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vesna_86 [32]
3 years ago
5

Calculate the factor of safety against bending failure of a gear with a given tooth bending stress of 117MPa and a corrected ben

ding strength of 354MPa.
Engineering
1 answer:
Deffense [45]3 years ago
3 0

Answer:

3

Explanation:

Bending stress = 117 MPa

Yield stress = 354 MPa

Factor of safety is defined as the ratio of yield stress to the allowable stress (tensile, compressive or bending) of a given material

FS= \frac{\text{Yield stress}}{\text{Allowable stress}}\\\Rightarrow FS=\frac{354}{117}\\\Rightarrow FS=3.025

So, factor of safety is almost equal to 3.

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Explanation:

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An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in
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Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, T_{a} = 140^{\circ}C = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, T_{i} = 60^{\circ}C = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

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Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)

Thus

T_{e} = 109.71^{\circ}C = 273 + 109.71 = 382.71 K

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T_{e} = Equilibrium temperature

Now,

To calculate the changer in entropy:

\Delta s = \Delta s_{a} + \Delta s_{i}

Now,

For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K

For Iron:

\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

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