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Chapter 10: Problem 9

Determine the critical load of a pin-ended steel tube that is \(5 \mathrm{m}\) long and has a 100 -mm outer diameter and a 16 -mm wall thick. ness. Use \(E=200 \mathrm{GPa}\).

Short Answer

Expert verified
The critical load is approximately 18.27 kN.

Step by step solution

01

Calculate the inner diameter

To calculate the inner diameter of the tube, subtract twice the wall thickness from the outer diameter: \[ d_{ ext{inner}} = d_{ ext{outer}} - 2 imes ext{wall thickness} = 100 ext{ mm} - 2 imes 16 ext{ mm} = 68 ext{ mm} \]
02

Convert dimensions to meters

Convert both the outer and inner diameter from millimeters to meters for ease in further calculations. \[ d_{ ext{outer}} = 100 ext{ mm} = 0.1 ext{ m} \] \[ d_{ ext{inner}} = 68 ext{ mm} = 0.068 ext{ m} \]
03

Calculate the cross-sectional area and moment of inertia

For a hollow cylindrical section, the moment of inertia is calculated as follows: \[ I = \frac{\pi}{64} (d_{ ext{outer}}^4 - d_{ ext{inner}}^4) = \frac{\pi}{64} (0.1^4 - 0.068^4) = 4.62 imes 10^{-6} ext{ m}^4 \]
04

Use Euler's Formula for critical load

Euler's formula for the critical load of a column with pinned ends is given by: \[ P_{ ext{cr}} = \frac{\pi^2 EI}{L^2} \] Substitute the known values: \[ E = 200 imes 10^9 ext{ Pa} \] \[ I = 4.62 imes 10^{-6} ext{ m}^4 \] \[ L = 5 ext{ m} \] \[ P_{ ext{cr}} = \frac{\pi^2 imes (200 imes 10^9 ) imes (4.62 imes 10^{-6})}{5^2} = 18.27 imes 10^3 ext{ N} \]
05

Present the calculated critical load

The critical load for the given pin-ended steel tube, as calculated using Euler's formula, is approximately 18,270 N (or 18.27 kN).

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Key Concepts

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

Understanding Critical Load
The concept of critical load is a fundamental aspect in the study of structures, particularly when analyzing columns. Critical load represents the maximum load a column can bear before it becomes unstable and buckles. This is especially important in engineering, as stability is crucial in construction and load-bearing designs.
When a column reaches its critical load, it may buckle sideways, which can lead to potential failure. Engineers must calculate this value accurately to ensure safety and reliability.
  • Critical load depends on factors such as material, geometric properties, and constraints.
  • It is typically calculated using Euler's Formula, which is applied in various scenarios, including pin-ended columns.
This formula helps predict the collapse behavior of structures and ensure they can sustain applied loads without risk of buckling.
Exploring Moment of Inertia
Moment of inertia is a concept that refers to a body's resistance to bending or twisting. It plays a critical role in determining a column's capacity to withstand applied loads. In the context of critical load, moment of inertia is a crucial component in the calculation.
For a hollow cylindrical section, the moment of inertia (I) is defined mathematically:
\[ I = \frac{\pi}{64} (d_{\text{outer}}^4 - d_{\text{inner}}^4) \]
Where:
  • \(d_{\text{outer}}\) and \(d_{\text{inner}}\) are the outer and inner diameters of the tube respectively.
This formula highlights how the geometry of a structure directly influences its mechanical properties. A larger moment of inertia indicates a greater ability to resist bending, contributing to the column's overall stability.
Significance of Cross-Sectional Area
Cross-sectional area is vital in assessing how well a column and other structural elements handle stress. It is the area of the column perpendicular to its length. For a hollow tube, this area is determined by both the outer and inner radii.
The cross-sectional area is considered when calculating both strength and stiffness properties. A larger area tends to correlate with higher load-carrying capacity.
  • In hollow structures like tubes, material is distributed around the perimeter, maximizing both strength and efficient use of material.
By understanding the relationships between cross-sectional area, material properties, and applied forces, engineers can design safer and more efficient structures.

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Most popular questions from this chapter

A laminated column of 2.1 -m effective length is to be made by gluing together wood pieces of \(25 \times 150-\mathrm{mm}\) cross section. Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is \(\sigma_{C}=7.7 \mathrm{MPa}\) and the adjusted modulus is \(E=5.4 \mathrm{GPa}\), determine the number of wood pieces that must be used to support the concentric load shown when \((a) P=52 \mathrm{kN},(b) P=108 \mathrm{kN}\).

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 sawn-lumber column of \(5.0 \times 7.5\) -in. cross section has an effective length of \(8.5 \mathrm{ft}\). The grade of wood used has an adjusted allowable stress for compression parallel to the grain \(\sigma_{C}=\) 1180 psi and an adjusted modulus \(E=440 \times 10^{3}\) psi. Using the allowable-stress method, determine the largest eccentric load \(\mathbf{P}\) that can be applied when \((a) e=0.5\) in., \((b) e=1.0\) in.

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 column of 17 -ft effective length must carry a centric load of 235 kips. Using allowable stress design, select the wide-flange shape of 10 -in. nominal depth that should be used. Use \(\sigma_{Y}=\) \(36 \mathrm{ksi}\) and \(E=29 \times 10^{6} \mathrm{psi}\).
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