  − M0 = 0
 L  2 

ΣFY = 0:
M 0 3M 0
−
+ RB = 0
2L
L
RC =
3M 0
↓ 
L
RA =
RB =
M0
↑ 
2L
5M 0
↑ 
2L
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Esercizio 9.49
SOLUTION
Beam is indeterminate to first degree. Consider RA as
redundant and replace the given loading by loadings I, II,
and III.
Loading I:
Case 1 of Appendix D.
( yA)I =
Loading II:
RA L3
3EI
Case 2 of Appendix D.
( y A ) II = −
Loading III:
wL4
8EI
Case 2 of Appendix D (portion CB).
(θC ) III = −
( yC ) III =
w( L/2)3
1 wL3
=−
6 EI
48 EI
1 wL4
w( L/2)4
=
8EI
128 EI
Portion AC remains straight.
( y A ) III = ( yC ) III +
Superposition and constraint:
(a)
L
7 wL4
(θC ) III =
2
384 EI
y A = ( y A ) I + ( y A ) II + ( y A ) III = 0
1 RA L3 1 wL4
7 wL4
1 RA L3
41 wL4
−
+
=
−
=0
3 3EI
8 EI
384 EI
3 EI
384 EI
RA =
41
wL ↑ 
128
RB =
23
wL ↑ 
128
Statics:
(b)
ΣFy = 0 :
41
1
wL − wL + RB = 0
128
2
 41

1
 3L 
wL  L −  wL   − M B = 0
ΣM B = 0 : − 
128
2



 4 
MB =
7
wL2
128

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Esercizio 9.50
SOLUTION
Consider RA as redundant and replace loading system by I and II.
y ′A =
Loading I:
(Case 1 of Appendix D.)
Loading II:
Portion BC. (Case 3 of Appendix D.)
yC′′ = −
M 0 ( L − a)2
2 EI
θC′′ =
M 0 ( L − a)
EI
y ′′A = yC′′ − (a)θC
Portion AC is straight.
=−
(a)
RA L3
3EI
Superposition and constraint:
M 0 ( L − a) 2 aM 0 ( L − a )
−
2 EI
EI
y A = y ′A + y ′′A = 0
RA L3 M 0 ( L − a) 2 aM 0 ( L − a)
−
−
=0
3EI
2 EI
EI
2
RA L3 − M 0 ( L − a)( L − a + 2a ) = 0
3
2
RA L3 = M 0 ( L2 − a 2 )
3
(b)
ΣFy = 0: RA + RB = 0
RB = − RA
ΣM B = 0: M B + M 0 − RA L = 0
MB =

RA =
3M 0  2
2 
L − a 2 − L2 
2 
3 
2L 
RB =
MB + M0 −
3M 0
2 L3
3M 0
2 L3
( L2 − a 2 ) ↑ 
( L2 − a 2 ) ↓ 
3 M0 2
(L − a2 ) = 0
2 L2
MB =
M0
2 L2
( L2 − 3a 2 )


PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.51
SOLUTION
Beam is second degree indeterminate. Choose RB and M B as
redundant reactions.
Loading I:
Case 1 of Appendix D.
( yB ) I =
Loading II:
RB L3
3EI
(θ B ) I =
RB L2
2EI
Case 3 of Appendix D.
( yB ) II = −
M B L2
2 EI
(θ B ) II = −
M BL
EI
Loading III: Case 2 of Appendix D.
( yB ) III = −
wL4
8EI
(θ B ) III = −
wL2
6 EI
Superposition and constraint:
yB = ( yB ) I + ( yB )II + ( yB ) III = 0
L3
L2
wL4
RB −
MB −
=0
3EI
2EI
8EI
(1)
θ B = (θ B ) I + (θ B ) II + (θ B ) III = 0
L2
L
wL3
=0
RB −
MB −
2EI
6EI
EI
(2)
Solving Eqs. (1) and (2) simultaneously,
RB =
MB =
1
wL ↑ 
2
1
wL2
12

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.52
SOLUTION
Beam is second degree indeterminate. Choose RB and M B as redundant reactions.
Loading I:
Case 1 of Appendix D.
( yB ) I = −
Loading II:
RB L3
R L2
, (θ B ) I = − B
3EI
2 EI
Case 3 of Appendix D.
( yB ) II =
M B L2
M L
, (θ B )II = B
EI
2 EI
Loading III: Case 3 applied to portion AC.
M 0 ( L/2) 2 M 0 L2
=
2 EI
8 EI
M 0 ( L/2) M 0 L
=
=
2 EI
EI
( yC )III =
(θC )III
Portion CB remains straight.
3 M 0 L2
L
(θC ) III =
2
8 EI
2
1 M0L
=
2 EI
( yB ) III = ( yC ) III +
(θ B ) III = (θC )III
Superposition and constraint:
yB = ( yB ) I + ( yB )II + ( yB ) III = 0
−
L3
L2
3 M 0 L2
RB +
MB +
=0
3EI
2 EI
8 EI
(1)
θ B = (θ B ) I + (θ B ) II + (θ B ) III = 0
−
L2
L
1 M0L
RB +
MB +
=0
EI
2 EI
2 EI
(2)
Solving Eqs. (1) and (2) simultaneously,
RB =
3 M0
↓ 
2 L
MB =
1
M0
4

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Esercizio 9.53
SOLUTION
Let
a = 0.4 m.
Cantilever beams AB and CD.
Cases 1 and 2 of Appendix D.
yB = yD = −
( wa) a3 wa 4
11 wa 4
−
=−
3EI
8EI
24 EI
Beam BCD, with L = 0.8 m, assuming that points B and D do not move.
Case 6 of Appendix D.
yC′ = −
5wL4
384 EI
Additional deflection due to movement of points B and D.
yC′′ = yB = yD = −
Total deflection at C.
yC = yC′ + yC′′
yC = −
Data:
11 wa 4
24 EI
w
EI
4
4
 5L 11a 
+


24 
 384
E = 200 × 109 Pa,
1
I = (24)(12)3 = 3.456 × 10−3 mm 4 = 3.456 × 10−9 m 4
12
EI = (200 × 109 )(3.456 × 10−9 ) = 691.2 N ⋅ m 2
yC = −3 × 10−3 m
−3 × 10−3 = −
w  (5)(0.8) 4 (11)(0.4) 4 
−6
+

 = −24.69 w
691.2  384
24

w = 121.5 N/m 
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Esercizio 9.56
SOLUTION
Units:
Forces in kN; lengths in m.
Using free body ABC,
 M A = 0: 4.4 RC − (4.4)(30)(2.2) = 0
RC = 66.0 kN
E = 200 × 109 Pa
I = 125 × 106 mm 4 = 125 × 10−6 m 4
EI = (200 × 109 )(125 × 10−6 ) = 25.0 × 10−6 N ⋅ m 2
= 25,000 kN ⋅ m 2
For slope and defection at C, use Case 1 of Appendix D
applied to portion CE of beam DCE.
θC =
RC L2
(66.0)(2.2) 2
=
= 6.3888 × 10−3 rad
2EI
(2)(25, 000)
yC = −
RC L3
(66.0)(2.2)3
=
= −9.3702 × 10−3 m
3EI
(3)(25, 000)
Defection at B, assuming that point C does not move.
Use Case 6 of Appendix D.
( yB )1 = −
5WL4
(5)(30)(4.4) 4
=−
= −5.8564 × 10−3
384EI
(384)(25, 000)
Additional defection at B due to movement of point C:
(a)
Total deflection at B.
( yB ) 2 =
1
yC = −4.6851 × 10−3 m
2
yB = ( yB )1 + ( yB )2 = −10.54 × 10−3 m
yB = 10.54 mm ↓ 
Portion DC of beam DCB remains straight.
(b)
Deflection at D. yD = yC − aθC = −9.3702 × 10−3 − (2.2)(6.3888 × 10−3 ) = −23.4 × 10−3 m
yD = 23.4 mm ↓ 
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Esercizio 9.58
SOLUTION
Let P be the tension developed in member AB and δ B be the elongation of that member.
A = 255 mm 2 = 255 × 10−6 m 2
Cable AB:
δB =
PL
( P)(3)
=
EA (200 × 109 )(255 × 10−6 )
= 58.82 × 10−9 P
I = 156 × 106 mm 4 = 156 × 10−6 m 4
Beam BC:
EI = (200 × 109 )(156 × 10−6 )
= 31.2 × 106 N ⋅ m 2
Loading I:
20 kN/m downward.
Refer to Case 2 of Appendix D.
( yB )1 = −
wL4
(20 × 103 )(6) 4
=−
8EI
(8)(31.2 × 106 )
= −103.846 × 10−3 m
Loading II:
Upward force P at Point B.
Refer to Case 1 of Appendix D.
( yB ) 2 =
PL3
P(6)3
=
= 2.3077 × 10−6 P
3EI (3)(31.2 × 106 )
yB = ( yB )1 + ( yB ) 2
By superposition,
Also, matching the deflection at B,
yB = −δ B
−103.846 × 10−3 + 2.3077 × 10−6 P = −58.82 × 10−9 P
2.3666 × 10−6 P = 103.846 × 10−3
P = 43.9 × 103 N
P = 43.9 kN 
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Esercizio 9.59
SOLUTION
Let length AB = L = 0.5 m
length BC = a = 0.2 m
Consider portion AB of beam ABC.
The loading becomes forces P and RB at B plus the couple Pa.
The deflection at B is δ 0. Using Cases 1 and 3 of Appendix D,
δ0 =
( P − RB ) L3 PaL2
+
3EI
2 EI
 L3 L2a 
L3
+
RB = EI δ 0

 P −
2 
3
 3
(1)
The deflection at C depends on the deformation of beam ABC
subjected to loads P and RB . For loading I, using Case 1 of
Appendix D,
(δ C )1 =
P ( L + a )3
3EI
For loading II, using Case 1 of Appendix D,
yB =
RB L3
3EI
θB =
RB L2
2EI
Portion BC remains straight.
 L3 L2a  RB
+
yC = yB + aθ B = 

2  EI
 3
By superposition, the downward deflection at C is
δC =
P( L + a)3  L3 L2a  RB
− 
+

3EI
2  EI
 3
 L3 L2a 
( L + a)3
P − 
+
 RB = EI δ C
3
2 
 3
(2)
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PROBLEM$ 9. (ContinuD
Data: E = 200 × 109 Pa
I =
1
(60)(60)3 = 1.08 × 106 mm 4 = 1.08 × 10−6 m 4
12
EI = 216 × 103 N ⋅ m 2
δ 0 = 0.5 × 10−3 m
δ C = 1.0 × 10−3 m
Using the data, eqs (1) and (2) become
0.06667 P − 0.04167 RB = 108
(1)′
0.11433 P − 0.06667 RB = 216
(2)′
Solving simultaneously,
P = 5.63 × 103 N
P = 5.63 kN ↓ 
RB = 6.42 × 103 N
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.62
SOLUTION
Let 200 N = P.
Consider torsion of rod AB.
TL ( PL) L PL2
=
=
JG
JG
JG
PL3
yC′ = − LφB = −
JG
φB =
Consider bending of AB. (Case 1 of Appendix D.)
yC′′ = yB = −
PL3
3EI
Consider bending of BC. (Case 1 of Appendix D.)
yC′′′ = −
PL3
3EI
Superposition:
yC = yC′ + yC′′ + yC′′′
=−
PL3 PL3 PL3
PL3  EI 2 
−
−
=−
+
JG 3EI 3EI
EI  JG 3 
Data:
G = 80(109 ) Pa
E = 200(109 ) Pa
EI = 643.40 N ⋅ m 2
yC = −

1
J = π (0.008) 4 = 6.4340(10−9 ) m 4
2
1
I = J = 3.2170(10−9 ) m 4
2
JG = 514.72 N ⋅ m 2
(200)(0.25)3  643.40 2 
+  = −9.3093(10−3 ) m

643.40  514.72 3 
yC = 9.31 mm ↓ 
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Esercizio 9.63
SOLUTION
w( x) =
2w0
x − w0
L
w
 2w

V ( x) = − w( x)dx = −  0 x − w0  dx = − 0 x 2 + w0 x + C1
L
 L



[ x = 0, V = 0] 0 = 0 + 0 + C1 ∴ C1 = 0
w
w
 w

M ( x) = V ( x)dx =  − 0 x 2 + w0 x  dx = − 0 x3 + 0 x 2 + C2
L
3
L
2




[ x = 0, M = 0] 0 = 0 + 0 + C2 ∴ C2 = 0
w
w
d2y
= M = − 0 x3 + 0 x 2
2
3
2
L
dx
w
w
dy
EI
= − 0 x 4 + 0 x3 + C3
dx
12 L
6
EI

 x = L,

w L3 w L3
dy

= 0 0 = − 0 + 0 + C3
dx
12
6

EIy = −
Elastic curve.
(b)
y at x = 0.
(c)
dy
at x = 0.
dx
w0 L4 w0 L4 w0 L4
+
−
+ C4
60
24
12
y=−
yA = +
dy
dx
=−
A
w0 L3
12
w0 5 w0 4 w0 L3
x +
x −
x + C4
60 L
24
12
[ x = L, y = 0] 0 = −
(a)
∴ C3 = −
7 w0 L4
120 EI
w0 L3
12 EI
∴ C4 =
7 w0 L4
120
w0
(2 x5 − 5 Lx 4 + 10 L4 x − 7 L5 ) 
120 EIL
yA =
θA =
7 w0 L4
↑ 
120 EI
w0 L3
12 EI

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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Esercizio 9.68
SOLUTION
M C = 0:
− 4.5RA + (20)(3)(1.5) − (60)(1.5) = 0
RA = 0
Forces in kN; lengths in m.
Units:
1
d2y
1
0
2
= M = 60 x − 1.5 − 90 x − 1.5 − (20) x − 1.5
2
2
dx
dy
1
EI
= 30 x − 1.5 2 − 90 x − 1.51 −   (20) x − 1.5 3 + C1
dx
6
EI
EIy = 10 x − 1.5 3 − 45 x − 1.5 2 −
1
(20) x − 1.5 4
24
+ C1x + C2
Boundary conditions:
Data:
[ x = 0, y = 0]
0 + 0 + 0 + 0 + C2 = 0
[ x = 4.5, y = 0]
(10)(3)3 − (45)(3) 2 −
1
(20)(3)4 + 4.5C1 + 0 = 0
24
C2 = 0
C1 = 45 kN ⋅ m 2
E = 200 × 109 Pa, I = 316 × 106 mm 4 = 316 × 10−6 m 4
EI = (200 × 109 )(316 × 10−6 ) = 63.2 × 106 N ⋅ m 2 = 63, 200 kN ⋅ m 2
(a)
Slope at A.
 dy

 dx at x = 0 


EIθ A = C1 = 45 kN ⋅ m 2
θA =
(b)
45
= 0.712 × 10−3 rad
63, 200
θ A = 0.712 × 10−3 rad

( y at x = 1.5)
Deflection at B.
EIyB = (C1)(1.5) = (45)(1.5) = 67.5 kN ⋅ m3
yB =
67.5
= 1.068 × 10−3 m
63, 200
yB = 1.068 mm ↑ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.69
SOLUTION
1
(50)(50)3 = 520.833 × 103 mm3 = 520.833 × 10−9 m
12
EI = (105 × 109 )(520.833 × 10−6 ) = 54.6875 × 103 N ⋅ m 2
I=
= 54.6875 kN ⋅ m 2
Units: Forces in kN; lengths in meters.
Compute deflection at B due to w. Case 8 of Appendix D.
( yB )1 = −
wL4
(30)(0.400) 4
=−
8 EI
(8)(54.6875)
= −1.75543 × 10−3 = −1.7553 mm
The displacement is more than δ 0 , the gap closes.
Let P be the contact force between points B and C.
Compute deflection of B due to P. Use Case 1 of Appendix D.
( yB )2 =
PL3
P(0.4)3
=
3EI (3)(54.6875)
= 390.095 × 10−6 P
m
Compute deflection of C due to P.
yC = −
PL3
P (0.25)3
=−
= −95.238 × 10−6 P
3EI
(3)(54.6875)
m
Displacement condition:
yB + δ 0 = yC
Using superposition,
( yB )1 + ( yB )2 − δ 0 = yC
−1.75543 × 10−3 + 390.095 × 10−6 P + 1.2 × 10−3 = −95.238 × 10−6 P
485.333 × 10−6 P = 0.55543 × 10−3
P = 1.14443 kN
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEM$ 9. (ContinuD
(a)
Reaction at A.
ΣFy = 0: RA − 12 + 1.14443 = 0
RA = 10.86 kN ↑ 
ΣM A = 0: M A − (0.2)(12) + (0.4)(1.14443) = 0
M A = 1.942 kN ⋅ m
(b)
Reaction at D.

ΣFy = 0: RD − 1.14443 = 0
RD = 1.144 kN ↑ 
ΣM D = 0: −M D + (0.25)(1.14443) = 0
M D = 0.286 kN ⋅ m

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.70
SOLUTION
Units:
Forces in kN; lengths in m.
E = 200 × 109 Pa
I = 40.1 × 106 mm 4 = 40.1 × 10−6 m 4
EI = (200 × 109 )(40.1 × 10−6 ) = 8.02 × 106 N ⋅ m 2
= 8020 kN ⋅ m 2
Draw M/EI diagram by parts.
M1
(18)(2.2)
=
= 4.9377 × 10−3 m −1
EI
8020
1
A1 = (4.9377 × 10 −3 )(2.2) = 5.4315 × 10−3
2
1
x1 = (2.2) = 0.7333 m
3
M2
(26)(2.7) 2
=−
= −11.8167 × 10−3 m −1
EI
(2)(8020)
1
(−11.8167 × 10−3 )(2.7) = −10.6350 × 10−3
3
1
x2 = (2.7) = 0.675 m
4
A2 =
Draw reference tangent at C.
θC = θ A + θC / A = θ A + A1 + A2 = 0
(a)
Slope at A.
θ A = − A1 − A2 = −5.4315 × 10−3 + 10.6350 × 10−3
= 5.20 × 10−3 rad
θ A = 5.20 × 10−3 rad




PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
PROBLEM$ 9. (ContinuD
(b)
Deflection at A.
y A = yC − θC L + t A / C
= 0 − 0 + A1x1 + A2 x2
= 0 − 0 + (5.4315 × 10−3 )(1.9667) − (10.6350 × 10−3 )(2.025)
= −10.85 × 10−3 m

y A = 10.85 mm ↓ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.71
SOLUTION
For W130 × 23.8, I x = 8.91 × 106 mm 4
EI = (200 × 106 kPa)(8.91 × 10−6 m 4 ) = 1782 kN ⋅ m 2
Let RB be the jack force in kN.
1
(1.8 RB )(1.8) = 1.62 RB
2
1
A2 = (−60)(3) = −90 kN ⋅ m 2
2
EI tC / A = (2.4) A1 + (2) A2
A1 =
0 = (2.4)(1.62 RB ) + (2)(−90)
RB = 46.296 kN
A1 = 75 kN ⋅ m 2
1
(−60)(1.8) = −54 kN ⋅ m 2
2
1
A4 = (−24)(1.8) = −21.6 kN ⋅ m 2
2
EIt B/A = (1.2) A1 + (1.2) A3 + (0.6) A4
A3 =
= (1.2)(75) + (1.2)(−54) + (0.6)(−21.6)
= 12.24 kN ⋅ m 2
(a)
Deflection at B.
(b)
Reaction at B.
yB = tB / A =
EI t B / A 12.24
=
= 6.8687 × 10−3 m
EI
1782
yB = 6.87 mm ↑ 
RB = 46.3 kN ↑ 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
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F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
Esercizio 9.74
SOLUTION
Symmetric beam and loading:
Spring force:
RC = RA
1
2
F = (2wL) = wL
3
3
ΣFy = 0: RA + F − 2wL + RC = 0
RA = RC =
2
wL
3
Draw M/EI diagram by parts.
A1 =
1  2 wL2 
1 wL3

 L =
2  3 EI 
3 EI
1  1 wL2 
1 wL3
A2 = − 
 L = −
3  2 EI 
6 EI
Place reference tangent at B.
θB = 0
yB = −t A/B
2
3 

= −  A1 ⋅ L + A2 ⋅ L 
3
4 

=−
7 wL4
72 EI
F = − kyB
k=−
F
=
yB
2
wL
3
7 wL4
72 EI
k=
48 EI

7 L3
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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