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DESIGN OF MACHINERY SOLUTION MANUAL 10-5-1

Rectangular prism: ma := a·b·c·p

ma ( 2 2)+ b

ma ( 2lay := 12· a

2)e

r r

d 1 1 1

e

1

JiS PROBLEM 10-5

Statement: Figure P l 0-l shows a bracket made of steel.a. Find the location of its centroid referred to point B.b. Find its mass moment of inertia /:a about the x axis through point B.

c. Find its mass moment of inertia lyy about the y axis through point B.

Given: Dimensions in the figures below. Density: p := 7800kg·m-3

Solution: See Figure PI0-1 and Mathcad file P1005.

1. Divide the bracket into five volumes, find the location ofthe CG and the mass moments for each ofthem and then add the results to get the CG and mass moments for the entire bracket.

i := 1,2.. 5

2. Volume 1 is the rectangular prism with two negative cylinders shown below.

Dimensions:

a:= 64·mm

b := 38-

mme:= l9·mm

a ---------¡t y !

! I 1··---iT r-(

"'+/ J - -4-- + l

d := 20-mm

X

e:= l9·mm

r:= 5·mm

Determine the location ofthe CG in global coordinates.

z 1 .· :1 1 . i

i "--+ /

L_ L_t- -ª -+--! ,

Xcgt := 2 Xcg 1

= 9.500mm Zcg1 :=O-mm

b 2a·b·- - 2·1t·r ·d2

Ycg1 := -----a·b - 2·1t·r

-- d

2

Ycg1 = -1.069mm

Determine the volume and mass ofthis segment.2 4 3

V 1

:= a·b·c - 2·1t·r

·eV

1 = 4.322 x 10 mm M

1 := p·V

1M = 0.337kg

Determine the mass moments about the local axes through the CGs.

fax := 12· a

ma = 0.360kg

fax= l.664 x

lay= 1.339 x

-4 210 kg·m

-4 210 kg·m


+a

1ma { 2 2) - 5 2

:=-· b12

faz= 5.421 x 10 kg·m

2 2)

2 2)

=

2

=

DESIGN OF MACHINERY

1

SOLUTION MANUAL 10-5-3

1

z

a

l¡

1

11

Cylinder (one): mb :=2

-n·r ·C·p mb = -O.Ol2kg

2mb·r - 7 2

lbx:=--2

mb {lby := -·3·r +e

12

mb {lbz := -·3·r + e

12

lbx = -1.455 x 10 kg·m

- 7 2lby -4.229 X 10 kg·m

- 7 2lbz = -4.229 x 10 kgm

Determine the mass moments about the global axes.2 2 2 -4 2

l.u1 := lax + ma{(Ycg

1)+ {Zcg1) ] + 2.(fbx +

mb·e )

/_u] = 1.58) X 10 kg·m

2I.w I := lay + ma{( Xcg

1)

+ (Zcg 1) J + 2 .[fby +

mbür]IY.Y 1

-4 21.634 x 10 kgm

2Jzz 1 :=faz+ m0{(xcg1)

2+ {Ycg1) ] +

2·(1bz)

/zz 1 = 8.631 x- 5

kg·m2

3. Volume 2 is the bendjust above segment 1. lt is a quarter hollow cylinder with dimensions shown below.

Dimensions: e ·1a:= 13·mm

b := 32·mm y / \J r J1

e:= 64·mm *Y,..d := 18·mm i

e:= 32·mm

J-1z 1

1

. .1

e


4 b3 - ª3 Xcg := e - -·---

2 3·1t b2 -Xcg

2 = 16.825 mm

4 b3 - ª3 Ycg := d + -

·--- 2 3·1t b22

-aYcg

2= 33.175 mm

Determine the volume and mass of this segment.

2 2

1t·b


2

·C - 1t·a ·Cv2 := 4 M

= 0.335kg

Determine the mass moments about the local axes noted on the drawing of the segment.

lx:= M2 (3·a

2 + 3·b

2 2) lx = 2.144 x -4 2kg·m

-· +e 1012

L . _ty

3

1

=

1


ly:= M2 (3·a

2 + 3·b

2 2) fy = 2.144 X -4 2kg·m

-· +e12

M2 ( 2 2)lz:=-· a + b

2

10

-4 2lz = 2.000 x 10 kg·m

Detennine the mass moments about the global axes.

2Iwl :=lay+ Mfe

2

-4 2Ixx2 = 3.230 x 10 kg·m

-4 2Iw

1 = 4.771 x 10 kg·m

-4 2lzz 1 := laz + M2.

{d

+ /)

Izz 1 = 5.061 x 10 kg·m

4. Volume 3 is a rectangular prism with dimensions shown below.

Dimensions: 1· a

------··---·- ---·-·-·---------"1

a:= 64·mm

b := 19·mm

.-- -1

1_J

e:= 38·mm1

d := 40.5·mm

e:= 51·mm

z 1


Xcg 3 :=e

Zcg3

:= O·mm

Xcg3 = 51.000 mm Ycg := d

Ycg3

= 40.500mm

Detennine the volume and mass ofthis segment.4 3

v3 := a·b·c v3 = 4.621X10 mm M3 := p·V3

Detennine the mass moments about the local axes through the CGs.

-4 2lx = 1.339 x 10 kg·m

-4 2/y = 1.664 X 10 kg· m

M 3 = 0.360kg

lz = 5.421 x

Detennine the mass moments about the global axes.

-5kg·m

2

-4 2Ixx3 7.250 x 10 kg·m

DESIGN OF SOLUTION MANUAL 10-

2 2 - 2

1

1

4

2

2 2)

Iw3 := Iy + M 3 {(xcg3)

+ (Zcg3)

]

J_w3

= 1.104 x 10 kg·m

2Izz 3 := lz + M 3{(xcg3)

2+ (Ycg3) ]

lzz = 5.061 x-4

kg·m 2

1

5. Volume 4 is a half cylinder with dimensions shown below.

Dimensions:

a:= l9·mm

b :=

40.5·mm e:=

10·mm r:=

32·mm

X

c----------<-i

------.a

_l

Determine the location ofthe CG in global coordinates .

4·rXcg :=c+-

4 3·1t

Zcg4 :=O-mm

Xcg = 83.581 mm Ycg3

= 40.500 mm

Determine the volume and mass ofthis segment.

x·r2

4 3

V4

:=a·-- V4

= 3.056 x 10 mm M 4 := p· V

4

M 4

= 0.238kg

Determine the mass moments about the local axes ofthe segment.

- 5 2lx = 6.820 x 10 kg·m

M4 2ly:=-·r

2

M4 (lz:= -·3·r +a

12

-4 2fy = 1.221 X 10 kgm

- 5 2lz = 6.820 x 10 kg·m


2 2 - 2Determine the mass moments about the global axes.


2 2 - 2

21

5

Ixx4 := lx + M4{(Ycg4)

+ (zcg4)

]

Ixx4 = 4.592 x 10 kg·m

2Iw4 := Iy+ M 4{(xcg4)

2

+ (Zcg4)

]

2

Iw4

Izz

= l.787 x

= 2.124 x

-3kg· m

2

-3kg·m

2

Izz4 := lz + M 4{(

Xcg4)

+ (Ycg4) ] 4 10

6. Volume 5 is a negative cylinder with dirnensions shown below.

Dimensions:

r a:= 10·mm

b :=

40.5·mm e:=

19·mm r:=

13·mm

z

Determine the location of the CG in global coordinates.

---.-1 ,_J j

Zcg 5 :=

O·mm

Xcg 5 = 70.000 mm Ycg = 40.500 mm

Determine the volume and mass ofthis segment.

2V 5 := c·n·r

4 3

V 5 = 1.009 x 10 mm M 5 := -p· V 5 M 5 = -0.079 kg

Determine the mass moments about the local axes ofthe segment.

M5 ( 2 2) -6 2

lx:= -·3-r +e

12

M5 2

ly:=-·r2


2 2 - 2lx = -5.691 x 10kg·m -6 2

fy = -6.649 X 10 kg·m

lz := M5 .(3.r2 +

/)12

lz = -5.691 x-6 2

10 kg·m


25

1

=

l

l

1

1

Determine the mass moments about the global axes.

2Ixx 5 := lx + M 5 {(Ycg5) + (Zcg5)

]Ixx = -1.348 x

-4kg·m

2

Iyys := Iy + Ms{( Xcgs)2 +

(Zcgs)2]Iyy

5= -3.922 x

-4 kg·m

2

2Izz5 := lz + M 5 {(xcg5)

2+ (Ycg5) ]

-4 2lzz

5-5.203 x 10 kg·m

a. Find the location ofits centroid referred to point B.

L""."J' Xcg. ·M.

LY.Jcg.·M.

Xcg = 34.919mm

YCg = 26.688 mm

Zcg = 0.000 mm

b. Find its mass moment of inertia lxx about the X axis through point B.

Ixx = l.531 x

c. Find its mass moment ofinertia lyy about the Y axis through point B.

-3kg·m

2

lyy:= L Iyyi

fyy = 2.976 X

-3kg·m

2

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