Openings in Beams

CE5510 Advanced Structural Concrete Design - Design & Detailing of Openings in RC Flexural MembersAssoc Prof Tan Kiang Hwee Department of Civil Engineering National University of Singapore

DEPARTMENT OF CIVIL ENGINEERING

In this lecture We will explore !Types

of openings !Behaviour of beams with large openings !Design for ultimate strength !Crack Control !Deflection Calculation A/P Tan K H, NUS

A/P Tan K H, NUS

2  Tan K H, NUS

DEPARTMENT OF CIVIL ENGINEERING

At the end of the lecture You should be able to !Understand

the behaviour of 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 A/P Tan Kserviceability H, NUS A/P Tan K H, NUS 3 and ultimate limit states  Tan K H, NUS

National University of Singapore

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

Additional references !

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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 of the American Concrete Institute, Proceedings, Vol. 82, No. 4, USA, July/August 1985, pp. 517-524. Tan, K.H. and Mansur, M.A., "Design Procedure for Reinforced Concrete Beams with Web Openings", ACI Structural Journal, Vol. 93, No. 4, USA, JulyAugust 1996, pp. 404-411. Mansur, M.A., Huang, L.M., Tan, K.H. and Lee, S.L., "Analysis and Design of R/C Beams with Large Web Openings", Journal of the Institution of Engineers, Singapore, Vol. 31, No. 4, July/August 1991, pp. 59-67.

A/P Tan K H, NUS

A/P Tan K H, NUS

5  Tan K H, NUS

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Opening in beam National University of Singapore

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Classification of Openings National University of Singapore

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By geometry: – Small opening if Depth or diameter ≤ 0.25 x beam depth and Length ≤ depth ≤d d ≤ 0.25h

≤ 0.25h

– Large opening otherwise. © Tan KH NUS Page 9

h

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By structural response: – Small opening if Beam-type behaviour persists.

– Large opening otherwise.

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TS

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T2

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DESIGN TOOLS National University of Singapore

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Finite element analysis: • elastic stress state • area of stress concentration

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Truss analysis (Strut-and-tie model): • lower bound approach • detailing for crack control and strength

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Tests: • specific cases • serviceability & strength aspects

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BEHAVIOUR of BEAMS WITH LARGE OPENINGS

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Principal tension stress contours

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Mt +Mb + Nz = M =M Nt + Nb = 0 Vt + Vb = V

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Inadequately reinforced openings © Tan KH NUS Page 17

actual National University of Singapore

1

2

idealised

1 3

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2 4

3

4

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Shear carried by chords © Tan KH NUS Page 24

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Contraflexural points © Tan KH NUS Page 25

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DESIGN & DETAILING of 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 of chord member • deflection requirement of beam

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Crack control • principal tensile stress direction • stress concentration • congestion of reinforcement

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⇒shear concentration factor of 2 (i.e. 2Vu) ⇒diagonal bars to carry 50~75% of total shear; vertical stirrups to carry the rest

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Strength requirement – 2 general approaches: •

mechanism approach ◊ failure at an opening is due to the formation of four plastic hinges, one at each corner of 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

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Mechanism Approach National University of Singapore

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Rigorous method 1. Calculate Mu and Vu at centre of opening. 2. Assume suitable reinft. for chord members.

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3. Determine Nu in chord members and hinge moments (Mu)*,* from National University of Singapore

Nu = [Mu - (Ms)t - (Ms)b]/z

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where (Ms)t = [(Mu)t,1 +(Mu)t,2]/2 (Ms)b = [(Mu)b,4 +(Mu)b,3]/2 and (Mu)*,* are related to Nu by the respective M-N interaction diagrams. © Tan KH NUS Page 32

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4. Calculate Vu in chord members from (Vu)t = [(Mu)t,2 - (Mu)t,1]/ll (Vu)b = [(Mu)b,4 - (Mu)b,3]/ll where l is the opening length. 5. Check that (Vu)t+(Vu)b ≅ Vu.

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Simplified method – Unknowns: Nt, Nb, Mt, Mb, Vt, Vb – 3 equations: Mt +Mb + Nz = M Nt + Nb = 0 Vt + Vb = V ⇒ Assume

where © Tan KH NUS Page 35

Mt = 0 Mb = 0 Vb =kVt k=0 k = Ab/At

or

k = Ib/It

EXAMPLE National University of Singapore

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A simply supported reinforced concrete beam, 300 mm wide and 600 mm deep, contains a rectangular opening and is subjected to a series of point loads as shown. Design and detail the reinforcement for the opening if the design ultimate load Pu is 52.8 kN. Material properties are: fc’=30 MPa, fyv=250 MPa, and fy=460 MPa.

Solution by Mechanism Approach National University of Singapore

Assume point of contraflexure at midpoint of chord members and that 40% of shear is carried by bottom chord member. At centre of opening, Mu = 171.6 kNm Vu = 79.2 kN

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1. Bottom chord member (Nu)b = Mu/z = 408.6 kN (Vu)b = 0.4 Vu = 31.7 kN (Mu)b =(Vu)bl/2 = 15.1 kNm 2As = 1954 mm2 (3T20 top + 4T20 bottom) Asv/s = 400 mm2/m (minimum) (M6 @ 100 mm spacing)

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2. Top chord member (Nu)t=408.6 kN; (Vu)t=47.5 kN; (Mu)t=22.6 kNm 2As = 516 mm2 (2T12 +1T10 top & bottom) Asv/s = 525 mm2/m (M6 @ 100 mm spacing)

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2T12 + 1T10 2T12 + 1T10

3T20

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Truss Model Approaches National University of Singapore

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Plasticity truss (AIJ approach)

As As

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Vu = 2 bdvsρvfyv cot φs

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where νfc’/ρ ρvfyv - 1) ≤ 2 cot φs = √(ν ρvfyv ≥ 1 νfc’/ρ ρv = Avfyv/bs ν = 0.7 - fc’/200 "

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Shear capacity

Longitudinal reinforcement Tsn = Asnfy = Vul/2dvs Tsf = Asffy = Vu(ll +dvs cot φs )/2dvs

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Vd = Asd fyd sin θd

Solution by Plasticity Truss Method (AIJ) National University of Singapore

1. Calculate v. v = Vu/(2bdvsνfc’)=0.089 2. Determine shear reinforcement. From v = √ψ(1-ψ √ψ ψ), obtain ψ ≡ ρvfyv/ν νfc’ = 0.008 → ρv = 0.00053 < ρv,min = 1/3fy = 0.01 Asv/s = 400 mm2/m (M6 @ 100 mm spacing)

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3. Determine longitudinal reinforcement. Asn = Vul/2fydvs = 957 mm2 Asf = Vu(ll +dvs cot φs )/2fydvs = 1080 mm2 ⇒ 3T20 (near) + 4T20 (far)

4T20 3T20

3T20 4T20

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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 of strut and tie forces • deformation capacity of concrete • angle between struts and ties

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2. Calculate forces in struts and ties from equilibrium. • •

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check stresses in struts (< 0.6fcd) detail reinforcement according to calculated tie forces check for anchorage of reinforcement

Solution by Strut-and-Tie Method 1. Postulate strut-and-tie model. National University of Singapore

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

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3. Longitudinal reinforcement. Top chord Top steel: 1298 mm2 (4T20) Bot. steel: 1741 mm2 (4T25) Bottom chord Top steel: 171 mm2 (2T12) Bot. steel: 590 mm2 (2T20)

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4. Transverse reinforcement. National University of Singapore

Top chord Asv/s= 1567 mm2/m (T8 @ 65 mm) Bottom chord Asv/s= 284 mm2/m (M6 @ 65 mm)

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CALCULATION OF DEFLECTIONS Rigid Abutment

It Stirrup ds

M-∆M

Hinges

V

V

M+∆M

Ib ∆d

P

P

le / 2

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le / 2

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δ δV Relative displacement of hinge w.r.t. one end of opening due to V (It based on gross concrete section; Ib based on a fully cracked section)

Relative displacement of one end of opening w.r.t. the other end

δv = 2δ =

3

 le  V   2 δ= 3 Ec ( It + Ib

)

V l3e

12 E c ( I t + I b )

Midspan deflection of beam

δ = δ w + (δ v A/P Tan K H, NUS

)opening 1 + (δ v )opening 2 + ...... 54

Calculate the midspan service load deflection of the beam described in the Example. SOLUTION Service load, Ps = Pu / 1.7 = 52.8 / 1.7 = 31.1 kN; V = 1.5 Ps = 46.6 kN. Effective length of chord members, le = (950 + 50) = 1000 mm. Assume It = Ig = 300 x 1803/12 = 146 x 106 mm4; Ib = 0.1 x 146 x 106 mm4 ≈ 15 x 106 mm4 Also, Ec = 4730√30 = 26 x 103 MPa Hence, δv = V le3 / [12 Ec( It + Ib)] = 46.6 x 1012 / [12 x 26 x (146 + 15) x 109] = 0.93 mm A/P Tan K H, NUS

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Midspan deflection, δw, of the beam is 11PL3/144EcI, where I can be conservatively estimated as the moment of inertia of the beam at a section through the opening. That is, I = 2 x [300 x 1803 / 12 + 300 x 180 x 2102] = 5054 x 106 mm4 As the span L is 6 m, therefore, δw =11x 31.1 x 103 x (6000)3 / [144 x 26 x 5054 x 109] = 3.91 mm Hence, the total midspan deflection is calculated as δ = δv + δw = 0.93 + 3.91 = 4.84 mm < L/360 = 16.7 mm A/P Tan K H, NUS

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Summary National University of Singapore

Deflection

Add deflection due to shear at each opening

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