8/2/2024 Pre-stressed Concrete Design (Homogeneous Beam Concept) ENGR. ALVIN C. CARIAGA, RCE, RMP, SO-2 Why use PRESSTRESSED CONCRETE instead of conventional RC? βπππ₯ = 5π€πΏ4 384 πΈπΌ πππ‘ 5π€ =k 384 πΈπΌ β = ππΏ4 Let’s have 2 scenarios, if L = 4m and if L =8m β = π 4 4 = 256π β = π 8 4 = 4096π As we have noticed, if we double our length the deflection will 16 times greater than the other Based on NSCP, The allowable deflection is L/240. 4000 Assume that Δπππ‘π’ππ(4π) = 10 ππ πππ βπππππ€ = 240 = 16.67ππ. We can say that our member is safe. But at 8m, 8000 the actual deflection is 16 times so we have 160 mm and our allowable is βπππππ€ = = 33.33 ππ, we can see that 240 we will be having an unsafe member. 1 8/2/2024 TYPES PRESTRESSED CONCRETE PRE-TENSIONING POST-TENSIONING Methods of Prestressing • Homogeneous Beam Concept • Load Balancing Concept Step 1: Transform the beam section into an equivalent homogeneous section π = ππ π ππ΄ππ If “n” is given: π΄ = π΄π + π − 1 π΄ππ π If “n” is not given: π΄ = π΄π π Step 2: Locate the Neutral Axis πππππ£π = ππππππ€ If “n” is given: π π π΄ π ππ πΌ πππ πΌ π=− Step 3: Analyze using Combined Stress π= π ππ ± π΄ πΌ π πππ ππ ± ± πΌ πΌ π΄ } πΆ π Due to Prestressing force Due to Loadings (+) Tension (-) Compression ππππ‘ = − π πππ ππ − + π΄ πΌ πΌ ππ‘ππ = − π πππ ππ + − π΄ πΌ πΌ 2 8/2/2024 ACI Maximum Permissible Stresses CONCRETE • Initial Stage – the prestressing force is called the initial force or the force immediately after transfer *- Compression ≤ 0.60f’ci (f’c is the ultimate compressive stress) *- Tension ≤ 0.25 f′ci (within the beam except from the supports) - Tension ≤ 0.50 f′ci (from supports) (f’ci is the initial compressive stress at initial stage) • Final or Service Stage – the force after all time dependent effects including shrinkage, creep, and steel relaxation that’s why the initial prestress force will be reduced, and the remaining force will be called the effective prestressing force. *- Compression ≤ 0.45f’c (extreme fibers in compression due to sustained dead load and live load) - Compression ≤ 0.60f’c (transient live load) *- Tension ≤ 0.50 f′c (short-term deflection) - Tension ≤ 1.0 f′c (long-term deflection) STEEL • Jacking Force • Initial Force • Post-tensioning 0.94fpy but ≤ 0.80fpu 0.82fpy but ≤ 0.74fpu 0.70fpu fpy & fpu are yield and ultimate stress of tendons, respectively. Problem No. 1 A rectangular beam (200mm by 400mm) is prestressed with a final prestressing force of 500 KN at an eccentricity of 100mm below the neutral axis. Determine the maximum moment that will produce no stress at the bottom fiber. π=− 400ππ 100ππ π πππ ππ ± ± π΄ πΌ πΌ ππππ‘ = − 0= − 200ππ π = 500 ππ π πππ ππ − + π΄ πΌ πΌ 500,000π 200ππ 400ππ − π(200ππ) 500,000π(100ππ)(200ππ) + 3 200 400 3 200 400 ππ4 ππ4 12 12 π = 83.33π₯106 π − ππ π = 83.33 ππ − π 3 8/2/2024 Problem No. 2 A 6-m long cantilever beam, 250 mm wide x 600 mm deep, carries a uniformly distributed dead load (beam weight included) of 5 kN/m throughout its length. To prevent excessive deflection of the beam, it is pretensioned with 12 mm diameter strands causing a final prestress force of 540 kN. Use fc’ = 27 MPa. 1. Determine the resulting stress (MPa) at the top fiber of the beam at the free end if the center of gravity of the strands coincide with the centroid of the section. 2. Determine the uniform eccentricity of the strands (mm) such that the resulting stress at the top fiber at the fixed end is zero. 3. Determine the maximum concentrated live load (kN) that maybe applied at the free end of the beam so that the stresses in the extreme fibers at the fix end will not exceed 0.45 fc’ for compression and 0.5 ππ′ for tension if the strands are placed at a uniform eccentricity of 150 mm above the centroid of the section Problem No. 2 A 6-m long cantilever beam, 250 mm wide x 600 mm deep, carries a uniformly distributed dead load (beam weight included) of 5 kN/m throughout its length. To prevent excessive deflection of the beam, it is pretensioned with 12 mm diameter strands causing a final prestress force of 540 kN. Use fc’ = 27 MPa. 1. Determine the resulting stress (MPa) at the top fiber of the beam at the free end if the center of gravity of the strands coincide with the centroid of the section. 5 ππ/π π=0 πππππ−πππ = 0 600ππ 6π π=− π πππ ππ ± ± π΄ πΌ πΌ π=− 540 000π 250 600 ππ2 250 ππ π = 540 ππ π = −3.6 πππ → #1 4 8/2/2024 Problem No. 2 A 6-m long cantilever beam, 250 mm wide x 600 mm deep, carries a uniformly distributed dead load (beam weight included) of 5 kN/m throughout its length. To prevent excessive deflection of the beam, it is pretensioned with 12 mm diameter strands causing a final prestress force of 540 kN. Use fc’ = 27 MPa. 2. Determine the uniform eccentricity of the strands (mm) such that the resulting stress at the top fiber at the fixed end is zero. ππππππ‘ @ πππ₯ππ πππ 5 ππ/π π€πΏ2 π= 2 600ππ π 6π π πΆ π πππ ππ ± π=− ± πΌ πΌ π΄ 250 ππ π = 540 ππ ππ‘ππ = 0 (@πππ₯ππ πππ) ππ‘ππ = − π= 5ππ/π 6π 2 2 ππΉπΈ = 90 ππ − π ππΉπΈ = 90 π₯106 π − ππ π πππ ππ − + π΄ πΌ πΌ 540 000π 540 000π(π)(300ππ) (90,000π₯106 π − ππ)(300ππ) − + 250 600 ππ 250 600 3 250 600 3 4 ππ ππ4 12 12 π = 66.67 ππ ππππ£π π‘βπ ππ΄ → #2 0=− Problem No. 2 A 6-m long cantilever beam, 250 mm wide x 600 mm deep, carries a uniformly distributed dead load (beam weight included) of 5 kN/m throughout its length. To prevent excessive deflection of the beam, it is pretensioned with 12 mm diameter strands causing a final prestress force of 540 kN. Use fc’ = 27 MPa. 3. Determine the maximum concentrated live load (kN) that maybe applied at the free end of the beam so that the stresses in the extreme fibers at the fix end will not exceed 0.45 fc’ for compression and 0.5 ππ′ for tension if the strands are placed at a uniform eccentricity of 150 mm above the centroid of the section. π 5 ππ/π π πππ ππ π=− ± ± π΄ πΌ πΌ 150 ππ 600ππ π πππ ππ 6π ππππ‘ = − + − π π΄ πΌ πΌ πΆ 250 ππ π = 540 ππ ππππ‘ = ππ‘ππ = 0.45ππ −0.45(27) = − 540,000π 540,000π (150ππ)(300ππ) π(300ππ) + − 250ππ 600ππ 3 250ππ(600ππ) 250ππ 600ππ 3 12 12 ππππ‘ = 209. 25 π₯106 π − ππ 5 8/2/2024 Problem No. 2 A 6-m long cantilever beam, 250 mm wide x 600 mm deep, carries a uniformly distributed dead load (beam weight included) of 5 kN/m throughout its length. To prevent excessive deflection of the beam, it is pretensioned with 12 mm diameter strands causing a final prestress force of 540 kN. Use fc’ = 27 MPa. 3. Determine the maximum concentrated live load (kN) that maybe applied at the free end of the beam so that the stresses in the extreme fibers at the fix end will not exceed 0.45 fc’ for compression and 0.5 ππ′ for tension if the strands are placed at a uniform eccentricity of 150 mm above the centroid of the section. π 5 ππ/π π πππ ππ π=− ± ± π΄ πΌ πΌ 150 ππ 600ππ π πππ ππ 6π ππ‘ππ = − − + π π΄ πΌ πΌ πΆ 0.5 27 = − 250 ππ 540,000π 250ππ(600ππ) π = 540 ππ ππππ‘ = ππ‘ππ = 0.45ππ − 540,000π (150ππ)(300ππ) π(300ππ) + 250ππ 600ππ 3 250ππ 600ππ 3 12 12 ππ‘ππ = 173. 96 π₯106 π − ππ < ππππ‘ ∴ π = 173. 96 ππ − π Problem No. 2 A 6-m long cantilever beam, 250 mm wide x 600 mm deep, carries a uniformly distributed dead load (beam weight included) of 5 kN/m throughout its length. To prevent excessive deflection of the beam, it is pretensioned with 12 mm diameter strands causing a final prestress force of 540 kN. Use fc’ = 27 MPa. 3. Determine the maximum concentrated live load (kN) that maybe applied at the free end of the beam so that the stresses in the extreme fibers at the fix end will not exceed 0.45 fc’ for compression and for 0.5 ππ′ tension if the strands are placed at a uniform eccentricity of 150 mm above the centroid of the section. π 5 ππ/π π = 173. 96 ππ − π 150 ππ 600ππ 6π π πΆ 250 ππ π= π€πΏ2 + ππΏ 2 173. 96 = 90 + ππΏ π = 13. 99 ππ → #3 π = 540 ππ ππππ‘ = ππ‘ππ = 0.45ππ 6 8/2/2024 Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. a. Determine the resulting stress at the bottom fibers of the DT at L/4 from the center of bearings or support. b. Compute the resulting stress at the bottom fibers of the DT at midspan due to the service loads and prestress force. c. What additional super imposed load the DT can carry such that the resulting stress at the bottom fibers at midspan is zero. Load imposed on the joists are: Dead load = 2.3 kPa Live Load = 6 kPa Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m π = 2.4 π 88 ππ π. π΄. 192 ππ 267 ππ 75 ππ Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. a. Determine the resulting stress at the bottom fibers of the DT at L/4 from the center of bearings or support. b. Compute the resulting stress at the bottom fibers of the DT at midspan due to the service loads and prestress force. c. What additional super imposed load the DT can carry such that the resulting stress at the bottom fibers at midspan is zero. Load imposed on the joists are: Dead load = 2.3 kPa Live Load = 6 kPa Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m π = 2.4 π 88 ππ π. π΄. 192 ππ 267 ππ 75 ππ 7 8/2/2024 π€ 7.5 π π = 2.4 π ππ πππ = 2 π π€= 88 ππ ππ π π. π΄. 192 ππ 267 ππ 75 ππ π€ 7.5 π π = 2.4 π ππ πππ = 2 π π€= ππ π 88 ππ π. π΄. 192 ππ 267 ππ Load imposed on the joists are: Dead load = 2.3 kPa Live Load = 6 kPa 75 ππ πππ‘ππ πΏπππ = 2.3 + 6 = 8.3 πππ π€π = 8.3 ππ 2.4 π = 19.92 ππ/π π2 8 8/2/2024 Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. a. Determine the resulting stress at the bottom fibers of the DT at L/4 from the center of bearings or support. b. Compute the resulting stress at the bottom fibers of the DT at midspan due to the service loads and prestress force. c. What additional super imposed load the DT can carry such that the resulting stress at the bottom fibers at midspan is zero. Load imposed on the joists are: π = 2.4 π wt = 19.92 kN/m Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m 88 ππ π. π΄. 192 ππ 267 ππ 75 ππ Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. a. Determine the resulting stress at the bottom fibers of the DT at L/4 from the center of bearings or support. π = 2.4 π 19.92 ππ/π 88 ππ π. π΄. 192 ππ 75 ππ π πππ ππ π=− ± ± π΄ πΌ πΌ ππππ‘ = − 7.5 π π = 74.7 ππ π πππ ππ − + → πππππ π΄ πΌ πΌ πΏ/4 ΰ· π πΏ/4 = 0; β· + 19.92 ππ/π π ππ ππ ππ ππ ππππ‘(πππππ) = − − + π΄ πΌ πΌ 74.7 ππ ππ = 745 1 − 0.18 2 = 1221.8ππ 19.92 7.5 2 π = 74.7 ππ π = 267 ππ 7.5π 4 7.5 7.5 π = 74.7 − 19.92 ∗ 4 4 7.5 4 2 π = 105. 0468 ππ − π 9 8/2/2024 Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. a. Determine the resulting stress at the bottom fibers of the DT at L/4 from the center of bearings or support. π = 2.4 π 88 ππ ππππ‘(πππππ) = − π. π΄. ππ ππ ππ ππ − + π΄ πΌ πΌ 192 ππ 267 ππ 75 ππ ππππ‘(πππππ) = − ππ = 1221.8ππ ππΏ/4 = 105. 0468 ππ − π + Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m 1221.8π₯103 π 192ππ (267 ππ) 1221.8π₯103 π − 1880 π₯106 ππ4 200,000 ππ2 105. 0468π₯106 π − ππ (267 ππ) 1880 π₯106 ππ4 ππππ‘(πππππ) = −24. 5062 πππ πΆ Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. b. Compute the resulting stress at the bottom fibers of the DT at midspan due to the service loads and prestress force. . π = 2.4 π 88 ππ ππππ‘(πππππ) = − π. π΄. ππ ππ ππ ππ − + π΄ πΌ πΌ 192 ππ 267 ππ 75 ππ π= ππ = 1221.8ππ Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m π€πΏ2 19.92 7.5 2 = 8 8 ππππ = 140.0625 ππ − π ππππ‘(πππππ) = − + 1221.8π₯103 π 192ππ (267 ππ) 1221.8π₯103 π − 1880 π₯106 ππ4 200,000 ππ2 140.0625π₯106 π − ππ (267 ππ) 1880 π₯106 ππ4 ππππ‘(πππππ) = −19.533 πππ πΆ 10 8/2/2024 Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. c. What additional super imposed load the DT can carry such that the resulting stress at the bottom fibers at midspan is zero. π = 2.4 π π€πππ 88 ππ } π€ π. π΄. 192 ππ π€π 267 ππ 75 ππ ππππ‘(πππππ) = − ππ = 1221.8ππ Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m 0=− ππ ππ ππ ππ − + π΄ πΌ πΌ 1221.8π₯103 π 1221.8π₯103 π 192ππ (267 ππ) − 2 200,000 ππ 1880 π₯106 ππ4 + ππ π₯106 (267 ππ) 1880 π₯106 ππ4 ππ = 277.6 ππ − π Problem No. 3: CE BOARD MAY 2012/2013/NOV 2015 The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m and are pre-tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. c. What additional super imposed load the DT can carry such that the resulting stress at the bottom fibers at midspan is zero. π = 2.4 π π€πππ 88 ππ π. π΄. π€ 192 ππ } π€π 267 ππ 75 ππ ππ = 1221.8ππ Properties of DT A = 200,000 mm2 I = 1880 x106 mm4 Yt = 88 mm Yb = 267 mm a = 2.4 m ππ = 277.6 ππ − π ππ = 277.6 = π€π πΏ2 8 π€π = π€πππ + π€ π€π 7.5 2 8 39. 48 = π€πππ + 19.92 π€πππ = 19.56 ππ/π π€π = 39. 48 ππ/π 11 8/2/2024 Deflection of Prestressed Concrete Beam ENGR. ALVIN C. CARIAGA, RCE, RMP, SO-2 12
