Concept explainers
A centric load P must be supported by the steel bar AB. Using allowable stress design, determine the smallest dimension d of the cross section that can be used when (a) P = 108 kN, (b) P = 166 kN. Use σY = 250 MPa and E = 200 GPa.
Fig. P10.80
(a)
Find the smallest dimension d of the cross section.
Answer to Problem 80P
The smallest dimension d of the cross section is
Explanation of Solution
Given information:
The length of the column is
The allowable yield strength of the steel is
The modulus of elasticity of the steel is
The centric load acting in the column is
Calculation:
The effective length of the column
Find the cross sectional area (A) using the equation.
Here, the width of the column is b and the depth of the column is d.
Substitute 3d for b.
Find the moment of inertia (I) of the cross section using the equation.
Substitute 3d for b.
Find the minimum radius of gyration (r) using the relation.
Substitute
Find the slenderness ratio
Here, the modulus of elasticity of the material is E and the allowable yield strength is
Substitute 200 GPa for E and 250 MPa for
Find the ratio of effective length
Consider
Find the effective stress
Substitute 200 GPa for E and
Find the critical stress
Substitute
Calculate the allowable stress
Substitute
Calculate the allowable load
Substitute
Consider the allowable load is equal to the centric load.
Substitute 108 kN for
Check:
Substitute 30.1 mm for d in Equation (1).
The assumed condition is correct.
Therefore, the smallest dimension d of the cross section is
(b)
Find the smallest dimension d of the cross section.
Answer to Problem 80P
The smallest dimension d of the cross section is
Explanation of Solution
Given information:
The length of the column is
The allowable yield strength of the steel is
The modulus of elasticity of the steel is
The centric load acting in the column is
Calculation:
The effective length of the column
Find the cross sectional area (A) using the equation.
Substitute 3d for b.
Find the moment of inertia (I) of the cross section using the equation.
Substitute 3d for b.
Find the minimum radius of gyration (r) using the relation.
Substitute
Find the slenderness ratio
Here, the modulus of elasticity of the material is E and the allowable yield strength is
Substitute 200 GPa for E and 250 MPa for
Find the ratio of effective length
Consider
Find the effective stress
Substitute 200 GPa for E and
Find the critical stress
Substitute
Calculate the allowable stress
Substitute
Calculate the allowable load
Substitute
Consider the allowable load is equal to the centric load.
Substitute 166 kN for
Check:
Substitute 33.5 mm for d in Equation (2).
Therefore, the smallest dimension d of the cross section is
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Chapter 10 Solutions
Mechanics of Materials, 7th Edition
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