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Chapter 10: Problem 1

A steel beam is placed vertically in the basement of a building to keep the floor above from sagging. The load on the beam is $5.8 \times 10^{4} \mathrm{N},\( the length of the beam is \)2.5 \mathrm{m},$ and the cross- sectional area of the beam is \(7.5 \times 10^{-3} \mathrm{m}^{2}\) Find the vertical compression of the beam.

Short Answer

Expert verified
Answer: The vertical compression of the steel beam is approximately \(0.000096625\) meters.

Step by step solution

01

Write down the given information

We know the following values: Load on the beam (F) = \(5.8 \times 10^{4} \mathrm{N}\) Length of the beam (L) = \(2.5 \mathrm{m}\) Cross-sectional area of the beam (A) = \(7.5 \times 10^{-3} \mathrm{m}^{2}\)
02

Find the stress

Using the formula for stress, we have: Stress (σ) = \(\frac{F}{A}\) Plug in the values: σ = \(\frac{5.8 \times 10^{4} \mathrm{N}}{7.5 \times 10^{-3} \mathrm{m^2}}\) σ = \(7.73 \times 10^6 \mathrm{Pa}\)
03

Find the strain

Using Hooke's Law, we have: Strain (ε) = \(\frac{σ}{E}\) The modulus of elasticity of steel (E) is approximately \(2 \times 10^{11} \mathrm{Pa}\). Plug in the values: ε = \(\frac{7.73 \times 10^6 \mathrm{Pa}}{2 \times 10^{11} \mathrm{Pa}}\) ε = \(3.865 \times 10^{-5}\)
04

Calculate the vertical compression

Using the formula for strain, we have: ε = \(\frac{ΔL}{L}\) We need to find ΔL (the vertical compression), so we rearrange the formula: ΔL = ε × L Plug in the values: ΔL = \(3.865 \times 10^{-5} \times 2.5 \mathrm{m}\) ΔL = \(9.6625 \times 10^{-5} \mathrm{m}\) Therefore, the vertical compression of the steel beam is approximately \(9.6625 \times 10^{-5}\) meters or \(0.000096625\) meters.

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Most popular questions from this chapter

An object of mass \(m\) is hung from the base of an ideal spring that is suspended from the ceiling. The spring has a spring constant \(k .\) The object is pulled down a distance \(D\) from equilibrium and released. Later, the same system is set oscillating by pulling the object down a distance \(2 D\) from equilibrium and then releasing it. (a) How do the period and frequency of oscillation change when the initial displacement is increased from \(D\) to $2 D ?$ (b) How does the total energy of oscillation change when the initial displacement is increased from \(D\) to \(2 D ?\) Give the answer as a numerical ratio. (c) The mass-spring system is set into oscillation a third time. This time the object is pulled down a distance of \(2 D\) and then given a push downward some more, so that it has an initial speed \(v_{i}\) downward. How do the period and frequency of oscillation compare to those you found in part (a)? (d) How does the total energy compare to when the object was released from rest at a displacement \(2 D ?\)
Show that the equation \(a=-\omega^{2} x\) is consistent for units, and that \(\sqrt{k / m}\) has the same units as \(\omega\).
A 91 -kg man's thighbone has a relaxed length of \(0.50 \mathrm{m},\) a cross- sectional area of \(7.0 \times 10^{-4} \mathrm{m}^{2},\) and a Young's modulus of \(1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) By how much does the thighbone compress when the man is standing on both feet?
The maximum height of a cylindrical column is limited by the compressive strength of the material; if the compressive stress at the bottom were to exceed the compressive strength of the material, the column would be crushed under its own weight. (a) For a cylindrical column of height \(h\) and radius \(r,\) made of material of density \(\rho,\) calculate the compressive stress at the bottom of the column. (b) since the answer to part (a) is independent of the radius \(r,\) there is an absolute limit to the height of a cylindrical column, regardless of how wide it is. For marble, which has a density of $2.7 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\( and a compressive strength of \)2.0 \times 10^{8} \mathrm{Pa},$ find the maximum height of a cylindrical column. (c) Is this limit a practical concern in the construction of marble columns? Might it limit the height of a beanstalk?
(a) Given that the position of an object is \(x(t)=A\) cos \(\omega t\) show that \(v_{x}(t)=-\omega A\) sin \(\omega t .\) [Hint: Draw the velocity vector for point \(P\) in Fig. \(10.17 \mathrm{b}\) and then find its \(x\) component.] (b) Verify that the expressions for \(x(t)\) and \(v_{x}(t)\) are consistent with energy conservation. [Hint: Use the trigonometric identity $\sin ^{2} \omega t+\cos ^{2} \omega t=1.1$.]
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