HW09 - Eighth Homework Assignment PDF

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Eighth Homework Assignment...

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5/4/2019

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

50/50

Points

100

%

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1.

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Award:





The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

The first moment (Qx)1 of component area A1 with respect to the x axis is

25

in3.

The first moment (Qx)2 of component area A2 with respect to the x axis is

-25

in3.

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The above results are expected since x

is a centroidal axis and yˉ =

0

in.



Hints Hint #1

References Worksheet

Problem 05.021 - First moment of a composite area

Difficulty: Easy



The horizontal x axis is drawn through the centroid C of the area shown, and it divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained.

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25 ± 2% in3. The first moment (Qx)1 of component area A1 with respect to the x axis is The first moment (Qx)2 of component area A2 with respect to the x axis is -25 ± 2% in3. 0 ± 2% in. is a centroidal axis and yˉ = The above results are expected since x

 Explanation:

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Note that Q x = ΣˉyA

(

Then, Q x) 1 =

( )( 5

3

in.

1 2

× 6 in. × 5 in.

)

(

or Q x) 1 = 25 in 3

(

and Q x) 2 =

(

-2 3

× 2.5 in.

)(

1 2

) (

1

× 9 in. × 2.5 in. + - × 2.5 in. 3

)(

1 2

× 6 in. × 2.5 in.

)

(

or Q x) 2 = - 25 in 3

(

(

Now, Q x = Q x) 1 +  Q x) 2 = 0 This result is expected since x is a centroidal axis, and thus, (yˉ = 0)

(

ˉ and Q x = ΣˉyA = YΣA yˉ = 0 ⇒ Q x = 0

)



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2.

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



Award:

Determine the centroid of the area shown by direct integration.

2

xˉ = ( yˉ = (

1

/ /

5 2

)a. )b.



Hints Hint #1

References Worksheet

Problem 05.035 - Area centroid

Difficulty: Medium

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

Determine the centroid of the area shown by direct integration.

xˉ = ( yˉ = (

2/ 1/

5 )a. 2 )b.

 Explanation:

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b = ma, but y 1 = mx, thus y 1 = 1

b x a

1

1

b

b = ka 2 , but y 2 = kx 2 , thus y 2 =

1

x2

a2

¯ Now x EL = x

¯ y EL =

( ) ( ) ( ) (( ) ) ( ) ( ) ( ( )) ( )( ) ( ) ( ( 1

1 2

(

)

y2 + y1 =

b

x2

2

1 a2

+

x

a

1

(

)

x2

dA = y 2 - y 1 dx = b

-

1 a2

x

a

dx

1

¯ ∫ x EL dA = ∫0axb

¯ b ∫ y EL dA = ∫0a 2

¯ ˉxA = ∫ x EL dA: xˉ

3

x2

Then, A = ∫ dA = ∫0a b

-

1 a2

x

a

dx = b

3

1

x2 1 a2

x

-

a

dx = b∫ a0

a

1 a2

+

x

( ) 1

6

1 2

-

x2

3

1 a2

x2 a

-

a

ab =

b

1 15

x2

1 a2

-

x

a

dx =

a

x2

=

2a

1

6

5

dx = b

b2 a ∫ 2 0

ab

0

1

1

x2

x2

2

x

a

-

2

x2

5

1 a2

x2 a

-

dx =

x3

a

=

3a

b2 2

1 15

a 2b

0

x2 2a

-

x3 3a 2

))

a

= 0

1 12

ab 2

a 2b

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xˉ =

2 a 5

¯ ˉyA = ∫ y EL dA: yˉ

yˉ =

1 2

( ) 1

6

ab =

1 12

ab 2

b



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3.

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Award:





For the machine element shown, locate the x coordinate of the center of gravity. Given: r = 1.4 in.

The x coordinate of the center of gravity of the machine element is

.643

in.



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References Worksheet

Problem 05.102 - Centroid of a composite body

Difficulty: Easy



For the machine element shown, locate the x coordinate of the center of gravity. Given: r = 1.4 in.

The x coordinate of the center of gravity of the machine element is

0.6431 ± 2% in.

 Explanation: ezto.mheducation.com/hm.tpx?todo=C15PrintView&wid=13252710750162999&role=student&sid=13252710775950179

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Divide the element into four sections consisting of half cylinder, minus a whole cylinder, a vertical plate, and a horizontal plate.

1

Half Cylinder

2

-Whole Cylinder

3 4

Vertical Plate Horizontal Plate Σ

V, in 3

xˉ , in.

π

0

2

(3) 2(1)

-π(1.4) 2

= 14.1372

= - 6.1575 (1)(6)(5) = 30 (3)(6)(1) = 18 55.9797

xˉ V, in 4 0

0

0

0 2

0 36 36

Xˉ ΣV = ΣxV ˉ

(

)

Xˉ 55.9797 in. 3 = 36 in. 4 Xˉ = 0.6431 in.



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4.

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



A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity if h = 2 m.

The center of gravity of the steel wire is: ˉ X: m .842 ˉ Y: m .4537 ˉ Z: m .765 

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References Worksheet

Problem 05.114 - Centroid of a shape formed by a thin wire

Difficulty: Easy



A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity if h = 2 m.

The center of gravity of the steel wire is: ˉ 0.842 ± 2% m X: ˉ 0.454 ± 2% m Y: ˉ 0.765 ± 2% m Z:  Explanation:

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First assume that the sheet metal is homogenous so that the center of gravity of the form will coincide with the centroid of the corresponding area.

−

−

2 × 2.4

x2 = z2 =

π

=

4.8 π

m

L, m 1 2 3 4 Σ

3.124 3.7699 2.4 2.000 11.294 ˉ

ˉ X =

∑xL

ˉ Y =

∑yL

ˉ Z =

∑L

=

ˉ

∑L

=

ˉ

∑ zL ∑L

=

9.509 11.294 5.124 11.294 8.640 11.294

ˉ x, m 1.2 1.5279 0 0

ˉ y, m 1.000 0 0 1.000

ˉ z, m 0 1.5279 1.2 0

xˉ L, m 2

yˉ L, m 2

zˉ L, m 2

1.2 × 3.124 1.5279 × 3.7699 0 × 2.4 0 × 2.000 9.509

1.000 × 3.124 0 × 3.7699 0 × 2.4 1.000 × 2.000 5.124

0 × 3.124 1.5279 × 3.7699 1.2 × 2.4 0 × 2.000 8.640

= 0.842 m = 0.454 m = 0.765 m



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5.

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Award:





Identify the centroid of the volume obtained by rotating the shaded area about the x-axis.

 xˉ  =  1 h, ˉy = 0, andˉz = 0. 7 ✓

 xˉ  =  1 h, ˉy = 0, andˉz = 0. 6  xˉ  =  1 h, ˉy = 0, andˉz = 0. 8  xˉ  =  1 h, ˉy = 0, andˉz = 0. 9

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y  = 0andˉz = 0 First, notethatsymmetryimpliesˉ Wehavey = k( x − h ) Atx = 0,  y = a ,

2

a = k (− h) 2 ork = 

a h2

Chooseastheelementofvolumea diskofradius randthicknessdx.  ThendV = πr 2dx, ˉ x EL = x a

Nowr =  2 ( x − h ) 2 h sothatdV = π h

ThenV =  ∫ π 0

a2 h4

a2 h4

(x − h) 4dx

( x −  h) 4dx

1

 =  5 πa 2h h

[

and∫xˉ ELdV = ∫ x π 0

a2 h4

]

(x − h ) 4dx

1

 =  30 πa 2h2 NowˉxV = ∫ ˉxELdV 1

ˉx =  6 h 1

x =  6 h, ˉy = 0, andˉz = 0. Hence, ˉ

Hint #1

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Problem 05.126 - Integrate for the centroid of an area

Difficulty: Hard

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