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2 years ago

12

Calculate the second moment of area of a 4-in-diameter shaft about the x-x and yy axes, as shown.

1 answer:

Keith_Richards [23]2 years ago

6 0

Second moment of area about an axis along any diameter in the plane of the cross section (i.e. x-x, y-y) is each equal to (1/4)pi r^4.
The second moment of area about the zz-axis (along the axis of the cylinder) is the sum of the two, namely (1/2)pi r^4.

The derivation is by integration of the following:
int int y^2 dA
over the area of the cross section, and can be found in any book on mechanics of materials.

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ALGEBRAIC EXPRESSION 11. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7. 12.

Naily [24]

Answer:

Explained below.

Step-by-step explanation:

(11)

Subtract the sum of (13x - 4y + 7z) and (- 6z + 6x + 3y) from the sum of (6x - 4y - 4z) and (2x + 4y - 7z).

$[(6x - 4y - 4z) +(2x + 4y - 7z)]-[(13x - 4y + 7z) + (- 6z + 6x + 3y) ]\\=[6x-4y-4z+2x+4y-7z]-[13x-4y+7z-6z+6x+3y]\\=6x-4y-4z+2x+4y-7z-13x+4y-7z+6z-6x-3y\\=(6x+2x-13x-6x)+(4y-4y+4y-3y)-(4z+7z+7z-6z)\\=-11x+y-12z$

Thus, the final expression is (-11x + y - 12z).

(12)

From the sum of (x² + 3y² - 6xy), (2x² - y² + 8xy), (y² + 8) and (x² - 3xy) subtract (-3x² + 4y² - xy + x - y + 3).

$[(x^{2} + 3y^{2} - 6xy)+(2x^{2} - y^{2} + 8xy)+(y^{2} + 8)+(x^{2} - 3xy)] - [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[x^{2} + 3y^{2} - 6xy+2x^{2} - y^{2} + 8xy+y^{2} + 8+x^{2} - 3xy]- [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[4x^{2}+3y^{2}-xy+8]-[-3x^{2} + 4y^{2} - xy + x - y + 3]\\=4x^{2}+3y^{2}-xy+8+3x^{2}-4y^{2}+xy-x+y-3\\=7x^{2}-y^{2}-x+y+5$

Thus, the final expression is (7x² - y² - x + y + 5).

(13)

What should be subtracted from (x² – xy + y² – x + y + 3) to obtain (-x²+ 3y²- 4xy + 1)?

$A=(x^{2} - xy + y^{2} - x + y + 3) - (-x^{2}+ 3y^{2}- 4xy + 1)\\=x^{2} - xy + y^{2} - x + y + 3 +x^{2}- 3y^{2}+ 4xy -1\\=2x^{2}-2y^{2}+3xy-x+y+2$

Thus, the expression is (2x² - 2y² + 3xy - x + y + 2).

(14)

What should be added to (xy – 3yz + 4zx) to get (4xy – 3zx + 4yz + 7)?

$A=(4xy-3zx + 4yz + 7)-(xy - 3yz + 4zx) \\=4xy-3zx + 4yz + 7 -xy + 3yz - 4zx\\=3xy-7zx+7yz+7$

Thus, the expression is (3xy - 7zx + 7yz + 7).

(15)

How much is (x² − 2xy + 3y²) less than (2x² − 3y² + xy)?

$A=(2x^{2} - 3y^{2} + xy)-(x^{2} - 2xy + 3y^{2})\\=2x^{2} - 3y^{2} + xy-x^{2} + 2xy - 3y^{2}\\=x^{2}-6y^{2}+3xy$

Thus, the expression is (x² - 6y² + 3xy).

7 0

2 years ago

(9,5) and (4,-7) find the distance between two points

Pani-rosa [81]

Answer:

13 units

Step-by-step explanation:

Use the distance formula, d = $\sqrt{(x2 - x1)^2 + (y2 - y1)^2}$ , where (x2, y2) and (x1, y1) are two different points on the line.

Plug in the values:

d = $\sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

d = $\sqrt{(9 - 4)^2 + (5 + 7)^2}$

d = $\sqrt{5^2 + 12^2}$

d = $\sqrt{25 + 144}$

d = $\sqrt{169}$

d = 13

6 0

3 years ago

Which are not the dimensions

Paul [167]

It would be B. 5cm 12cm and 13cm

4 0

1 year ago

Describe how the graph of y=x^2 can be transformed to the graph of the given equation y=x^2+13. A) shift the graph of y=x^2 righ

stiks02 [169]

Answer:

D

Step-by-step explanation:

when a number is added at the end it results in a vertical transformation

8 0

2 years ago

A box is x to inches high,(2x+3)inches wide, and (2x+5) of inches long. In terms of x, what is the volume (v) of the box?

Advocard [28]

Answer:

The volume of the box is $V=(4x^3+16x^2+15x)\ in^3$

Step-by-step explanation:

we know that

The volume of the box is equal to

$V=LWH$

where

L is the length of the box in inches

W is the width of the box in inches

H is the height of the box in inches

we have

$L=(2x+5)\ in\\W=(2x+3)\ in\\H=x\ in$

substitute in the formula

$V=(2x+5)(2x+3)x$

$V=(4x^2+6x+10x+15)(x)$

$V=(4x^2+16x+15)(x)$

$V=(4x^3+16x^2+15x)\ in^3$

6 0

2 years ago