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Lecture Goals

Lecture Goals. Columns Long Column Design. Introduction.

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Lecture Goals

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  1. Lecture Goals • Columns • Long Column Design

  2. Introduction • Columns are classified as long or short depending on the length effect considerations. In the long column the deflection of the longitudinal axis by the first order moment (primary moment) creates additional secondary moment which is significant. While in stocky columns (short), bending deformation will be very small due to its large bending stiffness, resulting small secondary moment which can be neglected. • Example: member P-delta and structure P-delta

  3. “Long” Columns Slender Columns: ACI Slender = Column

  4. “Long” Columns Less than 10 % of columns in “braced” or “non-sway” frames and less than half of columns in “unbraced” or “sway” frames would be classified as “slender” following ACI Code Procedure.

  5. Effective Length The effective length - Klu lu - It measures the clear distance between floors. K - a factor, which represents the ratio of the distance between points of zero moments in the columns Examples: pin-pin, pin-fixed, fixed-fixed, cantilever, …

  6. (a) (b) Pinned-Pinned Connection Fixed-Fixed Connection “Long” Column - Slenderness Ratio Slenderness Ratio for columns

  7. (c) (d) Fixed-Pinned Connection Partial restrained Connection “Long” Column - Slenderness Ratio Slenderness Ratio for columns

  8. “Long” Column - Slenderness Ratio Slenderness Ratio for columns in frames

  9. “Long” Column - Slenderness Ratio Slenderness Ratio for columns in frames

  10. K Factor YA and YB are the top and bottom factors of the column. For a hinged end Y is infinite or 10 and for a fixed end Y is zero or 1

  11. K Factor The general assumptions are 1. The girders are rigidly connected to the columns at all joints. 2. All columns of a story buckle simultaneously. 3. All members are prismatic and the behavior is elastic. 4. At the onset of buckling, the rotations of the ends of girders within a story are equal in magnitude and direction for unbraced frames (double curvature) and the rotations are equal and opposite for braced frames (single curvature).

  12. K Factor Use the Y values to obtain the K factors for the columns.

  13. “Long” Column Eccentrically loaded pin-ended column: member P-delta. Lateral deflection - increases moment M = P*( e + D )

  14. “Long” Column Eccentrically loaded pin-ended column. OA - curve for end moment OB - curve for maximum column moment @ mid-height) Axial capacity is reduced from A to B due to increase in maximum moment due to D’s (slenderness effects)

  15. Long columns

  16. “Long” Column lu = Unsupported height of column from top of floor to bottom of beams or slab in floor Radius of gyration = 0.3* overall depth of rectangular columns = 0.25* overall depth of circular columns r =

  17. M1/M2 = Ratio of moments at two column ends, where M2 > M1 (-1 to 1 range) “Long” Column double curvature singular curvature

  18. General Formulation Do a second order analysis: P-delta where A. Modulus of Elasticity B. Moment of inertia is given by ACI 10.10.4.1:

  19. General Formulation

  20. Discuss M2min • Solve example 1

  21. Design of columns (short) 1. Calculate factored loads (Pu , Mu ) and e for relevant load combinations. Select material properties and trial size. 2. compute g. 3. Use interaction diagram to determine 4. Select the reinforcement 5. Check maximum load capacity 6. Design the lap splices, select the ties or spiral.

  22. Design of columns (general) • Calculate factored loads (Pu , Mu ) and e for relevant load combinations. Select material properties and trial size. • Check slenderness: if short go to 4 • Compute magnified moment • 4. Select column reinforcement

  23. Example 1 • A hinged end column 6m tall supports unfactored loads of 410kN dead load and 340kN live load. These loads are applied at an eccentricity of 0.05m at bottom and 0.075m at the top. Both eccentricities are on the same side of the centerline of the column. Design a tied column with 28MPA concrete and 420MPa steel.

  24. Example 2 Figure next slide shows a column in a sway frame. The dimensions are center-to-center of the joints. The beams are 0.4m wide by 0.8m over-all depth. The floor slab is 15cm thick (effective flange width for the beam is 1.5m). Use 28MPa concrete and 420MPa reinforcement. The unfactored loads for column AB are: PD=273kN, PL=52kN, PW=2.3kN and the unfactored moments for the column are (all clockwise): At A: MD=460kN.m, ML=110kN.m, MW=17kN.m At B: MD=224kN.m, ML=53kN.m, MW=32kN.m Using the load combination: U=1.2D+L+1.6W, design the column reinforcement given that the column is 0.4m width by 0.8m depth?

  25. Remark • You are a nice class. • I wish you a successful career

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