Technical Note Structural Concrete Software System

ADAPT Corporation Redwood City, CA, USA

ADAPT International Kolkata, India

ADAPT Latin America Miami, FL, USA

ADAPT Europe Zurich, Switzerland

www.adaptsoft.com Tel: +1 (650) 306 2400 Fax: +1 (650) 306 2401

TN386_HK_code_in_ABI_092910

HONG KONG CODE IMPLEMENTATION IN ADAPT ABI SOFTWARE1

ADAPT-ABI is a computer program developed for the analysis and design of segmentally constructed structures with detailed allowance for the following time-dependent effects of its concrete material:

Creep of concrete Shrinkage of concrete Aging of concrete Relaxation of prestressing

In addition to the time-dependency, the program can handle temperature effects, impact of construction equipment, base reinforcement, and explicit representation of pre- or post-tensioned strands. For bridge structures, in addition to static loads, a flexible implementation of moving loads in the program covers practically the load train configuration and application of all major codes. The material properties used in the construction can be defined by the users, based on laboratory measurements, or taken from the major building/bridge codes. The list of the building codes already implemented in the program are given in the appendix of this Technical Note. The following lists the implementation of the material properties of Hong Kong Code of Practice for Concrete Structures 2004, extended where necessary for program implementation. The description of the material implementation if followed by the verification of its implementation in the program. MATERIAL PROPERTIES OF HK CODE IMPLEMENTED IN ABI

CONCRETE STRENGTH The variation of concrete strength with time is shown in Table 1:

Table 1

Days Fcu/Fcu28

1.6 0.55 7 0.8

28 1 60 1.08

720 1.2

1 Copyright ADAPT Corporation 2010

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CONCRETE MODULUS OF ELASTICITY Modulus of elasticity of concrete is shown in Table 2: Table 2

Strength (MPa)

E (MPa)

20 18900 25 20200 30 21700 40 24000 45 26000 50 27400 55 28800 60 30200

CONCRETE CREEP STRAIN Creep coefficient and its variation with time is defined as: C(t) = CrKLKmKcKeKjKs (1) where,

Cr = Ultimate creep coefficient for age at loading 28 days; KL = depends on the relative humidity (Figure 1); Km = depends on the hardening of the concrete at the age of loading (Figure 2); Kc = depends on the composition of concrete (Figure 3); Ke = depends on the effective thickness of the member (Figure 4); Kj = defines the development of the time-dependent deformation with time (Figure 5); and Ks = depends on the percent of reinforcement :

Where,

= steel ratio As/Ac As = total area of longitudinal reinforcement Ac = gross cross-sectional concrete area Es = modulus of elasticity of the reinforcement Ec = short-term modulus of concrete

For plain concrete For reinforced concrete

=

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CONCRETE SHRINKAGE STRAIN Shrinkage strain of concrete is calculated using the following equation:

S(t) = CsSrKLKcKeKjKs (2) where, Cs = shrinkage coefficient for Hong Kong; = 4.0; Sr = coefficient by which users can scale the code stipulated factor for specific conditions; KL = depends on the relative humidity (Figure 6); Kc = depends on the composition of concrete (Figure 3); Ke = depends on the effective thickness of the member (Figure 7); Kj = defines the development of the time-dependent deformation with time (Figure 5); and Ks = depends on the percent of reinforcement

Figure 1. Coefficient KL (environmental thickness)

Figure 2. Coefficient Km (hardening at the age of loading)

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Figure 3. Coefficient Kc (composition of the concrete)

Figure 4. Coefficient Ke (effective thickness)

Figure 5. Coefficient Kj (variation as a function of time)

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Figure 6. Coefficient KL for shrinkage (environment)

Figure 7. Coefficient Ke for shrinkage (effective thickness)

STRESS RELAXATION IN PRE-STRESSING REINFORCEMENT The user can specify relaxation coefficient directly or indirectly through the use of initial stress to ultimate stress ratio, number of hours of duration of test, and observed stress loss ratio in the test. The formulation used for loss of prestressing is: fs/fsi = 1 – log(t)/c * (fsi/fpy – 0.55) (3) where, fs = steel stress at time t; fsi = initial steel stress; fpy = 0.001 offset yield stress; c = constant, 10 for stress relieved strand, 45 for low relaxation strand t = time in hours after stressing

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VALIDATION EXAMPLES – CREEP STRAIN

Description This set of examples validates the implementation of the Hong Kong standard time dependent creep model in ADAPT-ABI and illustrates how the associated hand calculations can be carried out. A rectangular specimen is axially loaded at different loading ages. The resulting creep of the specimen due to the applied axial loading is monitored at a number of observation times using ADAPT-ABI. The corresponding total creep strains are hand calculated using Hong Kong standard relationships as described in previous chapter of this technical note. The hand calculations are compared with the values obtained from ADAPT-ABI program. The agreement is found to be good.

Structure The dimensions and loading of the axially loaded column of uniform cross-section are shown in Figure 8. Other particulars of the specimen are:

• 28 days concrete cube strength fcu = 30 MPa; • Ultimate creep coefficient Cr = 1; • Humidity H = 80% • Temperature T = 20 ºC • Cement content c = 300 kg/m3 • Cement type is ordinary Portland cement • Effective thickness h = 200 mm • Water-cement ratio w/c= 0.60 • Percent of reinforcement p = 0%

Figure 8 Three cases A through C are considered:

• In Cases A the specimen is loaded at age 28 days. The loading is sustained and the shortening of the specimen is observed in time.

• In Case B the specimen is loaded with two concentrated axial compressive forces applied at different ages. The objective of this case is to test the formulation for multiple loading. The first loading is applied on day 28. With the first loading retained, the second loading is applied on day 60. Both loadings are kept on the specimen while its shortening is monitored with lapse of time.

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• In Case C the specimen is loaded on day 28 and the load was removed on day 60. The shortening was observed in time.

The column is discretized into one finite element only. The shrinkage of concrete is neglected.

Results The sample hand calculations are presented in the following.

Example 1 - Case A Verify the shortening (δ) of the specimen on day 180. The specimen is loaded on day 28. Verification:

• First, calculate the instantaneous elastic shortening for loading age of 28 days:

δi28 = P x L / (A x E(28))

Concrete strength at time of loading is:

fcu(28) = 30 MPa

Modulus of elasticity at time of loading is:

E(28) = 21700 MPa

Therefore, elastic shortening is:

δi28 = 100000*2000/(10000*21700) = 0.9216 mm (Solution from ADAPT-ABI 0.9242 mm, OK)

• Second, calculate the creep coefficient.

The creep coefficient for observation on day 180 is: C(180) = CrKLKmKcKeKjKs Cr = 1 KL = 1.9 Km(28) = 1 Kc = 1.18 Ke = 0.85 Kj(180-28) = 0.53 Ks = 1 C(180) = 1*1.9*1*1.18*0.85*0.53*1 = 1.0100

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• Total shortening on day 180 is: δ = [1 + C(180)]*( δi28 ) = [1 + 1.0100]* 0.9216 = 1.8524 mm (Solution from ADAPT-ABI 1.8429 mm, OK)

Example 2 – Case B Verify the shortening (δ) of the specimen on day 180. The specimen is loaded on day 28 with P=100 kN. Subsequently on day 60 a second load P=100 kN is added. Both loads are sustained on the specimen. Verification: For multiple loading, the creep strain is obtained using the principle of superposition. That is to say, the creep on day 180 is the sum of creeps on day 180 due to the application of the first load on day 28 and subsequently the application of the second load on day 60. The creep as well as the instantaneous displacement due to each application is treated independently from the other applications. (i) Shortening on day 180 due to the application of the first load

The total shortening on day 180, due to the application of the first load on day 28, is identical to the solution obtained for Example 1 and verified in the preceding. It is equal to 1.8349 mm.

(ii) Shortening on day 180 due to the application of the second load

• First, calculate the instantaneous shortening for loading age of 60 days: δi60 = P x L / (A x E(60)) Concrete strength at time of loading is: fcu(60) = 1.08 *30 = 32.4 MPa Modulus of elasticity at time of loading is: E(60) = 22252 MPa Therefore, instantaneous shortening is: δi60 = 100000*2000/(10000*22252) = 0.8988 mm (Solution from ADAPT-ABI 0.8997 mm, OK)

• Second, calculate the creep shortening. The creep coefficient for observation on day 180 is: C(180) = CrKLKmKcKeKjKs Cr = 1

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KL = 1.9 Km(60) = 0.85 Kc = 1.18 Ke = 0.85 Kj(180-60) = 0.50 Ks = 1 C(180) = 1*1.9*0.85*1.18*0.85*0.50*1 = 0.8099

• Total shortening is: δ = [1 + C(180)]*(instantaneous shortening, δi ) = [1 + 0.8099]* 0.8988 = 1.6258 mm

(iii) Shortening on day 180 due to the application of both loads

δ = sum of deflections due to each application = 1.8524 + 1.6258 = 3.4792 mm

(Solution from ADAPT-ABI 3.4497 mm, OK) Example 3 – Case C Verify the shortening (δ) of the specimen on day 180. The specimen is loaded on day 28 with P=100 kN and unloaded on day 60. Verification: (i) Shortening on day 180 due to the application of load

The total shortening on day 180, due to the application of the load on day 28, is identical to the solution obtained for Example 1. It is equal to 1.8524 mm.

(ii) Shortening on day 180 due to the removal of the load Total shortening on day 180 due to removal of load on day 60 is equivalent to shortening due to application of load at day 60 and is identical to the solution obtained in Example 2 but with opposite sign. It is equal to -1.6258 mm.

(iii) Shortening on day 180 δ = sum of deflections due to each application

= 1.8524 + (-1.6258) = 0.2257 mm

(Solution from ADAPT-ABI 0.2361 mm, OK)

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Table 3 Creep coefficient calculation summary Loading

age [days]

Observation day

Time since loading, t

[days]

Kj (t) Creep coefficient

C(t)=CrKLKmKcKeKjKs

Shortening =(1+C(t))*δi

[mm]

28

Fc = 30MPa, E = 21700 MPa, δi = 0.9216 mm Cr =1, KL = 1.9, Km(28) = 1, Kc= 1.18, Ke = 0.85, Ks = 1

28 0 0 0.0000 0.9216 35 7 0.1 0.1906 1.0972 46 18 0.15 0.2859 1.1850 60 32 0.24 0.4574 1.3431 85 57 0.34 0.6479 1.5187

100 72 0.38 0.7242 1.5890 180 152 0.53 1.0100 1.8524 260 232 0.65 1.2387 2.0632 350 322 0.7 1.3340 2.1510 470 442 0.75 1.4293 2.2388 630 602 0.8 1.5246 2.3266 750 722 0.82 1.5627 2.3618

1100 1072 0.89 1.6961 2.4847 2000 1972 0.95 1.8104 2.5901

60

Fc = 32.4MPa, E = 22252 MPa, δi = 0.8988 mm Cr =1, KL = 1.9, Km(60) = 0.85, Kc= 1.18, Ke = 0.85, Ks = 1

60 0 0 0.0000 0.8988 85 25 0.18 0.2916 1.1609

100 40 0.28 0.4536 1.3065 180 120 0.5 0.8099 1.6268 260 200 0.62 1.0043 1.8015 350 290 0.68 1.1015 1.8888 470 410 0.73 1.1825 1.9616 630 570 0.78 1.2635 2.0344 750 690 0.81 1.3121 2.0781

1100 1040 0.88 1.4255 2.1800 2000 1940 0.93 1.5065 2.2528

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Figure 10 Case A – Creep Shortening due to Load Applied

on Day 28

Figure 11 Case B – Creep Shortening due to Load Applied

on Day 28 and Day 60

Figure 12 Case C – Creep Shortening due to Load Applied

on Day 28 and Removed on Day 60

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VALIDATION EXAMPLES - SHRINKAGE STRAIN

Description This example validates the implementation of the Hong Kong time dependent shrinkage model in ADAPT-ABI and illustrates how the associated hand calculations can be carried out. Structure The dimensions of the column of uniform cross-section are shown in Figure 9. Other particulars of the specimen are:

• 28 days concrete cube strength fcu = 30 MPa; • Unit weight W = 2.4019 kg/mm3; • Ultimate shrinkage strain εshu = 0.03 • Relative humidity H = 80% • Water-cement ratio = 0.60 • Effective thickness h = 200 mm • Cement content c = 300 kg/m3 • Percent of reinforcement p = 0%

Figure 9

The column is discretized into one finite element only.

Results The sample hand calculations are presented in the following.

Example 1 Verify the shortening (δ) of the specimen at age of 100 days. Verification: Shrinkage strain at day 100 is calculated using following equation: S(t) = CsSrKLKcKeKjKs Where, Cr = 4 Sr = 1 KL = 200 x 10-6 Kc = 1.18 Ke = 0.80 Kj(100) = 0.44

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Ks = 1 Shrinkage strain is: S(100) = 4*1*(200*10-6)*1.18*0.80*0.44*1 = 0.000332 Shortening at day 100 is: δ = S(100) * L = 0.000332* 2000 = 0.6645 mm (Solution from ADAPT-ABI 0.6501 mm, OK)

Table 4 Shrinkage Strain Calculation Summary

Observation day Kj (t) Shrinkage strain S(t)=CsSrKLKcKeKjKs

Shortening =S(t) x L

[mm]

L = 2000 mm Cs =4, Sr = 1, KL = 200 x 10-6, Kc=1.18, Ke = 0.80 Ks = 1

28 0.21 1.59E-04 0.3172 35 0.27 2.04E-04 0.4078 46 0.29 2.19E-04 0.4380 60 0.32 2.42E-04 0.4833 85 0.39 2.95E-04 0.5891

100 0.44 3.32E-04 0.6646 180 0.58 4.38E-04 0.8760 260 0.63 4.76E-04 0.9516 350 0.7 5.29E-04 1.0573 470 0.75 5.66E-04 1.1328 630 0.8 6.04E-04 1.2083 750 0.82 6.19E-04 1.2385

1100 0.9 6.80E-04 1.3594 2000 0.95 7.17E-04 1.4349

Figure 13 Shrinkage Shortening

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APPENDIX Other material codes implemented in the program are:

ACI 1978 ACI 1992 IS 1343-1980 AASHTO 1994 AASHTO 2007 IRC 18-2000 BS8110 CEB/FIP 1978 CEB1 1978 EC 2004 Hong Kong code 2004 User Defined