HOLLOWCORE WORKED EXAMPLE - Concrete Society

1 ASSESSMENT OF EXISTING PRECAST

CONCRETE FLOORS Hollowcore Worked Example

HOLLOWCORE WORKED EXAMPLE

The following worked example uses the November 2018 draft C5 Appendix C5E to assess a

1981 Wellington Hollowcore floor.

The following assumptions have been made:

Hollowcore self-weight + topping G 4 kPa

Floor superimposed dead load SDL 0.5 kPa

Floor live load Q 3 kPa

Elastic storey drift at system yield δe 0.60% (from C5 Assessment)

Beam concrete probable compressive strength f'c 30 MPa

Hollowcore concrete compressive strength f’c 42 MPa

Hollowcore concrete elastic modulus Ec 30460 MPa

Hollowcore section area AHC 0.101 m2

Mild streel reinforcement yield strength fy 324 MPa

Mesh reinforcement “yield” strength fy 600 MPa

Beam distance to CL longitudinal reinforcement d’ 65 mm

Beam plastic hinge length Lp,b 335 mm

Initial Hollowcore prestressing force Fps 852 kN

le =

36

00

mm

D12-300

600 lap

D12-300

600 lap

500

W x

600

D

500W x 800D

500

W x

800

D

8000mm (typ.)

U

R

200 Hollowcore

65mm topping

665 mesh

50mm seating

6

70

mm

550mm

53

5m

m



In accordance with Appendix C5E, a number of locations and scenarios will need to be

assessed, even for this comparatively simplistic example. These include the following:

A. Units adjacent corner columns (inside elongation affected length le)

A.1 Loss of support (unseating) check at free end using unrestrained elongation (U)

A.2 Negative moment check at the end of the starter bars

A.3 Positive moment check noting potential for plan diagonal cracking

B. Units adjacent elongating beam away from corners (inside elongation affected length le)

B.1 Loss of support (unseating) check using restrained elongation (R)

B.2 Negative moment check at the end of the starter bars

B.3 Positive moment check using restrained elongation

C. Internal units away from elongating beam (outside elongation affected length le)

C.1 Loss of support (unseating) check using starter bar elongation

C.2 Negative moment check at the end of the starter bars

C.3 Positive moment check using starter bar elongation

This example will look at ‘Type A’ floor units that are within the plastic hinge elongation

affected zone adjacent the perimeter seismic frame, but outside of the predominant corner

cracking region. The length of this elongation affected zone is called the “elongation

length”, le, and is considered to be half of the beam span when adjacent to an unrestrained

(U) plastic hinge.

In real buildings a number of units will need to be checked throughout the floor plate due

to varying beam spans, unit spans, starter bar conditions, number/location of plastic hinges,

seismic system types and deflection profiles etc.

Note that even when a building does not have elongating hinges, elongation and rotation in

accordance with ‘Type C’ units will need to be assessed.

A.

C.

B.

le

U R R

le



A. UNITS ADJACENT CORNER COLUMNS

A.1 Loss of support (C5E.5.2)

Loss of support need not be checked if the hollows at the supports are reinforced in

accordance with the detail provided within NZS 3101:2006 A3. Because this is a 1980s

building its detailing does not satisfy this provision.

Loss of seating needs to allow for the following movements and allowances:

Initial seating reduction due to:

- Construction tolerance of 20mm unless comprehensively measured

- Initial spalling of 15mm, typically to back face of unit

- Required bearing with a minimum of 5mm

Further seating reduction and unit movement due to:

- Spalling of seating due to rotation of unit

- Sliding under rotation of unit

- Sliding under elongation of either adjacent beam hinge or starter bar hinge at

end of unit

Two things should be clarified in relation to the illustration above. Construction tolerance

results from a number of factors including beam face alignment, beam position, ledge

width, unit length, unit face alignment and unit position. This might be reduced by taking

onsite measurements using destructive testing; however, an intensive testing regime is

required due to the consequence of error. Initial spalling is considered to occur regardless

of face armouring and drift related spalling can occur in the seating and/or at the rear face

of the unit. Ledge armouring, therefore, has no effect on spalling reduction calculated in

accordance with the Appendix.

Construction tolerance

(portion of)

Initial spalling

(example)

Drift related spalling (example)

Elongation +

Rotation



Through spread sheet iteration of the formulas below, it can be shown that the 50mm

seating can tolerate a total inter-storey drift of 1.15% as follows:

Construction tolerance = 20 mm

Total spalling = 15 mm + 3 mm

= 18 mm

Load on seating w* = (G + SDL)*LHC / 2

= (4.2+0.5)*(8-0.5)/2

= 18 kN/m

Ava

ilab

le s

eati

ng

(mm

)

Interstorey Drift (%)

Initial seating = 50mm

Construction tolerance = -20mm

Initial spalling = -15mm

Additional spalling with drift

Seating required to permit hinge

elongation and support rotation

LOS = 1.15%

Elastic elongation

Bearing = -5mm



Bearing length required Lbear = w*/(0.85 f’c) (NZS3101 cl16.3)

= 18*(1200/885)/(0.85*30)

= 1 mm < 5 mm

= 5 mm

Remaining seating capacity = 50 – 20 – 18 – 5

= 7 mm

Frame elongation

𝛿𝑒𝑙 = 2.6𝜃𝑝

2(𝑑 − 𝑑′) ≤ 0.036ℎ𝑏 (Eqn C5E.1)

Plastic beam rotation θp = θpcol

* L / (L - hcol - Lp)

= (1.15-0.6)*8000/(8000-800-335)

= 0.6%

Beam elongation δel = 2.6*0.006/2*(800-65*2)

= 5 mm > 0.5hb

Beam rotation

𝛿𝑟1 = (ℎ𝑏

2 − ℎ𝐿)𝜃 (Eqn C5E.3a)

Support rotation θ = 1.15%

δr1 = (800/2-(200+65))*0.0115

= 2 mm

Seating demand = 5 + 2

= 7 mm

Therefore, the total seating demand at an interstorey drift of 1.15% is:

= 20 + 18 + 5 + 2 + 5

= 50 mm

= Total seating

Therefore, loss of support (LOS) of the Hollowcore floor unit adjacent the seismic frame is Loss of support

assessed to occur at an inter-storey drift of 1.15%. Max drift = 1.15%



Note that a check that rotation and elongation of the unit itself without frame elongation

should be completed to ensure this is not a critical mechanism. This check is critical for

“Type C” units, which are assessed to not experience the effects of beam elongation.

Check at the previous limiting drift of 1.15% to prove that the mechanism is not critical:

Unit elongation

𝛿𝑒𝑙_𝑢𝑛𝑖𝑡 = 1.3𝜃𝑝

2(ℎ𝐿 − 𝑑′) ≤ 0.018ℎ𝐿 (Eqn C5E.4)

Unit rotation θp = (δ – δe)* L / (L – 2s)

= (1.15-0.6)*7500/(7500-2*300)

= 0.6%

Unit elongation δel_unit = 1.3*0.006/2*(265-30)

= 1 mm

Beam rotation

𝛿𝑟2 = ℎ𝐿𝜃 (Eqn C5E.3b)

δr2 = 265*0.0115

= 3 mm

δr2 + δel_unit = 4 mm < 7 mm

Therefore, unit elongation mechanism is not critical over frame elongation.



A.2 Negative moment (C5E.5.3)

Two negative moment scenarios exist within the elongation affected zone and both need to

be assessed to determine the governing loading condition.

The first scenario is a negative moment induced at the support due to frame elongation

causing the starter bar above the Hollowcore unit to yield. The eccentricity of the starter

bar about the section results in a constant negative moment applied throughout the length

of the unit.

The first loading condition only requires a check of the moment demand to capacity at the

end of the starter bars.

Loading condition i

With the second scenario, negative moment may be induced at the support as a result of

rotation causing the back face of the unit to compress against the support beam creating a

resistance couple with the starter bars. This can result in the formation of a mechanism

which will either take the form of the opening of the back face of the unit and yielding of

the starter bar, or a flexural shear mechanism propagating from the end of the starter bars.

The critical mechanism will depend on the relative ratios of moment demand to capacity at

the two sections.

The second loading condition requires a check of the moment demand to capacity both at

the support of the hollowcore and at the end of the starter bars.

Loading condition ii

Loading condition one considers the case of tension and rotation within the plastic hinge

elongation affected length while loading condition two considers the effects of rotation,

anywhere in the building. Both scenarios need to be assessed within the elongation

affected length.



A.2.1 Loading condition i

As the adjacent beam elongates a perpendicular crack is opened at the end of the floor,

causing the starter bars (continuity reinforcement) to yield in tension. The eccentricity of

these bars about the centroid of the floor section results in a constant negative bending

demand over the length of the floor.

When the gravity demand is draped below the line of constant bending, it can be compared

to the bending capacity to assess whether the critical section occurs at the end of the unit

or within the unit. If the bending strength of the mesh plus the prestressing strands is much

lower than the strength of the starter bars alone, negative moment failure may be induced

within the span of the floor:

For this example, the following Hollowcore construction properties will be assumed:

HC elastic modulus Ec 30460 MPa

Prestressed strand elastic modulus Eps 200000 MPa

Stress in strand after losses fpsi 967 MPa

No. of strands Nse 7

Strand area Ap 93 mm2

Height of strand from base ht 35 mm

Moment capacity

Moment demand = M*G - M*

N

M*N = A

sf

oe

600mm



Loading condition i - Negative moment capacity at end of starter bars

𝑆𝑐𝑓 = 4 (ℎ

375) (Eqn C5E.6)

Strain concentration factor Scf = 4*(200+65)/375

= 2.83

ε𝑦𝑚 = 0.002 + 𝑓𝑦𝑝𝑚

𝐸𝑚 (Eqn C5E.7)

Mesh yield strain εym/Scf = (0.002 + 600/200000)/2.83

= 0.00177

Knowing the strain within the mesh, the stress and strain diagrams may be used to iterate

for the neutral axis position, therefore giving the bending capacity of the section. A

spreadsheet or section analysis tool may be used to find the bending capacity.

𝑓𝑝 = 𝑓𝑠𝑒 − ε𝑝𝐸𝑝𝑠 (Eqn C5E.8)

fp = 967 – 0.00021*200000

= 925 MPa

fc = εc*Ec

= 0.00056*30460

= 17 MPa

Mesh contribution Mym = 23 kNm

Prestress contribution Mps = 11 kNm

Axial tension force N = Asfo

= 113*1.25*324*1200/300* 10-3

= 183 kN

Axial load contribution M = -22 kNm

Therefore, the total negative moment capacity at the end of the starter bars is:

Mn_span = 23 + 11 – 22

= 12 kNm



To assess the moment demand at this location, the constant moment due to the axial

tension imparted by the yielding starter bars must be calculated.

M*N = N.e

= 183 * [(265-140-65/2]* 10-3

= 17 kNm

M*G = (4+0.5)*1200*600/2*(7500-600)* 10-9

= 11 kNm

Therefore M*_span = M*N – M*G

= 17 – 11

= 6 kNm

Mn_span > M*_span

12 kNm > 6 kNm

The moment capacity within the span is greater than the moment demand; therefore,

loading condition one is not assessed to result in negative moment failure.

Ybl

e

N

Mo

me

nt

cap

acit

y (k

Nm

)

Distance (m)

Mn = 12kNm

M* = 6kNm



A.2.2 Loading condition ii

Loading condition two is similar to loading condition one, except that it considers the case

of the starter bars yielding due to rotation induced bending rather than tension from

outside of the floor system.

Assuming that positive moment resistance is not achieved at the opposite end, the gravity

demand may be draped below a sloped rotation demand.

It should be noted that in many cases load case two is more critical than load case one and

so both checks must be performed within the elongation affected regions.

Loading condition ii - Negative moment capacity at end of starter bars

Moment capacity

Moment demand = M*G – Mo(L – x)/L

Mo

x



ε𝑦𝑚 = 0.002 + 𝑓𝑦𝑝𝑚

𝐸𝑚 (Eqn C5E.7)

Mesh yield strain εym/Scf = (0.002 + 600/200000)/2.83

= 0.00177

Knowing the strain within the mesh, the stress and strain diagrams may be used to iterate

for the neutral axis position, therefore giving the bending capacity of the section. A

spreadsheet or section analysis tool may be used to find the bending capacity.

𝑓𝑝 = 𝑓𝑠𝑒 − ε𝑦𝑚𝐸𝑝𝑠 (Eqn C5E.8)

fp = 967 – 0.00021*200000

= 903 MPa

fc = εc*Ec

= 0.0007*30460

= 21 MPa

Mesh contribution Mym = 23 kNm

Prestress contribution Mps = 10 kNm

Therefore, the total negative moment capacity at the end of the starter bars is:

Mn_span = 23 + 10

= 33 kNm



Loading condition ii - Negative moment capacity at support

Assuming the mechanism is governed by concrete crushing, the stress and strain diagrams

may be used to iterate for the neutral axis position, therefore giving the bending capacity of

the section. A spreadsheet or section analysis tool may be used to find the bending

capacity.

Ultimate concrete strain εc = 0.003

Starter bar contribution Mo_support = 46 kNm

Taking the moment demand equal to the moment capacity at the support, the

corresponding moment demand within the span at the end of the starter bars may be

calculated.

M*_span = M* – M*G

= 46*(7500-600)/7500 – 11

= 31 kNm

Mn_span > M*_span

33 kNm > 31 kNm

The moment capacity within the span is greater than the moment demand; therefore, Negative Moment

loading condition two is not assessed to result in negative moment failure. Failure is precluded

εc

εs fs

fc



Note that if the moment demand was found to exceed the moment capacity within the

span at the end of the starter bars, the limiting drift for NMF would be set to 1%.

In the example above, NMF would have been triggered if any of the following

alterations to the assumed arrangement was made:

Grade 380 or 430 starter bars (HD12-300)

D12 starter bars, 300mm long (D12-300)

D16 starter bars at 400mm centres (D16-400)

D12 starter bars at 200mm centres (D12-200)

At the limiting drift of 1%, this would become the governing mechanism.

Prestress

Mo

me

nt

cap

acit

y (k

Nm

)

Distance (m)

Mesh

Starters

Mn = 33kNm

M* = 31kNm



A.3 Positive moment (C5E.5.4)

A3.1 Transverse cracking due to positive moment failure near a support (C5E.5.5)

Through spread sheet iteration of the formulas below, it can be shown that positive

moment failure will ensue under a total inter-storey drift of 1.7% as follows:

Maximum permissible crack width = Strand diameter

Δtot = 12.5 mm

Plastic beam rotation θp = θpcol

* L / (L – hcol - Lp)

= (1.7-0.6)*8000/(8000-800-335)

= 1.3%

Beam elongation δel = 2.6*0.013/2*(800-65*2) (Eqn C5E.1)

= 11 mm

Beam rotation δr1 = (800/2-(200+65))*0.016 (Eqn C5E.3a)

= 2 mm

Total crack width = 11 + 2

= 13 mm

= Strand diameter

Therefore, positive moment failure (PMF) due to transverse cracking is assessed to ensue at

an inter-storey drift of 1.7%.

Cra

ck w

idth

(m

m)

Interstorey Drift (%)

Strand dia. = critical crack width

Crack due to plastic hinge

elongation and support rotation

PMF = 1.7%

Elastic elongation



A.3.2 Web cracking

Web cracking can occur due to the following three mechanisms:

Deformation incompatibility between hollowcore unit and adjacent beam

Torsion

Shear induce from support rotation

A.3.2.1 Deformation incompatibility

Through spread sheet iteration of the formulas below, it can be shown that web cracking

due to deformation incompatibility is possible under a total inter-storey drift of 1.8% as

follows:

Max diff displacement δd max = 8 mm

Hollowcore displacement

𝛿𝑟 = 𝜃𝑐𝑠(

𝐿

2−𝑠−0.9ℎ𝑏)

(𝐿

2−𝑠)

(Eqn C5E.9)

Ledge to col CL s = 300 mm

Hollowcore displacement δr = 0.018*300*(8000/2-300-0.9*800)

(8000/2-300)

= 4 mm

Beam displacement

From Table C5E.1: hc/hb = 1.0

L/hb = 10

Beam displacement δb = [0.017 - (2.0-1.8)*(0.017-0.013) ]*800

(2.0-1.5)

= 12 mm



Differential displacement δd = δb - δr

= 12 – 4

= 8 mm = δd max

Therefore, web cracking is assessed to ensue at an inter-storey drift of 1.8%. This is greater

than transverse cracking at 1.7%, and so transverse cracking is the critical PMF mechanism.

A.3.2.2 Torsion

Although torsion does not necessarily need to be checked for this example, as the torsional

demands on the unit are likely to be low, a check will be completed to both demonstrate

the process and show how sensitive Hollowcore units are to torsion.

Hollowcore prestress assuming an 82% reduction due to losses

fps = 0.82*Fps/AHC

= 0.82 * 852 /0.101* 10-3

= 6.9 MPa

fpc = fps/3

= 2.3 MPa

Web tensile strength fct = 0.33*sqrt(f’c)

= 2.14 MPa

𝑣𝑡𝑛 = 𝑚𝑖𝑛 {𝑓𝑐𝑡√1 +𝑓𝑝𝑐

𝑓𝑐𝑡; 0.2𝑓′𝑐; 10𝑀𝑃𝑎} (Eqn C5E.11)

Limiting shear stress Vtn = 2.14*sqrt(1+2.3/2.14)

= 3.1 MPa

Max differential displacement δmax = 7 mm



Torsional shear flow q = vtn*t

= 3.1 MPa*25 mm

= 77 N/mm

Aco = (1200-35)*(200+65-25/2-(65+32)/2)* 10-6

= 0.238 m2

Nominal torsional cracking strength

2𝐴𝑐𝑜𝑞 = 2𝑞𝑏𝑡𝑑𝑡 = 𝑇𝑐𝑟𝑎𝑐𝑘 (Eqn C5E.12)

Tcrack = 2*Aco*q

= 2*0.238*77

= 36.7 kNm

Therefore, the internal strain energy may be calculated for each of the top flange, sides and

bottom flange making up the equivalent wall, using the upperbound wall thickness for

stiffness:

Noting that γi = vi/(0.4Ec)

and eεi = γi*Vi/2

Location of wall in tube vi = q/ti Vi = q*wi γi eεi

(MPa) (kN) (N)

Top 0.80 90 7.72E-05 3.5

Webs 2.20 31 1.81E-04 2.8

Soffit 2.20 90 1.81E-04 8.1

Sum 14.4



𝐸ε =1

2𝐿 ∑ (

𝑞

𝑡𝑖) 𝑡𝑖𝑤𝑖𝛾𝑖 (Eqn C5E.13)

Internal strain energy Eε = Σeεi*L

= 14.4*7500* 10-6

= 0.108 kNm

Angle of twist at cracking = 2*Eε/Tcrack

= 2*0.108/36.7

= 0.0059 rads

Therefore, the maximum torsional displacement across the unit prior to cracking is:

δmax = 0.0059*1200mm

= 7 mm

The torsional displacement capacity of 7mm may be compared to the torsional Positive Moment

displacement demand imposed by the supports to assess the drift capacity of the floor in Max drift = 1.7%

torsion. This should be calculated in accordance with the assessment procedures outlined

in the Guidelines.



SUMMARY

The following limiting drifts were assessed to be critical for unit ‘Type A’:

Loss of support at a limiting drift of 1.15%

Negative moment failure is precluded

Positive moment failure due to transverse cracking at a limiting drift of 1.7%

Therefore, floor system ‘Type A’ is assessed to possess a limiting total inter-storey drift of Unit ‘Type A’

1.15%, at which point a loss of support ensues due to insufficient seating. Max drift = 1.15%

The same assessment procedure may be completed for floor unit ‘Types B and C’. Note that

floor unit ‘Type C’ will not require a check for negative moment loading condition one;

however, loading condition two will need to be assessed. Similarly, web cracking due to

incompatibility between frame and floor will not need to be assessed; however transverse

cracking due to support rotation and elongation of the starter bars will need to be assessed.

The %NBS of the floor may be obtained by comparing the limiting inter-storey drift with the

drift demands on the primary lateral system assessed in accordance with the main body of

section C5, increased by a factor of 2.

HOLLOWCORE WORKED EXAMPLE - Concrete Society

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