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victus00 [196]
3 years ago
14

Asolid rectangular rodhas a length of 90mm, made of steel material (E =207,000 MPa, Syield= 300 MPa), the cross section of the r

od is shownbelow.If the boundary conditions are fixed-freeand a safety factor of 2 on loadis required, (1)Find themaximumallowable load (Pallowable)on the rod. [9points]{hint: Pallowable= Pcr/required safety factor}(2)If the same critical load (Pcr) calculated from (1)above(Pcrisalready taken into accounta safety factor of 2)is applied to a solid round rodunderthe same boundary conditions, what is the minimum required diameter of the rod against buckling? [9points]

Engineering
1 answer:
erastova [34]3 years ago
3 0

Answer:

13.6mm

Explanation:

We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.

See attachment for the step by step solution of the problem

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D. left foot for the accelerator and your right foot for the brake.

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For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

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5 0
3 years ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0
3 years ago
The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

Losses=K\dfrac{V^2}{2g}

8 0
3 years ago
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