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The following will be discussed in this chapter: Stability of structures, euler’s formula for pin-ended beams, extension of euler’s formula, eccentric loading, the secant formula, design of columns under centric load, design of columns under an eccentric load.
Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P Beer E Russell Johnston, Jr John T DeWolf Columns Lecture Notes: J Walt Oler Texas Tech University © 2002 The McGraw-Hill Companies, Inc All rights reserved Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stability of Structures • In the design of columns, cross-sectional area is selected such that - allowable stress is not exceeded σ= P ≤ σ all A - deformation falls within specifications δ= PL ≤ δ spec AE • 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 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stability of Structures • Consider model with two rods and torsional spring After a small perturbation, K (2∆θ ) = restoring moment L L P sin ∆θ = P ∆θ = destabilizing moment 2 • Column is stable (tends to return to aligned orientation) if L P ∆θ < K (2∆θ ) P < Pcr = © 2002 The McGraw-Hill Companies, Inc All rights reserved 4K L 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 θ = K (2θ ) PL P θ = = K Pcr sin θ • Noting that sinθ < θ , the assumed configuration is only possible if P > Pcr © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 = π EI L2 ( ) π E Ar π 2E P = σ = > σ cr = A L A (L r )2 © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Euler’s Formula for Pin-Ended Beams • The value of stress corresponding to the critical load, π EI P > Pcr = σ= σ cr = = L2 P P > σ cr = cr A A ( ) π E Ar L2 A π 2E (L r ) = critical stress L = slenderness ratio r Preceding analysis is limited to centric loadings â 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 = σ cr = π EI L2e π 2E (Le r )2 Le = L = equivalent length © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Extension of Euler’s Formula © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 L = 20 in E = 10.1 x 106 psi b) Design the most efficient cross-section for the column P = kips FS = 2.5 © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 10 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.1 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 a2 z 12 rz = = = 12 A ab Le, z rz = rz = a 12 L a 12 • Most efficient design: Le, z rz • Buckling in xz Plane: ry2 = Le, y ry Iy A = = ab3 12 ab b2 = 12 ry = b 12 2L b / 12 © 2002 The McGraw-Hill Companies, Inc All rights reserved = Le, y ry L 2L = a 12 b / 12 a = b a = 0.35 b 10 - 11 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.1 • Design: Le 2L 2(20 in ) 138.6 = = = ry b 12 b 12 b Pcr = ( FS )P = (2.5)(5 kips ) = 12.5 kips σ cr = σ cr = Pcr 12500 lbs = (0.35b )b A π 2E = (Le r ) ( ( π 10.1 × 106 psi (138.6 b )2 E = 10.1 x 106 psi 12500 lbs π 10.1 × 106 psi = (0.35b )b (138.6 b )2 P = kips b = 1.620 in L = 20 in FS = 2.5 ) ) a = 0.35b = 0.567 in a/b = 0.35 © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 12 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 ⎡ ⎛π P ⎞ ⎤ ⎟ − 1⎥ ymax = e ⎢sec⎜⎜ ⎟ P cr ⎠ ⎦ ⎝ ⎣ Pcr = π EI L2e • Maximum stress σ max = P ⎡ ( ymax + e )c ⎤ 1+ ⎢ ⎥ A⎣ ⎦ r2 P ⎡ ec ⎛ P Le ⎞⎤ ⎟⎟⎥ = ⎢1 + sec⎜⎜ A⎣ r EA r ⎠⎦ ⎝ © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 13 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Eccentric Loading; The Secant Formula σ max = σ Y = P ⎡ ec ⎛ P Le ⎞⎤ ⎟⎥ ⎢1 + sec⎜ A ⎣ r ⎜⎝ EA r ⎟⎠⎦ © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 14 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 E = 29 × 106 psi 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 © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 15 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.2 SOLUTION: • Maximum allowable centric load: - Effective length, Le = 2(8 ft ) = 16 ft = 192 in - Critical load, Pcr = π EI = Le ( )( π 29 × 106 psi 8.0 in (192 in )2 ) = 62.1 kips - Allowable load, 62.1 kips P Pall = cr = FS σ= © 2002 The McGraw-Hill Companies, Inc All rights reserved Pall 31.1 kips = A 3.54 in Pall = 31.1 kips σ = 8.79 ksi 10 - 16 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.2 • Eccentric load: - End deflection, ⎡ ⎛π P ⎞ ⎤ ⎟ − 1⎥ ym = e ⎢sec⎜⎜ ⎟ ⎣ ⎝ Pcr ⎠ ⎦ ⎡ ⎛ π ⎞ ⎤ = (0.075 in )⎢sec⎜ ⎟ − 1⎥ 2 ⎠ ⎦ ⎣ ⎝ ym = 0.939 in - Maximum normal stress, P ⎡ ec ⎛ π P ⎞⎤ ⎟⎥ σ m = ⎢1 + sec⎜⎜ ⎟ A⎣ = r ⎝ Pcr ⎠⎦ 31.1 kips ⎡ (0.75 in )(2 in ) ⎛ π ⎞⎤ 1+ sec⎜ ⎟⎥ ⎢ 2 ⎝ ⎠⎦ 3.54 in ⎣ (1.50 in ) σ m = 22.0 ksi © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 17 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 10 - 18 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design of Columns Under Centric Load Structural Steel American Inst of Steel Construction • For Le/r > Cc π 2E σ cr = σ all = (Le / r ) σ cr FS FS = 1.92 • For Le/r > Cc ⎡ (Le / r )2 ⎤ σ cr = σ Y ⎢1 − ⎥ C ⎢⎣ c ⎥⎦ σ all = L / r 1⎛ L / r ⎞ FS = + e − ⎜⎜ e ⎟⎟ Cc ⎝ Cc ⎠ σ cr FS • At Le/r = Cc σ cr = © 2002 The McGraw-Hill Companies, Inc All rights reserved 1σ Y Cc2 = 2π E σY 10 - 19 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design of Columns Under Centric Load Aluminum Aluminum Association, Inc • Alloy 6061-T6 Le/r < 66: σ all = [20.2 − 0.126(Le / r )] ksi = [139 − 0.868(Le / r )] MPa Le/r > 66: σ all = 51000 ksi (Le / r ) = 351 × 103 MPa (Le / r )2 • Alloy 2014-T6 Le/r < 55: σ all = [30.7 − 0.23(Le / r )] ksi = [212 − 1.585( Le / r )] MPa Le/r > 66: σ all = © 2002 The McGraw-Hill Companies, Inc All rights reserved 54000 ksi (Le / r ) = 372 × 103 MPa (Le / r )2 10 - 20 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.4 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 © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 21 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.4 • For L = 750 mm, assume L/r > 55 • Determine cylinder radius: P 372 × 103 MPa σ all = = A (L r )2 60 × 103 N πc c = cylinder radius r = radius of gyration = πc 4 c I = = A πc = 372 × 103 MPa ⎛ 0.750 m ⎞ ⎜ ⎟ ⎝ c/2 ⎠ c = 18.44 mm • Check slenderness ratio assumption: L L 750mm = = = 81.3 > 55 r c / (18.44 mm ) assumption was correct d = 2c = 36.9 mm © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 22 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.4 • For L = 300 mm, assume L/r < 55 • Determine cylinder radius: σ all = P ⎡ ⎛ L ⎞⎤ = ⎢212 − 1.585⎜ ⎟⎥ MPa A ⎣ ⎝ r ⎠⎦ 60 × 103 N πc ⎡ ⎛ 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 © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - 23 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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, σ = σ centric + σ bending σ max = P Mc + A I • Allowable stress method: P Mc + ≤ σ all A I • Interaction method: P A + Mc I (σ all )centric (σ all )bending © 2002 The McGraw-Hill Companies, Inc All rights reserved ≤1 10 - 24 ... reserved 10 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Extension of Euler’s Formula © 2002 The McGraw-Hill Companies, Inc All rights reserved 10 - Third Edition MECHANICS OF MATERIALS. .. 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 10 - Third Edition MECHANICS OF MATERIALS. .. McGraw-Hill Companies, Inc All rights reserved 10 - 14 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 10.2 The uniform column consists of an 8-ft section of structural