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elena-14-01-66 [18.8K]
3 years ago
15

Which of the following terms is defined as small bumps and slashes within a fluid power system?

Engineering
2 answers:
-BARSIC- [3]3 years ago
7 0

Answer:

friction

Explanation:

''.''

Firdavs [7]3 years ago
5 0

Answer:

the answer is friction

Explanation:

because

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A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough
jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection  

Explanation:

We are given:

Plane strain fracture toughness K = 98.9 MPa \sqrt{m}

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

oc = \frac{K}{Y \sqrt{pi * a}}

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

a= \frac{1}{pi} * [\frac{k}{y*a}]^2

Substituting figures in the expression above, we have:

= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection  

3 0
2 years ago
Read 2 more answers
True or False: Galvanized steel pipe can withstand moderate freezing.
yaroslaw [1]

Answer:

True

Explanation:

Older galvanized steel pipes, which have a tendency to freeze, are a bit more forgiving and will likely not burst. They can withstand extreme cold and warm temperatures.

5 0
3 years ago
A CL soil is being used for compacted fill on a project. A sample of the compacted soil with a total volume of 1/30 ft3 weighs 4
Genrish500 [490]

Answer:

A. 0.4

B. 1.003

C. 0.83

Explanation:

The void ratio of a mixture is defined as the ratio of the volume of voids to volume of solids.

Total volume of soil = 1/30 ft3

= 1 ft3/30 * 0.0283 m3/1 ft3

= 9.43 x 10^-4 m3

Mass of water is in the soil = 20% * 4.8

= 0.96 pounds of water

= 0.96 * 0.454

= 0.44 kg

SG = density of substance/density of water

= 2.66 * 1 kg/l

Density of the soil = 2.660 kg/l

Mass of solid = 80 %

= 80% * 4.8 * 0.454

= 1.74 kg

Volume of solids = mass/density

= 1.74/2.66

= 6.63 l

= 6.63 x 10^-4 m3.

The volume of voids is found by adding the volume of water and the volume of air.

Total volume of soil = volume of (solids + voids)

9.43 x 10^-4 = 6.63 x 10^-4 + voids

Volume of voids = 2.8 x 10^-4 m3

A.

Void ratio = volume of void : volume of solids

= 2.8 : 6.63

= 0.4

B.

Y = (1 + w) * Gs * Yw * (1 + e)

Y = moist unit weight

Yw = unit weight of water

w = moisture content of the material

Gs = pecific gravity of the solid

e = void ratio

= (1 + 0.2) * 2.66 * 0.44 * (1 + 0.4)

= 1.003.

C.

gd = Y/(1 + w)

Or

= Gs * Yw * 1/(1 + e)

= 0.83.

6 0
3 years ago
A. Paleozoic
garri49 [273]
Pretty sure it’s Cenozoic
5 0
2 years ago
Read 2 more answers
A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The
Minchanka [31]

Answers:

31.7 inches

Explanation:

Given:

Diameter = 3ft

Let D = Diameter

So, D = 3ft. (Convert to inches)

D = 3 * 12in = 36 inches

Coss-sectional area of the steel = 0.2in²

Gauge Pressure (P) = 4psi

Stress in Steel (σ)= 11.4ksi

Force in steel = ½ (Pressure * Projected Area)

Area (A) = 2 * Force/Pressure

Also, Area (A) = Spacing (S) * Wood Pipe Diameter

Area = Area

2*Force/Pressure = Spacing * Diameter

Substitute values I to the above expression

2 * Force / 4psi = S * 36 inches

Also

Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel

So, F = 11.4ksi * 0.2in²

F = 11.4 * 10³psi * 0.2in²

F = 2.28 * 10³ psi.in²

So, 2 * Force / 4psi = S * 36 becomes

2 * 2.28 * 10³/4 = S * 36

S = 2 * 2.28 * 10³ / (4 * 36)

S = 4560/144

S = 31.66667 inches

S = 31.7 inches (approximated)

5 0
3 years ago
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