Chapter 9 - Hibbert Statics solution PDF

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Hibbert Statics solution...

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9–1. Locate the center of mass of the homogeneous rod bent into the shape of a circular arc.

y

30

300 mm x

SOLUTION dL = 300 d u 30

' x = 300 cos u ' y = 300 sin u

x =

L

2p 3

L-2p3

' x dL

L

=

300 cos u (300du) 2p 3

L-2p3

dL

300d u

(300)2 C sin u D-32p3 2p

=

4 300a pb 3

= 124 mm y = 0

Ans. (By symmetry)

Ans.

Ans: x = 124 mm y = 0 8 79

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9–2. y

Determine the location (x, y) of the centroid of the wire.

2 ft

4 ft

SOLUTION Length and Moment Arm: The length of the differential element is dL = 2dx2 + dy2 = ¢

B

1 + a

b ≤ dx and its centroid is ∼ y = y = x2. Here, 2

dy dx

y = x2

dy = 2x. dx

x

Centroid: Due to symmetry ∼

x = 0

Ans.

Applying Eq. 9–7 and performing the integration, we have

∼

y =

LL

2 ft

∼

ydL

LL

=

L-2 ft

x 21 + 4x2 dx

2 ft

L-2 ft

dL

=

2

21 + 4x2 dx

16.9423 = 1.82 ft 9.2936

Ans.

Ans: x = 0 y = 1.82 ft 88 0

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9–3. y

Locate the center of gravityx of the homogeneous rod. If >m, determine the rod has a weight per unit length of 100 N the vertical reaction at A and the x and y components of reaction at the pin B.

1m B

1m y  x2 A

x

SOLUTION Arm. The length of the differential element is dy dy 2 dL = 2dx + dy = c 1 + a b d dx and its centroid is ~ x = x. Here = 2x. A dx dx Perform the integration Length

And

Moment

2

2

L =

LL

dL =

L0

= 2

1m

L0

21 + 4x2 dx

1m

A

x2 +

1 dx 4

1 1 1 1m + ln ax + x2 + b d = c x x2 + A 4 4 A 4 0 = 1.4789 m LL

~ x dL =

L0

1m

x21 + 4x2 dx 1m

= 2 = c

L0

x

A

x2 +

1 dx 4

1 3>2 1 m 2 2 ax + b d 4 3 0

= 0.8484 m2 Centroid. x =

~ 0.8484 m2 1L xdL = 0.5736 m = 0.574 m = 1.4789 m 1L dL

Ans.

Equations of Equilibrium. Refering to the FBD of the rod shown in Fig. a + Σ F S

x

= 0;

a+Σ MB = 0;

Bx = 0

Ans.

100(1.4789) (0.4264) - A y(1) = 0 Ay = 63.06 N = 63.1 N

MA = 0; a+Σ

Ans.

B y(1) - 100(1.4789) (0.5736) = 0 By = 84.84 N = 84.8 N

Ans. Ans: x = 0.574 m Bx = 0 Ay = 63.1 N By = 84.8 N 8 81

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*9–4. y

Locate the center of gravityy of the homogeneous rod.

1m B

1m y  x2 A

x

SOLUTION Arm. The length of the differential element is dy dy 2 dL = 2dx + dy = c 1 + a b d dx and its centroid is ~ y = y. Here = 2x. A dx dx Perform the integration, Length

And

Moment

2

L =

2

1m

LL

L0

dL =

= 2

L0

= cx

21 + 4x2 dx

1m

A

A

x2 +

x2 +

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 4 0 A

= 1.4789 m LL

~ y dL =

L0

= 2

1m

L0

= 2c

x2 21 + 4x2 dx

1m

x

2

A

x2 +

1 dx 4

1 x 1 1 1m 1 3 1 x x2 + ln ax + x2 + b d ax2 + b A 4 0 4A 32 A 4 128 4

= 0.6063 m2 Centroid. y =

~ 0.6063 m2 1L y dL = 0.40998 m = 0.410 m = 1.4789 m 1L dL

Ans.

Ans: y = 0.410 m 88 2

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9–5. Determine the distance y to the center of gravity of the homogeneous rod.

y 1m

2m

SOLUTION Length

And

y  2x3

x

Arm. The length of the differential element is

Moment

dy 2 dL = 2 dx + dy = a 1 + a b b dx and its centroid is at ~ y = y. Here A dx dy = 6x2. Evaluate the integral numerically, dx 2

L = LL

2

LL

dL =

~ y dL =

L0

L0

1m

21 + 36x4 dx = 2.4214 m

2x 21 + 36x4 dx = 2.0747 m2

1m

3

Centroid. Applying Eq. 9–7, y =

~ 2 1L ydL = 2.0747 m = 0.8568 = 0.857 m 2.4214 m 1L dL

Ans.

Ans: y = 0.857 m 8 83

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9–6. Locate the centroid y of the area.

y

y

1 – 1– x2 4

1m x 2m

SOLUTION Area and Moment Arm: The area of the differential element is y 1 1 1 ' dA = ydx = a 1 - x2 b dx and its centroid is y = = a 1 - x2 b. 4 4 2 2 Centroid: Due to symmetry x = 0

Ans.

Applying Eq. 9–4 and performing the integration, we have

y =

LA

2m

' y dA

dA L A

=

=

1 1 2 1 ¢ 1 - x ≤ ¢ 1 - x2 ≤ dx 4 4 L- 2m 2 2m

L- 2m

¢

¢1 -

1 2 x ≤dx 4

2m x3 x5 x + ≤` 160 - 2m 2 12

x ¢x ≤` 12 - 2m 3

2m

=

2 m 5

Ans.

Ans: y =

88 4

2 m 5

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9–7. Determine the area and the centroid x of the parabolic area.

y

h

SOLUTION

y

Differential Element:The area element parallel to the x axis shown shaded in Fig. a will be considered. The area of the element is dA = x dy =

a h1>2

h 2 –– a2 x x

a

y 1>2 dy

x a ' ' Centroid: The centroid of the element is located at x = = y 1>2 and y = y. 2 2h1>2 Area: Integrating, A =

x =

L A

dA LA

LA L0

' x dA =

h

dA =

h

¢

L0 h

a 1>2

2h

a 1>2

y 1> 2 ≤ ¢ 2 ah 3

y 1>2 dy = a 1>2

h

Ay 3>2 B 2 = 1> 2 h

2a 3h

y 1>2 dy≤

0

2 ah 3

Ans.

h 2 a2 y a ¢ ≤` y dy 2h 2 0 3 L 0 2h = = = a Ans. 2 2 8 ah ah 3 3 h 2

Ans: x =

8 85

3 a 8

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*9–8. Locate the centroid of the shaded area.

y

y  a cos px L a

L 2

x

L 2

SOLUTION Area And Moment Arm. The area of the differential element shown shaded in y a p p Fig.a is dA = ydx = a cos x dx and its centroid is at y~ = = cos x. L 2 2 2 Centroid. Perform the integration

y =

L>2

p p a a cos x b aa cos x dx b 2 L L L-L>2

~ 1A y dA

1A dA

=

L>2

L-L>2 L>2

L-L>2 =

a cos

p x dx L

a2 2p acos x + 1 b dx 4 L L>2

L- L>2

a cos

p x dx L

L>2 a2 L 2p a x + xb ` sin 4 2p L -L>2

=

=

L>2 aL p a sin x b ` p L -L> 2

a 2 L >4 p = a 2aL> p 8

Ans.

Due to Symmetry, x = 0

Ans.

Ans: p a 8 x = 0 y =

88 6

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9–9. Locate the centroid x of the shaded area.

y

4m y  1 x2 4 x 4m

SOLUTION Area And Moment Arm. The area of the differential element shown shaded in Fig.a 1 is dA = x dy and its centroid is at ~x = x. Here, x = 2y 1>2 2 Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

L0

4m1

2

a 2y1>2 b a 2y1>2 dyb

L0 =

4m

2y1>2 dy

3 m 2

Ans.

Ans: x =

8 87

3 m 2

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9–10. Locate the centroid y of the shaded area.

y

4m y  1 x2 4 x 4m

SOLUTION Area And Moment Arm. The area of the differential element shown shaded in Fig.a ∼ is dA = x dy and its centroid is at y = y . Here, x = 2y1>2. Centroid. Perform the integration

y =

~ 1A y dA

1A dA

=

L0

4m

y a 2y1> 2 dyb

4m

L0

=

=

2y1>2 dy

4m 4 a y 5>2 b ` 5 0 4m 4 a y3>2 b ` 3 0

12 m 5

Ans.

Ans: y =

88 8

12 m 5

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9–11. y

Locate the centroid x of the area.

h y  —2 x2 b

SOLUTION

h

dA = y dx ' x = x

x =

x

A L

B

b

' x dA

LA

= dA

h 3 2 x dx b L0 b

L0 b2 h

2

x dx

=

h 4b2

x4 R

b

b 0

b h B 2 x3 R 3b 0

=

3 b 4

Ans.

Ans: x =

8 89

3 b 4

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*9–12. y

Locate the centroid y of the shaded area.

h y  —2 x2 b

SOLUTION

h

dA = y dx y ' y = 2

y =

LA

x b

' y dA

LA

= dA

4 L0 2b b

h

2 L 0 b

h

x4 dx = 2

x dx

b

b h2 B 4 x5 R 10b 0

2

b h B 2 x3 R 3b 0

=

3 h 10

Ans.

Ans: y =

89 0

3 h 10

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9–13. Locate the centroid x of the shaded area.

y

SOLUTION

4m y

1 dA = 14 - y2dx = a x2 b dx 16 ' x = x

x =

L A

' x dA

LA

= dA

L0

8

xa

L0

8

a

4

1 2 –– 16 x 8m

x2 b dx 16

1 2 x b dx 16

x = 6m

Ans.

Ans: x = 6m 8 91

x

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9–14. Locate the centroid yof the shaded area.

y

SOLUTION

4m y

dA = 14 - y2dx = a y =

1 2 x b dx 16

4

1 2 –– 16 x 8m

4 + y 2 x2 1 x2 ¢8 b a b dx 2 L0 16 16 8

y =

LA

' y dA

LA

= dA

L0

8

a

1 2 x b dx 16

y = 2.8 m

Ans.

Ans: y = 2.8 m 89 2

x

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9–15. Locate the centroid x of the shaded area.Solve the problem by evaluating the integrals using Simpson’s rule.

y

y = 0.5e x

2

SOLUTION At x = 1 m 2

y = 0.5e1 = 1.359 m A L

L0

dA =

1

L 0

(1.359 - y) dx =

1

x

a 1.359 = 0.5 ex b dx = 0.6278 m 2 2

1m

x = x LA

x dA =

L0

1

x a 1.359 - 0.5 ex b dx 2

= 0.25 m3

x =

LA

x dA

LA

= dA

0.25 = 0.398 m 0.6278

Ans.

Ans: x = 0.398 m 8 93

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*9–16. Locate the centroid y of the shaded area.Solve the problem by evaluating the integrals using Simpson’s rule.

y

y = 0.5ex

2

SOLUTION L0

LA

dA =

y =

1

L0

(1.359 - y) dx =

1

a 1.359 - 0.5ex b dx = 0.6278 m 2

2

1.359 + y

x

2

1m 1

2 1.359 + 0.5 ex b A 1.359 - 0.5 ex B dx y dA = a 2 L0 LA 2

1

=

y =

1 2 a 1.847 - 0.25 e2x b dx = 0.6278 m3 2 L0

y dA L A LA

dA

=

0.6278 = 1.00 m 0.6278

Ans.

Ans: y = 1.00 m 89 4

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9–17. Locate the centroid y of the area.

y

SOLUTION

2 ––

y 4 in.

Area: Integrating the area of the differential element gives

A =

LA

dA =

L0

8 in.

x3

8 in.

3 x2>3 dx = c x5>3 d 2 0 5

x

= 19.2 in.2

8 in.

1 Centroid: The centroid of the element is located at y = y>2 = x2>3. Applying 2 Eq. 9–4, we have '

A y = L

'

y dA

LA

=

L0

dA

c =

3 7>3 2 x d 14 0

8 in.

1 2> 3 2> 3 x A x B dx 2 19.2

=

L0

8 in.

1 4>3 x dx 2 19.2

8 in.

19.2

Ans.

= 1.43 in.

Ans: y = 1.43 in. 8 95

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9–18. y

Locate the centroid x of the area.

y

SOLUTION

h

h xn — an

h

dA = y dx x

' x = x

x =

a

LA

' x dA

LA

= dA

B x2 h 2

=

0 L

Bhx -

¢ hx -

h n+1 x ≤ dx an

a

h n x ≤ dx an

a

L0

¢h -

h1xn + 22

a 1n + 22 n

h1x

n+1

2

an1n + 12

R

R

a 0

a 0

a

h h b a2 n + 2 2 a(1 + n) = x = 2(2 + n) h ba ah n + 1

Ans.

Ans: x =

89 6

a(1 + n) 2(2 + n)

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