Ch. 07_Columns and Struts.pdf .
- 1. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 1 Chapter 7 Columns and Struts MME 2154: STRENGTH OF MATERIALS Subraya Krishna Bhat, PhD Assistant Professor Dept. of Mech & Mfg. Engg. MIT Manipal sk.bhat@manipal.edu
- 2. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 2 Introduction to Columns/Struts • A bar or member of a structure in any position acted upon by a compressive load is known as a strut. • However, when the compressive member is in a vertical position and is liable to fail by bending or buckling, it may be referred to as column or strut, for example a vertical pillar between the roof and floor. Failure of columns: • The failure of a column takes place due to the anyone of the following stresses set up in the columns: – Direct compressive stresses – Buckling stresses – Combined of direct compressive and buckling stresses
- 3. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 3 Introduction to Columns/Struts Failure of short columns: • A short column of uniform cross-sectional area A, subjected to an axial compressive load P. The compressive stress induced is given by σ=P/A. • If the compressive load on the short column is gradually increased, a stage will reach when the column will be on the point of failure by crushing. • The stress induced in the column corresponding to this load is known as crushing stress and the load is called crushing load. • AII short columns fail due to crushing.
- 4. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 4 Buckling of Columns • The buckling load at which the column just buckles, is known as buckling or crippling load. • The buckling load is less than the crushing Ioad for a long column. • Actually the value of buckling load for long columns is low whereas for short columns the value of buckling load is relatively high.
- 5. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 5 Buckling of Columns L = Length of a long column P = Load (compressive) at which the column has just buckled A = Cross-sectional area of the column e = Maximum bending of the column at the centre 𝜎𝑜 = Stress due to direct load = P/A 𝜎𝑏 = Stress due to bending at the centre of the column = (P x e)/Z Z = Section modulus about the axis of bending.
- 6. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 6 Buckling of Columns The extreme stresses on the mid-section are given by: 1. Maximum stress = 𝜎𝑜 + 𝜎𝑏 and 2. Minimum stress = 𝜎𝑜 − 𝜎𝑏 ❑ The column will fail when maximum stress (i.e., 𝜎𝑜 + 𝜎𝑏 ) is more than the crushing stress 𝜎𝑐. But in case of long columns, the direct compressive stresses are negligible as compared to buckling stresses. ❑ Hence very long columns are subjected to buckling stresses only.
- 7. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 7 Euler’s Column Theory Assumptions made in the Euler’s Column theory: 1. The column is initially perfectly straight and the load is applied axially. 2. The cross-section of the column is uniform throughout its length. 3. The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke's law. 4. The length of the column is very large as compared to its lateral dimensions. 5. The direct stress is very small as compared to the bending stress. 6. The column will fail by buckling alone. 7. The self-weight of column is negligible.
- 8. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 8 End Conditions for Long Columns: ▪ In case of long columns, the stress due to direct load is very small in comparison with the stress due to buckling. ▪ Hence the failure of long columns takes place entirely due to buckling (or bending). ▪ The following four types of end conditions of the columns are important: 1. Both the ends of the column are hinged (or pinned). 2. One end is fixed and the other end is free. 3. Both the ends of the column are fixed. 4. One end is fixed and the other is pinned. ▪ Note: For a hinged (or pinned) end, the deflection is zero. For a fixed end the deflection and slope are zero. For a free end the deflection is not zero.
- 9. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 9 Effective Length (or Equivalent Length) of Columns ▪ The effective length of a given column with given end conditions is the length of an equivalent column of the same material and cross-section with hinged ends and having the value of the crippling load equal to that of the given column. ▪ Effective length is also called equivalent length. ▪ Let Le = Effective length of a column, L = Actual length of the column, PE = Crippling load for the column, and E = Young’s modulus. ▪ Then the crippling load for any type of end condition is given by 𝑃𝐸 = 𝜋2𝐸𝐼 𝐿𝑒 2 ▪ There are two values of moment of inertia i.e., 𝐼𝑥𝑥 and 𝐼𝑦𝑦. The value of 𝐼 (moment of inertia) in the expressions of crippling load (𝑃𝐸) should be taken as the least value of the two moments of inertia as the column will tend to bend in the direction of least moment of inertia.
- 10. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 10 Effective Length (or Equivalent Length) of Columns
- 11. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 11 Effective Length (or Equivalent Length) of Columns
- 12. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 12 Slenderness Ratio and Limitation of Euler’s Theory ▪ The ratio of the actual length of the column to the least radius of gyration of the column is known as slenderness ratio. ▪ If the slenderness ratio i.e. (𝑙/𝑘) is small the crippling stress (or the stress at failure) will be high. ▪ But for the column material the crippling-stress cannot be greater than the crushing stress. 𝑃𝐸 = 𝜋2 𝐸𝐼 𝐿𝑒 2 = 𝜋2 𝐸𝐴𝑘2 𝐴𝐿𝑒 2 = 𝜋2 𝐸 Τ 𝐿𝑒 𝑘 2
- 13. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 13 Slenderness Ratio and Limitation of Euler’s Theory ▪ Hence when the slenderness ratio is less than a certain limit Euler's formula gives a value of crippling stress greater than the crushing stress. ▪ In the limiting case we can find the value of 𝑙/𝑘, for which the crippling stress is equal to crushing stress. ▪ If OD represents the yield stress of the material, obviously the Euler formula cannot be applied if 𝑙/𝑘 is less than OE as below this value the material will become plastic and will not follow Hooke’s law. 𝑃𝐸 = 𝜋2 𝐸𝐼 𝐿𝑒 2 = 𝜋2 𝐸𝐴𝑘2 𝐴𝐿𝑒 2 = 𝜋2 𝐸 Τ 𝐿𝑒 𝑘 2
- 14. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 14 Rankine’s Formula ▪ We have learnt that Euler's formula gives correct results only for very long columns. ▪ But what happens when the column is a short or the column is not very long? ▪ On the basis of results of experiments performed by Rankine, he established an empirical formula which is applicable to all columns whether they are short or long. ▪ For a given column material the crushing stress σc is a constant. Hence the crushing load Pc (which is equal to σc x A) will also be constant for a given cross-sectional area of the column.
- 15. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 15 Rankine’s Formula ▪ In equation (𝑖), PC is constant and hence value of P also depends upon the value of PE. ▪ But for a given column material and given cross-sectional area, the value of PE depends upon the effective length of the column. 1 𝑃 = 1 𝑃𝑐 + 1 𝑃𝐸 • P is the Rankine’s crippling load. The above relation is also known as the Rankine-Gorden formula.
- 16. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 16 Rankine’s Formula • If the column is a short, which means the value of Le is small, then the value of PE will be large. Hence the value of 1/PE will be small enough and is negligible as compared to the value of 1/PC. Neglecting the value of 1/PE, we get, 1 𝑃 ≈ 1 𝑃𝑐 𝑃 ≈ 𝑃𝑐
- 17. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 17 Rankine’s Formula • If the column is long, which means the value of Le is large. Then the value of PE will be small and the value of 1/PE will be large enough compared with 1/PC. Hence the value of 1/PC may be neglected. 1 𝑃 ≈ 1 𝑃𝐸 𝑃 ≈ 𝑃𝐸
- 18. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 18 Rankine’s Formula • The crippling load by Rankine's formula for long columns is approximately equal to crippling load given by Euler's formula. • Hence the Rankine’s formula gives satisfactory results for all lengths of columns, ranging from short to long columns. Rewriting the eqn., 1 𝑃 = 1 𝑃𝑐 + 1 𝑃𝐸 = 𝑃𝑐 + 𝑃E 𝑃𝑐 ∙ 𝑃𝐸 𝑃 = 𝑃𝑐 1 + 𝑃𝑐 𝑃𝐸 𝑃 = 𝜎𝑐 ∙ 𝐴 1 + 𝜎𝑐 ∙ 𝐴 𝜋2𝐸𝐼 𝐿𝑒 2 𝑃 = 𝜎𝑐 ∙ 𝐴 1 + 𝜎𝑐 ∙ 𝐿𝑒 2 𝜋2𝐸𝑘2 𝑘 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑔𝑦𝑟𝑎𝑡𝑖𝑜𝑛
- 19. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 19 Rankine’s Formula • Rankine’s constant 𝑎 which is equal 𝜎𝑐 𝜋2𝐸 to can be introduced in the equation as 𝑃 = 𝜎𝑐 ∙ 𝐴 1 + 𝑎 𝐿𝑒 𝑘 2
- 20. Department of Mechanical & Manufacturing Engineering, MIT, Manipal 20 Example 1
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