Physics, 23.01.2020 17:31 mamieengler

Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. knowing that the average normal stress must not exceed 216 mpa in the bolts and 143 mpa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design. the outer diameter of the spacers that yields the most economical and safe design is 34.854 mm

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Answer from: dootdootkazoot

$d_s = 16 mm \sqrt{1+\frac{216 Mpa}{143 Mpa}} =25.35 mm$

So then the outer diameter for the spaces should be 25.35 in order to have the most economical and safe design.

Explanation:

Let's define some notation first:

$\sigma_b =216 Mpa$ represent the tensile stress for the bolt

$\sigma_s = 143Mpa$ represent the tensile stress for the spacer

d_b = 16 mm the bolt diameter

d_s the value that we want to find.

For this case we need to remember the definition of tensile stress given by:

$\sigma = \frac{F}{A} = \frac{P}{ \frac{\pi}{4} d^2}$

And if we solve for P we got:

$P = \frac{\pi}{4} d^2 \sigma_b$ where P represent the normal stress

For this case we want that the most economical space and we can do this setting equal the both values of P for this case like this:

P_b =P_s

And if we replace we got:

$\frac{\pi}{4} \sigma_s (d^2_s - d^2_b) = \frac{\pi}{4} \sigma_b d^2_b$

And if we cancel the constant terms we got:

$d^2_s - d^2_b = \frac{\sigma_b}{\sigma_s} d^2_b$

And if we solve for the variable of interest we got:

$d^2_s = d^2_b +\frac{\sigma_b}{\sigma_s} d^2_b$

And taking common factor we got:

$d^2_s = d^2_b (1+\frac{\sigma_b}{\sigma_s})$

And if we take the square root on both sides we got:

$d_s = d_b \sqrt{1+\frac{\sigma_b}{\sigma_s}}$

And if we replace we got:

$d_s = 16 mm \sqrt{1+\frac{216 Mpa}{143 Mpa}} =25.35 mm$

So then the outer diameter for the spaces should be 25.35 in order to have the most economical and safe design.

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a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm. the wire diameter is d = 0.9 mm. the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

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d = 0.9 mm

i = 9 a

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calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

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