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2 years ago

Jason, your client, is developing a subdivision of 140 houses. He may place deed restrictions on as many as ______ properties.

Business

2 answers:

Andrei [34K]2 years ago

7 0

Jason may place deed restrictions on 140 (100%) properties.

<h3>Deed of restrictions</h3>

A deed of restriction is an action that is recognized by law which gives the property owner or agent the right to restrict, how the properties are to be used. For example, Jason's deed of restriction could contain restrictions against carrying out certain types of businesses on all 140 properties.

You can learn more about deed of restrictions here brainly.com/question/4510408

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kolbaska11 [484]2 years ago

4 0

Jason, your client, is developing a subdivision of 140 houses. He may place deed restrictions on as many as 140 (100%) properties.

<h3>What is a deed restriction?</h3>

This is the term that is used to refer to the written agreements that are done in order to restrict and limit activities that may go on in a property.

These are private agreements that are made. It has to be 100 percent on the properties.

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3 years ago

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of pya, where a 5 pd2y4. if the load is kno

raketka [301]

Here is the correct question.

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of P/A, where A= πd²/4. if the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answer:

the minimum design factor that will guarantee that the part will not fail. = 1.434

Explanation:

Looking at the uncertainty; loss of strength must be raised to $\dfrac{1}{0.85}$ due to the stress that causes the failure (strength) is known within ±15% uncertainty.

Looking at the uncertainty; the maximum allowable load must be reduced to $\dfrac{1}{1.1}$ because the load is known with an uncertainty of ±10.

Looking at the uncertainty; the diameter must be raised to $\dfrac{1}{0.95}$ because the diameter is known within an uncertainty of ±5.

The decrease in the maximum allowable stress can be estimated as:

$\sigma' = \dfrac{P'}{A'}$

where,

$\sigma =$ stress

P = load

A = cross-sectional area of the cylinder

∴

$\sigma' = \dfrac{P'}{\dfrac{\pi}{4}(d')^2}$

replacing P' with $\dfrac{1}{1.1}P$ and d' with $\dfrac{1}{0.95}d$ , we have:

$\sigma' = \dfrac{(\dfrac{1}{1.1})\times p }{\dfrac{\pi}{4}(\dfrac{1}{0.95 } d)^2 }$

$\sigma' =\dfrac{P}{\dfrac{\pi}{4}d^2} (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times 0.82045$

$\dfrac{\sigma' }{\sigma } =0.82045$

Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045

Now, the minimum design factor that will ascertain that the part will not fail can be computed as:

$n_d = \dfrac{loss \ of \ function \ parameter }{maximum \ allowable \ parameter}$

where;

the design factor = $n_d$

$n_d =\dfrac{\dfrac{1}{0.85} }{0.82045}$

the design factor $n_d$ = 1.434.

Thus, the minimum design factor that will guarantee that the part will not fail. = 1.434

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