A cylindrical specimen of a brass alloy having a length of 104 mm (4.094 in.) must elongate only 5.20 mm (0.2047 in.) when a tensile load of 101000 N (22710 lbf) is applied. Under these circumstances what must be the radius of the specimen? Consider this brass alloy to have the stress–strain behavior

Answers

Answer 1

Answer:

The radius of the specimen is assumed to be 9.724 mm

Explanation:

Given that:

For a cylindrical specimen of a brass alloy;

The length = 104 mm, Elongation = 5.20 mm and the tensile load = 101000 N

Let's first determine the radius of the cylindrical brass alloy from the knowledge of the cross-sectional area of a cylinder.

[tex]A_0 = \pi r ^2[/tex]

[tex]r = \sqrt{\dfrac{A_o}{\pi}}[/tex]

[tex]r = \sqrt{\dfrac{\bigg (\dfrac{F}{\sigma} \bigg )}{\pi}}[/tex]

[tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]

To estimate the tensile stress:

We need to first determine the strain relating to elongation at 5.20 mm

[tex]Strain \ \ \varepsilon= \dfrac{\Delta l}{l_o}[/tex]

[tex]Strain \ \ \varepsilon= \dfrac{5.20}{104}[/tex]

Strain ε = 0.05

Using the stress-strain plot; let assume that under the circumstances; [tex]\sigma[/tex] = 340 MPa for stress corresponding to 0.05 strain

Thus;

The cylindrical brass alloy radius [tex]r = \sqrt{\dfrac{F}{ \sigma \pi}}[/tex]

[tex]r =\sqrt{ \dfrac{101000}{(340\times 10^{6})\pi}[/tex]

r = 0.009724 m

r = 9.724 mm


Related Questions

Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 2.6 kg/s. If the mixture leaves the mixing chamber at 60°C.

Required:
Determine the mass flow rate of the superheated steam required.

Answers

Answer:

0.154kg/s

Explanation:

From this question we have the following information:

P1 = 300kpa

T1 = 20⁰c

M1 = 2.6kg/s

For superheated system

P2 = 300kpa

T2 = 300⁰c

M2 = ??

T2 = 60⁰c

From saturated water table

h1 = 83.91kj/kg

h3 = 251.18kj/kg

From superheated water,

h2 = 3069.6kj/kg

The equation of energy balance

m1h1 + m2h2 = m3h3

When we input all the corresponding values:

We get

m2 = -434.902/-2818.42

m2 = 0.15430

m2 = 0.154kg/s

This is the mass flow rate of the superheated steam

Please check attachment for more detailed explanation.

thank you!

This question involves the concepts of energy balance and mass flow rate.

The mass flow rate of the superheated steam required is "0.15 kg/s".

Applying the energy balance in this situation, we get:

[tex]m_1h_1+m_2h_2=m_3h_3[/tex]

where,

m₁ = mass flow rate of liquid water at 300 KPa and 200°C = 2.6 kg/s

m₂ = mass flow rate of superheated at 300 KPa and 300°C = ?

h₁ = enthalpy of liquid water at 300 KPa and 200°C = 83.91 KJ/kg (from saturated steam table)

h₂ = enthalpy of superheated at 300 KPa and 300°C = 3069.6 KJ/kg (from superheated steam table)

h₃ = enthalpy of exiting fluid at 60°C = 251.18 KJ/kg (from saturated steam table)

m₃ = mass flow rate of exiting fluid = 2.6 kg/s + m₂

Therefore,

[tex](2.6\ kg/s)(83.91\ KJ/kg)+(m_2)(3069.6\ KJ/kg)=(2.6\ kg/s+m_2)(251.18\ KJ/kg)\\m_2(3069.6\ KJ/kg-251.18\ KJ/kg)=(2.6\ kg/s)(251.18\ KJ/kg-83.91\ KJ/kg)\\\\m_2=\frac{434.902\ KW}{2818.42\ KJ/kg}[/tex]

m₂ = 0.15 kg/s

Learn more about energy balance here:

https://brainly.com/question/9839609?referrer=searchResults

A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?

Answers

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate of 1.3 m3/min. The work input to the control volume is 240 kJ per kg of air flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW.

Answers

Answer:

Heat transfer = 2.617 Kw

Explanation:

Given:

T1 = 300 k

T2 = 440 k

h1 = 300.19 KJ/kg

h2 = 441.61 KJ/kg

Density = 1.225 kg/m²

Find:

Mass flow rate = 1.225 x [1.3/60]

Mass flow rate = 0.02654 kg/s

mh1 + mw = mh2 + Q

0.02654(300.19 + 240) = 0.02654(441.61) + Q

Q = 2.617 Kw

Heat transfer = 2.617 Kw

Match the use of the magnetic field to its respective description.​

Answers

oooExplanation:

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Write a program that asks the user to enter a list of numbers. The program should take the list of numbers and add only those numbers between 0 and 100 to a new list. It should then print the contents of the new list. Running the program should look something like this:

Please enter a list of numbers: 10.5 -8 105 76 83.2 206

The numbers between 0 and 100 are: 10.5 76.0 83.2

Answers

In python 3.8

nums = input("Please enter a list of numbers: ").split()

new_nums = [x for x in nums if 0 < float(x) < 100]

print("The numbers between 0 and 100 are: " + " ".join(new_nums))

When you said numbers between 0 and 100, I didn't know if that was inclusive or exclusive so I made it exclusive. I hope this helps!

Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.

Answers

Hope this helps...........


A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
of the water when the specific heat capacity of water is 4200 J/kg °C.​

Answers

Answer:

  2.728 kg

Explanation:

The units help you keep the calculation straight.

  [tex]\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}[/tex]

1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?​

Answers

Answer:

  a) up

  b) down

Explanation:

When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).

a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.

__

b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.

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