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Columns

10. Columns. Columns. Stability of Structures Euler’s Formula for Pin-Ended Beams Extension of Euler’s Formula Sample Problem 10.1 Eccentric Loading; The Secant Formula Sample Problem 10.2 Design of Columns Under Centric Load Sample Problem 10.4

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Columns

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  1. 10 Columns

  2. Columns Stability of Structures Euler’s Formula for Pin-Ended Beams Extension of Euler’s Formula Sample Problem 10.1 Eccentric Loading; The Secant Formula Sample Problem 10.2 Design of Columns Under Centric Load Sample Problem 10.4 Design of Columns Under an Eccentric Load

  3. Pont du Gard (France) Roman times The bridge is constructed from limestone blocks fitted together without mortar and secured with iron clamps. The three tiered structure avoids the need for long compressive members.  

  4. Royal Border Bridge (England) Opened in 1850, the bridge continues being used today. The increased slenderness of the columns compared to the Pont du Gard reflect technological improvements over many centuries.

  5. Crymlyn Viaduct (U.K.) The advance from masonary to the slender metal compressive members which make up each column requires substantial bracing to prevent buckling . Opened in 1857 ; Closed in 1964.

  6. Humber road bridge Opened in 1981, comprises a continuously welded closed box road deck suspended from catenary cables supported on reinforced concrete towers. Suspension bridges eliminate the need for struts other than the two towers, however avoiding buckles in other slender components becomes an issue  

  7. The column behind the steering wheel is designed to fail: it is mean to buckle during a car crash to prevent impaling the driver.

  8. In the design of columns, cross-sectional area is selected such that • allowable stress is not exceeded • deformation falls within specifications • After these design calculations, may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles. Stability of Structures

  9. When a structure ( subjected usually to compression ) undergoes visibly large displacements transverse to the load then it is said to   buckle. Buckling is an instability of equilibrium.

  10. Buckling and bending are similar in that they both involve bending moments. In bending these moments are substantially independent of the resulting deflections. In buckling the moments and deflections are mutually inter-dependent - so moments, deflections and stresses are not proportional to loads. If buckling deflections become too large then the structure fails - this is a geometric consideration, completely divorced from any material strength consideration.

  11. Consider model with two rods and torsional spring. After a small perturbation, • Column is stable (tends to return to aligned orientation) if Stability of Structures

  12. Assume that a load P is applied. After a perturbation, the system settles to a new equilibrium configuration at a finite deflection angle. Stability of Structures • Noting that sinq < q , the assumed configuration is only possible if P > Pcr.

  13. Consider an axially loaded beam. After a small perturbation, the system reaches an equilibrium configuration such that • Solution with assumed configuration can only be obtained if Euler’s Formula for Pin-Ended Beams

  14. The value of stress corresponding to the critical load, Euler’s Formula for Pin-Ended Beams • Preceding analysis is limited to centric loadings.

  15. The two rings are elastic supports which change the value of the critical load.

  16. A column with one fixed and one free end, will behave as the upper-half of a pin-connected column. • The critical loading is calculated from Euler’s formula, Extension of Euler’s Formula

  17. Extension of Euler’s Formula

  18. Sample Problem 10.1 An aluminum column of length L and rectangular cross-section has a fixed end at B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry but allow it to move in the other plane. • Determine the ratio a/b of the two sides of the cross-section corresponding to the most efficient design against buckling. • Design the most efficient cross-section for the column. L = 20 in. E = 10.1 x 106 psi P = 5 kips FS = 2.5

  19. SOLUTION: The most efficient design occurs when the resistance to buckling is equal in both planes of symmetry. This occurs when the slenderness ratios are equal. • Buckling in xy Plane: • Most efficient design: • Buckling in xz Plane: Sample Problem 10.1

  20. Sample Problem 10.1 • Design: L = 20 in. E = 10.1 x 106 psi P = 5 kips FS = 2.5 a/b = 0.35

  21. Eccentric loading is equivalent to a centric load and a couple. • The deflection becomes infinite when P = Pcr • Maximum stress Eccentric Loading; The Secant Formula • Bending occurs for any nonzero eccentricity. Question of buckling becomes whether the resulting deflection is excessive.

  22. Eccentric Loading; The Secant Formula

  23. Sample Problem 10.2 The uniform column consists of an 8-ft section of structural tubing having the cross-section shown. • Using Euler’s formula and a factor of safety of two, determine the allowable centric load for the column and the corresponding normal stress. • Assuming that the allowable load, found in part a, is applied at a point 0.75 in. from the geometric axis of the column, determine the horizontal deflection of the top of the column and the maximum normal stress in the column.

  24. SOLUTION: • Maximum allowable centric load: - Effective length, - Critical load, - Allowable load, Sample Problem 10.2

  25. Eccentric load: - End deflection, - Maximum normal stress, Sample Problem 10.2

  26. Previous analyses assumed stresses below the proportional limit and initially straight, homogeneous columns Design of Columns Under Centric Load • Experimental data demonstrate • for large Le/r, scr follows Euler’s formula and depends upon E but not sY. • for small Le/r, scr is determined by the yield strength sY and not E. • for intermediate Le/r, scr depends on both sY and E.

  27. For Le/r > Cc Structural Steel American Inst. of Steel Construction • For Le/r > Cc • At Le/r = Cc Design of Columns Under Centric Load

  28. Alloy 6061-T6Le/r < 66: Le/r> 66: • Alloy 2014-T6Le/r < 55: Le/r> 66: Design of Columns Under Centric Load Aluminum Aluminum Association, Inc.

  29. Sample Problem 10.4 • SOLUTION: • With the diameter unknown, the slenderness ratio can not be evaluated. Must make an assumption on which slenderness ratio regime to utilize. • Calculate required diameter for assumed slenderness ratio regime. • Evaluate slenderness ratio and verify initial assumption. Repeat if necessary. Using the aluminum alloy 2014-T6, determine the smallest diameter rod which can be used to support the centric load P = 60 kN if a) L = 750 mm, b) L = 300 mm

  30. For L = 750 mm, assume L/r > 55 • Determine cylinder radius: • Check slenderness ratio assumption: assumption was correct Sample Problem 10.4

  31. For L = 300 mm, assume L/r < 55 • Determine cylinder radius: • Check slenderness ratio assumption: assumption was correct Sample Problem 10.4

  32. Normal stresses can be found from superposing the stresses due to the centric load and couple, • Allowable stress method: • Interaction method: Design of Columns Under an Eccentric Load • An eccentric load P can be replaced by a centric load P and a couple M = Pe.

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