11 columns 2015 bách khoa

24 2 0
  • Loading ...
1/24 trang

Thông tin tài liệu

Ngày đăng: 27/01/2019, 14:56

Đây là tài liệu của các bạn sinh viện hiện tại đang học tại Đại học Bách Khoa TP HCM. Đồng thời cũng là giáo án của giảng viên tại Đại học Bách Khoa. Nó sẽ rất hữu ích cho công việc học tập của các Bạn. Chúc Bạn thành công. 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 11.1 STABILITY OF STRUCTURES STABILITY OF STRUCTURES • In the design of columns, cross-sectional area is selected such that - allowable stress is not exceeded  P   all A - deformation falls within specifications  PL   spec AE • After these design calculations, may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles 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   destabiliz ing moment 2 • Column is stable (tends to return to aligned orientation) if L P   K 2  P L L sin   P  2 P  Pcr  4K L 11.1 STABILITY OF STRUCTURES STABILITY OF STRUCTURES • Assume that a load P is applied After a perturbation, the system settles to a new equilibrium configuration at a finite deflection angle L P sin   K 2  PL P    K Pcr sin  • Noting that sin <  , the assumed configuration is only possible if P > Pcr P L sin  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 d2y M P    y EI EI dx d2y P  y0 EI dx • Solution with assumed configuration can only be obtained if P  Pcr   EI L2   P  E Ar  2E     cr   A L A L r 2 11.1 STABILITY OF STRUCTURES EULER’S FORMULA FOR PIN-ENDED BEAMS • The value of stress corresponding to the critical load,  EI P  Pcr    cr   L2 P P   cr  cr A A    E Ar L2 A  2E L r   critical stress L  slenderness ratio r • Preceding analysis is limited to centric loadings 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, Pcr   cr   EI L2e  2E Le r 2 Le  L  equivalent length 11.1 STABILITY OF STRUCTURES EXTENSION OF EULER’S FORMULA 2  1   0.7   0.5 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 E = 10.1 x 106 psi P = kips FS = 2.5 b) Design the most efficient cross-section for the column 11.1 STABILITY OF STRUCTURES EXAMPLE 11.01 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: ba I a rz  z  12  A ab 12 Le, z rz  rz  a 12 0.7 L a 12 • Most efficient design: Le, z • Buckling in xz Plane: b2 I y 12 ab ry    A Le, y ry  ab 2L b / 12 12 rz ry  b 12  Le, y ry 0.7 L 2L  a 12 b / 12 a 0.7  b a  0.35 b 11.1 STABILITY OF STRUCTURES EXAMPLE 11.01 • Design: Le 2L 220 in  138    ry b 12 b 12 b Pcr  FS P  2.55 kips   12.5 kips  cr   cr  Pcr 12500 lbs  0.35b b A  2E  Le r     10.1  10 psi 138 b 2 E = 10.1 x 106 psi 12500 lbs  10.1  10 psi  0.35b b 138 b 2 P = kips b  1.620 in L = 20 in FS = 2.5 a/b = 0.35 a  0.35b  0.567 in   11.2 ECCENTRIC LOADING; THE SECANT FORMULA ECCENTRIC LOADING; THE SECANT FORMULA • Eccentric loading is equivalent to a centric load and a couple • Bending occurs for any nonzero eccentricity Question of buckling becomes whether the resulting deflection is excessive • The deflection become infinite when P = Pcr d2y  dx  Py  Pe EI   P     1 ymax  e sec    Pcr   Pcr  • Maximum stress Remind: sec(x) = 1/cos(x) cse(x) = 1/sin(x)  max   P   ymax  e c  1   A  r2 P  ec  P Le   1  sec A r  EA r   EI L2e 11.2 ECCENTRIC LOADING; THE SECANT FORMULA ECCENTRIC LOADING; THE SECANT FORMULA  max   Y  P  ec  P Le   1  sec A  r  EA r  11.2 ECCENTRIC LOADING; THE SECANT FORMULA EXAMPLE 11.02 The uniform column consists of an 8-ft section of structural tubing having the cross-section shown a) Using Euler’s formula and a factor of safety of two, determine the allowable centric load for the column and the corresponding normal stress b) 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 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, Pcr   EI  Le    29  106 psi 8.0 in 192 in 2   62.1 kips - Allowable load, P 62.1 kips Pall  cr  FS  Pall 31.1 kips  A 3.54 in Pall  31.1 kips   8.79 ksi 11.2 ECCENTRIC LOADING; THE SECANT FORMULA EXAMPLE 11.02 • Eccentric load: - End deflection,   P     1 ym  e sec    Pcr         0.075 in sec   1  2 2  ym  0.939 in - Maximum normal stress, P  ec   P    m  1  sec  A r  Pcr  31.1 kips  0.75 in 2 in      1 sec   2   3.54 in  1.50 in   m  22.0 ksi 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 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD DESIGN OF COLUMNS UNDER CENTRIC LOAD Structural Steel American Inst of Steel Construction • For Le/r > Cc  cr   2E Le / r 2  all   cr FS FS  1.92 • For Le/r < Cc  Le / r 2   cr   Y 1   2Cc    all  L / r 1 L / r  FS   e   e  Cc  Cc  • At Le/r = Cc  cr  12  Y 2 2 E Cc  Y  cr FS 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD DESIGN OF COLUMNS UNDER CENTRIC LOAD Aluminum Aluminum Association, Inc • Alloy 6061-T6 Le/r < 66:  all  20.2  0.126 Le / r  ksi  139  0.868Le / r  MPa Le/r > 66:  all  51000 ksi Le / r   351  103 MPa Le / r 2 • Alloy 2014-T6 Le/r < 55:  all  30.7  0.23Le / r  ksi  212  1.585Le / r  MPa Le/r > 66:  all  54000 ksi Le / r   372  103 MPa Le / r 2 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD EXAMPLE 11.03 SOLUTION: • With the diameter unknown, the slenderness ration 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 alloy2014-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 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 c c  cylinder radius r  radius of gyration  I c 4 c   A c  372  103 MPa  0.750 m     c/2  c  18.44 mm • Check slenderness ratio assumption: L L 750 mm    81.3  55 r c / 18.44 mm assumption was correct d  2c  36.9 mm 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD EXAMPLE 11.03 • For L = 300 mm, assume L/r < 55 • Determine cylinder radius:  all  P   L   212  1.585  MPa A   r  60  103 N c   0.3 m   212  1.585   10 Pa  c /   c  12.00 mm • Check slenderness ratio assumption: L L 300 mm    50  55 r c / 12.00 mm assumption was correct d  2c  24.0 mm 11.3 DESIGN OF COLUMNS UNDER CENTRIC LOAD PRACTICAL METHOD FOR DETERMINING REQUIRED SECTION USING TABULATED SLENDERNESS RATIOS  L r  L I A From Lambda, based on material, tabulated slenderness ratios have been established They are called  Nz L L '    A  Iteration 1: assume   0.5 calculate calculate r    I A0  0' from table ' If error between  and  are smaller than 5% then A0 will be chosen If not, go interpolate to iteration Iteration 2: let 1     0' calculate A1  interpolate 1 from table ' Nz 1   ' calculate 1  L r  L I A1 If error between 1 and 1 are smaller than 5% then A1 will be chosen If not, go to the next iteration ' 11.4 DESIGN OF COLUMNS UNDER AN ECCENTRIC LOAD DESIGN OF COLUMNS UNDER AN ECCENTRIC LOAD • An eccentric load P can be replaced by a centric load P and a couple M = Pe • Normal stresses can be found from superposing the stresses due to the centric load and couple,    centric   bending  max  P Mc  A I • Allowable stress method: P Mc    all A I • Interaction method: P A  Mc I  all centric  all bending 1
- Xem thêm -

Xem thêm: 11 columns 2015 bách khoa, 11 columns 2015 bách khoa

Gợi ý tài liệu liên quan cho bạn

Nhận lời giải ngay chưa đến 10 phút Đăng bài tập ngay