worked example ashby part 2 vfinal - concrete society

Double Tee Section

Qualitative Assessment of your Floor before you Start

ASSESSMENT OF EXISTING PRECAST CONCRETE FLOORS

This example is for a Hollowcore floor building – but the same principles apply for all precast floors including Double Tees, Rib and Infill, and Flat Slab assessments.

Step 1 - Qualitative Review• Before you start your detailed

capacity calculations • Carry out a qualitative review of the

floor and potential damage areas obtained from your building analysis

• Check the drawings and details for what information you need to get from the site investigation

Step 2 - Visit the building• Review the parameters and details

on site. • Site measure and audit against

drawings• Is there some additional invasive

investigation needed• The better the information the more

accurate the resultStep 3 Detailed Check• Now start your detailed calculations


Design Drawings and Details

Double Tee Example – Ductile Frame Example Parameters

• Ductile frame building with 350 deep 2400 module Double Tee floors

• Beam span = 10,500 mm

• Beam depth = 800 mm

• Column width = 800 mm

• Floor unit seated 285 mm above the beam centreline

• D12-300 starter bars

• Elastic drift = 0.6%

• Detailed example calculations provided in the Appendix


le=

3600

mm

D12-300600 lap

D12-300600 lap

500W

x 6

00D

500W x 800D

500W

x 8

00D

10500mm (typ.)

U R

350 Double Tee75mm topping

665 mesh75mm seating

Double Tee Example – Required checks

Scenarios to check:

A. Unit adjacent corner columns

• Check using unrestrained hinge (U)

B. Unit adjacent elongating beam, away from corners

• Check using restrained hinge (R)

C. Internal units away from elongating beam


Note: Highlighted Units – the topping delaminates so is a special case A or B to consider

Double Tee Example – Required checks

Checks for each scenario:

1. Loss of support

– Spalling

– Elongation2. Birds Mouth Failure

– Review connection load transfer mechanism and probable failure modes of connection due to drift.

– Strut and Tie check of flexural capacity and failure modes

– Details on the assessment of the failure modes of flange hung double tees units is provided in Hare et al. (2009)


Loss of Support Example

Double Tee Floor Example – Loss of Support Review



• Tee Unit adjacent corner column

• Initial seating = 75 mm

• Construction tolerance = 20 mm

• Initial spalling = 10 mm

• Bearing (calc = 8mm) = 10 mm

• Therefore remaining seating to permit elongation, rotation + further spalling:

75 – 20 – 10 – 10 = 35 mm


Construction tolerance (portion of)

Initial spalling (example)

Drift related spalling (example)

Elongation + Rotation

Double Tee Unit

SUPPORT BEAM



Calculation of Elongation

Plastic beam rotation

θp = θpcol * L / (L - hcol - Lp)

= (1.39-0.6)*10500/(10500-800-335)

= 0.89%

Beam elongation

del = 2.6 * θp/2 * (d - d’) ≤ 0.036hb

= 2.6*0.0089/2*(800-65)

(U) = 8.5 mm => 9 mm ≥ 0.005hb

(R) = 4.2 mm => 4 mm

(For reversing plastic hinges)

(Restrained taken as ½ of reversing)

(Lp = 0.5*d/2)

Double Tee Example – 1. Loss of support review

Calculation of Beam Rotation

Similar to the hollowcore design example the support rotation is the maximum {S del + dr1 or dr2 + del unit}

qbeam = 1.39%

dr1 = ((hb/2)-hl)*θbeam

= (800/2-(115))*0.0139

= 4.2 mm

dr2 = hlθbeam

=115*0.0139

=1.4 mm



How is the spalling calculated ?

• Initial Spalling = 10 mm

+ Additional spalling at

limiting drift of 1.39%

• Unit (27 -10) = 17 mm

• Ledge = 5 mm

• Total spalling = 32 mm


L.O.S = 1.39%

5mm

27mm

10mm

Figure C5E.26 Spalling depths to be considered for Flange hung and web supported double tees

COVER TO FIRST BAR 25mm + 10mm

COVER TO FIRST BAR XX mm + 10mm


Through iteration 35mm of seating

is exceeded at a total inter-storey

drift of 1.4% components of loss

• Unit Spalling = 17 mm

• Ledge Spalling = 5 mm

• Beam elongation = 9 mm

• Beam rotation = 4 mm

35 mm


L.O.S = 1.4%

Initial seating = 75mm

Construction Tolerance = 20mm

Bearing = 10 mm

Initial Spalling = 10 mm


Through iteration 45mm of seating

is exceeded at a total inter-storey

drift of 1.7% components of loss

• Unit Spalling = 17 mm

• Ledge Spalling = 21 mm

• Beam elongation = 11 mm

• Beam rotation = 5 mm

45 mm

With suitable site investigation 20mm tolerance may be able to be reduced to actual site tolerance

e.g. for the above example with a site measured 10mm variance on seating (75mm +/- 10mm)


L.O.S = 1.7%

Initial seating = 75mm

Construction Tolerance = 20mm

Bearing = 10 mm

Initial Spalling = 10 mm

Extra investigation – tolerance reduced



Now repeat the L.O.S for rest of the floor

• The L.O.S support calculation is easily put into a spreadsheet and once set up with the project specific geometry and spalling parameters can be repeated for the rest of the floor and levels relatively quickly


Results for L.O.S for Building

Summary

Results repeated for the other units gives;

Case A Adjacent Unrestrained Hinge

• L.O.S = 1.4% drift

Case B Adjacent Restrained Hinge

• L.O.S = 1.5% drift

Case C Internal Unit

• L.O.S. = 1.7% drift

These are the limiting drifts for the loss of support due to spalling or elongation

Now you need to investigate the Birdsmouth Support detail and possible failures to see if these are critical for limiting drift*Note: Highlighted Units – the topping delaminates so is a special case A or B


A*

Double Tee Example – Loss of Support NBS Rating


How to use Limiting Drift for Rating of %NBSULS design drifts are factored by 2.0 in the Yellow Book to get drift demand

[Note: this is different to NZBC = dULS x 1/SP x 1.5]

Building Drift Demand dULS x 2 = 3.2%

Limiting Drift Calculated = 1.4%

Correct Rating Method Limiting Drift / Building Drift Demand

1.4 / 3.2 = 44% NBS

Incorrect Rating MethodSeating Provided / Seating Required

75 / 132 = 57% NBS

The NBS rating is based on a limiting drift to loss of support i.e. the methodology assesses the amount of drift required before loss of precast support – this is a bi-linear function and gives a different answer than simply comparing against the total seating that would be required for design.

Figure: Total required seating at calculated demand building drift using C5E Methodology

Total required seating at calculated building drift demand 3.2%

Summary

Elongation and Rotation 37.8 mmUnit Spalling 35.0 mmLedge Spalling 29.5 mmShrinkage 0.0 mmConstruction Tolerance 20.0 mmBearing Length 10.0 mmTotal 132.3 mm

TOTAL SEATING REQUIRED 132 mmAS PER C5 GUIDELINES

Seating Provided 75 mm


Effect of L.O.S Due to Spalling and Elongation


No Spalling and After Spalling

Birdsmouth / Loop Bar Review

Double Tee Example – Birdsmouth / Loop Bar


350x2400 Double Tee Example

• Span = 10000mm Unit Width = 2400mm

• Topping = 75mm f’c = 20MPa => f’cp = 30 MPa

• Double Tee f’c unit = 45MPa

• “Pigtail” Loop Bar Detail – 3 R12 Bars / fy = 275 MPa => fyp = 324MPa

Loading

Gravity [G unit = 2.5 KPa] + [Topping = 1.8 KPa] + [SDL = 0.5KPa] = 4.8KPa

Live Load = Q= 3.0 KPa , YE = 0.3

Total G + YE Q + Eu = 5.7 KPa

Load per Unit = 5.7*2.4*10 = 136 kN

Load per web / Pigtail = 136/4 = 34 kN



Loop Bar Hanger damage being investigated post Christchurch EQ



Possible Failure Modes

There are a number of possible failure modes / crack patterns that can occur at the birdsmouthconnection.

The failure in fig C5E 29 is a combination of – a, b, c, d, e & f. and the designer needs to be satisfied that they have investigated all likely modes in the failure analysis.

Note: For the double tee supported in the elongation zone (immediately adjacent column) topping delamination is likely to occur and the check is for the precast portion (t-ttopping) only

in fig C5E.29C5E 29 Example demonstrated



Tension Shift

e’ mm tension shift



Elongation Demand:

Solve for d to find limiting drift

M* = P (d + e + e’) + mPt < MBM.capacity

Friction Static (typical ranges)mf= 1.0 - 1.2 concrete to concrete

mf= 0.8 - 1.0 on soft mortar

mf= 0.8 - 1.0 on steel

mf= 0.6 - 1.0 on bearing strip

Note: Be careful some published values of m are lower bound or dependable values i.e. they have a f.o.s or reduction factor for design built in and may be un-conservative for calculation of demand.

75mm seating15mm gap10mm min seating



Scenario Review a) Flexure

a) Yes - check required

b) Topping delamination a) at loop bar – Yes likely - check

b) In unit – Yes - possible as the R6 spirals are unlikely to be sufficient

c) Shear Failurea) Yes - check required

d) Diagonal tension failurea) Yes - check required

e) Bond failure a) At Leg – <600mm anchorage for

plain R12 bar in tension – to short to develop

b) At bottom hangar bar – ? may be OK

f) Separation of Flange from Weba) 4 sets of R6 Stirups at 50mm crs at

end of unit so failure is likely suppressed but should check

Qualitative Review of Detailing

C5E 29 Example demonstrated

30O

6590

75

50



Elongation Demand:

Solve for d => to find limiting drift

Iteration 1 => try max d = (75/2-10) = 27.5mm

Design Actions for Review

P* = 34 kN (G + Qe + E)

P*m = 27 kN (assuming m=0.8)

M* = P (d + e + e’) + mPt

= 34*[(75/2-10)+(75/2+15)+65]mm + 0.8*34*125mm

= 8.3kNm (at max d of 27.5mm)

Iteration 2 => try min d = 0mm

M* = 7.4 kNm (at d of 0mm) i.e. At initiation of movement / crack

75

P



But why can’t I just use simple flexural theory?Solve for limiting d - take first trial as max = 27mm

fMn = fAsfy (d-a/2) f = 0.85

= 11.0 kNm

M* = P (d + e + e’) + mPt

= 8.3kNm < 11.0 kNm => OK

T*N = P / Cos qh

= 34 /Cos 60o

= 68kN < 110 kN => OK

This check may be simple but is un-conservative and does not consider the actual failure modes of the connection. The birds mouth is a disturbed region and you must use an appropriate method i.e. Strut and Tie to assess the failure modes.

Further details on the assessment of the failure modes is in Hare et al. (2009)

75mm seating15mm gap10mm min seating

65 mm tension shift => e’

q =60o

qh =60o

30O

6590

75

50



Review the Flexural MechanismHorizontal Reinforcing Tie

To develop Tjd of 11kNm the tension steel yeilds

=> for 3 R12 T* = C = Asfy = 110 kN

Compression Strut Check

Strut / Node Force = 110/Cos30 = 127 kN

Node width = 200mm x 20mm (Pigtail bar width)

Node Capacity

fFn = f*bn*f’c = 0.75*0.80*200*20*30 = 72kN < 127kN

=>Capacity Ratio = 0.56

Strut Capacity

fFs = f*bs*f’c = 0.75*1.0*200*20*30 = 90kN => Ratio = 0.70

Therefore node failure and concrete crushing may occur at loop bar node before you can fully develop tension steel capacity required for flexure theory.

P

T = ASfy

CC

ConcreteTension

30O

6590

75

50



Note: Caution Needed - Concrete Tension Capacity - NZBCUnreinforced Tie Check

Not permitted by B1 / VM1 - NZS3101 Cl 2.3.2.3

Equations for Concrete Tension – Use with Caution!

Tension capacity in slab

(NZS3101 5.2.4) gives ft = 0.38 l [f’c]1/2 = f x 2.08 MPa (for 30MPa)

Cold joint interface - for tension across slab/unit interface

(ACI) = 0.25 [f’c]1/2 = f x 1.36 MPa

P

T

C

ConcreteTension



Review the Flexural Mechanism Review slab in tension

Tension tie required = 127*Sin30 = 63.5kN

Assume 50mm at node x 200mm wide

Tie capacity = f * f’t*At = 15.6 kN << 63.5kN

Ratio = 0.25 …….

Iterating gives a max capacity = 22kN / 0.35

Slab Interface capacity = 10.2 kN << 63.5 kN

Ratio = 0.16…….

Iterating gives a max capacity = 23kN / 0.36

By iteration you can find S&T mechanisms which may give better results than this initial review example. A key consideration is the capacity of the tie is limited by the geometry of the joint including the pigtail compression node width + amount of spread of strut of 1:2.5 max.

Further details on the assessment of the double tee failure modes is in Hare et al. (2009)

Slab Interface

Simplified S&T Model

Reinforcing for Pigtail

Tension tie

Other failure mode?


Review the results for failure mode a) Flexure for this example

• Flexural Capacity derived by S&T < 40% of that provided by simple Flexure theory fMi= Tjd=>this is quite sensitive to geometry of loop bar and joint.

• Large increase in design actions due to tension from elongation Pmt => actions almost double – also quite sensitive to support load and friction coefficient chosen

• Tension Shift – once concrete cracks also accounts for another shift approximately additional 30% - 40% increase

• Offset of support due to elongation has less affect and accounts for about another 10% to 15% increase in load




Ideal Flexural Capacity fMi = 11kNm

MoR Flange = 5.5kNm

S&T Capacity = 4.5 kNm

2.7 kNm

27 mm Limit before L.O.S governs

1.8 kNm

7.1 kNm

8.3 kNmCrack may form due to tension Pmt from elongation

6.1 kNm

4.9 kNm

Birdsmouth / Loop Bar – Demand / Capacity Envelope

kNm

mm



Review - OutcomeStrut and Tie Review of Mechanism (a)

• Review of the flexural mechanism with an initial S&T model gives a lower bound limiting drift sensitive to cracking with d = <5 mm for the birdsmouth after onset of cracking.

• The S&T review indicated however that the unit it may be able to (just) carry gravity loads after elongation has finished i.e. (Pm loads from elongation no longer action).

• Could investigate geometry further and sliding capacity – important to understand site constraints and geometry of loop bar and joint

Other Failures – mechanism (b), (c), (d), (e), etc

• Need to now complete the review and check other modes to determine critical mode of failure has been determined.

The floor %NBS in this example (a) Flexure will be difficult to rate as a result of the birdsmouth / loop bar connection having insufficient capacity to accept movement that would result in a crack forming and tension (from elongation) across the birdsmouth connection. A detailed further investigation of the loop bar details should be undertaken to help determine the actual as built capacity.



Scenario Review a) Flexure

a) Completed – Probably OK

b) Topping delamination a) at loop bar – possible - check

b) In unit – possible as the R6 spirals are unlikely to be sufficient

c) Shear Failurea) OK

d) Diagonal tension failurea) Yes – further check required

e) Bond failure a) At Leg – <600mm anchorage for

plain R12 bar in tension – Ld x 2 = 610mm ~ 570 => OK

b) At bottom hangar bar – Ldh OK

f) Separation of Flange from Weba) 4 sets of R6 Stirups at 50mm crs

b) Capacity > 34kN => OK


C5E 29 Example demonstrated



Review of Double Tees - SummaryLoss of Support – Elongation

• Likely to govern most situations for most 1980s and early 90s buildings which have typical seating detailed between 30mm to 50mm.

• Failures observed to date Clarendon / Statistics were loss of support.

Birdsmouth Failure

• Likely to govern all buildings at PHZ where delamination of topping occurs

• Likely to govern newer buildings mid 1990s onwards using the loop bar detail that are provided with a more generous seating of 75 mm or more.

• Be aware of double tees that have the flange trimmed back locally (more common on steel beam support) with no distributed flange bearing means redundancy is reduced.

• Be aware of long span double tees – high shear load at the birds mouth is problematic

• It can be difficult to assess the capacity and point of failure of the failure of those modes relying on concrete in tension. Need to undertake a rational approach and look at likely scenarios.

• Hare et al 2009 covers this topic very well.

• No direct failures– but lots of in service evidence that birdsmouth cracking mechanisms are occurring even under service loads so may be problematic with the right earthquake conditions.



What happens at the Plastic hinge regions and corners ?

Topping Delaminates

tf

Section capacity is based solely on the double tee flange depth tf for these areas

Delamination of topping

4040ASSESSMENT OF EXISTING PRECAST CONCRETE FLOORS

Can not hang the unit from the topping

Concrete Tension at Interface(ACI) = 0.25 [f’c]1/2 = 1.36 MPaCapacity 10,000 x 2,400 x 1.36 MPa = 32,640 kNDouble Tee Weight10,000 x 2,400 x 2.4 kPa = 60 kN

F.O.S = > 500

Why did this fail ?

Not permitted by B1 / VM1 - NZS3101 Cl 2.3.2.3

Assess Birdsmouth and other failure mechanisms and using concrete tension with caution

Re-Cast Project => programmed to look at this loop bar issue in existing buildings in more detail with detailed FEM solid modelling and lab tests

Precast Rib and Infill

Failure of Rib and Timber Infill Floors

• Possible rib entrapment under positive moments – Casting of ribs into the beam

– Haunching final vertical form Seating on mortar

• Weak section forms along rib

• Follow

42


Assessment of Rib and Timber Infill Floors

• Similar Process to hollowcoreexample

• Checking for;

• Spalling and loss of support due to elongation

• Displacement / drift induced design actions on the precast member and floor

Flat Slab Floors

Assessment of Flat Slab Floors


• Similar Process to hollowcoreexample

• Checking for;

• Spalling and loss of support due to elongation

• Displacement / drift induced design actions on the precast member and floor

Questions

worked example ashby part 2 vfinal - concrete society

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