N.W.F.P. University of Engineering and Technology Peshawar 1 By: Prof Dr. Akhtar Naeem Khan Lecture 09: Compression Members

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 2 Effective length of columns in frames Rotation of the ends of the columns in building frames is usually limited by the beams connecting to them, while compression members in trusses may have restricted end rotations because of other members connecting at the joints.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 3 Effective length of columns in frames KL is called effective length of column and K effective length factor.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 4 Effective length of columns in frames

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 5 Effective length of columns in frames So far, we have looked at the buckling strength of individual columns. These columns had various boundary conditions at the ends, but they were not connected to other members with moment (fix) connections. The effective length factor K for the buckling of an individual column can be obtained for the appropriate end conditions from Table C-C2.1 of the AISC Manual

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 6 Effective length of columns in frames However, when these individual columns are part of a frame, their ends are connected to other members (beams etc.). These frames are sometimes braced and sometimes un braced.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 7 Effective length of columns in frames A Braced frame is one in which a sideway (joint translation) is prevented by means of bracing, shear walls, or lateral support from adjoining structure.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 8 Effective length of columns in frames A Un Braced does not have any bracing and must depend on stiffness of its own members and rotational rigidity of joints between frame members to prevent lateral buckling.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 9 Effective length of columns in frames Effective length coefficient increases with decreasing stiffness of the beam and becomes unity with zero stiffness. Critical loads for a column depends on: Its stiffness relative to that of beams framing into it and Presence or absence of restraint to lateral displacement of its ends. Conclusions

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 10 Effective length of columns in frames Braced and Un-braced Frames

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 11 Effective length of columns in frames Braced and Un-braced Frames Similarly you can analyze multi bay, multistory frames. Assumptions Subjected to vertical loads only All columns become unstable simultaneously All joint rotations at floor are equal Restraining moment distributed in proportion to stiffness.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 12 Effective length of columns in frames Method of Analysis First, you have to determine whether the column is part of a braced frame or an unbraced (moment resisting) frame. Then, you have to determine the relative rigidity factor G for both ends of the column

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 13 Effective length of columns in frames Method of Analysis G is defined as the ratio of the summation of the rigidity (EI/L) of all columns coming together at an end to the summation of the rigidity (EI/L) of all beams coming together at the same end. G = It must be calculated for both ends of the column.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 14 Effective length of columns in frames Method of Analysis Then, you can determine the effective length factor K for the column using the calculated value of G at both ends, i.e., G A and G B and the appropriate alignment chart. There are two alignment charts provided by the AISC manual.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 15 Effective length of columns in frames Method of Analysis One is for columns in braced (side sway inhibited) frames. See Figure C-C2.2a on page of the AISC manual. 0 < K ≤ 1 The second is for columns in un-braced (side sway uninhibited) frames. See Figure C-C2.2b on page of the AISC manual. 1 < K ≤ ∞

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 16 Effective length of columns in frames Alignment Chart

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 17 Effective length of columns in frames Method of Analysis: Inelastic Case G is a measure of the relative flexural rigidity of the columns (EI c /L c ) with respect to the beams (EI b /L b ) However, if column buckling were to occur in the inelastic range ( c < 1.5), then the flexural rigidity of the column will be reduced because I c will be the moment of inertia of only the elastic core of the entire cross-section.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 18 Effective length of columns in frames Method of Analysis: Inelastic Case

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 19 Effective length of columns in frames Method of Analysis: Inelastic Case The beams will have greater flexural rigidity when compared with the reduced rigidity (EI­ c ) of the inelastic columns. As a result, the beams will be able to restrain the columns better, which is good for column design.

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 20 Effective length of columns in frames Method of Analysis: Inelastic Case The ratio Fcr/ Fe is called Stiffness reduction factor

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 21 Effective length of columns in frames Method of Analysis: Inelastic Case The ratio Fcr/ Fe is called Stiffness reduction factor

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 22 Procedure for Column Design 1. Design Load 2.Assume F cr ØP n = ØA g F cr = P u Find Ag Select a section 3.Find 4.Find c =

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 23 Procedure for Column Design For c ≤ 1.5F cr = F y For c > 1.5 F cr = F y 5.Fcr Calculated > Fcr Assumed 6.Ø P > Pu…………………. Check

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 24 Procedure for Column Design LRFD mannual contains variety of Design aids, helpful in making original trial section. 1.Design load 2.Find section for Corresponding P & KL using table Calculate an equivalent (KL)eq = Using Design Aids

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 25 Procedure for Column Design 4.Use the calculated (KL)eq value to find (ØcPn) the column strength. 1. Select section and its properties 5.Find c 6.Find Fcr 7.Find ØP Using Design Aids

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 26 Problem

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 27 Problem

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 28 Problem

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 29 Example Problem 01 ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 30 Example Problem 01 ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 31 Example Problem 01 ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 32 Example Problem 01 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 33 Example Problem 01 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 34 Example Problem 01 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 35 Example Problem 01 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 36 Example Problem 02 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 37 Example Problem 02 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 38 Example Problem 02 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 39 Example Problem 02 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 40 Example Problem 02 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 41 Example Problem 02 LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 42 Column Bases

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 43 Column Bases

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 44 Example ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 45 Example ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 46 Example ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 47 Example ASD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 48 Example LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 49 Example LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 50 Example LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan 51 Example LRFD

CE-409: Lecture 09Prof. Dr Akhtar Naeem Khan THANKS