Theories of columns
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The document discusses column buckling and failure modes. It defines a strut and column, and describes failure due to direct stress, buckling stress, or a combination. It provides differential equations to model column buckling based on the column's end conditions, including fixed-fixed, fixed-free, and fixed-hinged. Solutions provide the critical buckling load. The document also discusses Rankine's formula which relates crushing load and buckling load for a column's total capacity.
Theories of columns
- 2. A member of a structure which carries an axial compressive load is called a strut A vertical strut which is liable for failure due to buckling or bending is called a column Failure of a STRUT occurs due to any one of the following stresses set up in the column Direct stress- Due to compressive stress (for short columns) Buckling stress or crippling stress- Due to lateral bending of columns Combined direct compressive stress and buckling stresses
- 3. Column is initially straight and the load is applied axially Cross section of the column is uniform through out its length The column material is isotropic and homogenous Length of the column is very large compared to its lateral dimensions Direct stress (compressive stress) is very small compared to buckling stress The column will fail by buckling alone Self weight of the column is negligible
- 6. M=EI d²y/dx² EI d²y/dx²= -Py d²y/dx² + (P/EI).y=0 Solution of the differential equation, d²y/dx² + m²y=0 is y= Acosmx + Bsinmx Hence solution of DE d²y/dx² + (P/EI).y=0 is y=Acos√(P/EI)x + Bsin√(P/EI)x
- 7. Boundary conditions are (i) When x=0, y=0 (ii) When x=L, y=0 Using BC (i), A=0 Hence y= Bsin √(P/EI)x Using BC (ii), B sin√(P/EI)L=0 √(P/EI)L=nΠ, where n=1,2,3...... Simplest case is when n=1, √(P/EI)L=Π P=Π²EI/L²
- 9. y= Bsin √(P/EI)x; This means y is a sin function of x. Hence, we can plot various column failure modes as follows:
- 10. Consider a column AB of length L and uniform cross sectional area, fixed at A and free at B. The free end will sway sideways when the load is applied and the curvature will be similar to that of upper half of the column subjected to compressive load whose both ends are hinged. Let P be the critical load, ie crippling load at which the column starts buckling
- 11. Let y be the deflection at a section which is x distance away from A.
- 12. Moment due to crippling load at the section, M= P(a-y) M=EI d²y/dx² EI d²y/dx²= -P(y-a) d²y/dx² + (P/EI).y - Pa=0 (1) Let λ = (P/EI).y – Pa; d λ/dx= (P/EI).dy/dx
- 13. d² λ/dx²= (P/EI).d²y/dx² d²y/dx²= (EI/P). d² λ/dx² Eqn (1) becomes (EI/P). d² λ/dx² + λ=0 d² λ/dx² + λ.(P/EI)=0 λ= Acos√(P/EI)x + Bsin√(P/EI)x (P/EI).y – Pa= Acos√(P/EI)x + Bsin√(P/EI)x y=(EI/P){ Acos√(P/EI)x + Bsin√(P/EI)x+Pa}
- 14. Boundary conditions: I) At x=0, y=0 II) At x=L, dy/dx=0 Using condition (i), y=0=A + Pa; A=-Pa y= (EI/P){-Pacos√(P/EI)x + Bsin√(P/EI)x+Pa} Using condition (ii), dy/dx=0= B√(P/EI).cos √(P/EI).L √(P/EI).L=(2n-1) Π/2, n=1,2,3.... Simplest case is when n=1; √(P/EI).L= Π/2 P=Π²EI/4L²
- 15. Consider a column AB of length L and uniform cross sectional area, fixed at both A and B. Let P be the critical load, ie crippling load at which the column starts buckling
- 16. Let y be the deflection at a section which is x distance away from B. Moment due to crippling load at the section, M= Mo-Py M=EI d²y/dx²
- 17. EI d²y/dx²= Mo-Py d²y/dx² + (P/EI).y – Mo/EI=0 (1) Let λ = (P/EI).y – Mo/EI; d λ/dx= (P/EI).dy/dx d² λ/dx²= (P/EI).d²y/dx² d²y/dx²= (EI/P). d² λ/dx² Eqn (1) becomes (EI/P). d² λ/dx² + λ=0
- 18. d² λ/dx² + λ.(P/EI)=0 λ= Acos√(P/EI)x + Bsin√(P/EI)x (P/EI).y – Mo= Acos√(P/EI)x + Bsin√(P/EI)x y=(EI/P){ Acos√(P/EI)x + Bsin√(P/EI)x+Mo} Boundary conditions: a) At x=0, y=0 b) At x=L, y=0 c) At x=0, dy/dx=0 Using condition (a), y=0=A + Mo; A=-Mo
- 19. y= (EI/P){-Mocos√(P/EI)x + Bsin√(P/EI)x+Mo} Using condition (c),B=0 y= (EI/P){-Mocos√(P/EI)x + Mo) Using condition (b), -Mocos√(P/EI)L + Mo=0 cos√(P/EI)L =1 √(P/EI)L=2nΠ, n=1,2,3,..... Simplest case is when n=1, √(P/EI)L=2Π P= 4Π²EI/L²
- 20. Consider a column AB of length L and uniform cross sectional area, fixed at end A and hinged at end B Let P be the critical load, ie crippling load at which the column starts buckling At the fixed end of the column, there will be a fixed end moment=Mo This Mo will tend to make the slop of deflection at fixed end zero. Inorder to balance the moment a reaction force H will be generated at B
- 21. Let y be the deflection at a section which is x distance away from A.
- 22. Hence the moment at section at a distance x from fixed end, Moment M=-Py + H(L-x)
- 23. M=EI d²y/dx² EI d²y/dx²= -Py + H (L-x) d²y/dx² + (P/EI).y –H (L-x)=0 (1) Let λ = (P/EI).y –H (L-x); d λ/dx=(P/EI)dy/dx + H d ²λ/dx²=(P/EI)d²y/dx² d²y/dx²= (EI/P) d ²λ/dx²
- 24. Hence eqn (1) becomes (EI/P) d ²λ/dx² + λ=0 λ= A cos √(P/EI) x + B sin√(P/EI) x (P/EI).y –H (L-x)= A cos √(P/EI) x + B sin√(P/EI) x Boundary conditions: a) At x=0, y=0 b) At x=L, y=0 c) At x=0, dy/dx=0
- 25. Using condition (a) A = -HL Using condition (b), 0= -HL cos {√(P/EI) L} + B sin{√(P/EI) L} (2) Using condition (c), -H= √(P/EI)B (3) Using condition eqns (2) and (3),√(P/EI)Lcos{√(P/EI)L} + sin√(P/EI)L=0
- 26. Tan √(P/EI)L=√(P/EI)L Solution of tanθ= θ is θ= 4.5 radian √(P/EI)L=4.5 P/EI=20.25/L²; 2π²=20.25 P=2π²EI/L²
- 27. The effective length (Le) of a given column with given end conditions is the length of an equivalent column of same material and cross section hinged at its either ends and having the value of Euler’s crippling load equal to that of the column. Crippling load for any type of end conditions is given by, P=π²EI/Le² The moment of inertia in the equation in Euler’s equation is the least among Ixx and Iyy
- 29. Moment of inertia I =Ak², where A is the cross sectional area and k is the least radius of gyration. Crippling load P=π²EI/Le²; Substituting I=Ak², P= π²E (Ak²)/Le² P= π²E (A)/(Le/k)² P/A= π²E /(Le/k)² Le/k = Equivalent length/least radius of gyration=Slenderness ratio
- 30. It is an empirical formula which is applicable to all columns whether short or long 1/P=1/Pc + 1/PE - P= Crippling load - Pc= Crushing load= σc x A - PE= Euler’s crippling load= π²EI/ Le² - σc= Ultimate crushing stress (=330 MPa for Mild steel) - A= Cross sectional area
- 31. P=Pc.PE/(Pc + PE) OR P=Pc/{(Pc/PE) + 1} P= σc.A/( σc.A.Le²/Π²EI + 1) P= σc.A/( σc /Π²E x (Le/k)² + 1) σc /Π²E=α=Rankine’s constant Hence P= σc.A/( α x (Le/k)² + 1) Case-1:- If a column is short Pc>>Pe; Hence P≈Pc Case-2:- If a column is long Pe>>Pc; Hence P≈PE
- 32. SL No Material Crushing stress in MPa α 1 Wrought iron 250 1/9000 2 Cast iron 550 1/1600 3 Mild steel 330 1/7500 4 Timber 50 1/750
- 33. Consider a column AB of length L and uniform cross section fixed at A and free at B subjected to a compressive load P at an eccentricity of amount e Let a be the lateral deflection at free end Consider a section at a distance x from A where the lateral displacement is y.
- 34. Bending moment at the section, M= P(a+e-y) EI d²y/dx²= P(a+e-y) d²y/dx² + (P/EI) x y – P (a + e)/EI=0
- 35. Let λ=(P/EI) x y – P (a + e)/EI Then d² λ/dx²=(P/EI) x d² y/dx² (EI/P) d² λ/dx² + λ= 0 λ= A cos {√(P/EI)x} + B sin {√(P/EI)x} (P/EI).y – P(a+e)/EI= A cos {√(P/EI)x} + B sin {√(P/EI)x}
- 36. Boundary conditions are A) x=0, y=0 B) x=0,y’=0 C) x=L, y=a Condition (A) A= -P(a+e)/EI Condition (B) B=0
- 37. Condition (c) e= (a + e) x cos{√(P/EI)L} (a+e)= e x sec {√(P/EI)L} Maximum stress induced at a section Maximum stress at any section= crushing stress + bending stress Crushing stress= P/A Bending stress= M/Z; where Z is the section modulus M= P (a+e)= P x e x sec {√(P/EI)L} Hence σmax= P/A + P x e x sec {√(P/EI)L}