Direct Calculation Method

This requires knowledge of the curvatures in the section in both a cracked and uncracked state. In a cracked section, the contribution of the concrete in tension is ignored, whereas in an uncracked section the concrete is assumed to act elastically in both the tension and compression zones with the maximum tensile capacity f_ctmof the concrete. For concrete grades not greater than C50, this is taken as

f_ctm¼0,3 fð _ckÞ²⁼³ ð5:11Þ

where f_ck is the characteristic strength of the concrete. The effective Young’s modulus for the concrete E_c,effis given by

E_c,eff¼ E_cm 1 þ ð1, t0Þ¼

22^fck₁₀^þ80,3

1 þ ð1, t0Þ ð5:12Þ

where (1, t₀) is the creep coefficient.

4 D 40

2 D 32 64

370 Beam cross section Beam data : Transient load 40 kN/m

Permanent load 30 kN/m

F^IGURE5.3 Beam data for Examples 5.1 and 5.2

74
Chapter 5 / Durability, Serviceability and Fire

The curvature (or other appropriate deformation parameter)
under any condition is given by

¼ 
_IIþ ð1  Þ
_I ð5:13Þ

where
_I is calculated on the uncracked section,
_II is calculated on the cracked section, and  is a distribution parameter which allows for tension stiffening and is given by

 ¼1   sr

s

2

ð5:14Þ

where  allows for the duration of the loading and is 1,0 for a single short term load and 0,5 for sustained or repetitive loading, _sis the stress in the steel calculated on a cracked section and _sris the stress in the steel as the section is just about to crack.

Beeby, Scott and Jones (2005) suggest that as the decay of tensile stress is relatively fast (i.e. a matter of hours), then  should always be taken as 0,5. In flexure, the ratio _s/_sr can be replaced by M/M_cr (where M_cr is the moment which just causes cracking) or in tension by N/N_cr(where N_cris the force which just causes cracking).

For an uncracked section  is zero.

The curvature due to shrinkage (1/r_cs) is given by 1

rcs

¼"cs
e

S

I ð5:15Þ

where "_cs is the shrinkage strain,
₁the modular ratio (defined as E_s/E_c,eff), S the first moment of area of the reinforcement about the centroidal axis and I the second moment of area. Both S and I should be calculated for the cracked and uncracked section and combined using Eq. (5.13) with
replaced by 1/r_cs.

The section properties required are determined as follows:

(a) Uncracked section (Fig. 5.4):

In an uncracked section the concrete is taken as linear elastic with equal Young’s moduli for both compression and tension. The reinforcement will be linear elastic.

FIGURE5.4 Cross section data

The second moment of area I_gross is given by

where x/d is given by x

d¼1=2 h=dð Þ²þ
eð ⁰ðd⁰=dÞ þ Þ

ðh=dÞ þ
eð ⁰þ Þ ð5:17Þ

The uncracked moment M_uncrthat can be taken by the section is given by

Muncr¼f_ctI_gross

h  x ð5:18Þ

where f_ctis the relevant tensile stress in the concrete, and the curvature 1/r_M,uncr is given by

The moment M_cr and curvature 1/r_cr at the point of cracking is obtained by setting f_ct¼f_ctmin Eq. (5.19).

(b) Cracked section:

In a cracked section, the concrete is taken as linear elastic in compression and cracked in tension. The reinforcement will be linear elastic.

The second moment of area I_cris given by I_cr

where x/d is given by, x

The moment M_crtaken by the section is given by

Mcr¼fcsIcr

x ð5:22Þ

where f_cs is the compressive stress in the concrete, and the curvature 1/r_cr is given by 76
Chapter 5 / Durability, Serviceability and Fire

Where there is no compression steel Eqs (5.20) and (5.21), reduce to

where x/d is given by,

x

The resultant deflections can be determined using two methods of deceasing accuracy:

 Integration of the curvatures calculated at discrete increments along the length of the member,

 direct use of the curvature at the point of maximum moment interpolating between the cracked and uncracked values section using Eq. (5.13).

In the exact method, the total curvature 1/r_totat any point is related to the deflection w by the following expression

1 r_tot¼d²w

dx² ð5:26Þ

The second order differential can be replaced by its finite difference form to give 1

rtot

¼w_xþ1þw_x12w_x

x

ð Þ² ð5:27Þ

where the deflections are determined at the point x and either side of it, and  x is the incremental length along the beam between points at which the curvature is calculated. For a simply supported (or continuous) beam the deflections are zero at the supports and for a cantilever the deflection and slope are zero at the (encastre´) support.

In the approximate method, the deflections are calculated assuming the whole member is either cracked or uncracked and the resultant deflection determined using the maximum curvature derived from substituting Eq. (5.4) into Eq. (5.3),

 ¼k₁

where k₃depends on the loading pattern and L is the span. For common load cases, values of k₃are given in Fig. 5.5.

E

^XAMPLE

5.2

Deflection calculations. Determine the actual deflection for the cantilever shown in Fig. 5.3.

Basic data: numerical values of M_A etc.

are used

k = 0.104 (1−b / 10 ) b = ( M_A + M_B ) / M_C

M_C may be either the midspan or maximum value, any error will be small

FIGURE5.5 Values of coefficient k3 for deflection calculations 78
Chapter 5 / Durability, Serviceability and Fire

Shrinkage strain (Table 3.2 EN 1992-1-1):

The shrinkage strain "_cs comprises two components: drying shrinkage ("_cd) and autogenous shrinkage ("_ca).

Drying shrinkage "_cd:

Assuming 40 per cent relative humidity, the nominal shrinkage strain "_cd,0 is 0,6010^3. Assume loading is at 28 days (i.e. t ¼ 28) and that curing finished at a time of 3 days (i.e. t_s¼3), then the shrinkage strain at time of loading is

"cdðtÞ ¼ dsðtÞkh"cd,0

where k_his a parameter dependant upon the notional size h₀given by 2 A_c/u, and

_ds(t,t_s) is given by

dsðt, t_sÞ ¼ t  t_s t  t_sþ0,04

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Ac

u

3

s

where A_c is the concrete gross sectional area and u is the perimeter exposed to drying.

h₀¼2  ð700  370Þ=ð2ð700 þ 370ÞÞ ¼ 242 mm:

From Table 3.3 (EN 1992-1-1), k_h¼0,808 (by linear interpolation)

dsðt, t_sÞ ¼ 28  3 28  3 þ 0,04 ffiffiffiffiffiffiffiffiffiffi

242³

p ¼0,142

So, "_cd(t) ¼ 0,142  0,808  0,60  10^3¼69  10^6 Autogenous shrinkage strain, "_ca:

"_caðtÞ ¼ _asðtÞ"_cað1Þ

where the limiting autogenous shrinkage strain "_ca(1) is given by

"_cað1Þ ¼2,5ð f_ck10Þ  10^6

¼2,5ð25  10Þ  10^6¼37,5  10^6 the loading correction factor _as(t) is given by

asðtÞ ¼ 1  e^0,2t0,5

¼1  e^0,2280,5 ¼0,653

Thus, the autogenous shrinkage strain "_ca,

"caðtÞ ¼ asðtÞ"cað1Þ

¼0,653  37,9  10^6 ¼24  10^6

The total shrinkage strain "_cs:

"cs¼"_cdþ"ca¼69  10^6þ24  10^6

¼93  10^6

Creep coefficient (t,t₀):

Use Annex B of EN 1992-1-1:

ðt,t0Þ ¼0cðt,t0Þ

where the basic creep coefficient ₀is given by 0¼RHðfcmÞðt0Þ

For concrete grade C25, _RHis given by

RH¼1 þ1  RH=100 0,1 ffiffiffiffiffi

h0

p3

¼1 þ1  40=100 0,1 ffiffiffiffiffiffiffiffi

3242

p ¼1,963

The modification factor for mean concrete strength, ( f_cm):

ðfcmÞ ¼ 16,8 ffiffiffiffiffiffi f_cm

p ¼ 16,8

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi 25 þ 8

p ¼2,92

The correction factor for the time of loading of the element (t₀):

ðt0Þ ¼ 1

0,1 þ t^0,20₀ ¼ 1

0,1 þ 3^0,20¼0,445

or ₀is given as

₀¼_RHðf_cmÞðt₀Þ

¼1,963  2,92  0,445 ¼ 2,55

The factor (t,t₀) to allow for the time relation between time at loading t₀and time considered, t, is given by

cðt,t0Þ ¼ t  t0

Hþt  t0

 0,3

80
Chapter 5 / Durability, Serviceability and Fire

where for concrete grade less than C35, _His given by

H¼1,51 þ ð0,012RHÞ¹⁸ h0

þ250

¼1,51 þ ð0,012  40Þ¹⁸

242 þ250 ¼ 613 Thus (t,t₀) is

cðt,t0Þ ¼ t  t0

_Hþt  t₀

 0,3

¼ 28  3 613 þ 28  3

 0,3

¼0,378 so (t,t₀) becomes

ðt,t₀Þ ¼₀_cðt,t₀Þ ¼2,55  0,378 ¼ 0,964

Ec,eff¼Ecm=ð1 þ ðt,t0ÞÞ ¼31,5=ð1 þ 0,964Þ ¼ 16,04 GPa:

_e¼200=16,04 ¼ 12,5

Determine the cracked and uncracked second moments of area:

⁰¼A⁰_s=bd ¼ 1608=635  370 ¼ 0,0068;

¼As=bd ¼ 5026=635  370 ¼ 0,0214;

d⁰=d ¼ 64=635 ¼ 0,101; h=d ¼ 700=635 ¼ 1,102

 ¼0, 5 ðsustained loadsÞ Uncracked section:

Use Eq. (5.17) to determine x/d:

x=d ¼ ð0,5  1,102²þ12,5ð0,0214 þ 0,0068  0,101ÞÞ= . . . ð1,102 þ 12,5ð0,0068 þ 0,0214ÞÞ ¼ 0,607

x ¼ 0; 607  635 ¼ 385 mm and Eq. (5.16) for I_gross,

Igross=bd³¼0,607³=3 þ ð1,102  0,607Þ³=3 þ 12,5  0,0068 ð0,607  0,101Þ²þ12,5  0,0214ð1  0,607Þ²¼0,178 I_gross¼0,178  370  635³¼16,86  10⁹mm⁴

Use Eq. (5.18) with f_ctreplaced by f_ctmto determine M_cr

M_cr¼f_ctmI_gross=ðh  xÞ ¼ 2,56  16,86  10⁹=ð700  385Þ Nmm ¼ 137 kNm:

Cracked section:

Use Eq. (5.21) to determine x/d

x=d ¼ 12,5ð0,0068 þ 0,0214Þ þ . . .

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð12,5²ð0,0068 þ 0,0214Þ²þ2  12,5ð0,0068  0,101 þ 0,0214ÞÞ q

¼0,470 x ¼ 0,470  635 ¼ 298 mm

and Eq. (5.20) to determine I_cr

I_cr=bd³¼0,47³=3 þ 12,5  0,0068ð0,101  0,47Þ²þ12,5  0,0214ð1  0,47Þ²¼0,121 I_cr¼0,121  370  635³¼11,46  10⁹mm⁴

Shrinkage parameters:

"cs
e¼93  10^612,5 ¼ 1163  10^6 Calculate S/bd²:

Uncracked section:

S=bd²¼0,0214ð1  0,607Þ  0,0068ð0,101  0,607Þ ¼ 0,0119 S ¼ 0,0119  370  635²¼1,775  10⁶mm³

S=Igross¼1,775  10⁶=16,86  10⁹¼0,1053  10^3=mm

Cracked section:

S=bd²¼0,0214ð1  0,47Þ  0,0068ð0,101  0,47Þ ¼ 0,0139 S ¼ 0,0139  370  635²¼2,074  10⁶mm³

S=I_cr¼2,074  10⁶=11,46  10⁹¼0,181  10^3=mm

Integration method:

Divide the beam into four segments each 1 m long (x ¼ 0 at free end)

Total UDL (assuming the whole of the transient load acts on a quasi-permanent basis) ¼ 40 þ 30 ¼ 70 kN/m

M_Sd¼70x²/2 ¼ 35x²kNm.

As the curvature 1/r_cson the cracked and uncracked sections is constant along the beam these may be determined first.

Shrinkage:

Uncracked section:

1=r_cs,unc¼ ð"cs
eÞðS=IÞ_unc¼0,001163  0,1053  10^3

¼0,1225  10^6=mm 82
Chapter 5 / Durability, Serviceability and Fire

Cracked:

1=r_cs,cr¼ ð"_cs
eÞðS=IÞ_cr¼0,001163  0,181  10^3

¼0,211  10^6=mm Point 1 (x ¼ 0): curvature is zero (zero moment) Point 2 (x ¼ 1)

M_Sd¼35 kNm 5 M_cr, thus  ¼ 0:

Bending curvature:

f_ct¼M_Sdðh  xÞ=I_gross¼35  10⁶ ð700  385Þ=16,89  10⁹¼0,653 MPa 1=r_M,unc¼ ðf_ct=E_c,effÞ=ðh  xÞ ¼ ð0,653=16,04  10³Þ=ð700  385Þ

¼0,129  10^6=mm

Total uncracked curvature, 1/r_I:

1=rI¼1=rcs,uncþ1=rM,unc¼ ð0,129 þ 0,1225Þ  10^6¼0,252  10^6=mm As the concrete is uncracked, 1/r_tot¼1/r_I

Point 3 (x ¼ 2)

MSd¼140 kNm 4 Mcr, thus  needs calculating:

_s=_sr¼M_Sd=M_cr¼140=137 ¼ 1,022 Use Eq. (5.14) to calculate :

 ¼1  0,5=1,022²¼0,478 Uncracked curvatures:

Bending:

f_ct¼M_Sdðh  xÞ=I_gross¼140  10⁶ ð700  385Þ=16,89  10⁹¼2,61 MPa 1=_rM,unc¼ ðfct=Ec,effÞ=ðh  xÞ ¼ ð2,61=16,04  10³Þ=ð700  385Þ

¼0,517  10^6=mm

Total uncracked curvature 1/r_I:

1=r_I¼1=r_cs,uncþ1=r_M,unc¼ ð0,517 þ 0,1225Þ  10^6 ¼0,640  10^6=mm Cracked curvatures:

Bending:

fcs¼MSdx=Icr¼140  10⁶298=11,46  10⁹¼3,64 MPa

1=rM,cr¼ ðfcs=Ec,effÞ=x ¼ ð3,64=16,04  10³Þ=298 ¼ 0,762  10^6=mm

Total cracked curvature, 1/r_II

1=rII¼1=rcs,crþ1=rM,cr¼0,762  10^6þ0,211  10^6 ¼0,973  10^6=mm Effective total curvature 1/r_tot is given by Eq. (5.13),

1=r_tot¼0,478  0,973  10^6þð1  0,478Þ  0,64  10^6¼0,799  10^6=mm Point 4 (x ¼ 3)

MSd¼315 kNm 4 Mcr, thus  needs calculating:

s=sr¼M_Sd=Mcr¼315=137 ¼ 2,3

Use Eq. (5.14) to calculate :

 ¼1  0,5=2,3² ¼0,906 Uncracked section:

Bending curvature:

f_ct¼M_Sdðh  xÞ=I_gross¼315  10⁶ ð700  385Þ=16,89  10⁹¼5,87 MPa:

1=r_M,unc¼ ðf_ct=E_c,effÞ=ðh  xÞ ¼ ð5,87=16,04  10³Þ=ð700  385Þ

¼1,162  10^6=mm Total uncracked curvature 1/r_I:

1=rI¼1=rcs,uncþ1=rM,unc¼1,162  10^6þ0,1225  10^6

¼1,285  10^6=mm Cracked section:

Bending:

fcs¼MSdx=Icr¼315  10⁶298=11,46  10⁹¼8,411 MPa

1=rM,cr¼ ðfcs=Ec,effÞ=x ¼ ð8,411=16,04  10³Þ=298 ¼ 1,760  10^6=mm Total cracked curvature, 1/r_II

1=r_II¼1=r_cs,crþ1=r_M,cr¼1,760  10^6þ0,211  10^6 ¼1,971  10^6=mm Effective curvature 1/r_totis given by Eq. (5.13),

1=rtot ¼0,906  1,971  10^6þ ð1  0,906Þ  1,285  10^6¼1,907  10^6=mm 84
Chapter 5 / Durability, Serviceability and Fire

Point 5 (x ¼ 4)

MSd¼560 kNm 4 Mcr, thus  needs calculating:

s=sr¼M_Sd=Mcr¼560=137 ¼ 4,088 Use Eq. (5.14) to calculate :

 ¼1  0,5=4,088²¼0,970 Uncracked section:

Bending curvature:

fct¼MSdðh  xÞ=Igross¼560  10⁶ ð700  385Þ=16,89  10⁹¼10,44 MPa 1=rM,cr¼ ðfct=Ec,effÞ=ðh  xÞ ¼ ð10,44=16,04  10³Þ=ð700  385Þ

¼2,066  10^6=mm Total uncracked curvature, 1/r_I:

1=rI¼1=rcs,uncþ1=rM,unc¼2,066  10^6þ0,1225  10^6

¼2,189  10^6=mm Cracked section:

fcs¼MSdx=Icr¼560  10⁶298=11,46  10⁹¼14,56 MPa 1=rII¼ ðfcs=Ec,effÞ=x ¼ ð14,56=16,04  10³Þ=298 ¼ 3,046  10⁶=mm Total cracked curvature 1/r_II

1=rII¼1=rcs,crþ1=rM,cr¼3,046  10⁶þ0,211  10^6

¼3,257  10⁶=mm Effective curvature 1/r_tot is given by Eq. (5.13),

1=r_tot¼0,970  3,257  10⁶þ ð1  0,970Þ2,189  10^6¼3,225  10^6=mm The results from determining ( x)²1/r_tot at each node point are given in Table 5.3.

To allow the boundary conditions to be evaluated at point 5 in the beam, a fictitious point 6 needs to be introduced (Fig. 5.6). The boundary conditions at node 5 are:

(a) deflection is zero, or w₅¼0

(b) slope is zero, or (w₆w₄)/(2 x) ¼ 0; or, w₆¼w₄.

Thus applying Eq. (5.27) at each node point, together the boundary conditions, gives in matrix form

AW ¼  x²1

r ð5:29Þ

or, solving

W ¼ A^1x²1

r ð5:30Þ

Substituting in numerical values gives, 2

Thus the deflection at the free end, w₁¼14,0 mm.

This is equivalent to span/285 which is acceptable.

Approximate method:

The deflection w is given by Eq. (5.28). The calculation must be determined for loading and shrinkage separately as each has different k₃factors.

For a UDL on a cantilever k₃¼0,25 (Fig. 5.5).

T^ABLE5.3 Deflection parameters for Example 5.2.

Node Point

1000 1000 1000 1000 1000

3

F^IGURE5.6 Deflection profile for Example 5.2 86
Chapter 5 / Durability, Serviceability and Fire

At the support, the loading curvature 1/r_I based on an uncracked section is 2,06610^6/mm, and 1/r_IIfor a cracked section is 3,04610^6/mm. Equation (5.13) may be used to determine the resultant deflection, so with  ¼ 0,97 (see earlier), the resultant curvature is given by

1=r_res¼0,97  3,046  10^6þ ð1  0,97Þ2,066  10^6¼3,02  10^6=mm

thus, the deflection due to the loading is given by w ¼ 0,25  4000²3,02  10^6¼12,1 mm

For shrinkage, 1/r_cs,uncr¼0,122510^6/mm and 1/r_cs,cr¼0,21110^6/mm, resultant shrinkage curvature 1/r_cs,res is given by

1=r_cs,res¼0,97  0,211  10^6þ ð1  0,97Þ0,1225  10^6¼0,208  10^6=mm

k₃¼0,5 for this case, so

w ¼ 0,5  4000²0,208  10^6¼1,7 mm

Total estimated deflection ¼ 12,1 þ 1,7 ¼ 13,8 mm.

The exact calculations and the approximate calculations give virtually identical results.

NOTE: Under normal circumstances, deflections due to shrinkage are not calculated, but in this example it should be noted that the shrinkage deflection is around 14 per cent of the total. Johnson and Anderson (2004) indicate that European Design Codes appear to give higher shrinkage strains than those traditionally from current British Codes, and thus neglect of shrinkage strains may no longer be justified.

In document Concrete Design to en 1992, Second Edition (2006)-Lawrence Martin, John a Purkiss- (Page 92-105)