Kiến trúc xây dựng - Chương 10: Mechanics of materials

MECHANICS OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER © 2002 The McGraw-Hill Companies, Inc. All rights reserved. Columns © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 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 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 3 Stability of Structures • In the design of columns, cross-sectional area is selected such that - allowable stress is not exceeded allA P σσ ≤= - deformation falls within specifications specAE PL δδ ≤= • After these design calculations, may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 4 Stability of Structures • Consider model with two rods and torsional spring. After a small perturbation, ( ) moment ingdestabiliz 2 sin 2 moment restoring 2 =∆=∆ =∆ θθ θ LPLP K • Column is stable (tends to return to aligned orientation) if ( ) L KPP KLP cr 4 2 2 =< ∆<∆ θθ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 5 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. ( ) θ θ θθ sin4 2sin 2 == = crP P K PL KLP • Noting that sinθ < θ , the assumed configuration is only possible if P > Pcr. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 6 Euler’s Formula for Pin-Ended Beams • Consider an axially loaded beam. After a small perturbation, the system reaches an equilibrium configuration such that 02 2 2 2 =+ −== y EI P dx yd y EI P EI M dx yd • Solution with assumed configuration can only be obtained if ( ) ( )2 2 2 22 2 2 rL E AL ArE A P L EIPP cr cr ππσσ π ==>= => © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 7 Euler’s Formula for Pin-Ended Beams ( ) ( ) s ratioslendernes r L tresscritical s rL E AL ArE A P A P L EIPP cr cr cr cr 2 2 2 22 2 2 = == = =>= => π πσ σσ π • The value of stress corresponding to the critical load, • Preceding analysis is limited to centric loadings. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 8 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, ( ) length equivalent 2 2 2 2 2 == = = LL rL E L EIP e e cr e cr πσ π © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 9 Extension of Euler’s Formula © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 10 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. a) Determine the ratio a/b of the two sides of the cross-section corresponding to the most efficient design against buckling. b) Design the most efficient cross-section for the column. L = 20 in. E = 10.1 x 106 psi P = 5 kips FS = 2.5 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 11 Sample Problem 10.1 • Buckling in xy Plane: 12 7.0 1212 , 23 12 1 2 a L r L ara ab ba A Ir z ze z z z = ==== • Buckling in xz Plane: 12/ 2 1212 , 23 12 1 2 b L r L brb ab ab A I r y ye y y y = ==== • Most efficient design: 2 7.0 12/ 2 12 7.0 ,, = = = b a b L a L r L r L y ye z ze 35.0= b a 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 12 Sample Problem 10.1 • Design: ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 62 2 62 2 2 cr cr 6.138 psi101.10 0.35 lbs 12500 6.138 psi101.10 0.35 lbs 12500 kips 5.12kips 55.2 6.138 12 in 202 12 2 bbb brL E bbA P PFSP bbb L r L e cr cr y e ×= ×== == === === π ππσ σ L = 20 in. E = 10.1 x 106 psi P = 5 kips FS = 2.5 a/b = 0.35 in.567.035.0 in.620.1 == = ba b © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 13 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. 2 2 max 2 2 1 2 sec e cr cr L EIP P Pey EI PePy dx yd ππ =⎥⎦ ⎤⎢⎣ ⎡ −⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= −−= • The deflection become infinite when P = Pcr • Maximum stress ( ) ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛+= ⎥⎦ ⎤⎢⎣ ⎡ ++= r L EA P r ec A P r cey A P e 2 1sec1 1 2 2 max maxσ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 14 Eccentric Loading; The Secant Formula ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛+== r L EA P r ec A P e Y 2 1sec1 2max σσ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 15 Sample Problem 10.2 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. .psi1029 6×=E © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 16 Sample Problem 10.2 SOLUTION: • Maximum allowable centric load: ( ) in.192ft16ft 82 ===eL - Effective length, ( )( ) ( ) kips 1.62 in 192 in 0.8psi 1029 2 462 2 2 = ×== ππ e cr L EIP - Critical load, 2in3.54 kips 1.31 2 kips1.62 == == A P FS PP all cr all σ kips1.31=allP ksi79.8=σ - Allowable load, © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 17 Sample Problem 10.2 • Eccentric load: in.939.0=my ( ) ⎥⎦ ⎤⎢⎣ ⎡ −⎟⎠ ⎞⎜⎝ ⎛= ⎥⎦ ⎤⎢⎣ ⎡ −⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛= 1 22 secin 075.0 1 2 sec π π cr m P Pey - End deflection, ( )( ) ( ) ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛+= ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛+= 22 sec in 1.50 in 2in 75.01 in 3.54 kips 31.1 2 sec1 22 2 π πσ cr m P P r ec A P ksi0.22=mσ - Maximum normal stress, © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 18 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, σcr follows Euler’s formula and depends upon E but not σY. - for small Le/r, σcr is determined by the yield strength σY and not E. - for intermediate Le/r, σcr depends on both σY and E. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 19 Design of Columns Under Centric Load Structural Steel American Inst. of Steel Construction • For Le/r > Cc ( ) 92.1 / 2 2 = == FS FSrL E cr all e cr σσπσ • For Le/r > Cc ( ) 3 2 2 / 8 1/ 8 3 3 5 2 /1 ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛−+= = ⎥⎥⎦ ⎤ ⎢⎢⎣ ⎡ −= c e c e cr all c e Ycr C rL C rLFS FSC rL σσσσ • At Le/r = Cc Y cYcr EC σ πσσ 2 2 2 1 2== © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 20 Design of Columns Under Centric Load Aluminum Aluminum Association, Inc. • Alloy 6061-T6 Le/r < 66: ( )[ ] ( )[ ]MPa /868.0139 ksi /126.02.20 rL rL e eall −= −=σ Le/r > 66: ( ) ( )2 3 2 / MPa 10513 / ksi 51000 rLrL ee all ×==σ • Alloy 2014-T6 Le/r < 55: ( )[ ] ( )[ ]MPa /585.1212 ksi /23.07.30 rL rL e eall −= −=σ Le/r > 66: ( ) ( )2 3 2 / MPa 10273 / ksi 54000 rLrL ee all ×==σ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 21 Sample Problem 10.4 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 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. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 22 Sample Problem 10.4 2 4 gyration of radius radiuscylinder 2 4 c c c A I r c === = = π π • For L = 750 mm, assume L/r > 55 • Determine cylinder radius: ( ) mm44.18 c/2 m 0.750 MPa 103721060 rL MPa 10372 2 3 2 3 2 3 = ⎟⎠ ⎞⎜⎝ ⎛ ×=× ×== c c N A P all π σ • Check slenderness ratio assumption: ( ) 553.81mm 18.44 mm750 2/ >=== c L r L assumption was correct mm 9.362 == cd © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 23 Sample Problem 10.4 • For L = 300 mm, assume L/r < 55 • Determine cylinder radius: mm00.12 Pa10 2/ m 3.0585.12121060 MPa 585.1212 6 2 3 = ×⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛−=× ⎥⎦ ⎤⎢⎣ ⎡ ⎟⎠ ⎞⎜⎝ ⎛−== c cc N r L A P all π σ • Check slenderness ratio assumption: ( ) 5550mm 12.00 mm 003 2/ <=== c L r L assumption was correct mm0.242 == cd © 2002 The McGraw-Hill Companies, Inc. All rights reserved. MECHANICS OF MATERIALS Third Edition Beer • Johnston • DeWolf 10 - 24 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, I Mc A P bendingcentric += += maxσ σσσ • Allowable stress method: allI Mc A P σ≤+ • Interaction method: ( ) ( ) 1≤+ bendingallcentricall IMcAP σσ