Unit5 Torsion in Shafts and Buckling of Axially
Unit-5. Torsion in Shafts and Buckling of Axially Loaded Columns Lecture Number-5 Mr. M. A. Mohite Mechanical Engineering S. I. T. , Lonavala
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. Est = 200 GPa.
Illustrative Example. 1…. . Contd Solution: • This force creates average compressive stress in column • Since cr < Y = 250 MPa means Column safe
Illustrative Example. 2 • The A-36 steel W 200 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 xsectional area and moments of inertia are A = 5890 mm 2, Ix = 45. 5 106 mm 4, and Iy = 15. 3 106 mm 4. By inspection, buckling will occur about the y-y axis.
Illustrative Example. 2…. Contd Critical Load, When fully loaded, average compressive stress in column is
Illustrative Example. 2…Contd Since this stress exceeds yield stress (250 N/mm 2), the load P is determined from simple compression:
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 k. N/mm 2. Hint for Solution 1. Calculate the moment of inertia of the I-section about X-X, IXX == 200 × 106 mm 4 and moment of inertia of the I-section about Y-Y, IYY =13. 36 × 106 mm 4 Take I as IYY =13. 36 × 106 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 k. N Ans.
- Slides: 7