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Chapter 10: Problem 24

Column \(A B C\) has a uniform rectangular cross section with \(b=12 \mathrm{mm}\) and \(d=22 \mathrm{mm} .\) The column is braced in the \(x z\) plane at its midpoint \(C\) and carries a centric load \(\mathbf{P}\) of magnitude \(3.8 \mathrm{kN} .\) Knowing that a factor of safety of 3.2 is required, determine the largest allowable length \(L\). Use \(E=200\) GPa.

Short Answer

Expert verified
The largest allowable length, L, is approximately 1.547 m.

Step by step solution

01

Calculate the Area Moment of Inertia

The area moment of inertia for a rectangle about its base is given by the formula:\[ I = \frac{{b imes d^3}}{12} \]Substitute the given values \( b = 12 \text{ mm} \) and \( d = 22 \text{ mm} \) into the formula:\[ I = \frac{{12 \times 22^3}}{12} = 10648 \text{ mm}^4 \]
02

Find the Critical Load using Euler's Formula

The critical load \(P_{cr}\) for a column braced at its midpoint in the \(x z\) plane is given by:\[ P_{cr} = \frac{{\pi^2 E I}}{{(KL)^2}} \]Since the column is braced at its midpoint, effectively it behaves like two columns of length \(L/2\). Therefore, the effective length \(KL = L/2\).Substituting \(E = 200 \times 10^9 \text{ Pa} \) and \( I = 10648 \times 10^{-12} \text{ m}^4 \):\[ P_{cr} = \frac{{\pi^2 \times 200 \times 10^9 \times 10648 \times 10^{-12}}}{{(L/2)^2}} \]
03

Apply the Factor of Safety

Given that the factor of safety \(FS = 3.2\), the allowable load \(P_{allow}\) on the column can be given by:\[ P_{allow} = \frac{P_{cr}}{FS} = \frac{3.8 \times 10^3}{3.2} \text{ N} \]Calculate \(P_{allow}\):\[ P_{allow} \approx 1187.5 \text{ N} \]
04

Solve for the Largest Allowable Length

Equating the allowable load \(P_{allow}\) with \(P_{cr}\) and solve for \(L\):\[ P_{allow} = \frac{{\pi^2 \times 200 \times 10^9 \times 10648 \times 10^{-12}}}{{(L/2)^2}} \]Rearrange to solve for \(L\):\[ 1187.5 = \frac{{\pi^2 \times 200 \times 10^9 \times 10648 \times 10^{-12}}}{{(L/2)^2}} \]After rearranging, solve for \(L\):\[ L = 2 \times \sqrt{\frac{{\pi^2 \times 200 \times 10^9 \times 10648 \times 10^{-12}}}{1187.5}} \]Carrying out the calculations will yield:\[ L \approx 1.547 \text{ m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factor of Safety
The factor of safety (FS) is an engineering principle used to ensure that structures are built with enough strength to withstand expected loads. It provides a margin for potential errors in calculations, unexpected loads, or material aging.
In essence, it is a measure of reliability for a given structure.
  • A higher factor of safety generally results in a more robust and reliable structure, as it can handle higher loads than anticipated.
  • The FS is calculated by dividing the structure's maximum capacity by the actual load it is subjected to.
  • A factor of safety of 3.2, as specified in the problem, indicates that the column should be able to support 3.2 times the anticipated load.
This ensures that even in unexpected situations, the column remains functional and safe, reducing the chances of failure.
The FS is important in ensuring public safety, especially in critical load-bearing structures like columns.
Area Moment of Inertia
The area moment of inertia, often symbolized as "I," is a geometric property of a cross-section that influences its resistance to bending.
It is crucial in determining a beam's or a column's ability to bear loads without undergoing significant bending or deflection.
  • The formula for calculating the area moment of inertia for a rectangular section is \(I = \frac{b \times d^3}{12} \\), where "b" is the width and "d" is the depth.
  • A larger value of "I" means greater resistance to bending; it is effectively a measure of a section's stiffness.
  • In our scenario, the area moment of inertia has been calculated as 10648 mm"); this means that the column's rectangular profile has a specific stiffness to support loads applied to it.
Choosing the correct cross-sectional dimensions based on the desired area moment of inertia is key in architectural and engineering designs, as it influences the column's structural capacity.
Critical Load
The critical load refers to the maximum load a column can carry before it becomes unstable and buckles.
This concept is incredibly important in the design of columns since exceeding the critical load can lead to sudden structural failure.
  • Euler's formula, \(P_{cr} = \frac{\pi^2 E I}{(KL)^2} \\), has been employed to determine this load. Here, "E" is the modulus of elasticity, "I" is the area moment of inertia, and "K" refers to the column's effective length factor.
  • The calculations in the solution show that knowing the critical load is essential to safely dimension the column length, ensuring safety beyond just basic loading conditions.
  • A column subject to a load above the critical threshold may experience bending or buckling, leading to potential catastrophic failure if unchecked.
Understanding the critical load is vital for designing columns that can withstand real-world loads without risk of sudden instability.
Effective Column Length
The effective column length is a concept used in the calculation of the critical load.
It accounts for how different end constraints and bracing positions impact the structural behavior of a column.
  • Effectively, it is the length of a column that is "free" to buckle. This could be less than the actual physical length of the column if it is braced or has different support conditions.
  • For example, a column that is pinned or firmly supported at its endpoints might act as though it is shorter, thanks to these constraints.
  • The effective length factor ("K") is used to adjust calculations to account for these differences. In our exercise, the mid-braced column behaves like two shorter columns, hence \(KL = L/2\).
  • Taking the effective length into account is crucial for correct sizing and safety assurance. The column's stability depends not just on the material or its dimensions but also on the effective length.
This allows engineers to accurately predict the performance of a column under various conditions, making it a fundamental part of structural design.

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Most popular questions from this chapter

A steel tube of \(80-\mathrm{mm}\) outer diameter is to carry a 93 -kN load \(\mathbf{P}\) with an eccentricity of \(20 \mathrm{mm}\). The tubes available for use are made with wall thicknesses in increments of 3 mm from 6 mm to \(15 \mathrm{mm} .\) Using the allowable-stress method, determine the lightest tube that can be used. Assume \(E=200 \mathrm{GPa}\) and \(\sigma_{Y}=\) \(250 \mathrm{MPa}\).

A rectangular column with a 4.4 -m effective length is made of glued laminated wood. Knowing that for the grade of wood used the adjusted allowable-stress for compression parallel to the grain is \(\sigma_{C}=8.3 \mathrm{MPa}\) and the adjusted modulus \(E=4.6 \mathrm{GPa}\) determine the maximum allowable centric load for the column.

A sawn lumber column with a \(7.5 \times 5.5\) -in. cross section has an 18-ft effective length. Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is \(\sigma_{C}=1200\) psi and that the adjusted modulus \(E=470 \times 10^{3}\) psi, determine the maximum allowable centric load for the column.

A sawn lumber column of rectangular cross section has a \(2.2-\mathrm{m}\) effective length and supports a 41 -kN load as shown. The sizes available for use have \(b\) equal to \(90 \mathrm{mm}, 140 \mathrm{mm}, 190 \mathrm{mm},\) and \(240 \mathrm{mm} .\) The grade of wood has an adjusted allowable stress for compression parallel to the grain \(\sigma_{C}=8.1 \mathrm{MPa}\) and an adjusted modulus \(E=8.3\) GPa. Using the allowable-stress method, determine the lightest section that can be used.

The steel tube having the cross section shown is used as a column of 15 -ft effective length to carry a centric dead load of 51 kips and a centric live load of 58 kips. Knowing that the tubes available for use are made with wall thicknesses in increments of \(\frac{1}{16}\) in. from \(\frac{3}{16}\) in. to \(\frac{3}{8}\) in., use load and resistance factor design to determine the lightest tube that can be used. Use \(\sigma_{Y}=36 \mathrm{ksi}\) and \(E=29 \times 10^{6}\) psi. The dead load factor \(\gamma_{D}=1.2,\) the live load factor \(\gamma_{L}=1.6,\) and the resistance factor \(\phi=0.90\).
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