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german
2 years ago
14

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

ded tractor-trailer crosses over the bridge, however, the stress in the beam increases to 45 ksi. The beam is fabricated from steel with an ultimate tensile strength of 76 ksi, a yield strength of 50 ksi, and an endurance limit of 38 ksi. Find the safety factor for an infinite fatigue life:
a. if the effect of mean stress on fatigue strength is ignored
b. when the effect of mean stress on fatigue strength is considered.
Engineering
1 answer:
zlopas [31]2 years ago
8 0

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

Determine the safety factor for an infinite fatigue life

a) If mean stress on fatigue strength is ignored

β = ( 45 - 14 ) / 2

  = 15.5 Ksi

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

                                                 = 38 / 15.5 = 2.452

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

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An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel castin
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Answer:

h = 375 KW/m^2K

Explanation:

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                                               E_in = E_out

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Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2

q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)

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victus00 [196]

Answer:

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Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

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32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

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The principal stress is:

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σp1 = 20 ksi

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Solving both equations:

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σp2 = -37 ksi

The shear stress on the vertical plane is zero

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3 years ago
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