Los mercados de locales industriales y de hotelería y sus perspectivas

The final task of this project is to propose a design methodology for repairing and retrofitting end damage girders with short shear spans. The proposed design methodology is discussed in the following sections along with two numerical examples. The layout of FRP laminate repair system is illustrated in Figure 7.1.

7.1 PROPOSED DESIGN STEPS

Figure 7.1. Proposed layout of FRP laminate repair system.

Step 1: Estimate the total loss of shear capacity, ∆𝐹𝐹 using any method or by using Equation 7-1,

which assumes that the loss of shear capacity is mainly due to the cracking of concrete in the web. ∆𝑭𝑭 = 𝒇𝒇𝒓𝒓 × 𝒍𝒍 × 𝒄𝒄 (7-1)

Where 𝑓𝑓𝑟𝑟 is the modulus of rupture of concrete, 𝑐𝑐 is the thickness of the damaged concrete, and 𝑙𝑙 is

the longitudinal projection of initial shear crack. The initial shear crack appears mainly in the web, and for simplicity, is assumed to be at an angle of approximately 45° with respect to longitudinal axis of the beam as shown in Figure 7.2.

Step 2: Determine the average longitudinal tensile strain in the web, 𝜺𝜺𝒎𝒎 (see Figure 7.3) using

Equation 7-2. It is noted that Equation 7-2 is used for the case where the shear reinforcement satisfies the minimum requirement of shear reinforcement specified in AASHTO (2017).

Table B5.2-1 from AASHTO-LRFD (2017) is used to perform iterations to obtain 𝜽𝜽 and 𝜷𝜷 that are used in Equation 7-2.

Figure 7.3. Average tensile strain in the web.

𝜺𝜺𝒎𝒎 =

|𝑴𝑴𝒖𝒖|

𝒅𝒅𝒗𝒗 +𝟎𝟎.𝟓𝟓𝑵𝑵𝒖𝒖+𝟎𝟎.𝟓𝟓�𝑽𝑽𝒖𝒖−𝑽𝑽𝒑𝒑�𝒄𝒄𝒄𝒄𝒄𝒄𝜽𝜽−𝑨𝑨𝒑𝒑𝒑𝒑𝒇𝒇𝒑𝒑𝟎𝟎

𝟐𝟐(𝑬𝑬𝒑𝒑𝑨𝑨𝒑𝒑+𝑬𝑬𝒑𝒑𝑨𝑨𝒑𝒑𝒑𝒑) (7-2)

Where 𝜽𝜽 is an angle between diagonal compressive stress and longitudinal axis of the beam as shown in Figure 7.4; 𝜷𝜷 is a factor relating to the effect of longitudinal strain on the shear capacity of

concrete. 𝑴𝑴𝒖𝒖 is factored moment and is not taken less than 𝑽𝑽𝒖𝒖𝒅𝒅𝒗𝒗; 𝑽𝑽𝒖𝒖 is factored shear force; 𝑵𝑵𝒖𝒖 is

factored axial force; 𝑽𝑽𝒑𝒑 is component in the direction of the applied shear of the effective

prestressing force; 𝒅𝒅𝒗𝒗 is effective shear depth. 𝑨𝑨𝒑𝒑 and 𝑨𝑨𝒑𝒑𝒑𝒑 are the area of non-prestressing tensile

reinforcement and area of prestressing steel, respectively; 𝑬𝑬𝒑𝒑 and 𝑬𝑬𝒑𝒑 are the Young’s modulus of

non-prestressing tensile reinforcement and prestressing steel, respectively. 𝒇𝒇𝒑𝒑𝟎𝟎 is a parameter taken

as modulus of elasticity of prestressing tendons multiplied by the locked-in difference in strain between the prestressing tendons and surrounding concrete, for the usual levels of prestressing, a value of 0.7𝒇𝒇𝒑𝒑𝒖𝒖 is appropriate (AASHTO 2017).

Step 3: Plug 𝜺𝜺𝒎𝒎, 𝜽𝜽 into the Equation 7-3 to compute principal tensile strain (𝜺𝜺𝟏𝟏).

𝜺𝜺𝟏𝟏 = 𝜺𝜺𝒎𝒎 + [𝜺𝜺𝒎𝒎+ 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟐𝟐(𝟏𝟏 − �𝟏𝟏 −_𝒇𝒇′𝝂𝝂𝒖𝒖_𝒄𝒄𝟎𝟎.𝟖𝟖+𝟏𝟏𝟏𝟏𝟎𝟎𝜺𝜺_{𝒑𝒑𝒔𝒔𝒔𝒔𝜽𝜽𝒄𝒄𝒄𝒄𝒑𝒑𝜽𝜽}𝟏𝟏)]𝒄𝒄𝒄𝒄𝒄𝒄𝟐𝟐𝜽𝜽 (7-3)

Where 𝝂𝝂𝒖𝒖 is average factored shear stress in the concrete.

Step 4: Obtain vertical component (𝜺𝜺𝒚𝒚) of 𝜺𝜺𝟏𝟏 by multiplying 𝜺𝜺𝟏𝟏 by 𝒄𝒄𝒄𝒄𝒑𝒑𝜽𝜽. The directions of 𝜺𝜺𝟏𝟏 and 𝜺𝜺𝒚𝒚

are shown in Figure 7.4. The FRP design strain (𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭) is computed by dividing 𝜺𝜺𝒚𝒚 by a factor 𝝁𝝁 (see

small-scale and full-scale beam tests, the value of 𝝁𝝁 was found to be on average equal to 6.0. If the computed 𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭 is found to be greater than 0.004, a value of 0.004 should be used to design FRP

laminate.

Figure 7.4. Direction of 𝜺𝜺𝟏𝟏 and 𝜺𝜺𝒚𝒚.

𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭= 𝜺𝜺_𝝁𝝁𝒚𝒚 = 𝜺𝜺𝟏𝟏 × 𝒄𝒄𝒄𝒄𝒑𝒑𝜽𝜽_𝝁𝝁 (7-4)

Step 5: Choose the material type of FRP system and calculate the amount of FRP material (thickness

of FRP layer, t) needed using Equation 7-5.

𝒄𝒄 = _𝑬𝑬 ∆𝑭𝑭

𝑭𝑭𝑭𝑭𝑭𝑭 × 𝑳𝑳 × 𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭 (7-5)

Where 𝑬𝑬𝑭𝑭𝑭𝑭𝑭𝑭 is the Young’s modulus of the selected FRP material, 𝑳𝑳 is the length of shear span as

shown in Figure 7.1.

Step 6: Determine the width of longitudinal strip (𝒘𝒘), spacing (𝒑𝒑) between panels and anchor length

(𝑳𝑳𝒎𝒎) (see Figure 7.1).

It is recommended to place 3 longitudinal strips: at the top of the web, the bottom of the girder, and the joint between web and bottom flange. Based on the testing of full-scale girders, the width of the strips is recommended to be at least 3 in. The spacing (𝒑𝒑) between FRP panels is suggested to be at least 1.0 in. to ensure sufficient bond between anchor strip and concrete. For girders with sufficient space around end region, longitudinal strip could be looped around the end of the girder similar to the case in the testing of small-scale beam in Chapter 4. For girders with inadequate space around end region, the longitudinal strips could be terminated at girder end with sufficient anchorage length using the same layout of full-scale girder testing in Chapter 5. For both cases, the anchorage length (𝑳𝑳𝒎𝒎) needs to be equal or greater than 6 in. It is important to note that detailing values (𝒘𝒘, 𝒑𝒑 and 𝑳𝑳𝒎𝒎)

recommended herein are primarily based on the limited tests that were carried out during this study. More tests are recommended to further optimize these values.

7.2 DESIGN EXAMPLE USING SMALL-SCALE BEAMS

The small-scale beam test conducted in Chapter 4 is used as a design example following the proposed design method. The cover of the small-scale beam is 12.7 mm (0.5 in.) in thickness and the concrete compressive strength is 49.4 MPa (7.16 ksi). The length of shear span 𝑳𝑳 equals to 508 mm (20 in.) and the height (𝒅𝒅) and thickness (𝒃𝒃𝒘𝒘) of the web is 190 mm (7.5 in.) and 76.2 mm (3 in.) respectively. The

area of prestressing strands (𝑨𝑨𝒑𝒑𝒑𝒑) is 296.1 mm2 (0.459 in2). The area of mild steel bars (𝑨𝑨𝒑𝒑) is 189.7

mm2 _{(0.294 in}2_{). The effective shear depth (𝒅𝒅}

𝒗𝒗) is calculated as 377 mm (14.83 in.). Factor 𝒇𝒇𝒑𝒑𝟎𝟎 is

taken as 0.7𝒇𝒇𝒑𝒑𝒖𝒖 which is 1303 MPa (189 ksi). The Young’s moduli of mild steel and prestressing strand

are 200 GPa (29000 ksi) and 196 GPa (28700 ksi), respectively. The FRP design process is shown below:

Step 1: calculate the tensile force loss (∆𝑭𝑭)

∆𝑭𝑭 = 𝒇𝒇𝒓𝒓 × 𝒍𝒍 × 𝒄𝒄 = 𝟏𝟏. 𝟓𝟓√𝟏𝟏𝟏𝟏𝟕𝟕𝟎𝟎_{𝟏𝟏𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟏𝟏. 𝟓𝟓 × 𝟎𝟎. 𝟓𝟓 = 𝟐𝟐. 𝟒𝟒 𝒌𝒌𝒔𝒔𝒑𝒑𝒑𝒑}

Step 2: compute average longitudinal tensile strain (𝜺𝜺𝒎𝒎) in the web

𝑴𝑴𝒖𝒖 and 𝑽𝑽𝒖𝒖 of the small-scale beams are calculated as 205.9 kN-m (151.8 kip-ft) and 94.3 kN

(21.2 kips). Through iterations using AASHTO Table B5.2-1, 𝜽𝜽 and 𝜷𝜷 are 36.4 and 2.23, respectively. Since there is no axial load or draped strand, 𝑵𝑵𝒖𝒖 and 𝑽𝑽𝒑𝒑 are zero.

𝜺𝜺𝒎𝒎 = |𝑴𝑴𝒖𝒖| 𝒅𝒅𝒗𝒗 + 𝟎𝟎. 𝟓𝟓|𝑽𝑽𝒖𝒖|𝒄𝒄𝒄𝒄𝒄𝒄𝜽𝜽 − 𝑨𝑨𝒑𝒑𝒑𝒑𝒇𝒇𝒑𝒑𝟎𝟎 𝟐𝟐�𝑬𝑬𝒑𝒑𝑨𝑨𝒑𝒑+ 𝑬𝑬𝒑𝒑𝑨𝑨𝒑𝒑𝒑𝒑� = 𝟏𝟏𝟓𝟓𝟏𝟏. 𝟖𝟖 × 𝟏𝟏𝟐𝟐 𝟏𝟏𝟒𝟒. 𝟖𝟖𝟖𝟖 + 𝟎𝟎. 𝟓𝟓 × 𝟐𝟐𝟏𝟏. 𝟐𝟐 × 𝒄𝒄𝒄𝒄𝒄𝒄𝟖𝟖𝟕𝟕. 𝟒𝟒° − 𝟎𝟎. 𝟒𝟒𝟓𝟓𝟒𝟒 × 𝟏𝟏𝟖𝟖𝟒𝟒 𝟐𝟐(𝟐𝟐𝟒𝟒𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟎𝟎. 𝟐𝟐𝟒𝟒𝟒𝟒 + 𝟐𝟐𝟖𝟖𝟏𝟏𝟎𝟎𝟎𝟎 × 𝟎𝟎. 𝟒𝟒𝟓𝟓𝟒𝟒) = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟏𝟏𝟏𝟏𝟕𝟕𝟒𝟒

Step 3: plug 𝜺𝜺𝒎𝒎 into Equation 7-3 and solve for principal tensile strain (𝜺𝜺𝟏𝟏), 𝜺𝜺𝟏𝟏= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟖𝟖𝟏𝟏𝟒𝟒𝟎𝟎.

Step 4: obtain FRP design strain (𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭)

𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭 = 𝜺𝜺_{𝝁𝝁 =} 𝒚𝒚 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟖𝟖𝟏𝟏𝟒𝟒𝟎𝟎 × 𝒄𝒄𝒄𝒄𝒑𝒑𝟖𝟖𝟕𝟕. 𝟒𝟒°_{𝟕𝟕. 𝟎𝟎} = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟓𝟓𝟎𝟎

Step 5: choose FRP material and compute the number of FRP layers

CFRP laminate is selected as repair material. From Table 4.2, the Young’s modulus 𝑬𝑬𝑪𝑪𝑭𝑭𝑭𝑭𝑭𝑭 is 86.9 GPa

(13000 ksi) and the thickness of each CFRP layer is 1.24 mm (0.049 in.).

𝒄𝒄𝑪𝑪𝑭𝑭𝑭𝑭𝑭𝑭= _𝑬𝑬 ∆𝑭𝑭

𝑪𝑪𝑭𝑭𝑭𝑭𝑭𝑭 × 𝑳𝑳 × 𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭=

𝟐𝟐. 𝟒𝟒

𝟏𝟏𝟖𝟖𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟐𝟐𝟎𝟎 × 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟓𝟓𝟎𝟎 = 𝟎𝟎. 𝟎𝟎𝟏𝟏𝟒𝟒 𝒔𝒔𝒔𝒔. Hence, one layer of CFRP laminate is selected.

7.3 DESIGN EXAMPLE USING FULL-SCALE GIRDERS

Full-scale girder tested in Chapter 5 is used as another design example following the proposed design method. The cover of the full-scale beam is 31.75 mm (1.25 in.) in thickness and the concrete

compressive strength is 60.3 MPa (8.75 ksi). The length of shear span 𝑳𝑳 equals to 1079 mm (42.5 in.) and the height (𝒅𝒅) and thickness (𝒃𝒃𝒘𝒘) of the web is 431.8 mm (17 in.) and 152.4 mm (6 in.)

respectively. The area of prestressing strands (𝑨𝑨𝒑𝒑𝒑𝒑) is 592.3 mm2 (0.918 in2). The effective shear

depth (𝒅𝒅𝒗𝒗) is calculated as 830 mm (32.7 in.). Factor 𝒇𝒇𝒑𝒑𝟎𝟎 could be taken as 0.7𝒇𝒇𝒑𝒑𝒖𝒖 which is 1303 MPa

(189 ksi). The Young’s moduli of mild steel and prestressing strand are 200 GPa (29000 ksi) and 186 GPa (27000 ksi), respectively. The FRP design process is shown below:

Step 1: calculate the tensile force loss (𝑭𝑭)

∆𝑭𝑭 = 𝒇𝒇𝒓𝒓 × 𝒍𝒍 × 𝒄𝒄 = 𝟏𝟏. 𝟓𝟓√𝟖𝟖𝟏𝟏𝟓𝟓𝟎𝟎_{𝟏𝟏𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟏𝟏𝟏𝟏 × 𝟏𝟏. 𝟐𝟐𝟓𝟓 = 𝟏𝟏𝟒𝟒. 𝟒𝟒 𝒌𝒌𝒔𝒔𝒑𝒑𝒑𝒑}

Step 2: compute average tensile strain (𝜺𝜺𝒎𝒎) in the web

𝑴𝑴𝒖𝒖 and 𝑽𝑽𝒖𝒖 of the full-scale beams are calculated as 797.4 kN-m (587.8 kip-ft) and 378.1 kN (85.0

kips). Through iterations using AASHTO Table B5.2-1, 𝜽𝜽 and 𝜷𝜷 are 36.4 and 2.23. Since there is no axial load or draped strand, 𝑵𝑵𝒖𝒖 and 𝑽𝑽𝒑𝒑 are zero.

𝜺𝜺𝒎𝒎 = |𝑴𝑴𝒖𝒖| 𝒅𝒅𝒗𝒗 + 𝟎𝟎. 𝟓𝟓|𝑽𝑽𝒖𝒖|𝒄𝒄𝒄𝒄𝒄𝒄𝜽𝜽 − 𝑨𝑨𝒑𝒑𝒑𝒑𝒇𝒇𝒑𝒑𝟎𝟎 𝟐𝟐�𝑬𝑬𝒑𝒑𝑨𝑨𝒑𝒑+ 𝑬𝑬𝒑𝒑𝑨𝑨𝒑𝒑𝒑𝒑� = 𝟓𝟓𝟖𝟖𝟏𝟏. 𝟖𝟖 × 𝟏𝟏𝟐𝟐 𝟖𝟖𝟐𝟐. 𝟏𝟏 + 𝟎𝟎. 𝟓𝟓 × 𝟖𝟖𝟓𝟓. 𝟎𝟎 × 𝒄𝒄𝒄𝒄𝒄𝒄𝟖𝟖𝟕𝟕. 𝟒𝟒° − 𝟎𝟎. 𝟒𝟒𝟏𝟏𝟖𝟖 × 𝟏𝟏𝟖𝟖𝟒𝟒 𝟐𝟐(𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟎𝟎. 𝟒𝟒𝟏𝟏𝟖𝟖) = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟐𝟐𝟎𝟎𝟏𝟏𝟓𝟓

Step 3: plug 𝜺𝜺𝒎𝒎 into Equation 7-3 and solve for principal tensile strain (𝜺𝜺𝟏𝟏), 𝜺𝜺𝟏𝟏= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟕𝟕𝟏𝟏𝟒𝟒.

Step 4: obtain FRP design strain (𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭)

𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭= 𝜺𝜺_{𝝁𝝁 =} 𝒚𝒚 𝜺𝜺𝟏𝟏× 𝒄𝒄𝒄𝒄𝒑𝒑𝜽𝜽_𝝁𝝁 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟕𝟕𝟏𝟏𝟒𝟒 × 𝒄𝒄𝒄𝒄𝒑𝒑𝟖𝟖𝟕𝟕. 𝟒𝟒°_{𝟕𝟕. 𝟎𝟎} = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟖𝟖𝟐𝟐

Step 5: choose FRP material and compute the number of FRP layers

CFRP laminate is selected as repair materials. From Table 4.2, the Young’s modulus 𝑬𝑬𝑪𝑪𝑭𝑭𝑭𝑭𝑭𝑭 is

86.9 GPa (13000 ksi) and the thickness of each CFRP layer is 1.24 mm (0.049 in.).

𝒄𝒄𝑪𝑪𝑭𝑭𝑭𝑭𝑭𝑭= _𝑬𝑬 ∆𝑭𝑭

𝑪𝑪𝑭𝑭𝑭𝑭𝑭𝑭 × 𝑳𝑳 × 𝜺𝜺𝑭𝑭𝑭𝑭𝑭𝑭 =

𝟏𝟏𝟒𝟒. 𝟒𝟒

𝟏𝟏𝟖𝟖𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟒𝟒𝟐𝟐. 𝟓𝟓 × 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟖𝟖𝟐𝟐 = 𝟎𝟎. 𝟎𝟎𝟖𝟖𝟖𝟖 𝒔𝒔𝒔𝒔. Hence, one layer of CFRP laminate is selected.

CHAPTER 8: SUMMARY, CONCLUSIONS AND