Concept explainers
Select the lightest W200 shape.
Answer to Problem 52P
The lightest W200 shape is
Explanation of Solution
Given information:
The magnitude of axial load is
The eccentricity of the load is
The modulus of elasticity of the shape is
The allowable stress in the column is
Calculation:
The effective length of the column
Consider the section
Refer to the Appendix C “Properties of Rolled-Steel Shapes” in the textbook.
For
The cross sectional area is
The minimum moment of inertia in the y-axis is
The minimum radius of gyration in the y-axis is
The width of the flange is
Find the distance between the neutral axis and the outermost fibre (c) using the relation.
Substitute 166 mm for
Determine the critical load
Here, the modulus of elasticity is E, the minimum moment of inertia is I, and the effective length is
Substitute 200 GPa for E,
Find the maximum stress
Here, the maximum stress in the column is
Substitute 345 kN for P, 1,375 kN for
Consider the section
Refer to the Appendix C “Properties of Rolled-Steel Shapes” in the textbook.
For
The cross sectional area is
The minimum moment of inertia in the y-axis is
The minimum radius of gyration in the y-axis is
The width of the flange is
Find the distance between the neutral axis and the outermost fibre (c) using the relation.
Substitute 133 mm for
Determine the critical load
Substitute 200 GPa for E,
Find the maximum stress
Substitute 345 kN for P, 505.66 kN for
Consider the section
Refer to the Appendix C “Properties of Rolled-Steel Shapes” in the textbook.
For
The cross sectional area is
The minimum moment of inertia in the y-axis is
The minimum radius of gyration in the y-axis is
The width of the flange is
Find the distance between the neutral axis and the outermost fibre (c) using the relation.
Substitute 165 mm for
Determine the critical load
Substitute 200 GPa for E,
Find the maximum stress
Substitute 345 kN for P, 1,161 kN for
The calculated maximum stress is closer to the allowable stress.
The shape
Therefore, the lightest W200 shape is
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