1. BUCKLING OF COLUMNS

2. AIM To study the failure analysis of buckling of columns

3. BASIC DEFINITIONS ASSUMPTIONS & SIGN CONVENTIONS FAILURE ANALYSIS OF COLUMN WITH EULER EQUATION EFFECTIVE LENGTH AND SUPPORT CONDITION GRAPHICAL ANALYSIS OF STRESS VS SLENDERNESS RATIO RANKINE’S FORMULA AND ITS IMPORTANCE NUMERICAL PREVIEW

4. BASIC DEFINITIONS COLUMN Compressive member in vert position Consist of Short column Long column STRUTS Bar or member of structure acted upon by compressive load Eg: strut in truss, piston rods P P Short column Long column Compression/Crushing Bending / Buckling

5. BASIC DEFINITIONS STABLE/UNSTABLE EQUILIBRIUM Stable equilibrium: body in static equilibrium on being displaced slightly, returns to its original position Unstable equilibrium : body in equilibrium on being slightly disturbed, moves away from its equilibrium position NEUTRAL EQUILIBRIUM Body in equilibrium on being slightly disturbed, moves away from its equilibrium position and remains in the same position on removal of load Load acting- crippling/buckling/critical load unstable stable neutral

6. ASSUMPTIONS & SIGN CONVENTIONS Column is initially perfectly straight and the load is applied axially Uniform cross-section area Perfectly elastic and obeys Hooke’s law. homogeneous and isotropic Length very large as compared to its lateral dimensions Direct stress <<<bending stress Self-weight of column is negligible +ve Sign Convention Assumptions

7. The failure analysis on columns by buckling has been carried out for the fwg conditions FAILURE ANALYSIS OF COLUMN Le=0.5L P Le=L P Le=2L P Le=0.7L P Both ends hinged One end fixed other end free One end fixed other end hinged Both ends fixed

8. CONDITION I : BOTH ENDS HINGED Consider a column AB of length l hinged at both its ends A and B carries an axial crippling load at A. Consider any section X-X at a distance of x from B. Let the deflection at X-X is y.  The bending moment at X-X due to the load P, M = (-)P*y Where( ) Solution of this differential equation is P P XX y L B A x

9. (Now taking the least significant value (i.e)  ) By using Boundary conditions, At B, (x = 0, y = 0  A = 0) and At A,(x = l, y = 0) CONDITION I : BOTH ENDS HINGED Le=L P

10. CONDITION II : ONE END FIXED ONE END FREE Consider a column AB of length l, fixed at B and free at A, carrying an axial rippling load P at D de to which it just buckles. The deflected form of the column AB is shown in fig. Let the new position of A is A 1. Let a be the deflection at the free end. Consider any section X-X at a distance x from B. Let the deflection at xx is y. Bending moment due to critical load P at xx, B a a-y x y P a x

11. The solution of the above differential equation is, Where A and B are constants of integration. At B, x = 0, y = 0  A = -a Differentiating the equation ( w.r.t to x) At the fixed end B, x = 0 and CONDITION II : ONE END FIXED ONE END FREE

12. Substitute A = -a and B = 0 in equation (i) we get, At the free end A, x = l, y = a, substitute these values in equation (ii) (ii)  The crippling load for the columns with one end fixed and other end free. Le=2L P CONDITION II : ONE END FIXED ONE END FREE

13. CONDITION III : BOTH ENDS FIXED Solution of this differential equation is P P XX y L B A x M M B = 0 … on applying bdy condition ….On substitution and applying 2 nd bdy condition Le=0.5L P

14. CONDITION IV : FIXED/HINGED Differential equation Solution of differential equation Applying bdy condition P P X X y L B A x M M H

15. EFFECTIVE LENGTH AND SUPPORT CONDITION Pinned/PinnedFixed/FreeFixed/FixedPinned/Fixed Le = LLe = 2LLe = L/2Le = 0.7L Le=0.5L P Le=L P Le=2L P Le=0.7L P Both ends hinged One end fixed other end free One end fixed other end hinged Both ends fixed 1 2 34

16. The Euler formula was derived with the assumption that load is always applied through the centroid of the column & column is straight. Manufactured columns are never perfectly straight & begin to bend slightly upon the application of the load. To investigate this effect load, P is applied to the column at a short eccentric distance, e from the centroid of the cross section. SECANT FORMULA P P P P M=Pe e l l

17. Secant Formula SECANT FORMULA

18. GRAPHICAL ANALYSIS OF STRESS VS SLENDERNESS RATIO For a steel column if (L/r) ≥ 89, Euler’s formula can be used to determine the buckling load since the stress in the column remains elastic. (L/r) - Slenderness ratio. Buckling will occur about the axis when the ratio gives the greatest value. Critical stress= P(e)/A= (Π^2 ) EI/A*l(e)^2= (Π^2 ) E/(l(e)/k)^2 L/r 89 250 FAILURE BY BUCKLING FAILURE BY CRUSHING

19. Euler’s formula long columns(buckling condition) Rankine’s formula applicable to all columns irrespective of whether they are short or long. If P is the crippling load by Rankine’s formula. P c is the crushing load of the column material P E is the crippling load by Euler’s formula. RANKINE’S FORMULA AND ITS IMPORTANCE 1/P= 1/Pe +1/Pc Also, 1/P=1/Pe(long columns) 1/P=1/Pc(short columns)

20. DERIVATION OF RANKINE GORDEN FORMULA Substitute the value of P c = f c A and where, f c =Ultimate crushing stress of the column material. A=Cross-sectional are of the column L=Effective length of the column I=Ak 2 Where k = Least radius of gyration. where  = Rankine’s constant P = Crushing Load

21. 21.A column with alone end hinged and the other end fixed has a length of 5m and a hollow circular cross section of outer diameter 100 mm and wall thickness 10 mm. If E = 1.60 x 10 5,stress =350N/mm 2 and crushing strength, Find the load that the column may carry with a factor of safety of 2.5 according to Euler theory and Rankine – Gordon theory. If the column is hinged on one side and fixed on other, find the safe load according to the two theories. NUMERICAL ON BUCKLING OF COLUMN Given: Length= 5 m = 5000 mm Outer diameter D = 100 mm Inner diameter d = D-2t = 100 – 2 (10) = 80 mm Thickness = 10 mm E=1.60 x 10 5 N/mm 2 FOS= 2.5 Stress =350N/mm 2

22. 1. Calculation of load by Euler’s theory: Column with one end fixed and other end hinged. (i) I = 28.96 x 10 5 mm 4 By putting the value of I in the eqn (I), we get P = 73.074 x 10 3 N NUMERICAL ON BUCKLING OF COLUMN

23. ii.Calculation of load by Rankine-Gordon theory: Rankine’s Constant (assume the column material is mild steel) K = least radius of gyration mKwhere Ans: Euler’s Theory P = 73.074 x 10 4 N Rankine’s Theory P = 60.94 x 10 4 N Safe Load= P/2.5 Euler’s Theory =29229.6N Rankine’s theory = 243760 N NUMERICAL ON BUCKLING OF COLUMN

24. BIBLIOGRAPHY https://web.mit.edu/16.unified https://wp.optics.arizona.edu/optomech/wp- content/uploads/sites NPTEL lectures

25. - END - Thank you