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4 years ago

A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.

Engineering

1 answer:

Tems11 [23]4 years ago

6 0

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

$S_{ut}=K \left(\cfrac ne \right)^n$

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

$K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}$

And we can replace values

$K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87$

Thus we get that the strength coefficient is K = 591.87 MPa

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The following displacement (cm), output (V) data have been recorded during a calibration of an LVDT. Displacement (cm), output (

nadya68 [22]

Answer:

2.08V/cm

Explanation:

Plot the points on a graph. Draw the line of best fit. Calculate the gradietn of line of best fit.

Attached is the graph plotted on excel.

The equation of the line is

Votlage= 2.08× distance + 0.276

4 0

4 years ago

The acceleration (in m/s^2) of a linear slider (undergoing rectilinear motion) within a If the machine can be expressed in terms

Inga [223]

Answer:

47.91 sec

Explanation:

it is given that $\alpha =\frac{1}{4v^{2}}$

at t=0 velocity =0 ( as it is given that it is starting from rest )

we have to find time at which velocity will be 3.3 $\frac{m}{sec^{2}}$

we know that $\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}$

$4v^{2}dv=dt$

integrating both side

$\frac{4v^{3}}{3}=t+c$ ---------------eqn 1

at t=o it is given that v=0 putting these value in eqn 1 c=0

so $\frac{4v^{3}}{3}=t$

when v= 3.3 $\frac{m}{sec^{2}}$

t= $\frac{4}{3}\times 3.3^{3}$

=47.91 sec

6 0

4 years ago

timofeeve [1]

A. The factor by which a machine multiplies a force

6 0

3 years ago

An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci

Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0

3 years ago

The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%

Aleonysh [2.5K]

Answer:

4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

4 0

3 years ago