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evablogger [386]
3 years ago
7

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Engineering
1 answer:
Kamila [148]3 years ago
7 0

Answer:

(b)False

Explanation:

I_{xy} defined as

      I_{xy} =\int \left (x\cdot y\right )dA

Where x is the distance from centroidal x-axis

           y is the distance from centroidal y-axis

          dA is the elemental area.

The product of x and y can be positive or negative ,so the value of  I_{xy} can be positive as well as negative .

So from the above expressions we can say that the product of I_{x},I_y is different from I_{xy} .

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2. There are three drawings that architects and designers use to indicate spaces. What are these drawing?
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Architectural plans.

Explanation:

An architectural plan is called the drawings made by architects, civil engineers or designers of spaces or interiors, in which these professionals capture their building projects, organizing the distribution of the spaces to be used, the elements to be located in them and, fundamentally, to give construction planning a projection into reality. Thus, the plans help professionals to have a better understanding of the expected end result of the projects they are carrying out.

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

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