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valentinak56 [21]
3 years ago
11

When preparing a foundation for a heavy duty machine tool, discuss any four (4) statics machine characteristics to be considered

.
Engineering
1 answer:
kozerog [31]3 years ago
5 0

Answer:

1 ) Accuracy of the Machine Tool

2) Load bearing capacity

3) Linearity in the product line

4) Torque of the machine

Explanation:

we know that machine tool is the permanent essential in manufacturing industries

it is a machine use for different form like cutting , grinding and boring etc

so 1st is

1 ) Accuracy of the Machine Tool

we know it is very important Characteristic of the machine tool because when we use it in manufacturing unit Accuracy of the Machine Tool should be higher concern

2) Load bearing capacity

we should very careful about Load bearing capacity because how much amount of load tool will bear check by some parameter like creep , shear stress and strength etc

3) Linearity in the product line

Linearity in the product line mean that it should be group of related product produced by the any one of the manufacturer otherwise it will take time or it may be intermixing

4) Torque of the machine

we know that Torque is a rotational force or a turning force so amount of force multiplied by the distance of the operation

and we know torque per second give the power rating of machine tool

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A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo
Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, e_{min} = 0.51

Void ratio in the loosest state, e_{max} = 0.87

Now,

Dry density, \gamma_d=\frac{\gamma_t}{1+w}

=\frac{18}{1+0.05}

= 17.14 kN/m³

Also,

\gamma_d=\frac{G\gamma_w}{1+e}

here, G = Specific gravity = 2.7 for sand

17.14=\frac{2.7\times9.81}{1+e}

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e = 0.545

Relative density = \frac{e_{max}-e}{e_{max}-e_{min}}

= \frac{0.87-0.545}{0.87-0.51}

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Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

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7 0
3 years ago
Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new
Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))     where i=1 to N-1 and N=6

C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))

A area of the polygon can be found by the following formulaA=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i}) where i=1 to N-1

A=\frac{1}{2}[ (x_{1}  y_{2} -x_{2}  y_{1})+ (x_{2}  y_{3} -x_{3}  y_{2})+(x_{3}  y_{4} -x_{4}  y_{3})+(x_{4}  y_{5} -x_{5}  y_{4})+(x_{5}  y_{6} -x_{6}  y_{5})]

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(x_{2}+x_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(x_{3}+x_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(x_{4}+x_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(x_{5}+x_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{x} =15.36

For Cy

C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1}  y_{2} -x_{2}  y_{1})]+ [(y_{2}+y_{3})(x_{2}  y_{3} -x_{3}  y_{2})]+[(y_{3}+y_{4})(x_{3}  y_{4} -x_{4}  y_{3})]+[(y_{4}+y_{5})(x_{4}  y_{5} -x_{5}  y_{4})]+[(y_{5}+y_{6})(x_{5}  y_{6} -x_{6}  y_{5})]]

putting the values of x's and y's you will get

C_{y} =22.55

So coordinates for the fire station should be (15.36,22.55)

5 0
3 years ago
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