Example 24.3 Flexural Strength of Prestressed Member Using Approximate Value for f ps

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Madeleine Miller
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1 Example 24.3 Flexural Strength of Prestressed Member Using Approximate Value for f ps Calulate the nominal moment strength of the prestressed member shown. f = 5000 psi f pu = 270,000 psi (low-relaxation strands; f py = 0.90f pu ) Calulations and Disussion Code Referene 1. Calulate stress in prestressed reinforement at nominal strength using approximate value for f ps. For a fully prestressed member, Eq. (18-3) redues to: f ps = f pu 1 - γ p f p pu β ρ Eq. (18-3) 1 f = where = 252 ksi γ p = 0.28 for f py fpu = 0.90 for low-relaxation strand β 1 = 0.80 for f = 5000 psi ρ p = Aps bdp = =

2 Code Example 24.3 (ont d) Calulations and Disussion Referene 2. Calulate nominal moment strength from Eqs. (9) and (10) of Part 24 Compute the depth of the ompression blok: a = A f ps ps 085. bf = = 4.54 in. Eq. (10) M n = A ps f a ps dp - 2 M n = Eq. (9) 454. = 4565 in- = 380 ft Chek to see if tension ontrolled /d p = (a/β 1 )/d p = 454. / /d p = < R Tension ontrolled φ =

3 Dut A onduit (plain or orrugated) to aommodate prestressing steel for post-tensioned installation. Requirements for post-tensioning duts are given in Sheathing A material enasing prestressing steel to prevent bonding of the prestressing steel with the surrounding onrete, to provide orrosion protetion, and to ontain the orrosion inhibiting oating. 270 E ps = 28,500 ksi 270 ksi x Per ASTM A 416 minimum yield strength at 1% elongation for 270 ksi = 243 ksi Stress f ps (ksi) Strain, ps (in./in.) The above urve an be approximated by the following equations: ps : f ps = 28,500 ps (ksi) ps > : f ps = ps (ksi) 18.2 GENERAL Figure 24-1 Stress-Strain Curve for Grade 270, Low Relaxation Strand (24.1) The ode speifies strength and servieability requirements for all onrete members, prestressed or nonprestressed. This setion requires that, for prestressed members, both strength and behavior at servie onditions must be heked. All load stages that may be ritial during the life of the struture, beginning with the transfer of the prestressing fore to the member and inluding handling and transportation, must be onsidered. This setion also alls attention to several strutural issues speifi to prestressed onrete strutures that must be onsidered in design: Stress onentrations. See for requirements for post-tensioned anhorages. 24-3

4 Example 24.4 Flexural Strength of Prestressed Member Based on Strain Compatibility The retangular beam setion shown below is reinfored with a ombination of prestressed and nonprestressed strands. Calulate the nominal moment strength using the strain ompatibility (moment-urvature) method. f = 5000 psi f pu = 270,000 psi (low-relaxation strand; f py = 0.9f pu ) E ps = 28,500 ksi jaking stress = 0.74f pu losses = 31.7 ksi (alulated by method of Ref See 18.6 Loss of Prestress for proedure.) Calulations and Disussion Code Referene 1. Calulate effetive strain in prestressing steel. ε = (0.74f pu - losses)/e ps = ( )/28,500 = Draw strain diagram at nominal moment strength, defined by the maximum onrete ompressive strain of and an assumed distane to the neutral axis,. For f = 5000, β 1 = Trials C a 2 a = /2"φ 1'-8" Prest. Nonprest. 2'-0" M n 12" 2" A T 1 = A s1 f 1 T 2 = A s2 f 2 3. Obtain equilibrium of horizontal fores. Strains Stresses The strain line drawn above from point 0 must be loated to obtain equilibrium of horizontal fores: C = T 1 + T 2 To ompute T 1 and T 2, strains ε1 and ε2 are used with the stress-strain relation for the strand to determine the orresponding stresses f 1 and f 2. Equilibrium is obtained using the following iterative proedure: 24-26

5 Code Example 24.4 (ont d) Calulations and Disussion Referene a. assume (loation of neutral axis) b. ompute ε 1 and ε 2. obtain f 1 and f 2 from the equations in Fig d. ompute a = β 1 e. ompute C = 0.85 ab f. ompute T 1 and T 2 g. hek equilibrium using C = T 1 + T 2 h. if C < T 1 + T 2, inrease, or vie versa and return to step b of this proedure. Repeat until satisfatory onvergene is ahieved. Estimate a neutral axis loation for first trial. Estimate stressed strand at 260 ksi, unstressed strand at 200 ksi. T = ΣA ps f s = (200) (260) = 220 = C a = C/(0.85 b) = 220/( ) = 4.32 in. = a/β 1 = 4.32/0.80 = 5.4 in. Use = 5.4 in. for first try The following table summarizes the iterations required to solve this problem: Trial No. in. ε 1 ε 2 f 1 ksi f 2 ksi a in. C T 1 T 2 T 1 + T O.K Calulate nominal moment strength. Using C = 228.5, T 1 = 67 and T 2 = 162, the nominal moment strength an be alulated as follows by taking moments about T 2 : M n = [((d 2 - a/2) C) - ((d 2 - d 1 ) T 1 )]/12 = [(22 - (4.48/2)) (22-20) 67)]/12 = 365 ft

3.1 Analysis of Members under Axial Load

3.1 Analysis of Members under Axial Load This setion overs the following topis. Introdution Analysis at Transfer Analysis at Servie Loads Analysis of Ultimate Strength Analysis of Behaviour Notations Geometri