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Two \(89 \times 64-\mathrm{mm}\) angles are bolted together as shown for use as a column of 2.4 -m effective length to carry a centric load of \(180 \mathrm{kN} .\) Knowing that the angles available have thicknesses of \(6.4 \mathrm{mm}, 9.5 \mathrm{mm},\) and \(12.7 \mathrm{mm},\) use allowable stress design to determine the lightest angles that can be used. Use \(\sigma_{Y}=250 \mathrm{MPa}\) and \(E=200 \mathrm{GPa}\).

Short Answer

Expert verified
Use the 6.4 mm thick angles for the lightest fit.

Step by step solution

01

Calculate Slenderness Ratio

Determine the slenderness ratio using the formula \[ \lambda = \frac{KL}{r} \]where \(K\) is the effective length factor (1 for a pin-pin column), \(L = 2.4\text{ m} = 2400\text{ mm}\), and \(r\) is the radius of gyration. Initially, assume one of the angles for further calculations.
02

Calculate Critical Stress \(\sigma_c\)

Once the slenderness ratio \(\lambda\) is found, compute the critical stress \(\sigma_c\) using Euler's formula for columns:\[ \sigma_c = \frac{\pi^2 E}{\lambda^2} \]. Use \(E = 200 \text{ GPa} = 200,000 \text{ MPa}\). Compare \(\sigma_c\) with the allowable yield stress \(\sigma_Y = 250 \text{ MPa}\).
03

Determine Required Area

Determine the required cross-sectional area \(A_{req}\) by setting the allowable axial load equal to the product of the critical stress and the area:\[ P = \sigma_c \times A_{req} \]where \(P = 180 \text{ kN} = 180,000 \text{ N}\). Solve for \(A_{req}\).
04

Check Angle Dimensions

With the determined \(A_{req}\), compare with the cross-sectional properties of the available angles (\(6.4 \text{ mm}, 9.5 \text{ mm}, 12.7 \text{ mm}\)). Find an angle pair whose combined area meets or exceeds \(A_{req}\), ensuring each pair's lesser sectional radius sustains calculated \(\sigma_c\).
05

Select Lightest Angle

Select the angle configuration that meets the requirements with the smallest mass, ensuring that its performance under load is adequate. Consider the cross-sectional properties listed and choose the lightest option.

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Key Concepts

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

Slenderness Ratio
When designing a column, one of the critical factors we consider is the slenderness ratio, which helps to understand its tendency to buckle. The slenderness ratio \( \lambda \) is calculated using the formula \( \lambda = \frac{KL}{r} \). In this formula:
  • \( K \) represents the effective length factor. For a pin-ended column, \( K \) is usually taken as 1.
  • \( L \) is the actual length of the column, and in this exercise, is given as 2400 mm.
  • \( r \) is the radius of gyration, found from the cross-sectional properties of the column.
The radius of gyration \( r \) helps to understand the distribution of the cross-sectional area around an axis and is fundamental to identify the column's resistance to buckling. Understanding and calculating the slenderness ratio allows engineers to predict and enhance the stability of columns used in construction.
Critical Stress
After determining the slenderness ratio, the next step in allowable stress design is to calculate the critical stress \( \sigma_c \). This is the stress level needed to cause the column to buckle, and it is computed using Euler's Formula: \[ \sigma_c = \frac{\pi^2 E}{\lambda^2} \]Where:
  • \( E \) is Young's modulus, the measure of the stiffness of a material (200 GPa or 200,000 MPa in this exercise).
  • \( \lambda \) is the slenderness ratio.
With the calculated \( \sigma_c \), engineers need to ensure it does not exceed the allowable yield stress \( \sigma_Y \), which is 250 MPa in this example. Balancing these stresses ensures the column remains safe under load, protecting against buckling before reaching yield stress limits.
Cross-sectional Area
The final step in assuring a column's stability is determining the required cross-sectional area \( A_{req} \). This ensures the column can handle the given load without exceeding the critical stress. The required area is calculated using the formula: \[ P = \sigma_c \times A_{req} \]With:
  • \( P \) being the load the column must support (180 kN or 180,000 N in this problem).
  • \( \sigma_c \) being the previously calculated critical stress.
To find an appropriate cross-sectional area, engineers compare \( A_{req} \) to available column sections. For this exercise, the thicknesses of the angles (6.4 mm, 9.5 mm, 12.7 mm) are considered, seeking the lightest configuration meeting or exceeding the required area, providing both safety and efficiency in design. Selecting the optimal cross-section ensures the structure can withstand the specified load without failure.

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Most popular questions from this chapter

A compression member of \(9-\mathrm{m}\) effective length is obtained by welding two 10 -mm-thick steel plates to a \(\mathrm{W} 250 \times 80\) rolled-steel shape as shown. Knowing that \(\sigma_{Y}=345 \mathrm{MPa}\) and \(E=200 \mathrm{GPa}\) and using allowable stress design, determine the allowable centric load for the compression member.

Axial loads of magnitude \(P=135\) kips are applied parallel to the geometric axis of the \(\mathrm{W} 10 \times 54\) rolled-steel column \(A B\) and intersect the \(x\) axis at a distance \(e\) from the geometric axis. Knowing that \(\sigma_{\text {all }}=12\) ksi and \(E=29 \times 10^{6}\) psi, determine the largest permissible length \(L\) when \((a) e=0.25\) in. \((b) e=0.5\) in.

A \(26-\) kip axial load \(\mathbf{P}\) is applied to a \(\mathrm{W} 6 \times 12\) 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 \(e=0.25\) in., determine the largest permissible length \(L\) if the allowable stress in the column is 14 ksi. Use \(E=29 \times 10^{6}\) psi.

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 \(\phi=0.90\).

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.

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