96/64 reduced to its lowest term

Rubber bushings are used on suspensions to

Harlamova29_29 [7]

D. All of the above

4 0

3 years ago

A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque

Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

τ = 16*T / pi*d^3

T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

T_c = τ_c*pi*d_c^3 / 16

T_c = 12*1000*pi*1^3 / 16

T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

T_s = τ_s*pi*d_s^3 / 16

T_s = 15*1000*pi*0.25^3 / 16

T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

T = min ( 2356.2 , 46.02 )

T = 46.02 lb-in

6 0

3 years ago

A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

PIT_PIT [208]

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0

2 years ago

The "Big Dig" was the nickname of the civil engineering project that redesigned the highway Infrastructure for the city of

zheka24 [161]

Geotechnical since it’s geologicaly based

4 0

3 years ago

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are

Vika [28.1K]

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle $W_{net$ = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ $^{Y-1$ = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = $($ T₂ / T₁ $)^{\frac{1}{Y-1}$

so we substitute

⇒ V₁ / V₂ = $($ 973 K / 303 K $)^{\frac{1}{1.4-1}$

= $($ 3.21122 $)^{\frac{1}{0.4}$

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c) the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂ = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

8 0

3 years ago