EFECTOS DEL CÁNCER Y DE SU TRATAMIENTO EN EL CRECIMIENTO Y DESARROLLO

Example 5.6: A simply supported concrete beam in an interior location (Figure 5.27) reinforced with four 25-mm bars is required to carry a 100% increase in its original design live load. Assuming that the beam is safe in shear for such a strength increase, and also assuming a deflection and crack width under 100% increased loads are acceptable, design a CFRP wrap system (low grade) to carry flexural loads for future live loads. Use the beam and CFRP properties given in Table 5.8 and Table 5.9.

Existing and new loadings and associated midspan moments for the beam are summarized in Table 5.10. Strengthening the existing reinforced concrete beam using FRP system is proposed having the properties shown in Table 5.9. Three 381-mm wide × 6.7-m long plies will be bonded at the bottom (soffit) of the beam.

Part 1: Design of FRP wrap system for flexural strengthening using CFRP wraps.

The level of strengthening is acceptable because the criterion of strength limit as FIGURE 5.27 Simply supported beam with FRP external reinforcement.

TABLE 5.8

Dimensions and Beam Properties

Length of the beam, l 6.7 m Width of the beam, w 381 mm

d 546 mm

h 610 mm

fc′ 38 N/mm²

f_y 414 N/mm²

φ(Mn) w/o FRP 386.34 kN-m

Bars Four 25mm

546 mm 6.7 m

381 mm

610 mm 4–25 mm bars

f_c′ = 38 N/mm²

f_y = 414 N/mm² 3–381 mm × 6400 mm FRP wrap plies

W_DL, W_LL

Simply supported interior beam Cross section FRP wrap

described by Equation 5.5 is satisfied. The existing moment capacity without wrap, (φMn)_{w/o FRP} = 386.34 kN-m, is greater than the unstrengthened moment limit, (1.2MDL

+ 0.85MLL)new = 282.58 kN-m.

Step 1. Analyze the current beam. The properties of the concrete are (SI units) from ACI 318, Section 10.2.7.3:

The properties of the existing reinforcing steel:

TABLE 5.9

Manufacturer’s Reported FRP-System Properties

Thickness per ply, t_f 0.89 mm

Ultimate tensile strength, f_fu* 0.65 kN/mm²

Rupture strain, εfu* 0.017 mm/mm

Modulus of elasticity of FRP laminates, E_f 38.6 kN/mm²

TABLE 5.10

Loadings and Corresponding Moments

Existing Loads Future Loads

Dead loads, w_DL 16.05 N/mm 16.05 N/mm

Live load, w_LL 18.24 N/mm 36.47 N/mm

Unfactored loads, w_DL + w_LL 34.28 N/mm 52.52 N/mm Unstrengthened load limit, 1.2w_DL + 0.85w_LL n/a 50.26 N/mm Factored loads, 1.2w_DL + 1.6w_LL 48.44 N/mm 77.62 N/mm

Dead-load moment, M_DL 90.16 kN-m 90.16 kN-m

Live-load moment, M_LL 102.53 kN-m 205.06 kN-m

Service-load moment, M_s 192.76 kN-m 295.29 kN-m

Unstrengthened moment limit, 1.2M_DL + 0.85M_LL n/a 282.58 kN-m

Factored moment, M_u 272.33 kN-m 436.38 kN-m

β1=1 09. −0 008. fc′

β1−1 09 0 008 38. − . ( )=0 786.

Ec=4750 fc′

Ec=4750 38 N/mm² =29 280, N/mm²

ρsn_s = (0.0097)(6.86) = 0.0668 The minimum reinforcement ratio:

The depth of neutral axis (NA):

c = a/β1

ρ^s A^s

≡bd

As≡( )( )( . ) = 4 25 4 .

4 2026 83

π 2mm² mm²

ρ^s≡ 2026 83 =

381 . 0 0097

( mm .

mm)(546 mm)

2

n E

s E

s c

≡

ns≡ 200 =

29 15 kN/mm 6 86 kN/mm

2

. 2 .

ρmin= . ′ 0 249 f

f

c y

ρmin≡0 249. 38 = . 414 0 0037 ρ^provided(=0 0097. )>ρ_min(=0 0037. )

a A f f b

s y c

=0 85. ′

a= ( . )( ) = ( .2026 83 414)( )( ) .

0 85 38 381 68 18 mm

c = 68.18/0.786 = 86.74 mm The ratio of the depth of the NA to effective depth:

The strain in steel:

The beam is ductile with the steel strain exceeding the yield value of 0.002 and the minimum limit of tension-controlled failure mode strain value of 0.005. A nominal strength reduction factor φ = 0.9 will be used.

M_n = (2026.83)(414) = 429.55 kN-m

φMn = 0.9 M_n

φMn = 0.9(429.55) = 386.6 kN-m

Step 2. Compute the FRP-system design material properties. The beam is in an interior location and CFRP material will be used. Therefore, an environmental-reduction factor CE of 0.95 is used as per Table 5.1:

ffu = CEffu* c

dt

=86 75= 546. 0 1588

.

c dt

=0 1588. ≤0 375.

ε^s d c

= ⎛ c−

⎝⎜ ⎞

0 003. ⎠⎟

ε^s= ⎛ −

⎝⎜ ⎞

⎠⎟ = >

0 003 546 86 74

86 74 0 0158 0 005

. .

. . .

ε^y=ε^sy= ⎛⎝⎜ ⎞

⎠⎟= 414

200 000 0 002

, . mm/mm

Mn=A f ds y( −a/ )2

546 68 18

m− 2 mm

⎛

⎝⎜ ⎞

⎠⎟ .

ffu = (0.95)(650 N/mm²) = 617.5 N/mm² εfu = CEεfu*

εfu = (0.95)(0.017 mm/mm) = 0.0161 mm/mm

Step 3. Preliminary calculations. The properties of the concrete are from ACI 318, Section 10.2.7.3:

β1 = 1.09 – 0.008(38) = 0.786

The properties of the existing reinforcing steel (calculated in Step 1):

A_s = 2026.83 mm² ρs = 0.0097

n_s = 6.86 ρsns = 0.0668

The properties of the externally bonded CFRP reinforcement:

Af = ntfwf

Af = (3 plies)(0.89 mm/ply)(381 mm) = 1017.27 mm² β1=1 09. −0 008. fc′

Ec=4750 fc′

Ec=4750 38 N/mm² =29 280, N/mm²

ρ^f A^f

≡ bd

ρ^f≡ 1017 27 =

381 . 0 0049

( mm .

mm)(546 mm)

2

n E

f E

f c

≡

Step 4. Determine the existing state of strain on the soffit. The existing state of strain is calculated assuming the beam is cracked and the only loads acting on the beam at the time of the FRP installation are dead loads. A cracked section analysis of the existing beam gives k = 0.2984 and I_cr = 2.827 × 10⁹ mm⁴.

(with steel bars)

c = kd

c = (0.3048)(546) = 166.42 mm

(with steel bars)

εbi = 0.00045

Step 5. Determine the bond-dependent coefficient of the FRP system. The dimen-sionless bond-dependent coefficient for flexure, κm, is calculated using Equation 5.6.

Compare nEftf to 1,000,000:

(3)(38,600 N/mm²)(0.89 mm) = 103,062 < 175,336

nf ≡ 38 6 =

29 28. 1 32

. kN/mm .

kN/mm

2 2

ρ^fn^f =( .0 0049 1 32)( . )=0 0065.

k= (ρs sn)²+2(ρs sn)−(ρs sn)

k= ( .0 0668)²+2 0 0668( . ) ( .− 0 0668)=0 3048.

I bc

n A d c

cr= ³+ s s − ²

3 ( )

Icr=( )( . ) + −

( . )( . )(

546 166 42

3 6 86 2026 83 564 163

3

..92)²=3 037 10. × ⁹ mm⁴

ε^{bi DL}

cr c

M h kd

= I E( − )

ε^bi=(90 16 10. × ⁶ N-mm)[610 mm−( .0 3048 546)( mm))]

(3.037×10⁹ mm⁴)(29 280, N/mm⁴)

Therefore,

κm = 0.74 ≤ 0.90

Step 6a. Estimate c, the depth to the neutral axis. A reasonable initial estimate of c is or c = 0.2d, as suggested by ACI 440.2R-02. The value of c is adjusted after checking equilibrium.

Step 6b. Determine the effective strain in the FRP reinforcement. The effective strain in the FRP is calculated from Equation 5.7:

Note that for the neutral axis depth selected, we can expect a tension- and compres-sion-controlled failure mode with steel reinforcement strain (0.0114 > 0.005, from Step 5c) without FRP rupture leading to secondary compression failure in the form of concrete crushing.

εfe = 0.0125 ≤ 0.0119 (not satisfied)

Note that the values are close but the condition is not satisfied; check again after obtaining final value of c.

Step 6c. Calculate the strain in the existing reinforcing steel. The strain in the reinforcing steel can be calculated using similar triangles according to Equation 5.33b:

Step 6d. Calculate the stress level in the reinforcing steel and FRP. The stresses are calculated using Equation 5.13a and b, Equation 5.14a, b, and c, and Equation 5.25.

fs = Esεs≤ fy

Step 6e. Calculate the internal force resultants and check equilibrium. Force equilibrium is verified by checking the initial estimate of c with Equation 5.35 by noting that c = a/β1.

= 137.49 mm

c = 137.49 mm ≠ 144.06 mm (assumed)

Therefore, revise the estimate of c and repeat Step 6a through Step 6e until equi-librium is achieved. Using spreadsheet programming for iterations is suggested to obtain a hassle-free solution within a few seconds. Results of the final iteration are:

= 129.84 mm c = 129.84 mm (as assumed)

Therefore, the value of c selected for the final iteration is correct.

a = β1c = 0.786(129.84 mm) = 102.05 mm c = 129.84 mm

εs = 0.0095 f_s = f_y = 414 N/mm² εfe = 0.0104 (see note below)

ffe = 407.9 N/mm² Note:

εfe = 0.0106 ≤ 0.0119 (satisfied)

Note that the alternate approach for Step 6a to Step 6e is shown in Example 5.8 and Example 5.9.

Step 7. Calculate the design flexural strength of the section. The design flexural strength is calculated using Equation 5.4 and Equation 5.36 with appropriate

sub-c=(2026 83. mm²)(414 N/mm²) (+ 1017 27. mm²)(4077 9

0 85 0 786 38 381

. )

( . )( . )( )(

N/mm

N/mm mm)

2 2

ε^fe h c ε^bi κ ε^{m fu}

= ⎛ c−

⎝⎜ ⎞

⎠⎟ − ≤ 0 003.

ε^fe= ⎛ −

⎝⎜ ⎞

⎠⎟− ≤

0 003 610 129 84

129 84 0 00045 0 7

. .

. . . 44 0 0161( . )

stitutions. An additional reduction factor, ψf = 0.85, is applied to the bending strength contributed by FRP system. Because εs = 0.0095 > 0.005, a strength-reduction factor of φ = 0.90 is used as per Equation 5.9:

φMn = 551.2 kN-m ≥ Mu = 436.38 kN-m

Therefore, the strengthened section is capable of sustaining the new required moment strength.

Step 8. Check the service stresses in the reinforcing steel and the FRP. Calculate the depth of the cracked neutral axis by summing the first moment of the areas of the elastic transformed section without accounting for the compression reinforce-ment as per Equation 5.28:

k = 0.3185

kd = (0.3185)(546 mm) = 173.9 mm

Calculate the stress level in the reinforcing steel using Equation 5.30h:

φM φ A f d a ψ

fs,s = 0.287 kN/mm²

Verify that the stress in steel is less than recommended limit as per Equation 5.10:

fs,s≤ 0.80fy

fs,s = 287 N/mm²≤ (0.80)(414 N/mm²) = 331.2 N/mm²

Therefore, the stress level in the reinforcing steel is within the recommended limit.

Calculate the stress level in the FRP due to maximum service loads (assume sustained) using Equation 5.31d and verify that it is less than the creep-rupture stress limit.

For a carbon FRP system, the creep-rupture stress limit is listed as per Table 5.2:

Ff,s = 0.55ffu

fs,s = 47.55 N/mm²≤ (339.625 N/mm²)

Therefore, the stress level in the FRP is within the recommended creep-rupture stress limit.

Example 5.7: If the beam in Example 5.6 was to be designed with a GFRP wrap system (normal grade) to carry the same increase in flexural loads due to future live loads, what changes in design are expected? Assume that the proposed GFRP wrap system (normal grade) has same properties (hypothetically) as those of the CFRP wrap system provided in Table 5.9.

Design: Though the properties of GFRP are hypothetically stated to be same as those of CFRP, design values with environmental reduction factor and creep-rupture stress limit values are lower for GFRP. Because it is tension-controlled failure without FRP rupture as noted in Example 5.1, we will verify the values of GFRP stress and strain at the ultimate and compare these with the design values (similar to Example 5.1). If stresses are within the design values, then check the creep-rupture strength, which is different for GFRP and CFRP systems.

Note that the procedures in Step 6 of Example 5.6 must be calculated with the new design values of the proposed GFRP because the design values of stress and strain are lower for the proposed GFRP as compared to CFRP in Example 5.6 due to lower value of CE.

Analysis and calculation details for GFRP wrap system. FRP properties:

ffu = CEffu*

ffu = (0.75)(0.65 kN/mm²) = 0.4875 kN/mm² εfu = C_Eεfu*

εfu = (0.75)(0.017 mm/mm) = 0.0127 mm/mm

From Step 6e of Example 5.6, the depth of the neutral axis and the stress and strain in steel and FRP:

= 129.84 mm c = 129.84 mm (as assumed)

c A f A f f b

s s f fe

c

= +

′ 0 85. β1

c=(2026 83. mm²)(414 N/mm²) (+ 1017 27. mm²)(4077 91

0 85 0 786 38 381

. )

( . )( . )( )(

N/mm

N/mm mm)

2 2

a = β1c = 0.786(129.84 mm) = 102.05 mm

Step 8. Check the service stresses in the reinforcing steel and the FRP.

For a carbon FRP system, the creep-rupture stress limit is listed as per Table 5.2:

Ff,s = 0.55ffu

ff,s = 44.76 N/mm²≤ (0.2)(487.5 N/mm²) ff,s = 44.76 N/mm²≤ (97.5 N/mm²)

Therefore, the stress level in the FRP is within the recommended creep-rupture stress limit.

Example 5.8: Analyze the design in Example 5.6 to determine the depth of the neutral axis for a balanced failure and the amount of FRP reinforcement needed for a balanced failure. Compare the calculated amount of FRP reinforcement for a balanced failure and compare it with the amount of FRP reinforcement provided to determine the possible failure mode for the FRP strengthened beam.

Analysis: The hypothetical balanced failure mode is assumed to occur when strains in extreme tension and compression fibers have reached their limit values simultaneously.

In document TÍTULO: Máster Universitario en Condicionantes Genéticos, Nutricionales y Ambientales del Crecimiento y Desarrollo (página 67-76)