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3 years ago

11

A 2.5-m-long tension member (E = 70 GPa) has cross-sectional dimensions of 25 mm x 35 mm. Determine the maximum load P that may

be supported by this member if the allowable normal stress may not exceed 90 MPa and the total elongation must not exceed 3 mm. State your answer in kN.

1 answer:

Alja [10]3 years ago

8 0

Answer:

Force = 7.3 kN

Explanation:

Elastic Modulus = Stress / Strain

Since the maximum stress applied can be 90 MPa, lets check the corresponding maximum strain using the above equation.

70 * 10^9 = 90 * 10^6 / Strain

Strain = 0.0012857

Strain = Change in length/total length

Change in Length = 0.0012857 * 2.5

Change in Length = 0.003214 m

Converted to mm this is: 0.003214 * 1000 = 3.214 mm

Thus we can see that strain is the limiting factor, and not stress. This is because with the maximum stress possible applied, we go over the strain limit.

Doing another quick calculation, we can take out the maximum stress with strain equal to 3mm/2.5m

Strain = (3/1000)/2.5

Strain = 0.0012

Modulus of elasticity = Stress/ Strain

Stress = 70 * 10^9 * 0.0012

Stress = 84 MPa

This stress is under both the strain and stress limit.

Taking out the force due to this stress:

Stress = force / area

Force = 84 * 10^6 * Area

Force = 84 * 10^6 * (25/1000) * (35/1000)

Force = 7350 N or 7.3 kN

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Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours

Nostrana [21]

Answer:

Approximately $116\; \text{miles}$ for the train from Boston to NYC Penn Station.

Approximately $105\; \text{miles}$ for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

$\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}$ .

$\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}$ .

Calculate average speed of each train:

$\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$ .

$\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$

Assume that it takes a time period of $t$ for the trains to pass by each other after departure. Distance each train travelled would be:

$s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t$ .

$s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t$ .

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

$v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}$ .

Rearrange and solve for $t$ :

$(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}$ .

$\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}$ .

Distance each train travelled in $t = (539 / 279)\; \text{hour}$ :

$\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}$ .

$\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}$ .

8 0

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Answer:

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Explanation:

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3 years ago

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4 years ago

Read 2 more answers

6. What is the change in temperature of a metal rod that is 55.0 cm long, decreases length by 0.20 cm, and that has a coefficien

MrMuchimi

Explanation:

We have,

Length of a metal rod is 55 cm or 0.55 m

Change in length is 0.2 cm or 0.002 m

It is required to find the change in temperature of a metal rod. The coefficient of linear expansion is given by :

$\alpha =\dfrac{\Delta L}{L_0\Delta T}$

$\Delta T$ is the change in temperature

$\Delta T =\dfrac{\Delta L}{L_0\alpha }\\\\\Delta T =\dfrac{0.002}{0.55\times 12\times 10^{-6}}\\\\\Delta T= 303.03^{\circ} C$

So, the change in temperature is 303.03 degrees Celsius.

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3 years ago

The moon Umbriel orbits Uranus (mass = 8.68 x 10^25 kg) at a distance of 2.66 x 10^8 m. What is Umbriel's orbital period (in hou

Whitepunk [10]

Answer:

T = 99.51 hour

Explanation:

Mass of Uranus, $M=8.68\times 10^{25}\ kg$

The moon Umbriel orbits Uranus at a distance of $2.66\times 10^8\ m$

We need to find Umbriel's orbital period. Let it is T. Using Kepler's third law of motion to find it.

$T^2\propto r^3\\\\T^2=\dfrac{4\pi^2r^3}{GM}\\\\T^2=\dfrac{4\pi^2\times (2.66\times 10^8)^3}{6.67\times 10^{-11}\times 8.68\times 10^{25}}\\\\T=358244.51\ s$

As 1 hour = 3600 s

358244.51 s = 99.51 hour

Hence, Umbriel's orbital period is 99.51 hour.

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3 years ago