Design of concrete structures based on ACI 318 Lecture note 5

Presented by Dr Nguyen Dai Minh PhD PEng IBST, Hanoi, Dec 2011

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Lecture note 5Analysis and design of columns under compression and axial bending

(based on ACI 318)

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Contents of the lecture1. 2. 3. 4. 5. 6. Short column under axial compression Short columns under compression and bending Short columns under biaxial bending Load contour method Reciprocal load method Summary

Actually, this content has been presented by Prof. Darwin in previous lectures. Here, I only present a summary of analysis and design of columns with more details on column under biaxial bending.

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1. Short columns under axial compressionColumns are defined as members that carry chiefly in compression. There are 3 types: - members reinforced longitudinal rebars and lateral ties - members with rebars and continuous spirals - members reinforced longitudinally with structural steel shapes, pipes, with or without additional longitudinal rebars, and various types of stirrups Sections of columns can be rectangular, circular or L shapes depending of the architectural or production/mechanical requirements.4

Figure 1: Reinforcement of the columns (Nelson & Darwin 2001)

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The ultimate strength of the short column (exactly of the column section) is reached when the concrete crushes while the rebars yield, or: Pn = 0.85*fc*Ac + fy*Ast (1)

Numerous careful tests have shown the reliability of eq. (1) in predicting the ultimate/nominal strength of a concentrically loaded RC column, provided its slenderness ratio is small so that buckling will not reduce its strength.

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However, taken into account the accidental eccentricities of loading not considered in the analysis, the upper limit is taken as 0.85 for design strength for spiral reinforced columns, and 0.8 for the calculated strength for tied columns, or: for spiral reinforced columns: *Pn(max) = 0.85*[0.85fc(Ag - Ast) + fyAst] where * = 0.7 for spiral columns for tied columns: *Pn(max) = 0.85*[0.85fc(Ag - Ast) + fyAst] where F = 0.7 for tied columns (3) (2)

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2. Short columns under compression and bending

Figure 1: Equivalent eccentricity of column load

General equations are: *Mn >= Mu * Pn >= Pu (4a) (4b)

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a. Strain compatibility analysis and interaction diagrams

Iu = 0.003 (crushing)

Figure 5: Column subject to eccentric compression: (a) loaded columns, (b) strain distribution, (c) stresses and forces at nominal ultimate strength (already lectured by Prof. Darwin)

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Figure 6: From actual to rectangular equivalent stress block10

Value of Eand F in ACI 318-0211

Figure 7: Strain compatibilities, at ultimate state: strain in concrete at compression zone is Iu, strain in steel is Is and Is.

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Strain compatibility conditions:

For tension steel:

d c Is ! Iu cf s ! I u Es d c c and e f y

(5a) (5b)

For compression steel:

c d' I 's ! I u cf 's ! I u E s

(6a)

c d' (6b) and e f y c The concrete stress block has depth: a = F1 c and e h (7) and consequently the concrete compressive resultant is C = 0.85 fc a b (8)13

Equilibrium of external and internal forces requires that: Pn = 0.85 fc a b + As fs As fs (9a) (9b)

h h a h M n ! Pn e ! 0.85 f 'c ab A's f 's d ' As f s d 2 2 2 2

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Interaction diagram (Mn-Pn curve): For given section: b, d, h, As, As, d, d, Es, fc, fy to construct M-P curve? The interaction-strength diagram can be built as follows: - for given c: from eqs.: (5a), (5b) can determine fs from (6a), (6b) can calculate fs from (7) can calculate a from (8) can calculate C from (9a) can determine Pn from (9b) can calculate Mn Then point (Mn, Pn) can be determined based on given c. - Processes are carried out for various values of c. - Finally curve of Mn and Pn relation can be constructed.15

Example of (Mn, Pn) curve is given in Fig 9:

Figure 8: Interaction diagram for nominal column strength in combined bending and axial load.16

EXAMPLE

Figure 9 (balanced section when Is = Iy)17

b. ACI code safety provisions: Design requirements: * Pn >= Pu (10a) * Mn >= Mu (10b) * = 0.65 for tied columns * = 0.70 for spirally reinforced columns

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Figure 10: ACI safety provisions superimposed on column strength interaction diagram

More ductility

The provisions of ACI 318-02 are shown in Fig 10.

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c. Design aids

Can use computer program for design (PCACOLUMN, HBCOLUMN etc.) or using EXCEL sheet to establish the M-P interaction diagrams Can also use the graphs from the technical books as a tool for design

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Pu

Figure 11: Graph for designVg = 0.045

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Mu

Given cross section with b * h, concrete fc, factored moment Pu and Mu=Pu e, to calculate the total steel area Ast: h (d c d c' ) (1) Calculate the ratio K ! , and select the design chart h Pu Pu e (2) Calculate values K n ! and Rn ! where Ag = b*h ' ' * f c Ag * f c Ag h (3) In the graph: from Kn and Rn, to find the required steel ratio V g (4) Calculate the total steel ratio Ast ! V g bh

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3. Short columns under biaxial bendingGeneral: There are situations in which axial compression is accompanied by simultaneous bending about both principal axes of the section, for examples: the corner column between 2 frames. Figure 12 shows the column under biaxial bending. It is called column as the compression load is dominated.

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Figure 12: Interaction diagram for compression plus biaxial bending: (a) uniaxial bending about Y axis; (b) uniaxial bending about X axia; (c) biaxial bending about diagonal axis; (d) interaction surface

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Failure surface: Failure curve for each angle Ucan construct the strength interaction curve P-M as shown in Fig 12d. The series of these curves shall be the failure surface. If the point of (Pu, Mux, Muy) outside of this failure surface the column will fail. Constructing of the failure surface is an extension of uniaxial bending of the column. In Fig 12c, for a given angle U, the strength interaction curve of compression and uniaxial bending can be built using for various value of c, the strain compatibility and equilibrium equations can be established, then ultimate capacities of the column (Pn and Mn) can be determined for the given angle U. Hence, the failure surface can be constructed.

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The main difficulty is that the neutral axis will not be perpendicular to the resultant eccentricity, drawn from the column center to the load Pn (Fig 12c). Therefore, points on the failure surface established in this way will not actually represent a plane interaction as shown for case c in Fig 12. Hence, for design practice, more simple methods are used in the analysis for columns under biaxial bending.

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4. Load contour method The load contour method is based on the representing the failure surface by a family of curves corresponding to constant values of Pn.

Figure 13: Load contour, plane of constant Pn.

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General form of these curves can be approximated by a nondimensional interaction equation:

M ny M nx M M nx 0 ny 0

E1

E2

!1

(8.18(11) )

where E1 and E2 are exponents depending on column dimensions, amount and distribution of steel reinforcement, stress-strain characteristics of rebars and concrete, concrete cover and size of stirrups. When E1 = E2 = E, the shapes of such interaction contours are as shown in Fig 14 for specific a values. ACI recommends value of E is between 1.15 and 1.55 (E 1.19).

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Figure 14: Interaction contours at constant Pn for varying29

By using ACI * factors, equation (11) becomes:

J M nx J M ny J M J M ! 1.0 nx 0 ny 0

E

E

(12)

However, equation (12) is the same as equation (11). Hence, Fig 14 can also be used to describe the load contours for surface of design strength of equation (12).

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Design procedure: For given Pu, Mux, Muy , b, h, fc and Ast to check the column will be failed or not ? - Let Pn = Pu => determine Mnx0 and Mny0 based on the uniaxial bending strength curve. - Choose value of E between 1.15 and 1.55 (E 1.19). - Check equation (12) to be satisfied or not?

JM nx J M ny !1 JM J M nx 0 ny 0

E

E

(8.18)

This method is of course an approximate method and trial design needs to be carried out.

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BS 8110 for biaxial bending of columns (similar to load contour method)

Figure 15: after Kong and Evans (1995)32

Let: Mx = N ey My = N ex M = (M2x + M2y)0.5Kong and Evans (1995)

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Figure 16: N-M interaction surface

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Figure 17

(d) N=0- 0.2Nuz:

Mx My !1 M ux M uy2 2

Mx My !1 (e) N/Nuz approaches 0.8-1: M ux M uy (c) At 0.2Nuz