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4 years ago

14

The container that holds the water for the football team is 7/10 full. After pouring out 5 gallons of water, it is 1/2 full. How

many gallons can the container hold?

2 answers:

Shtirlitz [24]4 years ago

6 0

7/10x - 5 = 1/2x
7/10x - 1/2x = 5
7/10x - 5/10x = 5
2/10x = 5
x = 5 / (2/10)
x = 5 * 10/2
x = 50/2
x = 25 gallons <===

expeople1 [14]4 years ago

3 0

(7/10)X - 5 - (1/2)X
(7/10)X - (5/10)X = 5
(2/10)X = 5
X = 25 gallons capacity

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please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

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

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

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+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point A'(0+<em>(0.03)</em><em>(0),0+</em><em>(-0.01)</em><em>(0))</em>

Point A'(0<em>,0)</em>

Point B(x=1,y=0)

Point B'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point B'(1+<em>(0.03)</em><em>(1),0+</em><em>(-0.01)</em><em>(0))</em>

Point <em>B</em>'(1.03<em>,0)</em>

Point C(x=1,y=1)

Point C'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point C'(1+<em>(0.03)</em><em>(1),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>C</em>'(1.03<em>,0.99)</em>

Point D(x=0,y=1)

Point D'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point D'(0+<em>(0.03)</em><em>(0),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>D</em>'(0<em>,0.99)</em>

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.

<em>Part b: Calculate the six strain components.</em>

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: <em>Find the volume change</em>

<em></em> $\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$ <em></em>

<em>Also the change in volume is 0.0197</em>

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.

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