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____ [38]
3 years ago
7

A structural component in the shape of a flat plate 25.0 mm thick is to be fabricated from a metal alloy for which the yield str

ength and plane strain fracture toughness values are 545 MPa and 29.6 MPa-m1/2, respectively. For this particular geometry, the value of Y is 1.3. Assuming a design stress of 0.3 times the yield strength, calculate the critical length of a surface flaw.
Engineering
2 answers:
taurus [48]3 years ago
6 0

Answer:

The critical length of a surface flaw is 6.17 mm

Explanation:

according to the exercise:

flat plate size=25 mm

σ=545 MPa

plane strain fracture toughness=29.6 MPa-m1/2

Y=1.3

Design stress = 0.3*σ = 163.5 MPa

For plain structure fracture toughness is:

K_{IC} =Y*o\sqrt{\pi a_{c}  }

Where ac is the maximum allowable flat size. Clearing ac:

a_{c} =\frac{1}{\pi } (\frac{K_{IC} }{o*Y} )^{2} =\frac{1}{\pi } (\frac{29.6}{163.5*1.3} )^{2} =6.17x10^{-3}m=6.17mm

balandron [24]3 years ago
5 0

Answer:

The critical length of surface flaw = 6.176 mm

Explanation:

Given data-

Plane strain fracture toughness Kc = 29.6 MPa-m1/2

Yield Strength = 545 MPa

Design stress. =0.3 × yield strength

= 0.3 × 545

= 163.5 MPa

Dimensionless parameter. Y = 1.3

The critical length of surface flaw is given by

= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2

Now putting values in above equation we get,

= 1/3.14( 29.6 / 1.3 × 163.5)^2

=6.176 × 10^-3 m

=6.176 mm

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Answer:

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The maximum shear strain is 1790 μrad.

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Answer:

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
Bad White [126]

Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

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length of the stroke = 3.4 in

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the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature T_1 = 60 F = 519.67 R    

and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

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The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

V_1-V_2 = \dfrac{\pi}{4}D^2L

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V_1-0.16V_1= 36.55714291

0.84 V_1 =36.55714291

V_1 =\dfrac{36.55714291}{0.84 }

V_1 =43.52040823 \ in^3 \\ \\  V_1 = 43.52 \ in^3

V_1 = 0.02518 \ ft^3

the mass in air ( lb) can be determined by using the formula:

m = \dfrac{P_1V_1}{RT}

where;

R = 53.3533 ft.lbf/lb.R°

m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R  \times 519 .67 ^0 R}

m = 0.0018962 lb

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v_{r1} =158.58

u_1 = 88.62 Btu/lb

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v_{r2} = v_{r1}  \times \dfrac{V_2}{V_1}

v_{r2} = 158.58 \times 0.16

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The specific energy u_2 at v_{r2} = 25.3728 is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

v_{r3} = 0.1828

u_3 = 1098 \ Btu/lb

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v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}

v_{r4} =0.1828 \times \dfrac{1}{0.16}

v_{r4} =1.1425

The specific energy u_4 at v_{r4} =1.1425 is 591.84 Btu/lb

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W_{net} = Heat  \ supplied - Heat  \ rejected

W_{net} = m(u_3-u_2)-m(u_4 - u_1)

W_{net} = m(u_3-u_2- u_4 + u_1)

W_{net} = m(1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)

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W = 4 \times N'  \times W_{net

where ;

N' = \dfrac{2400}{2}

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec ×  0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

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