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mylen [45]
4 years ago
9

Assuming the torsional yield strength of a compression spring is 0.43Sut and the maximum shear stress is equal to 434MPa. What i

s the factor of safety if the spring is a music wire (A=2170MPa and m=0.146) with a wire diameter of 4mm?
Engineering
1 answer:
pav-90 [236]4 years ago
4 0

Answer:

factor of safety is 1.756

Explanation:

given data

torsional yield strength = 0.43 sut  

maximum shear stress = 434 MPa

A = 2170 MPa

m = 0.146

diameter = 4 mm

to find out

factor of safety

solution

we know 1 sut is equal to 1772.39 MPa

so 0.43 sut = 0.43 ×1772.39 =  762.128 MPa

so here

factor of safety formula is

factor of safety = yield strength / shear stress   ............1

put here value

factor of safety = 762.128 / 434

factor of safety is 1.756

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3 years ago
Two added to four times a number, minus 3 times the number, equals 5.
vladimir1956 [14]
<h2>Answer:</h2>

<u>x= 3</u>.

<h2>Explanation:</h2>

<em>What is presented in this problem is basically an equation in verbal form.</em>

<em />

<h3>1. Write the equation.</h3>

2+4x-3x=5

<h3>2. Solve for x.</h3>

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ale4655 [162]

Answer:

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Explanation:

For this exercise let's use ohm's law

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the third way of calculation

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