The density of a certain material is such that it weighs 9 pounds per cubic foot of
1 answer:
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Answer:
F = 8849 N
Explanation:
Given:
Load at a given point = F = 4250 N
Support span = L = 44 mm
Radius = R = 5.6 mm
length thickness of tested material = 12 mm
First compute the flexural strength for circular cross section using the formula below:
σ
σ = FL / π R³
Putting the given values in the above formula:
σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³
= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³
= 187 / 3.141593 (0.0056)³
= 338943767.745358
= 338.943768 x 10⁶
σ = 338 x 10⁶ N/m²
Now we compute the load i.e. F from the following formula:
= 2 σ
d³/3 L
F = 2σd³/3L
= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)
= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)
= 146016 / 125 / 3 ( 44 x 1 / 1000 )
= ( 146016 / 125 ) / (3 ( 11 / 250 ))
= 97344 / 11
F = 8849 N
This question is incomplete, the missing image in uploaded along this answer below.
Answer:
The required stress is 200 Mpa
Explanation:
Given the data in the question;
diameter D = 12 mm = 12 × 10⁻³ m
Length L = 188 mm = 188 × 10⁻³ m
Poisson's ratio v = 0.34
Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m
The transverse strain will;
εˣ = Δd / D
εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m
εˣ = -0.00088
The longitudinal strain will be;
= - ( εˣ / v )
= - ( -0.00088 / 0.34 )
= - ( - 0.002588 )
= 0.0026
Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.
From the graph, in the Second image;
The stress is 200 Mpa
Therefore, The required stress is 200 Mpa
