Openings in Beams

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DEPARTMENT OF CIVIL ENGINEERING

In t h is le c t u re

We will explore Types of openings Behavi viou ourr of b bea eams ms wi with th larg large e !Beha 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


In t h is le c t u re

We will explore Types of openings Behavi viou ourr of b bea eams ms wi with th larg large e !Beha 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


At the e nd o f the le le c tur ture e

You should be able to Understand the be beha hav vio iour ur of be beam ams s 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 Aand /P Tan K H , NUS 3 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


Ad d iti tio o na l re f e re nc e s !

!

!

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


© Tan KH NUS

Opening in beam National University of Singapore

© Tan KH NUS


© Tan KH NUS

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

h

"


By structural response: – Small opening if

Beam-type behaviour persists.

– Large opening otherwise.

© Tan KH NUS


TS

© Tan KH NUS

T2


© Tan KH NUS

DESIGN TOOLS National University of Singapore

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"

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© Tan KH NUS

Finite element analysis: • elastic stress state • area of stress concentration Truss analysis (Strut-and-tie model): • lower bound approach • detailing for crack control and strength Tests: • specific cases • serviceability & strength aspects


© Tan KH NUS

BEHAVIOUR of BEAMS WITH LARGE OPENINGS


Principal tension stress contours

© Tan KH NUS


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

© Tan KH NUS


Inadequately reinforced openings © Tan KH NUS

actual National University of Singapore

1

2

idealised

1 3

© Tan KH NUS

2 4

3

4


© Tan KH NUS


© Tan KH NUS


© Tan KH NUS


© Tan KH NUS


© Tan KH NUS


Shear carried by chords © Tan KH NUS


Contraflexural points © Tan KH NUS


DESIGN & DETAILING of OPENINGS - General guidelines •

•

• •

© Tan KH NUS

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


© Tan KH NUS

" National University of Singapore

© Tan KH NUS

Crack control

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


© Tan KH NUS

2

⇒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

© Tan KH NUS

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.

© Tan KH NUS

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

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

Slide 15

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


© Tan KH NUS


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

© Tan KH NUS

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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

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

or

k = Ib/It

EXAMPLE National University of Singapore

© Tan KH NUS

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

© Tan KH NUS


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)

© Tan KH NUS


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)

© Tan KH NUS


2T12 + 1T10 2T12 + 1T10

3T20

© Tan KH NUS

Truss Model Approaches National University of Singapore

© Tan KH NUS

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Plasticity truss (AIJ approach)

As As

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Vu = 2 bdvsρvfyv cot φs


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

© Tan KH NUS

Shear capacity

Longitudinal reinforcement

Tsn = Asnfy = Vul/2dvs Tsf = Asffy = Vu(l +dvs cot φs )/2dvs


© Tan KH NUS

Vd = Asd fyd sin θd

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

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

© Tan KH NUS


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

4T20 3T20

3T20 4T20

© Tan KH NUS

" National University of Singapore

© Tan KH NUS

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


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

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© Tan KH NUS

check stresses in struts (< 0.6f cd ) 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

© Tan KH NUS


© Tan KH NUS

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


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)

© Tan KH NUS

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)

© Tan KH NUS


© Tan KH NUS

C ALC ULATIO N O F DEFLEC TIO NS Rigid Abutment

It Stirrup ds

M-∆M

Hinges

V

V

M+∆M

Ib ∆d

P

P l e / 2

A/P Tan K H, NUS

l e / 2

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

δ =

 l e  V    2 

3

3 E c ( I t + I b )

Relative displacement V l3e of one end of opening δ v = 2 δ = 12 E c ( I t + I b ) w.r.t. the other end

Midspan deflection of beam

= δ w + (δ v )opening 1 + (δ v )opening 2 + ...... A/P Tan K H, NUS

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Calculate the midspan service load deflection of 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 of chord members, le = (950 + 50) = 1000 mm. Assume I t = I g = 300 x 1803 /12 = 146 x 106 mm4; I b = 0.1 x 146 x 106 mm4 ≈ 15 x 106 mm4 Also, E c = 4730√30 = 26 x 103 MPa Hence, δ v = V le3 / [12 E c ( I t + I b )] = 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, δ w , of the beam is 11PL3 /144E c I , 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, δ w =11x 31.1 x 103 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 = 0.93 + 3.91 = 4.84 mm < L /360 = 16.7 mm A/P Tan K H, NUS

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Openings in Beams

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