Capitolo 5
ANALISI E PROGETTO
DI TRAVI INFLESSE
F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, © 2014 McGraw-Hill Education (Italy) srl
ESERCIZIO 5.1
SOLUTION
Reactions:

From A to B:
ΣM C = 0: LA − bP = 0
A=
Pb
L
ΣM A = 0: LC − aP = 0
C =
Pa
L
0< x<a
ΣFy = 0:
Pb
−V = 0
L

Pb

L
V =
ΣM J = 0: M −
Pb
x=0
L
M =
From B to C:
Pbx

L
a< x< L
ΣFy = 0: V +
Pa
=0
L
V =−
ΣM K = 0: − M +
Pa
( L − x) = 0
L
M =
At section B:
Pa

L
Pa( L − x)

L
M =
Pab

L2
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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 5.2
SOLUTION
Reactions:
ΣM B = 0: − AL + wL ⋅
L
=0
2
A=
wL
2
ΣM A = 0:
L
=0
2
B=
wL
2
BL − wL ⋅
Free body diagram for determining reactions.
Over whole beam,
0< x< L
Place section at x.
Replace distributed load by equivalent concentrated load.
ΣFy = 0:
wL
− wx − V = 0
2
L

V = w − x  
2

ΣM J = 0: −
M =
wL
x
x + wx + M = 0
2
2
w
( Lx − x 2 )
2
M =
Maximum bending moment occurs at x =
w
x( L − x) 
2
L
.
2
M max =
wL2

8
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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 5.3
SOLUTION
From A to B (0 < x < a) :
 Fy = 0 : − wx − V = 0
V = −wx 
 M J = 0 : (wx)
x
+M =0
2
M =−
wx 2

2
From B to C (a < x < L) :
 Fy = 0 : − wa − V = 0
a

 M J = 0 : (wa)  x −  + M = 0
2

V = − wa 
a

M = −wa  x −  
2

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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 5.4
SOLUTION
ΣFy = 0:
ΣM J = 0:

−
1 w0 x
⋅ x −V = 0
2 L
V =−
w0 x 2

2L
M =−
w0 x3

6L
1 w0 x
x
⋅x⋅ +M = 0
2 L
3
At x = L,
V =−
w0 L
2
M =−
w0 L2
6
|V |max =
|M |max =
w0 L

2
w0 L2

6
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 5.5
SOLUTION
Reactions:
A = D = wa
From A to B:
0< x<a
 Fy = 0 :
wa − wx − V = 0
V = w(a − x) 
MJ = 0 :
− wax + (wx)
x
+M =0
2

x2 
M = w  ax −
 
2 

From B to C:
a< x< L−a
 Fy = 0 :
wa − wa − V = 0
V =0 
MJ = 0 :
From C to D:
 Fy = 0:
a

− wax + wa  x −  + M = 0
2

M =
1 2
wa 
2
L−a< x< L
V − w( L − x) + wa = 0
V = w( L − x − a) 
 M J = 0:
L − x
− M − w( L − x) 
 + wa( L − x) = 0
 2 
M = wa[( L − x) −
1
( L − x) 2 ] 
2
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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 5.6
SOLUTION
Calculate reactions after replacing distributed load by an equivalent
concentrated load.


Reactions are

A= D=


From A to B:

0< x<a
ΣFy = 0:

1
w ( L − 2a)
2
1
w ( L − 2a) − V = 0
2
V =
1
w ( L − 2a) 
2
1
ΣM = 0: − w ( L − 2a) + M = 0
2
M =
From B to C:
a< x< L−a
b=


ΣFy = 0:

x−a
2
Place section cut at x. Replace distributed load by equivalent concentrated
load.


1
w ( L − 2a ) x 
2
1
w ( L − 2a) − w ( x − a) − V = 0
2
L

V = w − x  
2

1
x − a
M J = 0: − w ( L − 2a) x + w ( x − a) 
+M =0
2
 2 


M =


1
w [( L − 2a) x − ( x − a) 2 ] 
2
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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 5.6 (ContinuD



From C to D:

L−a< x< L
ΣFy = 0: V +

1
w ( L − 2a ) = 0
2

V =−


ΣM J = 0: −M +




1
w ( L − 2a)( L − x) = 0
2
M =

At x =
L
,
2
w
( L − 2a ) 
2
1
w ( L − 2a)( L − x) 
2
 L2 a 2 
M max = w 
−
 
2 
 8

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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 5.8
SOLUTION
V = 200N, M = 0
At B,
At E + ,
 Fy = 0 :
V − 200 = 0 V = 200N
ME = 0 :
− M − (0.225)(200) = 0
M = −45 N ⋅ m
At D + ,
 Fy = 0 :
V + 500 − 200 = 0
(a) V = −300 N 
MD = 0 :
− M + (0.3)(500) − (0.525)(200) = 0
M = 45 N ⋅ m

At C + ,
 Fy = 0 :
 MC = 0 :
V − 200 + 500 − 200 = 0 V = −100N
− M − (0.225)(200) + (0.525)(500) − (0.75)(200) = 0
(b) M = 67.5 N ⋅ m 
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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 5.8 (Continued)
At A,
 Fy = 0 :
MA = 0 :
V − 200 − 200 + 500 − 200 = 0
V = 100 N
− M − (0.3)(200) − (0.525)(200) + (0.825)(500) − (1.05)(200) = 0
M = 37.5 N ⋅ m
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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 5.9
SOLUTION
Reactions:
 M C = 0 : − 2 A + (1)(24) − (1)(40) = 0
A = −8kN = 8kN ↓
MA = 0 :
2C − (1)(24) − (3)(40) = 0
C = 72 kN = 72 kN ↑
A to C.
0 < x < 2m
 Fy = 0 : − 8 − 12 x − V = 0 V = (−8 − 12 x) kN
MJ = 0 :
x
− 8x − (12 x)   − M = 0
2
M = (−8 x − 6 x 2 ) kN ⋅ m
C to B.
2m < x < 3m
 Fy = 0 : V − 40 = 0
V = 40 kN
MK = 0 :
− M − (3 − x)(40) = 0
M = (40 x − 120) kN ⋅ m
From the diagrams,
(a) V
(b)
M
max
max
= 40.0 kN 
= 40.0 kN ⋅ m 
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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 5.12
SOLUTION
M B = 0: − 0.750 RA + (0.550) (75) + (0.300) (75) = 0
Reaction at A:
RA = 85 N ↑
RB = 65 N ↑
Also,
A to C :
V = 85 N
C to D :
V = 10 N
D to B :
V = −65 N
At A and B,
M =0
Just to the left of C,
Σ M C = 0: − (0.25) (85) + M = 0
M = 21.25 N ⋅ m
Just to the right of C,
Σ M C = 0:
− (0.25) (85) + (0.050)(75) + M = 0
M = 17.50 N ⋅ m
Just to the left of D,
Σ M D = 0:
−(0.50) (85) + (0.300)(75) + M = 0
M = 20 N ⋅ m
Just to the right of D,
Σ M D = 0:
−M + (0.25) (65) = 0
M = 16.25 kN
(a)
(b)
|V |max = 85.0 N 
|M |max = 21.25 N ⋅ m 

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ESERCIZIO 5.14
SOLUTION
Over the whole beam,
ΣFy = 0: 1.5w − 1.5 − 1.5 = 0
A to C:
w = 2 kN/m
0 ≤ x < 0.3 m
ΣFy = 0: 2 x − V = 0
V = (2 x) kN
 x
ΣM J = 0: −(2 x)   + M = 0
2
At C −,
M = ( x 2 ) kN ⋅ m
x = 0.3 m
V = 0.6 kN, M = 0.090 kN ⋅ m
= 90 N ⋅ m
C to D:
0.3 m < x < 1.2 m
ΣFy = 0: 2 x − 1.5 − V = 0
V = (2 x − 1.5) kN
x
ΣM J = 0: − (2 x)   + (1.5)( x − 0.3) + M = 0
2
M = ( x 2 − 1.5x + 0.45) kN ⋅ m
At the center of the beam: x = 0.75 m
V =0
M = −0.1125 kN ⋅ m
= −112.5 N ⋅ m
At C +,
x = 0.3 m,
V = −0.9 kN
(a) Maximum |V | = 0.9 kN = 900 N 
(b)
Maximum |M | = 112.5 N ⋅ m 
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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 5.15
SOLUTION
Reaction at A:
M B = 0:
− 4.5 A + (3.0)(3) + (1.5)(3) + (1.8)(4.5)(2.25) = 0
A = 7.05 kN ↑
Use AC as free body.
ΣM C = 0: M C − (7.05)(1.5) + (1.8)(1.5)(0.75) = 0
M C = 8.55 kN ⋅ m = 8.55 × 103 N ⋅ m
I =
1 3
1
(80)(300)3 = 180 × 106 mm 4
bh =
12
12
= 180 × 10−6 m 4
c=
1
(300) = 150 mm = 0.150 m
2
σ =
Mc (8.55 × 103 )(0.150)
=
= 7.125 × 106 Pa
−6
I
180 × 10
σ = 7.13 MPa 
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ESERCIZIO 5.16
SOLUTION
Use CB as free body.
6
M C = 0: −M − (200)(6)   = 0
2
M = −3600 lb ⋅ ft
= −43.2 × 103 lb ⋅ in
For rectangular section,
I =
1 3
1
bh =
(4)(8)3 = 170.667 in 3
12
12
c=
1
h = 4 in.
2
σ =
|M |c (43.2 × 103 )(4)
=
= 1.0125 × 103 psi
I
170.667
σ = 1.013 ksi 
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ESERCIZIO 5.17
SOLUTION
Use portion CB as free body.
 MC = 0 :
− M + (3)(2.1)(1.05) + (8)(2.1) = 0
M = 23.415 kN ⋅ m = 23.415 × 103 N ⋅ m
For W310 × 60 : S = 844 × 103 mm3
= 844 × 10−6 m3
Normal stress: σ =
M
23.415 × 103
=
= 27.7 × 106 Pa
S
844 × 10−6
σ = 27.7 MPa 

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 5.18
SOLUTION
Reactions:
A= B
By symmetry,
 Fy = 0 :
A = B = 80 kN
Using left half of beam as free body,
MJ = 0 :
−(80)(2) + (30)(1.2) + (50)(0.4) + M = 0
M = 104 kN ⋅ m = 104 × 103 N ⋅ m
For
W310 × 52 : S = 747 × 103 mm3
= 747 × 10−6 m3
Normal stress:
σ =
M
104 × 103
=
= 139.2 × 106 Pa
−6
S
747 × 10
σ = 139.2 MPa 

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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 5.20
SOLUTION
Use entire beam as free body.
MB = 0 :
−4.8 A + (3.6)(216) + (1.6)(150) + (0.8)(150) = 0
A = 237 kN
Use portion AC as free body.
 MC = 0 :
M − (2.4)(237) + (1.2)(216) = 0
M = 309.6 kN ⋅ m
For W460 × 113, S = 2390 × 106 mm3
Normal stress:
σ =
309.6 × 103 N ⋅ m
M
=
S
2390 × 10−6 m3
= 129.5 × 106 Pa
σ = 129.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.22
SOLUTION
Reactions:
 M D = 0 : 4 A − 64 − (24)(2)(1) = 0
A = 28 kN
 Fy = 0 : − 28 + D − (24)(2) = 0 D = 76 kN
A to C:
0 < x < 2m
 Fy = 0 : − V − 28 = 0
V = −28 kN
 M J = 0 : M + 28x = 0
M = (−28 x) kN ⋅ m
C to D:
2m < x < 4m
 Fy = 0 : − V − 28 = 0
V = −28 kN
MJ = 0 :
M + 28x − 64 = 0
M = (−28x + 64) kN ⋅ m
D to B:
4m < x < 6m
 Fy = 0 :
V − 24(6 − x) = 0
V = (−24 x + 144) kN
MJ = 0 :
6 − x
− M − 24(6 − x) 
=0
 2 
M = −12(6 − x) 2 kN ⋅ m
max M = 56 kN ⋅ m = 56 × 103 N ⋅ m
S = 482 × 103 mm3
For S250 × 52 section,
Normal Stress: σ =
M
S
=
56 × 103 N ⋅ m
482 × 10 −6 m3
= 116.2 × 106 Pa
σ = 116.2 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.23
SOLUTION
Statics: Consider portion AB and BE separately.
Portion BE:
ME = 0 :
(96)(3.6) + (48)(3.3) − C (3) + (160)(1.5) = 0
C = 248kN ↑
E = 56 kN ↑
MA = MB = ME = 0
At midpoint of AB:
 Fy = 0 : V = 0
 M = 0 : M = (96)(1.2) − (96)(0.6) = 57.6 kN ⋅ m
Just to the left of C:
 Fy = 0 : V = −96 − 48 = −144 kN
 M C = 0 : M = −(96)(0.6) − (48)(0.3) = −72 kN
Just to the left of D:
 Fy = 0 : V = 160 − 56 = +104 kN
MD = 0 :
M = (56)(1.5) = +84 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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PROBLEM 5.23 (Continued)

From the diagram:

M
max
= 84 kN ⋅ m = 84 × 103 N ⋅ m 
For W310 × 60 rolled steel shape,
S x = 844 × 103 mm3
= 844 × 10−6 m3
Stress: σ m =
σm =
M
max
S
84 × 103
= 99.5 × 106 Pa
844 × 10−6
σ m = 99.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.24
SOLUTION
Reaction at A:
 M B = 0 : − 4.8 A + 40 + (25)(3.2)(1.6) = 0
A = 35 kN
A to C:
0 < x < 1.6 m
 Fy = 0 : 35 − V = 0 V = 35 kN
 M J = 0 : M + 40 − 35x = 0
M = (30 x − 40) kN ⋅ m
C to B:
1.6 m < x < 4.8m
 Fy = 0 : 35 − 25( x − 1.6) − V = 0
V = (−25 x + 75) kN
 M K = 0 : M + 40 − 35x
 x − 1.6 
+ (25)( x − 1.6) 
=0
 2 
M = (−12.5x 2 + 75x − 72) kN ⋅ m
Normal stress:
σ =
For W200 × 31.3, S = 298 × 103 mm3
M
40.5 × 103 N ⋅ m
=
= 135.9 × 106 Pa
S
298 × 10−6 m3
σ = 135.9 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.26
SOLUTION
By symmetry, A = B
Σ Fy = 0: A − 8 + 12 − 8 + B = 0
A = B = 2 kN
Shear:
A to C −:
V = 2 kN


C + to D −:
V = −6 kN 


D + to E −:
V = 6 kN 


E + to B :
V = −2 kN 

Bending moment:
Σ M C = 0: M C − (1)(2) = 0
At C,
M C = 2 kN ⋅ m 
At D,
+Σ M D = 0: M D − (2)(2) + (8) (1) = 0
M D − 4 kN ⋅ m 
By symmetry,
M = 2 kN ⋅ m at E.
M E = 2 kN ⋅ m 
max|M | = 4 kN ⋅ m occurs at E.
For W310 × 23.8,
Normal stress:
S x = 280 × 103 mm3 = 280 × 10−6 m3
σ max =
|M |max
4 × 103
=
Sx
280 × 10−6
= 14.29 × 106 Pa
σ max = 14.29 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.27
SOLUTION
By symmetry,
A=B
Σ Fy = 0: A − 8 + W − 8 + B = 0
A = B = 8 − 0.5W
Bending moment at C:
Σ M C = 0: −(8 − 0.5W )(1) + M C = 0
M C = (8 − 0.5W ) kN ⋅ m
Bending moment at D:
Σ M D = 0: − (8 − 0.5W )(2) + (8) (1) + M D = 0
M D = (8 − W ) kN ⋅ m
−M D = M C
Equate:
W − 8 = 8 − 0.5W
W = 10.67 kN 
(a)
W = 10.6667 kN
M C = −2.6667 kN ⋅ m
M D = 2.6667 kN ⋅ m = 2.6667.103 N ⋅ m
|M |max = 2.6667 kN ⋅ m
For W310 × 23.8 rolled steel shape,
S x = 280 × 103 mm3 = 280 × 10−6 m3
(b)
σ max =
|M |max
2.6667 × 103
=
= 9.52 × 106 Pa
Sx
280 × 10−6
σ max = 9.52 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.30
SOLUTION
P = 480 N
Q = 480 N
Σ M D = 0: − Aa + 480(a − 0.5)
Reaction at A:
− 480(1 − a) = 0
720 

A =  960 −
N
a 

Bending moment at C:
Σ M C = 0: − 0.5 A + M C = 0
360 

M C = 0.5 A =  480 −
N⋅m
a 

Bending moment at D:
Σ M D = 0: − M D − 480 (1 − a) = 0
M D = −480 (1 − a) N ⋅ m
(a)
Equate:
−M D = M C
480 (1 − a) = 480 −
360
a
a = 0.86603 m
A = 128.62 N
(b)
For rectangular section, S =
S =
M C = 64.31 N ⋅ m
a = 866 mm 
M D = −64.31 N ⋅ m
1 2
bh
6
1
(12)(13)2 = 648 mm3 = 648 × 10−9 m3
6
σ max =
|M |max
64.31
=
= 99.2 × 106 Pa
S
6.48 × 10−9
σ max = 99.2 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.31
SOLUTION
P = 480 N
Reaction at A:
Q = 320 N
Σ M D = 0: Aa + 480(a − 0.5) − 320(1 − a) = 0
560 

A =  800 −
N
a 

Bending moment at C:
Σ M C = 0: − 0.5 A + M C = 0
280 

M C = 0.5 A =  400 −
 N⋅m
a 

Bending moment at D:
Σ M D = 0: −M D − 320(1 − a) = 0
M D = (−320 + 320 a) N ⋅ m
(a)
−M D = M C
Equate:
320 a 2 + 80a − 280 = 0
320 − 320 a = 400 −
280
a
a = 0.81873 m, − 1.06873 m
a = 819 mm 
Reject negative root.
A = 116.014 N
(b)
M C = 58.007 N ⋅ m
M D = −58.006 N ⋅ m
1 2
bh
6
1
S = (12)(18) 2 = 648 mm3 = 648 × 10−9 m3
6
For rectangular section, S =
σ max =
|M |max
58.0065
=
= 89.5 × 106 Pa
S
648 × 10−9
σ max = 89.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.32
SOLUTION
Weight density: γ = ρ g
Let L = total length of beam.
W = AL ρ g = b 2 L ρ g
Reactions at C and D:
C = D=
W
2
Bending moment at C:
 L  W 
Σ M C = 0:    + M = 0
 6  3 
WL
M =−
18
Bending moment at center of beam:
 L  W   L  W 
Σ M E = 0:    −    + M = 0
 4  2   6  2 
max|M | =
S =
For a square section,
Normal stress:
Solve for b:
Data:
σ =
M =−
WL
24
WL b 2 L2 ρ g
=
18
18
1 3
b
6
|M | b 2 L2 ρ g /18 L2 ρ g
=
=
S
3b
b3 /6
b=
L = 3.6 m ρ = 7860 kg/m3
L2 ρ g
3σ
g = 9.81 m/s 2
(a) σ = 10 × 106 Pa
(b) σ = 50 × 106 Pa
(a)
b=
(3.6) 2 (7860) (9.81)
= 33.3 × 10−3 m
6
(3) (10 × 10 )
b = 33.3 mm 
(b)
b=
(3.6) 2 (7860) (9.81)
= 6.66 × 10−3 m
(3) (50 × 106 )
b = 6.66 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 5.34
SOLUTION
At A+,
ΣM C = 0: LA − bP = 0
A=
Pb
L
ΣM A = 0: LC − aP = 0
C =
Pa
L
V = A=
A to B − :
Pb
L
M =0
0< x<a
x
0 wdx = 0
w=0
V − VA = 0
a Pb
a
M B − M A = 0 Vdx =  0
At B +,
V = A−P =
B + to C:
Pb

L
V =
L
dx =
Pba
L
MB =
Pba

L
Pb
Pa
−P=−
L
L
a< x< L
x
a wdx = 0
w=0
VC − VB = 0
V =−
Pa

L
Pa
Pab
( L − a) = −
L
L
Pab
Pba Pab
= MB −
=
−
=0
L
L
L
L
M C − M B = a Vdx = −
MC
|M | max =
Pab

L
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ESERCIZIO 5.35
SOLUTION
Σ M B = 0: −AL + wL ⋅
Σ M A = 0: BL − wL ⋅
L
=0
2
L
=0
2
A=
wL
2
B=
wL
2
dV
= −w
dx
x
V − VA = −0 wdx = − wx
V = VA − wx = A − wx
V =
wL
− wx 
2
dM
=V
dx
x
x  wL

M − M A = 0 Vd x = 0 
− wx  dx
2


=
wLx wx 2
−
2
2
M = MA +
Maximum M occurs at x =
V =
wLx wx 2
−
2
2
M =
w
( Lx − x 2 ) 
2
1
, where
2
dM
=0
dx
|M |max =
wL2

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 5.36
SOLUTION
Over AB:
VA = 0 M A = 0
x
V = − 0 wdx = −wx
dM
= V = −wx
dx
x
Vdx
0
M =
At B:
x=a
Over BC:
w=0
dV
=0
dx
wx 2
=−
2
x
M =−
wx 2

2
MB = −
wa 2

2
0
VB = −wa
V = constant = VB
V = − wa 
dM
= V = − wa
dx
x
x
M − M B = a Vdx = − wax a = −wa( x − a)
M = − wa( x − a) −
wa 2
2
a

M = −wa  x −  
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 5.37
SOLUTION
x
L
w = w0
VA = 0,
MA = 0
dV
Wx
= −w = − 0
dx
L
x
V − VA = −0
w0 x
w x2
=− 0
2L
L





V =−
w0 x 2

2L
M =−
w0 x3

6L
dM
w x2
=V = − 0
dx
2L
x
x
M − M A = 0 V dx = −0


w0 x 2
dx
2L

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.38
SOLUTION
Reactions:
A = D = wa
A to B:
0< x<a
w=w
VA = A = wa,
MA = 0
x
V − VA = − 0 w dx = −wx
V = w(a − x) 
dM
= V = wa − wx
dx
M − MA =
x
x
0 Vdx = 0 (wa − wx)dx
M = wax −
VB = 0
B to C:
MB =
1 2
wx 
2
1 2
wa
2
a< x< L−a
V =0 
dM
=V =0
dx
M − MB =
M = MB
x
a V dx = 0
M =
1 2
wa 
2






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PROBLEM 5.38 (Continued)
C to D:
x
V − VC = −L − a w dx = −w[ x − ( L − a)]
V = −w[ x − ( L − a)] 
M − MC =
x
x
L − a V dx = L − a −[wx − (L − a)]dx
 x2

= − w  − ( L − a) x 
2

x
L−a
 x2

( L − a) 2
= − w  − ( L − a) x −
+ ( L − a) 2 
2
2

 x2
( L − a)2 
= − w  − ( L − a) x +

2
2

M =
 x2
1 2
( L − a) 2 
wa − w  − ( L − a) x +

2
2
2


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ESERCIZIO 5.39
SOLUTION
Reactions.
A= D =
1
w( L − 2a)
2
At A.
VA = A =
1
w( L − 2a), M A = 0
2
A to B.
0< x<a












VB = VA =
1
w( L − 2a)
2
MB =

B to C.
a
VB − VA = −
MB − MA =


w=0
0

a
0
w dx = 0
V dx =
0
1
w( L − 2a)dx
2
1
w( L − 2a)a
2
a< x< L−a
V − VB = −
V =

a

x
a
w=w
w dx = − w( x − a)
1
1
w( L − 2a) − w( x − a) = w( L − 2 x)
2
2
dM
1
= V = w( L − 2 x)
2
dx
M − MB =
=
M =
=
1
x
x
2
a V dx = 2 w ( Lx − x ) a
1
w( Lx − x 2 − La + a 2 )
2
1
1
w( L − 2a)a + w( Lx − x 2 − La + a 2 )
2
2
1
w( Lx − x 2 − a 2 )
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 5.39 (Continued)
1
1
VC = − w( L − 2a) M C = ( L − 2a)a
2
2
At C.
x= L−a
C to D.
1
V = VC = − w( L − 2a)
2
MD = 0
At x =
L
,
2
 L2 a 2 
M max = w 
−
 
2 
 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 5.41
SOLUTION
 M A = 0 : − M A − (0.3)(200) − (0.525)(200)
+ (0.825)(500) − (1.05)(200) = 0
M A = 37.5 N ⋅ m
 Fy = 0 : VA − 200 − 200 + 500 − 200 = 0
VA = 100 N
Shear:
A to C:
V = 100 N
C to D:
V = 100 − 200 = −100 N
D to E:
V = −100 − 200 = −300 N
E to B:
V = −300 + 500 = 200 N
Areas under shear diagram:
A to C:
 Vdx = (100)(0.3) = 30 N ⋅ m
C to D:
 Vdx = (−100)(0.225) = −22.5 N ⋅ m
D to E:
 Vdx = (−300)(0.3) = −90 N ⋅ m
E to B:
 Vdx = (200)(0.225) = 45 N ⋅ m
Bending moments:
M A = 37.5 N ⋅ m
MC = M A +
C
A V dx = 37.5 + 30 = 67.5 N ⋅ m
M D = MC +
D
C V dx = 67.5 − 22.5 = 45 N ⋅ m
ME = MD +
E
D V dx = 45 − 90 = −45 N ⋅ m
MB = ME +
D
E V dx = −45 + 45 = 0
(a) Maximum V = 300 N 
(b) Maximum M = 67.5 N ⋅ 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 5.42
SOLUTION
Reactions:
 M C = 0 : 2 A + (12)(2)(1) − (40)(1) = 0
A = 8kN ↓
 M A = 0 : 2C − (12)(2)(1) − (40)(3) = 0
C = 72 kN ↑
Shear diagram: VA = −8 kN
A to C:
0 < x < 2 m w = 12kN/m
VC − VA = −

2
0

2
wdx = − 12dx = −24 kN
0
VC = −24 − 8 = −32 kN
C to B:
VB = −32 + 72 = 40 kN
Areas of shear diagram:
1
( −8 − 32)(2) = −40 kN ⋅ m
2
A to C:
 Vdx =
C to B:
 Vdx = (1)(40) = 40 kN ⋅ m
Bending moments:
MA = 0
M C = M A +  Vdx = 0 − 40 = −40 kN ⋅ m
M B = M C +  Vdx = −40 + 40 = 0
(a) Maximum V = 40.0 kN 
(b) Maximum M = 40.0 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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ESERCIZIO 5.44
SOLUTION
Reaction at A:
ΣM B = 0: −3.0 A + (1.5)(3.0)(3.5) + (1.5)(3) = 0
A = 6.75 kN ↑
B = 6.75 kN ↑
Reaction at B:
Beam ACB and loading: (See sketch.)
Areas of load diagram:
A to C:
(2.4)(3.5) = 8.4 kN
C to B:
(0.6)(3.5) = 2.1 kN
Shear diagram:
VA = 6.75 kN
VC − = 6.75 − 8.4 = −1.65 kN
VC + = −1.65 − 3 = −4.65 kN
VB = −4.65 − 2.1 = −6.75 kN
Over A to C,
V = 6.75 − 3.5x
At G,
V = 6.75 − 3.5xG = 0 xG = 1.9286 m
Areas of shear diagram:
A to G:
1
(1.9286)(6.75) = 6.5089 kN ⋅ m
2
G to C:
1
(0.4714)(−1.65) = −0.3889 kN ⋅ m
2
C to B:
1
(0.6)(−4.65 − 6.75) = −3.42 kN ⋅ m
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 5.44 (Continued)
Bending moments:
MA = 0
M G = 0 + 6.5089 = 6.5089 kN ⋅ m
M C − = 6.5089 − 0.3889 = 6.12 kN ⋅ m
M C + = 6.12 − 2.7 = 3.42 kN ⋅ m
M B = 3.42 − 3.42 = 0
(a)
(b)
|V |max = 6.75 kN 
|M |max = 6.51 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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ESERCIZIO 5.45
SOLUTION
 M B = 0:
− 3 A + (1)(4) + (0.5)(4) = 0
A = 2 kN ↑
 M A = 0: 3B − (2)(4) − (2.5)(4) = 0
B = 6 kN ↑
Shear diagram:
A to C:
V = 2 kN
C to D:
V = 2 − 4 = −2 kN
D to B:
V = −2 − 4 = −6 kN
Areas of shear diagram:
 Vdx = (1)(2) = 2 kN ⋅ m
 Vdx = (1)(−2) = −2 kN ⋅ m
 Vdx = (1)(−6) = −6 kN ⋅ m
A to C:
C to D:
D to E:
Bending moments:
MA = 0
M C − = 0 + 2 = 2 kN ⋅ m
M C + = 2 + 4 = 6 kN ⋅ m
M D − = 6 − 2 = 4 kN ⋅ m
M D + = 4 + 2 = 6 kN ⋅ m
MB = 6 − 6 = 0
(a)
(b)
V
M
max
max
= 6.00 kN 
= 6.00 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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ESERCIZIO 5.46
SOLUTION
By symmetry, A = B.
ΣFy = 0: A + B − 3 − 3 − (4.5)(1.8) = 0
A = B = 7.05 kN
VA = 7.05 kN
Shear diagram:
A to C − :
w = 1.8 kN/m
−
V = 7.05 − (1.8)(1.5) = 4.35 kN
At C ,
At C +,
+
V = 4.35 − 3 = 1.35 kN
−
w = 1.8 kN/m
C to D :
−
At D ,
V = 1.35 − (1.5)(1.8) = −1.35 kN
At D +,
V = −1.35 − 3 = −4.35 kN
D + to B:
w = 1.8 kN
At B,
V = −4.35 − (1.5)(1.8) = −7.05 kN
Draw the shear diagram:
V = 0 at point E, the midpoint of CD.
Areas of the shear diagram:
A to C:
1
(7.05 + 4.35)(1.5) = 8.55 kN ⋅ m
2
C to E:
1
(1.35)(0.75) = 0.50625 kN ⋅ m
2
E to D:
1
(−1.35)(0.75) = −0.50625 kN ⋅ m
2
D to B:
1
(−4.35 − 7.05)(1.5) = −8.55 kN ⋅ m
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
PROBLEM 5.46 (Continued)
For a rectangular section,
Bending moments:
MA = 0
M C = 0 + 8.55 = 8.55 kN ⋅ m
M E = 8.55 + 0.50625 = 9.05625 kN ⋅ m
M D = 9.05625 − 0.50625 = 8.55 kN ⋅ m
M B = 8.55 − 8.55 = 0
3
M C = 8.55 × 10 N ⋅ m
1
1
S = bh 2 =   (80)(300) 2
6
6
= 1.2 × 106 mm3 = 1.2 × 10−3 m3
Maximum normal stress at C:
M C 8.55 × 103
=
S
1.2 × 10−3
= 7.125 × 106 Pa
σ=
σ = 7.13 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.48
SOLUTION
Reactions:
By symmetry, A = B.
 Fy = 0 : A = B = 80 kN ↑
Shear diagram:
A to C:
V = 80 kN
C to D:
V = 80 − 30 = 50 kN
D to E:
V = 50 − 50 = 0
Areas of shear diagram:
A to C:
 Vdx = (80)(0.8) = 64 kN ⋅ m
C to D:
 Vdx = (50)(0.8) = 40 kN ⋅ m
D to E:
 Vdx = 0
Bending moments:
MA = 0
M C = 0 + 64 = 64 kN ⋅ m
M D = 64 + 40 = 104 kN ⋅ m
M E = 104 + 0 = 104 kN ⋅ m
M
max
= 104 kN ⋅ m = 104 × 103 N ⋅ m
For W310 × 52, S = 747 × 103 mm3 = 747 × 10−6 m3
Normal stress:
σ =
M
104 × 103
=
= 139.2 × 106 Pa
S
747 × 10−6
σ = 139.2 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.50
SOLUTION
w0 x kw0 ( L − x)
wx
−
= (1 + k ) 0 − kw.
L
L
L
w0 x
= − w = kw0 − (1 + k )
L
2
wx
= kw0 x − (1 + k ) 0 + C1
2L
= 0 at x = 0 C1 = 0
w=
dV
dx
V
V
w x2
dM
= V = kw0 x − (1 + k ) 0
dx
2L
kw0 x 2
w x3
− (1 + k ) 0 + C2
2
6L
M = 0 at x = 0 C2 = 0
M=
M=
(a)
w0 x 2

L
w x 2 w x3

M= 0 − 0
2
3L
k = 1.
V = w0 x −
x = L.
Maximum M occurs at
(b)
kw0 x 2 (1 + k ) w0 x3
−
2
6L
k=
M
1
.
2
V = 0 at

At
2
x = L,
3
At
x = L,
M=
x=
w0 ( 32 L )
4
2
−
max
=
w0 L2

6
V=
w0 x 3w0 x 2
−

2
4L
M=
w0 x 2 w0 x3
−

4
4L
2
L
3
w0 ( 32 L )
4L
3
=
w0 L2
= 0.03704 w0 L2
27
M =0
|M |max =
w0 L2

27
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.51
SOLUTION
dV
x
= − w = − w0
dx
L
2
1
x
dM
V = − w0
+ C1 =
2
L
dx
3
1
x
M = − w0
+ C1 x + C2
6
L
(a)
M = 0 at
x=0
C2 = 0
M = 0 at
x=L
1
0 = − w0 L2 + C1 L
6
1
x2 1
V = − w0
+ w0 L2
2
L 6
M max occurs when
1
w0 L
6
V=
1
x3 1
M = − w0
+ w0 Lx
6
L 6
(b)
C1 =
1
w0 ( L2 − 3x 2 )/L 
6
M=
1
w0 ( Lx − x3 /L) 
6
dM
= V = 0. L2 − 3 xm2 = 0
dx
xm =
L
3
M max =
1  L2
L2 
−
w0 

6  3 3 3 
M max = 0.0642w0 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 5.52
SOLUTION
πx
dV
= − w = − w0 sin
dx
L
w0 L
dM
πx
V=
cos
+ C1 =
L
dx
π
2
wL
πx
M = 0 2 sin
+ C1 x + C2
L
π
M = 0 at x = 0
C2 = 0
M = 0 at x = L
0 = 0 + C1 L + 0
C1 = 0
V=
(a)
M=
dM
= V = 0 at
dx
(b)
M max =
w0 L2
π2
sin
x=
π
2
w0 L
π
w0 L2
π
2
cos
πx
sin
πx
L
L


L
2
M max =
w0 L2
π2

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.55
SOLUTION
 M C = 0 : (2)(1) − (3)(4)(2) + 4B = 0
B = 5.5 kN
 M B = 0 : (5)(2) + (3)(4)(2) − 4C = 0
C = 8.5 kN
Shear:
A to C:
V = −2 kN
C+ :
V = −2 + 8.5 = 6.5 kN
B:
V = 6.5 − (3)(4) = −5.5 kN
Locate point D where V = 0.
d
4−d
=
6.5
5.5
d = 2.1667 m
12d = 26
4 − d = 3.8333 m
Areas of the shear diagram:
A to C:
 Vdx = (−2.0)(1) = −2.0 kN ⋅ m
C to D:
 Vdx =
1
(2.16667)(6.5) = 7.0417 kN ⋅ m
2
D to B:
 Vdx =
1
(3.83333)(−5.5) = −5.0417 kN ⋅ m
2
Bending moments:
MA = 0
M C = 0 − 2.0 = −2.0 kN ⋅ m
M D = −2.0 + 7.0417 = 5.0417 kN ⋅ m
M B = 5.0417 − 5.0417 = 0
Maximum M = 5.0417 kN ⋅ m = 5.0417 × 103 N ⋅ 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
PROBLEM 5.55 (Continued)
For pipe:
I =
Normal stress:
1
1
1
1
do = (160) = 80 mm, ci = di = (140) = 70 mm
2
2
2
2
co =
π
(c
4
4
o
)
− ci4 =
π
4
4
6
4
(80) − (70)  = 13.3125 × 10 mm
4
S =
I
13.3125 × 106
=
= 166.406 × 103 mm3 = 166.406 × 10−6 m3
co
80
σ =
M
5.0417 × 103
=
= 30.3 × 106 Pa
S
166.406 × 10−6
σ = 30.3 MPa 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.57
SOLUTION
w=0
MD = 0 :
− 4 RA + (2)(250) − (2)(150) = 0
RA = 50 kN ↑
MA = 0 :
4RD − (2)(250) − (6)(150) = 0
RD = 350 kN ↑
Shear:
VA = 50 kN
A to C:
V = 50 kN
C to D:
V = 50 − 250 = −200 kN
D to B:
V = −200 + 350 = 150 kN
Areas of shear diagram:
A to C:
 Vdx = (50)(2) = 100 kN ⋅ m
C to D:
 Vdx = (−200)(2) = −400 kN ⋅ m
D to B:
 Vdx = (150)(2) = 300 kN ⋅ m
Bending moments: M A = 0
M C = M A +  Vdx = 0 + 100 = 100 kN ⋅ m
M D = M C +  Vdx = 100 − 400 = −300 kN ⋅ m
M B = M D +  Vdx = −300 + 300 = 0
Maximum M = 300 kN ⋅ m = 300 × 103 N ⋅ m
For W410 × 114 rolled steel section,
σm =
M
max
Sx
=
S x = 2200 × 103 mm3 = 2200 × 10−6 m3
300 × 103
= 136.4 × 106 Pa
2200 × 10−6
σ m = 136.4 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.58
SOLUTION
Reaction:
 M B = 0 : − 4 A + 60 + (80)(1.6)(2) − 12 = 0
A = 76 kN ↑
Shear:
VA = 76 kN
V = 76 kN 
A to C :
VD = 76 − (80)(1.6) = −52 kN
V = −52 kN
D to C :
Locate point where V = 0:
V ( x) = −80 x + 76 = 0
x = 0.95 m
Areas of shear diagram:
A to C:  Vdx = (1.2)(76) = 91.2 kN ⋅ m
C to E:  Vdx =
1
(0.95)(76) = 36.1 kN ⋅ m
2
E to D:  Vdx =
1
(0.65)(−52) = −16.9 kN ⋅ m
2
D to B:  Vdx = (1.2)(−52) = −62.4 kN ⋅ m
Bending moments:
M A = −60 kN ⋅ m
M C = −60 + 91.2 = 31.2 kN ⋅ m
M E = 31.2 + 36.1 = 67.3 kN ⋅ m

M D = 67.3 − 16.9 = 50.4 kN ⋅ m
M B = 50.4 − 62.4 = −12 kN ⋅ m
For W250 × 80, S = 983 × 103 mm3
Normal stress:
σ max =
M
67.3 × 103 N ⋅ m
=
= 68.5 × 106 Pa
−6 3
S
983 × 10 m
σ m = 68.5 MPa 
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 5.60
SOLUTION
(a)
Replace distributed load by equivalent concentrated load at the centroid of the area of the load diagram.
For the triangular distribution, the centroid lies at x =
(a)
 Fy = 0 : RD − W = 0
RD =
1
w0 L
2
2L
.
3
W =
1
w0 L
2
 M C = 0 : 0 = 0 equilibrium

V = 0, M = 0, at x = 0
0< x<
2L
,
3
dV
wx
= −w = − 0
dx
L
dM
w x2
w x2
= V = − 0 + C1 = − 0
dx
2L
2L
M =−
V =−
Just to the left of C,
w0 x3
w x3
+ C2 = − 0
6L
6L
w0 (2 L / 3)2
2
= − w0 L
2L
9
2
5
V = − w0 L + RD =
w0 L
9
18
Just to the right of C,
Note sign change. Maximum M occurs at C.
Maximum M =
σm =
M
max
I
w0 (2L / 3)3
4
= − w0 L2
6L
81
4
w0 L2
81
For square cross section,
(b)
MC = −
c
I =
1 4
a
12
c=
1
a
2
3
=
4 w0 L2 6
8 w0 L2  2  w0 L2
=
= 
3
3
81 a
27 a3
3 a
σm =
w0 L2

(1.5a)3
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 5.61
SOLUTION

1
(18)(36)3 = 69.984 × 103 mm 4
12
1
c = d = 18 mm
2
I
S = = 3.888 × 103 mm3 = 3.888 × 10−6 m3
c
I=



At A,
M A = Sσ A = (3.888 × 10−6 )(−56.9) = −221.25 N ⋅ m
At C,
M C = Sσ C = (3.888 × 10−6 )( −29.9) = −116.25 N ⋅ m
ΣM A = 0: 221.23 − (0.1)(400) − 0.2 P − 0.325Q = 0
0.2 P + 0.325Q = 181.25

(1)
ΣM C = 0: 116.25 − (0.05)(200) − 0.1P − 0.225Q = 0

0.1P + 0.225Q = 106.25
Solving (1) and (2) simultaneously,
(2)
P = 500 N 
Q = 250 N 
RA − 400 − 500 − 250 = 0 RA = 1150 N ⋅ m
Reaction force at A:
VA = 1150 N
VD = 250
M A = −221.25 N ⋅ m
M C = −116.25 N ⋅ m
M D = −31.25 N ⋅ m
|V |max = 1150 N 
|M |max = 221 N ⋅ m 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.62
SOLUTION
At D,
1
(24)(60)3 = 432 × 103 mm 4 c = 30 mm
12
I
S = = 14.4 × 103 mm3 = 14.4 × 10−6 m3 M = Sσ
c
M D = (14.4 × 10−6 )(55 × 106 ) = 792 N ⋅ m
At F,
M F = (14.4 × 10−6 )(37.5 × 106 ) = 540 N ⋅ m
(a)
I=
Using free body FB,
ΣM F = 0: −540 + 0.3B = 0
B=
540
= 1800 N
0.3
ΣM D = 0: −792 − 3Q + (0.8)(1800) = 0
Using free body DEFB,
Q = 2160 N
Using entire beam,
ΣM A = 0: − 0.2 P − (0.7)(2160) + (1.2)(1800) = 0
P = 3240 N
ΣFy = 0: A − 3240 − 2160 + 1800 = 0
A = 3600 N
Shear diagram and its areas:
A to C − :
+
−
C to E :
+
E to B:
V = 3600 N
AAC = (0.2)(3600) = 720 N ⋅ m
V = 3600 − 3240 = 360 N
ACE = (0.5)(360) = 180 N ⋅ m
V = 360 − 2160 = −1800 N
AEB = (0.5)(−1800) = −900 N ⋅ m
Bending moments:
MA = 0
M C = 0 + 720 = 720 N ⋅ m
M E = 720 + 180 = 900 N ⋅ m
|M |max = 900 N ⋅ m
M B = 900 − 900 = 0
(b)
Normal stress.
σ max =
|M |max
900
=
= 62.5 × 106 Pa
S
14.4 × 10−6
σ max = 62.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.64
SOLUTION
Reactions:
ΣM D = 0: − 2.4 A + (1.6)(1.8) + (0.8)(3.6) = 0
A = 2.4 kN
ΣM A = 0: − (0.8)(1.8) − (1.6)(3.6) + 2.4 D = 0
D = 3 kN
Construct shear and bending moment diagrams:
|M |max = 2.4 kN ⋅ m = 2.4 × 103 N ⋅ m
σ all = 12 MPa
= 12 × 106 Pa
Smin =
|M |max
σ all
=
2.4 × 103
12 × 106
= 200 × 10−6 m3
= 200 × 103 mm3
1
1
S = bh 2 = (40)h 2
6
6
= 200 × 103
(6)(200 × 103 )
40
= 30 × 103 mm 2
h2 =
h = 173.2 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 5.65
SOLUTION
 Fy = 0 : A − (1.2)(18) = 0
Reactions:
A = 21.6 kN ↑
 M A = 0 : − M A − (1.8)(1.2)(18) = 0
M A = −38.88 kN ⋅ m
VA = VB = 21.6 kN
Shear diagram:
VC = 21.6 − (1.2)(18) = 0
Areas of shear diagram:
A to B : (1.2)(21.6) = 25.92 kN ⋅ m
B to C :
1
(1.2)(21.6) = 12.96 kN ⋅ m
2
Bending moments: M A = −38.88 kN ⋅ m
M B = −38.88 + 25.92 = −12.96 kN ⋅ m
M C = −12.96 + 12.96 = 0
M
max
= 38.88 kN ⋅ m = 38.8 × 103 N ⋅ m
σ max =
S =
M
max
σ max
=
6S
=
b
max
S
38.8 × 103 N ⋅ m
= 3240 × 10−6 m3 = 3240 × 103 mm3
6
12 × 10 Pa
For a rectangular section, S =
h=
M
1 2
bh
6
6(3240 × 103 )
= 394 mm
125
h = 394 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 5.66
SOLUTION
By symmetry, A = D.
Reactions:
1
1
(3)(1.5) − (6)(1.5) − (3)(1.5) − D = 0
2
2
A = D = 6.75 kips ↑
 Fy = 0 : A −
Shear diagram:
VB = 6.75 −
VA = 6.75 kips
1
(3)(1.5) = 4.5 kips
2
VC = 4.5 − (6)(1.5) = −4.5 kips
VD = −4.5 −
1
(3)(1.5) = −6.75 kips
2
Locate point E where V = 0 :
By symmetry, E is the midpoint of BC.
Areas of the shear diagram:
A to B : (3)(4.5) +
2
(3)(2.25) = 18 kip ⋅ ft
3
B to E :
1
(3)(4.5) = 6.75 kip ⋅ ft
2
E to C :
1
(3)(−4.5) = −6.75 kip ⋅ ft
2
C to D : By antisymmetry, − 18 kip ⋅ ft
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 5.6 (ContinuD)
Bending moments: M A = 0
M B = 0 + 18 = 18 kip ⋅ ft
M E = 18 + 6.75 = 24.75 kip ⋅ ft
M C = 24.75 − 6.75 = 18 kip ⋅ ft
M D = 18 − 18 = 0
σ max =
M
max
S
S =
M
max
σ max
For a rectangular section,
=
(24.75 kip ⋅ ft)(12 in/ft)
= 169.714 in 3
1.750 ksi
S =
1 2
bh
6
h=
6S
=
b
6(169.714)
= 14.27 in.
5
h = 14.27 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.68
SOLUTION
By symmetry,
B=C
ΣFy = 0: B + C + 2.5 + 2.5 − (3)(6) = 0
B = C = 6.5 kN
Shear:
V = 2.5 kN
VB+ = 2.5 + 6.5 = 9 kN
VC − = 9 − (3)(6) = −9 kN
A to B:
V = −9 + 6.5 = −2.5 kN
C to D:
Areas of the shear diagram:
 Vdx = (0.6)(2.5) = 1.5 kN ⋅ m
1
 Vdx =  2  (1.5)(9) = 6.75 kN ⋅ m
A to B:
B to E:
 Vdx = −6.75 kN ⋅ m
 Vdx = −1.5 kN ⋅ m
E to C:
C to D:
Bending moments:
MA
MB
ME
MC
MD
=0
= 0 + 1.5 = 1.5 kN ⋅ m
= 1.5 + 6.75 = 8.25 kN ⋅ m
= 8.25 − 6.75 = 1.5 kN ⋅ m
= 1.5 − 1.5 = 0
Maximum |M | = 8.25 kN ⋅ m = 8.25 × 103 N ⋅ m
σ all = 12 MPa = 12 × 106 Pa
Smin =
For a rectangular section,
|M |max
σ all
=
8.25 × 103
= 687.5 × 10−6 m3 = 687.5 × 103 mm3
6
12 × 10
1
S = bh 2
6
1

687.5 × 103 =   (100) h 2
6
(6)(687.5 × 103 )
= 41.25 × 103 mm 2
h2 =
100
h = 203 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 5.69
SOLUTION
Shear:
M B = 0: − 2.4 A + (0.6)(3.6)(3) = 0
A = 2.7 kN
M A = 0: −(1.8)(3.6)(3) + 2.4 B = 0
B = 8.1 kN
VA = 2.7 kN
VB − = 2.7 − (2.4)(3) = −4.5 kN
VB+ = −4.5 + 8.1 = 3.6 kN
VC = 3.6 − (1.2)(3) = 0
d
2.4 − d
=
2.7
4.5
d = 0.9 m
Locate point D where V = 0.
7.2 d = 6.48
2.4 − d = 1.5 m
Areas of the shear diagram:
1
A to D:
 Vdx =  2  (0.9)(2.7) = 1.215 kN ⋅ m
D to B:
 Vdx =  2  (1.5)(−4.5) = −3.375 kN ⋅ m
B to C:
 Vdx =  2  (1.2)(3.6) = 2.16 kN ⋅ m
Bending moments:
1
1
MA = 0
M D = 0 + 1.215 = 1.215 kN ⋅ m
M B = 1.215 − 3.375 = −2.16 kN ⋅ m
M C = −2.16 + 2.16 = 0
Maximum |M | = 2.16 kN ⋅ m = 2.16 × 103 N ⋅ m
Smin =
For rectangular section,
|M |
σ all
=
σ all = 12 MPa = 12 × 106 Pa
2.16 × 103
= 180 × 10−6 m3 = 180 × 103 mm3
6
12 × 10
1
1
S = bh 2 = b(150)2 = 180 × 103
6
6
b=
(6)(180 × 103 )
1502
b = 48.0 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 5.72
SOLUTION
18 − 6 

w = 6 +
x  = (6 + 2 x) kN/m
6


dV
= −w = −6 − 2 x
dx
V = −6 x − x 2 + C1
V = 0 at
x = 0, C1 = 0
dM
= V = −6 x − x 2
dx
1
M = −3x 2 − x3 + C2
3
M = 0 at x = 0, C2 = 0
M = −3x 2 −
M
M
max
max
1 3
x
3
occurs at x = 6 m.
1
= −(3)(6)2 −   (6)3 = 80 kN ⋅ m = 180 × 103 N ⋅ m
3
σ all = 160 MPa = 160 × 106 Pa
Smin =
Shape
W530 × 66
M
σ all
=
180 × 103
= 1.125 × 10−3 m3 = 1125 × 103 mm3
160 × 106
S, ( 103 mm3 )
W460 × 74
1340 ←
1460
W410 × 85
1510
W360 × 79
1270
W310 × 107
1600
W250 × 101
1240
Lightest acceptable wide flange beam: W530 × 66 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.73
SOLUTION
Shape
Section modulus
σ all = 160 Mpa
Smin =
M max
σ all
=
286 kN ⋅ m
= 1787 × 10−6 m3
160 MPa
= 1787 × 103 mm3
S, ( 103 mm3 )
W610 × 101
2520
W530 × 92
2080 ←
W460 × 113
2390
W410 × 114
2200
W360 × 122
2020
W310 × 143
2150

Use W530 × 92 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.74
SOLUTION
Reaction:
 M D = 0 : − 10 A + (7.5)(60) + (5)(40) = 0
A = 65 kN ↑
Shear diagram:
A to B:
V = 65 kN
B to C:
V = 65 − 60 = 5 kN
C to D:
V = 5 − 40 = −35 kN
Areas of shear diagram:
A to B: (2.5)(65) = 162.5 kN ⋅ m
B to C: (2.5)(5) = 12.5 kN ⋅ m
C to D: (5)(−35) = −175 kN ⋅ m
Bending moments: M A = 0
M B = 0 + 162.5 = 162.5 kN ⋅ m
M C = 162.5 + 12.5 = 175 kN ⋅ m
M D = 175 − 175 = 0
M
max
= 175 kN ⋅ m = 175 × 103 N ⋅ m
σ all = 160 MPa = 160 × 106 Pa
Shape
Sx, ( 103 mm3 )
S610 × 119
2870
S510 × 98.2
1950
S460 × 81.4
1460 ←
Smin =
M
σ all
=
175 × 103
= 1093.75 × 10−6 m3
160 × 106
= 1093.75 × 103 mm3
Lightest S-section:
S460 × 81.4 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.75
SOLUTION
Reactions:
By symmetry,
B = C.
 Fy = 0 : − 70 + B − (9)(45) + C − 70 = 0
B = C = 272.5 kN ↑
VA = −70 kN
Shear:
VB − = −70 + 0 = −70 kN
VB + = −70 + 272.5 = 202.5 kN
VC − = 202.5 − (9)(45) = −202.5 kN
VC + = −202.5 + 272.5 = 70 kN
VD = 70 kN
Draw shear diagram. Locate point E where V = 0 .
E is the midpoint of BC.
Areas of the shear diagram:
A to B:
 Vdx = (3)(−70) = −210 kN ⋅ m
B to E:
 Vdx = 2 (4.5)(202.5) = 455.625 kN ⋅ m
E to C:
 Vdx = 2 (4.5)(−202.5) = −455.625 kN ⋅ m
C to D:
 Vdx = (3)(70) = 210 kN ⋅ m
1
1

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 5.7 (ContinuD)
Bending moments: M A = 0
M B = 0 − 210 = −210.5 kN ⋅ m
M E = −210 + 455.625 = 245.625 kN
M C = 245.625 − 455.625 = −210 kN
M D = −210 + 210 = 0
M
max
S =
S( 103 mm3 )
S610 × 119
2870
S510 × 98.2
1950 ←
S460 × 104
1690
= 245.625kN ⋅ m = 245.625 × 103 N ⋅ m
σ all = 160 MPa = 160 × 106 Pa
σ =
Shape
Lightest S-shape S510 × 98.2 
M
S
M
σ
=
245.625 × 103
= 1.5352 × 10−3 m3
6
160 × 10
= 1535.2 × 103 mm3
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.80
SOLUTION
 M A = 0 : − M A − (1)(1.5) − (1.5)(1.5) − (2)(1.5) = 0
M
max
M A = −6.75 kN ⋅ m
= M A = 6.75 kN ⋅ m
Smin =
Smin =
I m in =
M
max
σ all
=
6.75 × 103 N ⋅ m
= 45 × 10−6 m3 = 45 × 103 mm3
6
150 × 10 Pa
I min
c2
π
4
I min = c2 Smin = (50)(45 × 103 ) = 2.25 × 106 mm 4
(c
4
2
4
c1max
= c24 −
4
− c1max
4
π
)
I min = (50) 4 −
4
π
(2.25 × 106 ) = 3.3852 × 106 mm 4
c1max = 42.894 mm
tmin = c2 − c1max = 50 − 42.894 = 7.106 mm
t = 9 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 5.83
SOLUTION
By symmetry, B = C
Reactions.
 Fy = 0 : B + C − 0.9w = 0
B = C = 0.45w ↑
VA = 0
Shear:
VB − = 0 − 0.2w = −0.2w
VB + = −0.2w + 0.45w = 0.25w
VC − = 0.25w − 0.5w = −0.25w
VC + = −0.25w + 0.45w = 0.2w
VD = 0.2w − 0.2w = 0
Areas:
A to B.
1
(0.2)(−0.2 w) = −0.02w
2
B to E
1
(0.25)(0.25w) = 0.03125w
2
Bending moments:
MA = 0
M B = 0 − 0.02 w = −0.02 w
M E = −0.02w + 0.03125w = 0.01125w
Centroid and moment of inertia:
Part
A, mm 2
y , mm
Ay (103 mm3 )
d , mm.
Ad 2 (103 mm 4 ) I (103 mm 4 )

1200
70
84
20
480
40

1200
30
36
20
480
360
Σ
2400
960
400
Y =
120 × 103
= 50 mm
2400
I =
 Ad 2 +  I = 1360 × 103 mm 4
120
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 5.8 (Continued)
Top:
I /y = (1360 × 103 )/30 = 45.333 × 103 mm3 = 45.333 × 10−6 m3
Bottom:
I /y = (1360 × 103 ) / (−50) = −27.2 × 103 mm3 = −27.2 × 10−6 m3
Bending moment limits ( M = −σ I / y ) and load limits w.
Tension at B and C:
−0.02 w = −(80 × 106 ) (45.333 × 10−6 )
w = 181.3 × 103 N/m
Compression at B and C:
−0.02 w = −(−130 × 106 ) (27.2 × 10−6 )
w = 176.8 × 103 N/m
Tension at E:
0.01125 w = −(80 × 106 ) (27.2 × 10−6 )
Compression at E:
0.01125 w = −(−130 × 10) (45.333 × 10−6 ) w = 523.8 × 103 N/m
The smallest allowable load controls:
w = 193.4 × 103 N/m
w = 176.8 × 103 N/m
w = 176.8 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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ESERCIZIO 5.84
SOLUTION
By symmetry, B = C
Reactions:
Σ Fy = 0: B + C − 0.9 w = 0
B = C = 0.45 w ↑
VA = 0
Shear:
VB − = 0 − 0.2 w = −0.2 w
VB + = −0.2 w + 0.45 w = 0.25 w
VC − = 0.25 w − 0.5 w = −0.25 w
VC + = −0.25 w + 0.45 w = 0.2 w
VD = 0.2 w − 0.2 w = 0
Areas:
1
(0.2)(−0.2 w) = −0.02 w
2
1
(0.25) (0.25 w) = 0.03125 w
2
A to B:
B to E:
MA = 0
Bending moments:
M B = 0 − 0.02 w = −0.02 w
M E = −0.02 w + 0.03125 w = 0.01125 w
Centroid and moment of inertia:
Part
A, mm 2
y , mm
Ay , (103 mm3 )
d , mm
Ad 2 (103 mm 4 )
I , (103 mm 4 )

1200
50
60
20
480
360

1200
10
12
20
480
40
Σ
2400
960
400
72
Y =
72 × 103
= 30 mm
2400
I = Σ Ad 2 + ΣI = 1360 × 103 mm3
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.8 (Continued)
Top:
I /y = (1360 × 103 ) / (50) = 27.2 × 103 mm3 = 27.2 × 10−6 m3
Bottom:
I /y = (1360 × 103 ) / (−30) = −45.333 × 108 mm3 = −45.333 × 10−6 m3
Bending moment limits ( M = −σ I / y ) and load limits w.
Tension at B and C:
−0.02 w = −(80 × 106 ) (27.2 × 10−6 )
w = 108.8 × 103 N/m
Compression at B and C:
−0.02 w = −( −130 × 106 ) (−45.333 × 10−6 )
w = 294.7 × 103 N/m
Tension at E:
0.01125 w = −(80 × 106 ) (−45.333 × 10−6 )
w = 322.4 × 103 N/m
Compression at E:
0.01125 w = −( −130 × 106 ) (27.2 × 10−6 )
w = 314.3 × 103 N/m
The smallest allowable load controls:
w = 108.8 × 103 N/m
w = 108.8 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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ESERCIZIO 5.85
SOLUTION
For W460 × 74,
S = 1460 × 103 mm3 = 1460 × 10−6 m3
σ all = 140 MPa = 140 × 106 Pa
M all = Sσ all = (1460 × 10−6 )(140 × 106 )
= 204.4 × 103 N ⋅ m = 204.4 kN ⋅ m
A = B,
Reactions: By symmetry,
C=D
+↑ ΣFy = 0: A + B − (6)(66) = 0
A = B = 198 kN = 198 × 103 N
+ΣFy = 0: C + D − 66l = 0
C = D = (33l ) kN
(1)
Shear and bending moment in beam AB:
0 < x < a,
V = 198 − 66 x kN
M = 198 x − 33x 2 kN ⋅ m
At C, x = a.
M = M max
M = 198a − 33a 2 kN ⋅ m
Set M = M all .
198a − 33a 2 = 204.4
33a 2 − 198a + 204.4 = 0
a = 4.6751 m ,
(a)
By geometry,
From (1),
1.32487 m
l = 6 − 2a = 3.35 m
l = 3.35 m 
C = D = 110.56 kN
Draw shear and bending moment diagrams for beam CD. V = 0 at point E, the midpoint of CD.
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5. (Continued)
1 
1
 Vdx = 2 (110.560)  2 l  = 92.602 kN ⋅ m
Area from A to E:
M E = 92.602 kN ⋅ m = 92.602 × 103 N ⋅ m
Smin =
ME
σ all
=
92.602 × 103
= 661.44 × 10−6 m3
140 × 106
= 661.44 × 103 mm3
Shape
S (103 mm3 )
W410 × 46.1
774
W360 × 44
693
W310 × 52
748
W250 × 58
693
W200 × 71
709
←
(b) Use W360 × 44. 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.87
SOLUTION
M B = MC = 0
1
VB = −VC =   (7.2) w = 3.6w
2
Area B to E of shear diagram:
1
 2  (3.6) (3.6 w) = 6.48 w
 
M E = 0 + 6.48 w = 6.48 w
Centroid and moment of inertia:
Part
A (mm 2 )
y (mm)
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )

2500
156.25
390625
34.82
3.031 × 106
0.0326 × 106

1875
140625
46.43
4.042 × 106
3.516 × 106
Σ
4375
7.073 × 106
3.548 × 106
75
531250
531250
= 121.43 mm
4375
I = ΣAd 2 + ΣI = 10.621 × 106 mm 4
Y =
Location
Top
Bottom
y (mm)
41.07
−121.43
I / y (103 mm3 )
← also (10−6 m3 )
258.6
−87.47
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 5. (Continued)
M = −σ I / y
Bending moment limits:
Tension at E:
− (110 × 106 ) (−87.47 × 10−6 ) = 9.622 × 103 N ⋅ m
Compression at E:
− (−150 × 10−6 )(258.6 × 10−6 ) = 38.8 × 103 N ⋅ m
Tension at A and D:
− (110 × 106 ) (258.6 × 10−6 ) = −28.45 × 103 N ⋅ m
Compression at A and D: − (−150 × 106 )(−87.47 × 10−6 ) = −13.121 × 103 N ⋅ m
(a)
Allowable load w:
Shear at A:
6.48 w = 9.622 × 103
w = 1.485 × 103 N/m
w = 1.485 kN/m 
VA = (a + 3.6) w
Area A to B of shear diagram:
1
1
a (VA + VB ) = a(a + 7.2) w
2
2
1
Bending moment at A (also D): M A = − a(a + 7.2) w
2
1
− a (a + 7.2)(4.485 × 103 ) = −13.121 × 103
2
(b)
Distance a:
1 2
a + 3.6a − 8.837 = 0
2
a = 1.935 m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.88
SOLUTION
M B = MC = 0
VB = −VC = P
Area B to E of shear diagram: 2.4 P
M E = 0 + 2.4 P = 2.4 P = M F
Centroid and moment of inertia:
Part
A (mm 2 )
y (mm)
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )

2500
156.25
390625
34.82
3.031 × 106
0.0326 × 106

1875
140625
46.43
4.042 × 106
3.516 × 106
Σ
4375
7.073 × 106
3.548 × 106
75
531250
531250
= 121.43 mm
4375
I = ΣAd 2 + ΣI = 10.621 × 106 mm 4
Y =
Location
Top
Bottom
y (mm)
I / y (103 mm3 )
41.07
258.6
−121.43
← also (10−6 m3 )
−87.47
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5. (Continued)
M = −σ I / y
Bending moment limits:
− (110 × 106 ) ( −87.47 × 10−6 ) = 9.622 × 103 N ⋅ m
Tension at E and F:
Compression at E and F:
−(110 × 106 ) (258.6 × 10−6 ) = −28.45 × 103 N ⋅ m
Tension at A and D:
Compression at A and D:
(a)
Allowable load P:
−( −150 × 106 )( −87.47 × 10−6 ) = −13.121 × 103 N ⋅ m
2.4 P = 9.622 × 103
Shear at A:
(b)
−(−150 × 106 )(258.6 × 10−6 ) = 38.8 × 103 N ⋅ m
P = 4.01 × 103 N
VA = P
Area A to B of shear diagram:
aVA = aP
Bending moment at A:
M A = −aP = −4.01 × 103 a
Distance a:
P = 4.01 kN 
−4.01 × 103 a = −13.121 × 103
a = 3.27 m 

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ESERCIZIO 5.89
SOLUTION
L = 48 ft a = 14 ft
P1 = 24 kips
P2 = 6 kips W = 0.75 kip/ft
Dead load:
1
RA = RB =   (48)(0.75) = 18 kips
2
Area A to E of shear diagram:
1
 2  (8)(18) = 216 kip ⋅ ft
 
M max = 216 kip ⋅ ft = 2592 kip ⋅ in at point E.
Live load:
u=
aP2
(14)(6)
=
= 1.4 ft
2( P1 + P2 ) (2)(30)
L
− u = 24 − 1.4 = 22.6 ft
2
x + a = 22.6 + 14 = 36.6 ft
L − x − a = 48 − 36.6 = 11.4 ft
x=
ΣM B = 0: −48 RA + (25.4)(24) + (11.4)(6) = 0
RA = 14.125 kips
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 5. (Continued)
Shear:
A to C:
V = 14.125 kips
C to D:
V = 14.125 − 24 = −9.875 kips
D to B:
V = −15.875 kips
Area:
(22.6)(14.125) = 319.225 kip ⋅ ft
A to C:
M C = 319.225 kip ⋅ ft = 3831 kip ⋅ in
Bending moment:
Design:
γ D M D + γ L M L = ϕ M U = ϕσ U Smin
Smin =
=
γ DM D + γ LM L
ϕσ U
(1.25)(2592) + (1.75)(3831)
(0.9)(60)
= 184.2 in 3

Shape
S (in 3 )
W 30 × 99
269
W 27 × 84
213
W 24 × 104
258
W 21 × 101
227
W18 × 106
204
←
Use W 27 × 84. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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ESERCIZIO 5.90
SOLUTION
L = 48 ft
a = 14 ft
P1 = 24 kips
P2 = 6 kips W = 0.75 kip/ft
See solution to Prob. 5.94 for calculation of the following:
M D = 2592 kip ⋅ in
M L = 3831 kip ⋅ in
For rolled steel section W27 × 84, S = 213 in 3
Allowable live load moment M L* :
γ D M D + γ L M L* = ϕ M U = ϕσ U S
ϕσ U S − γ D M D
M L* =
γL
(0.9)(60)(213) − (1.25)(2592)
1.75
= 4721 kip ⋅ in
=
Ratio:
M L* 4721
=
= 1.232 = 1 + 0.232
M L 3831
Increase 23.2%. 
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ESERCIZIO 5.91
SOLUTION
L = 16 m, wD = 350 N/m = 0.35 kN/m
wL = 600 N/m = 0.6 kN/m, P = 6 kN
Dead load:
1
RA =   (16)(0.35) = 2.8 kN
2
Area A to C of shear diagram:
1
 2  (8)(2.8) = 11.2 kN ⋅ m
 
Bending moment at C:
11.2 kN ⋅ m = 11.2 × 103 N ⋅ m
Live load:
1
RA = [(16)(0.6) + 6] = 7.8 kN
2
Shear at C −:
V = 7.8 − (8)(0.6) = 3 kN
Area A to C of shear diagram:
1
 2  (8)(7.8 + 3) = 43.2 kN ⋅ m
 
Bending moment at C:
43.2 kN ⋅ m = 43.2 × 103 N ⋅ m
Design:
γ D M D + γ L M L = ϕ M U = ϕσ U S
S=
γ D M D + γ L M L (1.2)(11.2 × 103 ) + (1.6)(43.2 × 103 )
=
ϕσ U
(0.9)(50 × 106 )
= 1.8347 × 10−3 m3 = 1.8347 × 106 mm3
For a rectangular section,
1
S = bh 2
6
h=
6S
(6)(1.8347 × 106 )
=
b
75
h = 383 mm 
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ESERCIZIO 5.93
SOLUTION
RA = RB =
ΣM J = 0: −
M=
S=
For a rectangular cross section,
Equating,
(a)
At x =
L
,
2
For x >
(b)
P
↑
2
P
x+M =0
2
L

0 < x < 2 


Px
2
M
σ all
=
Px
2σ all
1
S = bh 2
6
1 2
Px
bh =
6
2σ all
h=
h = h0 =
3PL
2σ all b
3Px
σ allb
h = h0
2x
L
, 0< x< 
L
2
L
, replace x by L − x.
2
Solving for P,
P=
2σ all bh02 (2)(72 × 106 )(0.025)(0.200)2
=
= 60 × 103 N
3L
(3)(0.8)
P = 60 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 5.94
SOLUTION
A= B
By symmetry,
ΣFy = 0: A −
For 0 ≤ x ≤
L
,
2
At x = 0,
w=
2w0 x
L
V=
1
w0 L :
4
1
w0 L + B = 0
2
2w x
dV
= −w = − 0
dx
L
C1 =
M = 0:
M=
(a)
L
At x = ,
2
1
w0 L ↑
4
V = C1 −
w0 x 2
L
1
w0 L
4
w x2
dM
1
= V = w0 L − 0
dx
4
L
At x = 0,
A=B=
M = C2 +
1
1 w0 x3
w0 Lx −
4
3 L
C2 = 0
1 w0
(3L2 − 4 x3 )
2 L
3
1 w0  2  L 
L  1
M = MC =
3L   − 4    = w0 L2
12 L 
2
 2   12
M
For constant strength,
S=
For a rectangular section,
1
S = bh 2
6
σ all
,
S0 =
M0
σ all
MC
=
1
S
M
=
= 3 (3L2 x − 4 x3 )
S0 M 0 L
σ all
1
S0 = bh02
6
S  h 
= 
S0  h0 
2
h = h0
(b)
Data:
L = 800 mm
h0 = 200 mm
b = 25 mm
3L2 x − 4 x3

L3
σ all = 72 MPa
1
1
S0 = bh02 = (25)(200)2 = 166.667 × 103 mm3 = 166.667 × 10−6 m3
6
6
M C = σ all S0 = (72 × 106 )(166.667 × 10−6 ) = 12 × 103 N ⋅ m
w0 =
12 M C
L2
=
(12)(12 × 103 )
= 225 × 103 N/m
2
(0.800)
w0 = 225 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 5.97
SOLUTION
wx
dV
= −w = − 0
dx
L
2
wx
dM
V =− 0 =
2L
dx
3
wx
M =− 0
6L
w x3
|M |
= 0
S=
σ all 6 Lσ all
For a rectangular cross section,
At x = L,
w0 x3
6L
1
S = bh 2
6
w x3
1 2
bh = 0
6
6 Lσ all
Equating,
|M |=
h = h0 =
h=
w0 x3
σ allbL
w0 L2
σ allb
x
h = h0  
L
(a)
Data:
3/2

L = 750 mm = 0.75 m, b = 30 mm = 0.030 m
w0 = 300 kN/m = 300 × 103 N/m, σ all = 200 MPa = 200 × 106 Pa
(b)
h0 =
(300 × 103 )(0.75) 2
= 167.7 × 10−3 m
6
(200 × 10 )(0.030)
h0 = 167.7 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 5.98
SOLUTION
dV
πx
= − w = − w0 sin
dx
2L
2w L
πx
V = 0 cos
+ C1
2L
π
V = 0 at
x = 0 → C1 =
2 w0 L
π
2w L 
dM
πx
= V = − 0 1 − cos

dx
2L 
π 
2 w0 L 
2w L 
2L
2L
πx
πx
sin
|M | = 0  x −
sin
x−


L 
2L 
π 
π
π 
π
|M | 2 w0 L 
2L
πx
=
S=
x−
sin
2 L 
σ all πσ all 
π
M =−
1
S = bh 2
6
For a rectangular cross section,
πx
1 2 2 w0 L 
2L
bh =
x−
sin
πσ all 
π
6
2 L 
Equating,
1/2
12w0 L 
2L
π x 
h=
sin
x−

2L 
π
 πσ all b 
At x = L,
2
12 w0 L
h = h0 = 
 πσ all b
1/ 2
2  

1 −  
 π  
= 1.178
w0 L2
σ allb
1/2
 x 2
π x   2 
h = h0  − sin
 1 − 
2 L   π  
 L π
(a)
Data:
1/2
πx
x 2
h = 1.659 h0  − sin
2 L 
L π

L = 750 mm = 0.75 m, b = 30 mm = 0.030 m
w0 = 300 kN/m = 300 × 103 N/m, σ all = 200 MPa = 200 × 106 Pa
(b)
h0 = 1.178
(300 × 103 )(0.75) 2
= 197.6 × 10−3 m
6
(200 × 10 )(0.030)
h0 = 197.6 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 5.99
SOLUTION
P
2
RA = RB =
0< x<
1
2
P
x+M =0
2
M x
or M = max
1.2
ΣM J = 0:
M=
−
Px
2
Bending moment diagram is two straight lines.
At C,
SC =
1 2
bhC
6
M C = M max
Let D be the point where the thickness changes.
At D,
SD =
1 2
bhD
6
MD =
M max xD
1.2
2
S D hD2  100 mm 
1 M D xD
= 2 =
=
 = =
SC hC  200 mm 
4 M C 1.2
l
= 1.2 − xD = 0.9
2
xD = 0.3 m
l = 1.800 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 5.100
SOLUTION
RA = RB =
0.8 N
= 0.4 w
2
Shear:
A to C:
V = 0.4 w
D to B:
V = −0.4 w
Areas:
A to C:
(0.8)(0.4) w = 0.32 w
C to E:
1
 2  (0.4)(0.4) w = 0.08 w
 
Bending moments:
M C = 0.40 w
At C,
M = 0.40 wx
A to C:
At C,
SC =
1 2
bhC
6
M C = M max = 0.40 w
Let F be the point were the thickness changes.
At F,
SF =
1 2
bhF
6
M F = 0.40 wxF
2
S F hF2  100 mm 
1 M F 0.40 wxF
= 2 =
=
 = =
SC hC  200 mm 
4 MC
0.40 w
xF = 0.25 m
l
= 1.2 − xF = 0.95 m
2
l = 1.900 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 5.105
SOLUTION
A
=
B
= 160RkN ↑R
x
+M =0
2
M = 160 x − 20 x 2 kN ⋅ m
ΣM J = 0: −160 x + (40 x)
At center of beam,
x = 4m
At D,
x=
At center of beam,
M C = 320 kN ⋅ m
1
(8 − l ) = 1.5 m
2
M D = 195 kN ⋅ m
I = I beam + 2 I plate
2


1
 457 7.5 
= 333 × 106 + 2 (200)(7.5) 
+
+ (200)(7.5)3 

2  12
 2


= 494.8 × 106 mm 4
457
+ 7.5 = 236 mm
2
I
S = = 2097 × 103 mm3 = 2097 × 10−6 m3
c
c=
(a)
Normal stress:
σ=
320 × 103
M
=
= 152.6 × 106 Pa
−6
S
2097 × 10
σ = 152.6 MPa 
(b)
At D,
S = 1460 × 103 mm3 = 1460 × 10−6 m3
Normal stress:
σ=
195 × 103
M
=
= 133.6 × 106 Pa
S 1460 × 10−6
σ = 133.6 MPa 
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 5.107
SOLUTION
1
wL ↑
2
1
x
Σ M J = 0: − wLx + wx + M = 0
2
2
w
M = ( Lx − x 2 )
2
w
= x( L − x)
2
RA = RB =
where w = 160 kN/m and L = 1.2 m.
h = a + kx
For the tapered beam,
a = 120 mm
300 − 120
k=
= 300 mm/m
0.6
1
1
For a rectangular cross section, S = bh 2 = b(a + kx) 2
6
6
σ=
Bending stress:
M 3w Lx − x 2
=
S
b (a + kx)2
dσ
To find location of maximum bending stress, set dx = 0.
dσ 3w d  Lx − x 2  3w  (a + kx) 2 ( L − 2 x) − ( Lx − x 2 )2(a + kx)k 
=

=


dx
b dx  (a + kx) 2  b 
(a + kx) 4

=
3w  (a + kx)( L − 2 x) − 2k ( Lx − x 2 ) 


b 
(a + kx)3

=
3w  aL + kLx − 2ax − 2kx2 − 2kLx + 2kx2

b 
( a + kx)3
=
3w  aL − 2ax + kLx 

=0
b  (a + k x)3 



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 5.1 (Continued)
(a)
xm =
aL
(120) (1.2)
=
2a + kL (2) (120) + (300) (1.2)
xm = 0.240 m 
hm = a + k xm = 120 + (300)(0.24) = 192 mm
1
1
Sm = bhm2 = (20) (192)2 = 122.88 × 103 mm3 = 122.88 × 10−6 m3
6
6
w
160 × 103
M m = xm ( L − xm ) =
(0.24) (0.96) = 18.432 × 103 N ⋅ m
2
2
(b)
Maximum bending stress:
σm =
M m 18.432 × 103
=
= 150 × 106 Pa
−6
S m 122.88 × 10
σ m = 150.0 MPa 
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 5.115
SOLUTION
ΣM B = 0: −2.5 A + (1.75)(1.5)(16)
0
=
A = 16.8 kN
ΣM A = 0: −(0.75) + (1.5)(16) + 2.5B = 0
B = 7.2 kN
Shear diagram:
VA = 16.8 kN
VC = 16.8 − (1.5)(16) = −7.2 kN
VB = −7.2 kN
Locate point D where V = 0.
1.5 − d
d
24d = 25.2
=
16.8
7.2
d = 1.05 m 1.5 − d = 0.45 m
Areas of the shear diagram:
1
A to D:
 Vdx =  2  (1.05)(16.8) = 8.82 kN ⋅ m
D to C:
 Vdx =  2  (0.45)(−7.2) = −1.62 kN ⋅ m
C to B:
 Vdx = (1)(−7.2) = −7.2 kN ⋅ m
1
Bending moments:
MA = 0
M D = 0 + 8.82 = 8.82 kN ⋅ m
M C = 8.82 − 1.62 = 7.2 kN ⋅ m
M B = 7.2 − 7.2 = 0
Maximum |M | = 8.82 kN ⋅ m = 8.82 × 103 N ⋅ m
For S150 × 18.6 rolled steel section, S = 120 × 103 mm3 = 120 × 10−6 m3
Normal stress:
σ=
|M | 8.82 × 103
=
= 73.5 × 106 Pa
S
120 × 10−6
σ = 73.5 MPa 
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 5.118
SOLUTION
ΣM D = 0: −3.2 B + (24)(3.2)(50) = 0
B = 120 kN
ΣM B = 0: 3.2 D − (0.8)(3.2)(50) = 0
D = 40 kN
VA = 0
Shear:
VB− = 0 − (0.8)(50) = −40 kN
VB+ = −40 + 120 = 80 kN
VC = 80 − (2.4)(50) = −40 kN
VD = −40 + 0 = −40 kN
Locate point E where V = 0.
e 2.4 − e
=
80
40
e = 1.6 m
Areas:
120e = 192
2.4 − e = 0.8 m
1
A to B :
 Vdx =  2  (0.8)(−40) = −16 kN ⋅ m
B to E :
 Vdx =  2 (1.6)(80) = 64 kN ⋅ m
E to C :
 Vdx =  2  (0.8)(−40) = −16 kN ⋅ m
C to D :
 Vdx = (0.8)(−40) = −32 kN ⋅ m
Bending moments:
1
1
MA = 0
M B = 0 − 16 = −16 kN ⋅ m
M E = −16 + 64 = 48 kN ⋅ m
M C = 48 − 16 = 32 kN ⋅ m
M D = 32 − 32 = 0
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.1 (Continued)
Maximum |M | = 48 kN ⋅ m = 48 × 103 N ⋅ m
σ all = 160 MPa = 160 × 106 Pa
Smin =
Shape
S (103 mm3 )
W 310 × 32.7 415
W 250 × 28.4 308 ←
W 200 × 35.9 342
|M |
σ all
=
48 × 103
= 300 × 10−6 m3 = 300 × 103 mm3
160 × 106
Lightest wide flange beam:
W 250 × 28.4 @ 28.4 kg/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
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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 5.122
SOLUTION
V = −P =
dM
dx
Let
d = d0 + k x
For a solid circular section,
I=
π
4
c4 =
π
64
M = − Px
d3
d
I π 3 π
S= =
d = ( d 0 + k x )3
2
c 32
32
dS 3π
3π 2
=
(d 0 + k x) 2 k =
d k
32
dx 32
c=
Stress:
At H,
σ=
|M | Px
=
S
S
dσ
1 
dS 
= 2  PS − PxH
=0
dx S 
dx 
dS π 3
3π 2
S − xH
d − xH
d k
=
dx 32
32
1
1
1
k xH = d = ( d 0 + k H x H ) k xH = d 0
3
3
2
d = d0 +
1
3
d0 = d0
2
2
d = 1.5d0 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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