Post-tensioned prestressed concrete bridge - assignment

Post-tensioned prestressed concrete bridge - assignment Design a post-tensioned prestressed concrete bridge of a three-span arrangement. The construction is prestressed at the age of 7 days and put into operation at the age of 100 days. The durability is expected to be 100 years. The structure is loaded by the dead load g 0 + g add and live load q. Individually assigned input parameters are: L [m] b [m] a [-] h [m] g add,k [kn/m ] q k [kn/m ] length of the middle span cross-sectional width ratio of outer span to middle span cross-sectional height additional dead load except own weight characteristic value live load characteristic value Prestressing reinforcement: 7-wire tendons, diameter 15 mm; f pk = 1770 MPa; f p0.1k = 1560 MPa Cross-section Tasks: - cross-section geometry characteristics (area A; position of centre of gravity - cg; moment of inertia I) - internal forces (N, V, M), extreme values of internal forces caused by live load - number of tendons - losses of prestress - rigorous SLS assessment in decisive cross-sections (stress limit) - ULS assessment in one of the decisive cross-section - structural drawing, prestressing reinforcement drawing 1

Following model homework uses model input parameters. In your homework, use your own parameters obtained in the first lesson. 1. Dimensions, cross-section geometry, material characteristics Define the geometry of the beam based on your own input parameters. The positions of the decisive cross-sections are marked as 5, 10 and 15. Calculate following cross-sectional characteristics: Area A = 1.738 m Distance between cg and bottom surface e b = 0.78 m Distance between cg and top surface e t = 0.518 m Moment of inertia in vertical axis I y = 0.76 m 4 Define the properties of used materials: Concrete: C 30/37, f cd = f ck = 0 MPa 1,5 Prestressing steel: 7-wire tendons, diameter 15 mm, area of one tendon A p1 = 150 mm f pk = 1770 MPa (ultimate stress) f p0.1k = 1560 MPa (conventional yield stress)

. Loading Dead load g k [kn.m -1 ] γ f [-] g d [kn.m -1 ] Weight of structure g 0 43.45 1.35 58.66 Other dead load g-g 0 7.56 1.35 10.1 Total 51.01-68.87 Imposed load q k [kn.m -1 ] γ f [-] q d [kn.m -1 ] Live load q 16.5 1.5 4.75 Total 16.5-4.75 Loading cases listed below must be considered: LC1 Own weight LC Additional dead load LC3 Live load I LC4 Live load II LC5 Live load III LC6 Live load IV 3. Internal forces Use any software for structural analysis to calculate internal forces on the structure (N, V, M). Calculate the internal forces in combinations listed in the table below. Following table summarizes obtained internal forces in the decisive cross-sections (5, 10, 15) caused by loading cases and their combinations. Loading Force, cross-section g 0 g add,k g q min q max g + q min g + q max M 5 617.74 109.6 77.34-64.91 504.11 46.43 131.45 M 10-773.09-491.98-365.07-1159.89 86.11-444.96-3178.96 M 15 00.78 358.5 379.3-190.74 973.1 188.56 335.51 Q 10, L -537.57-95.37-63.94-1.94 - -845.88 - Q 10, P 639.18 113.4 75.58-59.6-101.18 3.1 Result combination Prepare a result combination of maximal decisive values. 3

M 5,max = 131.45 knm M 10,min = -444.96 knm M 15,max = 335.51 knm (Notice: The position of the cross-section 5 is not in the centre of the outer span it is placed at the extreme value of bending moment.) 3. Estimated secondary prestressing moment As the structure is statically indefinite, secondary prestressing moments are created (moment diagram in the picture below). Firstly, its maximal value is estimated. M P = (10 15)% M 10,min Chosen estimation: 10 % M P = 0,1 M 10,min = 0.1 444.96 knm = 44.5 knm 3.3 Moment values for prestress design To obtain decisive moment values, both primary and secondary prestressing moment must be considered. M 5 = M 5,max + M P 5.4 18 5.4 = 131.45 knm + 44.5 knm = 1364.05 knm 18 M 10 = M 10,min + M P = 444.96 knm + 44.5 knm = 398.46 knm M 15 = M 15,max + M P = 335.51 knm + 44.5 knm = 3795.01 knm 4

4. Tendon design (position, prestressing force) 4.1 Eccentricity of tendon in cross-section 10 and 15 The tendon will be placed at the position of maximal possible eccentricity e 10 in cross-section 10. Maximal possible eccentricity e 5 or e 15 will be also used in cross-section 5 or 15, depending on whichever of these two has a higher value of bending moment. In this model homework, M 15 > M 5, so the maximal eccentricity is used for cross-section 15. The maximal possible eccentricity is then given by the thickness of the cover layer c of concrete. For this model homework, cover layer is given 100 mm. Do not forget to consider the diameter of the tendon duct, which is preliminarily designed to be 100 mm. Eccentricity of the tendon (cross-section 10): e P,10 = e t c d = 0.518 0.1 0.1 Eccentricity of the tendon (cross-section 15): e P,15 = e b c d = 0.78 0.1 0.1 = 0.368 m = 0.63 m (The eccentricity of tendon in cross-section 5 will be determined subsequently.) 4. Prestressing force design The aim of prestressing is to reduce the tension in concrete: THE REQUIRED STRESS AT THE TENSIONED SIDE IS ZERO (= the sum of stress created by load and prestressing is zero.) this requirement is a base for prestress design. 5 σ t f + σ t P = 0 [MPa]... for the cross-section 10 at the top of the beam σ b f + σ b P = 0 [MPa]... for the cross-section 15 at the bottom of the beam Cross-section 10: σ t f + σ t P = 0 where σ t f is a stress at the top caused by loading σ t P is a stress at the top caused by prestressing σ f t = M 10 e I t = 398.46 0.518 = 7.47 [MPa] (tension) y 0.76

σ p t = N P, 10 A N 10 P, ep,10 e I t y CM03 - Model homework Combining equations above, necessary prestressing force at the time of 100 years: N 10 P, = N 10 P, = f σ t 1 A +e P,10 e t Iy [kn] 7 470 000 1 1.738 +0.368 0.518 0.76 = 5900.8 [kn] Cross-section 15: (If M 15 < M 5, continue with cross-section 5 instead of 15) σ b f + σ b P = 0 where σ b f is a stress at the bottom caused by loading σ b P is a stress at the bottom caused by prestressing σ f b = M 15 e I b = 3795.01 0.78 = 10.75 [MPa](tension) y 0.76 σ p b = N P, 15 A N 15 P, ep,15 e I b y Combining equations above, necessary prestressing force at the time of 100 years: N 15 P, = N 15 P, = f σ b 1 A +e P,15 e b Iy [kn] 10 750 000 1 1,738 +0,63 0,78 0,76 = 4543.46 [kn] The decisive value of prestressing force: There can be only one value of prestressing force in the tendon it is necessary to choose the highest value for the design: N P, = max[n 10 P, ; N 15 P, ] = max [5900.8; 4543.46] = 5900.8 [kn] Due to losses of prestress, the force changes during the life of the structure. The prestressing force calculated above is designed for the age of 100 years of the construction. 4.3 Eccentricity of tendon in cross-section 5 To design the eccentricity in the cross-section 5, the ratio of primary moments will be used as a proportionality coefficient (This applies only for case M 15 > M 5, otherwise the process is reverse): M 5 = 131.45 M 15 335.51 = 0.37 e 5 = e 15 M 5 M 15 = 0.63 0.37 = 0.34 m (The position of the tendon peak in cross-section 5 is at the place of maximal bending moment, not in the mid-span). 6

Notice: If the extreme value of bending moment is higher in cross-section 5 than in cross-section 15, the prestressing force N P, should be determined based on the equation σ f b + σ P b = 0 [MPa] in cross-section 5 (not 15). Then the decisive value of prestressing force should be determined as: N P, = max[n 5 P, ; N 10 P, ]. 4.4 Geometry of tendon position Create points in the distance of 400 mm from the eccentricity of cross-section 10. Split the interval between this point and the mid-span into halves. Draw lines between created points. Parabolic shape of tendon is approximated using circular arcs. The lines drawn in the previous step are tangent to circular arcs of the tendon. Draw the upper (concave) circular arc from the cross-section 10 to the right side. The arc starts at the peak in the cross-section 10 and its length is up to the point placed 400 mm from the breaking point on the askew tangent. (In AutoCAD, use the circle described by two tangents and radius.) The ending point of the arc drawn in previous step is also the starting point of the lower (convex) arc. It is also the contact point of the lower arc with its askew tangent. Draw the lower arc using its tangents. Fill the distance between the end of the lower arch and the mid-span with a straight line. To draw the tendon shape to the left from the cross-section 10, use analogical procedure as in previous steps. The lowest point of the lower arch must be placed in the position of maximal bending moment M 5. Notice that the first 1500 mm of the tendon must be straight. 7

5. Preliminary prestressing design Number of tendons: Calculate the total cross-sectional area of all tendons A P necessary to carry the load: N P, = σ P, A p A P = N P, σ P, [mm ] A P = 5900.8 1053 10 6 = 5603 [mm ] where σ P, = σ P,max (1 5%) = 1404 0,75 = 1053 [MPa] The loss of prestress is estimated to be 5% at the end of the service life of the structure (100 years of age), the precise value will be determined in further calculation. σ P,max = min[0,8 f pk ; 0,9 f p,0,1,k ] = = min[0,8 1770; 0,9 1560] = = min[1416; 1404] = 1404 [MPa] Calculate the number of tendons into which the total cross-sectional area will be divided: n = A P = 5603 = 37.4 38 tendons A P,1 150 Two ducts of 19 tendons will be used. (The usual amount of tendons in a duct is 1, 15, 19 choose from these three options.) 8

6. Losses of prestress 9 CM03 - Model homework In previous step, the losses of prestress in the end of the service life were estimated to be 5% of the initial prestress. In further text, the losses of prestress are calculated precisely. 6.1 Immediate losses of prestress 6.1.1 Loss due to friction There is a friction between the tendon and the tendon duct in the curved parts. Also, there is friction in the straight tendon parts caused by presence of the duct supports. The friction forces along the length of the tendon are in opposite direction to the prestressing force in the tendon, decreasing it. This is the cause of the loss of prestress. σ P,1 = σ P,max (1 e μ (α+k l) ) where μ friction coefficient in curved part μ = 0.19 (for metal duct) k = 0.01 m -1 α central angle of circular arc in particular interval k friction coeficient in straight part (related to 1m of length) l length of particular interval angles: (cross-section 5) α = α 1 = 0.0657 [rad] lengths: (cross-section 10) α = α 1 + α + α 3 = 0.0657 + 0.10097 + 0.10097 = 0.671 [rad] (cross-section 15) α = α 1 + α + α 3 + α 4 + α 5 = 0.0657 + 0.10097 + 0.10097 + 0.11835 + 0.11835 = 0.50391 [rad] (cross-section 5) l = l 1 + l = 1.503 + 4.653 = 6.156 [m] (cross-section 10) l = l 1 + l + l 3 + l 4 + l 5 = 1.503 + 4.653 + 0.369 + 11.453 + 0.799 = 18.777 [m] (cross-section 15) l = l 1 + l + l 3 + l 4 + l 5 + l 6 + l 7 + l 8 = 1.503 + 4.653 + 0.369 + 11.453 + 0.799 + 0.799 + 13.887 + 0.349 = 33.81 [m] Loss of prestress due to friction: σ P,1,5 = 1404 (1 e 0.19 (0.0657+0.01 6.156) ) = 33.43 [MPa] σ P,1,10 = 1404 (1 e 0.19 (0.671+0.01 18.777) ) = 116.7 [MPa] σ P,1,15 = 1404 (1 e 0.19 (0.50391+0.01 33.81) ) = 07.57 [MPa] 6.1. Anchorage set loss At the moment of anchoring, the anchor wedge slips into the anchor head, causing length reduction of the tendon, which leads to the loss of prestress. The slip is decreased by the friction that means that the effect of slip decreases with the length of the tendon. Initial slip: w = 0.005 m

Hooke s law applies: σ = E p ε = E p w L w = σl E p = A w E p A w = w E P = 0.005 195 000 = 975 [MPa] where A w is the area under the stress-length curve changes of stress related to 1m of length in straight tendon: p P = σ P,0,max (1 e μ ( α+k l) ) = 1404 (1 e 0.19 (0+0.01 1) ) =.665 [MPa] changes of stress related to 1m of length in curved part no. 1: p 1 = σ P,0,max (1 e μ ( α 1+k l) ) = 1404 (1 e 0.19 (0.01403+0.01 1) ) = 6.395 [MPa] where: α 1 = a 1 L 1 = 0.0657 4.653 = 0.01403 [rad m 1 ] changes of stress related to 1m of length in curved part no. : p = σ P,0,max (1 e μ ( α +k l) ) = 1404 (1 e 0.19 (0.0088+0.01 1) ) = 5.011 [MPa] where: α = a L 1 = 0.10097 11.453 = 0.0088 [rad m 1 ] The effect of anchorage set loss is assumed to be eliminated in the curved part no. of the tendon. Area under the stress-length curve: (Notice: there is a straight part between curve no. 1 and curve no. make sure you omit that straight part, if you did not designed it in previous steps) A w = ( 1 x P,1 p P x P,1 + x P,1 (x 1 p 1 + x P, p P + x p )) + straight part + ( 1 x 1 p 1 x 1 + x 1 (x P, p P + x p )) + curve no. 1 10

+ ( 1 x P, x P, p P + x P, (x p P )) + (straight part) + 1 x x p + curve no. A w = ( 1 1.5.655 1.5 + 1.5 (4.65 6.395 + 0.369.665 + x 5.011)) + + ( 1 4.65 6.395 4.65 + 4.65 (0.369.665 + x 5.011)) + + ( 1 0.369 0.369.665 + 0.369 (x 5.011)) + + 1 x x 5.011 A w = 5.011x + 65.333x + 45.936 = 975 The solution of x (only one reasonable): x = 7.19 [m] Length of the tendon where the anchorage set loss is effective x w : x w = x P,1 + x 1 + x P, + x = 1.5 + 4.65 + 0.369 + 7.19 = 13.711 [m] The result shows that the effect of anchorage set loss will affect only the cross-section 5. Total anchorage set loss: σ P,w = (x P,1 p P + x 1 p 1 + x P, p P + x p ) = (1.5.665 + 4.65 6.395 + 0.369,665 + 7.19 5.011) = 141.51 [MPa] Anchorage set loss in cross-section 5: σ P,,5 = (141.51 (x P,1 p P + x 1 p 1 )) = (141.51 (1.5.665 + 4.65 6.395)) = 74.04 [MPa] Stresses in tendons after immediate losses summary: σ P0 = σ P,0,max + σ P,1 + σ P, Cross-section σ P,0,max σ P,1 σ P, σ P0 μ 5 1404-33.43-74.04 196.53 0.733 10 1404-116.7 0 187.73 0.78 15 1404-07.57 0 1196.43 0.676 (Notice: μ in this table does not stand for the friction coefficient) μ = σ P0 f Pk, where f pk = 1770 MPa 11

6. Long-term losses of prestress 6..1 Loss due to relaxation of prestressing reinforcement Metal reinforcement bars have the ability to relax when elongated. This means, that the stress in the tendon decreases over time even though their elongation was not reduced. σ p,r = σ P0 0.66.5 e 9.1 μ ( t 0.75 (1 μ) 1000 ) 10 5 where t is the time after prestressing [hours] Loss of prestress at the time of 100 days: Cross-section 5 σ 100 P,r.5 = 196.53 0.66.5 e 9.1 0.733 ( 400 0.75 (1 0.733) 1000 ) = 0.10 [MPa] Cross-section 10 100 σ P,r,10 = 187.73 0.66.5 e 9.1 0.78 ( 400 0.75 (1 0.78) 1000 ) = 19.14 [MPa] Cross-section 15 100 σ P,r,15 = 1196.43 0.66.5 e 9.1 0.676 ( 400 0.75 (1 0.676) 1000 ) = 11.46 [MPa] 10 5 10 5 10 5 Loss of prestress at the end of the service life (100 years): Total relaxation σ capacity Pr for t = 500 000 h (approx. 57 years) it is assumed that after this time the relaxation has no effect: Cross-section 5 0.75 (1 0.733) σ P,r,5 = 196.53 0.66.5 e 9,1 0,733 500 000 ( 1000 ) = 58.56 [MPa] Cross-section 10 0.75 (1 0.78) σ P,r,10 = 187.73 0.66.5 e 9.1 0.78 500 000 ( 1000 ) = 56.89 [MPa] Cross-section 15 0.75 (1 0.676) σ P,r,15 = 1196.43 0.66.5 e 9.1 0.676 500 000 ( 1000 ) = 41.96 [MPa] 10 5 10 5 10 5 1

6.. Loss due to creep CM03 - Model homework Concrete as a material is slightly viscous. As a result, its deformation increases even though the loading conditions are constant. As the concrete creeps, there is loss of prestress in the prestressing reinforcement. Force in the tendon after prestressing: = 5 σp,0 n AP1 = 196.53 38 150 = 7390. [kn] 5 N P,0 = 10 σp,0 n AP1 = 187.73 38 150 = 7340.06 [kn] 10 N P,0 = 15 σp,0 n AP1 = 1196.43 38 150 = 6819.65 [kn] 15 N P,0 Bending moments caused by dead load and secondary prestressing moment: M g 5 = 77.34 + 13.75 = 860.09 [knm] M g 10 = 365.07 + 44.50 = 8.57 [knm] M g 15 = 379.30 + 44.50 = 81.8 [knm Elastic stress in concrete at the place of cg of the tendon: σ g+p cp = N P,0 A i (g+p) 7.390 σ cp,5 = 1.757 N P,0 e P,i e I P,i + M g e y,i I P,i y,i 7.390 (0.5 (0.78 0.78)) 0.77 (0.5 (0.78 0.78)) = 5.08 [MPa] + 0.860 0.77 (g+p) 7.340 σ cp,10 = 1.757 7.340 (0.368 (0.786 0.78)) 0.78 (0.368 (0.786 0.78)) = 3.98 [MPa] +.83 0.78 (g+p) 6,80 σ cp,15 = 1.757 Loss of prestress due to creep: E p 6.80 (0.5 (0.78 0.777)) 0.80 (0.5 (0.78 0.777)) = 4.86 [MPa] σ Pc = E c (τ) σ g+p cp φ(t; τ) where τ time of prestressing τ = 7 day +.8 0.80 t time in the life of the beam t 1 = 100 days, t = 100 years The values of the Young s modulus at the time of 7 days (E c (τ)) and the creep coefitient of the concrete (φ(t; τ)) are usually determined using software. For the purpose of this homework, the values are given defaultly: 13

E c (7) = 1 700 MPa φ(100 days) = 0.8 φ(100 years) =.8 Loss of prestress due to creep at the time of 100 days: σ 100 P,4,5 = 195000 5.08 0.83 = 37.81 [MPa] 1748 100 σ P,4,10 = 195000 3.98 0.83 = 9.6 [MPa] 1748 100 σ P,4,15 = 195000 4.86 0.83 = 36.17 [MPa] 1748 Loss of prestress due to creep at the time of 100 years: σ P,4,5 = 195000 5.08.81 = 17.99 [MPa] 1748 σ P,4,10 = 195000 3.98.81 = 100.8 [MPa] 1748 σ P,4,15 = 195000 4.86.81 = 1.45 [MPa] 1748 6..3 Loss due to shrinkage of concrete Concrete shrinks over time, as the water leaves the matrix. As a result of shrinkage, there is a loss of prestress in the tendons. σ PS (t) = E p (ε c S (t) ε c S (τ)) The shrinkage coefficient is usually determined using software. For the purpose of this homework, the values are given defaultly: ε c S (τ = 7days) = 8. 10 6 ε c S (t = 100 days) = 56.33 10 6 ε c S (t = 100 years) = 439.51 10 6 Loss of prestress due to shrinkage at the time of 100 days: σ 100 P,5 = E p (ε S c (100) ε S c (7)) = 195000 (56.33 10 6 8. 10 6 ) = 9.38 [knm] Loss of prestress due to shrinkage at the time of 100 years: σ P,5 = E p (ε S c (36500) ε S c (7)) = 195000 (439.51 10 6 8. 10 6 ) = 84.10 [knm] Stresses in tendons after long-term losses summary: σ P = σ P0 + σ P,3 + σ P,4 + σ P,5 14

100 years Long-term losses 100 days Immedi ate losses Summary of stresses in cross-sections 5, 10, 15 at different times of life: Stresses and forces CM03 - Model homework Cross-section 5 10 15 Input stress σ P,0,max [Mpa] 1404.00 1404.00 1404.00 Friction σ P.1 [Mpa] -33.43-116.7-07.57 Anchorage set σ P. [Mpa] -74.04 0.00 0.00 Stress after immediate losses σ P,0 [Mpa] 196.53 187.73 1196.43 Force after immediate losses N P,0 [kn] 7390. 7340.06 6819.65 Relaxation σ P.3 [Mpa] -0.10-19.14-11.46 Creep σ P.4 [Mpa] -37.81-9.6-36.17 Shrinkage σ P.5 [Mpa] -9.38-9.38-9.38 Stress at the time of 100 days σ P.100 [Mpa] 19.4 19.59 1139.45 Force at the time of 100 days N P.100 [kn] 7006.67 7008.66 6494.86 Relaxation σ P.3 [Mpa] -58.56-56.89-41.96 Creep σ P.4 [Mpa] -17.99-100.8-1.45 Shrinkage σ P.5 [Mpa] -84.10-84.10-84.10 Stress at the end of service life σ P, [Mpa] 105.88 1046.46 947.9 Force at the end of service life N P, [kn] 5847.5 5964.8 5403.14 15