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Svetradugi [14.3K]
4 years ago
6

A load P is applied horizontally while the other end is fixed to a structure. A load P is applied horizontally while the other e

nd is fixed to a structure. How many bolts are required to make the assembly safe? Use 8 mm bolts (M8-1.25) that yield at 0.5 MPa. (You must provide an answer before moving to the next part.)
Engineering
1 answer:
sveta [45]4 years ago
8 0

Answer:

The number of bolt is 4.

Explanation:

Given that,

\tau = 0.5\ MPa

distance = 8 mm

Suppose, Load P = 205 N

We need to calculate the shear area of bolt

Using formula of shear stress

\tau=\dfrac{P}{A}

Here is double shear.

So, \tau will be double.

A=\dfrac{P}{2\tau}

Put the value into the formula

A=\dfrac{205}{2\times0.5\times10^{6}}

A=205\times10^{-6}\ m^2

We need to calculate the area of one bolt

Using formula of area

A'=\dfrac{\pi}{4}d^2

Put the value into the formula

A'=\dfrac{\pi}{4}\times(8\times10^{-3})^2

A'=50.26\times10^{-6}\ m^2

We need to calculate the number of bolt

Using formula of number of bolt

n=\dfrac{A}{A'}

Put the value into the formula

n=\dfrac{205\times10^{-6}}{50.26\times10^{-6}}

n=4.0

Hence, The number of bolt is 4.

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Answer:

Explanation:

Fist you need to identify where the leak is coming from. You can do this by either listening for the leak or spraying soapy water on the exhaust to look for air bubbles coming out of the exhaust. Depending on the spot of the leak there are many ways you can fix this leak.

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4 0
2 years ago
You must create a Low Pass Filter for an audio amplifier. You must pass 10khz and block
Katena32 [7]

Answer:

) You must create a Low Pass Filter for an audio amplifier. You must pass 10khz and block

60khz. In designing this, you must choose a cutoff frequency exactly half way between

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Given these criteria, analyze this circuit, and determine the necessary resistor value in

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“Insert”--- “Equation”). (6 marks)

a. Solve for fc

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Explanation:

6 0
3 years ago
A metal having a cubic structure has a density of , an atomic weight of , and a lattice parameter of Å One atom is associated wi
Veronika [31]

Answer:

Explanation:

Answer: The crystal structure of the metal is BCC

Explanation:

we first calculate the volume of the unit cell.

Volume of unit cell= (a°)^3.

The lattice parameter here is a°.

Substitute (6.13 * 10^-8)cm for a°.

Volume of unit cell = (6.13 * 10^-8)^3 = 2.3034 * 10^-22 cm^3/cell.

To determine the crystal structure we use

Density (p) = {(Number of atoms per cell) (Atomic mass)} / {(volume of unit cell)(Avogrado constant)}.

Substitute 1.892g/cm^3 for p (6.02*10^23) atoms/mol for Avogrado constant 1.3921g/mol.

For atomic mass and (2.3034 * 10^-22) cm^3/cell for unit cell.

1.892g/cm^3 = {(Number of atoms per cell) (1.3291g/mol)} / {(2.3034 * 10^-22) (6.02 * 10^23 atoms/mol)}.

Changing the subject of formula we have :

Number of atoms per cell = {(2.3034 * 10^-22) * (6.02 * 10^23) * 1.892} / 132.91

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4 0
3 years ago
A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b
Mamont248 [21]

Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

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Let,

critical stress required for initiating crack propagation Cc = 112MPa

plain strain fracture toughness = 27.0MPa

surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

Also for 6.2mm length of surface crack;

Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m

The dimensionless parameter

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Y = Kic/(Cc*√pia*a)

Y = 27/(112*√pia*4.4*10-³)

Y = 2.05

Now,

Cc = Kic/(Y*√pia*a)

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The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

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