Chapter 10: Problem 91
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.
Short Answer
Step by step solution
Calculate the Cross-Sectional Area
Determine the Moment of Inertia
Calculate the Radius of Gyration
Compute the Euler Critical Load
Find Allowable Eccentric Load when e=0.5 in.
Find Allowable Eccentric Load when e=1.0 in.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Allowable Stress Method
- \( \sigma_{allowable} = \frac{P}{A} + \frac{Pe}{r^2} \)
Cross-Sectional Area Calculation
- Cross-sectional area, \( A = b \times d = 5.0 \times 7.5 = 37.5 \text{ in}^2 \)
Moment of Inertia
- \( I = \frac{b \cdot d^3}{12} \)
- \( I = \frac{5.0 \times 7.5^3}{12} = 175.78 \text{ in}^4 \)
Euler Critical Load
- \( P_{cr} = \frac{\pi^2EI}{L_{eff}^2} \)
- \( P_{cr} \) is the critical load.
- \( E \) is the modulus of elasticity.
- \( I \) is the moment of inertia.
- \( L_{eff} \) is the effective length of the column.
- \( P_{cr} \approx 92661.57 \text{ lb} \)