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Anna007 [38]
3 years ago
13

*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t

hat θ = 89.3 ° . Determine the strains ε x , ε y and γ x y . Take ν r = 0.5.

Engineering
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

Explanation:

We are given;

δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

Let's calculate ε_x in the direction of axis x

Thus, ε_x = δ/L = 0.03/4 = 0.0075

Let's calculate ε_y in the direction of axis y;

ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375

Now, shear strain is angle between π/2 rad surfaces at that point.

Thus,

γ_xy =  π/2 - θ = π/2 - 89.3π/180

γ_xy = π(0.003889) = 0.0122 rad

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A) Consider an air standard otto cycle that has a heat addition of 2800 kJ/kg of air, a compression ratio of 8 and a pressure an
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Answer:

a) i) The maximum pressure is approximately 122.37 bar

ii) The thermal efficiency is approximately 56.47%

iii) The mean effective pressure is approximately 20.974 bar

b) (b) Four types of internal combustion engine includes;

1) The diesel engine

2) The Otto engine

3) The Brayton engine

4) The Wankel engine

Explanation:

The parameters of the Otto cycle are;

The heat added, Q_{in} = 2,800 kJ/kg

The compression ratio, r = 8

The beginning compression pressure, P₁ = 1 bar

The beginning compression temperature, T₁ = 300 K

Cp = 1.005 kJ/kg·K

Cv = 0.718 kJ/kg·K

R = 287 kJ/kg·K

K = Cp/Cv = 1.005 kJ/kg·K/(0.718 kJ/kg·K) ≈ 1.4

T₂ = T₁×r^(k - 1)

∴ T₂ = 300 K×8^(1.4 - 1) ≈ 689.219 K

\dfrac{P_1\cdot V_1}{T_1}  = \dfrac{P_2\cdot V_2}{T_2}

P_2 = \dfrac{P_1\cdot V_1 \cdot T_2}{T_1 \cdot V_2}  = \dfrac{V_1}{V_2} \cdot  \dfrac{P_1 \cdot T_2}{T_1 } = r \cdot  \dfrac{P_1 \cdot T_2}{T_1 }

∴ P₂ = 8 × 1 bar × (689.219K)/300 K ≈ 18.379 bar

Q_{in} = m·Cv·(T₃ - T₂)

∴ Q_{in} = 2,800 ≈ 0.718 × (T₃ - 689.219)

T₃ = 2,800/0.718 + 689.219 = 4588.94 K

P₃ = P₂ × (T₃/T₂)

P₃ = 18.379 bar × 4588.94K/(689.219 K) = 122.37 bar

The maximum pressure = P₃ ≈ 122.37 bar

(ii) The thermal efficiency, \eta_{Otto}, is given as follows;

\eta_{Otto} = 1 - \dfrac{1}{r^{k - 1}}

Therefore, we have;

\eta_{Otto} = 1 - \dfrac{1}{8^{1.4 - 1}} \approx 0.5647

The thermal efficiency, \eta_{Otto} ≈ 0.5647

Therefore, the thermal efficiency ≈ 56.47%

(iii) The mean effective pressure, MEP is given as follows;

MEP = \dfrac{\left(P_3 - P_1 \cdot r^k \right) \cdot \left(1 - \dfrac{1}{r^{k-1}} \right)}{(k -1)\cdot (r - 1)}

Therefore, we get;

MEP = \dfrac{\left(122.37 - 1 \times 8^{1.4} \right) \cdot \left(1 - \dfrac{1}{8^{1.4-1}} \right)}{(1.4 -1)\cdot (8 - 1)} \approx 20.974

The mean effective pressure, MEP ≈ 20.974 bar

(b) Four types of internal combustion engine includes;

1) The diesel engine; Compression heating is the source of the ignition, with constant pressure combustion

2) The Otto engine which is the internal combustion engine found in cars that make use of gasoline as the source of fuel

The Otto engine cycle comprises of five steps; intake, compression, ignition, expansion and exhaust

3) The Brayton engine works on the principle of the steam turbine

4) The Wankel it follows the pattern of the Otto cycle but it does not have piston strokes

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A cross beam in a highway bridge experiences a stress of 14 ksi due to the dead weight of the bridge structure. When a fully loa
zlopas [31]

Answer:

a) 2.452

b) 1.256

Explanation:

Stress due to dead weight. = 14 Ksi

Stress due to fully loaded tractor-trailer = 45Ksi

ultimate tensile strength of beam = 76 Ksi

yield strength = 50 Ksi

endurance limit = 38 Ksi

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β = ( 45 - 14 ) / 2

  = 15.5 Ksi

hence FOS ( factor of safety ) = endurance limit / β

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b) When mean stress on fatigue strength is considered

β2 = 45 + 14 / 2

    = 29.5 Ksi

Ratio  = β / β2 = 15.5 / 29.5 = 0.5254

Next step: applying Goodman method

Sa =  [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]

     = 19.47 Ksi

hence the FOS ( factor of safety ) = Sa / β

                                                      = 19.47 / 15.5 = 1.256

8 0
3 years ago
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