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Aleksandr-060686 [28]
3 years ago
6

A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e

xperience either plastic deformation or a diameter reduction of more than 7.5×10-3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).
Material Modulus of Elasticity(GPa) Yield Strength(MPa) Poission's Ratio
Aluminum alloy 70 200 0.33
Brass alloy 101 300 0.34
Steel alloy 207 400 0.3
Titanium alloy 107 650 0.34
Engineering
1 answer:
mash [69]3 years ago
7 0

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first  criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is  applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³  mm), it is necessary to  calculate the change in diameter Δd for these four alloys.

Then we use the equation   υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒  υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.

For the Steel alloy

∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm

0.005 mm < 7.5×10 ⁻³ mm

Therefore, the steel is a candidate.

For the Titanium alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm

0.0111 mm > 7.5×10 ⁻³ mm

Hence, the Titanium alloy is not a candidate.

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All MOS devices are subject to damage from:________
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Answer:

  d. all of these

Explanation:

Electrostatic discharge will generally produce excess voltage in a local area that results in excessive current and excessive heat. It will blast a crater in an MOS device, or melt bond wires, or cause damage of other sorts. In short, MOS devices are subject to damage from "all of these."

6 0
3 years ago
Royce has a Bevel protractor with a damaged vernier scale. What limitation will Royce face if he uses this protractor to measure
liraira [26]

Answer:

Limitation in the level of possible accuracy of the that can be obtained to mainly whole degrees

Explanation:

The angle measurement values are located around the circularly shaped bevel protractor to which a Vernier scale can be attached to increase the accuracy of the angle measurement reading

The accuracy of the bevel protractor is up to 5 arc minutes or 1/12° with the space on the Vernier scale having graduations of 1/12°, such that two spaces on the main scale is 5 arc minutes more than a space on the Vernier scale and the dimensions of the coincidence of the two scale will have an accuracy of up to 1/12°

Therefore, whereby the Vernier scale is damaged, the accuracy will be limited to the accuracy main scale reading in whole degrees.

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3 years ago
A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D
zhuklara [117]

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

d = \sqrt{4*P / (\pi*\sigma)}

d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm

7 0
3 years ago
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

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