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Ratling [72]
3 years ago
9

A crystalline grain of aluminum in a metal plate is situated so that a tensile load is oriented along the [1 1 1] direction. Wha

t tensile stress is required (to 3 significant figures) to produce a critical resolved shear stress of 0.242 MPa along [1 0 1] in the (1 1 -1) plane
Engineering
1 answer:
Ivenika [448]3 years ago
3 0

Answer: required tensile stress is 0.889 MPa

Explanation:

Given that;

tensile load is oriented along the [1 1 1] direction

shear stress is 0.242 MPa along [1 0 1] in the (1 1 -1) plane

first we determine

λ which is Angle between  [1 1 1]  and  [1 0 1]

so

cosλ = [ 1(1) +  1(0) + 1(1) ] / [ √(1² + 1² + 1²) √(1² + 0² + 1²)]

= 2 / √3√2 =  2/√6

Next, we determine ∅ which is angle between [1 1 1]  and  [1 1 -1]

so,

cos∅ = [ 1(1) +  1(1) + 1(-1) ] / [ √(1² + 1² + 1²) √(1² + 1² + (-1)²)]

cos∅ = [ 2-1] / [√3√3 ]

cos∅ = 1/3

Now, we know that;

σ = T_stress/cosλcosθ

so we substitute

σ = 0.242 / ( 2/√6 × 1/3 )

σ = 0.242 / 0.2721

σ = 0.889 MPa

Therefore, required tensile stress is 0.889 MPa

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The population size would be p' = 5000

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So in this problem we are going to be examining the application of a  population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield,  so basically we would be applying an engineering solution to fishing

 

    So the current  yield which is mathematically represented as

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            and dt is the change in time

So in order to obtain the solution we need to obtain the  rate of growth

    For this we would be making use of the growth rate equation which is

                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

  Where N is the population of the fish which is given as 4,000 fishes

          and  K is the carrying capacity which is given as 10,000 fishes

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                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

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                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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