Slender Columns and Two-way Slabs

Lecture Goals Slender Column Design One-way and two-way slab Slab thickness, h

Design of Long Columns- Example A rectangular braced column of a multistory frame building has floor height lu =25 ft. It is subjected to service dead-load moments M2= 3500 k-in. on top and M1=2500 k-in. at the bottom. The service live load moments are 80% of the dead-load moments. The column carries a service axial dead-load PD = 200 k and a service axial live-load PL = 350 k. Design the cross section size and reinforcement for this column. Given YA = 1.3 and YB = 0.9. Use a d’=2.5 in. cover with an sustain load = 50 % and fc = 7 ksi and fy = 60 ksi.

Design of Long Columns- Example Compute the factored loads and moments are 80% of the dead loads

Design of Long Columns- Example Compute the k value for the braced compression members Therefore, use k = 0.81

Design of Long Columns- Example Check to see if slenderness is going to matter. An initial estimate of the size of the column will be an inch for every foot of height. So h = 25 in. We need to be concerned with slender columns

Design of Long Columns- Example So slenderness must be considered. Since frame has no side sway, M2 = M2ns, ds = 0 Calculate the minimum M2 for the ratio computations.

Design of Long Columns- Example Compute components of concrete The moment of inertia of the column is

Design of Long Columns- Example Compute the stiffness, EI

Design of Long Columns- Example The critical load (buckling), Pcr, is

Design of Long Columns- Example Compute the coefficient, Cm, for the magnification d coefficient

Design of Long Columns- Example The magnification factor

Design of Long Columns- Example The design moment is Therefore, the design conditions are

Design of Long Columns- Example Assume that the r = 2.0 % or 0.020 Use 14 # 9 bars or 14 in2

Design of Long Columns- Example The column is compression controlled so c/d > 0.6. Check the values for c/d = 0.6

Design of Long Columns- Example Check the strain in the tension steel and compression steel.

Design of Long Columns- Example The tension steel is

Design of Long Columns- Example Combined forces are

Design of Long Columns- Example Combined force is

Design of Long Columns- Example Moment is

Design of Long Columns- Example The eccentricity is Since the e = 11.28 in. < 13.62 in. The section is in the compression controlled region f = 0.65. You will want to match up the eccentricity with the design.

Design of Long Columns- Example We need to match up the eccentricity of the problem. This done varying the c/d ratio to get the eccentricity to match. Check the values for c/d = 0.66

Design of Long Columns- Example Check the strain in the tension steel and compression steel.

Design of Long Columns- Example The tension steel is

Design of Long Columns- Example Combined forces are

Design of Long Columns- Example Combined force is

Design of Long Columns- Example Moment is

Design of Long Columns- Example The eccentricity is Since the e 11.28 in. The reduction factor is equal to f = 0.65. Compute the design load and moment.

Design of Long Columns- Example The design conditions are The problem matches the selection of the column.

Design of Long Columns- Example Design the ties for the column Provide #3 ties, spacing will be the minimum of: Therefore, provide #3 ties @ 18 in. spacing.

Using Interaction Diagrams Determine eccentricity. Estimate column size required base on axial load. Determine e/h and required fPn/Ag, fMn/(Agh) Determine which chart to use from fc, fy and g. Determine r from the chart. Select steel sizes. Check values. Design ties by ACI code Design sketch

Two-way Slabs

Comparison of One-way and Two-way slab behavior One-way slabs carry load in one direction. Two-way slabs carry load in two directions.

Comparison of One-way and Two-way slab behavior One-way and two-way slab action carry load in two directions. One-way slabs: Generally, long side/short side > 1.5

Comparison of One-way and Two-way slab behavior Two-way slab with beams Flat slab

Comparison of One-way and Two-way slab behavior For flat plates and slabs the column connections can vary between:

Comparison of One-way and Two-way slab behavior Flat Plate Waffle slab

Comparison of One-way and Two-way slab behavior The two-way ribbed slab and waffled slab system: General thickness of the slab is 2 to 4 in.

Comparison of One-way and Two-way slab behavior Economic Choices Flat Plate suitable span 20 to 25 ft with LL= 60 -100 psf Advantages Low cost formwork Exposed flat ceilings Fast Disadvantages Low shear capacity Low Stiffness (notable deflection)

Comparison of One-way and Two-way slab behavior Economic Choices Flat Slab suitable span 20 to 30 ft with LL= 80 -150 psf Advantages Low cost formwork Exposed flat ceilings Fast Disadvantages Need more formwork for capital and panels

Comparison of One-way and Two-way slab behavior Economic Choices Waffle Slab suitable span 30 to 48 ft with LL= 80 -150 psf Advantages Carries heavy loads Attractive exposed ceilings Fast Disadvantages Formwork with panels is expensive

Comparison of One-way and Two-way slab behavior Economic Choices One-way Slab on beams suitable span 10 to 20 ft with LL= 60-100 psf Can be used for larger spans with relatively higher cost and higher deflections One-way joist floor system is suitable span 20 to 30 ft with LL= 80-120 psf Deep ribs, the concrete and steel quantities are relative low Expensive formwork expected.

Comparison of One-way and Two-way slab behavior ws =load taken by short direction wl = load taken by long direction dA = dB Rule of Thumb: For B/A > 2, design as one-way slab

Two-Way Slab Design Static Equilibrium of Two-Way Slabs Analogy of two-way slab to plank and beam floor Section A-A: Moment per ft width in planks Total Moment

Two-Way Slab Design Static Equilibrium of Two-Way Slabs Analogy of two-way slab to plank and beam floor Uniform load on each beam Moment in one beam (Sec: B-B)

Two-Way Slab Design Static Equilibrium of Two-Way Slabs Total Moment in both beams Full load was transferred east-west by the planks and then was transferred north-south by the beams; The same is true for a two-way slab or any other floor system.

General Design Concepts (1) Direct Design Method (DDM) Limited to slab systems to uniformly distributed loads and supported on equally spaced columns. Method uses a set of coefficients to determine the design moment at critical sections. Two-way slab system that do not meet the limitations of the ACI Code 13.6.1 must be analyzed more accurate procedures

General Design Concepts (2) Equivalent Frame Method (EFM) A three-dimensional building is divided into a series of two-dimensional equivalent frames by cutting the building along lines midway between columns. The resulting frames are considered separately in the longitudinal and transverse directions of the building and treated floor by floor.

Equivalent Frame Method (EFM) Transverse equivalent frame Longitudinal equivalent frame

Equivalent Frame Method (EFM) Perspective view Elevation of the frame

Method of Analysis (1) Elastic Analysis Concrete slab may be treated as an elastic plate. Use Timoshenko’s method of analyzing the structure. Finite element analysis

Method of Analysis (2) Plastic Analysis The yield method used to determine the limit state of slab by considering the yield lines that occur in the slab as a collapse mechanism. The strip method, where slab is divided into strips and the load on the slab is distributed in two orthogonal directions and the strips are analyzed as beams. The optimal analysis presents methods for minimizing the reinforcement based on plastic analysis

Method of Analysis (3) Nonlinear analysis Simulates the true load-deformation characteristics of a reinforced concrete slab with finite-element method takes into consideration of nonlinearities of the stress-strain relationship of the individual members.

Column and Middle Strips The slab is broken up into column and middle strips for analysis

Minimum Slab Thickness for Two-way Construction The ACI Code 9.5.3 specifies a minimum slab thickness to control deflection. There are three empirical limitations for calculating the slab thickness (h), which are based on experimental research. If these limitations are not met, it will be necessary to compute deflection.

Minimum Slab Thickness for Two-way Construction (a) For fy in psi. But not less than 5 in.

Minimum Slab Thickness for Two-way Construction (b) For fy in psi. But not less than 3.5 in.

Minimum Slab Thickness for Two-way Construction (c) For Use the following table 9.5(c)

Minimum Slab Thickness for Two-way Construction Slabs without interior beams spanning between supports and ratio of long span to short span < 2 See section 9.5.3.3 For slabs with beams spanning between supports on all sides.

Minimum Slab Thickness for two-way construction The definitions of the terms are: h = Minimum slab thickness without interior beams ln = b = am= Clear span in the long direction measured face to face of column the ratio of the long to short clear span The average value of a for all beams on the sides of the panel.

Definition of Beam-to-Slab Stiffness Ratio, a Accounts for stiffness effect of beams located along slab edge reduces deflections of panel adjacent to beams.

Definition of Beam-to-Slab Stiffness Ratio, a With width bounded laterally by centerline of adjacent panels on each side of the beam.

Beam and Slab Sections for calculation of a

Beam and Slab Sections for calculation of a

Beam and Slab Sections for calculation of a Definition of beam cross-section Charts may be used to calculate a

Minimum Slab Thickness for Two-way Construction Slabs without drop panels meeting 13.3.7.1 and 13.3.7.2, tmin = 5 in Slabs with drop panels meeting 13.3.7.1 and 13.3.7.2, tmin = 4 in

Example - Slab A flat plate floor system with panels 24 by 20 ft is supported on 20 in. square columns. Determine the minimum slab thickness required for the interior and corner panels. Use fc = 4 ksi and fy = 60 ksi

Example - Slab Slab thickness, from table 9.5(c) for fy = 60 ksi and no edge beams

Example - Slab Slab thickness, from table 9.5(c) for fy = 60 ksi and no edge beams for a = am = 0 (no beams)

Example – a Calculations The floor system consists of solid slabs and beams in two directions supported on 20-in. square columns. Determine the minimum slab thickness, h, required for the floor system. Use fc = 4 ksi and fy = 60 ksi

Example – a Calculations The cross-sections are:

Example – a Calculations To find h, we need to find am therefore Ib, Islab and a for each beam and slab in long short direction. Assume slab thickness h = 7 in. so that x = y < 4 tf

Example – a Calculations Compute the moment of inertia and centroid

Example – a Calculations Compute the a coefficient for the long direction Short side of the moment of inertia

Example – a Calculations Compute the a coefficient for short direction The average am for an interior panel is

Example – a Calculations Compute the b coefficient Compute the thickness for am > 2 Use slab thickness, 6.5 in. or 7 in.

Example – a Calculations Compute the moment of inertia and centroid for the L-beam

Example – a Calculations Compute the am coefficient for long direction Short side of the moment of inertia

Example – a Calculations Compute the am coefficient for the short direction

Example – a Calculations Compute the am coefficient for the edges and corner

Example – a Calculations Compute the am coefficient for the edges and corner

Example – a Calculations Compute the largest length ln of the slab/beam, edge to first interior column.

Example – a Calculations Compute the thickness of the slab with am > 2 The overall depth of the slab is 7 in. Use slab thickness, 6.5 in. or 7 in.