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chap11 slides fm M Vable Mechanics of Materials Chapter 11 Pr in te d fr om h ttp // w w w m e m tu e du /~ m av ab le /M oM 2n d ht m Stability of Columns • Bending due to a compressive axial load is[.]
M Vable Mechanics of Materials: Chapter 11 Stability of Columns • Bending due to a compressive axial load is called Buckling • Structural members that support compressive axial loads are called Columns • Buckling is the study of stability of a structure’s equilibrium Learning objectives • Develop an appreciation of the phenomena of buckling and the various types of Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm structure instabilities • Understand the development and use of buckling formulas in analysis and design of structures August 2012 11-1 M Vable Mechanics of Materials: Chapter 11 Buckling Phenomenon Energy Approach Unstable Equilibrium Stable Equilibrium Potential Energy is Minimum Neutral Equilibrium Bifurcation/Eigenvalue Problem P P PL ⁄ K θ = θ ⁄ sin θ Lsinθ θ Unstable Unstable Stable Stable (c) L O R Kθθ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm O August 2012 11-2 M Vable Mechanics of Materials: Chapter 11 Snap Buckling Problem θ = 0 < θ < 45 P o P Lsin(45-θ) Fs L o 45 (45−θ) Fs L cos(45-θ) P P = ( cos ( 45 – θ ) – cos 45 ) tan ( 45 – θ ) KL L < θ < 45 P = ( cos ( θ – 45 ) – cos 45 ) tan ( θ – 45 ) KL L θ > 45 θ > 45 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Lsin(θ−45) L cos(θ−45) Fs P (θ−45) P Fs August 2012 11-3 0 M Vable Mechanics of Materials: Chapter 11 Local Buckling Axial Loads Torsional Loads crinkling Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Compressive August 2012 11-4 M Vable Mechanics of Materials: Chapter 11 C11.1 Linear deflection springs and torsional springs are attached to rigid bars as shown.The springs can act in tension or compression and resist rotation in either direction Determine Pcr, the critical load value P k = lbs/in 30 in k = lbs/in 30 in K = 2000 in-lbs/rads O Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C11.1 August 2012 11-5 M Vable Mechanics of Materials: Chapter 11 Euler Buckling y P Mz x A N=P L v(x) A P Boundary Value Problem Differential Equation: EI d v + Pv = dx Boundary conditions: v ( ) = v(L) = Solution Trivial Solution: v = Non-Trivial Solution: v ( x ) = A cos λx + B sin λx where: λ = P -EI Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Characteristic Equation: sin λL = 2 n π EI P n = -2 L n = 1, 2, ⋅ ⋅ π EI Euler Buckling Load: P cr = -2 L • Buckling occurs about an axis that has a minimum value of I August 2012 11-6 M Vable Mechanics of Materials: Chapter 11 x L⎠ Buckling Mode: v = B sin ⎛ nπ -⎞ ⎝ Mode shape Pcr Pcr π EI P cr = -2 L L Pcr Mode shape Pcr Pcr I L/2 L/2 Pcr 4π EI P cr = -2 L Mode shape Pcr L/3 Pcr I I L/3 L/3 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Pcr August 2012 11-7 9π EI P cr = -2 L M Vable Mechanics of Materials: Chapter 11 Effects of End Conditions P B L x Differential Equation Boundary Conditions One end fixed, other end free EI dv dx = One end fixed, other end pinned + Pv v(0) = v( L) = EI x y A A A L x y Pinned at both Ends B L L x P B B Case P P y A Fixed at both ends dv EI + Pv dx = Pv ( L ) dv EI + Pv dx = RB ( L – x ) dv + Pv dx = RB ( L – x ) + MB v( 0) = v( 0) = v( 0) = dv (0) = dx dv (0) = dx v( L) = dv (0) = dx v( L) = dv (L) = dx Characteristic Equation Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm λ = sin λL = cos λL = tan λL = λL ( – cos λL ) – λL sin λL = P -EI Critical Load Pcr π EI L π EI π EI - = -22 4L ( 2L ) 20.13EI π EI = -22 L ( 0.7L ) 4π EI π EI = -2- [ L ( 0.5L ) Effective Length— Leff L 2L 0.7L 0.5L 2 2 π EI P cr = -2 L eff August 2012 11-8 2 M Vable Mechanics of Materials: Chapter 11 P cr Axial Stress: σ cr = -A Radius of gyration: r = I A P cr π E σ cr = = A ( L eff ⁄ r ) Slenderness ratio: (Leff/r) Failure Envelopes A A A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Short columns: Designed to prevent material elastic failure • Long columns: Designed to prevent buckling failure August 2012 11-9 M Vable Mechanics of Materials: Chapter 11 C11.2 Columns made from an alloy will be used in a construction of a frame The cross-section of the columns is a hollow-cylinder of thickness 10 mm and outer diameter of ‘d’ mm The modulus of elasticity is E = 200 GPa and the yield stress is σyield = 300 MPa Table below shows a list of the lengths ‘L’ and outer diameters ‘d’ Identify the long and the short columns Assume the ends of the column are built in d (mm) 60 80 100 150 200 225 250 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm L (m) August 2012 11-10 M Vable Mechanics of Materials: Chapter 11 C11.3 A force F= 750 lb is applied to the two bars structure as shown Both bars have a diameter of d = 1/4 inch, modulus of elasticity E = 30,000 ksi, and yield stress σyield = 30 ksi Bar AP and BP have lengths of LAP= inches and LBP= 10 inches, respectively Determine the factor of safety for the two-bar structures B A 60o P 25o F Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C11.3 August 2012 11-11 M Vable Mechanics of Materials: Chapter 11 Class Problem Identify the members in the structures that you would check for buckling Circle the correct answers F P B 40o A 110o Structure Structure o 110 A B F P B B o Structure 60 P 75o 30o Structure P F F Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm A Structure Structure Structure Structure August 2012 AP AP AP AP BP BP BP BP 11-12 Both Both Both Both None None None None A M Vable Mechanics of Materials: Chapter 11 C11.4 A hoist is constructed using two wooden bars to lift a weight of kips The modulus of elasticity for wood is E = 1,800 ksi and the allowable normal stress 3.0 ksi Determine the maximum value of L to the nearest inch that can be used in constructing the hoist B A L ft C in 30o A A W A A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C11.4 August 2012 11-13 in Cross-section AA ... http://www.me.mtu.edu/~mavable/MoM2nd.htm • Short columns: Designed to prevent material elastic failure • Long columns: Designed to prevent buckling failure August 2012 11- 9 M Vable Mechanics of Materials: Chapter 11 C11.2 Columns made... August 2012 11- 10 M Vable Mechanics of Materials: Chapter 11 C11.3 A force F= 750 lb is applied to the two bars structure as shown Both bars have a diameter of d = 1/4 inch, modulus of elasticity... BP BP 11- 12 Both Both Both Both None None None None A M Vable Mechanics of Materials: Chapter 11 C11.4 A hoist is constructed using two wooden bars to lift a weight of kips The modulus of elasticity