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3 years ago

Engineering Specificaitons is ...

1 answer:

5 0

"Engineering specifications" means those records which provide detailed documentation of the construction, wiring, arrangement and related engineering details of the information processing equipment.

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

3. In order to obtain your commercial driver's license (CDL) you must first:

Murljashka [212]

A and C is the answer to the question. Be 15 years old & get a permit

8 0

3 years ago

A 100 meter dash is run on a track in the direction of the vector = 2 + 7 . The wind velocity is = 5 + km/hr. The rules say that

Scrat [10]

Answer:joe mamma

Explanation:

6 0

3 years ago

You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,

nadezda [96]

Answer:

Explanation:

cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

= 1.075 x 10⁻³

Young's modulus = stress / strain

= 83.32 x 10⁶ / 1.075 x 10⁻³

= 77.5 x 10⁹ N/m²

5 0

3 years ago

Which items are NOT found on a

Alja [10]

Answer:

None of the above cause thats what i put

3 0

3 years ago