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Zigmanuir [339]
2 years ago
10

Match the words with their definitions

Engineering
1 answer:
Pavlova-9 [17]2 years ago
5 0

Wooden 2 X 8's, 10's, or 12's that run parallel to one another and support a floor or ceiling, and supported in turn by larger beams, girders, or bearing walls- <u>Joist</u>

Explanation:

  1. Wooden 2 X 8's, 10's, or 12's that run parallel to one another and support a floor or ceiling, and supported in turn by larger beams, girders, or bearing walls- <u>Joist</u>
  2. A point where a bearing/structural weight is concentrated and transferred to the foundation-<u>Point Load</u>
  3. Includes all exterior walls and any interior wall that is aligned above a support beam or girder-<u>Load Bearing Wall</u>
  4. Plywood substitute made of coarse sawdust that is mixed with resin and pressed into sheets-<u>Particleboard</u>
  5. The replacement of excavated earth into a trench around or against a basement/crawl space foundation wall- <u>Backfill</u>
  6. Continuous 8m" or 10" thick concrete pad installed before and supports the foundation wall or monopost-<u>Footing</u>
  7. Ribbed steel bars installed in foundation concrete walls, footers and poured in place concrete structures designed to strengthen concrete-<u>Rebar</u>
  8. The end, upper, triangular area of a home, beneath the roof-<u>Gable</u>
  9. A manufactured 4X8 wood panel made out of 1-2 wood chips and glue. Often used as a substitute for plywood-<u>OSB</u>
  10. The incline slope of a roof or the ratio of the total width of a house-<u>Pitch</u>
  11. A manufactured structual wood beam. It is constructed of pressure and adhesive wood strands of wood-<u>Microlam</u>

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Read 2 more answers
For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

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