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Tiêu đề Eurocode 2-04 Pt-Sl-001 Post-Tensioned Slab Design
Trường học University of Engineering and Technology
Chuyên ngành Civil Engineering
Thể loại bài báo
Năm xuất bản 2023
Thành phố Hanoi
Định dạng
Số trang 6
Dung lượng 209,68 KB

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EXAMPLE Eurocode 2-04 PT-SL-001

Post-Tensioned Slab Design

P ROBLEM D ESCRIPTION

The purpose of this example is to verify the slab stresses and the required area of mild steel strength reinforcing for a post-tensioned slab

A one-way simply supported slab is modeled in SAFE The modeled slab is 254

mm thick by 914 mm wide and spans 9754 mm as shown in shown in Figure 1

Length, L = 9754 mm

Prestressing tendon, Ap Mild Steel, As

914 mm

25 mm

229 mm

254 mm

Length, L = 9754 mm

Prestressing tendon, Ap Mild Steel, As

914 mm

25 mm

229 mm

254 mm

Figure 1 One-Way Slab

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A 254 mm wide design strip is centered along the length of the slab and has been defined as an A-Strip B-strips have been placed at each end of the span, perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile)

A tendon with two strands, each having an area of 99 mm2, was added to the A-Strip The self weight and live loads have been added to the slab The loads and post-tensioning forces are as follows:

Loads: Dead = self weight, Live = 4.788 kN/m2 The total factored strip moments, required area of mild steel reinforcement and slab stresses are reported at the midspan of the slab Independent hand calculations were compared with the SAFE results and summarized for verification and validation of the SAFE results

G EOMETRY , P ROPERTIES AND L OADING

Yield strength of steel f y = 400 MPa Prestressing, ultimate f pu = 1862 MPa Prestressing, effective f e = 1210 MPa Area of Prestress (single strand) A p = 198 mm2 Concrete unit weight w c = 23.56 KN/m3 Modulus of elasticity E c = 25000 N/mm3 Modulus of elasticity E s = 200,000 N/mm3

T ECHNICAL F EATURES OF SAFE T ESTED

¾ Calculation of the required flexural reinforcement

¾ Check of slab stresses due to the application of dead, live, and post-tensioning loads

R ESULTS C OMPARISON

Table 1 shows the comparison of the SAFE total factored moments, required mild steel reinforcing, and slab stresses with independent hand calculations

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Table 1 Comparison of Results

RESULTS

SAFE

Factored moment,

Area of Mild Steel req’d,

Transfer Conc Stress, top

Transfer Conc Stress, bot

Normal Conc Stress, top (D+L+PT F ), MPa −10.460 −10.465 0.05%

Normal Conc Stress, bot

Long-Term Conc Stress, top (D+0.5L+PT F(L) ), MPa −7.817 −7.817 0.00%

Long-Term Conc Stress, bot (D+0.5L+PT F(L) ), MPa 5.759 5.759 0.00%

C OMPUTER F ILE : EUROCODE 2-04PT-SL-001.FDB

C ONCLUSION

The SAFE results show a close comparison with the independent results

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H AND C ALCULATIONS :

Design Parameters:

Mild Steel Reinforcing Post-Tensioning

Stressing Loss = 186 MPa Long-Term Loss = 94 MPa

f i = 1490 MPa

f e = 1210 MPa

γm, steel = 1.15

γm, concrete = 1.50

0 1

=

η for f ck ≤ 50 MPa 8

0

=

λ for f ck ≤ 50 MPa

Length, L = 9754 mm

Prestressing tendon, Ap Mild Steel, As

914 mm

25 mm

229 mm

254 mm

Length, L = 9754 mm

Prestressing tendon, Ap Mild Steel, As

914 mm

25 mm

229 mm

254 mm

Loads:

Dead, self-wt = 0.254 m x 23.56 kN/m3 = 5.984 kN/m2 (D) x 1.35 = 8.078 kN/m2 (Du) Live, = 4.788 kN/m2 (L) x 1.50 = 7.182 kN/m2 (Lu)

Total = 10.772 kN/m2 (D+L) = 15.260 kN/m2 (D+L)ult

ω=10.772 kN/m2 x 0.914 m = 9.846 kN/m, ωu= 15.260 kN/m2 x 0.914 m = 13.948 kN/m Ultimate Moment,

2 1 8

U wl

M = = 13.948 x (9.754)2/8 = 165.9 kN-m

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Ultimate Stress in strand, 7000 1 1.36 PU P /

PS SE

CK

f A

f bd

1210 7000(229) 1 1.3630(914) 2291862(198)( ) /(9754)

1361 MPa

=

Ultimate force in PT, F ult PT, = A P(f PS) 2(99)(1362) /1000 269.6 kN= =

Compression block depth ratio:

cd

f bd

M m

η 2

=

165.92 0.1731

(0.914)(0.229) (1)(30000 /1.50)

Required area of mild steel reinforcing,

m

2 1

1− −

=

ω = 1− 1 2(0.1731) 0.1914− =

2

1(30 /1.5)(914)(229)

400 /1.15

cd EquivTotal

yd

f bd A

f

η

2

1365

2303 mm

400 /1.15

2

1361

400

S

Check of Concrete Stresses at Midspan:

Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI

Tendon stress at transfer = jacking stress − stressing losses =1490 − 186 = 1304 MPa

The force in the tendon at transfer = 1304(197.4) /1000 257.4 kN=

Moment due to dead load, M D =5.984(0.914)(9.754) / 8 65.04 kN-m2 =

Moment due to PT, M PT =F PTI(sag) 257.4(102 mm) /1000 26.25 kN-m= =

0.254(0.914) 0.00983

f

where S=0.00983m3

f = −1.109 3.948 MPa±

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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF

Tendon stress at normal = jacking − stressing − long-term=1490 − 186 − 94 = 1210 MPa The force in tendon at normal = 1210(197.4) /1000 238.9 kN=

Moment due to dead load M D =5.984(0.914)(9.754) / 8 65.04 kN-m2 =

Moment due to live load M L =4.788(0.914)(9.754) / 8 52.04 kN-m2 =

Moment due to PT, M PT =F PTI(sag) 238.9(102 mm) /1000 24.37 kN-m= = Stress in concrete for (D+L+PTF),

238.8 117.08 24.37

0.254(0.914) 0.00983

PTI D L PT

f

f = −1 029 9 431. ± .

f = −10.460(Comp) max, 8.402(Tension) max

Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF

Tendon stress at normal = jacking − stressing − long-term = 1490 − 186 − 94 = 1210 MPa The force in tendon at normal, = 1210(197.4) /1000 238.9 kN=

5.984(0.914)(9.754) / 8 65.04 kN-m

D

4.788(0.914)(9.754) / 8 52.04 kN-m

L

Moment due to PT, M PT =F PTI(sag) 238.9(102 mm) /1000 24.37 kN-m= =

Stress in concrete for (D+0.5L+PTF(L)),

0.254(0.914) 0.00983

F f

f = −1 029 6 788. ± .

f = −7.817(Comp) max, 5.759(Tension) max

Ngày đăng: 18/09/2023, 20:42

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