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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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Answer: Symbol rate, Fs = 0.846

Explanation:

The attachment below shows the calculations

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Explain why different types of equipment are required for proper conditioning of air
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Answer:

 Different types of equipment are required for proper conditioning of air because every air conditional space faces some geometrical and environmental issues or problems. There are some different types of equipment used for conditioning of air that are air system, water system and air-water system. In many cases the air conditioning of the system varies with size of the equipment.  

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3 years ago
Describe how you would control employee exposure to excessive noise in a mining environment
Alexandra [31]

Answer:

1. Buy Quiet – select and purchase low-noise tools and machinery

2. Maintain tools and equipment routinely (such 3. as lubricate gears)

3. Reduce vibration where possible

4. Isolate the noise source in an insulated room or enclosure

5. Place a barrier between the noise source and the employee

6. Isolate the employee from the source in a room or booth (such as sound wall or window

Explanation:

Hope my answer will help u.

7 0
2 years ago
Witch measuring tool would be used to determine the diameter of a crankshaft journal
djverab [1.8K]
Answer =

dial bore gauge

a “dial bore gauge” measures the inside of round holes, such as the bearing journals . can mesure up to 2” and 6” diameter holes .

when ( “ ) is next to a number it means inches fwi - but hope this helped have a good day :)
3 0
3 years ago
The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
3 years ago
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