# Copy of M.Sc-RC-12

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### slide 1:

Load Contour Method If the interaction diagram is horizontally sliced at a particular load level as done in Fig. 14.27 the resulting slice is actually a graph between M nx and M ny at a constant load and is called a load contour. The values of the moments M nx and M ny are usually normalized with M nxo when M ny 0 and M nyo when M nx 0 respectively. In this method a curve of the following form is passed through the failure line of a load contour:

### slide 2:

P n M ny M nx Load Contour M ny M nx a Isometric View b Plan View Fig. 14.27. A Typical Load Contour.

### slide 3:

0 . 1 2 1         +         α α nyo ny nxo nx M M M M The constants α 1 and α 2 are the exponents depending on • column dimensions • amount and distribution of steel reinforcement • stress-strain characteristics of steel and concrete • amount of concrete cover and • size of lateral ties or spirals. Generally the values of α 1 and α 2 equal to a constant value α give satisfactory results reducing the equation to: 0 . 1         +         α α nyo ny nxo nx M M M M

### slide 4:

The range of values of α for square and rectangular columns is between 1.15 and 1.55 and the lower values of α are more conservative. A value of α 1.5 is reasonably accurate for the most square and rectangular sections having uniformly distributed reinforcement. According to some researchers the value of α may be more accurately obtained in the following form: α log 0.5 / log β

### slide 5:

Values of β are given in the form of charts on pages 7-17 to 7-19 of PCA Notes on ACI-02 against the values of P u / P o and the reinforcement index. Reinforcement index Separate curves are available for various bar arrangements. The column must be rectangular with larger to shorter sides ratio of less than 4.0 f c ′ must be between 12 and 41 MPa and γ must be between 0.6 and 1.0. c y t f f ′ ρ

### slide 6:

Example 14.5: Select a square tied column cross- section to resist P u 1600 kN M ux 95 kN-m and M uy 110 kN-m. f c ′ 20 MPa f y 300 MPa and clear cover 40 mm. Use the following two methods: a Equivalent uniaxial eccentricity method b Reciprocal load method. Solution: a Equivalent Uniaxial Eccentricity Method y c uy ux u g f f M M P A 008 . 0 43 . 0 2 2 trial + ′ + + 300 008 . 0 20 43 . 0 1000 110 2 95 2 1600 × + × × + × + 182727 mm 2

### slide 7:

Try 450 × 450 mm column. 450 40 2 25 10 2 450 × − − × − γ 0.72 e x 68.8 mm u uy P M 1600 1000 110 × e y 59.4 mm u ux P M 1600 1000 95 × e x / b 68.8 / 450 ≥ e y / h 59.4 / 450 0.395 ≤ 0.4 g c u A f P ′ 2 450 20 1000 1600 × ×

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α 6 . 0 720 300 5 . 0 ≥ +         ′ + y g c u f A f P 0.746 720 300 300 395 . 0 5 . 0 + + e ox e x + α b h e y 68.8 + 0.746 × 59.4 × 450 / 450 113.1 mm Equivalent uniaxial moment is: M oy P u e ox 1600 × 113.1 / 1000 180.98 kN-m Use uniaxial interaction diagrams with bars in all the four faces to determine the total steel ratio ρ t .

### slide 9:

1600000 / 450 2 7.9 MPa g u A P 180.98 ×10 6 / 450 3 2.0 MPa h A M g oy For γ 0.6 ρ t 0.023 For γ 0.75 ρ t 0.018 For γ 0.72 ρ t 0.018 + 0.005 / 0.15 × 0.75 − 0.72 0.019 A st ρ t × A g 0.019 × 450 2 3848 mm 2 Use 8 − 25 bars

### slide 10:

Reciprocal Load Method Trial Size: 450 × 450 mm column Use 8 − 25 bars as the first try. ρ t 8 × 510 / 450 2 0.02 110 ×10 6 / 450 3 1.21 MPa h A M g uy h A M g ny φ 1600000 / 450 2 7.90 MPa g n A P φ Join this point with the origin and extend to the 2 steel curve to get the value of the required capacity.

### slide 11:

For γ 0.60 9.8 MPa g nx A P φ For γ 0.75 10.2 MPa g nx A P φ For γ 0.72 10.12 MPa g nx A P φ ⇒ φP nx 2049 kN 95 ×10 6 / 450 3 1.04 MPa h A M g ux h A M g nx φ 1600000 / 450 2 7.90 MPa g n A P φ

### slide 12:

For γ 0.60 10.6 MPa g ny A P φ For γ 0.75 10.75 MPa g ny A P φ 10.72 MPa for γ 0.72 ⇒ φP ny 2171 kN g ny A P φ The point to calculate φ P no is located on the diagram where the interaction curve for ρ 0.02 intersects the vertical load line. 14.70 MPa ⇒ φP no 2977 kN g no A P φ As the loads φ P nx and φ P ny are quite closer to φ P no φ- factor of 0.65 seems reasonable for all the loads.

### slide 13:

o ny nx ni P P P P φ φ φ φ 1 1 1 1 − + 2977 1 2171 1 2049 1 − + φ P ni 1632 kN P u 1600 kN Design is OK according to the Reciprocal Load Method. Example 14.6: Check the adequacy of a rectangular tied column X-section of size 300 × 450 mm to resist P u 1000 kN acting at e x 125 mm and e y 50 mm as shown in Fig. 14.28. f c ′ 25 MPa f y 420 MPa and cover to centroid of bars 60 mm. The reinforcement is arranged around the perimeter of the column consisting of 8 − 25 bars.

### slide 14:

Use the following two methods: a Reciprocal load method b Load contour method. P n x y Fig. 14.28.Column For Example 14.28. Solution: ρ t 4080 / 300 × 450 0.03 a Reciprocal Load Method i Considering bending about y-axis: γ 330 / 450 0.73 ≅ 0.75 1000 × 1000 / 300 × 450 7.41 MPa g u A P

### slide 15:

1000 × 1000 × 125 / 300 × 450 2 2.06 MPa h A M g u h A e P g x u × Join the point P u /A g M u /A g h with the origin and extend to ρ 0.03 to get the following: 10.8 MPa ⇒ φP nx 1458 kN g nx A P φ ii Considering bending about x-axis: γ 180 / 300 0.60 1000 × 1000 / 300 × 450 7.41 MPa g u A P

### slide 16:

1000 × 1000 × 50 / 450 × 300 2 1.23 MPa h A M g u h A e P g y u × Join the point P u /A g M u /A g h with the origin and extend to ρ 0.03 to get the following: 13.6 MPa ⇒ φP ny 1836 kN g ny A P φ iii No eccentricity case: φP no 21.4 × 300 × 450 2889 kN Assuming same φ-factors for all the loads we have

### slide 17:

o ny nx ni P P P P φ φ φ φ 1 1 1 1 − + 2889 1 1836 1 1458 1 − + φ P ni 1131 kN P u 1000 kN OK a Load Contour Method i Considering bending about x-axis: γ 180 / 300 0.60 1000 × 1000 / 300 × 450 7.41 MPa g u A P ρ t 0.03

### slide 18:

2.92 MPa ⇒ h A M g nx φ φM nxo 2.92 × 450 × 300 2 / 10 6 118.26 kN-m ii Considering bending about y-axis: γ 320 / 450 0.71 ≅ 0.75 1000 × 1000 / 300 × 450 7.41 MPa g u A P ρ t 0.03 3.47 MPa ⇒ h A M g ny φ φM nyo 3.47 × 300 × 450 2 / 10 6 210.80 kN-m

### slide 19:

M ux P u × e y 1000 × 50 / 1000 50 kN-m M uy P u × e x 1000 × 125 / 1000 125 kN-m Selecting the conservative value of α 1.15 we have α α         +         nyo ny nxo nx M M M M α α φ φ         +         nyo uy nxo ux M M M M 15 . 1 15 . 1 80 . 210 00 . 125 26 . 118 00 . 50       +       0.92 1.0 OK Hence the column is safe according to the load contour method. This method is relatively more exact for portion of curve below the balanced condition.

### slide 20:

SLENDER COLUMNS There are three methods of design of slender columns described below: 1. Perform exact P −∆ analysis to find P u and M umax and then use the standard interaction diagram for the short columns. This method is called the second order analysis of frames. 2. Find P u and approximately magnified moment M u and then use the standard interaction diagram for short columns. This method is called Moment Magnification Method. In this method the moments obtained from the first order analysis are multiplied with an empirical moment magnifier.

### slide 21:

3. Modify the interaction diagram to account for the slenderness effects having the coordinates P u and unmagnified moment M u just like a regular diagram. Elastic Buckling Load For Concentrically Loaded Columns P c 2 2 u t k I E l π 2 2 / r k A E u t l π Concrete is an inelastic material and hence the modulus of elasticity varies all along the stress-strain curve as shown in Fig. 14.29. By replacing the E-value with the tangent modulus of elasticity E t Euler’s formula may be used for materials like concrete.

### slide 22:

e f E t Fig. 14.29.Modulus of Elasticity For Concrete.

### slide 23:

The tangent modulus of elasticity is different at all points of the stress-strain curve and is difficult to estimate precisely. Hence approximate methods are used to calculate the effective E-value along with the reduced moment of inertia due to cracking and long-term effects. A typical load − slenderness ratio curve is shown in Fig.14.30 to represent the buckling behavior. For concrete columns the chances of elastic buckling are usually very less and only moment magnification and at the most inelastic buckling are the important parameters.

### slide 24:

Elastic Buckling Inelastic Buckling With Crushing r k u l P Short Column Intermediate Column Long Column Fig. 14.30. Buckling Load Versus Slenderness Ratio For Columns.

### slide 25:

Effective Length Factor This factor gives the ratio of length of half sine wave portion of defected shape after buckling distance between two points of contra-flexure to full-unsupported length of column. This depends upon the end conditions of the column and the fact that whether side-sway is permitted or not. Greater the k-value greater is the effective length and slenderness ratio and hence smaller is the buckling load. The value of k-factor in case of no side-sway is between 0.5 and 1.0 whereas in case of appreciable side-sway it is always greater than or equal to 1.0.

### slide 26:

Any appreciable lateral or sideward movement of top of a vertical column relative to its bottom is called side- sway sway or lateral drift. If side-sway is possible k-value increases by a greater degree and column buckles at a lesser load. Side-sway in a frame takes place due to the following factors shown in Fig. 24.31:- 1. Lengths of different columns are unequal. 2. Sections of columns have different cross-sectional properties. 3. Loads are un-symmetrical. 4. Lateral loads are acting.

### slide 27:

2I I I Fig. 14.31. Chances of Side-sway.

### slide 28:

Side-sway can be prevented in a frame by:- 1. Providing shear or partition walls. 2. Fixing the top of frame with adjoining rigid structures. 3. Provision of properly designed lift well in a building which may act like backbone of the structure reducing the lateral deflections. 4. Provision of lateral bracing which may be of following two types: A. Diagonal bracing. B. Longitudinal bracing. Effective length factor and the buckled shape of columns having well-defined end conditions are given in Fig. 14.32.

### slide 29:

Theoretical k 0.5 Practical k 0.65 No side-sway Theoretical k 1.0 Practical k 1.0 No side-sway Inflection points l e k l l e k l

### slide 30:

Theoretical k 1.0 Practical k 1.2 Side-sway present Theoretical k 0.7 Practical k 0.8 No side-sway l e k l l e k l

### slide 31:

Theoretical k 2.0 Practical k 2.10 Side-sway present Theoretical k 2.0 Practical k 2.0 Side-sway present l e k l l e k l Fig. 14.32. Values of k-factor For Various End Conditions.

### slide 32:

Consider the example of column AB shown in Fig. 14.33. The ends are not free to rotate. However these ends are also not perfectly fixed. Instead the ends are partially fixed with the fixity determined by the ratio of relative flexural stiffness of columns meeting at a joint to the flexural stiffness of beams meeting at the joint. This ratio is denoted by ψ or G using the expression given below:-

### slide 33:

A B B A G A or ψ A G B or ψ B Columns Beams Fig. 14.33. Partially Fixed Column Ends.

### slide 34:

ψ or G at each end EI of columns EI of beams l l ∑ ∑ This value is calculated at both ends of the columns denoted by points A and B and summation is taken for all members meeting at a particular end. The lower columns of Fig. 14.33 have their top ends partially fixed but the bottom ends may have well-defined end conditions. The ψ or G value at these ends are decided as follows:

### slide 35:

Hinged Support G or Ψ 10.0 for braced columns 20.0 for sway columns Fixed Support G or Ψ 0.5 for braced columns 1.0 for sway columns To find out the k-value the alignment charts given in Figs. 14.34 and 14.35 are used. The alignment chart of Fig. 14.34 is for columns without any side-sway and the alignment chart of Fig. 14.35 is for columns having appreciable side-sway.

### slide 36:

∞ 10.0 1.0 0.8 0.6 0.3 0.5 0.4 0.2 0.1 0 2.0 3.0 5.0 ∞ 10.0 1.0 0.8 0.6 0.3 0.5 0.4 0.2 0.1 0 2.0 3.0 5.0 1.0 0.8 0.6 0.5 0.6 0.5 Ψ A Ψ B k Fig. 14.34. Effective Length Factor For Braced Columns.

### slide 37:

∞ 100.0 10.0 8.0 6.0 3.0 5.0 4.0 2.0 1.0 0 20.0 30.0 50.0 ∞ 100.0 10.0 20.0 30.0 50.0 8.0 6.0 3.0 5.0 4.0 2.0 1.0 0 ∞ 10.0 3.0 2.0 5.0 4.0 1.5 1.0 Ψ A Ψ B k Fig. 14.35. Effective Length Factor For Unbraced Columns.

### slide 38:

Restraint Provided By Footings For Calculation Of Ψ Factor In case of footing resting on soil the Ψ value is calculated by using the foundation stiffness K f in place of the beam stiffness in the usual expression as under: ψ ∑ ∑ footing of columns of l l EI EI f c K K To calculate the stiffness of the footing a moment M is applied to the footing of Fig. 14.36 and the corresponding rotation θ f settlement at extreme compression end due to moment alone leaving the uniform downward settlement ∆ and the contact stress under the footing due to the rotation σ are observed.

### slide 39:

θ f ∆ y Fig. 14.36. Rotation and Stiffness of a Foundation. The extreme edge contact stress may be calculated as follows: σ M / S f M y / I f where S f Section modulus of the contact surface and I f Moment of inertia of the contact surface.

### slide 40:

θ f y ∆ y K s / σ s f K I M where K s is the modulus of sub-grade reaction pressure corresponding to 1mm settlement which may approximately be found from Table 14.1. Table 14.1. Soil Sub-grade Reaction. Allowable Soil Bearing Capacity Sub-grade Modulus kN/m 2 or kPa N/mm 2 /mm 60 0.0136 100 0.0272 180 0.0450 240 0.0504 360 0.0654 480 0.0736

### slide 41:

∴ K f f M θ s f K I The value of Ψ at a footing to column joint is estimated as follows: ψ f c K K s f c c c K I I E l / 4 After calculation of Ψ factors at both ends of the column the value of k may also be calculated from the following ACI equations: For braced members k should be the lesser of the following two values:

### slide 42:

k 0.7 + 0.05 Ψ A + Ψ B ≤ 1.0 k 0.85 + 0.05 Ψ min ≤ 1.0 where Ψ min smaller of Ψ A and Ψ B For unbraced compression members k may be evaluated as follows: For Ψ m 2k 2 B A ψ ψ + m m ψ ψ + + 1 20 20 For unbraced compression members hinged at one end and partially fixed at the other end k-value is obtained as follows: k 2.0 + 0.3 Ψ