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3 years ago

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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 210 MPa in the bolts and 145 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design.

1 answer:

dusya [7]3 years ago

5 0

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

For steel bolt

Stress = 210 MPa or 210 N/mm2

Pressure = Stress* Area

Pbolt = 210 N/mm2 * 16^2 *(pi)/4

Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N

For Brass spacer

Pressure = 42201.6 N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2

d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758

d^2 = 370.758 + 16^2

d^2 = 626.758

d = 25.03 mm

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

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A light pressure vessel is made of 2024-T3 aluminum alloy tubing with suitable end closures. This cylinder has a 90mm OD, a 1.65

Andru [333]

Given:

Outer Diameter, OD = 90 mm

thickness, t = 0.1656 mm

Poisson's ratio, $\mu$ = 0.334

Strength = 320 MPa

Pressure, P =3.5 MPa

Formula Used:

1). $Axial Stress_{max}$ = $P[\frac{r_{o}^{2} + r_{i}^{2}}{r_{o}^{2} - r_{i}^{2}}]$

2). factor of safety, m = $\frac{strength}{stress_{max}}$

Explanation:

Now, for Inner Diameter of cylinder, ID = OD - 2t

ID = 90 - 2(1.65) = 86.7 mm

Outer radius, $r_{o}$ = 45mm

Inner radius, $r_{i}$ = 43.35 mm

Now by using the given formula (1)

$Axial Stress_{max}$ = $3.5[\frac{45^{2} + 43.35^{2}}{45^{2} - 43.35^{2}}]$

$Axial Stress_{max}$ = $3.5\times 26.78$ =93.74 MPa

Now, Using formula (2)

factor of safety, m = \frac{320}{93.74} = 3.414

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3 years ago

Please help ASAP!!

masha68 [24]

Answer:

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Explanation:

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A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when

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Answer:

1. $3.4^{o}$

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

= (50)(1000/3600) m/s

= 13.89 m/s

The acceleration due to gravity, $g = 9.81 m/s^{2}$

The frictional force, f = μsN ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

ΣF = mg sinθ-f = 0 ...... (2)

Substitute the equation (1) in equation (2), we get

ΣF = mgsinθ-μsN = 0

mgsinθ-μsmgcosθ =0

μs = sinθ/cosθ

tanθ = μs

θ = tan-1( μs) = tan-1(0.06) = $3.4^{o}$

(b)The vertical component of the force is

N cosθ = fsinθ+mg

N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma (Centripetal acceleration, $a = \frac {v^{2}}{r}$

Nsinθ+fcosθ = $m(\frac {v^{2}}{r})$

Nsinθ+μsNcosθ = $m(\frac {v^{2}}{r})$

N[sinθ+μs cosθ] = $m(\frac {v^{2}}{r})$ ...... (4)

Dividing the equation (4) with equation (3),

[sinθ+μscosθ]/[cosθ- μs sinθ] = $\frac {v^{2}}{rg}$

cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] = $\frac {v^{2}}{rg}$

[tanθ+μs]/[1-μs tanθ] = $\frac {v^{2}}{rg}$

From part (1), tanθ = μs

Then the above equation becomes

$\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

$\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

Therefore, the minimum radius of the curvature of the curve is

$r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}$

= $\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}$

= $\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}$

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3 years ago

In fully developed laminar flow in a circular pipe the velocity at R/2 (mid-way between the wall surface and the centerline) is

Butoxors [25]

Answer:

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Explanation:

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$u_{max} = 2\cdot U_{avg}$

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Please answer fast. With full step by step solution.

lina2011 [118]

Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

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8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

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$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

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$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

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$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

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$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

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