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3 years ago

Various factors to be considered in deciding the factor of safety?

Engineering

1 answer:

Eduardwww [97]3 years ago

4 0

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

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A civil engineer is analyzing the compressive strength of concrete. The compressive strength is approximately normal distributed

hram777 [196]

Answer:

See explanation

Explanation:

Solution:-

- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).

- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.

- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:

3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,

3286, 3210, 3265, 3273

- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.

Null hypothesis: μ = 3500 psi

Alternate hypothesis: μ ≠ 3500 psi

- The type of test performed on the sample data will depend on the application of Central Limit Theorem.

- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )

- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.

- The mean of the sample ( x ) is calculated as follows:

$x = \frac{Sum ( x_i )}{n} \\\\x = \frac{Sum ( 3273+ 3229+ 3256+ 3272+ 3201+ 3247+ 3267+ 3237+ 3286+ 3210+ 3265+3273 )}{12} \\\\x = \frac{39016}{12} \\\\x = 3251.3333$

- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.

- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:

The population variance is known OR sample size ( n ≥ 30 )

- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:

$Z-test = \frac{x - u}{\sqrt{\frac{sigma^2}{n} } } \\\\Z-test = \frac{3251.333 - 3500}{\sqrt{\frac{1000^2}{12} } } \\\\Z-test = -27.24$

- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:

Z-critical = Z_α/2 = Z_0.005

- Use the list of correlation of significance level ( α ) and critical values of Z to determine:

Z-critical = Z_0.005 = ± 2.576

- Compare the Z-test value against the rejection region defined by the Z-critical value.

Rejection region: Z > 2.576 or Z < -2.576

- The Z-test value lies in the rejection region:

Z-test < Z-critical

-27.24 < -2.576 .... Null hypothesis rejected

Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.

- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.

- The margin of error (ME) is defined by the following formula:

$ME = Z^*. \frac{sigma}{\sqrt{n} }$

Where,

- The ( Z* ) is the critical value for the defined confidence level ( CI ):

- The confidence interval and significance level are related and critical value Z* is as such:

α = 1 - CI , Z* = Z_α/2

- The critical values for ( CI = 99% & 95% ) are evaluated:

α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05

Z* = Z_0.005 , Z* = Z_0.025

Z* = ± 2.58 , Z* = ± 1.96

- The formulation of Confidence interval is given by the following inequality:

[ x - ME < μ < x + ME ]

[ x - Z*√σ^2 / n < μ < x + Z*√σ^2 / n ]

- The CI of 95% yields:

[ 3251.33 - 1.96*√(1000 / 12) < μ < 3251.33 + 1.96*√(1000 / 12) ]

[ 3251.333 - 17.89227 < μ < 3251.33 + 17.89227 ]

[ 3233.44 < μ < 3269.23 ]

- The CI of 99% yields:

[ 3251.33 - 2.58*√(1000 / 12) < μ < 3251.33 + 2.58*√(1000 / 12) ]

[ 3251.333 - 23.552 < μ < 3251.33 + 23.552 ]

[ 3227.78 < μ < 3274.88 ]

- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.

7 0

3 years ago

2 times 4 - 3 times 5

Papessa [141]

Answer:

<h3>The answer is - 7</h3>

Explanation:

The statement above is written as

2 × 4 - 3 × 5

8 - 15

= - 7

Hope this helps you

4 0

4 years ago

Read 2 more answers

The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False

Alex

Answer:

(B) FALSE

Explanation:

view factor $F_{ij}$ depends on the surface emissivity and the surface of geometry view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by $F_{ij}$

4 0

3 years ago

1. When should a synthetic oil change be completed? *

Roman55 [17]

Answer:

A range from 5,000 miles to 7,500 miles, on average. Some recommended intervals may be shorter or longer.

4 0

3 years ago

Read 2 more answers

Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo

noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

3 0

3 years ago