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3 years ago

(1st pic) What is the equation of the line shown in this graph?

Mathematics

1 answer:

vredina [299]3 years ago

4 0

Answer:

For figure 1: The equation of a line is y=1

For figure 2: The equation of a line is y=(-4)x+1

For figure 3:The equation of a line is y= $\frac{-5}{2}x+5$

Step-by-step explanation:

The equation of line slope-intercept form is given by y=mx+c

Where m is the slope of the line and c is the y-intercept.

For figure 1:

Here, Line is parallel to x-axis

Hence, Slope m=0

Also, Line passing to y axis at (0,1)

Y-intercept is c=1

Therefore,

The equation of line is

y=0x+1

y=1

For figure 2:

Figure show a line passing through point (1,-3) and (-1,5)

The slope of the line is given by m= $\frac{Y2-Y1}{X2-X1}$

Using given points to find out the slope of a line

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{5-(-3)}{(-1)-1}$

m= $\frac{8}{-2}$

m=(-4)

Also, Line is intersecting y-axis at (0,1)

Hence, c=1

We can write the equation of line as

y=mx+c

y=(-4)x+1

Thus, The correct option is D). y=(-4)x+1

For figure 3:

From the figure, a line is passing through points (-2,0) and (0,5)

The slope of the line is given by m= $\frac{Y2-Y1}{X2-X1}$

Using given points to find out the slope of a line

m= $\frac{Y2-Y1}{X2-X1}$

m= $\frac{5-0}{0-(-2)}$

m= $\frac{-5}{2}$

Also, Line is intersecting y-axis at (0,5)

Hence, c=5

We can write the equation of line as

y=mx+c

y= $\frac{-5}{2}x+5$

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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

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