Pont 1

Institute of Technology of Cambodia

F. PretensioningTo estimate the number minimum of strands, Service Limit III should be usedIn this case , only 80% of live load plus impact (LL+IM) is used in tension controlleddesign of prestressed concrete.

Service III = 1.0(DC + DW) + 0.8(LL + IM)Note that: only moment at midspan is used to design minimum number of strandIn Service Limit State III, the moment due to precast beam, cast in place slab and prestressing forceis used for non-composite section and moment due to wearing surface, parapet, and Live Load.Limit tensile stress in service III

σts-

= 3.054132 MPa

σts = -Fpe/Ag – Fpee/Sb + MNDC/Sb + MDC/Sbc + MLLC/Sbc

Fpe = η.Fpi : Effective prestressing force

Fpi : Prestressing force at transfer

η = 0.8e : Excentricty maximum at midspan

Assume e 4.64058 mmMNDC 1034.208 KN.m :Moment due to precast beam, cast in place slab

MDC 319.316 KN.m :Moment due to wearing surface and parapet

MLLC 1181.8 KN.m

Ag 0.36 m2

Eq. VI.1 we can solve for effective prestressing force :Fpe 2252.06 KN


ηFpi 100.44 KN

minimum number of strand 22.42194We try 26 strands

calculate the distance from center of gravity of strands to bottom fiber ybs 93.784615 mm

excentricity ec = yb - ybs 421.073 mm

1 Prestressing Losses

Total prestress loss:

ΔfpT = ΔfpES + ΔfpSR + ΔfpCR + ΔfpR2where ΔfpES = loss of prestress due to elastic shortening

ΔfpSR = loss of prestress due to concrete shrinkageΔfpSR = loss of prestress due to concrete shrinkageΔfpSR = loss of prestress due to concrete shrinkage

In pretensioned members, there are no losses due to frottening and to sliping of anchorage

1.1 Elastic Shortening loss

fcgp sum of concrete stresses at the center of gravity of the prestressing steel due

to prestressing force at transfer and self weight of the member at sectionsof maximum moment

Pi pretension force after allowing for the initial losses

assume 8% losed after jackingPi (number of strand)(area of one strand)(0.92*fpi) 3003.156 KN

Mg momen t midspan du to self weight 650.3616 KN.m

fcgp 13.26701 MPa

ΔfpES 82.19145 MPa

Calculate prestressing force after transferPt = Atot(fpi - ΔfpES) 2789.458 KN

1.2 Shinkage LossesThe expression for prestress loss due to shrinkage is a function of the average annualambient relative humidity, H, and is given as Equation 5.9.5.4.2-1 for pretensioned membersΔfpSR = 117 – 1.03 HMPaWhere H relatve humidity

ΔfpSR

1.3 Creep Losses ΔfpCR = 12 fcgp - 7Δfcdp

fcgp concrete stress at the center of gravity of the prestressing steel at transferΔfcdp change in concrete stress at center of gravity of prestressing steel

due to permanent loads, except the load acting at the time the prestressing force is appliedValues of Dfcdp should be calculated at the same section or at sections for which fcgpis calculated


Δfcdp 1.859079 MPa

Creep loss

ΔfpCR 146.1905 MPa

1.4. Steel RelaxationFor stress relieved strands after transfer :

ΔfpR2 64.84532 MPa

For low relaxation strands after transfer, ΔfpR2 equal that of stress relieved strands time 30%

ΔfpR2 19.4536 MPa

Total LossTotal prestress loss:ΔfpT = ΔfpES + ΔfpSR + ΔfpCR + ΔfpR2 303.0356 MPaTotal prestressing force at service state Fpe 2476.408 KN

G. Investisgate Limit State

In this section we check the stress at midspan both transfer and service stage.

Allowable stress at transfer:The allowable compression stress limit for pretensioned concrete components is

fa_ci = 0,6.fc' = 21 MPa

The allowable tension stress limit for pretensioned concrete components is

fa_ti = 0.58.(fc')0.5 = -3.43133 MPa

Allowable stress at service:Compression due to the sum of effective prestress and permanent load

fa_cs1 = 0,45.fc' = 20.25 MPaCompression due to one half Live load an the sum of effective prestress and permanent load

fa_cs2 = 0,4.fc' = 18 MPaAllowable tensile stress with moderate corrosion condition

fa_ts = -3.3551 MPa

At trasfer midspan

Concrete stress at bottom fiber at midspan due to prestress and permanent load, fb

fb 1.8961 MPa < fa_ci Ok!Concrete stress at top fiber at midspan due to prestress and permanent load, ft

ft 5.54675 MPa < fa_ci Ok!

Stress at service midspan

Concrete stress at bottom fiber:


fb -2.7557 MPa > fa_ts OK !!!Concrete stress at top fiber case 1:

ft 7.60725 MPa < fa_cs1 OK !!!Concrete stress at top fiber case 2:ft 7.62076 MPa < fa_cs2 OK !!!

All the stress in concrete are all verified with stress limitH. Strength Limit State I

η = 1.05

1.05[1.25DC+1.5DW+1.75(LL+IM)] ≤ φMn

DC moment due to dead load of components, it includes also nonstructural attachementDW moment due to future wearing surfaceLL+IM moment due to live load and impact

Ultimate moment Mu

Mu 4134.88 KN.m

c distance between the neutral axis and thecompressive face (mm)

Suppose that c is in the table, so the section is considered as rectangular section

A ps Area of prestressing steel

A s Area of tension steel = 0

A s' Area of compression steel = 0

f y Yield strength of non prestressing reinforcement

f pu Yield strength of prestressing steel

d p Distance between center of prestressing and compressive face 1239.215 mm

k = 0.28c = 115.878641 mm < 190 mma = 98.4968451 mmNominal moment


f ps 1811.3

M n 5604.007 KN.m

φM n 5604.007 KN.m (φ = 1)

M u = 4134.88 KN.m < φM n 5604.007 KN.m OK!!!!

Maximum reinforcementThe amount of prestress and nonprestress reinforcement shall be such that :

in which :

In this case d p = d e

0.09350969 < 0.42 OK!!

Minimulm reinforcementThe amount of prestress and nonprestress reinforcement shall be adequate to develop flexural strength.φM n > 1.2M cr

f cpe compressive stress in concrete due to effective prestress force only at bottom fiber

f cpe 17.1463952 MPa

f r = 0.97(f c' ) 0.5 = 6.50695781 MPa modulus of rupture for normal weight concrete

Mcr 2040.95941 KN.m

φM n = 4134.88 KN.m > 1.2M cr = 2449.1513 KN.m

I. Contol of crackingSection is uncracked when:f c < 0.8f r

f c = 2340.44453 KPa

0.8f r = 0.8*0.63(f c' ) 0.5 = 4494.49663 KPa

f c < 0.8f r the section is uncrack

J. Deflection1. The deflection due to the prestressing force and the self-weight of precast beam

: due to self-weight

(Δ i ) G 22.55236 mm

(Δ i ) Fi -54.86022 mm

Thus, (Δ i ) 1 = (Δ i ) G + (Δi) Fi -32.30786 mmIt's acamber !!!


2.Additional long-term deflection until addition of slab (using λ1=1)

(Δ add ) 1 = λ1 (Δ i ) 1 = -32.30786 mm

3. Instantaneous deflction in precast beam due to weight of cast in place slab

(Δ i ) 3 = 0.545211 mm

4. Resulting deflection(Δ i ) 4 = (Δ i ) 3 + (Δadd) 1 + (Δ i ) 1 = -64.07051 mm

5. Instantaneous deflection in composite beam due to asphalt weight

(Δ i ) 5 = 2.128053916 mm6. Resulting deflection in compositeΔ 6 = Δ 4 + (Δ i ) 5 = -61.94246 mm

7. Additional long-term deflection in the composite beam (use λ 2 =1.2) :

(Δ add ) 7 = λ 2 .Δ 6 = -74.33095 mmThus the long-term deflection under sustained loads will be a camber of :

(-61,9425-74.3309) = -136.2734 mm Check deflection criteria

In order to determine the deflection due to live load plus impact, an equivalent uniformload is assumed to lead to the same moment as that of the live load ous impact. Thus :

w e = 16.4138889 KN/mThe corresponding deflection will be :

Δ LL+I 16.559237 mm < L/800 = 30 mm Ok!

Thus the deflection limit state is satisfactory.

Strands at section L/4ultime Moment: MStreng I = 1.05.(1.25.MDC + 1.5.MDW + 1.75.MLL+IM) = 2959.118

kN.m

k = 0.28

We debound 10 Strands at Section of L/4 from support: ec = 69.84 mm

dp = 1263.16 mm

Where: Aps = 1600 mm2

fpu = 1860 MpaAs = 0

A,s = 0

So: c = 72.05758 mm < hf = 190 mm b= 1.25 m


= 1830.29071 Mpa

a = 0.85.c = 61.24894398 mm

= 3609.43732 kN.m

MStrength I < φ Mn φ = 1

2959.118 < 3609.43732 Ok!!

K. Shear Design:

Critical distance for Shear:

Where: de = dp = 1263.16 mm

So: dv = 1.232535528 m = 0.004044 ft

M(x) = 0.256624 kips-ft = 0.35479554 kN.m

we consider that The moment of Dead Load at Section of dv from Support equal to 0

Mu = MLL+IM = 0.471878074 kN.m

V p = Fpesinα 131.832965 kN.m Vu = 494.163262 kN.m V = 1.903934266 MPa

0.04230965

2max 0.9

0.72

e

v e

ad

d d

h

u p

V v

V VV

b d

'c

v

f


We choose: θ = 22.3degree

fpo= 0.75fpu = 1395 Mpa

εx = -0.00568

1000.εx = -5.6778 < -0.20 and 0.04231 < 0.075

By table 5.8.3.4.2-1 we have θ = 22.3o and β = 6.32

Vs = 392.8518 kN

we use # 4 bar: Av = 253.2253 mm2

So: Smax = 0.774847784 m

S'max = 1.023177496 m

S = min(Smax , S'max ) = 0.77484778 m

Therefore: We use # 4 bar: with Spacing S = 50 cm

/ 0.5 0.5 cot

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xs s p ps

M d N V V A f

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v

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'0.083us p c v v

u

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cotv y v

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Pont 1

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