Strength of materials - Chapter 11: Columns - Nguyễn Sỹ Lâm

Nội dung text: Strength of materials - Chapter 11: Columns - Nguyễn Sỹ Lâm

1. CHAPTER 11: COLUMNS 11.1 Stability of Structures 11.2 Eccentric Loadings; The Secant Formula 11.3 Design of Columns Under Centric Load 11.4 Design of Columns Under An Eccentric Load
2. 11.1 STABILITY OF STRUCTURES STABILITY OF STRUCTURES • Consider model with two rods and torsional spring (stiffness is K). After a small perturbation, K 2  restoring moment L L P sin  P  destabilizing moment 2 2 • Column is stable (tends to return to aligned orientation) if L P  K 2  2 4K P P L L cr P sin  P  L 2 2
3. 11.1 STABILITY OF STRUCTURES EULER’S FORMULA FOR PIN-ENDED BEAMS • Consider an axially loaded beam. After a small perturbation, the system reaches an equilibrium configuration such that d 2 y M P y dx2 EI EI d 2 y P y 0 dx2 EI • Solution with assumed configuration can only be obtained if 2EI P Pcr L2 P 2E Ar 2 2E   cr A L2A L r 2
4. 11.1 STABILITY OF STRUCTURES EXTENSION OF EULER’S FORMULA • 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, 2EI P cr 2 Le 2E  cr 2 Le r Le 2L equivalent length
5. 11.1 STABILITY OF STRUCTURES EXAMPLE 11.01 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. a) Determine the ratio a/b of the two sides of the cross-section corresponding to the most efficient design against buckling. L = 20 in. b) Design the most efficient cross-section for the column. E = 10.1 x 106 psi P = 5 kips FS = 2.5
6. 11.1 STABILITY OF STRUCTURES EXAMPLE 11.01 • Design: L 2L 2 20 in 138.6 e ry b 12 b 12 b Pcr FS P 2.5 5 kips 12.5 kips P 12500 lbs  cr cr A 0.35b b 2E 2 10.1 106psi  cr 2 2 Le r 138.6 b L = 20 in. 12500 lbs 2 10.1 106psi 0.35b b 2 E = 10.1 x 106 psi 138.6 b P = 5 kips b 1.620 in. a 0.35b 0.567 in. FS = 2.5 a/b = 0.35
7. 11.2 ECCENTRIC LOADING; THE SECANT FORMULA ECCENTRIC LOADING; THE SECANT FORMULA P ec 1 P L   1 sec e max Y 2 A r 2 EA r
8. 11.2 ECCENTRIC LOADING; THE SECANT FORMULA EXAMPLE 11.02 SOLUTION: • Maximum allowable centric load: - Effective length, Le 2 8 ft 16 ft 192 in. - Critical load, 2EI 2 29 106 psi 8.0 in 4 P cr 2 2 Le 192 in 62.1 kips - Allowable load, Pcr 62.1 kips P Pall 31.1 kips all FS 2 P 31.1 kips  all  8.79 ksi A 3.54 in 2
9. 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD DESIGN OF COLUMNS UNDER CENTRIC LOAD • Previous analyses assumed stresses below the proportional limit and initially straight, homogeneous columns • Experimental data demonstrate - for large Le/r, cr follows Euler’s formula and depends upon E but not Y. - for small Le/r, cr is determined by the yield strength Y and not E. - for intermediate Le/r, cr depends on both Y and E.
10. 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD DESIGN OF COLUMNS UNDER CENTRIC LOAD Aluminum • Alloy 6061-T6 Le/r 66: 51000 ksi 351 103 MPa  all 2 2 Le / r Le / r • Alloy 2014-T6 Le/r 66: 54000 ksi 372 103 MPa  all 2 2 Le / r Le / r
11. 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD EXAMPLE 11.03 • For L = 750 mm, assume L/r > 55 • Determine cylinder radius: P 372 103 MPa  all A L r 2 60 103 N 372 103 MPa c 18.44 mm 2 2 c 0.750 m c/2 • Check slenderness ratio assumption: c cylinder radius L L 750mm 81.3 55 r radius of gyration r c / 2 18.44 mm I c4 4 c assumption was correct A c2 2 d 2c 36.9 mm
12. 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD PRACTICAL METHOD FOR DETERMINING REQUIRED SECTION USING TABULATED SLENDERNESS RATIOS L L  r I A From Lambda, based on material, tabulated slenderness ratios have been established. They are called N L L z ' Iteration 1: assume 0 0.5 calculate A0 calculate 0 0   r I ' A0 interpolate 0 from table If error between 0 and are smaller than 5% then A0 will be chosen. If not, go to iteration 2. ' 0 0 N z ' L L Iteration 2: let 1 calculate A1 calculate 1 2 1  r I ' A1 interpolate 1 from table ' If error between 1 and 1 are smaller than 5% then A1 will be chosen. If not, go to the next iteration