EMA 370 Mechanics & Materials Science (Mechanics of Materials) Chapter 10 Columns
Columns
Introduction Columns are vertical prismatic members subjected to compressive forces Goals: 1. Study the stability of columns: resistance to buckling under axial loadings. Determine critical load, stress, and/or column dimension information (length/cross-section area)
Column Subjected to Vertical Load Example: a vertical column with pin connections at both ends subject to centric axial compressive load P Concerns addressed before (chapters 1-): (Internal compressive) stress is within allowable limits σ = P A < σ allowable (Axial or normal) deformation is within the allowable limit δ = PL AE < δ allowable Loads/length should not be too large while transverse section not too small
Concern with Stability for a Column Additional concern for columns under compression: Due to unavoidable imperfect alignment for loads and/or environmental disturbance, axial compressive force/stress could lead to transverse deformation or buckling, often in a sudden way, causing loss of stability for a column. Consideration of a column s stability puts additional constraints on compressive load and part dimension
Pin & Spring Connected Rigid Rods (1) Consider two rigid rods of AC and BC pin connected at A and B and also connected at C by a pin and a torsional spring of constant K Perfect alignment After slight disturbance
Pin & Spring Connected Rigid Rods () To determine whether the system is stable or not against a small disturbance, consider the top rod AC at an angle and draw FBD If reaching equilibrium, balance of moment: When disturbance ~ 0 (very small) We have P cr L sin K( ) P cr 4K / sin θ θ L Implication: When actual load P < P cr, system will return to straight system stable
Pin & Spring Connected Rigid Rods (3) When P > P cr, system moves further away from straight/vertical state, if it settles into a new equilibrium state with a finite angle, P L sin K( ) Because not close to 0, equation cannot be simplified, and we have PL 4K sin
Pin & Spring Connected Rigid Rods (4) For in the range of 0 - /, > sin Therefore: For P < P cr No such exist system stable and remains in the straight state For P > P cr (but not too large) may be found numerically However, system unstable because disturbance for slight increase in P necessitates further increase in, which eventually lead to collapse or loss of stability PL 4K P 4K / L P < P cr P P cr sin P > P cr 1
Euler s Formula for Pin-Ended Columns (1) Similarly, to determine critical load P cr for a pin-ended column AB, the FBD for the entire column is drawn For part AQ near top draw FBD: Bending moment M = -Py From chapter 9 d y dx = M EI = P EI y Rewrite, we have d y dx + P EI y = 0
Euler s Formula for Pin-Ended Columns () From previous Therefore, d y dx + P EI y = 0 d y dx + p y = 0 The general solution to this differential equation is: y = Asinpx + Bcospx Consider boundary conditions: x = 0, y = 0 B = 0 x = L, y = 0 AsinpL = 0 Define p = P EI
Euler s Formula for Pin-Ended Columns (3) AsinpL = 0 Two possibilities: 1. A = 0 y 0 the column remains straight! or. sin pl = 0 pl = n We have, p = nπ L Recall p = P EI Therefore, P satisfy P = n π EI L Smallest P cr EI P occurs when n = 1 cr L Euler s formula
Euler s Formula for Pin-Ended Columns (4) For previous, if A 0 P cr When P = P cr p = P cr EI = π EI L EI = π L y = Asinpx + Bcospx becomes y = Asin πx L This is the elastic curve when P = P cr EI L Hence p = π L Since B = 0, the general solution
Euler s Formula for Pin-Ended Columns (5) If 0 < P < P cr pl < sin pl = 0 cannot be satisfied To satisfy boundary condition of x = L, y = Asin (x/l) = 0 0 P < P cr P P cr A = 0 and y = Asin (x/l) 0, The system remains stable (i.e., column remains straight and stable against small disturbance) If P P cr the system becomes unstable: a small disturbance (e.g., misalignment of force) may cause the system to suddenly change from straight to a buckled state
Euler s Formula for Pin-Ended Columns (4) EI Critical load P cr L Notes: If a column has circular or square crosssection, it may buckle in different directions If a column has other geometry, it tends to buckle in the orientation for which it has the smallest I. Therefore, to determine critical load P cr, I = I min for the cross-section should be used. For the rectangle as illustrated, I min = I y y x I y < I x
Euler s Formula for Pin-Ended Columns (5) Critical load Critical Stress P cr EI L σ cr = P cr A = π EI L A Setting I Ar g where r g is radius of gyration We have cr E L / r ) ( g L/r g Slenderness ratio of the column If a column has different directions with different moment of inertia, the minimum radius of gyration should be used
Euler s Formula for Pin-Ended Columns (6) Critical load P cr Notes about critical stress Proportional to E Decreases as slenderness ratio L/r g increases Ceases to be meaningful when cr > yield strength Y EI E Critical Stress cr L L / r ) ( g
Column w/ One End Fixed & One End Free For column AB with B side fixed and A side free & subjected to compressive load P: It can be treated as top half of a pin connected vertical column as discussed before, P cr E EI cr L ( L / ) e e rg with effective length L e = L Applied force/stress has to be even lower to avoid loss of stability
Column w/ Both Ends Fixed (1) For column AB with both A and B side fixed subjected to compressive load P. Draw FBD for AB and top half AC Symmetry of loads & supports dictates Zero horizontal reactions at A & B Zero shear at C Center symmetric for AC around D D is an inflection point (i.e., infinite curvature) with zero bending moment
Column w/ Both Ends Fixed () For column AB with both A and B sides fixed and subjected to compressive load P, symmetry considerations enable to treat DE section in the center as pin-ended column, with effective length L e = L/ P cr EI L e cr E L / r ) ( e g Applied force/stress can be a bit higher due to way of fixing at two ends
Summary Less stable More stable P cr EI E cr Le ( Le / rg )
Empirical Test Data for Critical Stress 1. Long columns: obey Euler s Equation. Short columns: dominated by y 3. Intermediate columns: mixed behavior
Design of Columns under an Eccentric Load Eccentric loads are treated as a centric load P superposed with bending moment M = Pe σ = σ centric + σ bending Maximum compressive stress σ max = P A + Mc I Two approaches: Allowable Stress Method P Over A + Mc σ I allowable conservative Interaction Method P/A Mc/I + 1 σ allowable(centric) σ allowable(bending)
EMA 370 Mechanics & Materials Science Zhe Cheng (018) 1 Introduction Homework 10.0 Read section 10.1, 10.3 and 10.4 and give a statement confirm the reading
EMA 370 Mechanics & Materials Science Zhe Cheng (018) 1 Introduction Homework 10.1 Knowing the spring constant at D is of constant k and the column DE is rigid. Please calculate the critical load P cr k P D L E
EMA 370 Mechanics & Materials Science Zhe Cheng (018) 1 Introduction Homework 10. A column EH is hang horizontally by two springs with equal length and spring constant k symmetrically at F and G, as illustrated. Please calculate the critical load P cr k P E F b G H P L
Homework 10.3 EMA 370 Mechanics & Materials Science Zhe Cheng (018) 1 Introduction A long cylinder shaped solid column has length L and outside diameter (OD) d. It is made of a material with elastic modulus E. If a separate cylinder-shaped hollow column is to be made with the same material and same length L and O.D. d, but has inner diameter of 0.5d, please calculate (a) the saving in weight for the hollow column versus the original solid column, and (b) the relative reduction in critical load for the hollow column versus the original solid cylinder.
EMA 370 Mechanics & Materials Science Zhe Cheng (018) 1 Introduction Homework 10.4 Members EF and GH are 3 mm diameter and 4 m long steel rod and members of EH and FG are 4 mm diameter and 3 m long steel road. When rope between EG and FH are tightened, and if the factor of safety is, please calculate the largest allowable tension force in EG. Knowing E = 00 GPa and consider buckling only in the plane of the structure. 4 m F G E 3 m H