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Ket [755]
3 years ago
11

A structural component in the shape of a flat plate 29.6 mm thick is to be fabricated from a metal alloy for which the yield str

ength and plane strain fracture toughness values are 535 MPa and 38.5 MPa-m1/2, respectively. For this particular geometry, the value of Y is 1.5. Assuming a design stress of 0.5 times the yield strength, calculate the critical length of a surface flaw. mm
Engineering
1 answer:
slega [8]3 years ago
8 0

Answer:

2.93 mm

Explanation:

Plane strain fracture toughness K_{IC}=38.5 Mpa\sqrt{m}

Yield strength, \sigma=535 Mpa

Design stress=0.5*535=267.5 Mpa

Dimensionless parameter Y=1.5

Critical length of surface flaw is given by

a_c=\frac {1}{\pi}\times (\frac {K_{IC}}{Y\sigma})^{2}=\frac {1}{\pi}\times (\frac {38.5}{1.5*267.5})^{2}= 0.00293\approx 2.93 mm

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Which of the following has special properties that allow forces and pressure to be distributed evenly?
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The y component of velocity in a steady, incompressible flow field in the xy plane is v = -Bxy3, where B = 0.4 m-3 · s-1, and x
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A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob
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The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

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12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

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$N_{avg}=\frac{9.86+14.8+14.8}{3}$

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