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A vertical steel beam in a building supports a load of 6.0×10⁴. If the length of the beam is 4.0m and it's cross-sectional area is 8.0×10^-3m². Find the diameter of the beam which is compressed along its length

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rafaleo84

Answer:

DL = 1.5*10^-4[m]

Explanation:

First we will determine the initial values of the problem, in this way we have:

F = 60000[N]

L = 4 [m]

A = 0.008 [m^2]

DL = distance of the beam compressed along its length [m]

With the following equation we can find DL

[tex]\frac{F}{A} = Y*\frac{DL}{L} \\where:\\Y = young's modulus = 2*10^{11} [Pa]\\DL=\frac{F*L}{Y*A} \\DL=\frac{60000*4}{2*10^{11} *0.008} \\DL= 1.5*10^{-4} [m][/tex]

Note: The question should be related with the distance, not with the diameter, since the diameter can be found very easily using the equation for a circular area.

[tex]A=\frac{\pi}{4} *D^{2} \\D = \sqrt{\frac{A*4}{\pi} } \\D = \sqrt{\frac{0.008*4}{\\pi } \\\\D = 0.1[m][/tex]

fichoh

Using the Young's modulus relationship, tbe diameter of the beam compressed along it's length is [tex]1.5 \times 10^{-4}[/tex]

Using the Young's Modulus relation :

  • Young's modulus = Stress / Strain

  • Stress = Force, F / Area, A
  • Strain = Extension, e / Length, L

HENCE,

  • [tex]\frac{F}{A} \times \frac{L}{e} [/tex]
  • Force = 6 × 10⁴
  • Length, L = 4 meters
  • Area = 8 × 10¯³ m²
  • Young's modulus = 2 × 10^11

[tex]\frac{6 \times 10^{4}}{8 \times 10^{-3}} \times \frac{4}{e} [/tex]

Using cross multiplication :

[tex] e = \frac{4 \times 6 \times 10^{4} \times 4 }{8 \times 10^{(3} \times 2 \times 10^{-11}} = 1.5 \times 10^{-4} [/tex]

The diameter of the beam compressed along its length is [tex]1.5 \times 10^{-4}[/tex]

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