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lesantik [10]
3 years ago
8

A bronze bushing 60 mm in outer diameter and 40 mm in inner diameter is to be pressed into a hollow steel cylinder of 120-mm out

er diameter. Calculate the tangential stresses for steel and bronze at the boundary between the two parts
Engineering
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

The tangential stress for the steel at 30mm = 88.8MPa.

The tangential stress for the steel at 60mm = 35.5 MPa.

The tangential stress for the bronze at 20mm =  - 191.9 MPa.

The tangential stress for the bronze at 30mm =  - 138.6 MPa.

Explanation:

The outer radius bronze bushing = 60 mm/2 = 30 mm, the inner radius for bronze bushing = 40mm/2 = 20mm and the cylinder radius = 120mm/2 = 60mm.

Step one: The first step is to calculate or Determine the interface pressure.

The interface pressure = 0.05/ [ [ 30 × ( 1/210 ×10⁹) ] ×  [( 60² + 30²/ ( 60² - 30²) + 0.3 ] + [ 1/105 × 10⁹× [ ( 20² + 30²)/( 30² - 20²) - 0.3] ].

The interface pressure = 53.3 MPa

Step two: Determine the tangential stresses for steel and bronze as given below:

The tangential stress for the steel at 30mm = 10⁶ × 53.3 [ ( 0.06)² + (0.03)²/ ( 0.06)² - (0.03)² ] = 88.8 MPa.

The tangential stress for the steel at 60mm = 10⁶ × 53.3 × 2 × [0.03]²/  ( 0.06)² - (0.03)² ] = 35.5 MPa.

The tangential stresses for the bronze is calculated below;

=> The tangential stress for the bronze at 20mm = - [ 10⁶ × 53.3 × 2 × [0.03]² ] /  ( 0.03)² - (0.02)² ] = - 191.9 MPa.

The tangential stress for the bronze at 30mm = - [ 10⁶ × 53.3  × [ 0.03]² + (0.03)² ] /  ( 0.03)² - (0.02)² ] = - 138.6 MPa.

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rate of insulation, Q = 100 Btu/h

To get the minimum electrical power required, use the relation below:

\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt

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3 years ago
What is the minimum efficiency of a functioning current-model catalytic converter? a. 60% b. 75% c. 80% d. 90%
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Answer:

d. 90%

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A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W
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Answer:

It will not  experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

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Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:

<em>σc = KIc / (Y*√(π*a))</em>

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

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3 years ago
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A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
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Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

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Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

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Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

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c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

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Tau ts = Kb {(8FD)/(Πd^3)}

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A = 147 kpsi.inm^3

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= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
3 years ago
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