ME 2406: Strength of Materials Buckling of Columns Dr. Faraz Junejo Objectives • Discuss the behavior of columns. • Discuss the columns. buckling of • Determine the axial load needed to buckle an ideal column. Critical Load • Long slender members subjected to axial compressive force are called columns. • The lateral (or sideways ) deflection that occurs is called buckling. • The maximum axial load a column can support when it is on the verge of buckling is called the critical load, Pcr. • Any additional loading will cause the column to buckle and therefore deflect laterally as shown in Figure b. IDEAL COLUMN WITH PIN SUPPORTS • An ideal column is perfectly straight before loading, made of homogeneous material, and upon which the load is applied through the centroid of the xsection. • We also assume that the material behaves in a linear-elastic manner and the column buckles or bends in a single plane. IDEAL COLUMN WITH PIN SUPPORTS • Smallest value of P is obtained for n = 1, so critical load for column is 2 EI Pcr 2 L • This load is also referred to as the Euler load. The corresponding buckled shape is defined by x C1 sin L • C1 represents maximum deflection, max, which occurs at midpoint of the column. IDEAL COLUMN WITH PIN SUPPORTS • A column will buckle about the principal axis of the x-section having the least moment of inertia (weakest axis). • For example, the meter stick shown will buckle about the a-a axis and not the b-b axis. • Thus, circular tubes made excellent columns, and square tube or those shapes having Ix ≈ Iy are selected for columns. IDEAL COLUMN WITH PIN SUPPORTS • Buckling eqn for a pin-supported long slender column, 2 Pcr EI L2 Pcr = critical or maximum axial load on column just before it begins to buckle. This load must not cause the stress in column to exceed proportional limit. E = modulus of elasticity of material I = Least moment of inertia for column’s x-sectional area. L = unsupported length of pinned-end columns. IDEAL COLUMN WITH PIN SUPPORTS • Expressing I = Ar2 where A is x-sectional area of column and r is the radius of gyration of x-sectional area. 2E cr L r 2 cr = critical stress, an average stress in column just before the column buckles. This stress is an elastic stress and therefore cr Y E = modulus of elasticity of material L = unsupported length of pinned-end columns. r = smallest radius of gyration of column, determined from r = √(I/A), where I is least moment of inertia of column’s x-sectional area A. IDEAL COLUMN WITH PIN SUPPORTS • The geometric ratio L/r in Eqn is known as the slenderness ratio. • It is a measure of the column’s flexibility and will be used to classify columns as long, intermediate or short. Critical load vs Slenderness ratio • Euler load equations are hyperbolic equations and are valid for critical stresses below the materials yield point. IDEAL COLUMN WITH PIN SUPPORTS IMPORTANT • Columns are long slender members that are subjected to axial loads. • Critical load is the maximum axial load that a column can support when it is on the verge of buckling. • This loading equilibrium. represents a case of neutral IDEAL COLUMN WITH PIN SUPPORTS IMPORTANT • An ideal column is initially perfectly straight, made of homogeneous material, and the load is applied through the centroid of the x-section. • A pin-connected column will buckle about the principal axis of the x-section having the least moment of intertia. • The slenderness ratio L/r, where r is the smallest radius of gyration of x-section. Buckling will occur about the axis where this ratio gives the greatest value. EXAMPLE : 1 A 7.2-m long A-36 steel tube having the x-section shown is to be used a pin-ended column. Determine the maximum allowable axial load the column can support so that it does not buckle. Est = 200 GPa and Y = 250 MPa EXAMPLE : 1 (contd.) Use Equation 13-5 i.e. Euler Formula to obtain critical load with Est = 200 GPa. For hollow circular x-section, I = π/64 (do4 - di4 )= π/4 (ro4 - ri4 ) Pcr 2 EI L2 20010 2 228.2 kN 9 1 N/m 0.0754 0.0704 m 4 4 7.2 m 2 2 EXAMPLE : 1 (contd.) This force creates an critical compressive stress in the column of Pcr 228.2 103 N cr 2 2 2 A 0.075 0.070 m 100.2 MPa 100 MPa Since cr < Y = 250 MPa, application of Euler’s equation is appropriate to compute largest axial load i.e.228.2 kN in this case. EXAMPLE : 2 The A-36 steel W20046 member shown is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields. Est = 200 GPa and Y = 250 MPa EXAMPLE : 2 (contd.) From table in Appendix B, column’s x-sectional area and moments of inertia are A = 5880 mm2, Ix = 45.8106 mm4,and Iy = 15.4106 mm4. By inspection, buckling will occur about the y-y axis. Applying Euler Equation, we have Pcr 2 EI L2 2 200109 N/m 2 15.410 6 m 4 1900.4 kN 4 m 2 EXAMPLE : 2 (contd.) When fully loaded, average compressive critical stress in column is Pcr 1900.4 kN cr A 5880 10 6 m 2 323.19 106 N/m 2 323.2MPa Since this stress exceeds yield stress (250 MPa), the load P is determined from simple compression formula: P 250 10 N/m 5880 10 6 m 2 6 2 P 1470 kN Exercise: 13.5 An A-36 steel (E = 200GPa) column has a length of 4 m and is pinned at both ends. If the cross sectional area has the dimensions shown, determine the critical load. Y = 250 MPa Answer: Since cr = 20.65 MPa < Y = 250 MPa, application of Euler’s equation is appropriate to compute largest axial load i.e. 22.7 kN in this case. Exercise: 13.7 A column is made of A-36 steel (E = 29000 ksi), has a length of 20 ft, and is pinned at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. Y = 36 ksi Answer: Since cr = 27.4 ksi < Y = 36 ksi, application of Euler’s equation is appropriate to compute largest axial load i.e. 157.7 kip in this case. Exercise: 13.12 An A-36 (E = 29000 ksi), steel column has a length of 15 ft and is pinned at both ends. If the crosssectional area has the dimensions shown, determine the critical load. Y = 36 ksi Answer: Since cr = 34.32 ksi < Y = 36 ksi, application of Euler’s equation is appropriate to compute largest axial load i.e. 377.5 kip in this case. Effective Lengths of Columns • The critical loads for columns with various support conditions can be related to the critical load of a pinned-end column through the concept of an effective length. • To demonstrate this idea, consider the deflected shape of a column fixed at the base and free at the top (Fig. 11-16a) with deflection curve for a pinned-end column top (Fig. 11-16b). • Consider the deflected shape of a column fixed at the base and free at the top(Fig. 11-16a). • This column buckles in a curve that is one-quarter of a complete sine wave. • If we extend the deflection curve (Fig. 11-16b), it becomes one half of a complete sine wave, which is the deflection curve for a pinnedend column. Therefore, the effective length Le for any column is the length of the equivalent pinned-end column, that is, it is the length of a pinned-end column having a deflection curve that exactly matches all or part of the deflection curve of the original column. Effective Lengths of Columns (contd.) • Another way of expressing this idea is to say that the effective length of a column is the distance between points of inflection (that is; points of zero moment) in its deflection curve, assuming that the curve is extended (if necessary) until points of inflection are reached. Thus, for a fixed-free column (Fig. 11-16), the effective length is Le = 2L • Because the effective length is the length of an equivalent pinned end column, we can write a general formula for critical loads as follows: Effective Lengths of Columns (contd.) • If we know the effective length of a column (no matter how complex the end conditions may be), we can substitute into the preceding equation and determine the critical load. • For instance, in the case of a fixed-free column, we can substitute Le = 2L • The effective length is often expressed in terms of an effective length factor (K) i.e. Le = KL where L is the actual length of the column. • Thus, the critical load is given by Pcr 2 EI KL 2 13 -11 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length • If a column is not supported by pinned-ends, then Euler’s formula can also be used to determine the critical load. • “L” must then represent the distance between the zero-moment points. • This distance is called the columns’ effective length, Le. COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length • Many design codes provide column formulae that use a dimensionless coefficient K, known as the effective-length factor. 13 - 10 Le KL • Thus, Euler’s formula can be expressed as Pcr cr 2 EI 13 - 11 KL 2 2E KL r 2 13 - 12 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS Effective length • Here (KL/r) is the column’s effective-slenderness ratio. Exercise: 13.6 An A-36 steel (E = 200GPa) column has a length of 4 m and is fixed at its bottom and pinned at its top. If the cross sectional area has the dimensions shown, determine the critical load. Y = 250 MPa Hint: k = 0.7 Answer: Since cr = 42.15 MPa < Y = 250 MPa, application of Euler’s equation is appropriate to compute largest axial load i.e. 46.4 kN in this case. Exercise: 13.8 A column is made of A-36 steel (E = 29000 ksi), has a length of 20 ft, and is fixed at its bottom and pinned at its top. If the cross-sectional area has the dimensions shown, determine the critical load. Y = 36 ksi Answer: Since cr = 9.1 ksi < Y = 36 ksi, application of Euler’s equation is appropriate to compute largest axial load i.e. 52.3 kip in this case. Exercise: 13.9 The W 14 X 38 column is made of A-36 (E = 29000 ksi), steel and is fixed supported at its base. If it is subjected to an axial load of determine the factor of safety with respect to buckling. Y = 36 ksi Exercise: 13.9 (contd.) • From the table in appendix, the cross-sectional area and moment of inertia about weak axis (y-axis) for W 14 X 38 are: A = 11.2 in2 Iy = 26.7 in4 • The column is fixed at its base and free at top (i.e. k = 2). Note that the column will buckle about the weak axis (y axis) in this case. Using Euler’s formula yiels Pcr = 33.17 kip; hence the factor of safety with respect to buckling is n = 33.17/15 = 2.21 (Ans) Since cr = 2.96 ksi < Y = 36 ksi, application of Euler’s equation is appropriate Exercise: 13.13 An A-36 steel column has a length of 5 m and is fixed at both ends. If the cross-sectional area has the dimensions shown, determine the critical load. Hint: k = 0.5 Answer: Since cr = 105 MPa < Y = 250 MPa, application of Euler’s equation is appropriate to compute largest axial load i.e. 272.1 kN in this case.