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Tiêu đề Is 456 00 pt sl 001
Trường học Standard University
Chuyên ngành Civil Engineering
Thể loại Bài Tốt Nghiệp
Năm xuất bản 2023
Thành phố Hanoi
Định dạng
Số trang 6
Dung lượng 154,73 KB

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EXAMPLE IS 456-00 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, has been 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 the independent hand calculations

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Table 1 Comparison of Results

FEATURE TESTED INDEPENDENT

RESULTS

SAFE RESULTS DIFFERENCE

Factored moment,

Area of Mild Steel req’d,

Transfer Conc Stress, top

Transfer Conc Stress, bot

Normal Conc Stress, top

Normal Conc Stress, bot

C OMPUTER F ILE : IS456-00PT-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

f ck = 30MPa f pu = 1862 MPa

fy = 400MPa f py = 1675 MPa

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

f i = 1490 MPa

f e =1210 MPa

γs = 1.15

γc = 1.50

α = 0.36

165

y max

y

f x

= max

u,

x

d 0.484

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 23.56kN/m3 = 5.984 kN/m2 (D) x 1.50 = 8.976 kN/m2 (Du) Live, = 4.788 kN/m2 (L) x 1.50 = 7.182 kN/m2 (Lu)

Total = 10.772 kN/m2 (D+L) = 16.158 kN/m2 (D+L)ult

ω=10.772 kN/m2 x 0.914 m = 9.846 kN/m, ωu= 16.158 kN/m2 x 0.914 m = 14.768 kN/m Ultimate Moment,

2 1 8

U

wl

M = = 14.768 x (9.754)2/8 = 175.6 kN-m

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Ultimate Stress in strand, f PS =from Table 11: f p = 1435 MPa

Ultimate force in PT, F ult PT, =A P(f PS)=197.4(1435) /1000=283.3 kN

Compression block depth ratio: 2

ck

M m

bd α f

=

2

175.6

0.3392 (0.914)(0.229) (0.36)(30000)

Required area of mild steel reinforcing,

β β

=1 1 4 =1 1 4(0.42)(0.3392) =0.4094

x

d

The area of tensile steel reinforcement is then given by:

= ⎧⎨ −β ⎫⎬= ( − )=

u

x

(1 6) 2663 mm

400 /1.15 189.6 /

u NET

M

400

P

y

f

f

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

D

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

Stress in concrete, 257.4 65.04 26.23

0.254(0.914) 0.00983

f

where S=0.00983m3

f = −1.109 3.948 MPa±

f = −5.058(Comp) max, 2.839(Tension) max

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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

D

L

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

f

f = −1 029 9 431. ± .

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

Ngày đăng: 18/09/2023, 20:43

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