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riadik2000 [5.3K]
3 years ago
8

Three 1.83 in. diameter bolts are used to connect the axial member to the support in a double shear connection. The ultimate she

ar strength of the bolts is 60 ksi, and a factor of safety of 3.9 is required with respect to fracture. Determine the allowable load P that can be applied to the axial member based on the shear strength of the bolts. Give your answer in kips. Enter a positive number.
Engineering
1 answer:
seraphim [82]3 years ago
3 0

Answer: the allowable load P is 242.7877 kips

Explanation:

Given that;

diameter bolts d = 1.83 in

ultimate shear strength of the bolts = 60 ksi

we know that

shear area = 2×(π/4)d²

= 2×(π/4)×(1.83)² = 5.2604 in²

so

p/3(5.2604) = 60000/3.9

p/15.7812 = 15384.6153

p = 15.7812 × 15384.6153

p = 242787.691 lb

p = 242.7877 kips

therefore the allowable load P is 242.7877 kips

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Signal generator‘s internal impedance is purely resistive and has an open-circuit voltage of 3.5 V. When the generator is loaded
Bezzdna [24]

Answer:

r = 5.5 ohms

Explanation:

Given:-

- The open circuit voltage, Vo = 3.5 V

- The terminal voltage, Vt = 2.8 V

- The load, R = 22 ohms

- The internal resistance = r

Find:-

What is the generator’s output impedance (pure resistance)?

Solution:-

- We see that the source Voltage (Vo) is not entirely used for the attached load. Some of the source voltage is dropped within due to the source internal resistance or impedance.

- The terminal voltage (Vt) is the amount of Voltage drop across the load. The current drawn by the load I can be determined by Ohm's Law:

                            Vt = I*R

                            I = Vt / R

                            I = 2.8 / 22

                            I = 0.12727 Amps

- Since, the attached load (R) and the pure impedance (r) of the source are in series. The current  ( I ) is constant across both. The potential drop across the pure resistance (r) can be determined from Ohm's law:

                          Vo - Vt = I*r

                          r = ( Vo - Vt ) / I

                          r = ( 3.5 - 2.8 ) / 0.12727

                          r = 5.5 ohms

4 0
3 years ago
Could I please get help with this​
alex41 [277]

Answer:

1.I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = 36.661458\overline 3 in.⁴

Iₓ = 28.6458\overline 3 in.⁴

I_y = 138.6548\overline 3 in.⁴

2. I_{xc} = 114.\overline 3 in.⁴

I_{yc} = 37.\overline 3 in.⁴

Iₓ = 457.\overline 3 in.⁴

I_y = 149.\overline 3 in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458\overline 3

I_{xc} = 7.161458\overline 3 in.⁴

I_{yc} = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458\overline 3

I_{yc} = 36.661458\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458\overline 3

Iₓ = 28.6458\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548\overline 3

I_y = 138.6548\overline 3 in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

I_{xc} = b·h³/12 = 4 × 7³/12 = 114.\overline 3

I_{xc} = 114.\overline 3 in.⁴

I_{yc} = h·b³/12 = 7 × 4³/12 = 37.\overline 3

I_{yc} = 37.\overline 3 in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457.\overline 3

Iₓ = 457.\overline 3 in.⁴

I_y = h·b³/3 = 2.5 × 5.5³/3 = 149.\overline 3

I_y = 149.\overline 3 in.⁴

3. The deflection, \delta _{max}, of a simply supported beam having a point load at the center is given as follows;

\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, I_{xc} = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552

The maximum deflection of the beam, \delta _{max} = 2.55552 inches

5 0
3 years ago
A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated pow
Anna [14]

Answer:

Rated power = 1345.66 W/m²

Mechanical power developed = 3169035.1875 W

Explanation:

Wind speed, V = 13 m/s

Coefficient of performance of turbine, C_p = 0.3

Rotor diameter, d = 100 m

or

Radius = 50 m

Air density, ρ = 1.225 kg/m³

Now,

Rated power = \frac{1}{2}\rho V^3

or

Rated power = \frac{1}{2}\times1.225\times13^3

or

Rated power = 1345.66 W/m²

b) Mechanical power developed =  \frac{1}{2}\rho AV^3C_p

Here, A is the area of the rotor

or

A = π × 50²

thus,

Mechanical power developed = \frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3

or

Mechanical power developed =  3169035.1875 W

8 0
3 years ago
For a 20 ohm resistor R, the current i = 2 A. What is the voltage V? Submit your answer as a number without units. ​
svetoff [14.1K]

Answer:

20*2=40

Explanation:

7 0
3 years ago
A driver complains that his front tires are wearing
Margarita [4]

Answer:

The correct option is;

Neither A nor B

Explanation:

The location of the where the thread wears in tire that has too high inflation is at the thread pattern center due to the reduced size of the contact patch with the load of the car resting on the central portion of the tire's contact surface

When the wear occurs at the outer edges of the tire, the load of the car rests on the outer edges as the contact patch increases due to the tire being under-inflated

Camber is the slope provided in road pavement to drain off water from the road

Roads with camber has a raised middle portion and wear due to camber includes outer-edge tread wear, inner-edge tread wear and tire feathering

8 0
2 years ago
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