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3 years ago

What is the radius of 9.5in

Mathematics

2 answers:

fredd [130]3 years ago

8 0

I'm not sure you are asking the right question? it tells you what the radius is right there. What is the question asking

lara [203]3 years ago

7 0

The radius is half the circle are you sure your not asking for the diameter? If so do 9.5x2.

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What does y equal y-16-3y=0

Afina-wow [57]

Answer:

y = - 8

Step-by-step explanation:

y - 16 - 3y = 0

Group like terms

y - 3y - 16 = 0

Add similar elements: y - 3y = - 2y

- 2y - 16 = 0

Add 16 to both sides

- 2y - 16 + 16 = 0 + 16

Simplify

- 2y = 16

Divide both sides by - 2

$\frac{-2y}{-2}$ = $\frac{16}{-2}$

Simplify $\frac{-2y}{-2}$ : y

$\frac{-2y}{-2}$

Apply the fraction rule: $\frac{-a}{-b} = \frac{a}{b}$

= $\frac{2y}{2}$

Divide the numbers: $\frac{2}{2} = 1$

= y

Simplify $\frac{16}{-2}$ : - 8

$\frac{16}{-2}$

Apply the fraction rule: $\frac{-a}{-b} = \frac{a}{b}$

$-\frac{16}{2}$

Divide the numbers: $\frac{16}{2}$ = 8

= - 8

y = - 8

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3 years ago

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aleksandrvk [35]

Answer:

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Step-by-step explanation:

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Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

2 years ago