Lecture # 3

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# Lecture # 3

## Lecture # 3

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##### Presentation Transcript

1. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Lecture # 3 Instructor: Eng. Mazen Alshorafa

2. Design of Concrete Structure II University of Palestine Page 1 Columns According to ACI Code a structural element with a ratio of height-to least lateral dimension exceeding three used primarily to support compressive loads is defined as column. P h b l h b Instructor: Eng. Mazen Alshorafa

3. Design of Concrete Structure II University of Palestine Sec A-A Page 2 Columns Columns are vertical compression members of a structural frame intended to support the load-carrying beams. They transmit loads from the upper floors to the lower levels and then to the soil through the foundations. Loads Column Column Beam Sec A Sec A Main beam Instructor: Eng. Mazen Alshorafa

4. Design of Concrete Structure II University of Palestine Page 3 Columns Usually columns carry bending moment as well, about one or both axes of the cross section, and the bending action may produce tensile forces over a part of the cross section The main reinforcement in columns is longitudinal, parallel to the direction of the load and consists of bars arranged in a square, rectangular, or circular Instructor: Eng. Mazen Alshorafa

5. Design of Concrete Structure II University of Palestine Page 4 Types of Columns 1- Form and arrangement of reinforcement Columns are divided into three types 1- Tied Columns It is a column in which the longitudinal reinforcement bars are tied together with separate smaller diameter transverse bars (ties) spaced at some interval along the column height. (Figure a) 2- Spirally-Reinforced Columns It is a column in which the longitudinal bars are arranged in a circle surrounded by a closely spaced continuous spiral. (Figure b) 3- Composite Columns It is a column made of structural steel shapes or pipes surrounded by or filled by concrete with or without longitudinal reinforcement. (Figure c) Instructor: Eng. Mazen Alshorafa

6. Design of Concrete Structure II University of Palestine Page 5 Types ofColumns Instructor: Eng. Mazen Alshorafa

7. Design of Concrete Structure II University of Palestine Page 6 Types of Columns 2- Length of the column in relation to its lateral dimensions. Columns may be divided into two categories 1- Short Columns,for which the strength is governed by the strength of the materials and the geometry of the cross section 2- Slender columns, for which the strength may be significantly reduced by lateral deflections. 3- Position of the load on the cross-section Columns can be classified as 1-Concentrically loaded columns, are subjected to axial force only 2-Eccentrically loaded columns, are subjected to moment in addition to the axial force. Instructor: Eng. Mazen Alshorafa

8. Design of Concrete Structure II University of Palestine Page 7 Columns Behavior of Tied and Spirally-Reinforced Columns Failure of a spiral column Failure of a tied column Deformation Instructor: Eng. Mazen Alshorafa

10. Design of Concrete Structure II University of Palestine Page 9 Columns Strength Reduction Factors ACI Code specifiesΦ values or strength reduction factors for most situations as in the following table Tension-controlled sections (εt ≥ 0.005) 0.90 Compression-controlled sections (εt ≤ 0.002) Members with spiral reinforcement 0.70 Other reinforced members 0.65 Φ Strength condition Instructor: Eng. Mazen Alshorafa

11. Design of Concrete Structure II University of Palestine lc P M ∆ M P Page 10 Sway and Nonsway Frames 1- Nonsway Frames (braced) It is a structural frames whose joints are restrained against lateral displacement by attachment to rigid elements or bracing According to ACI Code a column in a structure is nonsway if Columns Shear wall Columns Brace X Beams Beams Instructor: Eng. Mazen Alshorafa

12. Design of Concrete Structure II University of Palestine Page 11 Sway and Nonsway Frames 1- Nonsway Frames (braced) Moreover, ACI Code assumes a story within a structure is nonsway if: Where, Q is the stability index which is the ratio of secondary moment due to lateral displacement and primary moment, ΣPu is the total vertical load in the story, Vuis the story shear in the story under consideration, Lc is length of column measured center-to center of the joints in the frame, and Δ is the first-order relative deflection between the top and bottom of that story. Instructor: Eng. Mazen Alshorafa

13. Design of Concrete Structure II University of Palestine Page 12 Sway and Nonsway Frames 2- Sway Frames (Unbraced) Structural frames, not attached to an effective bracing element, but depend on the bending stiffness of the columns and girders to provide resistance to lateral displacement are called “sway frames” Instructor: Eng. Mazen Alshorafa

14. Design of Concrete Structure II University of Palestine Page 13 Slenderness effect The slenderness of columns is based on their geometry and on their lateral bracing. As their slenderness increases, their bending stresses increase, and thus buckling may occur. Several items involved in the calculation of slenderness ratios, these item unsupported column lengths, effective length factors and radii of gyration. Unsupported lengths (lu) It is clear distance between floor slabs, beams, or other members capable of providing lateral support as shown in figure. Instructor: Eng. Mazen Alshorafa

15. Design of Concrete Structure II University of Palestine Page 14 Slenderness effect Effective length factors K The effective length is the distance between points of zero moment in the columnThus, the effective length factor k, is the ratio of the effective length to the original length of column. Typical cases illustrating the buckled shape of the column for several end conditions and the corresponding length factor K Points of inflection Instructor: Eng. Mazen Alshorafa

16. Design of Concrete Structure II University of Palestine Page 15 Slenderness effect Effective length factors K For members in a structural frame, the end restraint lies between the hinged and fixed conditions. The actual k value can be estimated from the Jackson and Moreland alignment charts The effective length factor k is a function of the relative stiffness at each end of the column. In these charts, k is determined as the intersection of a line joining the values of ψ at the two ends of the column. The relative stiffness of the beams and columns at each end of the column ψ is given by the following equation Instructor: Eng. Mazen Alshorafa

17. Design of Concrete Structure II University of Palestine Page 16 Braced frame Instructor: Eng. Mazen Alshorafa

18. Design of Concrete Structure II University of Palestine Page 17 Unbraced frame Instructor: Eng. Mazen Alshorafa

19. Design of Concrete Structure II University of Palestine Page 18 Slenderness effect Effective length factors K where, lc= length of column center-to-center of the joints lb= length of beam center-to-center of the joints Ec= modulus of elasticity of column concrete Eb= modulus of elasticity of beam concrete Ic =moment of inertia of column cross section about an axis perpendicular to the plane of buckling being considered. Ib=moment of inertia of beam cross section about an axis perpendicular to the plane of buckling being considered. Σ indicates a summation of all member stiffness connected to the joint and lying in the plane in which buckling of the column is being considered Instructor: Eng. Mazen Alshorafa

20. Design of Concrete Structure II University of Palestine F C I h2 E B H h1 A D G L1 L2 Page 19 Slenderness effect Effective length factors K Consider the two-story frame shown in Figure. To determine the effective length factor k for column EF and For ψ = ------ and for ψ = ------ ACI Code specifies that for columns in nonsway frames, the effective length factor k should be taken as 1.0 Instructor: Eng. Mazen Alshorafa

21. Design of Concrete Structure II University of Palestine Page 20 Slenderness effect Effective length factors K To calculate the ψ values it is necessary to use realistic moments of inertia. Usually, the girder will be appreciably cracked on their tensile sides, whereas the columns will probably have only a few cracks. In the ACI code, it is stated that for determining ψ values for use in evaluating K factors, the rigidity for beams = 0.35 Ig and for columns= 0.7 Ig as follows Where, Ig is the gross moment of inertia Instructor: Eng. Mazen Alshorafa

22. Design of Concrete Structure II University of Palestine Page 21 Slenderness effect Effective length factors K ACI Code provides the following simplified equations for computing the effective length factors for nonsway and sway frame members For Nonsway frames, K is the smaller of Where, ψA and ψB are the values of ψ at the two ends of the column, ψmin is the smaller of the two values. Instructor: Eng. Mazen Alshorafa

23. Design of Concrete Structure II University of Palestine Page 22 Slenderness effect Effective length factors K For Sway frames, a) Restrained at both end For ψm < 2.0 , For ψm ≥ 2.0 , Where,ψmis the average of ψ at the two ends of the column b) Hinged at one end Where, ψ is the values at the restrained end of the column Instructor: Eng. Mazen Alshorafa

24. Design of Concrete Structure II University of Palestine Page 23 The ACI Procedure for Classifying Short and Slender Columns According to ACI Code, columns can be classified as short when their effective slenderness ratios satisfy the following criteria: For Nonsway frames For sway frames Where, k = effective length factor lu= unsupported length of member r = radius of gyration, for rectangular cross sections r = 0.30 h, and for circular sections, r = 0.25 h h = column dimension in the direction of bending. Instructor: Eng. Mazen Alshorafa

25. Design of Concrete Structure II University of Palestine Page 24 The ACI Procedure for Classifying Short and Slender Columns M1= smaller factored end moment on column, positive if member is bent single curvature, negative if bent in double curvature. M2= larger factored end moment on column, always positive. [M1/M2] = ratio of moments at two column ends [Range -1 to 1] Single curvature Double curvature Instructor: Eng. Mazen Alshorafa

26. Design of Concrete Structure II University of Palestine Column Design Non-sway frame Non-sway frame Neglect Slenderness [ Short ] Moment magnification [ long ] Exact P ∆ analysis [ long ] Page 25 Chart summarizes the process of column design as per the ACI Code Instructor: Eng. Mazen Alshorafa

27. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Example # 1 Instructor: Eng. Mazen Alshorafa

28. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Example # 1 • The frame shown in Figure consists of members with rectangular cross sections, made of the same strength concrete. Considering buckling in the plane of the figure. • Categorize column bc as long or short if the frame is: • Nonsway • Sway 270 kN.m 0.6x0.3 0.6x0.3 4.0 m 0.3x0.35 0.6x0.3 0.6x0.3 400 kN.m 4.5 m 0.3x0.35 9.0 m 7.5 m Instructor: Page Ex1-1 Eng. Mazen Alshorafa

29. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution a- Nonsway For a column to be short, Lu = 4-0.3-0.3=3.40 m k is conservatively taken as 1.0 Instructor: Page Ex1-2 Eng. Mazen Alshorafa

30. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution b- Sway For a column to be short, Using the appropriate alignment chart, k = 1.21, and i.e., column is classified as being slender Instructor: Page Ex1-3 Eng. Mazen Alshorafa

31. Design of Concrete Structure II University of Palestine Page 26 Short axially loaded columns For tied reinforced columns For spirally reinforced columns Instructor: Eng. Mazen Alshorafa

32. Design of Concrete Structure II University of Palestine Page 27 Design Considerations Maximum and Minimum Reinforcement Ratios Minimum Number of Reinforcing Bars Clear Distance between Reinforcing Bars Concrete Protection Cover Minimum Cross Sectional Dimensions Lateral Reinforcement Bundled Bars Instructor: Eng. Mazen Alshorafa

33. Design of Concrete Structure II University of Palestine Page 28 Short Columns Subjected to Axial force and Bending Generally, columns are subjected to axial forces in addition to some bending moments. These moments are generally due to: 1- Lateral loading such as wind or earthquake loads. 2- Loads from eccentric loading such as crane loads acting on corbels. 3- End restraints resulting from monolithic action between floor beams and columns. P M Instructor: Eng. Mazen Alshorafa

34. Design of Concrete Structure II University of Palestine Page 29 Behavior under Combined Bending and Axial Loads P P P P e e M A A = = Plastic centroid Axial load and bending moment Eccentrically loaded member P e=M/P e Section A-A Instructor: Eng. Mazen Alshorafa

35. Design of Concrete Structure II University of Palestine Page 30 Behavior under Combined Bending and Axial Loads Nominal strength interaction diagram Concrete crushes before steel yields Steel yields before concrete crushes Note: Any combination of P and M outside the envelope will cause failure. Instructor: Eng. Mazen Alshorafa

36. Design of Concrete Structure II University of Palestine Page 31 Interaction Diagram Design interaction diagram Instructor: Eng. Mazen Alshorafa

37. Design of Concrete Structure II University of Palestine Page 32 Point 1 [axial compression at zero moment] Point 2 [maximum permissible axial compression at zero eccentricity] Point 3 [maximum moment strength at maximum permissible axial compression] Instructor: Eng. Mazen Alshorafa

38. Design of Concrete Structure II University of Palestine Page 33 Point 4 [axial compression and moment strength at zero strain] 0.003 d’ Cs’ εs’ As’ a Cc h c=d As T=0 0.0 b stress strain Instructor: Eng. Mazen Alshorafa

39. Design of Concrete Structure II University of Palestine Page 34 Point 5 [axial compression and moment strength at 50% strain] 0.003 d’ Cs’ εs’ As’ a Cc c h T As εs =0.5 εy b stress strain Instructor: Eng. Mazen Alshorafa

40. Design of Concrete Structure II University of Palestine Page 35 Point6 [axial compression and moment strength at balanced conditions] 0.003 d’ Cs’ εs’ As’ a Cc c h T As εs = εy b stress strain Instructor: Eng. Mazen Alshorafa

41. Design of Concrete Structure II University of Palestine Page 36 Point7 [moment strength at zero axial force] 0.003 a c Cc d’ εs’~0 As’ h As T εs > εy b stress strain Instructor: Eng. Mazen Alshorafa

42. Design of Concrete Structure II University of Palestine Page 37 Point 8 [axial compression and moment strength at 100% strain plus 50% strain εs for point 7] 0.003 Cc a d’ εs’ c Cs’ As’ h As T εs > εy+0.5 εs b stress strain Instructor: Eng. Mazen Alshorafa

43. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Example # 2 Instructor: Eng. Mazen Alshorafa

44. Design of Concrete Structure II University of Palestine 60mm 4Φ20 P.C 0.5 m 4Φ20 60mm 0.30m بسم الله الرحمن الرحيم Example # 2 For the column cross section shown in the following Figure, plot a Design interaction diagram. Use fc’= 25 MPa fy =420 MPa, and Es= 2.04 (10)5MPa Instructor: Page Ex1-1 Eng. Mazen Alshorafa

45. Design of Concrete Structure II Design of Concrete Structure II University of Palestine University of Palestine 60mm 1257 P.C 0.5 m 1257 60mm 0.30m بسم الله الرحمن الرحيم بسم الله الرحمن الرحيم Solution Ag=150 (10)3 mm2, As =2(1257)=2513mm2 d = 500-60=440 mm, and d’ = 60mm Point #1 Point #1 [0, 2723] Point #2 Point #2 [0, 2179] Instructor: Page Ex1-1 Eng. Mazen Alshorafa

46. Design of Concrete Structure II Design of Concrete Structure II University of Palestine University of Palestine بسم الله الرحمن الرحيم بسم الله الرحمن الرحيم Solution 0.003 Point # 4 60 Cs’ εs’ 1257 Cc C=d=440 0.5 m 1257 T=0 0.0 0.30m Instructor: Page Ex1-1 Eng. Mazen Alshorafa

47. Design of Concrete Structure II Design of Concrete Structure II University of Palestine University of Palestine بسم الله الرحمن الرحيم بسم الله الرحمن الرحيم Solution Point # 4 [contd.] Point #4 [159, 1875] Instructor: Page Ex1-1 Eng. Mazen Alshorafa

48. Design of Concrete Structure II Design of Concrete Structure II University of Palestine University of Palestine 0.003 60 Cs’ εs’ 281 1257 330 Cc 0.5 m 1257 T εs =0.001 0.30m بسم الله الرحمن الرحيم بسم الله الرحمن الرحيم Solution Point # 5 Instructor: Page Ex1-1 Eng. Mazen Alshorafa

49. Design of Concrete Structure II Design of Concrete Structure II University of Palestine University of Palestine بسم الله الرحمن الرحيم بسم الله الرحمن الرحيم Solution Point # 5 [contd.] Point #5 [222, 1318] Instructor: Page Ex1-1 Eng. Mazen Alshorafa