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Design of Concrete Structure II. University of Palestine. Page. Long Columns. Moment Magnification for Sway Frames. The design moments M 1,max and M 2,max at the ends of a compression member are taken as

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  1. Design of Concrete Structure II University of Palestine Page Long Columns Moment Magnification for Sway Frames The design moments M1,max and M2,max at the ends of a compression member are taken as M1ns = factored end moment at the end M1acts due to loads that cause no sway calculated using a first-order elastic frame analysis M2ns = factored end moment at the end M2acts due to loads that cause no sway calculated using a first-order elastic frame analysis M1s = factored end moment at the end M1acts due to loads that cause substantial sway calculated using a first-order elastic frame analysis M2s = factored end moment at the end M2acts due to loads that cause substantial sway calculated using a first-order elastic frame analysis Instructor: Eng. Mazen Alshorafa

  2. Design of Concrete Structure II University of Palestine Page Long Columns Moment Magnification for Sway Frames [Contd.] • δs = moment magnification factor for sway frames to reflect lateral drift resulting from lateral and gravity loads • Computation of δs Ms • The magnified sway moments δsMsare computed in one of three methods. • By Using Second-Order Elastic Frame analysis • 2. By Using Sway-Frame Moment Magnifier • Where, • ΣPu = the summation of all vertical loads in a sto • ΣPcr = the summation of critical buckling loads for all sway resisting columns in a story • For flexural stiffness calculations, Instructor: Eng. Mazen Alshorafa

  3. Design of Concrete Structure II University of Palestine Page Long Columns Moment Magnification for Sway Frames [Contd.] • 3. By Using Direct P-∆ analysis • If δs exceeds 1.5, it has to be calculated using one of the two methods described in 1 and 2. • Heavily Loaded Slender Columns • If the slender column is heavily loaded, it is checked to see whether the moments at points between the ends of the column exceed those at the ends • of the column. ACI Code states that if a compression members has • it is to be designed for the factored axial load Puand Mmax = δns M2 , where • M2=M2ns+ δs M2s Instructor: Eng. Mazen Alshorafa

  4. Design of Concrete Structure II University of Palestine wD=60 kN/m , wL=40 kN/m B F PW=40 kN D 0.6m 0.6m 0.5m 5 m 0.5m 0.5m C E A 8 m 8 m بسم الله الرحمن الرحيم Example # 1 For the frame shown in Figure, design column EFit supports a uniform gravity load wu and a short-term concentrated lateral load Pw Use fc’ = 28 MPa and fy = 420 MPa Columns 0.5x0.3 and Beams 0.6x0.3 Instructor: Page Ex1-1 Eng. Mazen Alshorafa

  5. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution • Evaluate internal forces in story members • Effective moments of inertia are given by • Ibeam=0.35(0.3)(0.6)3/12=1.89x10-3 • Icolumn=0.7(0.3)(0.5)3/12=2.188x10-3 • The modules of elasticity of concrete is • Using SAP 2000, the normal forces and bending moments for Wu=1.2(60)+1.6(40)=136kN/m Case 1 U = 1.2D.L+1.6L.L B F D 423.6 kN.m C E A 1200.5 kN 487.8 kN 487.8 kN Instructor: Page Ex1-2 Eng. Mazen Alshorafa

  6. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution Case 2 U = 1.2D.L+1.0L.L+1.6W.L 1.2(60)+1.0(40)=112kN/m 1.6(40)=64kN B F F D D B 348.8 kN.m 89.2 kN.m C E C E A A 988.6 kN 20.0 kN 401.7 kN 401.7 kN 20.0 kN (b) Loading 1.6W.L (a) Loading 1.2D.L+1.0L.L The horizontal displacement at point B = 29.4 mm, as evaluated from SAP2000 Instructor: Page Ex1-3 Eng. Mazen Alshorafa

  7. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution Case 3 U = 0.9D.L+1.3W.L 0.9(60) =54kN/m 1.3(40)=52 kN B F F D D B 84.09 kN.m 72.5 kN.m C E C E A A 476.7 kN 16.25 kN 193.7 kN 193.7 kN 16.25 kN (b) Loading 1.3W.L (a) Loading 0.9 D.L The horizontal displacement at point B = 23.8 mm, as evaluated from SAP2000 Instructor: Page Ex1-4 Eng. Mazen Alshorafa

  8. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution 2. Check whether columns on the floor are sway or nonsway Case (1) 1.2D+1.6L In this case, the story is braced (nonsway). Case (2) 1.2D+1.0L+1.6W The stability index, i.e., the story is unbraced (sway). Case (3) 0.9D+1.3W The stability index, i.e., the story is unbraced (sway). Instructor: Page Ex1-5 Eng. Mazen Alshorafa

  9. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution 3. Check whether the column is short or long For column EF, Case (1) 1.2D+1.6L Using the appropriate alignment chart (Nonsway), k = 0.9, and Lu = 5.0-0.3= 4.7m For column to be short, i.e column is short. Instructor: Page Ex1-6 Eng. Mazen Alshorafa

  10. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution Case (2) 1.2D+1.0L+1.6W and Case (3) 0.9D+1.3W Using the appropriate alignment chart (sway), k = 2.1, and For column to be short, i.e column is long. 4. Sway and nonsway moments Mns =423.6 kN.m , Ms =0 kN.m and Pu = 487.8 kN Mns =348.8 kN.m , Ms =89.2 kN.m and Pu = 421.7 kN Mns =84.09 kN.m , Ms =72.5 kN.m and Pu = 210 kN Case (1) 1.2D+1.6L Case (2) 1.2D+1.0L+1.6w Case (3) 0.9D+1.3w Instructor: Page Ex1-7 Eng. Mazen Alshorafa

  11. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution Instructor: Page Ex1-8 Eng. Mazen Alshorafa

  12. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution Case (2) 1.2D+1.0L+1.6w Case (3) 0.9D+1.3w 5. Evaluate the magnified moments Case (1) 1.2D+1.6L Mmax =423.6 kN.m and Pu = 487.8 kN Case (2) 1.2D+1.0L+1.6W Mmax =348.8+1.3(89.2)=464.76 kN.m and Pu = 421.7 kN Case (3) 0.9D+1.3W Mmax =84.09+1.13(72.5)=166.01 kN.m and Pu = 210 kN Instructor: Page Ex1-9 Eng. Mazen Alshorafa

  13. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution 6. Determine if the maximum moment occurs at the end of the Member i.e., maximum moment occurs at one of the column ends, and the total moment M2does not have to be further magnified by δns . Instructor: Page Ex1-10 Eng. Mazen Alshorafa

  14. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Solution 7- Design the reinforcement Case (1) 1.2D+1.6L Case (2) 1.2D+1.0L+1.6W Case (3) 0.9D+1.3W Instructor: Page Ex1-11 Eng. Mazen Alshorafa

  15. Design of Concrete Structure II University of Palestine بسم الله الرحمن الرحيم Instructor: Page Ex1-12 Eng. Mazen Alshorafa

  16. Design of Concrete Structure II University of Palestine 0.30 12Φ25 Φ8@300 0.50 بسم الله الرحمن الرحيم Solution • Using the interaction diagram given for fc′ = 28 MPa, fy = 420 MPa and γ = 0.77, one gets ρreq,max = 0.034. • As = 0.034(500)(300) = 5100 mm2 , Use 12Φ25 • The Spacing of ties is the smallest of: • 48 (8) = 384mm • 16 (25) = 400mm • 300 mm • Use φ 8mm ties @ 300mm. Instructor: Page Ex1-13 Eng. Mazen Alshorafa

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