Calculate the Euler buckling load for an axially loaded, pin-connected W14 × 22 structural steel wide-flange column. The column length is 12 ft, as shown.
Calculate the Euler buckling load for an axially loaded, pin-connected
W14
×
22
structural steel wide-flange column. The column length is 12 ft, as shown.
Expert Solution & Answer
To determine
The Euler buckling load for the given member.
Answer to Problem 18.1P
Pe=100kips
Explanation of Solution
Given:
W14×22 Structural steel wide flange column
Column length is 12 ft
For structural steel modulus of elasticity is E=30000ksi
Calculation:
For the wide flange section W14×22
Least moment of inertia of the cross section is Iy=7.0in4
The Euler’s buckling for axially loaded member can be
Pe = π2EIL2
= π2×30000×7(12×12)2=99.95kips
Conclusion:
Therefore, the Euler’s buckling for axially loaded member is, Pe=100kips .
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The steel (E = 200GPa ) column hinged on both sides. the height of the column is 1 m and cross-section is 10mm× 20mm . Calculate the lowest Euler critical buckling load?
Derive the Euler Buckling formula for column having fixed support at both ends.
(eXPLAIN IN DETAIL IN HAND WRITTEN )
An aluminum tube of length 0.01R (m) is used as column (fixed at both ends) carrying an
axial load of 1.333R (N). If the outer diameter of the tube is 0.06R (mm),compute the inner
diameter that would provide a factor of safety of 2 against buckling. Use E = 70 GPa for aluminum.
Note R=993
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