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PRECAST CONCRETE HOLLOWCORE SLAB DESIGN (EN1992)

In accordance with EN1992-1-1:2004 + A1:2014 incorporating corrigenda January 2008, November 2010 and January
2014 and the UK national annex

Summary table
Description Unit Capacity Applied Utilisation Result
Bending moment kNm 75.3 20.1 0.267 Pass
Shear force kN 82.3 23.0 0.279 Pass
Spalling stress N/mm2 1.436 0.646 0.450 Pass
Compressive stress N/mm2 12.735 10.290 0.808 Pass
Initial camber mm 17.5 2.6 0.147 Pass
Camber after installation mm 11.7 4.0 0.340 Pass
Long term deflection mm 7.0 -1.7 0.247 Pass
Slab geometry
Nominal width of slab bnom = 1000 mm
Width of slab b = 1000 mm
Width of top flange btf = 1000 mm
Depth of slab h = 150 mm
Thickness of top flange htf = 28 mm
Thickness of bottom flange hbf = 28 mm
Width of inner ribs bw,i = 34.5 mm
Width of end ribs bw,e = 38 mm
Total width of ribs bw,t = (N - 1)  bw,i + 2  bw,e = 352 mm
Number of cores N=9
Width of cores bc = 72 mm
Height of cores hc = h - htf - hbf = 94 mm
Total width of cores bc,t = N  bc = 648 mm
1000
28
150

38 72 34.5
94
28

1000

Geometric properties
Cross-sectional area Ac = 146610 mm2
Height of neutral axis hna = 98 mm
First moment of area Sc = 6575578 mm3
Moment of Inertia Ic = 672726664 mm4
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Slab details
Effective span Le = 3500 mm
Bearing length lb = 100 mm

Loading details
Live load Flive = 4 kN/m2
Partitions Fpart = 1 kN/m2
Finishes Ffin = 0.5 kN/m2
Density of concrete yconc = 25 kN/m3
Self weight Fswt = Ac  yconc / bnom = 3.665 kN/m2

Partial load factors

Permanent loads yG = 1.35
Variable loads yQ = 1.5
Quasi-permanent variable load factor ψ2 = 0.3

Concrete details
Concrete class C50/60
Cement strength class N
Aggregate type Quartzite
Maximum aggregate size hagg = 20 mm
Partial factor for concrete - Table 2.1 yC =
1.5 Concrete properties at 28 days - Table 3.1
Characteristic compressive cylinder strength fck = 50 N/mm2
Compressive strength coefficient for flexure and axial loading - cl.3.1.6(1)
αcc = 0.85
Design compressive strength for flexure and axial loading - exp.3.15 fcd = αcc  fck / yC = 28.333
N/mm2 Comp.strength coefficient for shear - cl.3.1.6(1) αccv = 1
Design compressive strength for shear - exp.3.15 fcdv = αccv  fck / yC = 33.333
N/mm2 Mean value of concrete cylinder comp.strength fcm = fck + 8
N/mm2 = 58 N/mm2
Mean value of axial tensile strength fctm = 0.3 N/mm2  (fck / 1 N/mm2)2/3 = 4.072
N/mm2 Secant modulus of elasticity Ecm = 22 kN/mm2  (fcm / 10 N/mm2)0.3 = 37278
N/mm2 Characteristic axial tensile strength - 5% Fractile fctk,0.05 = 0.7  fctm = 2.85 N/mm2
Tensile strength coefficient - cl.3.1.6(2) αct = 1
Design tensile strength - exp.3.16 fctd = αct  fctk,0.05 / yC = 1.9 N/mm2
Ultimate compressive strain scu3 = (2.6 + 35  ((90 - fck / 1 N/mm2) / 100)4) / 1000 =
0.003496 Concrete properties at release
Time of release 2 days
Cement type coefficient - cl.3.1.2(6) s = 0.25
Time dependent comp.strength coefficient - exp.3.2 βcc,t = exp(s  (1 - (28 / t)0.5)) = 0.504
Mean value of comp.strength - exp.3.1 fcm,t = βcc,t  fcm = 29.225 N/mm2
α=1
Mean value of axial tensile strength - exp.3.4 fctm,t = βcc,tα  fctm = 2.052 N/mm2
Characteristic axial tensile strength - 5% Fractile fctk,0.05,t = 0.7  fctm,t = 1.436
N/mm2 Char.compressive cylinder strength - cl.3.1.2(5) fck,t = fcm,t - 8 N/mm2
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= 21.225 N/mm2
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Secant modulus of elasticity - exp.3.5 Ecm,t = (fcm,t / fcm)0.3  Ecm = 30349 N/mm2

Prestressing strand details

Strand type BS 5896 standard strands
Strand class 2
Modulus of elasticity Ep = 195 kN/mm2
Bond condition Good
Strand selected 10 x 9.3ф strands
Area per strand Ap,s1 = 52 mm2
Nominal tensile strength fpk,s1 = 1770 N/mm2
Characteristic 0.1% proof load Fp0.1,s1 = 78 kN
Cover from soffit; cs1 = 30 mm
Total area of prestressing strands Ap = 10  Ap,s1 = 520 mm2
Height to steel centroid hcp = cs1 + фs1 / 2 = 34.7 mm
Eccentricity of prestressing strands zcp = hna - hcp = 63.4 mm
Characteristic tensile strength fpk = fpk,s1 = 1770 N/mm2
Characteristic 0.1% proof-stress fp0.1k = (10  Fp0.1,s1) / Ap = 1500 N/mm2
Maximum stress factors - cl. 5.10.2.1(1) k1 = 0.8
k2 = 0.9
Max.stress applied to the tendon - cl.5.10.2.1(1) σp,max = min(k1  fpk, k2  fp0.1k) = 1350
N/mm2 Force applied to tendons - exp.5.41 Pmax = Ap  σp,max = 702 kN
Limiting prestressing force - cl.3.3.2(5) Plim = 0.7  fpk  Ap = 644.3 kN
Absolute value of initial prestress σpi = min(Pmax, Plim) / Ap = 1239
N/mm2

Final prestressing force after losses

Initial losses
a. Short term relaxation loss (48 hours)
Time (hours) t = 48
Relaxation percentage loss at 1000 hours after prestressing - cl.3.3.2(6)
p1000 = 2.5
Stress ratio µ = σpi / fpk = 0.7
Absolute value of the relaxation losses of the prestress - exp.3.29
σpr = σpi  0.66  p1000  exp(9.1  µ)  (t / 1000)0.75  (1 - µ)  10-5 = 6.03 N/mm2
Prestress just before release σpr = σpi - σpr = 1233 N/mm2

b. Elastic deformation loss due to prestress

Force in tendons P = σpr  Ap = 641.1 kN
Stress in concrete at tendons fc = P / Ac + P  zcp2 / Ic = 8.2
N/mm2 Elastic deformation loss due to prestress Pel = (Ep / Ecm,t)  fc = 52.7
N/mm2 Total of initial losses
σpr,el = σpr + Pel = 58.7 N/mm2
Prestress after initial losses σpm0 = σpi - σpr - Pel = 1180.3 N/mm2
Tendon stress factors - cl.5.10.3(2) k7 = 0.75
k8 = 0.85
Maximum value of prestress after initial losses - cl.5.10.3(2)
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σpm0,max = min(k7  fpk, k8  fp0.1k) = 1275

N/mm2 Prestressing force at release - exp.5.43 Pm0 = min(σpm0, σpm0,max)  Ap = 613.8
kN

Time dependent losses

Stress in concrete adjacent to tendons at 48 hours σc,QP = Pm0 / Ac + Pm0  zcp2 / Ic = 7.848 N/mm2

a. Loss due to creep

Perimeter in contact with atmosphere - Annex B.1(1)
u = 2  b = 2000 mm
Notional size of member - exp.B.6 h0 = 2  Ac / u = 146.6 mm
Concrete strength coefficients - exp.B.8c α1 = (35 N/mm2 / fcm)0.7 =
0.702
α2 = (35 N/mm2 / fcm)0.2 = 0.904
α3 = (35 N/mm2 / fcm)0.5 = 0.777
Relative humidity at release RH = 70
Relative humidity factor - exp.B.3b ф70 = (1 + (1 - RH / 100) / (0.1  (h0 / 1 mm)1/3)  α1)  α2 = 1.265
Relative humidity and size coefficient - exp.B.8b
βH,release = min(1.5  (1 + (0.012  RH)18)  h0 / 1 mm + 250 αα 3, 1500  α3) =
423.7
Ambient relative humidity RH = 50
Relative humidity factor - exp.B.3b ф50 = (1 + (1 - RH / 100) / (0.1  (h0 / 1 mm)1/3)  α1)  α2 = 1.506
Relative humidity and size coefficient - exp.B.8b
βH,final = min(1.5  (1 + (0.012  RH)18)  h0 / 1 mm + 250  α3, 1500  α3) =
414.1
Concrete strength factor - exp.B.4 βfcm = 16.8 / (fcm / 1 N/mm2) = 2.206
Initial time (days) t0 = 2
Concrete age factor - exp.B.5 βt0,release = 1 / (0.1 + t00.2) = 0.801
Notional creep coefficients - exp.B.2 ф0,release = ф70  βfcm  βt0,release =
2.235 Final time (days) t0 = 28
Concrete age factor - exp.B.5 βt0,final = 1 / (0.1 + t00.2) =
0.488 Notional creep coefficients - exp.B.2 ф0,final = ф50  βfcm  βt0,final =
1.622 Loss due to creep
σp,c = ((Ep / Ecm)  ф0,release  σc,QP) / (1 + (Ep  Ap / (Ecm  Ac))  (1 + Ac / Ic  zcp2)  (1 + 0.8  ф0,release)) = 83.6 N/mm2

b. Loss due to shrinkage

Ambient relative humidity RH = 50
RH0 =
100
Relative humidity ratio - exp.B.12 βRH = -1.55  [1 - (RH / RH0)3] = -1.356
Compressive strength fcmo = 10 N/mm2
Cement type coefficients - Appendix B.2 αds1 = 4
αds2 = 0.12
Basic drying shrinkage strain - exp.B.11
scd,0 = 0.85  ((220 + 110  αds1)  exp(-αds2  fcm / fcmo))  10-6  βRH = -0.000379
Notional size coefficient - Table 3.3 kh = 0.93
At infinte time, from exp.3.10 βds,t,ts = 1
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Drying shrinkage strain at infinite time - exp.3.9 scd,t = βds,t,ts  kh  scd,0 = -0.000353
Autogenous shrinkage strain at infinite time - exp.3.12
sca,inf = 2.5  (fck / 1 N/mm2 - 10)  10-6 = 0.0001
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At infinte time,from exp.3.13 βas,t = 1

Autogenous shrinkage strain - exp 3.11 sca,t = βas,t  sca,inf =
0.0001 Total shrinkage strain - exp.3.8 scs = -scd,t + sca,t =
0.000453 Loss due to shrinkage
σp,s = scs  Ep / (1 + Ep  Ap / (Ecm  Ac)  (1 + Ac  zcp2 / Ic)  (1 + 0.8  ф0,release)) = 80.5 N/mm2

c. Loss due to relaxation

Maximum tensile stress minus immediate losses - cl.5.10.4(1)
σpi = σpm0 = 1180.3 N/mm2
Initial prestress ratio - cl.3.3.2(7) µ = σpi / fpk = 0.667
Time for long term relaxation losses (hours) - cl.3.3.2(8)
t = 500000
Absolute value of the relaxation losses of the prestress - exp.3.29
σpr = σpi  0.66  p1000  exp(9.1  µ)  (t / 1000)0.75  (1 - µ)  10-5 = 39.7 N/mm2
Loss due to relaxation
σp,r = 0.8  σpr / (1 + Ep  Ap / (Ecm  Ac)  (1 + Ac  zcp2 / Ic)  (1 + 0.8  ф0,release)) = 29 N/mm2

Total of time dependent losses

σp,c,s,r = σp,c + σp,s + σp,r = 193.1 N/mm2
Prestress after final losses σp0 = σpm0 - σp,c,s,r = 987.2 N/mm2
Final prestressing force Pp0 = σp0  Ap = 513.3 kN

Stresses caused by prestress and resistance moment in service

Stress in bottom fibre fbottom = Pp0 / Ac + Pp0  zcp  hna / Ic = 8.239 N/mm2
Stress in top fibre ftop = Pp0 / Ac - Pp0  zcp  (h - hna) / Ic = 0.988
N/mm2 Quasi-permanent stress factor - cl.7.2(3) k2 = 0.45
Maximum permissible compress.stress - cl.7.2(3) σc,max = k2  fck =
22.5 N/mm2 Section modulus top Ztop = Ic / (h - hna) = 12937051
mm3
Resistance moment top MR,top = (k2  fck - ftop)  Ztop = 278.3 kNm
Section modulus bottom Zbottom = Ic / hna = 6864558 mm3
Resistance moment bottom MR,bottom = (fctm + fbottom)  Zbottom = 84.5 kNm
Serviceability load Fsls = Fswt + Ffin + Flive + Fpart = 9.165 kN/m2
Serviceability bending moment Msls = Fsls  bnom  Le2 / 8 = 14
kNm Msls / min(MR,bottom, MR,top)
= 0.166
PASS - Resistance moment is greater than serviceability bending moment
Ultimate moment of resistance
Characteristic strain of prestressing steel at maximum load - cl.3.3.6(7)
suk = 0.022222
Strain limit - cl.3.3.6(7) sud = 0.9  suk = 0.02
Strain due to prestressing sp0 = σp0 / Ep = 0.005062
Design tensile strength of prestressing steel - cl.3.3.6(6)
fpd = fp0.1k / yS = 1304.3 N/mm2
Strain at elastic limit - Fig 3.10 sel = fpd / Ep = 0.006689
Stress at maximum load - Fig 3.10 fpd,uk = fpk / yS = 1539.1 N/mm2
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Stress at limiting strain - Fig 3.10 fp,max = fpd + (fpd,uk - fpd)  (sud - sel) / (suk - sel) = 1505.5 N/mm2
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Depth to prestressing steel d = h - hcp = 115.4 mm

Stress distribution factor - exp.3.19 λ = 0.8
Stress block factor - exp.3.21 η=1
Depth of neutral axis, by iteration x = 32.5 mm
Depth to centroid of concrete in compression dn = 0.5  λ  x = 13 mm
Change in stress sp = scu3  (h - hcp) / x - scu3 = 0.009
Final strain in tendons sp = sp + sp0 = 0.013992
Final stress in tendons fp = min(fpd + (fpd,uk - fpd)  (sp - sel) / (suk - sel), fp,max) = 1414.7 N/mm2
Force in tendons Fp = fp  Ap = 735.7 kN
Area of concrete Ac,tf = btf  λ  x = 25964 mm2
Force in concrete Fc = Ac,tf  η  fcd = 735.7 kN
Ultimate moment of resistance MR = Fc  (d - dn) = 75.3 kNm
Ultimate load Fuls = Fswt  yG + Ffin  yG + Flive  yQ + Fpart  yQ = 13.123 kN/m2
Ultimate bending moment Muls = Fuls  bnom  Le2 / 8 = 20.1
kNm Muls / MR = 0.267
PASS - Resistance moment is greater than ultimate bending moment
Shear resistance
a. Cracked sections
Shear stress constant CRd,c = 0.18 / yC = 0.12
Reinforcement depth factor - cl.6.2.2(1) k = min(1 + (200 mm / d), 2.0)
= 2 Reinforcement ratio - cl.6.2.2(1) pl = min(Ap / (bw,t  d), 0.02) =
0.013
Axial force in cross-section NEd = Ap  σp0 = 513.3 kN
Stress in concrete section σcp = min(NEd / Ac, 0.2  fcdv) = 3.501 N/mm2
Minimum shear stress vmin = 0.035 N/mm2  k3/2  (fck / 1 N/mm2)1/2 = 0.7 N/mm2
Maximum stress factor - cl.6.2.2(1) k1 = 0.15
Shear resistance for cracked section - exp.6.2a & 6.2b
VRd,c,c = (max(CRd,c  k  (100  pl  fck / 1 N/mm2)1/3 1 N/mm2, vmin) + k1  σcp)  bw,t  d = 60.3 kN
Ultimate shear force Vuls = Fuls  bnom  Le / 2 = 23 kN
Extent of cracked section, point at which bottom fibre stress equals design tensile strength, by iteration
x = 1750 mm
Moment at x; Mx = Vuls  x - Fuls  bnom  x2 / 2 = 20.1 kNm
Shear at x Vx = Vuls - Fuls  bnom  x = 0 kN
Bottom fibre stress at x
fb = Mx / Zbottom + Vx  cos(45) / Ac - Pp0 / Ac - Pp0  zcp / Zbottom + Vuls / (0.7  Ac) + Vuls  hna / (0.7  Zbottom) = -4.619
N/mm2
Vx / VRd,c,c = 0
PASS - Section is uncracked throughout its length
b. Uncracked sections
Distance of section from starting point of transmission length - cl.6.2.2(2)
lx = lb + hna = 198 mm
Tendon release factor - cl.8.10.2.2(2) α1 = 1
Cross-section factor - cl.8.10.2.2(2) α2 = 0.19
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Average strand diameter ф s1) / (Ns1) = 9.3 mm
фav = (10 ф
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Tendon coefficient - cl.8.10.2.2(1) ηp1 =

3.2 Bond condition coefficient - cl.8.10.2.2(1) η1 = 1
Design value of tensile strength at release -
cl.8.10.2.2(1)
fctd,t = αct  0.7  fctm,t / yC = 0.957 N/mm2
Bond stress at transfer - exp.8.15 fbpt = ηp1  η1  fctd,t = 3.064 N/mm2
Basic value of transmission length - exp.8.16 lpt = α1  α2  фav  σpm0 / fbpt = 680.7
mm Upper bound value of transmission length - exp 8.18
lpt2 = 1.2  lpt = 816.9 mm
Length ratio - cl.6.2.2(2) αl = lx / lpt2 = 0.242
Shear resistance for uncracked section - exp.6.4 VRd,c,u = (Ic  bw,t / Sc)  (fctd2 + αl  σcp  fctd) = 82.3 kN
Vuls / VRd,c,u = 0.279
PASS - Shear resistance is greater than ultimate shear force
Resistance to spalling for prestressed hollowcore slabs - EN 1168 cl.4.3.3.2.1
Section modulus of bottom fibre Wb = Ic / hna = 6864558 mm3
Core radius k = Wb / Ac = 46.8 mm
Initial prestressing force in a single web P0 = σpm0  Ap,s1 = 61.4 kN
Eccentricity of prestressing steel e0 = hna - cs1 - фs1 / 2 = 63.4
mm
Alpha factor αe = max((e0 - k) / h , 0) = 0.110187
Basic value of transmission length - exp.8.16 lpt = α1  α2  фs1  σpm0 / fbpt = 680.7
mm Lower bound value of transmission length lpt1 = 0.8  lpt = 544.6 mm
Spalling stress
σsp = (P0 / (bw,i  e0))  (15  αe2.3 + 0.07) / (1 + (lpt1 / e0)1.5  (1.3  αe + 0.1)) = 0.646
N/mm2
σsp / fctk,0.05,t = 0.45
PASS - Axial tensile strength is greater than spalling stress
Serviceability
Stress limitations
Max.compressive stress at transfer - exp.5.42 σc = 0.60  fck,t = 12.735 N/mm2
Compressive stress at transfer fc,t = P / Ac + P  zcp / Zbottom = 10.29
N/mm2
fc,t / σc = 0.808
PASS - Maximum compressive stress is greater than compressive stress at transfer
Maximum tensile stress at transfer σt = fctm,t = 2.052 N/mm2
Self-weight moment Mswt = Fswt  bnom  Le2 / 8 = 5.612 kNm
Tensile stress at transfer ft,t = max(P  zcp / Ztop - P / Ac - Mswt / Ztop, 0 N/mm2) = 0
N/mm2 ft,t / σt = 0
PASS - Maximum tensile stress is greater than tensile stress at transfer
Max.compressive stress due to characteristic loads σc,c = 0.45  fck =
22.5 N/mm2 Compressive stress due to characteristic loads fc,c = ftop +
Msls / Ztop = 2.072 N/mm2
fc,c / σc,c = 0.092
PASS - Maximum compressive stress is greater than compressive stress under characteristic loads
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Max.tensile stress due to char.loads - cl.7.3.2(4) σt,c = fct,eff = fctm = 4.072 N/mm2
Tensile stress due to characteristic loads ft,c = max(Msls / Zbottom - fbottom, 0 N/mm2) = 0 N/mm2
ft,c / σt,c = 0
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PASS - Maximum tensile stress is greater than tensile stress under characteristic loads
Deflection
a. Initial camber
Camber due to prestress δ0,p = Ap  σpm0  zcp  Le2 / (8  Ecm,t  Ic) = 2.916 mm
Deflection due to self weight δ0,G = 5  Fswt  bnom  Le4 / (384  Ecm,t  Ic) = 0.351 mm
Initial camber δ0 = δ0,p - δ0,G = 2.565 mm
Deflection limit δ0,limit = Le / 200 = 17.5 mm
δ0 / δ0,limit = 0.147
PASS - Initial camber is within acceptable limits
b. Camber after installation
Variation in initial camber due to prestress over time
Coefficient of development of viscous effect in time
Initially, at 0 days αt0 = 0.1
At installation, after 1 month αt1 = 0.4
фtinf,t0 =
2.5
Concrete viscosity coefficient between time t0
and t1
фt0,t1 = фtinf,t0  (αt1 - αt0) = 0.75

Variation in initial camber due to prestress

over time
δ1,p = δ0,p  (1 + Ecm,t  2 / (Ecm,t + Ecm)  фt0,t1) + (Pp0 - Pm0)  zcp  Le2 / (8  Ecm  Ic) = 4.491 mm
Deflection due to self-weight δ1,G = 5  Fswt  bnom  Le4 / (384  Ecm  Ic) = 0.286 mm
Variation of deflection due to self-weight over time δ1,фG = δ0,G  (Ecm,t  2 / (Ecm,t + Ecm)  фt0,t1) =
0.236 mm Camber after installation δ1 = δ1,p - δ1,G - δ1,фG = 3.969 mm
Deflection limit δ1,limit = Le / 300 = 11.667 mm
δ1 / δ1,limit = 0.34
PASS - Deflection after installation is within acceptable limits
c. Long term deflection
Coefficient of development of viscous effect in time
At installation, after 1 month αt1 = 0.4
Application of live loads, after 3 months αt2 = 0.6
Finally αt,inf = 1
Concrete viscosity coefficient between time t1 and t2
фt2,t1 = фtinf,t0 * (αt2 - αt1) = 0.5
Concrete viscosity coefficient between time t2 and tinf
фtinf,t2 = фtinf,t0 * (αt,inf - αt2) = 1
Concrete viscosity coefficient between time t1 and tinf
фtinf,t1 = фtinf,t0 * (αt,inf - αt1) = 1.5
Concrete aging coefficient p = 0.8
Variation in initial camber due to prestress between time t1, installation and time tinf
δinf,фp = p  фtinf,t1  Ap  σp0  zcp  Le2 / (8  Ecm  Ic) = 2.383 mm
Viscous effects of self-weight from time t1, installation to time t2, application of live loads
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Calc. by Date Chk'd by Date App'd by Date
Abdullahi
Yusuf
δ2,фG,G1 = 5  Fswt  bnom  Le4  фt2,t1 / (384  Ecm  Ic) = 0.143 mm
Elastic deflection for permanent and long-term variable loads
 Project Job Ref.

Section Sheet no./rev.

15
Calc. by Date Chk'd by Date App'd by Date
Abdullahi
Yusuf

δ2,Qperm,Qinf = 5  (Ffin + ψ2  (Fpart + Flive))  bnom  Le4 / (384  Ecm  Ic) = 0.156
mm Viscous effects of all permanent loads from time t2, application of live loads to time tinf
δinf,фG,Q = 5  bnom  (Fswt + Ffin + ψ2  (Fpart + Flive))  Le4  p  фtinf,t2 / (384  Ecm  Ic) = 0.353 mm
Long term deflection δinf = δ2,фG,G1 + δ2,Qperm,Qinf + δinf,фG,Q - δinf,фp = -1.731 mm
Deflection limit δinf,limit = Le / 500 = 7
mm abs(δinf) / δinf,limit
= 0.247
PASS - Long term deflection is within acceptable limits
Final deflection δfinal = δ1 + δinf = 2.238 mm