Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
9th Edition
ISBN: 9781337594301
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 7, Problem 7.5.14P

A circle of a diameter d = 200 mm is etched on a brass plate (see figure). The plate has dimensions of 400 x 400 x 20 mm. Forces are applied to the plate, producing uniformly distributed normal stressescr^ =59 MPaander^ = —17 MPa. Calculate the following quantities: (a) the change in length Aac of diameter at: (b) the change in length Abd of diameter bd; (c) the change At in the thickness of the plate; (d) the change AV in the volume of the plate; (e) the strain energy U stored in the plate; (f) the maximum permissible thickness of the plate when strain energy £/must be at least 784 J; and (g) the maximum permissible value of normal stress axwhen the change in volume of the plate cannot exceed 0.015% of the original volume. (Assume E = 100 GPa and v = 0.34Chapter 7, Problem 7.5.14P, A circle of a diameter d = 200 mm is etched on a brass plate (see figure). The plate has dimensions

(a)

Expert Solution
Check Mark
To determine

The change in the length Δ a c of diameter a c .

Answer to Problem 7.5.14P

The change in the length Δ a c is 0.129 mm .

Explanation of Solution

Given information:

The normal stress acting on the x-direction is 59 MPa , the normal stress acting along y-direction is 17 MPa , modulus of elasticity is 100 GPa , diameter of the circle is 200 mm , the plate dimensions is 400 mm × 400 mm × 20 mm , and the Poisson s ratio is 0.35

Explanation:

The figure below shows the element.

  Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card, Chapter 7, Problem 7.5.14P

     Figure (1)

Write the expression of the strain along x-direction.

   ε x = 1 E ( σ x ν σ y ) ....... (I)

Here, the normal stress along x-direction is σ x ,the normal stress along y-direction is σ y , the strain along x-direction is ε x ,and the modulus of elasticity is E .

Write the expression of the change in the length.

   ε x = Δ a c a c (II)

Here, the change in the length is Δ a c , the length of the diameter is a c , and the strain along x-direction is ε x .

Calculation:

Substitute 59 MPa for σ x , 17 MPa for σ y , 100 × 10 3 MPa for E , and 0.34 for ν , in Equation (I).

   ε x = 1 100 × 10 3 MPa ( 59 MPa ( 0.34 × 17 MPa ) ) = 64.78 MPa 100 × 10 3 MPa = 0.64 × 10 3

Substitute 0.64 × 10 3 for ε x , and 200 mm for a c , in Equation (II).

   0.64 × 10 3 = Δ a c 200 mm Δ a c = 128 × 10 3 mm Δ a c = 0.129 mm

Conclusion:

The change in the length Δ a c is 0.129 mm .

(b)

Expert Solution
Check Mark
To determine

The change in the length Δ b d of the diameter b d .

Answer to Problem 7.5.14P

The change in the length Δ b d of the diameter b d is 0.07412 mm .

Explanation of Solution

Given information:

The length of the b d is 200 mm .

Write the expression of the strain along y-axis.

   ε y = 1 E ( σ y ν σ x ) ....... (III)

Write the expression of the change in the length Δ b d .

   Δ b d b d = ε y ....... (IV)

Calculation:

Substitute 59 MPa for σ x , 17 MPa for σ y , 100 × 10 3 MPa for E , and 0.34 for ν , in Equation (III).

   ε Y = 1 100 × 10 3 MPa ( 17 MPa - 0 .34 × 59 MPa ) = 37.6 MPa 100 × 10 3 MPa = 0.370 × 10 3

Substitute 0.370 × 10 3 for ε x , and 200 mm for b d , in Equation (IV).

   Δ b d 200 mm = 0.370 × 10 3 Δ b d = 0.370 × 10 3 × 200 mm Δ b d = 74.12 × 10 3 mm

Conclusion:

The change in the length Δ b d of the diameter b d is 0.07412 mm .

(c)

Expert Solution
Check Mark
To determine

The change in thickness of the plate.

Answer to Problem 7.5.14P

The change in the thickness is 2.856 × 10 3 mm .

Explanation of Solution

Given information:

Write the expression of the strain along z-direction.

   ε z = ν E ( σ x + σ y ) ....... (V)

Write the expression of the change in thickness.

   Δ t = ν E ( σ x + σ y ) t ....... (VI)

Here, the change in the thickness is Δ t , and the thickness is t .

Calculation:

Substitute 59 MPa for σ x , 17 MPa for σ y , 100 × 10 3 MPa for E ,and 0.34 for ν , in Equation (V).

   ε z = 0.34 100 × 10 3 MPa ( 59 MPa + ( 17 MPa ) ) = 14.28 MPa 100 × 10 3 MP = 0.1428 × 10 3

Substitute 0.1428 × 10 3 for ε z , and 20 mm for t , in Equation (VI).

   Δ t = 0.1428 × 10 3 × 20 mm = 2.856 × 10 3 mm

Conclusion:

The change in the thickness is 2.856 × 10 3 mm .

(d)

Expert Solution
Check Mark
To determine

The change in the volume of the plate.

Answer to Problem 7.5.14P

The change in the volume is 407 mm 3 .

Explanation of Solution

Write the expression for the volumetric strain.

   Δ V = V ο ( ε x + ε y + ε z ) ....... (VII)

Here, the change in the volume is Δ V , the volume is V ο ,the strain along x-direction is ε x , the strain along y-direction is ε y , and the strain along z-direction is ε z .

Calculation:

Substitute 0.64 × 10 3 for ε x , 0.370 × 10 3 for ε y , 0.1428 × 10 3 for ε z , and 400 mm × 400 mm × 20 mm for V ο , in Equation (VII).

   Δ V = 400 mm × 400 mm × 20 mm ( 0.64 × 10 3 0.370 × 10 3 0.1428 × 10 3 ) = 3200000 mm 3 ( 0.1272 × 10 3 ) = 407 mm 3

Conclusion:

The change in the volume is 407 mm 3 .

(e)

Expert Solution
Check Mark
To determine

The strain energy stored in the plate.

Answer to Problem 7.5.14P

The strain energy stored in the plate is 70.480 N m .

Explanation of Solution

Write the expression for the strain energy stored in the plate.

   U = V ο 2 ( σ x ε x + σ y ε y + σ z ε z ) ....... (VII)

Here, the energy in the plate is U .

Calculation:

Substitute 59 MPa for σ x , 17 MPa for σ y , 100 × 10 3 MPa for E , and 0.34 for ν , 0.64 × 10 3 for ε x , 0.370 × 10 3 for ε y , 0.1428 × 10 3 for ε z , and 400 mm × 400 mm × 20 mm for V ο , in Equation (VIII).

   U = 400 mm × 400 mm × 20 mm 2 ( 59 MPa ( 0.64 × 10 3 ) + 17 MPa × 0.370 × 10 3 + 0 MPa × ( 0.1428 × 10 3 ) ) = 1600000 mm 3 ( 0.04405 MPa ) ( 1 m 3 10 9 mm 3 ) ( 10 6 N / m 2 1 MPa ) = 70.480 N m The strain energy stored in the plate is 70.480 N m .

(f)

Expert Solution
Check Mark
To determine

The maximum permissible thickness of the plate.

Answer to Problem 7.5.14P

The maximum permissible thickness is 22.26 mm .

Explanation of Solution

Given information:

The strain energy stored must be at least 78.4 J ,the initial strain energy is 70.4 J ,and the initial thickness of the plate is 20 mm .

Write the expression of the strain energy directly proportional to the thickness.

   U final U initial = t final t initial ....... (IX)

Calculation:

Substitute 78.4 J for U final , 70.4 J for U initial , and 20 mm for t initial in Equation (IX).

   78.4 J 70.7 J = t final 20 mm t final = 20 mm × 1.1136 = 22.26 mm

Conclusion:

The maximum permissible thickness is 22.26 mm .

(g)

Expert Solution
Check Mark
To determine

The maximum permissible volume of the plate.

Answer to Problem 7.5.14P

The maximum permissible volume is . 63.75 MPa .

Explanation of Solution

Write the expression for the volumetric strain.

   Δ V V ο = ( 1 2 ν ) E ( σ x + σ y ) ....... (X)

Calculation:

Substitute 59 MPa for σ x , 17 MPa for σ y , 100 × 10 3 MPa for E , and 0.34 for ν ,and 0.015 % for Δ V V in Equation(X).

   0.015 100 = ( 1 0.68 ) 100 × 10 3 MPa ( σ x 17 MPa ) 1.5 × 10 4 MPa = 3.2 × 10 6 σ x 5.4 × 10 5 MPa σ x = 63.75 MPa

Conclusion:

The maximum permissible volume is . 63.75 MPa .

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Chapter 7 Solutions

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card

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