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3 years ago

14

A main cable in a large bridge is designed for a tensile force of 2,600,000 lb. The cable consists of 1470 parallel wires, each

0.16 in. in diameter. The wires are cold-drawn steel with an average ultimate strength of 230,000 psi. What factor of safety was used in the design of the cable

1 answer:

NISA [10]3 years ago

7 0

Answer:

the factor of safety was used in the design of the cable is 2.6146

Explanation:

Given the data in the question;

Load on the main capable $P_{initial$ = 2600000 lb

number of parallel wires n = 1470

Diameter d = 0.16 in

average ultimate strength $S_{ultimate$ = 230000 psi

First we calculate the Load acting on each cable;

$P_{initial$ = P × n

P = $P_{initial$ / n

we substitute

P = 2600000 lb / 1470

P = 1768.70748 lb

Next we determine the working stress acting in a member;

$S_{working$ = P/A

{ Area A = $\frac{\pi }{4}$ d² }

$S_{working$ = P / $\frac{\pi }{4}$ d²

we substitute

$S_{working$ = 1768.70748 / $\frac{\pi }{4}$ (0.16)²

$S_{working$ = 1768.70748 / 0.02010619298

$S_{working$ = 87968.29 psi

Now we calculate the factor of safety F.S

F.S = $S_{ultimate$ / $S_{working$

we substitute

F.S = 230000 psi / 87968.29 psi

F.S = 2.6145785 ≈ 2.6146

Therefore, the factor of safety was used in the design of the cable is 2.6146

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Answer:

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print(potato)

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Explanation:

The complete question is as follows

Create a variable called potato whose value corresponds to the number of potatoes you’ve eaten in the last week. Or something equally ridiculous. Print out the value of potato.

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The question was answered using R programming language.

At line 1, I assumed that I ate 100 potatoes in the previous week.

So, potato = 100

At line 2, the value of potato is printed as 100.

At line 3, the square root of potato is calculated using sqrt function: Square for of 100 = 10

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At line 5, the new value of potato is printed: 200.

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4 years ago

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

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Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

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Explanation:

From this question we are told that

The efficiency is $\eta = 33$ % = 0.33

Temperature for the hot day is $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

Temperature after cooling is $T_c = 5^oC = 278K$

The input power is $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

$Q_h = \frac{P_{in}}{\eta}$

$= \frac{2}{0.33} = 6.061 kW$

From the question the air condition is assumed to be half as a Carnot refrigeration unit

This can be Mathematically interpreted in terms of COP(coefficient of performance) as

$\beta_{air} = 0.5 \beta$

where $\beta$ denotes COP and is mathematically represented as

$\beta = \frac{Q_c}{P_{in}}$

= > $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if the COP of a Carnot refrigerator is having this value

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$= \frac{278}{308-278}$

$\beta_{Carnot} = 9.267\\$

Then

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$= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

$Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

$\r m = \frac{Q_c}{c_p \Delta T}$

$= \frac{9.2667}{1.004}* (308 - 278)$

$= 0.30765 kg/s$

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