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3 years ago

Why is the uniaxial tension test commonly used to evaluate the mechanical properties of metals?

Engineering

1 answer:

inn [45]3 years ago

3 0

Answer:

Because with this test you can determine complex material parameters like Young’s modulus, yield strength, ultimate strength and elongation at break. This is important because it provides us with the factor of safety that needs to be built-in the products using these materials.

Explanation:

This test consists in place the material between to tweezers to subdued the material into a stress-strain test. The figure shows the procedure.

In the x axe the strain and the y axe the stress.

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____ grinders are used to grind diameters, shoulders, and faces much like the lathe is used for turning, facing, and boring oper

skelet666 [1.2K]

Answer:

Cylindrical

Explanation:

A cylindrical grinder is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this. Cylindrical grinding machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.

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2 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

What is the mass of a brass axle that has a volume of 0.318 cm?

NeX [460]

Answer:

2.7g

Explanation:

the mass of a brass axle that has a volume of 0.318 cm is 2.7g.

8 0

3 years ago

A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut for the s

Xelga [282]

Answer:

See explanation

Explanation:

Given The bar is square and has a hot-rolled finish. The loading is fully reversed bending.

Tensile Strength

Sut: 600 MPa

Maximum temperature

Tmax: 500 °C

Bar side dimension

b: 150 mm

Alternating stress

σa: 100 MPa

Reliability

R: 0.999 Note 1.

Assumptions Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

Solution See Excel file Ex06-01.xls.

1 Since no endurance-limit or fatigue strength information is given, we will estimate S'e based on the ultimate tensile strength using equation 6.5a.

S'e: 300 MPa = 0.5 * Sut

2 The loading is bending so the load factor from equation 6.7a is

Cload: 1

3 The part size is greater than the test specimen and the part is not round, so an equivalent diameter based on its 95% stressed area must be determined and used to find the size factor. For a rectangular section in nonrotating bending, the A95 area is defined in Figure 6-25c and the equivalent diameter is found from equation 6.7d

A95: 1125 mm2 = 0.05 * b * b Note 2.

dequiv: 121.2 mm = SQRT(A95val / 0.0766)

and the size factor is found for this equivalent diameter from equation 6.7b, to be

Csize: 0.747 = 1.189 * dequiv^-0.097

4 The surface factor is found from equation 6.7e and the data in Table 6-3 for the specified hot-rolled finish.

Table 6-3 constants

A: 57.7

b: -0.718 Note 3.

Csurf: 0.584 = Acoeff * Sut^bCoeff

5 The temperature factor is found from equation 6.7f :

Ctemp: 0.710 = 1 - 0.0058 * (Tmax - 450)

6 The reliability factor is taken from Table 6-4 for R = 0.999 and is

Creliab: 0.753

7 The corrected endurance limit Se can now be calculated from equation 6.6:

Se: 69.94 MPa = Cload * Csize * Csurf * Ctemp *

Creliab * Sprme

Let

Se: 70 MPa

8 To create the S-N diagram, we also need a value for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading.

Sm: 540 MPa = 0.9 * Sut

9 The estimated S-N diagram is shown in Figure 6-34 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles.

b: -0.2958 Note 4.

a: 4165.7

Plotting Sn as a function of N from equation 6.10a

N Sn (MPa)

1000 540 =aa*B73^bb

2000 440

4000 358

8000 292

16000 238

32000 194

64000 158

128000 129

256000 105

512000 85

1000000 70

FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point

10 The number of cycles of life for any alternating stress level can now be found from equation 6.10a by replacing σa for Sn.

At N = 103 cycles,

Sn3: 540 MPa = aa * 1000^bb

At N = 106 cycles,

Sn6: 70 MPa = aa * 1000000^bb

The figure above shows the intersection of the alternating stress line (σa = 100 MPa) with the failure line at N = 3.0 x 105 cycles.

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3 years ago

Nêu đặc điểm của tín hiệu PAM rời rạc dạng lưỡng cực NRZ, RZ

zhuklara [117]

Answer:

yes it is certainly good ice cream

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3 years ago