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The bar has a cross-sectional area of 400×10^-6 m². If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads determine the average normal stress in the bar as a function of X for 0.5 m < x <= 1.25m

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ajeigbeibraheem

Answer:

the average normal stress in the bar is : [tex]\mathbf{\sigma = (32.5 - 20x )MPa}[/tex]

Explanation:

The free body flow of the missing diagram is attached to the answer below.

From the information given:

Let consider the sum of forces along horizontal direction to be equal to zero.

[tex]\begin{array}{l}\\\sum {{F_x}} = 0\\\\3 + 8\left( {1.25 - x} \right) - N = 0\\\\N = \left( {13 - 8x} \right){\rm{ kN}}\\\end{array}[/tex]

The average normal stress in the bar can be calculated by the formula:

[tex]\sigma = \dfrac{N}{A}[/tex]

where;

[tex]\sigma =[/tex] average normal stress in the bar

A = cross sectional area in the bar and it is given by: [tex]400*10^{-6 } m^2[/tex]

​N = (13- 8x) kN

∴ [tex]\sigma = \dfrac{(13-8)x}{400*10^{-6} m^2}[/tex]

[tex]\sigma = (32.5 - 20x )*10^3 kPa[/tex]

[tex]\sigma = (32.5 - 20x )MPa[/tex]

Thus; the average normal stress in the bar is : [tex]\mathbf{\sigma = (32.5 - 20x )MPa}[/tex]

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Lanuel

The average normal stress in the bar as a function of x is equal to 32.5 - 20x MPa.

Given the following data:

  • Cross-sectional area of bar = [tex]400 \times 10^{-6}\;m^2[/tex]
  • Range of x = 0.5 m < x ≤ 1.25 m.
  • Force A = 3 kN.
  • Force B = 8(1.25-x) kN.

To calculate the average normal stress in the bar as a function of x.

How to calculate average normal stress.

First of all, we would determine the sum of the forces acting on the bar in the horizontal direction;

[tex]\sum F_x=0\\\\3+8(1.25-x)-N=0\\\\3+10-8x-N=0\\\\13-8x-N=0\\\\N=(13-8x)\; kN[/tex]

For the average normal stress:

Mathematically, the average normal stress is given by this formula:

[tex]\sigma = \frac{N}{A}[/tex]

Where:

  • A is the cross-sectional area.
  • N is the resultant force.

Substituting the parameters into the formula, we have;

[tex]\sigma = \frac{(13-8x) \times 10^3}{400 \times 10^{-6}}\\\\\sigma = \frac{(13-8x) \times 10^{3+6}}{400}\\\\\sigma =(32.5-20x) \times 10^{9}[/tex]

Note: 1 MPa = [tex]1\times 10^9\;Pa[/tex]

Average normal stress = 32.5 - 20x MPa.

Read more on average normal stress here: https://brainly.com/question/7958709

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