1. Unit-5. Torsion in Shafts and Buckling of Axially Loaded Columns Lecture Number-5 Mr. M. A.Mohite Mechanical Engineering S.I.T., Lonavala

2. Illustrative Example.1 A 7.2-m long 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. E st = 200 GPa.

3. Illustrative Example.1…..Contd This force creates average compressive stress in column Since  cr <  Y = 250 MPa means Column safe Solution:

4. The A-36 steel W200  46 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. Soln : From table in Appendix B, column’s x- sectional area and moments of inertia are A = 5890 mm 2, I x = 45.5  10 6 mm 4,and I y = 15.3  10 6 mm 4. By inspection, buckling will occur about the y-y axis. Illustrative Example.2

5. When fully loaded, average compressive stress in column is Illustrative Example.2….Contd Critical Load,

6. Since this stress exceeds yield stress (250 N/mm 2 ), the load P is determined from simple compression: Illustrative Example.2…Contd

7. Practice Example-1 An I-section 400 mm × 200 mm × 10 mm and 6 m long is used as a strut with both ends fixed. Find Euler’s crippling load. Take Young’s modulus for the material of the section as 200 kN/mm 2. Hint for Solution- 1.Calculate the moment of inertia of the I-section about X-X, I XX == 200 × 10 6 mm 4 and moment of inertia of the I-section about Y-Y, I YY =13.36 × 10 6 mm 4 Take I as I YY =13.36 × 10 6 mm 4, Which is less column is fixed at its both ends- Le = l / 2 = 6000 / 2 = 3000 mm Calculate the crippling load using, = 2930 kN Ans.