CE5510 Advanced Structural Concrete Design - Design & Detailing of Openings in RC Flexural Members-

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1 CE5510 Advanced Structural Concrete Design - Design & Detailing Openings in RC Flexural Members- Assoc Pr Tan Kiang Hwee Department Civil Engineering National

2 In this lecture DEPARTMENT OF CIVIL ENGINEERING We will explore!types openings!behaviour beams with large openings!design for ultimate strength!crack Control!Deflection Calculation A/P Tan K H, A/P Tan K H, 2 Tan K H,

3 At the end the lecture DEPARTMENT OF CIVIL ENGINEERING You should be able to!understand the behaviour beams with openings under bending and shear!design the opening using Mechanism Approach, Plasticity Truss Method or Strut-and-Tie Method!Detail the reinforcement to satisfy serviceability and ultimate limit states A/P Tan K H, A/P Tan K H, 3 Tan K H,

4 Beams with Openings: and Design. Mansur, M.A. and Kiang-Hwee Tan. CRC Press Ltd, 1999 Page 4

5 Additional references DEPARTMENT OF CIVIL ENGINEERING! Mansur, M.A., Tan, K.H. and Lee, S.L., "A Design Method for Reinforced Concrete Beams with Large Openings", Journal the American Concrete Institute, Proceedings, Vol. 82, No. 4, USA, July/August 1985, pp ! Tan, K.H. and Mansur, M.A., "Design Procedure for Reinforced Concrete Beams with Web Openings", ACI Structural Journal, Vol. 93, No. 4, USA, July- August 1996, pp ! Mansur, M.A., Huang, L.M., Tan, K.H. and Lee, S.L., "Analysis and Design R/C Beams with Large Web Openings", Journal the Institution Engineers,, Vol. 31, No. 4, July/August 1991, pp A/P Tan K H, A/P Tan K H, 5 Tan K H,

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7 Opening in beam National Page 7

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9 Classification Openings National " By geometry: Small opening if Depth or diameter 0.25 x beam depth and Length depth d d 0.25h 0.25h h Page 9 Large opening otherwise.

10 " By structural response: National Small opening if Beam-type behaviour persists. Large opening otherwise. Page 10

11 TS T2 Page 11

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13 DESIGN TOOLS National " Finite element analysis: elastic stress state area stress concentration " Truss analysis (Strut-and-tie model): lower bound approach detailing for crack control and strength Page 13 " Tests: specific cases serviceability & strength aspects

14 BEHAVIOUR BEAMS WITH LARGE OPENINGS Page 14

15 Principal tension stress contours Page 15

16 M t +M b + Nz = M = M N t + N b = 0 V t + V b = V Page 16

17 Inadequately reinforced openings Page 17

18 1 actual 2 idealised Page 18

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24 Shear carried by chords Page 24

25 Contraflexural points Page 25

26 Page 26 DESIGN & DETAILING OPENINGS - General guidelines sufficient area to develop ultimate compression block in flexure openings to be located more than D/2 from supports, concentrated loads and adjacent openings opening depth D/2 opening length stability chord member deflection requirement beam

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28 " Crack control National principal tensile stress direction stress concentration congestion reinforcement Page 28

29 2 Page 29 shear concentration factor 2 (i.e. 2V u ) diagonal bars to carry 50~75% total shear; vertical stirrups to carry the rest

30 " Strength requirement 2 general approaches: mechanism approach failure at an opening is due to the formation four plastic hinges, one at each corner the opening truss model approach plasticity truss method strut-and-tie method applied loads are transferred across opening to supports by a truss system lower bound approaches Page 30

31 Mechanism Approach " Rigorous method 1. Calculate M u and V u at centre opening. 2. Assume suitable reinft. for chord members. Page 31

32 3. Determine N u in chord members and hinge moments (M u ) *,* from N u = [M u -(M s ) t -(M s ) b ]/z Slide 15 where (M s ) t = [(M u ) t,1 +(M u ) t,2 ]/2 (M s ) b = [(M u ) b,4 +(M u ) b,3 ]/2 and (M u ) *,* are related to N u by the respective M-N interaction diagrams. Page 32

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34 4. Calculate V u in chord members from (V u ) t = [(M u ) t,2 -(M u ) t,1 ]/l (V u ) b = [(M u ) b,4 -(M u ) b,3 ]/l where l is the opening length. 5. Check that (V u ) t +(V u ) b V u. Page 34

35 " Simplified method National Unknowns: N t, N b, M t, M b, V t, V b 3 equations: M t +M b + Nz = M N t + N b = 0 V t + V b = V Assume M t = 0 M b = 0 V b =kv t where k = 0 Page 35 or k = A b /A t k = I b /I t

EXAMPLE A simply supported reinforced concrete beam, 300 mm wide and 600 mm deep, contains a rectangular opening and is subjected to a series point loads as shown.

36 EXAMPLE A simply supported reinforced concrete beam, 300 mm wide and 600 mm deep, contains a rectangular opening and is subjected to a series point loads as shown. Design and detail the reinforcement for the opening if the design ultimate load P u is 52.8 kn. Material properties are: f c =30 MPa, f yv =250 MPa, and f y =460 MPa. Page 36

37 Solution by Mechanism Approach National Assume point contraflexure at midpoint chord members and that 40% shear is carried by bottom chord member. At centre opening, M u = knm V u = 79.2 kn Page 37

38 1. Bottom chord member (N u ) b = M u /z = kn (V u ) b = 0.4 V u = 31.7 kn (M u ) b =(V u ) b l/2 = 15.1 knm 2A s = 1954 mm 2 (3T20 top + 4T20 bottom) A sv /s = 400 mm 2 /m (minimum) 100 mm spacing) Page 38

39 2. Top chord member (N u ) t =408.6 kn; (V u ) t =47.5 kn; (M u ) t =22.6 knm 2A s = 516 mm 2 (2T12 +1T10 top & bottom) A sv /s = 525 mm 2 /m 100 mm spacing) Page 39

40 2T12 + 1T10 2T12 + 1T10 3T20 Page 40

41 Truss Model Approaches National " Plasticity truss (AIJ approach) A s A s Page 41

42 " Shear capacity National V u = 2 bd vs ρ v f yv cot φ s where cot φ s = (νf c /ρ v f yv - 1) 2 νf c /ρ v f yv 1 ρ v = A v f yv /bs ν = f c /200 " Longitudinal reinforcement Page 42 T sn = A sn f y = V u l/2d vs T sf = A sf f y = V u (l +d vs cot φ s )/2d vs

43 Page 43 V d = A sd f yd sin θ d

44 Solution by Plasticity Truss Method (AIJ) National 1. Calculate v. v = V u /(2bd vs νf c )= Determine shear reinforcement. From v = ψ(1-ψ), obtain ψ ρ v f yv /νf c = ρ v = < ρ v,min = 1/3f y = 0.01 A sv /s = 400 mm 2 /m 100 mm spacing) Page 44

45 3. Determine longitudinal reinforcement. A sn = V u l/2f y d vs = 957 mm 2 A sf = V u (l +d vs cot φ s )/2f y d vs = 1080 mm 2 3T20 (near) + 4T20 (far) 4T20 3T20 3T20 4T20 Page 45

46 " Strut-and-tie method General Principles 1. Idealize structure as comprising concrete struts and reinforcement ties, joined at nodes. follow stress path given by elastic theory; refine as necessary equilibrium strut and tie forces deformation capacity concrete angle between struts and ties Page 46

47 2. Calculate forces in struts and ties from equilibrium. check stresses in struts (< 0.6f cd ) detail reinforcement according to calculated tie forces " check for anchorage reinforcement Page 47

48 Solution by Strut-and-Tie Method 1. Postulate strut-and-tie model. Page 48

49 2. Calculate member forces. Top chord takes 91% total shear. Stirrup force 3.3 times shear (Model may be simplified.) Page 49

50 3. Longitudinal reinforcement. Top chord Top steel: 1298 mm 2 (4T20) Bot. steel: 1741 mm 2 (4T25) Bottom chord Top steel: 171 mm 2 (2T12) Bot. steel: 590 mm 2 (2T20) Page 50

51 4. Transverse reinforcement. National Top chord A sv /s= 1567 mm 2 /m 65 mm) Bottom chord A sv /s= 284 mm 2 /m 65 mm) Page 51

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53 CALCULATION OF DEFLECTIONS Stirrup I t Rigid Abutment d s M- M V Hinges V M+ M I b d P l / 2 / 2 e l e P A/P Tan K H, 53

54 δ δ V Relative displacement hinge w.r.t. one end opening due to V (I t based on gross concrete section; I b based on a fully cracked section) δ = V 3 E c l e 2 I t 3 ( + ) I b Relative displacement one end opening w.r.t. the other end δ v = 2 δ = 12 E V c ( + ) I l t 3 e I b Midspan deflection beam δ = δ w + ( ) ( )... δ v + δ + opening1 v opening 2 A/P Tan K H, 54

55 Calculate the midspan service load deflection the beam described in the Example. SOLUTION Service load, P s = P u / 1.7 = 52.8 / 1.7 = 31.1 kn; V = 1.5 P s = 46.6 kn. Effective length chord members, l e = ( ) = 1000 mm. Assume I t = I g = 300 x /12 = 146 x 10 6 mm 4 ; I b = 0.1 x 146 x 10 6 mm 4 15 x 10 6 mm 4 Also, E c = = 26 x 10 3 MPa Hence, δ v = V l e3 / [12 E c ( I t + I b )] = 46.6 x / [12 x 26 x ( ) x 10 9 ] = 0.93 mm A/P Tan K H, 55

56 Midspan deflection, δ w, the beam is 11PL 3 /144E c I, where I can be conservatively estimated as the moment inertia the beam at a section through the opening. That is, I = 2 x [300 x / x 180 x ] = 5054 x 10 6 mm 4 As the span L is 6 m, therefore, δ w =11x 31.1 x 10 3 x (6000) 3 / [144 x 26 x 5054 x 10 9 ] = 3.91 mm Hence, the total midspan deflection is calculated as δ = δ v + δ w = = 4.84 mm < L/360 = 16.7 mm A/P Tan K H, 56

57 Summary National Deflection Add deflection due to shear at each opening Page 57

Lecture-04 Design of RC Members for Shear and Torsion

Lecture-04 Design of RC Members for Shear and Torsion By: Prof. Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk www.drqaisarali.com 1 Topics Addressed Design of