Columns and struts_-_solved
- 1. 80 CHAPTER 9 Columns and Struts Problem 1. Compare the ratio of the strength of solid steel column to that of the hollow steel column of the same cross-sectional area. The internal diameter of the hollow column is 3/4th of the external diameter. The columns have the same length and are pinned at both ends. Use Euler’s theory. Solution: Let ‘d’ be the diameter of the solid strut and D be the outer diameter of the hollow strut. As the cross-sectional areas are same, 2 4 d π = 2 2 3 4 4 D D π − d2 = 2 2 29 7 16 16 D D D− = Let Pcrh = buckling load of the hollow column Pcrs = buckling load of solid column Ih = least moment of inertia of hollow column = 4 4 43 175 64 4 64 256 D D D π π − = × Is = Least moment of the solid column = 4 64 d π Now Pcrh = 2 2 hEI L π and Pcrs = 2 2 sEI L π ∴ crh crs P P = 4 4 175 64 256 64 h s D I I d π × = π × = 4 4 4 2 2 175 175 256 7 256 16 D D d D = = 25 7 Thus crh crs P P = 3.571. Ans. Problem 2. A solid round bar of 60 mm diameter and 2.5 m long is used as a strut. Find the safe compressive load for the strut using Euler’s formula if (a) both ends are hinged (b) both ends are fixed. Take E = 2 × 105 N/mm2 and factor of safety = 3. Solution: l = 2.5 m = 2500 mm d = 60 mm, E = 2 × 105 N/mm2 I = 4 4 60 64 64 d π π × = × = 636172.51 mm4
- 2. 81 Factor of safety = 3. (a) Both ends are hinged: Pcr = 2 2 5 2 2 2 10 636172.51 (2500) EI l π π × × × = = 200920 N = 200.92 kN ∴ Safe load = 200.92 Factor of safety 3 crP = = 66.97 kN. (b) Both ends are fixed: Pcr = 2 2 5 2 2 4 4 2 10 636172.51 (2500) EI l π × π × × × = = 803682 N = 803.682 kN ∴ Safe load = 803.682 Factor of safety 3 crP = = 262.89 kN. Ans. Problem 3. What is the ratio of the strength of a solid steel column of 150 mm diameter to that of a hollow circular steel column of the same cross-sectional area and a wall thickness of 15 mm? The two columns have the same length and similar end conditions. Solution: Diameter of circular column d = 150 mm ∴ C.S. Area = 2 150 4 π × Let the thickness of circular hollow column be t = 15 mm Let external diameter of hollow circular column be D = mm ∴ Its internal diameter = D – 2t = D – 2 × 15 = (D – 30) mm ∴ C.S. area = 4 π {D2 – (D – 30)2} This area is same as that of solid column ∴ 2 2 { ( 30) } 4 D D π − − = 2 150 4 π × D2 – {D2 – 60D + 900} = 1502 60D = 22500 + 900 = 23400 ∴ D = 390 mm ∴ Internal diameter of hollow column = 390 – 30 = 360 mm. Least moment of inertia: Is = 4 150 64 π × = 24850488.7 mm4 Ih = 4 4 (390 360 ) 64 π − = 311128119.5 mm4 Pcrh = 2 2 h e EI l π Pcrs = 2 2 s e EI l π
- 3. 82 ∴ crh crs P P = 311128119.5 24850488.7 h s I I = = 12.52. Ans. Problem 4. Find the Euler’s crushing load for a hollow cylindrical cast iron column 120 mm external diameter and 20 mm thick, if it is 4.2 m long and is hinged at both ends. Take E = 80 kN/mm2. Compare this load with the crushing load as given by Rankine’s formula using constants fc = 550 N/mm2 and a = 1/1600. For what length of strut does the Euler’s formula cease to apply? Solution: External diameter = 120 mm Thickness = 20 mm Internal diameter = 120 – 2 × 20 = 80 mm Least moment of inertia = 4 4 (120 80 ) 64 π − = 8168140.89 mm4 Column is hinged at both ends. ∴ le = 4.2 m = 4200 mm Euler’s buckling load = 2 2 3 2 2 80 10 8168140.89 (4200) EI l π π × × × = = 365606.89 N. Ans. A = 2 2 (120 80 ) 4 π − = 6283.18 mm2 K2 = 8168140.89 1300 6283.18 I A = = ∴ K2 = 36.05 mm ∴ Rankine’s critical load PR = 2 2 550 6283.18 1 4200 1 1 1600 36.05 cf A l a K × = + + = 364415.16 N ∴ E R P P = 365606 89 364415 16 . . = 1.003. Ans. Now, PE = 2 2 EI l π Equating it to crushing load, we get 2 2 EI l π = fc A 2 2 2 E K l π = fc 2 3 2 80 10 1300 l π × × × = 550 l 2 = 2 80 1000 1300 550 π × × × l = 1366.108 mm. Ans.
- 4. 83 Problem 5. An ISLB 300 section is provided with a flange plate 200 mm × 12 mm for each flange. The composite member is used as a column with one end fixed and the other end hinged. Calculate the length of the column for which, crippling loads given by Rankine’s formula and Euler’s formula will be the same. Take E = 210 kN/mm2, fc = 330 N/mm2, a = 1/7500 Properties of ISLB 300 section are: Overall width = 150 mm, Overall depth = 300 mm, Thickness of flange = 9.4 mm, Thickness of web = 6.7 mm Ixx = 73.329 × 106 mm4 Iyy = 3.762 × 106 mm4 A = 4808 mm2 . Solution: fc = 330 N/mm2, a = 1 , 7500 E = 210 × 103 N/mm2 Area A = 4808 mm2 Ixx = 73.329 × 106 mm4, Iyy = 3.762 × 106 mm4 Sectional area of ISLB 300 column. A = 4808 + 2 × (200 × 12) = 9608 mm2 Moment of inertia about x – x axis. Ixx = 73.329 × 106 + 2 3 2200 12 (200 12) 156 12 × + × = 190199406 mm4 Moment of inertia about y-y axis. Iyy = 3 6 12 200 3.762 10 2 12 × × + = 19762000 mm4 Since Iyy < Ixx, the column buckles about y-y axis. ∴ I = Imin = 19762000 mm4 Least radius of gyration = K = 19762000 9608 I A = = 45.35 mm. Let L = effective length 2 2 EI L π = 2 2 1 cf A L a K + 2 3 2 2 210 10 19762000 330 9608 1 1 7500 2056.82 L L π × × × × = + 12918229.65 = 2 2 8 1 6.4825 10 L L − + × × 12918229.65 + 0.8374 L2 = L2 ∴ L = 8914 mm
- 5. 84 Therefore, required actual length for one end hinged and other end fixed column for which critical load by Rankine’s formula and Euler’s formula will be equal is l = 2 2 8914L = × = 12606 mm = 12.606 m. Ans. Problem 6. A built up steel column, 8 m long and ends firmly fixed is having cross-section as shown in Fig. 1. The properties of I-section are Area = 9300 mm2, Ixx = 3 × 106 min4, Iyy = 8.4 × 106 mm4. Determine: (i) The safe axial load the column can carry with a factor of safety of 3.5 using (a) Euler’s Formula, (b) Rankine’s Formula. (ii) The length of the column for which both formulae give the same crippling load. (iii) The length of the column for which the Euler’s formula ceases to apply. Take E = 2 × 105 N/mm2, fc = 330 N/mm2, a = 1/7500 Fig. 1 Solution: (i) Length of the column = 8 m = 8000 mm Factor of safety = 3.5, fc = 330 N/mm2, a = 1 7500 , E = 2 × 105 N/mm2 A = 2 (9300 + 350 × 25) = 36100 mm2 Moment of inertia of column section about x-x axis: Ixx = 2 × 3 × 106 + 2 3 2350 25 350 25 237.5 12 × + × × = 994020833.5 mm4 Moment of inertia of the column section about y-y axis: Iyy = 2 × 8.4 × 106 + 2 3 225 350 9300 100 12 × + × = 381445833.3 mm4 Iyy < Ixx ∴ I = Imin = 381445833.3 mm4 K = 381445833.3 36100 I A = = 102.793
- 6. 85 Since column is fixed at both ends, PE = 2 2 5 2 2 4 4 2 10 381445833.3 (8000) EI l π × π × × × = = 47058993.44 N ∴ Safe Load P = 47058993.44 3.5 3.5 EP = = 13445.40 × 103 N = 13445.4 kN. Ans. (b) PR = 2 1 cf A L a K + where L = 8000 2 2 l = = 4000 mm = 2 330 36100 1 4000 1 7500 102.793 × + = 9911.806 × 103 N = 9911.806 kN. ∴∴∴∴ Safe load P = 9911.806 3.5 = 2831.95 kN. Ans. (ii) Let L1 be the effective length 2 2 1 EI L π = 2 1 2 1 cf A L a K + 2 5 2 1 2 10 381445833.3 L π × × × = 2 1 2 330 36100 1 1 7500 (102.793) L × + or 2 10.797 63203550.35L + = 2 1L or L1 = 17670 mm = 17.67 m. Ans. (iii) Let ‘l’ be the length of column for which Euler’s formula ceases to apply. Then PE = 2 2 4 EI l π fc A = 2 2 4 EI l π 330 = 2 2 2 5 2 2 2 4 4 2 10 (102.793)EK l l π π × × × = ∴ l 2 = 252815021.4 l = 15900 mm = 15.9 m. Ans. Problem 7. Find the safe load carrying capacity of column given in Problem 6 by Indian Standard code procedure. Given fc = 250 N/mm2.
- 7. 86 Solution: I = 381445833.3 mm4 A = 36100 mm2 K = 4 I = 102.793 mm Slenderness ratio λ = 4000 102.793 l k = = 38.914 From I.S. table, λ = 30 and fc = 250 N/mm2 σac = 145 N/mm2 and for λ = 40, fc = 250 N/mm2 σac = 139 N/mm2 Linearly interpolating between these two values of λ, σac = 8.914 145 (145 139) 10 − − = 139.65 N/mm2 Therefore, safe load carrying capacity of the column is, P = σac × A = 139.65 × 36100 = 5041444 N P = 5041.444 kN. Ans.