Compression member

design of compression member

1. Design of compression members As per IS 800 : 2007

2. Compression members Structural Members subjected to axial compression/compressive forces Design governed by strength and buckling Two types of compression member : Column Strut

3. Sections used for compression member In theory numerous shapes can be used for columns to resist given loads. However, from practical point of view, the number of possible solutions is severely limited by section availability, connection problems, and type of structure in which the section is to be used.

4. Effective length of compression member Column End Condition And Effective Length 1.Both end hinged. 2.Both end fixed. 3.One end fixed and other hinged. 4.One end fixed and other free.

5. Effective length

6. Slenderness ratio ( λ ) : Slenderness Ratio = effective length of column/Minimum radius of gyration λ = le/r If λ is more , its load carrying capacity will be less.

7. Short compression member L/r ≤ 88.85 for F y = 250 Mpa Failure stress equal to yield stress No buckling Long compression member 1. They will buckle elastically 2. Axial buckling stress is below proportional limit.

8. Compression Member Failure 10 There are three basic types of column failures. One, a compressive material failure( very short and fat). Two, a buckling failure,(very long and skinny). Three, a combination of both compressive and buckling failures.(length and width of a column is in between a short and fat and long and skinny column).

9. Compression Member Failure 11 Local Buckling This occurs when some part or parts of x-section of a column are so thin that they buckle locally in compression before other modes of buckling can occur

10. Compression Member Failure 12 Flexural Buckling (also called Euler Buckling) is the primary type of buckling members are subjected to bending or flexure when they become unstable

11. Compression Member Failure Squashing : When the length of column is relatively small and column is stocky and its component plates are prevented from local buckling, then column will be able to attain its full strength before failure.

12. Cross section classification

13. Design compressive strength

14. Clause 7.3.2

15. IS 800: 2007 Clause 7.1.2.1

16. Column buckling curves Classification of different sections under different buckling class a, b,c and d are given in Table 10 of IS 800: 2007 ( page 44). The stress reduction factor χ , and the design compressive stress fcd , for different buckling class, yield stress and effective slenderness ratio is given in table 8 ( page 37) Table 9( page 40) shows the design compressive stress, fcd for different buckling class a to d.

17. The curve corresponding to different buckling class are presented in non-dimensional form as shown in the figure below. Using this curve one can find the value of fcd ( design compressive stress) corresponding to non- dimensional effective slenderness ratio λ ( page 35)

18. Angle struts Single angle strut: ( IS : 800 cl. 7.5.1) The compression in single angle struts may be transfered either concentrically to its centroid through end gusset or eccentrically by connecting one of its legs to a gusset or adjacent member. Concentric loading Loaded through one leg √ k 1 + k 2 + ʎ 2 vv + k 3 ʎ 2 ɸ

19. Double angle struts Connected back to back on opposite sides of G.P. The effective length kL in plane of end gusset shall be taken as between 0.7 and 0.85 times the distance between intersections. 2. Connected back to back to one side of G.P.(cl.7.5.2.2) The outstanding legs shall be connected by tack bolting or tack welding spaced at a distance not exceeding 600 mm. (cl. 10.2.5.5)

20. Determine compressive strength of a single ISA 100×100×8 mm @ 12.1 kg/m with length of member 2.5 m . Ends of the members are hinged. Assume that the load is applied concentrically to the angle. F y = 250 MPa

21. For ISA 100×100×8 mm a = 1539 mm 2 ( Steel table ) r min = r vv = 19.5 mm Section Classification : b/t = 100/8 = 12.5 < 15.7 ε d/t = 100/8 = 12.5 < 15.7 ε ( IS : 800 , T – 2 , P.18) ( b+d )/ t = 200/8 = 25 ≤ 25 ε where ε = (250/ f y ) 0.5 = 1 Since all three criteria is satisfied , the section is semi compact. For the semi compact section, gross area is taken as effective area. A e = 1539 mm 2

22. Effective length : ( IS :800 , T-11 , P.45 ) Both end pinned, Effective length = kL = 1 × 2500 = 2500 mm As the member is concentrically loaded, the design compressive strength is calculated as per cl. 7.1.2 of code IS 800 – 2007. Euler’s buckling stress : ( cl. 7.1.2 P. 34 ) f cc = п 2 E / ( kL / r ) 2 = п 2 × 2 × 10 5 / (2500/19.5) 2 = 120.09 N/ mm 2

23. Non dimensional effective slenderness ratio : ( T.10 P.44 ) λ = √ f y / f cc = √ (250/ 120.09) = 1.44 α = Imperfection factor = 0.49 ( For a single ISA , buckling class is ‘c’ ) ( from T- 7 , α = 0.49) ɸ = 0.5[ 1 + α ( λ - 2) + λ 2 ] = 1.8406

24. Stress reduction factor χ : χ = 1 / [ ɸ + (ɸ 2 - λ 2 ) 0.5 ] = 1 / [ 1.8406 + (1.8406 2 – 1.44 2 ) 0.5 ] = 0.335 Design compressive stress ( f cd ) : f cd = ( f y / ɣ mo ) / [ ɸ + (ɸ 2 - λ 2 ) 0.5 ] = χ . f y / ɣ mo = 0.335 × 250/1.10 = 76.13 N/mm 2

25. Design compressive strength ( cl. 7.1.2) : P d = A e . F cd = 1539 × 76.13 = 117.16 kN

26. Design a double angle discontinuous strut to carry a factored load of 200 kN. The length of member is 3 m between intersections. The two angles are placed back to back on same side of gusset plate. Assume F e 410 steel with f y 250 Mpa . Angles are connected to the same side of gusset plate, as per IS : 800 cl.7.5.2.2 the member is designed as two single ISA. Load for single ISA = 200/2 = 100 kN Effective length = c/c length of member = 3000 mm Assume slenderness ratio kL /r = 120 F cd = 83.7 N/mm 2 ( T- 9(c) , P.42)

27. A g required = P / f cd = 100×10 3 / 83.7 = 1195 mm 2 Choose single ISA 90×90×8 mm a = 1379 mm 2 r min = r vv = 17.5 mm kL /r = 3000/17.5 = 171.42 < 180 ........OK Section Classification : b/t = 90/8 = 11.5 < 15.7 ε d/t = 90/8 = 11.5 < 15.7 ε ( T – 2 , P.18) ( b+d )/ t = 180/8 = 22.5 ≤ 25 ε where ε = (250/ f y ) 0.5 = 1 Since all three conditions are satisfied the section is semi compact.

28. A e = 1379 mm 2 Assume two bolts at each end with fixed end condition. k 1 = 0.20 k 2 = 0.35 IS : 800 , T – 12 , P. 48 k 3 = 20 λ vv = ( L / r vv ) / [ ε (√ п 2 E / 250) ] = (3000 / 17.5 ) / [1 √ ( п 2 2 × 10 5 / 250 )] = 1.929 λ ɸ =[ ( b 1 + b 2 ) / 2t] / [ ε √ п 2 E / 250 ] = ( 90 + 90/2 × 8 ) / 88.86 = 0.1266

29. λ e = √ k 1 + k 2 + λ 2 vv + k 3 λ 2 ɸ = √ 0.20 + 0.35(1.929) 2 + 20(0.1266) 2 = 1.35 Imperfaction factor = α = 0.49 ( T- 7 , P.35) ɸ = 0.5 [ 1+ α ( λ - 0.1) + λ 2 ] = 1.693 Stress reduction factor : χ = 1 / [ ɸ + (ɸ 2 - λ 2 ) 0.5 ] = 1 / 1.693 + (1.693 2 – 1.35 2 ) 0.5 = 0.368 Design compressive strength : (cl. 7.1.2.1 , P.34) f cd = χ . f y / γ mo = 0.368 × 250/1.10 = 83.64 N/mm 2

30. Design compressive strength ( P d ) : P d = A e f cd = 1379 × 83.64 = 115.34 kN For 2 – ISA , P d = 2× 115.34 = 230.67 > 200 kN .........SAFE Provide 2 – ISA, 90 × 90 × 8 mm on same side of the gusset plate.

31. Determine the design axial load on the column section of ISMB 350 having height 3 m , hinged at both ends. F y = 250 Mpa Properties of ISMB 350 ( steel table ) A = 6671 mm 2 b f = 140 mm , b = 70 mm t f = 14.2 mm t w = 8.1 mm h = 350 mm R = 14.0 mm r zz = 142.9 mm r yy = 28.4 mm

32. Section classification : ε = (250/ f y ) 0.5 = 1 Flange , b / t f = 70 / 14.2 = 4.93 < 9.4 ε ......plastic Web , d / t w = [ 350 – 2(14.2+14.0)] / 8.1 = 36.42 < 42 ε .....semi compact Flange and web both are fully effective. A e = A = 6671 mm 2 Effective length : ( IS : 800 , T-11 , P.45 ) (both end hinged) kL = 1.0 L = 3000 mm

33. Buckling curve classification : r min = r yy = 28.4 mm Failure is by buckling about minor axis y-y , h / b f = 350 / 140 = 2.5 > 1.2 ( T-10 , P.44 ) t f = 14.2 mm < 40 mm For failure about y-y axis use buckling curve ‘ b ‘ . kL / r yy = 3000/28.4 = 105.63 ( T- 9(B) , P.41 ) f y = 250 Mpa P.41 10 14 (difference) 5.63 ? ( 7.882 ) f cd = 118 – 7.882 = 110.12 N / mm 2

34. P d = A e × f cd = 6671 × 110.12 = 734.61 kN

35. Design a steel column to carry factored axial load of 1500 kN. The length of column is 3.6 m and hinged at both ends. Assume f y = 250 Mpa . P = 1500 kN Assume f cd = 0.6y = 0.6 × 250 = 150 N/mm 2 A g = P / f cd = 1500 × 10 3 / 150 = 100 cm 2

36. Choose ISHB 400@ 77.4 kg/ m , A= 9866 mm 2 h = 400 mm b f = 250 mm , b = 125 mm t f = 12.7 mm t w = 9.1 mm R = 14 mm r zz = 168.7 mm r yy = 52.3 mm Section classification : where ε = (250/ f y ) 0.5 ( T – 2 , P.18) = 1 Flange , b/ t f = 125/12.7 = 9.84 < 10.5 ε web , d/ t w = [ 400 – 2 (12.7+14)] / 9.1 = 38.08 < 42 ε .......semi compact

37. Full section is fully effective. A e = 9866 mm 2 Effective length : ( IS : 800 , T – 11 , P.45 ) column hinged at both ends kL = 1 . 0 × L = 3600 mm Buckling curve classification : ( IS : 800 , T – 10 , P.44) h / b f = 400 / 250 = 1.6 > 1.2 t f = 12.7 < 40 For buckling about y-y axis use curve – b. Since r yy is minimum failure is by buckling about y-y axis.

38. ( kL / r yy ) = 3600 / 52.6 = 68.44 < 180......OK F y = 250 Mpa ( IS : 800 , T – 9(b) , P.41) 10 15 (difference) 8.44 ? (12.66) F cd = 181 – 12.66 = 168.34 N/mm 2 Design compressive strength , P d = A e × f cd = 9866 × 168.34 = 1660.84 > 1500 kN ..........SAFE Provide ISHB 400 as a column.

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