Download Report

PROBLEM 5.100
For the machine element shown, locate the y coordinate of the
center of gravity.
SOLUTION
For half-cylindrical hole,
r = 1.25 in.
4(1.25)
3π
= 1.470 in.
yIII = 2 −
For half-cylindrical plate,
r = 2 in.
4(2)
zIV = 7 +
= 7.85 in.
3π
V, in 3
y , in.
z , in.
y V , in 4
z V , in 4
3.5
–7.875
73.50
I
Rectangular plate
(7)(4)(0.75) = 21.0
–0.375
II
Rectangular plate
(4)(2)(1) = 8.0
1.0
2
8.000
16.00
III
–(Half cylinder)
(1.25) 2 (1) = 2.454
1.470
2
–3.607
–4.908
IV
Half cylinder
(2) 2 (0.75) = 4.712
–0.375
–1.767
36.99
V
–(Cylinder)
−π (1.25) 2 (0.75) = −3.682
1.381
–25.77
Σ
27.58
–3.868
95.81
−
π
2
π
2
–0.375
–7.85
7
Y ΣV = Σ yV
Y (27.58 in 3 ) = −3.868 in 4
Y = −0.1403 in.
PROPRIETARY MATERIAL. © 2013 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. If you are a student using this Manual,
you are using it without permission.
672
PROBLEM 5.101
For the machine element shown, locate the y coordinate of the
center of gravity.
SOLUTION
First assume that the machine element is homogeneous so that its center of gravity will coincide with the
centroid of the corresponding volume.
V , mm3
x , mm
y , mm
x V, mm 4
y V, mm 4
I
(100)(18)(90) = 162, 000
50
9
8,100,000
1,458,000
II
(16)(60)(50) = 48, 000
92
48
4,416,000
2,304,000
III
π (12)2 (10) = 4523.9
105
54
475,010
244,290
IV
−π (13) 2 (18) = −9556.7
28
9
–267,590
–86,010
Σ
204,967.2
12,723,420
3,920,280
We have
Y ΣV = Σ yV
Y (204,967.2 mm3 ) = 3,920, 280 mm 4
or Y = 19.13 mm
PROPRIETARY MATERIAL. © 2013 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. If you are a student using this Manual,
you are using it without permission.
673
PROBLEM 5.118
Three brass plates are brazed to a steel pipe to form the flagpole
base shown. Knowing that the pipe has a wall thickness of 8 mm
and that each plate is 6 mm thick, determine the location of the
center of gravity of the base. (Densities: brass = 8470 kg/m3;
steel = 7860 kg/m3.)
SOLUTION
Since brass plates are equally spaced, we note that
the center of gravity lies on the y-axis.
x =z =0
Thus,
π
[(0.064 m) 2 − (0.048 m) 2 ](0.192 m)
4
= 270.22 × 10−6 m3
V=
Steel pipe:
m = ρ V = (7860 kg/m3 )(270.22 × 10−6 m3 )
= 2.1239 kg
Each brass plate:
1
(0.096 m)(0.192 m)(0.006 m) = 55.296 × 10−6 m3
2
m = ρ V = (8470 kg/m3 )(55.296 × 10−6 m3 ) = 0.46836 kg
V=
Flagpole base:
Σm = 2.1239 kg + 3(0.46836 kg) = 3.5290 kg
Σ y m = (0.096 m)(2.1239 kg) + 3[(0.064 m)(0.46836 kg)] = 0.29382 kg ⋅ m
Y Σm = Σ y m :
Y (3.5290 kg) = 0.29382 kg ⋅ m
Y = 0.083259 m
Y = 83.3 mm above the base
PROPRIETARY MATERIAL. © 2013 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. If you are a student using this Manual,
you are using it without permission.
694