Design of RC Columns CE A433 – RC Design T. Bart Quimby, P.E., Ph.D. Spring 2007 (Updated Spring 2009)
The Basic Design Inequality Pu<fPn or Mu<fMnAT THE SAME ECCENTRICTY
Pu, Mu vs. fPn, fMn • Pn, Mn • 1/e = Pn/Mn • fPn, fMn • 1/e = fPn/fMn • Pu, Mu • 1/e = Pu/Mu Compression Tension Failure Interaction Diagram
Pu<fPn is not sufficient • In this case Pu is less than fPn, however it falls outside of the curve • BOTH Pu<fPn and Mu<fMnat the same eccentricity must be true fPn, fMn Pu, Mu
Finding Pu, Mu • Pu is the computed internal axial force determined by structural analysis of the structure using factored loads • Mu is the computed internal moment determined by structural analysis of the structure using factored loads • If the moment is not about a principle axis then it can be expressed by it’s components about the principle axis, Mux and Muy
Plotting For Biaxial Bending + Axial Force • For the case of biaxial bending the condition is: Pu<fPnat same “e” AND at same rotation angle
Finding Pn, Mnx, and Mny • Use the principles of strain compatibility and equilibrium introduced at the start of the semester • A SURFACE of failure, or 3D interaction diagram, can be developed to for any RC section.
Pn Mnx Mny
Finding fPn, fMnx, and fMny • Each Pn, Mnx, and Mny combination (each is a unique point on the 3D interaction surface) is multiplied by f to generate a fPn, fMnx, and fMny surface. • ACI 318-08 9.3.2 lists applicable values of f
Determination of f(ACI 318-08 9.3.2) • For COMPRESSION CONTROLLED conditions: • Spirally Reinforced members: f = 0.75 • Other Reinforced members: f = 0.65 • For TENSION CONTROLLED conditions: • f = 0.90 • Between Tension and Compression controlled: • Linearly interpolate
Confinement Steel • Confinement steel is required to provide lateral support to rebar after the cover spalls off and to contain the rubble, thus increasing the ductility of the column • See pages 434-435 of text to see examples of column failure
Confinement Steel Types • Spirals (read text pg 442-444 for more detail) are like a big “slinky” • Spirals work for round reinforcing patterns • ACI 318-08 7.10.4 covers design requirements for spirals • Ties (read text pg 440-442 for more detail) are generally triangular or rectangular • Can be used with all reinforcing patterns • Review ACI 318-08 7.10.5 for rules pertaining to tie design
More on Confinement Steel • All RC columns MUST have either spirals or ties extending the full height of the column (ACI 318-08 18.104.22.168 & 22.214.171.124) • Confinement steel and the enclosed rubble provide lateral support for the longitudinal bars so that they can continue to support axial force after the cover spalls off • Keeps the slenderness (KL/r) small enough that the bars can still carry significant force
Tie Design • Must follow all of ACI 318-08 7.10.5 rules for layout and spacing. • There is frequently more than one way to achieve an acceptable design
Typical Tied Columns If you look real close you can see the cross ties Column “cages” can be pre-fabricated off site
Accidental Eccentricity • ACI 318-08 10.3.6 provisions recognize the fact that few columns ever have truly concentric loadings, even if analyzed that way. Consequently this section puts an upper limit on the axial force to be equal to that of a load at a small eccentricity. • For tied columns: Pnmax = 0.80Po • For spiral columns: Pnmax = .85Po • Where Po = maximum concentric capacity (i.e. the strain across the section = 0.003) • Po = .85f’c(Ag – As) + Asfy
3D Capacity Surface f modified Interaction Surface
Limits on Reinforcing • ACI 318-08 10.9.1: 0.01Ag< As< 0.08Ag • Limit on longitudinal steel • ACI 318-08 10.9.2: minimum number of bars is: • (3) if triangular ties are used • (4) if square or circular ties are used • (6) if spirals are used.
Summary • Design inequality: Pu<fPn at same “e” and angle of N.A. rotation • Determination of f: • Compression controlled: • f = 0.65 for “tied” or 0.75 for “spiral” • Tension controlled: f = 0.90 • Interpolate between tension & compression • Accidental eccentricity • fPnmax = f(0.80 or 0.85)Po • Limit on longitudinal reinforcing • 0.01Ag< As< 0.08Ag • Min # bars: (3) for triangular ties, (4) for square or circular ties, or (6) for spirals • Design Ties or Spirals to meet the requirements of ACI 318-08 7.10
Design Methods • Hand calculations • Always an option but are very time consuming if you have to look at multiple options • Make a spreadsheet • Takes the repetitiveness out of hand calculations • Good for well defined problems (rectangular sections with a given number of bar sets • Use Tables of problems already solved • CRSI produces some of these • Limited to the shapes analyzed • Extrapolation or interpolation not really an option • Limited to using their assumptions as part of your design • Write or purchase a program that can handle random shapes (both cross section and rebar layout) • The only practical option for random shapes. • Good for quick solution to any column problem
Use a Program • CONSTREN is an example of such a program • Can download a sample version of the program via a link on the class website
Example • Compute the 2D interaction curve for the given column (Muy = 0). • Assume: f’c = 6 ksi, fy = 60 ksi • All longitudinal bars are #7 • Are other applicable requirements met?
Other Requirements • 0.01 < As/Ag = 0.012 < 0.08 … OK • Tie layout meets requirements • tie spacing < min(16db,long, 48db,tie, least column dimension) tie spacing < min(16(7/8”), 48(3/8”), 20”) tie spacing < min(14”, 18”, 20”) = 14”