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lbvjy [14]
2 years ago
14

When bending metal, the material on the outside of the curve stretches while the material on the inside of the curve compresses.

That part of the material which is not affected by either stress is the g
Engineering
1 answer:
Nostrana [21]2 years ago
4 0

That part of the material which is not affected by either stress is the neutral line or axis.

<h3>What is the technique known as whilst you bend steel?</h3>

Sheet Metal Bending – Methods, Design Tips & K Factor. Bending is one of the maximum not unusual place sheet steel fabrication operations. Also referred to as press braking, flanging, die bending, folding and edging, this technique is used to deform a cloth to an angular shape. This is finished thru the utility of pressure on a workpiece.

The neutral axis is in which neither the fabric stretches nor is compressed. Hence, the duration of the nuetral axis lives the identical earlier than and after the bending operation.

Read more about the bending :

brainly.com/question/13637124

#SPJ1

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Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain
fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

N_{A}  = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1}  } ln(\frac{P_{T} -P_{A2}  }{P_{T} -P_{A1} })

Where

D_{AB} = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

P_{A1} = Pressure at first boundary

P_{A2} = Pressure at the destination boundary

T = System temperature

P_{T} = System pressure

Where P_{T} = 101.3 kPa P_{A2} =0, P_{A1} =y_{A}, P_{T} = 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R =  \frac{kJ}{(kmol)(K)} ,    T = 298 K   and  D_{AB} = 1.18 \frac{cm^{2} }{s} = 1.8×10⁻⁵\frac{m^{2} }{s}

N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65}) = 5.153×10⁻⁴\frac{kmol}{m^{2}s }

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0
3 years ago
Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
3 years ago
Calculate the density of the FCC nickel lattice with an interstitial hydrogen in the centered position of the unit cell. You may
Anastaziya [24]

Answer:

\rho=8907.94\ Kg/m^3

Explanation:

Given that

a=3.524 A

At.Wt. ,M= 58.7 g/mole,

 For FCC

 Z = 4

4r=\sqrt2\ a

The density given as

\rho=\dfrac{ZM}{N_Aa^3}

\rho=\dfrac{4\times 58.7\times 10^{-3} }{ 6.023\times 10^{23}\times (3.524\times 10^{-10})^3}

\rho=8907.94\ Kg/m^3

So the density is \rho=8907.94\ Kg/m^3

4 0
3 years ago
Why was information revolution different or unique <br>​
Xelga [282]

Answer:

It made information easily accessible and ensured individuals became more vast in subject topics of interest.

Explanation:

Information revolution is different and unique and it came with the advent of computers and the internet. A lot of information is stored there which is too large and complex for the human brain.This helped people to access information without much stress as informations about almost every subject is on the Internet.

Individuals can check the informations up and become more vast in interested topics.

8 0
3 years ago
Please answer the questions !
gizmo_the_mogwai [7]

Answer:

120

Explanation:

6 0
3 years ago
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