Chapter 10: Problem 60
A column having a 3.5 -m effective length is made of sawn lumber with a \(114 \times 140\) -mm cross section. Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is \(\sigma_{C}=7.6 \mathrm{MPa}\) and the adjusted modulus \(E=2.8 \mathrm{GPa},\) determine the maximum allowable centric load for the column.
Short Answer
Step by step solution
Analyze the Problem and Given Data
Calculate the Cross-Sectional Area
Convert Units
Calculate Maximum Load Using Allowable Stress
Consider Buckling (not needed if allowed stress governs)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Column Buckling
- \(P_{cr}\) = Euler's critical load
- \(E\) = Elastic modulus of the material
- \(I\) = Minimum moment of inertia of the cross section
- \(L_e\) = Effective length of the column
Compressive Stress
- \(P_{max}\) = Maximum allowable load
- \(\sigma_c\) = Allowable compressive stress
- \(A\) = Cross-sectional area
Elastic Modulus
Cross-Sectional Area
- \(b\) = Width of the cross section
- \(d\) = Depth of the cross section