PROBLEM 34. CE Board May 2009 540 π₯ 106 = 0.90(300)π2 (28)(0.21) π = 583.21 ππ A simply reinforced concrete beam reinforced for tension has a width of 300 mm and a total depth of 600 mm. It is subjected to an external moment Mu = 540 kN-m, fc’ = 28 MPa, fy = 280 MPa, Es = 200 GPa. β Which of the following gives the balance steel ration in percent. β‘ Which of the following gives the depth “a” in terms of “d” in percent using ρ = ½ ρb. β’ Which of the following gives the minimum effective depth. ππ’ = πΆ (π − π/2) Solution: ππ’ = 0.85 ππ′ ππ (π − π/2) β Balanced steel ratio: 0.85ππ′ π 600 ππ = π (600+ π ) π¦ π¦ 0.85(25)(0.85) ππ = 280 (600+ 280) ππ = 0.04926% ππ = π. πππ% 540 π₯ 106 = 0.85(28)(π)(300)(583.21 − π/ 2) 75630.25 = π(583.21 − 0.5π) 0.5π2 − 583.21π + 75630.25 = 0 π2 − 1166.42π + 151260.50 = 0 π= 1166.42±869.19 2 π = 148.61 ππ β‘ Depth “a” in terms of d in percent: ππ’ = ∅ππ 2 ππ′ π (1 − 0.59π) 1 π = 2 ππ 1 π = 2 (0.04926) π 148.61 = π 583.21 π = 0.2548 π π = ππ. ππ% π π = 0.0246 β’ Min. effective depth: π π = πππ. ππ ππ π = π ππ¦′ π π= 0.0246(280) = 0.246 28 π(1 − 0.59π) = 0.246(1 − 0.59 ∗ 0.246) ππ’ = ∅ππ2 ππ′ π (1 − 0.59π) PROBLEM 35. CE Board Nov. 2009 A rectangular beam has a width of 300 mm and an effective depth of 460 mm. The beam is reinforced with 2-28 mm ∅ at the top. fc’ = 35 MPa, fy = 350 MPa. β Compute the ratio of the depth of compression block to the distance of the top fiber to the neutral axis. ππ = π. ππππ% β’ Max. area of steel permitted: β‘ Compute the balanced steel ratio of the reinforcement. π΄π = π π π β’ Compute the max. area of steel permitted. π΄π = 0.75(0.0429)(300)(460) π΄π = 0.75 ππ π π π¨π = ππππ. ππ πππ Solution: β Ratio of the depth of compression block to the distance of the top fiber to the neutral axis” PROBLEM 36: A rectangular beam having a width of 300 mm and an effective depth of 450 mm. It is reinforced with 4-36 mm in diameter bars. fc’ = 28 MPa, MPa. fy = 270 MPa, Es = 200000 β Compute the depth of compression block for a balanced condition. β‘ Compute the nominal moment capacity of the beam. a = depth of compression block β’ If the value of fc’ is increased by 25%, compute the percentage of the increased nominal moment capacity of the beam. c = distance of top fiber to neutral axis π = π½π Solution: π π½=π π½ = 0.85 − 0.05(ππ′ −28) 7 π½ = 0.85 − 0.05(35−28) 7 β Depth of compression block for a balanced condition: π· = π. ππ β‘ Balanced steel ratio of the reinforcement: 0.85π′ π½ 600 π ππ = π (600+ π ) π¦ ππ = π¦ 0.85(35)(0.80)(600) 350 (600+ 350) 0.003 0.00135 = 450−π π β’ Percentage of the increased nominal moment capacity of the beam if fc’ is increased by 25%: 1.35 − 0.003π = 0.00135π 0.00435π = 1.35 π π = 310.34 π΄π = 4 (36)2 (4) π = π½π π΄π = 1296π π ππ = π΄π ππ¦ (π − 2 ) Note: π½ = 0.85 πππ ππ ′ = 28 πππ πΆ=π π = 0.85(310.34) 0.85ππ ′ ππ = π΄π ππ¦ π = πππ. ππ ππ ππ ′ = 1.28(28) ππ ′ = 35 πππ β‘ Nominal moment capacity: 0.85ππ ′ ππ = π΄π ππ¦ 0.85(35)(π)(300) = 1296π(270) π = 123.17 ππ π ππ = π΄π ππ¦ (π − 2 ) ππ = 1296π(270) (450 − 123.17 ) 2 ππ = 427 π₯ 106 π − ππ πΆ=π ππ = 427 ππ − π ′ 0.85ππ ππ = π΄π ππ¦ π 0.85(28)(π)(300) = 4 (36)2 (4)(270) Percentage increase in nominal moment: π = 153.96 ππ %πππππππ π = ( 427−410 ) π₯100 410 %ππππππππ = π. ππ% π ππ = π΄π ππ¦ (π − 2 ) π ππ = 4 (36)2 (4)(270) (450 − ππ = 410 π₯ 106 π − ππ π΄π = πππ ππ΅ − π 153.96 ) 2 PROBLEM 37: A reinforced concrete beam has a width of 400 mm and an effective depth of 600 mm. It is reinforced for tension with 4-28 mm ∅ bars. fc’ = 20.7 MPa, fy = 414.6 MPa. β Determine the percent increase in nominal moment if the depth is increased to 700 mm. β‘ Determine the percent increase in nominal moment if fc’ is increased to 27.6 MPa. β’ Determine the percent increase in nominal moment if the steel is changed to 4-32 mm ∅. ππ¦ = 0.00207 ππ > ππ¦ (π π‘πππ π¦πππππ ) Nominal moment if d = 600 mm: π ππ = ∅π΄π ππ¦ (π − 2 ) π ππ = 0.90 4 (28)2 (4)(414.6) (600 − Solution: β Percent increase in nominal moment if the depth is increased to 700 mm: 145.09 ) 2 ππ = 484755959 π − ππ ππ = 484756 ππ − π Nominal moment if d = 700 mm: π ππ = ∅π΄π ππ¦ (π − 2 ) ππ = 0.90 π 145.09 (28)2 (4)(414.6) (700 − ) 4 2 ππ = 576660663 π − ππ ππ = 576661 ππ − π πΆ=π 0.85ππ ′ ππ = π΄π ππ¦ π 0.85(20.7)(π)(400) = 4 (28)2 (4)(414.6) Percent increase in nominal moment: π = 145.09 ππ %πππππππ π = ( 576661−484756 ) π₯100 484756 %ππππππππ = ππ. ππ% π = π½π 145.09 = 0.85π π = 170.69 ππ β‘ Percent increase in nominal moment if fc’ is increased to 27.6 MPa: πΆ=π ππ 0.003 = 170.69 429.31 ππ = 0.00754 0.85ππ ′ ππ = π΄π ππ¦ π 0.85(27.6)(π)(400) = 4 (28)2 (4)(414.6) π = 108.82 ππ π ππ¦ = πΈπ¦ π 414.6 ππ¦ = 200,000 π ππ = ∅π΄π ππ¦ (π − 2 ) π ππ = 0.90 4 (28)2 (4)(414.6) (600 − 108.82 ) 2 ππ = 501423048.1 π − ππ β‘ Determine the moment capacity using moment reduction factor of 0.90. β’ Determine the super-imposed uniform live load it could carry in kPa besides a dead load of 20 kN/m including its own weight if it has a simple span of 6 m and a spacing of 1.8m. ππ = 501423 ππ − π Percentage increase in nominal moment: 501423−484756 ) π₯100 484756 %πππππππ π = ( Solution: β Depth of compression block: %ππππππππ = π. ππ % β’ Percent increase in nominal moment if the steel is change to 4-32 mm ∅ : πΆ=π 0.85ππ ′ ππ = π΄π ππ¦ π 4 0.85(20.7)(π)(400) = (32)2 (4)(414.6) π π΄π = 4 (32)2 (3) π = 189.51 ππ π΄π = 2413 ππ2 π ππ = ∅π΄π ππ¦ (π − 2 ) π ππ = 0.90 4 (32)2 (4)(414.6) (600 − π΄ π = πππ 189.51 ) 2 ππ = 606490 ππ − π 2413 π = 300(575) π = 0.014 606490−484756 ) π₯100 484756 %πππππππ π = ( %ππππππππ = ππ. π % ππππ₯ = 0.75ππ ππππ₯ = 0.75(0.02850 PROBLEM 38: A reinforced concrete beam has a width of 300 mm with an effective depth of 575 mm. It is reinforced with 3-32 mm ∅ at the bottom. fc’ = 27.6 MPa, fy = 414 MPa. Balanced steel ratio ρb = 0.0285. β Determine the depth of compression block. ππππ₯ = 0.021 > 0.014 ππ‘πππ π¦πππππ : ππ = ππ¦ πΆ=π 0.85ππ ′ ππ = π΄π ππ¦ π 0.85(27.6)(π)(300) = 4 (2413)(414) πΏπΏ = 47.98 ππ π π = πππ. ππ ππ 47.98 πΏπΏ = 1.8 β‘ Moment capacity using moment reduction factor of 0.90: π³π³ = ππ. ππ ππ·π π = π½π PROBLEM 39: 141.44 = 0.85π π = 166.99 ππ Check whether we could use ∅ = 0.90: π1 0.003 = 408.01 166.99 A fixed ended rectangular beam must support a uniform service dead and live loads of 220 kN/m and 182.6 kN/m respectively. It has a span of 6m. fc’ = 27.6 MPa, fy = 414.7 MPa, ρb = 0.028. β Effective depth of the beam using ω = 0.18. β‘ Compute the flexural reinforcements at the support. π1 = 0.0073 > 0.005 β’ Compute for the flexural reinforcements at the midspan. ππ π ∅ = 0.90 π 2 ππ’ = ∅π΄π ππ¦ (π − ) ππ’ = 0.90 (2413)(414) (575 − Solution: 141.44 ) 2 β Effective depth of the beam using ω = 0.18: ππ’ = 453.4 ππ − π β’Super-imposed uniform live load it could carry in kPa besides a dead load of 20 kN/m including its own weight if it has a simple span of 6 m and a spacing of 1.8m: ππ’ = 1.2π·πΏ + 1.6πΏπΏ π πΏ2 ππ’ = π’8 453.4 = ππ’ = 1.2(220) + 1.6(182.6) ππ’ (6)2 8 ππ’ = 556.2 ππ/π ππ ππ’ = 100.76 π Max. moment occurs at the fixed supports: ππ’ = 1.6πΏπΏ + 1.2π·πΏ −ππ’ = ππ’ πΏ2 12 100.76 = 1.6πΏπΏ + 1.2(20) −ππ’ = 556.2(6)2 12 −ππ’ = 1668.60 ππ − π π < ππππ₯ π΅π πππππππππππ ππππ πππ ππππ ππ Moment at midspan: +ππ’ = ππ’ πΏ2 24 +ππ’ = 556.2(6)2 24 π΄π = π ππ π΄π = 0.01198(600)(850) π¨π = ππππ πππ +ππ’ = 834.3 ππ − π β’ Flexural reinforcements at the midspan: +ππ’ = 834.3 ππ − π Effective depth of beam: ππ’ = ∅ππ′ ππ2 π (1 − 0.59π) ππ’ = ∅ππ′ ππ2 π (1 − 0.59π) 6 2 1668.6 π₯ 10 = 0.90(27.6)(600)π (0.18)[1 − 0.59(0.18)] π = 834.2 ππ π ππ¦ 850 ππ πΌππ π = πππ ππ 843.3 π₯ 106 = 0.90(27.6)(600)(850)2 (π)[1 − 0.59(π)] π[1 − 0.59π] = 0.0774 π2 − 1.6949π + 0.1313 = 0 π = 0.0814 β‘ Flexural reinforcements at the supports: π = 0.18 πππ ′ π = ππ¦ π= 0.18(27.6) 414.7 π = 0.01198 14 ππππ = 414.7 ππππ = 0.00338 π= πππ ′ ππ¦ π= 0.0814(27.6) 414.7 π = 0.00541 14 ππππ = 414.7 ππππ = 0.00338 ππ π < ππππ₯ 0.00541 < 0.021 ππππ₯ = 0.75ππ π΅π πππππππππππ ππππ πππ ππππ ππ ππππ₯ = 0.75(0.028) π΄π = π ππ ππ π π = 0.01198 π΄π = 0.00541(600)(850) ππππ₯ = 0.021 π¨π = ππππ πππ PROBLEM 40: A rectangular concrete beam has a width of 250 mm and a total depth of 450 mm. It is reinforced with a total steel area of 1875 mm2 placed at an effective depth of 375 mm. fc’ = 27.6 MPa, fy = 414.7 MPa. β Determine the depth of compression block. β‘ Determine the moment capacity reduction factor. β’ Determine the safe live load that the beam could carry in addition to a dead load of 20 kN/m if it has a span of 6m. ππ‘ 0.003 = 155.98 219.02 ππ‘ = 0.0042125 > 0.002 ππ’π‘ < 0.005 π πΈπ ππ¦ = π 414.7 ππ¦ = 200,000 Solution: β Depth of compression block: ππ¦ = 0.0020735 < 0.0042125 Steel yields: Assuming steel yields: πΆ=π 0.85ππ ′ ππ = π΄π ππ¦ 0.85(27.6)(π)(250) = 1875(414.7) π = πππ. ππ ππ Since π΄t is between 0.002 and 0.005, this value is within the transition range between compression controlled section and tension controlled section. 250 ππ π ∅ = 0.65 + (ππ‘ − 0.002) 3 250 ∅ = 0.65 + (0.0042125 − 0.002) 3 β‘ Moment capacity reduction factor: ∅ = π. πππ π = π½π 132.58 = 0.85π π = 155.98 ππ β’ Live load it could carry: π ππ’ = ∅π΄π ππ¦ (π − 2 ) ππ’ = 0.834 (1875)(414.7) (375 − 132.58 ) 2 Wu = 47.8 kN/m ππ’ = 200.2 π₯ 10 6 π − ππ ππ’ = 200.2 ππ − π ππ’ = ππ’ = ππ’ πΏ2 8 ππ’ = 47.8 (6)2 8 ππ’ = πππ. π ππ΅. π ππ’ πΏ2 8 200.2 = Wu = 1.2 (10.5) + 1.6 (22) ππ’ (6)2 8 2. Approximate flexural resistance factor: Mu = ø R b d2 ππ’ = 44.48 ππ/π d = 400 – 62.5 = 337.50 215.1 x 106 = 0.90 R (300) (337.5)2 ππ’ = 1.2π·πΏ + 1.6πΏπΏ R = 6.99 44.48 = 1.2(20) + 1.6πΏπΏ ππ΅ π³π³ = ππ. ππ π 3. Number of 32 mm ø bars: R = fc’ π (1 – 0.59 π) 6.99 = 27.6 π(1 – 0.59 π) PROBLEM 41: Architectural considerations limit the height of a 6 m. long simple span beam to 400 mm and width 0.59 π2 – 27.6 π + 6.99 = 0 π2 – 46.78 π + 11.85 = 0 of 300 mm. The following loads are material 46.78 ±46.27 = 0.25 2 properties are given: Use 62.5 mm ø as covering π= from center of reinforcing bars. π = π ππ¦′ π DL = 10.5 kN/m LL = 22 kN/m fc’ = 34.6 MPa fy = 414.7 MPa π 0.25 = π (414.7) 34.6 π = 0.0209 1. Determine the factored moment carried by the beam. 2. Determine the approximate flexural As = π b d As = 0.0209 (300) (337.50) As = 2112 mm2 resistance factor. Assume ø = 0.90 3. Determine the number of 32 mm ø bars needed for the beam. π = (32)2 π = 2112 4 π = 2.63 π ππ¦ 3 Solution: 1. Factored moment carried by the beam. Wu = 1.2 DL + 1.6 LL Use 3 – 32 mm ø bars π As = 4 = (32)2 (3) = 2413 ππ2 ππππ₯ = 0.75ππ 0.85 π′π½ 600 π ππ = π (600+π ) π¦ π¦ π½ = 0.85 − 0.05 (ππ′ −30) 7 π½ = 0.85 − 0.05 (34.6−30) 7 π½ = 0.817 ππ = 0.85 (34.6)(0.817)(600) 414.7 (600+414.7) ππ = 0.03426 ππππ₯ = 0.75ππ ππππ₯ = 0.75(0.03426) ππππ₯ = 0.0257 (250) 3 ø = 0.65 + (0.00426 – 0.002) (250) 3 ø = 0.838 (moment capacity reduction factor) π Mu = ø As fy (d - 2) Mu = 0.838 (2413) (414.7) (337.50 - 113.42 ) 2 Mu = 235.46 kN.m > 215.10 kN.m (safe) PROBLEM 42: A reinforced rectangular concrete beam has a width of 250 mm and an effective depth of 360 π΄ π = πππ π= Use ø = 0.65 + ( ππ‘ – 0.002) mm. It is reinforced for tension only at the bottom 2413 300(337.5) π = 0.0238 < 0.0257 ok with a total tension steel area of 600 mm2. fc’ = 40 MPa, fy = 400 MPa. 1. Determine the tension reinforcement index for this beam. 2. Determine the distance of the neutral axis below the compression surface. 3. Determine the ultimate flexural strength of the beam. Solution: C=T 0.85 fc’ ab = As fy 0.85 (34.6) (a) (300) = 2413 (414.7) a = 113.42 a = π½c 113.42 = 0.817 c c = 138.82 ππ‘ 0.003 = 138.82 198.68 ππ‘ = 0.00426 < 0.005 1. Tension reinforcement index for this beam. π π = π ππ¦′ π π΄ π = πππ 600 π = 250(360) π = 0.0067 ππ¦ π = ππ′ π 0.0067 (400) π= 40 π = π. πππ 2. Distance of the neutral axis below the π Mu = ø As fy (d - 2) ππ‘ 0.003 = 36.20 323.80 ππ‘ = 0.0026983 compression surface. π½ = 0.85 − 0.05 (ππ′ −30) 7 π½ = 0.85 − 0.05 (40−30) 7 π½ = 0.78 π ππ¦ = π¦ πΈπ ππ¦ = 400 200000 ππ¦ = 0.002 Assume steel yields: C=T ππ > ππ¦ steel yields (under reinforced) 0.85 fc’ ab = As fy ππ = ππ¦ 0.85 (40) (a) (250) = 600 (400) a = 28.23 mm ππ‘ = 0.0026983 < 0.005 ø = moment capacity reduction factor ø = 0.70 + ( ππ‘ – 0.002) a = π½c (250) 3 28.23 = 0.78 c c = 36.20 mm (neutral axis below compression surface) 3. Ultimate flexural strength of the beam. Use ø = 0.65 + ( ππ‘ – 0.002) (250) 3 ø = 0.65 + (0.0026983 – 0.002) Steel does not yield: fs ≠ fy (250) 3 Using Hookes Law, solve for the actual fs: ø = 0.71 fs = πΈπ ππ 0.003 π = π π 500−π π 2 Mu = ø As fy (d - ) Mu = 0.71 (600) (400) (360 - 28.23 ) 2 Mu = 58.93 kN.m ππ = (500−π)(0.003) π ππ = 200000(500−π)(0.003) π ππ = 600(500−π) π PROBLEM 43: A rectangular beam has a width of 280 mm and an effective depth of 500 mm. It is reinforced with 4 – 36 mm ø bars at the tension side of the beam placed 65 mm above the bottom of the beam. T=C As fy = 0.85 fc’ ab 4072 (600 (500−π) π ) = 0.85(25)(0.85)π(280) 500 – c = 0.00207 c2 1. Which of the following will give the c2 + 483.09 c – 241546 = 0 c = 306.08 mm location of the neutral axis from the top of the beam. 2. Which of the following give the stress of steel. 3. Which of the following will give the 2. Stress of steel. a = π½c a = 0.85 (306.08) a = 260.17 mm ultimate capacity of the beam. Solution: 1. Location of the neutral axis from the top of the beam . ππ = 600 (500−π) π ππ = 600 (500−306.08) 306.08 ππ = πππ. ππ ππ·π < ππ 3. Ultimate moment capacity of the beam π Mu = ø As fy (d - 2) Mu = 0.90 (4072) (380.14) (500 Mu = 515.3 kN.m 260.17 ) 2 ππ = PROBLEM 44: (500−π)(0.003) π π A rectangular beam has a width of 280 mm and an effective depth of 500 mm and is reinforced with steel area in tension equal to 4000 mm2. fc’ ππ = πΈπ π ππ (500−π)(0.003) = 200000 π ππ = 200000(500−π)(0.003) π ππ = 600(500−π) π = 25 MPa, fy = 400 MPa. 1. Compute the depth of compression strees block. 2. Compute the ultimate moment capacity C=T 0.85 fc’ ab = As fs of the beam. 3. What is the correct description of the beam? 0.85 fc’ π½cb = 40000 fs 0.85 (25) (0.85) c (280) = 400(600)(500−π) π a. Under reinforced c2 + 474.54c – 237271.4 = 0 b. Over reinforced c = 304.55 c. Balanced condition d. Reduction in depth of compression zone results in decrease in steel strain at failure. ππ = 600(500−π) π ππ = 600(500−304.55) 304.55 ππ = 385.06 πππ < ππ¦ = 400 πππ Solution: βΈ« Steel does not yield 1. Depth of compression stress block ok as assumed Assume steel does not yield a = π½c a = 0.85 (304.55) a = 258.87 mm depth of compression block 2. Ultimate moment capacity of the beam π 2 Mu = ø As fy (d - ) fs ≠ fy fs = πΈπ ππ 0.003 ππ = 500−π π Mu = 0.90 (4000) (385.06) (500 Mu = 513.7 kN.m 3. Description of beam 258.87 ) 2 ππ = (500−π)(0.003) π ππ = (500−304.55)(0.003) 304.55 1. Depth of compression block for a balanced condition. ππ = 0.00193 π πΈπ ππ¦ = π 400 ππ¦ = 200000 ππ¦ = 0.002 ππ < ππ¦ 0.00193 < 0.002 The beam is Over Reinforced. π ππ¦ = π¦ πΈπ 414 ππ¦ = 200000 ππ¦ = 0.002 PROBLEM 45: A beam has a width of 300 mm and an effective 0.003 0.002 = 500−π π c = 300 mm depth of 500 mm. fc’ = 28 MPa, fy =414 MPa, Es = 200,000 MPa 1. Determine the depth of compression block for a balanced condition. a = π½c a = 0.85 (300) a = 255 mm 2. Determine the balanced steel area required. 3. Determine the moment capacity for maximum steel area in a balanced condition. 2. Balanced steel area required. C=T 0.85 fc’ ab = As fy 0.85 (28) (255) (300) = Asb (414) Solution: Asb = 4398 mm2 3. Moment capacity for maximum steel area in a balanced condition. As = 0.75 Asb (max. steel area for balanced condition) As = 0.75 (4398) As = 3298.5 mm2 1. Total depth of the beam for a balanced condition. ππ¦ = ππ‘ = 0.002 < 0.005 Use ø = 0.65 π π Mu = ø As fy (d - 2) 255 Mu = 0.65 (3298.5) (414) (500 - 2 ) Mu = 330.64 kN.m ππ¦ = πΈπ¦ π 414 ππ¦ = 200000 ππ¦ = 0.0021 a = π½c 255 = 0.85 (c) c = 300 mm PROBLEM 46: The width of a rectangular beam is 300 mm. The depth of compression block for a balanced By ratio and proportion: condition is 255 mm. fc’ = 28 MPa, fy =414 MPa, 0.003 0.0021 = π−300 300 Es = 200,000 MPa. Use 70 mm as steel covering. d = 300 + 210 Unit weight of concrete is 24kN/m3. d = 510 mm 1. Determine the total depth of the beam for a balanced condition. Total depth = 510 + 70 = 580 mm 2. Determine the total area of reinforcement for a balanced condition. 3. Determine the factored support imposed 2. Area of reinforcement for a balanced condition. uniform load that a 6-m. simple span beam could support for a balanced T=C condition. Asb fy = 0.85 fc’ ab Asb (414) = 0.85 (28) (255) (300) Solution: Asb = 4398 mm2 3. Factored super imposed uniform load that a 6-m. simple span beam could support for a a balanced condition. Solution: 1. Depth of compression block for a balanced condition. π Mu = ø As fy (d - 2) 255 Mu = 0.90 (4398) (414) (500 - 2 ) Mu = 626.8 kN.m Mu = ππ’ πΏ2 8 626.8 = ππ’ (6)2 8 Wu = 139.29 kN/m ππ¦ ππ¦ = πΈ π 415 ππ¦ = 200000 Wt. of concrete = 0.3 (0.58) (24) ππ¦ = 0.0020750 Wt. of concrete = 4.18 kN/m Wu = 4.18 (1.2) + Ws 0.003 0.0020750 = π 500−π c = 295.57 mm 139.29 = 4.18 (1.2) + Ws Ws = 134.27 kN/m a = π½c PROBLEM 47: A reinforced concrete beam has a width of 250 a = 0.85 (295.57) a = 251.23 mm mm and an effective depth of 500 mm. The compression strength of concrete is 28 MPa and the yield strength of steel is fy = 415 MPa. 1. Determine the depth of compression block for a balanced condition. 2. Determine the steel area required for a balanced condition. 3. Determine the ultimate moment capacity to ensure that concrete fails in a ductile manner. 2. Steel area required for a balanced condition. Check for ø: C=T 0.85 fc’ ab = Asb fy 0.85 (28) (251.53) (250) = Asb (415) Asb = 3606 mm2 (balanced steel area) 3. Ultimate moment capacity to ensure that concrete fails in a ductile manner. Use As = 0.75 Asb a = π½c As = 0.75 (3606) 188.63 = 0.85 (c) As = 2704.50 mm2 c = 221.92 mm ππ‘ 0.003 = 221.92 278.08 ππ‘ = 0.00376 < 0.005 Use ø = 0.65 + (ππ‘ – 0.002) (250) 3 ø = 0.65 + (0.00376 – 0.002) (250) 3 ø = 0.80 π Mu = ø As fy (d - 2) C=T Mu = 0.80 (2704.05) (415) (500 - 0.85 fc’ ab = Asb fy Mu = 364.26 kN.m 188.63 ) 2 0.85 (28) (a) (250) = 2704.50 (415) a = 188.63 PROBLEM 48: A rectangular reinforced concrete beam has a π Mu = ø As fy (d - 2) width b = 300 mm and an effective depth d = 400 mm. If fc’ = 28 MPa, fy = 280 MPa, Es = 200,000 MPa. 1. Which of the following gives the nearest 2. Compressive force of concrete: vaule of the distance of the N.A. from the top of the beam so that the strain in concrete ππ = 0.003 will be attained at the same time with the yield strain of steel ππ¦ . 2. Which of the following gives the nearest value of the total compressive force of concrete. 3. Which of the following gives the nearest value of the balanced steel ratio. a = π½c Solution: a = 0.85 (272.73) 1. Distance from N.A. to the top of the a = 231.82 mm beam: C = 0.85 fc’ ab π ππ¦ = π¦ C = 0.85 (28) (231.82) (300) πΈπ 280 ππ¦ = 200000 C = 1655194.8 N C = 1655.2 kN ππ¦ = 0.0014 0.003 0.0014 = π 400−π c = 272.73 mm 3. Balanced steel ratio: 0.85 π′π½600 π ππ = π (600+π ) π¦ ππ = π¦ 0.85 (28)(0.85)600 280(600+280) ππ = 0.0493 ππ = π. ππ % PROBLEM 49: The beam has a cross section as shown in the figure. It carries an ultimate moment of 156 kN.m. Using fc’ = 20.7 MPa, fy = 414 MPa, Es = 200,000 MPa. T=C 509804 = 0.85 fc’ Ac 509804 = 0.85 (20.7) Ac Ac = 28974 mm2 Ac = 100 (a) (2) + 100(a – 75) 28974 = 200a + 100a – 7500 300a = 36474 a = 121.58 mm a = π½c 121.58 = 0.85 (c) 1. Compute the location of the neutral axis c = 143.04 mm measured from the top of the beam. 2. Compute the number of 20 mm ø bars needed. 2. Number of 20 mm ø A1 = 100 (121.58) (2) 3. Compute the actual strain of the steel reinforcements used. A1 = 24316 A2 = 100 (46.58) A2 = 4658 Solution: 1. Location of N.A. A π¦Μ = A1y1 + A2y2 (24316 + 4658) π¦Μ = 24316 (60.79) + 4658 (98.29) π¦Μ = 66.82 z = 400 – 66.82 = 333.18 mm Mu = ø T z 156 x 106 = 0.90 As (414) (333.18) As = 1256.62 mm2 Approximate z = 0.85 d z = 0.85 (400) z = 340 Mu = ø T z π (20)2 π = 1256.62 4 N = 4 bars Use 4 – 20 mm ø bars 156 x 106 = 0.90 T (340) T = 509804 3. Actual strain of steel bars Solution: ππ 0.003 = 143.04 256.96 ππ = π. ππππ (actual strain) 1. Location of neutral axis from the top of the beam for a balance condition. π ππ¦ = πΈπ¦ π ππ¦ = 414 200000 ππ¦ = 0.00207 ππ > ππ¦ steel yields ππ = ππ¦ PROBLEM 50: A symmetrical cross-section of a reinforced concrete shown has a value of fc’ = 24.13 MPa, fy = 482.7 MPa, Es = 200,000 MPa. ππ¦ ππ¦ = πΈ π 482.7 ππ¦ = 200000 ππ¦ = 0.00241 0.003 0.00241 = 681.25−π π c = 377.77 mm 2. Balanced steel area Asb 1. Which of the following gives the location of neutral axis from the top of the beam for a balance condition. 2. Which of the following gives the balanced steel area Asb. 3. Which of the following gives the max. area permitted by the code. a = π½c 3. Which of the following gives the max. a = 0.85 (377.77) area permitted by the code. a = 321.10 mm Solution: C = 0.85 fc’ ab C = 0.85 (24.13) [375(125) + 125(321.10)] 1. Location of neutral axis from the top of the beam for a balance condition. C = 1784670 N T=C Asb (482.7) = 1784670 Asb = 3697 mm2 3. Max. area permitted by the code Asmax = 0.75 (3697) Asmax = 2773 mm2 π ππ¦ = π¦ πΈπ PROBLEM 51: The hallow box beam in the figure must carry a factored moment of 540 kN.m. fc’ = 28 MPa, fy = ππ¦ = 345 200000 ππ¦ = 0.0017 345 MPa, Es = 200,000 MPa. 0.003 0.0017 = 725−π π c = 462.77 mm 2. Balanced steel area Asb a = π½c a = 0.85 (462.77) a = 393.35 mm C = 0.85 fc’ A 1. Which of the following gives the location of neutral axis from the top of the beam A = 125 (393.35)(2) + 250(150) A = 135,837.5 mm2 for a balance condition. 2. Which of the following gives the balanced steel area Asb. C = 0.85 (28) (135837.5) C = 3232933 N T=C T=C Asb fy = 0.85 fc’ Ac Asb (345) = 3232933 Asb = 9371 mm π (12)2 (3) (414) = 0.85 (20.7) (255) Ac 4 2 Ac = 7983.3 mm2 3. Max. area permitted by the code Asmax = 0.75 (9371) Asmax = 7028.25 mm 2 ππ₯ = 7983.3 2 π₯= 15966.7 π PROBLEM 52: A triangular beam having a base width of 300 By ration and proportion mm. has a total depth of 600 mm. It is reinforced π₯ 300 = 600 π with 3 – 12 mm ø bars placed at 70 mm above the π₯ = 0.5 π bottom of the beam. fc’ = 20.7 MPa, fy = 414 MPa, Es = 200,000 MPa. 0.5 π = π = 178.70 1. Compute the neutral axis of the beam from the apex of the section. 2. Compute the ultimate strength capacity of the beam. 3. What would be the steel area required for 15966.7 π a = π½c 178.70 = 0.85 (c) c = 210.24 mm a balanced condition? 2. Ultimate strength capacity Solution: 1. Neutral axis of the beam z = 530 – 2/3 (178.7) z = 410.87 Mu = ø T z π Mu = 0.90 4 (12)2 (3) (414) (410.87) Mu = 51.9 kN.m 3. Steel area required for a balanced condition. 3 As = 4 Asb ππ¦ ππ¦ = πΈ 3 As = 4 (755) π 414 ππ¦ = 200000 As = 566.25 mm2 ππ¦ = 0.00207 0.003 0.00207 = 530−π π PROBLEM 53: 1.59 – 0.003c = 0.00207 c = 313.61 A triangular beam has an effective depth of 687.50 mm and a base of 750 mm. The beam carries an ultimate moment of 197 kN.m. fc’ = 27.5 MPa, fy = 414 MPa 1. Compute the neutral axis of the beam 2. Compute the value of total compressive force of concrete. 3. Compute the streel area required. Solution: 1. Neutral axis of the beam. a = π½c a = 0.85 (313.61) a = 266.57 mm T=C Asb fy = 0.85 fc’ Ac 266.57 600 = 300 π₯ π 750 = 750 π x = 133.29 a=b Asb fy = 0.85 fc’ Ac Asb (414) = 0.85 (20.7) Asb = 755 mm2 (133.29)(266.57) 2 C=T ππ 0.85 fc’ 2 = As fy 0.85(27.5)(π)2 = π΄π (414.7) 2 As = 0.0282 a2 PROBLEM 54: Mu = ø T z The beam has a cross section shown in the figure. 2 197 x 106 = 0.90 As (414.7) (687.50 - 3 a) fc’ = 20.7 MPa, fy = 414.7 MPa 197 x 106 = 0.90 (0.282) a2 (414.7) (687.50 – 0.667a) 18717186.73 = 687.50a2 – 0.667a3 Solve for “a” by trial and error a = 181.8 a = π½c 181.8 = 0.85 (c) c = 213.88 mm 2. Compressive force of concrete ππ C = 0.85 fc’ 2 πΆ= 0.85 (27.5)(181.8)(181.8) 2 C = 386286 N C = 386.3 kN 1. Compute the minimum steel area permitted by the NSCP Specs. 2. Compute the flexural design strength ø Mn if it is reinforced with minimum steel area. 3. Compute the maximum area of flexural steel that can be used in reinforced the 3. Steel area required section. C=T 386286 = As fy 386286 = As (414.7) As = 931.5 mm2 Solution: 1. Min. steel permitted by the NSCP Specs. 1.4 ππππ = π π¦ ππππ = 1.4 414.7 ππππ = 0.00338 π΄π πππ = ππππ π π π΄π πππ = 0.00338 (125)(2)(437.5) π΄π πππ = πππ. π πππ 2. Flexural strength ø Mn C=T 0.85 fc’ (750) a = As fy 0.85 (20.7) (750) a = 369.2 (414.7) a = 11.60 mm π 2 ø Mn = 0.90 T (π − ) ø Mn = 0.90 (369.2) (414.7) (437.5 − 11.6 ) 2 ø Mn = 59.5 x 106 N.mm ø Mn = 59.5 kN.m 1. Determine the neutral axis of the section 3. Max. steel area that can be used 0.85 π′π½600 π ππ = π (600+π ) π¦ ππ = π¦ 0.85 (20.7)(0.85)600 414.7(600+414.7) from the top of the beam. 2. Determine the flexural capacity of the cross section. 3. Determine the strain in the steel at failure. ππ = 0.0213 Solution: ππππ₯ = 0.75 ππ 1. Neutral axis of the beam. ππππ₯ = 0.75 (0.0213) ππππ₯ = 0.016 π΄π πππ₯ = ππππ₯ π π π΄π πππ₯ = 0.016 (125)(2)(437.5) π΄π πππ₯ = ππππ πππ PROBLEM 55: Each leg of the cross section is reinforced with a 28 mm ø bar. C=T 0.85 fc’ Ac = As fy π 0.85 (24.8) (500) a = 4 (28)2 (2) (414.7) a = 48.45 mm a = π½c 48.45 = 0.85 (c) c = 57 mm 2. Flexural capacity ø Mn π ø Mn = 0.90 T (π − 2 ) π 4 ø Mn = 0.90 (28)2 (2) (414.7) (375 − 48.45 ) 2 ø Mn = 161.2 x 106 N.mm ø Mn = 161.2 kN.m 3. Strain in steel at failure: ππ 0.003 = 318 57 ππ = π. ππππ π΄π = 1963.5 ππ2 π΄ π = πππ 1963.5 PROBLEM 56: CE BOARD MAY 2010 π = 300(380) A 12 m simply supported beam is provided by an π = 0.017 additional support at midspan. The beam has a width of b = 300 mm and a total depth h = 450 If only tension bars are needed: mm. It is reinforced with 4 – 25 mm ø at the ππππ₯ = 0.75(ππ ) tension side and 2 – 25 mm ø at the compression ππππ₯ = 0.023 > 0.017 side with 70 mm cover to centroid of reinforcement. fc’ = 30 MPa, fy = 415 MPa. Use Therefore, the beam needs only tension 0.75 ππ = 0.023 bars a specified in the problem. 1. Determine the depth of the rectangular stress block. 2. Determine the nominal bending moment, Mn. C=T 0.85 fc’ ab = As fy π 4 0.85 (30) (a)(300) = (25)2 (4)(415) a = 106.52 mm 3. Determine the total factored uniform load including the beam weight considering moment capacity reduction of 0.90. Solution: 2. Nominal bending moment: π Mn = As fy (π − 2 ) π 1. Depth of the rectangular stress block: Mn = 4 (25)2 (4)(415) (380 − Mn = 266.2 x 106 N.mm Mn = 266.2 kN.m Check if compression bars is needed. π 4 π΄π = (25)2 (4) 106.52 ) 2 3. Total factored uniform load including beam weight: PROBLEM 57: A rectangular beam has a width of 300 mm and an effective depth of 537.50 mm to the centroid of tension steel bars. Tension reinforcement consists of 6 – 28 mm ø in two rows, compression reinforcement consists of 2 – 22 mm ø. fc’ = 27.6 MPa, fy = 414.7 MPa. Assume steel covering is 60 mm for compression bars. πΏ= 5 π€πΏ4 384 πΈπΌ 1. Compute the depth of compression ππΏ3 πΏ = 48 πΈπΌ block. 2. Compute the factored moment capacity 5 π€πΏ4 ππΏ3 = 384 πΈπΌ 48 πΈπΌ of beam. 5π€πΏ π= 8 3. Compute the maximum total tension steel 2R + P = wL 2R = wL 2R = allowed by specifications. 5π€πΏ 8 Solution: 3π€πΏ 8 1. Depth of compression block. 3 R = 16 π€πΏ πΏ 2 πΏ 2 πΏ 4 ππ΅ = π ( ) − π ( ) ( ) 3π€πΏ2 ππ΅ = 16(2) − π€πΏ2 8 π€πΏ2 ππ΅ = − 32 Mu = 0.90 Mn Mu = 0.90 (266.2) Mu = 239.58 kN.m 239.58 = π€πΏ2 32 239.58 = π€(12)2 32 w = 53.24 kN/m π 4 π΄π = (28)2 (6) π΄π = 3695 ππ2 π π΄π ′ = 4 (22)2 (2) π΄′π = 760 ππ2 π΄ π = πππ π= 3695 300(537.50) π = 0.0229 π΄ ′ π′ = πππ π′ = 760 300(537.50) π = 0.00471 As1 = As – As2 As2 = As’ when compression bars will yield As1 = As – As’ 0.85 fc’ ab = (As – As’) fy 0.85 (27.6) a (300) = (3695 – 760) 414.7 a = 172.94 mm Check if compression bar are really needed: ππ = 0.85 ππ′π½600 ππ¦ (600+ππ¦ ) 0.85 (27.6)(0.85)600 ππ = 414.7(600+414.7) 2. Factored moment capacity of the beam. a = π½c 172.94 = 0.85 (c) c = 203.46 mm ππ = 0.0284 When, π1 0.003 = 203.4 346.6 π > ππππ₯ compression bars are needed π1 = 0.00511 > 0.005 ππππ₯ = 0.75 ππ 0.0229 > 0.75(0.0284) Use capacity reduction factor: ø = 0.90 0.0229 > 0.0213 Check: if compression bars will yield π − π′ > 0.85 ππ′ π½ π ′ 600 π ππ¦ (600−ππ¦ ) π − π′ = 0.0229 − 0.00471 π − π′ = 0.0182 π 0.85 (27.6)(0.85)(60)600 = 537.5(414.7)(600−414.7) = 0.0173 M1 = ø As1 fy (π − 2 ) As2 = As’ As1 = As - As2 As1 = As – As’ 0.0182 > 0.0173 (compression bars will yield) π M1 = ø (As – As’) fy (π − 2 ) Check: π − π′ < ππππ₯ M1 = 0.90 (3695 – 760) 414.7 (537.50 − M1 = 494.07 x 106 N.mm 0.0182 < 0.0213 M2 = ø As2 fy (π − π′) C1 = T1 0.85 fc’ ab = As1 fy M2 = ø As’ fy (π − π′) 172.94 ) 2 M2 = 0.90 (760) (414.7) (537.5 – 60) 630 = 1.8 b M2 = 135.45 x 106 N.mm b = 350 mm Mu = M 1 + M 2 2. Reinforcement for compression Mu = 494.07 + 135.45 ππ = 0.85 ππ′π½600 ππ¦ (600+ππ¦ ) ππ = 0.85 (27.58)(0.85)600 413.4(600+413.4) Mu = 629.52 kN.m 3. Maximum total tension steel allowed by ππ = 0.0285 specifications. 0.85 fc’ ab = As1 fy ′ Max. As = ππππ₯ ππ + π ππ Max. As = 0.75ππ ππ + π′ ππ Max. As = ππ (0.75ππ + π′ ) Max. As = 300(537.5)(0.0213 + 0.00471) Max. As = ππππ πππ > ππππ πππ PROBLEM 58: A doubly reinforced concrete beam has a max. Assume effective depth of 630 mm and is subjected to a π1 = 0.85 ππ′ π½ (0.003) ππ¦ (0.008) π1 = 0.85 (27.58)(0.85)(0.003) 413.4 (0.008) total factored moment of 1062 kN.m including its own weight. fc’ = 27.58 MPa, fy = 413.4 MPa. Use 62.5 mm steel covering. π1 = 0.018 As1 = As – As2 1. Determine the width of the beam 2. Determine the reinforcement for compression 3. Determine the total reinforcement for As2 = As’ if compression steel will yield As1 = π1 bd As1 = 0.018 (350) (630) tension As1 = 3969 mm2 Solution: π 1. Width of the beam Approximate proportion of b and d d = 1.5 to 2 b Try d = 1.80 b M1 = ø As1 fy (π − 2 ) M1 = 0.90 (3969) 413.4 (630 − M1 = 782.7 kN.mm Mu = M 1 + M 2 199.97 ) 2 Use As = 5292 mm2 1062 = 494.07 + M2 M2 = 279.3 kN.m PROBLEM 59: Check whether compression bars will yield: π − π′ > 0.85 ππ′ π½ π ′ 600 π ππ¦ (600−ππ¦ ) As = As1 + As2 A rectangular beam has a width of 300 mm and an effective depth to the centroid of the tension reinforcement of 600 mm. The tension 2 reinforcement has an area of 4762 mm and the As2 = As’ area of compression reinforcement placed 62.50 As = As1 + As’ mm from the compression face to the beam is π bd = π1 bd + π′ bd ′ π − π = π1 987.5 mm2. fc’ = 34.56 MPa, fy = 414.6 MPa. Balanced steel ratio is 0.034. Assume that steel 0.85 (27.58)(0.85)(62.5)(600) 0.018 > 630 (413.4)(600−413.4) yields. 0.018 > 0.0154 1. Determine the depth of compression Therefore, compression bars will yield. fs’ = fy block 2. Determine the design strength using 0.90 As’ fy = As2 fy As’ = As2 as the reduction factor (ok as assumed) 3. Determine the concentrated live loads at M2 = ø As’ fy (π − π′) the midspan in addition to a dead load of 279.3 x 106 mm = 0.90 As’ (413.4) (630 – 62.5) 20 kN/m including the weight of the As’= 1323 mm2 3. Total reinforcement for tension. As = As1 + As2 As = 3969 + 1323 As = 5292 mm2 π′ = π΄π ′ ππ 1323 π′ = 350(630) π = 0.006 Max. As = bd (0.75 ππ + π′ ) Max. As = 350(630) [0.75(0.0285) + 0.00669)] Max. As = 6036 mm2 > 5292 mm2 ok beam if it has a span of 6 m. Solution: 1. Depth of compression block 0.85 π′π½600 π ππ = π (600+π ) π¦ π¦ π½ = 0.85 − 0.05 (ππ′ −28) 7 π½ = 0.85 − 0.05 (34.56−28) 7 π½ = 0.803 ππ = 0.85 (34.56)(0.803)(600) 414.6(600+414.6) ππ = 0.034 π΄ Since, compression bars will yield: 4762 π = πππ = 300 (600) fs’ = fy π = 0.0264 π΄ ′ T = C1 + C2 987.5 π′ = πππ = 300 (600) As fy = 0.85 fc’ ab + As’ fy π′ = 0.0054 π= Check the beam first as a singly reinforced beam to see if the compression bars can be disregarded. π= (π΄π −π΄′π )ππ¦ 0.85 ππ′ π (4762−987.5)414.6 0.85 (34.56)(300) π = πππ. ππ ππ ππππ₯ = 0.75 ππ ππππ₯ = 0.75 (0.034) 2. Design strength using 0.90 as the ππππ₯ = 0.0255 reduction factor. ππππ‘π’ππ = 0.0264 > 0.0255 Therefore, the beam must be analyzed as doubly reinforced beam. Check if the bars in compression will really yield, by computing the steel ratio that will ensure yielding of the compression bar at failure. ππππ‘π’ππ = 0.0264 C1 = T1 0.85 π′ π½ π ′ 600 π π − π′ > π π (600−π ) π¦ 0.85 fc’ ab = As1 fy π¦ 0.85 (34.56)(0.817)(62.5)(600) 0.0264 − 0.0054 > 414.6 (600)(600−414.6) 0.021 > 0.0195 bars will yield) (therefore, compression 0.85 (34.56) (177.57) (300) = As1 (414.6) As1 = 3774.50 mm2 As = As1 + As2 As2 = As – As1 PROBLEM 60: As2 = 4762 – 3774.50 A reinforced concrete beam has a width of 375 As2 = 987.50 mm2 ok mm and a total depth of 775 mm. Steel covering for both compression and tension bars is 75 mm. Area of compression bars is 1290 mm2 while that Mu = ø (M1 + M2) of the tension bars it is 6529 mm2. fc’ = 27.6 MPa, π M1 = T1 (π − 2 ) fy = 414.6 MPa. π M1 = As1 fy (π − 2 ) M1 = 3774.5 (414.6) (600 − 177.51 ) 2 M1 = 800 x 106 N.mm 1. Determine the depth of compression block. 2. Determine the capacity reduction factor M2 = ø As2 fy (π − π′) M2 = 987.50 (414.6) (600 – 62.5) M2 = 220 x 106 N.mm Mu = ø (M1 + M2) Mu = 0.90 (800 + 220) for moment. 3. Determine the ultimate moment capacity of the beam. Solution: 1. Depth of compression block. Mu = 918 kN.m 3. Concentrated live loads it could support at its midspan. 0.003 π ′ = π π π−75 ππ ′ = ππΏ π€πΏ2 Mu = (1.6) 4 + 8 (1.2) π(6) 20(6)2 918 = (1.6) 4 + 8 (1.2) P = 337.5 kN 0.003(π−75) π ππ ′ = ππ ′πΈπ ππ ′ = 0.003(π−75)(200000) π ππ ′ = 600(π−75) π T = C1 + C2 As fy = 0.85 fc’ ab + As’ fs’ 6529 (414.6) = 0.85(27.6) (0.85) c (375) + 1290 (600)(π−75) π 1932923.4 c = 7477.88c2 – 58050000 = 0 c2 – 258.49 c – 7762.90 = 0 c = 285.66 mm ππ ′ = 600 (π−75) π ππ ′ = 600 (285.66−75) 285.66 ππ ′ = 442.47 πππ > 414.6 πππ Note: When ππ < 0.005 Value of ø = 0.65 + ( ππ – 0.002) ø = 0.65 + (0.00435 – 0.002) Compression steel yield: (250) 3 (250) 3 ø = 0.846 Use fs’ = fy = 414.6 MPa a = π½c 3. Ultimate moment capacity of the beam. a = 0.85 (285.66) c = 242.81 mm 2. Capacity reduction factor for moment. π 0.003 = π π−π π= 0.003(π−π) π π= C2 = T2 0.003(700−285.66) 285.66 As’ fy = As2 fy π= 0.00435 < 0.005 As’ = As2 As2 = 1290 mm2 As1 = As – As2 As1 = 6529 – 1290 As1 = 5239 mm2 π ππ’ = ∅ [π1 (π − 2 ) + π2 (π − π′ )] π ππ’ = ∅ [π΄π 1 ππ¦ (π − 2 ) + π΄π 2 ππ¦ (π − π′ )] Solution: 1. Minimum tensile steel ratio that will ensure yielding of the compression steel 242.81 ππ’ = 0.846 [5239(414.6) (700 − 2 ) + at failure. 1290(414.6)(700 − 75)] ππππ = ππ’ = 1346 π₯ 106 π. ππ As = ππ’ = ππππ ππ΅. π 0.85 ππ′ π½ π ′ 600 + π′ ππ¦ π (600+ππ¦ ) π = (28)2 (2) = 1232 ππ2 4 0.85 π′ π½ π ′ 600 π¦ PROBLEM 61: A rectangular concrete beam has a width of 350 π΄ ′ π ππππ = π π (600+π + πππ ) ππππ = π¦ 0.85 (20.7)(0.85)(62.5)600 1232 + 414.6(537.5) (600+414.6) 350(537.5) ππππ = 0.020 mm and a total depth of 675 mm. It is reinforced for tension at the bottom with 4 – 36 mm ø bars at an effective depth of 537.5 mm and two 28 mm ø bars at the top placed at 62.5 mm from the top of the beam. fc’ = 20.7 MPa, fy = 414.6 MPa. Check: π΄π ππ π As = = (36)2 (4) = 4071.5 ππ2 4 π= 4071.5 π = 350(537.5) 1. Determine the minimum tensile steel ratio that will ensure yielding of the π = 0.0216 > 0.020 ππππ = π. πππ compression steel at failure. 2. Determine the total compressive force of 2. Total compressive force of concrete. concrete. 3. Determine the design moment capacity of the beam. π = (36)2 (4) = 4071.5 ππ2 4 π As’ = 4 = (28)2 (2) = 1232 ππ2 As = π= 4071.5 350(537.5) π = 0.0216 1232 π′ = 350(537.5) π′ = 0.0065 C1 = 0.85 (20.7) (188) (350) When: C1 = 1157751 N 0.85 ππ′ π½ π ′ 600 π − π′ > π π (600−π ) (compression bars will π¦ π¦ yield) C2 = As’ fs’ C2 = 1232 (414.6) π − π′ = 0.0216 − 0.0065 C2 = 510787 N π − π′ = 0.0151 0.85 (20.7)(0.85)(62.5)600 0.0151 > 414.6(537.5)(600−414.6) π M1 = C1 (π − 2 ) 0.0151 > 0.0136 M1 = 1157751 (537.5 − 188 ) 2 M1 = 513.46 x 106 N.mm Therefore, compression steel yields. M2 = C2 (π − π′) fs’ = fy = 414.6 MPa M2 = 510787 (537.5 – 62.5) C1 = 0.85 fc’ ab M2 = 242.62 x 106 N.mm a = π½c a = 0.85 (221.18) Mu = ø (M1 + M2) a = 188 mm Mu = 0.90 (513.46 + 242.62) C1 = 0.85 (20.7) (188) (350) Mu = 635.11 kN.m C1 = 1157751 N C1 = 1157.75 kN PROBLEM 3. Design moment capacity of the beam. 62: CE Board Nov. 2010, Nov.2012 A simply supported beam is reinforced with 4 – 28 mm ø at the bottom and 2 – 28 mm ø at the top ππ‘ 0.003 = 221.18 316.32 of the beam. Steel covering to centroid of ππ‘ = 0.00429 < reinforcement is 70 mm at the top and bottom of 0.005 the beam. The beam has a total depth of 400 mm and a width of 300 mm. fc’ = 30 MPa, fy = 415 Use ø = 0.65 + ( MPa. Balanced steel ratio ππ = 0.031. (250) ππ‘ – 0.002) 3 ø = 0.65 + (0.00429 – 0.002) ø = 0.84 (250) 3 1. Determine the depth of compression block. 2. Determine the design strength using 0.90 Mu = ø (M1 + M2) C1 = 0.85 fc’ ab as the reduction factor. 3. Determine the live load at the mid-span in addition to a DL = 20 kN/m including Therefore, compression bars will not yield. the weight of the beam if it has a span of fs’ ≠ fy 6 m. 0.003 ππ ′ = π−70 π ππ ′ = Solution: 1. Depth of compression block: 0.003(π−70) π ππ ′ = ππ ′πΈπ π = (28)2 (4) = 2463 ππ2 4 π As’ = = (28)2 (2) = 1231.5 ππ2 4 As = π΄ π = πππ π= 2463 350(330) π = 0.0248 T = C1 + C2 As fy = 0.85 fc’ ab + As’ fs’ ππππ₯ = 0.75 ππ 2463(415) ππππ₯ = 0.75 (0.031) 0.003(π−70)(200000) π ππππ₯ = 0.023 = (30) 1022145 = 7650a + (a) (300) 738900(π−70) π 738900(π−70) π π > ππππ₯ 1022145 = 7650 (0.85c) + Therefore, reinforcement for compression is 157.19 c = c2 + 113.63 (c – 70) needed. To ensure that compression bars will c2 + 43.56 c -7654.10 = 0 yield. π= 0.85 π′ π½ π ′ 600 43.56 ±183.61 2 π π − π′ > π π (600−π ) π¦ π − π′ > c = 113.59 π¦ 0.85 (30)(0.85)(70)600 415(330)(600−415) π − π′ > 0.036 π΄ ′ π′ = πππ a = π½c a = 0.85 (113.59) a = 96.55 mm 1231.5 π′ = 300(330) π′ = 0.0124 π − π′ > 0.036 0.0248 − 0.0124 < 0.036 0.0124 < 0.036 + 1231.5 2. Design strength using 0.90 as reduction factor. Mu = ø (M1 + M2) Mu = 0.90 (208 + 73.7) Mu = 253.53 kN.m 3. Concentrated live load at mid-span: ππΏ π€πΏ2 Mu = (1.6) 4 + 8 (1.2) 253.53 = (1.6) π(6) 20(6)2 + (1.2) 4 8 P = 60.64 kN C1 = 0.85 fc’ ab C1 = 0.85 (30) (96.55) (300) C1 = 738607.50 N PROBLEM 63: A reinforced concrete beam has a width of 350 mm and an effective depth of 562.5 mm. It is C2 = As’ fs’ reinforced for tension at the bottom of the section ππ ′ = ππ ′πΈπ having an area of 4896 mm2 and for compression ππ ′ = 0.003(π−70) π 0.003(113.59−70) ππ ′ = 113.59 at the top of the beam 62.5 mm below the extreme compression fibers of the beam, having an area of 1530 mm2. fc’ = 34.6 MPa, fy = 414.7 MPa. ππ ′ = 0.0011512 ππ ′ = 0.0011512 (200000) ππ ′ = 230.25 < 415 πππ 1. Determine the depth of compression block. 2. Determine the ultimate moment capacity. C2 = As’ fs’ 3. Determine the maximum total tension C2 = 1231.5 (230.25) area that could be used in this section. C2 = 283553 Solution: π 2 M1 = C1 (π − ) 96.55 M1 = 738607.5 (330 − 2 ) M1 = 208 x 106 N.mm 1. Depth of compression block. π΄ π = πππ 4896 π = 350(562.5) M2 = C2 (π − π′) π = 0.0249 M2 = 283553 (330 – 70) M2 = 73.7 x 106 N.mm π΄ ′ π′ = πππ 1530 π′ = 350(562.5) π′ = 0.0078 C1 + C2 = T1 + T2 C1 + C2 = T + As fy π − π′ = 0.0249 − 0.0078 π − π′ = 0.0171 8234.8 c + 918000 (π−62.5) = 4896 (414.7) π c2 + 111.48 (c – 62.5) = 246.56 c c2 – 135.08 c – 6967.5 = 0 0.85 ππ′ π½ π ′ 600 When, π − π′ > π π (600−π ) π¦ c = 174.91 mm π¦ (compression bar did not yield) a = π½c a = 0.85 (174.91) a = 139.93 mm 2. Ultimate moment capacity. C2 = As’ fs’ 0.003 ππ ′ = π−π′ π ππ ′ = ππ ′ = 600(174.91−62.5) 174.91 ππ ′ = 385.60 πππ 0.003(π−π′) π πΆ2 = ππ ′ = ππ ′πΈπ 91800(174.91−62.5) 174.91 πΆ2 = 589974 π ππ ′ = ππ ′ = 0.003(π−π ′ ) π (200,000) 600(π−62.5) π πΆ2 = 589.97 ππ C1 = 8234.8 c C1 = 8234.8 (174.91) C1 = 0.85 fc’ ab C1 = 1440349 N C1 = 0.85 fc’ π½ cb C1 = 0.85 (34.6) (0.8) c (350) C1 = 8234.8 c π M1 = C1 (π − 2 ) M1 = 1440349 (562.5 − C2 = As’ fs’ C2 = M1 = 709.42 x 106 N.mm 1530 (600)(π−62.5) π C2 = 918000 139.93 ) 2 (π−62.5) π M2 = C2 (π − π′) M2 = 589974 (562.5 – 62.5) M2 = 294.984 x 106 N.mm Mu = ø (M1 + M2) π΄ Mu = 0.90 (709.42 + 294.98) π = πππ Mu = 903.96 kN.m 4744 π = 375(500) π = 0.0253 3. Maximum total tension area that could be used in this section. π΄ ′ π′ = πππ π′ π Max. As = bd ( 0.75 ππ + π′ ππ¦ ) 1968 Max. As = 350(562.5) [0.75(0.03346) + π′ = 375(500) 0.0078 (385.60) ] 414.7 π′ = 0.010496 Max. As = 6369 mm2 ππππ₯ = 0.75 ππ ππ = 0.85 ππ′π½600 ππ¦ (600+ππ¦ ) mm and an effective depth of 500 mm. ππ = 0.85 (27.6)(0.85)(600) 414.6(600+414.6) compression bars has an area of 1968 mm2 ππ = 0.02844 PROBLEM 64: A rectangular concrete beam has a width of 375 located at 100 mm from the compression face of the beam. The tension bars have an area of 4744 ππππ₯ = 0.75(0.02844) mm2. fc’ = 27.6 MPa, fy = 414.6 MPa. ππππ₯ = 0.02133 1. Determine the depth of the compression 0.0253 > 0.02133 block. 2. Determine the max. steel ratio. Therefore, compression bars are really needed. 3. Determine the ultimate moment capacity of the beam. Check if compression bars will yield or not. 0.85 π′ π½ π ′ 600 π π − π′ > π π (600−π compression bars will ) Solution: 1. Depth of the compression block. π¦ π¦ not yield 0.85(27.6)(0.85)(100)600 π − π′ > 414.6 (500)(600−414.6) 0.0253 – 0.010496 < 0.03113 0.0148 < 0.03113 not yield fs’ ≠ fy ππ ′ = 600(π−π ′ ) π C1 + C2 = T1 + T2 compression bars will As = As1 + As2 3. Ultimate moment capacity of the beam. T1 + T2 = T As1 fy + As2 fy = T (As1 + As2) fy = T T = As fy C1 + C2 = T As’ fs’ = As2 fy 0.85 fc’ ab + As’ fs’ = As fy 0.85 (27.6) (0.85) c (375) + 1968 (600)(π−100) = π 4744 (414.6) c2 – 105.12 c – 15790.57 = 0 c = 188.77 1968 (282.15) = As2 (414.6) As2 = 1339 mm2 As = As1 + As2 As = 4744 + 1339 a = π½c As = 3405 mm2 a = 0.85 (188.77) a = 160.45 mm 2. Max. steel ratio. π′ ππππ₯ = 0.75ππ + π′ ππ π 2 M1 = As1 fy (π − ) M1 = 3405 (414.6) (500 − 160.45 ) 2 M1 = 704.44 x 106 N.mm π¦ ππ ′ = ππ ′ = 600(π−π ′ ) M2 = As2 fy (π − π′) π 600(188.77−100) 188.77 M2 = 1339 (414.6) (562.5 – 62.5) M2 = 222.06 x 106 N.mm ππ ′ = 282.15 πππ Mu = ø (M1 + M2) 282.15 ππππ₯ = 0.02844 + 0.010496 414.6 ππππ₯ = 0.02847 Check if we could use ø = 0.90 0.0253 < 002847 π1 0.003 = 188.77 311.23 π < ππππ₯ π1 = 0.005 ok ok π As = 4 = (32)2 (6) = 4825 ππ2 Use ø = 0.90 Mu = 0.90 (704.44 + 222.06) π΄ π = πππ Mu = 833.85 kN.m π= 4825 300(600) π = 0.0268 PROBLEM 65: A rectangular beam has a width of 300 mm and an effective depth to the centroid of the tension reinforcement of 600 mm. The ππππ₯ = 0.75 ππ 0.85 π′π½600 tension π ππ = π (600+π ) π¦ reinforcement consists of 6 – 32 mm ø bars placed π¦ reinforcement π½ = 0.85 − 0.05 (ππ′−28) 7 consisting of two 25 mm ø bars is placed 62.5 mm π½ = 0.85 − 0.05 (34.6−28) 7 in two rows. Compression from the compression face of the beam. fc’ = 34.6 π½ = 0.80 MPa, fy = 414.7 MPa. ππ = 0.85 (34.56)(0.80)(600) 414.7(600+414.7) ππ = 0.0335 1. Determine the depth of compression block. 2. Determine the maximum steel ratio. ππππ₯ = 0.75(0.0335) 3. Determine the design moment capacity ππππ₯ = 0.0252 of the beam. Compression bars is needed. Solution: π > ππππ₯ 1. Depth of compression block. Check whether compression bars is really 0.0268 > 0.0252 needed. Check if compression bars will yield or not. 0.85 π′ π½ π ′ 600 π π − π′ > π π (600−π ) π¦ π¦ π΄ ′ π′ = πππ π (25)2 (2) 4 π′ = 200(600) π′ = 0.00545 π − π′ > 0.85(34.6)(0.817)(62.5)600 414.7 (500)(600−414.7) 0.0268 – 0.00545 > 0.01954 0.02135 > 0.01954 fs’ = fy (steel compression yields) π As’ = 4 = (25)2 (2) = 981.75 ππ2 a = π½c 180.64 = 0.85 (c) c = 221.10 mm T1 + T2 = C1 + C2 ππ 0.003 = 378.30 221.10 T = C1 + C2 ππ = 0.0051 > ππ¦ As fy = 0.85 fc’ ab + As’ fy 4825 (414.7) = 0.85 (34.6) (a) (300) + 981.75 fs = fy Use ø = 0.90 π Mn = C1 (π − 2 ) + C2 (π − π′) π Mn = 0.85 fc’ ab (π − 2 ) + As’ fy (π − π′) C1 = 0.85 fc’ ab C1 = 0.85 (34.6) (180.64) (300) C1 = 1593787 N (414.7) a = 180.64 mm C2 = As’ fy C2 = 981.75 (414.7) C2 = 407132 N 2. Maximum steel ratio ππππ₯ = 0.75ππ + π′ ππππ₯ = 0.75 (0.0335) + (0.00545) π 2 Mn = C1 (π − ) + C2 (π − π′) ππππ₯ = π. ππππ Mn = 1593787 (600 − 3. Design moment capacity of the beam. 180.64 ) + 407132 (600 − 2 62.5) Mn = 1031 x 106 N.mm Mu = ø Mn Mu = 0.90 (1031) Mu = 927.9 kN.m PROBLEM 66: A rectangular beam reinforced for both tension and compression bars has an area of 1250 mm T = C1 + C2 As fy = 0.85 fc’ ab + As’ fs’ 2 and 4032 mm2 for tension bars. The tension bars 4032 (414.6) = 0.85(20.7) (0.85) c (350) 1250 (600)(π−62.5) π are placed at a distance of 75 mm from the bottom + of the beam while the compression bars are 1671667.2 c = 5234.5c2 – 750000 c - 46875000 placed 62.5 mm from the top of the beam. fc’ = c2 – 176.08 c – 8955 = 0 20.7 MPa, fy = 414.60 MPa. Width of beam is 350 c = 217.29 mm mm with a total depth of 675 mm. 1. Determine the depth of compression block. 2. Determine the ultimate moment of ππ ′ = 0.003(π−62.5) π ππ ′ = 0.003(217.29−62.5) 217.29 ππ ′ = 0.00214 capacity of the beam. 3. Determine the safe live concentrated load that the beam could support at its midspan if it has a span of 6-m. Assume ππ¦ ππ¦ = πΈ π 414.6 ππ¦ = 200000 ππ¦ = 0.00207 weight of concrete to be 23.5 kN/m. ππ ′ > ππ¦ ππ ′ > ππ¦ Solution: 1. Depth of the compression block. a = π½c a = 0.85 (217.29) a = 184.70 mm 2. Ultimate moment capacity of the beam 0.003 ππ ′ = π−62.5 π 0.003(π−62.5) ππ ′ = π ππ ′ = ππ ′πΈπ ππ ′ = 0.003(π−62.5)(200000) π ππ ′ = 600(π−62.5) π 0.003 ππ = π−π π ππ = 0.003(π−π) π ππ = 0.003(600−217.29) 217.29 ππ = 0.00529 > 0.005 Note: ππ’ = 0.90 [2782(414.6) (600 − When ππ < 0.005 Use ø = 0.65 + ( ππ‘ – 0.002) (250) 3 184.70 )+ 2 1250(414.6)(600 − 62.5)] ππ’ = 777.7 π₯ 106 π. ππ ππ’ = πππ. π ππ΅. π 3. Safe concentrated live load at mid span of the beam: ππΏ Mu = 4 + π€π·πΏ πΏ2 8 ππΏ Mu = (1.7) 4 + π€π·πΏ(πΏ)2 (1.4) 8 WDL = 23.5 (0.35) (0.675) WDL = 5.55 kN/m Use ø = 0.90 Mu = (1.7) ππΏ π€π·πΏ(πΏ)2 + (1.4) 4 8 777.7 = (1.7) π(6) 5.55(6)2 + (1.4) 4 8 P = 291.27 kN As’ fy = As2 fy As’ = As2 As = As1 + As2 As1 = As – As2 As1 = As – As’ As1 = 4032 – 1250 As1 = 2782 mm2 π ππ’ = ∅ [π1 (π − 2 ) + π2 (π − π′ )] π ππ’ = 0.90 [π΄π 1 ππ¦ (π − 2 ) + π΄π 2 ππ¦ (π − π′ )]
