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Analysis and Design of Precast Concrete Structures to Eurocode 2

Precast and Prestressed Floors

and Composite Floors

www.civil.utm.my

Precast & Prestressed Floors and Composite Floors

www.civil.utm.my

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• Hollow core floor units & slab fields

• Double tee units

• Half-slab (precast + in-situ topping)

• Composite floors

• Load vs. span data

Fixing Rates at 2000 m2 per week?

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40 11 feet Wide HCU onto Precast Walls at MGM Hotel, Las Vegas, 1992

Hollowcore Units

Daily production: 200 m2 – 1500 m2

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Double Tee Units

Twice the price but up to 4 capacity than hollow core

Prestressed Half Slab

Popular for housing and awkward shapes

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Prestressed Half Slab

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Prestressed Half Slab

Propping required for 5 m

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Prestressed Hollow Core Floor Units

• 400 – 3600 mm wide; typically 1200 mm

• 90 – 730 mm deep; typically 150, 200, 250, 300 mm

• Self weight 1.5 to 5.0 kN/m2

• Void ratio 40 – 60 % of solid section

• Spans 6 – 20 m (economical range)

Longitudinal pretensioning strand or wire

1195 mm

No shear or torsion links

30 mm flanges and webs

Shear key profile

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Splitting cracks due to sawing restraints as prestress in transferred

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Too much plasticiser!! in 450 deep units

Actually air-entrainment agent is used by several producers as a plasticiser

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Strand Pull-in:

An important indicator of success

Should be about 1 mm, irrespective of length

Theoretical pull-in limit for zero prestress:

𝑷𝑳

𝑨𝑬

* Use a linear scale between the extremes

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Bearing onto neoprene or mortar for spans more than about 15 m onto insitu or masonry

20 m long HCU

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Section for Analysis for Prestressed

• Pretension and losses (about 18 – 25%)

• Service moment (bottom tension critical)

• Ultimate moment (usually Msr 1.5)

• Ultimate shear uncracked & flexurally cracked

• End bearing and transmission length

• Deflection and camber (long-term, creep)

• Live load deflection after installation

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• 𝑀𝑠𝑟 = 𝑓𝑏𝑡 + 𝑓𝑐𝑡 𝑍𝑏

• 𝑀𝑢𝑟 = 0.87𝑓𝑝𝑏𝐴𝑝𝑠 𝑑 − 𝑑𝑛

• 𝑉𝑐𝑜 =𝐼∙𝑏𝑤

𝐴∙𝑦 𝑓𝑐𝑡𝑑

2 + 𝛼𝑙𝜎𝑐𝑝𝑓𝑐𝑡𝑑

Section for Analysis for Prestressed P

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Load vs Span Curve

Allowable span

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Load vs Span Curve

Allowable span

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Bearing limit

Handling limit span/depth = 50

Load vs Span Curve

Allowable span

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Bearing limit

Handling limit span/depth = 50

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Load vs Span Curve

Allowable span

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Shear

Service moment

Deflection Handling

Service moment control

Deflection control

Possibly shear ?

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Example 1 Calculate Msr for the 203 mm deep prestressed HCU shown below. The initial prestressing force may be taken as 70% of characteristics strength of the standard 7-wire helical strand. Take the final long-term losses as 24%. Geometric and data given by the manufacturer are as follows: Ac = 135 103 mm2, I = 678 106 mm4, yt = 99 mm, fck = 50 N/mm2, Ec = 30 kN/mm2, fck (t) = 35 N/mm2, Eci = 27 kN/mm2, fp0,1k = 1750 N/mm2, Eps = 195 kN/mm2, Aps = 94.2 mm2 per strand ( = 12.5 mm) and cover = 40 mm.

Section Properties

Zb = I/yb = 678 106/104 = 6.519 106 mm3

Zt = I/yt = 678 106/99 = 6.848 106 mm3

Eccentricity, e = 104 – 40 – 12.5/2 = 57.7 mm

Initial prestress in tendons, fpi = 0.70 1750 = 1225 N/mm2

Initial prestressing force, Pi = fpi Aps = 1225 7 94.2 10-3 = 807.8 kN

Example 1 - Solution

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Example 1 - Solution

Check Stress at Transfer

fbc = (807.8 103/135 103) + (807.8 103 57.75/6.519 106)

= +13.14 N/mm2 (compression) 0.6fck (t) (21 N/mm2)

ftt = (807.8 103/135 103) (807.8 103 57.75/6.848 106)

= 0.83 N/mm2 (tension) fctm (t) (3.20 N/mm2)

Final Prestress in Bottom and Top

fbc = +13.14 (1 – 0.24) = +9.97 N/mm2 (compression)

ftt = 0.83 (1 – 0.24) = 0.63 N/mm2 (tension)

OK

Service Moment

At the bottom fibre, Msr is limited by the tensile stress limit of 0 N/mm2

Msr = (fbc)Zb = (9.97) 6.519 106 10-6

= 65.0 kNm

At the top fibre, Msr is limited by the compressive stress limit of 0.6fck = 30 N/mm2

Msr = (ftt + 0.6fck)Zt = (0.63 + 30) 6.848 106 10-6

= 209.7 kNm 65.0 kNm

Example 1 - Solution

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Calculate Mur for the section given in Example 1. Manufacturer’s data gives the breadth of the top of the HCU as b = 1168 mm.

Effective depth, d = 203 – 40 – 12.5/2 = 156.75 mm

Stress in steel after losses = (1 – 0.24) 0.70 1750 = 931 N/mm2

Therefore, strain in steel after losses = 𝑓𝑠

𝐸𝑠=

931

205×103= 0.0045

which is less than y, the yield strain.

Example 2 - Solution

Example 2 - Solution

𝒇𝒑𝟎,𝟏𝒌

𝜸𝒎=𝟏𝟕𝟓𝟎

𝟏. 𝟏𝟓= 𝟏𝟓𝟐𝟐

𝜺𝒚 =𝟏𝟓𝟐𝟐

𝟐𝟎𝟓 × 𝟏𝟎𝟑= 𝟎. 𝟎𝟎𝟕𝟒𝟐

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As a first attempt, try x = 75 mm, approximately equal to 0.5d

(a) Steel Strains

Final stress strain, s = Prestress Strain + Bending Strain, ’s

Bottom layer at the tendon:

sb = 0.0045 + ’sb

= 0.0045 +𝑑−𝑥

𝑥𝜀𝑐𝑐

= 0.0045 +156.75−75

75× 0.0035 = 0.008315 y

Take sb = 0.00742

cc = 0.0035

’sb

x

d

Example 2 - Solution

(b) Steel Stresses Bottom layer at the tendon: fsb = sb Es

= 0.00742 205 103

= 1522 N/mm2 (c) Forces in Steel & Concrete Steel tensile forces, Fst =fs.Aps = (fsb) 7 94.2 = 1003 103 N Concrete compressive forces, Fcc= 0.567fckb 0.8x = 0.567 50 1168 0.8 75 = 1987 103 N Since Fcc Fst, a smaller depth of neutral axis, x must be tried.

Example 2 - Solution

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Example 2 - Solution Try x = 38 mm 28 mm (upper flange) and then carry out the previous analysis. Final Forces in Steel & Concrete Steel tensile forces, Fst =fs.Aps = (fsb) 7 94.2 = 1003 103 N Concrete compressive forces, Fcc= 0.567fckb 0.8x = 0.567 50 1168 0.8 38 = 1006 103 N Fst OK Ultimate Moment of Resistance: Mur = Fstz = Fst (d – 0.4x) = 1003 103 (156.75 – 0.4 38) 10-6 = 142 kNm

Shear searches out the weakest web

Shear failure in 400 deep units

Shear Analysis

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Calculate the uncracked shear capacity, Vco for the section in Example 1.

𝜎𝑐𝑝 =𝛾𝑐𝑝𝜂𝑃𝑖

𝐴=

0.9×(1−0.24)×0.70×1750×7×94.2

135×103 = 4.09 N/mm2 0.133fck (6.65 N/mm2)

𝑓𝑐𝑡𝑑 =𝑓𝑐𝑡𝑘,0.05𝛾𝑚

=2.90

1.5= 1.40 𝑁/𝑚𝑚2

𝛼1 =𝑙𝑥𝑙𝑝𝑡2

=800 + 203/2

961= 0.94 ≤ 1.0

Example 3 - Solution

𝒍𝒑𝒕𝟐 = 𝟏. 𝟐𝒍𝒑𝒕 = 𝟏. 𝟐𝜶𝟏𝜶𝟐∅𝝈𝒑𝒎𝟎

𝒇𝒃𝒑𝒕= 𝟏. 𝟐

𝟏. 𝟎 × 𝟎. 𝟐𝟓 × 𝟏𝟐. 𝟓 × 𝟏𝟐𝟐𝟓

𝟒. 𝟕𝟖

= 𝟗𝟔𝟏 𝒎𝒎 1 = 1.0 2 = 0.25 pm0 = 0.70 1750 = 1225 N/mm2

= 12.5 mm lpt (transmission length) = 800 mm

𝑓𝑏𝑝𝑡 = 𝜂𝑝1𝜂1𝑓𝑐𝑡𝑑 𝑡 = 𝜂𝑝1𝜂1𝑓𝑐𝑡𝑘,0.05 (𝑡)

𝛾𝑚= 3.2 × 1.0 ×

2.20

1.5= 4.78 𝑁/𝑚𝑚2

Example 3 - Solution

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Total breadth at centroid, bw = 1168 – (6 150) = 268 mm

Area above the centroidal axis, A = 67500 mm2

y’ = 64.5 mm (calculate from geometry)

𝑉𝑐𝑜 =𝐼 ∙ 𝑏𝑤𝐴 ∙ 𝑦

𝑓𝑐𝑡𝑑2 + 𝛼1𝜎𝑐𝑝𝑓𝑐𝑡𝑑

=678 × 106 × 268 × 10−3

67500 × 64.51.42 + 0.94 × 4.09 × 1.40

= 113.1 kN

Example 3 - Solution

Watch out for flexural-shear (Vcr) failure !

Shear Analysis

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Vco

Vcr

Decompression point

Vcr = [0.12k (100lfck)1/3 + 0.15cp] bwd

Vcr, min = [0.035k3/2 (fck)1/2 + 0.15cp] bwd

Flexural-Shear Formula, Vcr P

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SF diagram

Vcr failure here

Flexural-Shear Formula, Vcr

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Calculate the minimum value of Vcr for the slab in Example 1.

𝜎𝑐𝑝 =𝛾𝑐𝑝𝜂𝑃𝑖

𝐴=

0.9×(1−0.24)×0.70×1750×7×94.2

135×103 = 4.09 N/mm2 0.133fck (6.65

N/mm2)

𝑘 = 1 +200

𝑑= 1 +

200

156.75= 2.12 ≤ 2.0 (From Example 2, d = 156.75

mm) From Example 3, cp = 4.09 N/mm2

Vcr, min = [0.035k3/2 (fck)

1/2 + 0.15cp] bwd = [0.0035(2.0)3/2 (50)1/2 + 0.15 4.09] 268 156.75 10-3

= 28.7 kN

Example 4 - Solution

• Refer to Cl. 10.9.5

• Refers to contact bearing pressure at the bearing ledge

• Non-isolated – components are connected to other components with a secondary means of support

• Isolated – components rely entirely on their own bearing for total support

Ultimate bearing capacity, FEd = fRd b1 a1

Bearing Capacity

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𝒂 = 𝒂𝟏 + 𝒂𝟐 + 𝒂𝟑 + ∆𝒂𝟐𝟐 + ∆𝒂𝟑

𝟐

Bearing Definitions P

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a1 must NOT be less than the value in Table 10.2

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Net Bearing Length, a1 (Perpendicular to the Floor)

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Limiting Value of a2 P

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Limiting Value of a3

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Limiting Value of a2 and a3 P

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Net Bearing Width, b1

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Ultimate Bearing Strength, fRd P

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Calculate the bearing capacity of the HCU in Example 1. The unit has an actual 75 mm dry bearing length onto a reinforced concrete. Assumed that the unit is 6 m long and has secondary support on steel beams.

Ultimate bearing strength, fRd = 0.4fcd = 0.4 ×50

1.5 = 13.33 N/mm2

Net bearing width, b1 = 1200 mm b1, max = 600 mm Use b1 = 600 mm Net bearing length for non-isolated component, a1:

From a2 = 0 mm a2 = 10 mm a = 75 mm a3 = 5 mm a3 = 2.4 mm

Example 5 - Solution

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Refer to Table 10.2 for the minimum values of a1

The ultimate bearing capacity:

FEd = 13.33 60.3 600 10-3 = 482.3 kN > Vco (109.3 kN) OK

Note:

𝜎𝐸𝑑 =𝑉𝑐𝑜 (𝐹𝑟𝑜𝑚 𝐸𝑥𝑎𝑚𝑝𝑙𝑒 4)

1200 × 75= 1.25 𝑁/𝑚𝑚2

𝜎𝐸𝑑𝑓𝑐𝑑

=1.25

501.5

= 0.0375

Example 5 - Solution

𝒂𝟏 = 𝒂 − 𝒂𝟐 − 𝒂𝟑 − ∆𝒂𝟐𝟐 + ∆𝒂𝟑

𝟐

𝒂𝟏 = 𝟕𝟓 − 𝟎 − 𝟓 − 𝟏𝟎𝟐 + 𝟐. 𝟒𝟐 = 𝟔𝟎. 𝟑 𝒎𝒎 > 𝟐𝟓 𝒎𝒎

Composite Floors Required for Double Tee, but Optional for Hollowcore

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Surface Laitance due to Cutting Slurry

50 mm minimum at the highest point, increasing (with slab and beam cambers) to about 80 mm

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Composite Construction

• Minimum thickness of topping 40 mm

• Average depth of topping allowances for camber should be made – allowing span/300 will suffice

• Concrete grade – C25 to C30

• Minimum mesh reinforcement area = 0.13% Concrete area

Composite Construction

• Advantages to composite construction:

(a) Increase bending resistance and flexural stiffness

(b) Improve vibration, thermal & acoustic performance

(c) Provide horizontal diaphragm action

(d) Provide horizontal stability ties across floors

(e) Provide a continuous and monolithic floor finish

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with 75 mm topping

Composite Design

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• Selfweight of slab + in-situ concrete topping + construction traffic allowance (1.5 kN/m2)

• Section properties = Precast unit

• Selfweight of slab + in-situ concrete topping + construction traffic allowance (1.5 kN/m2)

• Section properties = Precast unit

Stage 1

• Superimposed load

• Section properties = Composite section

• Superimposed load

• Section properties = Composite section

Stage 2

Composite Design -2 Stage Approach Se

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– 2

Sta

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h

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_

+

_

+

_

+

+

+ =

_

Topping

Precast

Prestress Imposed Stage 1 stresses

Imposed Stage 2 stresses

Total stresses

0.6

f ck

limit

f ct l

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Stress Diagram at Serviceability

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Bottom fibre stress of the HCU:

𝒇𝒃𝟏 = 𝒇𝒃𝒄 −𝑴𝟏

𝒁𝒃𝟏< +𝟎.𝟔𝒇𝒄𝒌 (𝒕) 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 & 𝟎 (𝒔𝒆𝒓𝒗𝒊𝒄𝒆)

Top fibre stress of the HCU:

𝒇𝒕𝟏 = 𝒇𝒕𝒄 −𝑴𝟏

𝒁𝒕𝟏< 𝒇𝒄𝒕𝒎 𝒕 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 & + 𝟎. 𝟔𝒇𝒄𝒌 (𝒔𝒆𝒓𝒗𝒊𝒄𝒆)

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Stage 1 – Service Moment, M1

Bottom fibre stress of the HCU:

𝒇𝒃𝟐 = −𝑴𝟐

𝒁𝒃𝟐

Top fibre stress of the HCU:

𝒇𝒕𝟐 = +𝑴𝟐

𝒁𝒕𝟐

Top fibre stress of the in-situ topping:

𝒇′𝒕𝟐 = +𝑴𝟐

𝒁′𝒕𝟐

beff = b Ec’/Ec where b is the full breadth

Stage 2 – Service Moment, M2

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Adding Stage 1 + Stage 2:

fb =fb1 + fb2 = fbc – M1/Zb1 – M2/Zb2 0

ft = ft1 + ft2 = ftc + M1/Zt1 + M2/Zt2 +0.6fck

f’t = f’t2 = + M2/Z’t2 +0.6fck, in-situ

Service Moment, Msr

Bottom of the slab:

Ms2 = M2 (fbc)Zb2 – [M1(Zb2/Zb1)]

Top of the slab:

Ms2 = M2 (0.6fck – ftc)Zt2 – [M1(Zt2/Zt1)] Serv

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Total Service Moment, Ms

Calculate the Stage 2 service bending moment that is available if the HCU in Example 1 has a 50 mm minimum thickness structural topping. The floor is simply supported over an effective span of: (a) 4.0 m; (b) 8.0 m. The precamber of the HCU may be assumed as span/300 without loss of accuracy. Use fck, in-situ = 30 N/mm2 and Ec, in-

situ = 26 kN/mm2. Self weight of concrete = 25 kN/m3. Selfweight of HCU = 3.24 kN/m run. What is the minimum imposed loading for each span?

Example 6

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Example 6 - Solution

Effective breadth of topping, beff = 1200 26/30 = 1040 mm

Total depth of composite section = 203 + 50 = 253 mm

Depth to neutral axis from top of composite section, yt2

=1040×50×25 +(135000× 50+99 )

135000+(1040×50)= 114.5 mm

Second moment of area of composite section, I2 = 1266 106 mm4

Example 6 - Solution

𝑍𝑏2 =1266×106

(253−114.5)= 9.14 × 106 mm3

𝑍𝑡2 =1266×106

114.5= 19.63 × 106 mm3

Zb2/Zb1 = 9.14 106 /6.519 106 = 1.40

Zt2/Zt1 = 19.63 106 /6.848 106 = 2.86

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(a) 4.0 m

Precamber = 4000

300= 13 mm

Maximum depth of topping at supports = 50 + 13 = 63 mm

Average depth of topping = (50+63)

2= 57 𝑚𝑚

Self weight of topping = 0.057 25 1.2 = 1.71 kN/m run for 1.2 m wide unit

Self weight of HCU = 3.24 kN/m run

Stage 1 moment, M1 = (3.24+1.71)×42

8= 𝟗. 𝟗𝟎 𝒌𝑵𝒎

Example 6 - Solution

Bottom of the slab:

𝑀𝑠2 > 𝑓𝑏𝑐𝑍𝑏2 − 𝑀1

𝑍𝑏2𝑍𝑏1

𝑀𝑠2 > 9.97 × 9.14 × 10−6 − 9.90 × 106 1.40 × 10−6

= 77.46 kNm

Example 6 - Solution

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Top of the slab:

𝑀𝑠2 = 𝑀2 > 0.6𝑓𝑐𝑘 − 𝑓𝑡𝑡 𝑍𝑡2 − 𝑀1

𝑍𝑡2𝑍𝑡1

𝑀𝑠2 > 30 − (−0.63 ) × 19.63 × 106 − 9.90 × 106 2.86 × 10−6

= 573.35 kNm

The allowable maximum imposed load = 8 77.46/4.02 = 38.73 kN/m

* Additional notes:

Total Ms = Ms1 + Ms2 = 9.90 + 77.46 = 87.36 kNm

Example 6 - Solution

Ultimate Limit State - 2 Stage Approach

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Total area of reinforcement, Aps = Aps1 + Aps2

Effective breadth, beff = b fcu in-situ/fcu

Stage 1

fpdAps1 = 0.567fck b(0.8X) but dn1 = 0.4X or X = dn1/0.4

then dn1 = fpdAps1/1.134fckb

then Mu1 = fpdAps1(d – dn1) ---------------- (1)

Ultimate Limit State - 2 Stage Approach

Stage 2

Aps2 = Aps – Aps1

Mu2 = fpdAps2 (d + hs – dn2) ---------------- (2)

or by replacing (1) into (2):

Mu2 = fpd [Aps – Mu1/fpd(d – dn1)] (d + hs – dn2)

One Step Approach

Mu = fpdAps (d + hs – dn)

Ultimate Limit State - 2 Stage Approach

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Interface Shear Stress in Composite Slabs (Cl. 6.2.5)

In the neutral axis is below the interface, X hs

VEd = 0.567fckbeffhs

In the neutral axis is above the interface, X hs, VEd = 0.567fckbeff0.9X The shear stress at the interface should satisfy: vEdi vRdi

Interface Shear Stress in Composite Slabs (Cl. 6.2.5)

Design shear stress at the interface; 𝑣𝐸𝑑𝑖 =𝛽𝑉𝐸𝑑

𝑧𝑏𝑖

Design shear resistance at the interface; vRdi = cfctd +n + fyd ( sin + cos ) 0.5vfcd

If vEdi vRdi ; provide shear reinforcement (per 1 m run) projecting from the precast unit into the structural topping. The amount of steel required:

𝑨𝒇 =𝟏𝟎𝟎𝟎𝒃𝝉𝒉𝟎. 𝟖𝟕𝒇𝒚𝒌

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Interface Shear Stress in Composite Slabs (Cl. 6.2.5)

Interface Shear Stress in Composite Slabs (Cl. 6.2.5)

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Interface Shear Stress in Composite Slabs (Cl. 6.2.5)

Interface Shear Stress in Composite Slabs (Cl. 6.2.5)

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• 2400 – 3000 mm wide

• 300 – 2000 mm deep; typically 400 – 800 mm

• self weight 2.6 to 10 kN/m2

• void ratio 70 – 80 % of solid section

• spans 8 – 30 m (economical range)

Double Tee Floor Units

Mostly prestressed, but RC if manufacturer prefers

Bearing pads required 150 x 150 x 10 mm

Double Tee Floor Units