Assoc Prof Tan Kiang HweeDepartment of Civil EngineeringNational University of Singapore
CE5510 Advanced Structural Concrete Design
- Design & Detailing of Openings
in RC Flexural Members-
DEPARTMENT OF CIVIL ENGINEERING
Tan K H, NUSA/P Tan K H, NUS A/P Tan K H, NUS 2
In this lecture
We will explore
!Types of openings!Behaviour of beams with large
openings!Design for ultimate strength!Crack Control!Deflection Calculation
DEPARTMENT OF CIVIL ENGINEERING
Tan K H, NUSA/P Tan K H, NUS A/P Tan K H, NUS 3
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 serviceability and ultimate limit states
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Beams with Openings: and Design.Mansur, M.A. and Kiang-Hwee Tan.CRC Press Ltd, 1999
DEPARTMENT OF CIVIL ENGINEERING
Tan K H, NUSA/P Tan K H, NUS A/P Tan K H, NUS 5
Additional references
! 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, July-August 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.
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Opening in beam
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Classification of Openings" By geometry:
– Small opening ifDepth or diameter ≤ 0.25 x beam depthand Length ≤ depth
– Large opening otherwise.
h≤ 0.25hd ≤ 0.25h
≤ d
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" By structural response:
– Small opening ifBeam-type behaviour persists.
– Large opening otherwise.
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TS T2
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DESIGN TOOLS" 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
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BEHAVIOUR ofBEAMS WITH LARGE OPENINGS
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Principaltensionstress
contours
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Mt +Mb + Nz = M= MNt + Nb = 0Vt + Vb = V
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Inadequately reinforced openings
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actual
idealised
33
2
2
1
1
44
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Shear carried by chords
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Contra-flexural points
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DESIGN & DETAILINGof OPENINGS - General guidelines
• sufficient area to develop ultimate compression block in flexure
• openings to be located more than D/2from 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
2
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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
" 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
Nu = [Mu - (Ms)t - (Ms)b]/z
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.
Slide 15
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4. Calculate Vu in chord members from
(Vu)t = [(Mu)t,2 - (Mu)t,1]/llll(Vu)b = [(Mu)b,4 - (Mu)b,3]/llll
where llll 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 = 0Vt + Vb = V
⇒ Assume Mt = 0Mb = 0Vb =kVt
where k = 0k = Ab/At
or k = Ib/It
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EXAMPLEA 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.
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Solution by Mechanism Approach
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 kNmVu = 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)bllll/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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3T20
2T12 + 1T10
2T12 + 1T10
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Truss Model Approaches
" Plasticity truss (AIJ approach)
As
As
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" Shear capacity
Vu = 2 bdvsρρρρvfyv cot φφφφswhere
cot φφφφs = √√√√(ννννfc’/ρρρρvfyv - 1) ≤≤≤≤ 2ννννfc’/ρρρρvfyv ≥≥≥≥ 1ρρρρv = Avfyv/bsνννν = 0.7 - fc’/200
" Longitudinal reinforcement
Tsn = Asnfy = Vullll/2dvsTsf = Asffy = Vu(llll +dvs cot φφφφs )/2dvs
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Vd = Asd fyd sin θθθθd
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Solution by Plasticity Truss Method (AIJ)
1. Calculate v.v = Vu/(2bdvsννννfc’)=0.089
2. Determine shear reinforcement.From v = √ψ√ψ√ψ√ψ(1-ψψψψ), obtain
ψψψψ ≡≡≡≡ ρρρρvfyv/ννννfc’ = 0.008 →→→→ ρρρρv = 0.00053 < ρρρρv,min = 1/3fy = 0.01
Asv/s = 400 mm2/m(M6 @ 100 mm spacing)
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3. Determine longitudinal reinforcement.Asn = Vullll/2fydvs = 957 mm2
Asf = Vu(llll +dvs cot φφφφs )/2fydvs= 1080 mm2
⇒⇒⇒⇒ 3T20 (near) + 4T20 (far)
3T20
3T20
4T20
4T20
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" Strut-and-tie methodGeneral Principles
1. Idealize structure as comprising concrete struts and reinforcement ties, joined at nodes.
• follow stress path given by elastictheory; 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 fromequilibrium.
• check stresses in struts (< 0.6fcd)• detail reinforcement according to
calculated tie forces" check for anchorage of
reinforcement
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Solution by Strut-and-Tie Method1. Postulate strut-and-tie model.
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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 chordTop steel: 1298 mm2 (4T20)Bot. steel: 1741 mm2 (4T25)
Bottom chordTop steel: 171 mm2 (2T12)Bot. steel: 590 mm2 (2T20)
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4. Transverse reinforcement.
Top chordAsv/s= 1567 mm2/m (T8 @ 65 mm)
Bottom chordAsv/s= 284 mm2/m (M6 @ 65 mm)
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A/P Tan K H, NUS 53
CALCULATION OF DEFLECTIONS
It
Ib
RigidAbutment
M-∆M M+∆MV V
∆d
ds
P P
Hinges
Stirrup
2/el 2/el
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( ) I I E3
V
btc
e
+
=
3
2l
δ
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)
( ) ( ) ......21 +++= openingvopeningvw δδδδMidspan deflection of beam
( )I I E V
btc
ev +
==12
23lδδ
Relative displacement of one end of opening w.r.t. the other end
δVδ
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Calculate the midspan service load deflection of the beam described in the Example.
SOLUTIONService 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 le
3 / [12 Ec( It + Ib)] = 46.6 x 1012 / [12 x 26 x (146 + 15) x 109]= 0.93 mm
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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 asδ = δv + δw = 0.93 + 3.91 = 4.84 mm
< L/360 = 16.7 mm
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Add deflection due to shear
at each opening
Deflection
Summary