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Chapter 10: Problem 47

A 100 -kN axial load \(\mathbf{P}\) is applied to the \(\mathrm{W} 150 \times 18\) rolled-steel column \(B C\) that is free at its top \(C\) and fixed at its base \(B\). Knowing that the eccentricity of the load is \(e=6 \mathrm{mm}\), determine the largest permissible length \(L\) if the allowable stress in the column is 80 MPa. Use \(E=200 \mathrm{GPa}\).

Short Answer

Expert verified
The largest permissible length, \(L\), is approximately 4601 mm for the column.

Step by step solution

01

Understand the Problem

We have a steel column subject to an eccentric axial load causing both direct stress and bending stress. Our goal is to find the largest permissible length, \(L\), while ensuring the stress does not exceed the given allowable stress.
02

Calculate Direct Stress

Use the formula for direct stress, \( \sigma_d = \frac{P}{A} \), where \(P = 100\, \text{kN} = 100,000\, \text{N}\), and \(A\) is the cross-sectional area. For a W150 脳 18 section, \(A = 2290\, \text{mm}^2\). Thus, \( \sigma_d = \frac{100,000}{2290} \approx 43.67\, \text{MPa}\).
03

Calculate Bending Stress

The bending stress is given by \( \sigma_b = \frac{Mc}{I} \), where \(M = Pe\), \(c\) is the distance from the neutral axis to the outermost fiber, and \(I\) is the moment of inertia. \(e = 6\, \mathrm{mm}\), but to continue, \(c\) and \(I\) need to be obtained from standard values for W150 脳 18 section. Assume \(c = 75\, \mathrm{mm} / 2 = 37.5\, \mathrm{mm}\) and \(I=108.4 \times 10^6 \mathrm{mm}^4\). Then \( \sigma_b = \frac{100,000 \times 6 \times 10^{-3} \times 37.5}{108.4 \times 10^6} \approx 0.0208\, \text{MPa}\).
04

Combine Stresses and Compare to Allowable Stress

The total stress at the extreme fiber is \( \sigma = \sigma_d + \sigma_b \). Substitute to get \( \sigma = 43.67 + 0.0208 \approx 43.6908\, \text{MPa}\). The allowable stress is 80 MPa, so the combined stress is within limits.
05

Determine Effective Length for Buckling

Use the Euler buckling formula \( L = \pi \sqrt{\frac{EI}{P}} \) to find the largest permissible length. \(E = 200\, \text{GPa} = 200,000\, \text{MPa}\), \(P = 100\, \text{kN}\), and \(I = 108.4 \times 10^6\, \text{mm}^4\). Calculate \(L = \pi \sqrt{\frac{200,000 \times 108.4 \times 10^6}{100,000}}\) and evaluate to find \(L\).
06

Compute Permissible Length

Substitute values into the length formula to get \(L = \pi \sqrt{\frac{200,000 \times 108.4 \times 10^6}{100,000}} \approx 4601\, \text{mm}\). This is the largest permissible length for buckling to ensure that the stress does not exceed allowable limits.

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Key Concepts

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

Eccentric Load
The concept of an eccentric load is crucial when discussing column buckling and bending stress. In structural engineering, an eccentric load refers to a force applied at a distance away from the centroidal axis of a column. This is different from a concentric load which occurs directly along the axis. When a load is eccentric, it causes not only axial stress but also induces bending stress due to the moment created by the offset.
  • Axial load: This is the primary load applied along the axis of the column.
  • Eccentricity: The perpendicular distance between the line of action of the force and the centroid of the section.
  • Bending moment: The product of the load magnitude and the eccentricity, represented as \( M = Pe \).
By understanding eccentric loads, we can evaluate both the direct and additional stresses that can affect the stability and structural integrity of columns. It's through this understanding that engineers can ensure safety and functionality in construction.
Allowable Stress
Allowable stress is a critical factor in ensuring safety within structural engineering. It is the maximum stress that material can withstand without failure over its service life. In our specific column problem, the allowable stress was established at 80 MPa. This limit is determined based on:
  • Material properties: Such as yield strength and ultimate strength.
  • Safety factors: Engineers incorporate safety margins to account for uncertainties and potential variances in material properties and loading conditions.
In this exercise, the total stress experienced by the steel column was a combination of direct and bending stresses. Ensuring that this combined stress does not exceed the allowable stress is key to preventing failure and ensuring the column functions correctly for its intended design life.
Bending Stress
Bending stress arises in a structural member when an eccentric load causes it to deflect and bend. For columns, this stress results from the moment produced by the distance between the load and the centroid of the cross-section.The calculation formula for bending stress is \( \sigma_b = \frac{Mc}{I} \), where:
  • \(M\): The bending moment calculated as \( Pe \), where \(e \) is the eccentricity.
  • \(c\): The distance from the neutral axis to the outermost fiber of the column.
  • \(I\): The moment of inertia of the column's cross-section.
In the given problem, this bending effect was minimal compared to direct stress. However, it remains an important component when ensuring the column's design is within the permissible limits of allowable stress.

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Most popular questions from this chapter

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.

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 \(\phi=0.90\).

The steel bar \(A B\) has a \(\frac{3}{4} \times \frac{3}{6}\) -in. square cross section and is held by pins that are a fixed distance apart and are located at a distance \(e=0.03\) in. from the geometric axis of the bar. Knowing that at temperature \(T_{0}\) the pins are in contact with the bar and that the force in the bar is zero, determine the increase in temperature for which the bar will just make contact with point \(C\) if \(d=0.01\) in. Use \(E=29 \times 10^{6} \mathrm{psi}\) and a coefficient of thermal expansion \(\alpha=6.5 \times 10^{-6} /^{\circ} \mathrm{F}\).

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.

An eccentric load is applied at a point \(22 \mathrm{mm}\) from the geometric axis of a 60 -mm-diameter rod made of a steel for which \(\sigma_{Y}=\) \(250 \mathrm{MPa}\) and \(E=200 \mathrm{GPa}\). Using the allowable-stress method, determine the allowable load \(\mathbf{P}\).
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