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5. A non-cold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa. Estimate the yield strength o

Papessa [141]

Answer:

97.17 MPa

Explanation:

Given:-

- The nominal strength of the grain, σ0 = 25 MPa

- The average grain size of the brass specimen, d* = 0.01 m

- The yield strength of the non-cold worked specimen, σy = 150 MPa

- Conditions of cold-working: T = 500°C , t = 1000 s

Find:-

Estimate the yield strength of this alloy after cold - working process

Solution:-

- The nominal strength of the grain is a function of yield strength of the material, grain yield factor ( Ky ) and the grain size.

- the following relation is used to determine the grain strength:

σ0 = σy - ( Ky / √( d ) )

- We will use the above relation to determine the grain yield factor ( Ky ) for the alloy as follows. Note: here we will use the average value of grain size:

Ky = ( σy - σy )*√( d* )

Ky = ( 150 - 25 ) * √0.01

Ky = 12.5 MPa - √mm

- Now we will use the cold working conditions of T = 500 C and time of the process is t = 1000 s. We will look up the elongated size of the grain after the cold-working process in lieu with its yield factor ( Ky ). Use figure 7.25.

- The cold-worked grain size with the given conditions can be read off from the figure 7.25. The new size comes out to be d = 0.03 mm.

- We will again use the nominal grain strength relation expressed initially. And compute for the new yield strength of the cold-worked alloy.

σ0 = σy - ( Ky / √( d ) )

σy = σ0 + ( Ky / √( d ) )

σy = 25MPa + ( 12.5 / √( 0.03 mm ) )

σy = 97.17 MPa

- We see that the yield strength of the alloy decreases after cold-working process. This happens because the cold working process leaves with inter-granular strain ( dislocation of planes ) in the material structure which results from the increase in grain size.

6 0

3 years ago

The cold drawn AISI 1040 steel bar with 25-mm width and 10-mm thick has a 6- mm diameter thru hole in the center of the plate. T

4vir4ik [10]

Answer:

A) ( N ) = 1.54

B) N ( Goodman ) = 1.133, N ( Morrow) = 1.35

Explanation:

width of steel bar = 25-mm

thickness of steel bar = 10-mm

diameter = 6-mm

load on plate = between 12 kN AND 28 kN

notch sensitivity = 0.83

A ) Fatigue factor of safety based on yielding criteria

= δa + δm = $\frac{Syt}{n}$ = 91.03 + 227.58 = 490 / N

therefore Fatigue number of safety ( N ) = 1.54

δa (amplitude stress ) = kf ( Fa/A) = 2.162 * ( 8*10^3 / 190 ) = 91.03 MPa

A = area of steel bar = 190 mm^2 , Fa = amplitude load = 8 KN , kf = 2.162

δm (mean stress ) = kf ( Fm/A ) = (2.162 * 20*10^3 )/ 190 = 227.58 MPa

Fm = mean load = 20 *10^3

B) Fatigue factor of safety based on Goodman and Morrow criteria

δa / Se + δm / Sut = 1 / N

= 91.03 / 183.15 + 227.58 / 590 = 1 /N

Hence N = 1.133 ( based on Goodman criteria )

note : Se = endurance limit (calculated) = 183.15 , Sut = 590

applying Morrow criteria

N = 1 / ( δa/Se) + (δm/ δf )

= 1 / ( 91.03 / 183.15 ) + (227.58 / 935 )

= 1.35

6 0

3 years ago

g The parameters of a certain transmission line operating at 휔휔=6 ×108 [rad/s] are 퐿퐿=0.35 [휇휇H/m], 퐶퐶=75 [pF/m], 퐺퐺=75 [휇휇S/m],

yKpoI14uk [10]

Explanation:

$\begin{aligned}\gamma &=\sqrt{Z Y}=\sqrt{(R+j \omega L)(G+j \omega C)} \\&-\sqrt{|17|} j\left(6 \times 10^{8}\right)\left(0.35 \times 10^{-6}\right)|| 75 \times 10^{-6}\left|j\left(6 \times 10^{8}\right)\left(40 \times 10^{-12}\right)\right| \\&=0.094+j 2.25 \mathrm{m}^{-1}-\alpha+j \beta\end{aligned}$

Therefore,

$-\alpha-0.094 \mathrm{Np} / \mathrm{m} . \quad 3-2.25 \mathrm{rad} / \mathrm{m}, \text { and } \lambda-2 \pi / \beta-\underline{2.79} \mathrm{m}$

$Z_{0}-\sqrt{\frac{Z}{Y}}-\sqrt{\frac{R+j \omega L}{G+j \omega C}}-\sqrt{\frac{17+j 2.1 \times 10^{2}}{75 \times 10^{-6}+j 2.4 \times 10^{-2}}}-\frac{93.6-j 3.64 \Omega}{4}$

5 0

3 years ago

An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pres

My name is Ann [436]

Answer:

A) 222.58 kJ / kg

B) 0.8897 M^3/ kg

c) 0.7737 m^3/kg

D) 746.542 k

E) 536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) = 0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

$\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }$

hence T2 = $9^{1.4-1} * 310$ = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

3 0

3 years ago

Establishes general guidelines concerning licensing and vehicle

gavmur [86]

Answer:

A. National Highway Safety Act

Explanation:

The National Highway Safety Act establishes general guidelines concerning licensing, vehicle registration and inspection, and traffic laws for state regulations. The act was made in 1966 to reduce the amount of death on the highway as a result of increase in deaths by 30% between 1960 and 1965

National Traffic and Motor Vehicle Safety Act regulates vehicle manufacturers by ensuring national safety standards and issuance recalls for defective vehicles

Uniform Traffic Control Devices Act defines shapes, colors and locations for road signs, traffic signals, and road markings

5 0

3 years ago

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