On is viewed as as a linearly elastic material, As shown Figure 12a, the CFRP c,t = (3) t0 that is elastic up to its ultimatestrength and strain f fp and fp . As shown Figure 12b, t0 c,t tu ft the steel reinforcement is assumed to be ideal elastic-plastic [26,34], exactly where f y and y are the yielding pressure and strain, respectively. Furthermore, Figure 12c shows the anxiety train exactly where ft would be the axial tensile strength of UHPC, and ft = 0.668fr0 [37]; t0 could be the peak tension relationship of UHPC. For the compression branch, the initial curve terminates at the peak strain, and t0 = ft/Ec [36]; and cu would be the ultimate tension strain of UHPC, and tu = 30ft/Ec compression strain 0 , exactly where it joined a horizontal line [35]. For the tensile branch, the [36]. bilinear connection suggested by Graybeal [36] was adopted herein.(a)(b)(c)Figure 12. Stress train relationships ofof components: (a) pressure train partnership of CFRP; (b) stressFigure 12. Anxiety train relationships components: (a) strain train connection of CFRP; (b) stressstrain partnership of steel; (c) tension train relationship of UHPC. strain relationship of steel; (c) tension train relationship of UHPC.Appl. Sci. 2021, 11,14 ofFor the compression tension c,c -strain c,c partnership: c,c = fc fcn – two 1+(n-2)0 c,c 0 0 c,c cu(2)where f c may be the axial compressive strength of UHPC; n = Ec /Es , where Ec may be the elastic modulus of UHPC, Es will be the equivalent secant modulus at peak state; = cu /0 , exactly where cu will be the ultimate compression strain, cu = 4500 [30], and 0 could be the peak compression strain, which may be adopted as 3500 [30]. For the tensile anxiety c,t -strain c,t relationship: c,t =Appl. Sci. 2021, 11,ft ftt t0 c,t t0 t0 c,t tu(three)15 of 21 where f t will be the axial tensile strength of UHPC, and f t = 0.668f r0 [37]; t0 would be the peak tension strain, and t0 = f t /Ec [36]; and cu could be the ultimate tension strain of UHPC, and tu = 30f t /Ec [36].4.2. Determination of Stress Increment in CFRP Tendons 4.two. Determination of Strain Increment in CFRP Tendons Figure 13 shows the simplified calculation model of the just supported UHPC Figure 13 shows the simplified calculation model of your just supported UHPC beams prestressed with external CFRP tendons. beams prestressed with external CFRP tendons.Calculation Figure 13. Calculation model of specimens.The deflected length of the external tendon s+Ls could be determined by geometThe deflected length on the external tendon LsL+ Ls could be determined by aageometrical partnership, which might be written as: rical partnership, which could possibly be written as:( H + )two + Lw + k u ) two Ls +Ls +sLs a b = = ( H + ) two + (( Lw +ku )two L = = a’b'(four) (4)where Ls may be the initial length of your external tendon Ferritin light chain/FTL Protein Species involving the anchorage and the deviawhere Ls would be the initial length of the external tendon amongst the anchorage and the deviator; tor; Ls would be the elongation of CFRP tendon; H could be the Tartrazine Data Sheet vertical distance from the anchorage to Ls will be the elongation of CFRP tendon; H is the vertical distance in the anchorage towards the the deviator; would be the deflectionmidspan; L iswthethe horizontal distance in the anchordeviator; could be the deflection in in midspan; L is horizontal distance from the anchorage w age to deviator, and and =w = LLs22 -H 22;; k will be the actual distance from the anchorage for the the deviator, L L – H k would be the actual distance from the anchorage towards the to thew scentroid of cross section on the beam-end, which may be calculated employing Equation (five); centr.