IN SI UNITS
November 2017
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These solutions represent a preliminary version
of the Instructors' Solutions Manual (ISM). It is
possible and even likely that at this preliminary
stage of preparing the ISM there are some
omissions and errors in the draft solutions.
These will be corrected and this manual will be
republished.
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Spring force is given by F(x) = 157 x + 0.2 x3
(1)
F = 1000 N
With x in mm,
1000 = 157 x* + 0.2 (x*)3
or
0.2 (x*)3 + 157 x*1000 = 0
(2)
The roots of the cubic equation (2) can be found from MATLAB as
x* = 6.08,3 ± 28.5i
Thus x* is given by the real root of Eq (2) and is 6.08 mm.
(a)
Equivalent linear spring constant of rubber mounting at its static equilibrium
position, using Eq. (1.7), is
k eq 
dF
dx
 0.6 (x *)2  157
x*
 0.6(6.08)2  157  179.2 N mm
(b)
Deflection of rubber mounting corresponding to keq is
k
F
1000

 5.58 mm
Keq 179.2
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F(x) = 34.6 x + 0.34 x2 + 0.002 x3
The steady-state deflection is 12.7 mm = x*
The force extended on the spring is therefore
F = 34.6 (12.7) + 0.34 (12.7)2 + 0.002 (12.7)3 = 498.3 N
Equivalent linear spring constant at its steady deflection is given by Eq (1.7):
k eq 
dF
dx
 34.6  0.68 x  0.006 x 2
x  x*
 34.6  0.68(12.7)  0.68(12.7)2
 44.2 N mm
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k
rA2
where r  1.4 for air
v
12000 
(1.5  106 )(1.4) A2
v

A2
 5.71  103
v
Let diameter of piston = d = 6 cm; A 
A2  8  106 
(0.006)2
 2.82  103 m2
4
8  106
 v  .0014 m 3
3
5.71  10
d 2 l
 Al
4
2.82  10-3 l  0.0014 m 3
Volume of cylinder 
l  0.496 m
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For a helical spring, k 
Gd4
64nR 3
sinc e D 
k STEEL 
(80  109 )(0.05)4
 231.5 k N m
8  10  (0.3)3
k ALUM 
(24  109 )(0.025)4
 7.5 k N m
(8)(10)(0.25)3
R
2
k
Gd4
8nD3
(a)
Spring 2 inside spring 1: keq = k1 + k2 = 239 kN/m
(b)
Spring 2 on top of spring 1:
which gives k eq 
For a helical spring, k 
k1 k 2
k1  k 2

Gd4
64nR 3
1
1
1


k eq k1 k2
(231.5)(7.5)
 7.26 kN m
231.5  7.5
D
k1 
(80  109 )(0.025)4
 14.5 k N m
(8)(10)(0.3)3
k2 
(24  109 )(0.0125)4
 0.468 k N m
(8)(10)(0.25)3
(a)
keq = k1 + k2 = 14.968 kN/m
(b)
k eq 
k1 k 2
k1  k 2
R
Gd4
k
2
8nD3
 0.453 kN m
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x2  y sin 30
x1  y cos 30
Equivalence of strain energies:
1
1
1
k eq y2  k2 x 22  k1 x12
2
2
2
1
1
 k1 y2 cos2 30  k2 y2 sin2 30
2
2
i.e., k eq 
k1 
k2 
A1 E 1
l1
A2 E2
l2
3
1
k1  k 2
4
4

(0.252  0.242 )(207  109 )
4

 318.5 m N m
2.5

(0.182  0.1682 )(207  109 )
4

 357.1 m N m
1.9
Therefore, k eq 
3
1
(318.5)  (357.1)  328.2 m N m
4
4
Similarly, the equivalent dampting constant can be found as (using equivalence of kinetic
energies):
c eq 
3
1
3
1
c1  c2  (70)  (52)  65.5 N  s / m
4
4
4
4
Stainless steel: E = 200 × 109 N/m2, G = 86 × 109 N/m2
For each tube:
D = 8.25 mm = 0.00825 m
d = 8.00 mm = 0.008 m
l = 1.3 m
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Axial stiffness for a single tube
ka 
AE  (D2  d2 )E  (0.008252  0.0082 )(200  109 )


 490.6 kN / m.
l
4
l
4
13
Torsional stiffness for a single tube
kT 
G
32 l
(D4  d4 ) 
(86  104 )(0.008254  0.008 4 )
 3.48 N  m / rad.
(1.3)32
Total kA for 6 tubes = 6 KA = 6 × 490.6 = 2.94 mN/m
Total kT for 6 tubes = 6 KT = 6 × 3.48 = 20.88 N  m/rad
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 3 D3 l  
2d 
c  
 1 
 ; D=diameter of piston, l = axial length of piston, d = radial clearance
3
D
 4d  
µ = 0.30 Pa ∙ s
Let d = 0.02 mm, D = 70 mm, and above equation gives
18  106  0.30
 2  0.02  
3 l(703 ) 

1

 
3 
4(0.02) 
 70  
Therefore, l = 0.229 m = 22 cm
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x(t) = 0.4 sin 4t + 5.0 cos 4t cm
(1)
By expressing Eq. (1) as
x(t) = A sin(4t + ) = A (sin4t cos + cos4t sin)
we obtain
A cos = 0.4
(2)
A sin = 5.0
(3)
Eqs. (2) and (3) yield;
tan  
5.0
 12.5    85.4261  1.4910 rad
0.4
and
A
0.4
0.4

 5.0181
cos  0.0797
 x(t)  5.0181 sin(4t  1.4910)cm
Velocity:
x  20.0724 cos(4t  1.4910) cm/s
Acceleration:

x  80.2896 sin(4t  1.4910) cm s
2
Amplitudes of displacement, velocity and acceleration are:
x max  5.0181 cm, x max  20.0724 cm s, x max  80.2896 cm s
2
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Speed = 100 rpm
In a minute, a point will be subjected to the maximum pressure,
A = pmax = 690 kPa, 100 × 4 = 400 times. Hence
period   
60
 0.15 s
400
pressure, p(t)


A, 0  t 

4
p(t)  
0,   t  
 4

2
2
A = pmax
0

A
 345 kPa
a 0   p(t) dt  A(t) 
 0

2
4
0
τ
4
τ 5τ
4
2τ 9τ
4
time, t

2A  sin m t 
A
m

 
a m   p(t) cos m t dt 
sin
 0
  m 
m
2
0
2
4



2A  cos m t 
A 
m

 
 1
bm   p(t) sin m t dt  
cos

m 
 0
  m 
2

0
2
4

Evaluation of am and bm:
m=1
m=3
A
A
3
a3 
sin
sin   0
 73.2 kPa

3
2
2 
2
A

A
A
3
(cos
 1)  73.2 kPa
b1   (cos  1)  219.6 kPa b2  
(cos   1)  219.6 kPa b3  

3
2
2
2
a1 
A
sin
p(t) 
a0
2

m=2

A
 219.6 kPa
a2 

  (a m cos m t  bm sin m t) Pa
m 1
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Speed = 200 rpm
In a minute, a point will be subjected to the maximum pressure,
A = pmax = 690 kPa, 200 × 6 = 1200 times.
Hence
60
 0.05 s .
1200
period   
pressure, p(t)
A = pmax


A, 0  t 

4
p(t)  
0,   t  
 4

2
0

A
a 0   p(t) dt  A(t)04 
 345 kPa
2
 0

2
τ
4
τ 5τ
4
2τ 9τ
4
time, t

4
2A  sin m t 
2
A
m


sin
a m   p(t) cos m t dt 

2
m
 0
  m 
0


4

2A  cos m t 
A 
m
2
 

bm   p(t) sin m t dt  
cos
 1

2
m 
 0
  m 

0

Evaluation of am and bm:
m=1
a1 
A

sin


A
 219.5 kPa
2 

A

b1    cos  1   219.5 kPa
2
 

p(t) 
a0
2
m=2
m=3
A
A
3
 73.2 kPa
sin   0
a3 
sin
2
3
2

A
A 
3
 1   73.2 kPa
b2  
(cos   1)  219.5 kPa b3  
 cos
3 
2
2

a2 

  (a m cos m t  bm sin m t) Pa
m 1
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Goal: Mass to be maintained at 2.5 kg/min +/0.25 kg/min
Parameters to be determined: Angular velocity of crank (), lengths of crank and connecting
rod, dimensions of the wedge, dimensions of the orifice in the hopper, dimensions of the
actuating rod and dimensions of the lever arrangement.
Given: Density of the material in the hopper.
Procedure:
Select  based on available motor. Determine the dimensions of the orifice in the hopper which
delivers approximately 2.5 kg/min (assuming continuous flow of material). For trial dimensions
of the wedge, determine the increase/decrease in the size (diameter) of the orifice. Choose the
final dimensions of the wedge such that the material flow rate delivered by the orifice lies within
the specified range.
Force to be applied = 900 N, frequency = 50 Hz = 314.16 rad/s
Procedure:
1.
Select a motor that provides either directly or through a gear system the desired
frequency. Assume that it is connected to the cam.
2.
Determine the sizes and dimensions of the plate cam and roller.
3.
Choose the dimensions of the follower.
4.
Select the mass as 90 kg. From the geometry, determine the range of displacements
(vertical motion) of the mass.
5.
Determine the force exerted due to the falling mass.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
© 2018 Pearson Education Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.