Analysis And Design Of Beams Problems

ANALYSIS AND DESIGN OF BEAMS PROBLEMS PROBLEM 2.1 A reinforced concrete rectangular beam 300 mm wide has an effective depth of 460 mm and is reinforced for tension only. Assuming 𝑓′𝑐 = 21 𝑀𝑃𝑎 and 𝑓𝑦 = 345𝑀𝑃𝑎, determine the balance steel area in sq.mm. SOLUTION 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+𝑓𝑦 )

𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 < 30𝑀𝑃𝑎 𝜌𝑏 =

0.85(21)(0.85)(600) 345(600 + 345) 𝜌𝑏 = 0.02792

𝐴𝑠𝑏 = 𝜌𝑏 𝑏𝑑

PROBLEM 2.2 A rectangular beam has b = 300 mm and d =490 mm. Concrete compressive strength 𝑓′𝑐 = 27.6𝑀𝑃𝑎 and steel yield strength 𝑓𝑦 = 276 𝑀𝑃𝑎. Calculate the required tension steel area if the factored moment 𝑀𝑢 is (a) 20 kN-m, (b) 140 kN-m, (c) 485 kN-m, and (d)620 kN-m. SOLUTION Solve for 𝜌𝑚𝑎𝑥 𝑎𝑛𝑑 𝑀𝑢 𝑚𝑎𝑥 : 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+𝑓𝑦 )

𝜌𝑚𝑎𝑥 = 0.75 𝜌𝑏 𝜔𝑚𝑎𝑥 =

𝜌𝑚𝑎𝑥 𝑓𝑦 𝑓′ 𝑐

𝜌𝑏 =

0.85(27.6)0.85(600) 276(600+276)

𝜌𝑏 = 0.0495 𝜌𝑚𝑎𝑥 = 0.75(0.0495) 𝜌𝑚𝑎𝑥 = 0.0371 0.03711(276) 𝜔𝑚𝑎𝑥 = 27.6 𝜔𝑚𝑎𝑥 = 0.371

𝑅𝑛 𝑚𝑎𝑥 = 𝑓 ′ 𝑐 𝜔(1 − 0.59𝜔) 𝑅𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥 = 𝑅𝑛 𝑚𝑎𝑥 𝑏𝑑2

𝑅𝑛 𝑚𝑎𝑥 = 27.6(0.371)[1 − 0.59(0.37)] = 8.001𝑀𝑃𝑎

𝑀𝑛 𝑚𝑎𝑥 = 8.001(300)(490)2 𝑀𝑛 𝑚𝑎𝑥 = 576.279𝑥106 𝑁 − 𝑚𝑚 𝑀𝑛 𝑚𝑎𝑥 = 576.279𝑘𝑁 − 𝑚𝑚 𝑀𝑢 𝑚𝑎𝑥 = 0.90 𝑥 576.279 𝑀𝑢 𝑚𝑎𝑥 = 518.65 𝑘𝑁 − 𝑚

𝑀𝑢 𝑚𝑎𝑥 = 𝜑𝑀𝑛 𝑚𝑎𝑥

a) 𝑀𝑢 = 20𝑘𝑁 − 𝑚 < 𝑀𝑢 𝑚𝑎𝑥 = (𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑) 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑 2 20 x 106 = 0.90𝑅𝑛 (300)(490)2 𝑅𝑛 = 0.309 𝑀𝑃𝑎

𝜌=

𝜌=

0.85𝑓′𝑐 2𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.085𝑓′𝑐

0.85(27.6) 2(0.309 [1 − √1 − ] 276 0.85(27.6) 𝜌 = 0.00113 < 𝜌𝑚𝑖𝑛

𝜌𝑚𝑖𝑛 =

𝐴𝑠 = 𝜌𝑏𝑑

√𝑓′𝑐 1.4 𝑖𝑓 𝑓′𝑐 > 31.36𝑀𝑃𝑎, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝜌𝑚𝑖𝑛 = 4𝑓𝑦 𝑓𝑦 1.4 𝜌𝑚𝑖𝑛 = = 0.005072 𝑓𝑦 𝐴𝑠 = 0.00572(300)(490) 𝑨𝒔 = 𝟕𝟒𝟔𝒎𝒎𝟐

b) 𝑀𝑢 = 140𝑘𝑁 − 𝑚 < 𝑀𝑢 𝑚𝑎𝑥 (singly reinforced) 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2 140 x 106 = 0.90 𝑅𝑛 (300)(490)2 𝑅𝑛 = 2.16 𝑀𝑃𝑎

𝜌=

𝜌=

𝐴𝑠 = 𝜌𝑏𝑑

0.85𝑓′𝑐 2𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.85𝑓′𝑐

0.85(27.6) 2(2.16) [1 − √1 − ] 276 0.85(27.6)

𝜌 = 0.00822 > 𝜌𝑚𝑖𝑛 𝐴𝑠 = 0.00822(300)(490) 𝐴𝑠 = 𝟏, 𝟐𝟎𝟗𝒎𝒎𝟐

c) 𝑀𝑢 = 485 𝑘𝑁 − 𝑚 < 𝑀𝑢𝑚𝑎𝑥 (singly reinforced) 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2 485 x 102 = 0.90𝑅𝑛 (300)(490)2 𝑅𝑛 = 7.48 𝑀𝑃𝑎

𝜌=

𝜌=

0.85𝑓′𝑐 2𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.85𝑓′𝑐

0.85(27.6) 2(7.48) [1 − √1 − ] 276 0.85(27.6) 𝜌 = 0.03384 > 𝜌𝑚𝑖𝑛

𝐴𝑠 = 𝜌 𝑏 𝑑

𝐴𝑠 = 0.03384(300)(490) 𝐴𝑠 = 𝟒, 𝟗𝟕𝟓𝒎𝒎𝟐

d) 𝑀𝑢 = 600 𝑘𝑁 − 𝑚 > 𝑀𝑢𝑚𝑎𝑥 The beam will be doubly reinforced. See Chapter 3.

PROBLEM 2.3 (CE MAY 2012) A reinforced concrete beam has a width of 300 mm and an overall depth of 480 mm. The beam is simply supported over span of 5 m. Steel strength 𝑓𝑦 = 415 MPa and concrete𝑓′𝑐 = 28 𝑀𝑃𝑎. Concrete cover is 70 mm from the centroid of the steel area. Unit weight concrete is 23.5kN/𝑚3 .Other than the weight of the beam, the beam carries a superimposed dead of 18 kN/m and a live load of 14 kN/m. Use the strength design method. a) Determine the maximum factored moment on the beam. b) If the design ultimate moment capacity of the beam is 280 kN-m, determine the required number of 20 mm tension bars. c) If the beam will carry a factored load of 240 kN at midsPan, determine the required number of 20 mm tension bars.

SOLUTION Given:

b=300m 𝑓′𝑐 = 300 𝑀𝑃𝑎 d=480-70=410 mm 𝛽1 = 0.85 1.4 𝑓𝑦 = 415 𝑀𝑃𝑎 𝜌𝑚𝑖𝑛 = 𝑓 = 0.00337 𝑦

Bar diameter , 𝑑𝑏 = 20 𝑚 𝑘𝑁 Weight of beam, 𝑤𝑏 = 𝛾𝑐 𝐴𝑏 = 23.5(0.3 𝑥 0.48 ) = 3.384 𝑚 a) Maximum factored moment on the beam. Factored load, 𝑊𝑢 = 1.4(3.384 + 18) + .7 (14) Factored load, 𝑊𝑢 = 53.738 𝑘𝑁/𝑚 Maximum factored moment: 𝑊 𝐿2

53.738(5)2

𝑀𝑢 = 𝑢8 𝑀𝑢 = 8 𝑀𝑢 = 𝟏𝟔𝟕. 𝟗𝟑 𝒌𝑵 − 𝒎 b) 𝑀𝑢 = 280 𝑘𝑁 − 𝑚

Solve for 𝑀𝑢𝑚𝑎𝑥 to determine whether compression steel is needed 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+ 𝑓𝑦 )

𝜌𝑚𝑎𝑥 = 0.75 𝜌𝑏 𝜔𝑚𝑎𝑥 =

𝜌𝑚𝑎𝑥 𝑓𝑦 𝑓′𝑐

𝜌𝑏 =

0.85(28)(0.85)(600) 415(600+415)

𝜌𝑏 = 0.02881 𝜌𝑚𝑎𝑥 = 0.021261 𝜔𝑚𝑎𝑥 = 0.03203

𝑅𝑛 𝑚𝑎𝑥 = 𝑓 ′ 𝑐 𝜔𝑚𝑎𝑥 (1 − 0.59 𝜔𝑚𝑎𝑥 ) = 7.274 𝑀𝑢 𝑚𝑎𝑥 = 𝜑 𝑅𝑛𝑚𝑎𝑥 𝑏𝑑2 = 330.14 𝑘𝑁 − 𝑚 Required 𝑀𝑢 = 280 𝑘𝑁 − 𝑚<𝑀𝑢 𝑚𝑎𝑥 (singly reinforced) 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑 2 280 𝑥 106 = .90 𝑅𝑛 (300)(410)2 𝑅𝑛 = 6.169 𝑀𝑃𝑎 𝜌= 𝜌=

0.85𝑓′𝑐 2𝑅𝑛 [1 − ] 𝑓𝑦 0.85𝑓′𝑐

0.85(28) 2(6.619) [1 − √1 − ] = 0.01755 > 𝜌𝑚𝑖𝑛 415 0.85(28)

𝐴𝑠 = 𝜌 𝑏 𝑑 𝐴𝑠 = 0.01755(300)(410) 𝐴𝑠 = 2159𝑚𝑚2 𝜋 𝜋 𝐴𝑠 = 𝑑𝑏2 2159 = (20)2 𝑁 4 4 N=6.9 say 7 bars

1. 𝑃𝑢 = 240𝑘𝑁 𝑎𝑡 𝑚𝑖𝑑𝑠𝑝𝑎𝑛 𝑊_𝑑 = 3.384 𝑘𝑁/𝑚 (weight of beam) 𝑀𝑢 =

𝑃𝑢 𝐿 (1.4𝑊𝑑 )𝐿2 + = 314.805 𝑘𝑁 − 𝑚 < 𝑀𝑢 𝑚𝑎𝑥 4 8

(𝑠𝑖𝑛𝑔𝑙𝑦)

314.805 𝑥 106

𝑀

𝑅𝑛 = 𝜑𝑏𝑑𝑢 2

𝑅𝑛 = 0.90(300)(410)2 𝑅𝑛 = 6.936𝑀𝑃𝑎

𝜌=

0.85𝑓′𝑐 2𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.85𝑓′𝑐

𝜌=

0.85(28) 2𝑅𝑛 [1 − √1 − ] = 002031 > 𝜌𝑚𝑖𝑛 415 0.85𝑓′𝑐

𝐴𝑠 = 𝜌 𝑏 𝑑

𝐴𝑠 = 0.02031(300)(410) 𝐴𝑠 = 2498𝑚𝑚2 𝜋 𝜋 𝐴𝑠 = 4 𝑑𝑏2 𝑁 2498 = 4 (20)2 N 𝑁 = 7.95 𝑠𝑎𝑦 𝟖 𝒃𝒂𝒓𝒔

PROBLEM 2.4 (CE MAY 1993) A reinforced concrete beam has a width of 300 mm and an effective depth to tension bars of 600 mm. compression reinforcement if needed will be placed at a depth of 60 mm below the top. If 𝑓′𝑐 = 30 𝑀𝑃𝑎 and 𝑓𝑦 = 414 𝑀𝑃𝑎, determine the tension steel area if the beam is to resist an ultimate moment of 650 kN-m. SOLUTION Solve for 𝜌𝑚𝑎𝑥 and 𝑀𝑢𝑚𝑎𝑥 : 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

0.85(30)(0.85)(600) 414(600 + 414) 𝜌𝑏 = 0.031 𝜌𝑚𝑎𝑥 = 0.75(0.031) 𝜌𝑚𝑎𝑥 = 0.0232 𝜌𝑏 =

𝜌𝑚𝑎𝑥 = 0.15𝜌𝑏

𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 < 10 𝑀𝑃𝑎

𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

0.02323(414) 30 𝜔 = 0.3209

𝑀𝑢 𝑚𝑎𝑥 = 𝜑𝑓′𝑐 𝜔 𝑏 𝑑2 (1 − 0.59𝜔) 𝑀𝑢 𝑚𝑎𝑥 = 0.90(30)(0.3209)(300)(600)2 [1-0.59(0.309) 𝑀𝑢 𝑚𝑎𝑥 = 758.1 𝑘𝑁 − 𝑚 > 𝑀𝑢 Since𝑀𝑢 < 𝑀𝑢 𝑚𝑎𝑥 , the beam may be designed as singly reinforced. 𝑅𝑛 = 6.687 𝑀𝑃𝑎

650 x 106 = 0.90𝑅𝑛 (300)(600)2 𝑅𝑛 = 6.687𝑀𝑃𝑎

Solve for 𝜌:

𝜌=

0.85𝑓′𝑐 2𝑅𝑛 (1 − √1 − ) 𝑓𝑦 0.85𝑓′𝑐

𝜌=

0.85(30) 2(6.687) [1 − √1 − ] = 0.0191 > 𝜌𝑚𝑖𝑛 414 0.85(30)

𝜌𝑚𝑖𝑛 =

1.4 = 0.00338 𝑓𝑦

𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.0191(300)(600) 𝐴𝑠 = 3442 𝑚𝑚2

PROBLEM 2.5 (CE November 2000) A rectangular concrete beam has a width of 300 mm and an effective depth of 550 mm. The beam is simply supported over a span 6 m and is used to carry a uniform dead load of 25 kN/m and a uniform live load of 40 kN/m. Assume 𝑓′𝑐 = 21 𝑀𝑃𝑎 and 𝑓𝑦 = 312 𝑀𝑃𝑎. Compression

reinforcement if necessary shall be placed at a depth 80 mm from the outermost compression concrete. a) Determine 80 mm from the outermost compression concrete. b) Determine the required tension steel area. c) Determine the required number of 25-mm tension bars.

SOLUTION a) Maximum steel area: 𝜌𝑏 =

0.85 𝑓 ′ 𝑐 𝛽1 600

𝑓𝑦 (600 + 𝑓𝑦 ) 𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓𝑐 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 30 𝑀𝑃𝑎 𝜌𝑏 =

0.85(21)(0.85)(600) 312(312 + 600)

𝜌𝑏 = 0.03199 𝜌 𝑚𝑎𝑥 = 0.75𝜌𝑏

𝜌𝑚𝑎𝑥 = 0.75(0.03199) 𝜌𝑚𝑎𝑥 = 0.02399

𝐴𝑠 𝑚𝑎𝑥 = 𝜌 𝑚𝑎𝑥 𝑏𝑑

𝐴𝑠 𝑚𝑎𝑥 = 0.02399(300)(550) 𝐴𝑠 𝑚𝑎𝑥 = 𝟑, 𝟗𝟓𝟗 𝒎𝒎𝟐

b) Required tension steel area: Factored load: 𝑊𝑢 = 1.4 𝐷 + 1.7 𝐿

𝑊𝑢 = 1.4(25) + 1.7(40) 𝑊𝑢 = 103 𝑘𝑁/𝑚

Required strength: 𝑀𝑢 =

𝑊𝑢 𝐿2 8

Solve for 𝑀𝑢 𝑚𝑎𝑥

103(6)2 8 𝑀𝑢 =463.5kN-m

𝑀𝑢 =

𝜔=

𝜌 𝑚𝑎𝑥 𝑓𝑦 𝑓′𝑐

0.0299(312) 21 𝜔 = 0.356

𝜔=

𝑀𝑢 𝑚𝑎𝑥 = 𝜑𝑓 ′ 𝑐 𝜔𝑏𝑑2 (1 − 0.59𝜔) 𝑀𝑢 𝑚𝑎𝑥 = 0.90(30)(0.356)(300)(550)2 [1 − 0.59(0.356)] 𝑀𝑢 𝑚𝑎𝑥 = 536.5 𝑘𝑁 − 𝑚 > 𝑀𝑢 𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2

𝑀𝑢 = 0.39 𝑅𝑛 (300)(550)2 463.5 𝑥 106 = 0.9 𝑅𝑛 (300)(550)2 𝑅𝑛 = 5.67 𝑀𝑃𝑎

𝜌=

0.85𝑓′𝑐 2𝑅𝑛 (1 − √1 − ) 𝑓𝑦 0.85 𝑓,𝑐

𝜌=

0.85(21) 2(5.67) [1 − √1 − ] 312 0.85(21)

𝜌 = 0.02269 𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.002269(300)(550) 𝐴𝑠 = 3743 𝑚𝑚2

c) Number of 25 mm bars: Number of 25-mm bars=

𝐴𝑠 𝐴𝑠 25 3.743

Number of 25-mm bars=𝜋 4

(25)2

= 7.63 𝑠𝑎𝑦 8

PROBLEM 2.6 (CE MAY 2009) A reinforced concrete beam has a width of 300 mm and total depth of 600 mm. The beam will be design to carry a factored moment of 540kN-m.

Concrete strength 𝑓′𝑐 = 28 𝑀𝑃𝑎 and steel yield strength 𝑓𝑦 = 248 𝑀𝑃𝑎. Solve using the strength design method. a) Determine the balanced steel ratio in percent. b) Determine the minimum effective depth of the beam using a steel ratio 𝜌 equal to 0.5 of balanced steel ratio. c) Determine the minimum effective depth of the beam using the maximum allowable steel ratio. SOLUTION Given: 𝑓′𝑐 = 28 𝑀𝑃𝑎 𝑓𝑦 = 248 𝑀𝑃𝑎

b=300 mm h=600 mm 𝑀𝑢 = 540 𝑘𝑁 − 𝑚 𝛽1 = 0.85 a) Balanced steel ratio: 𝜌𝑏 =

0.85𝑓 ′ 𝑐 𝛽1 600

𝜌𝑏 =

𝑓𝑦 (600 + 𝑓𝑦 )

0.85(28)(0.85)600 248(600 + 248)

𝜌𝑏 = 0.0577 = 𝟓. 𝟕𝟕% b) Effective depth using 𝜌 = 0.5𝜌𝑏 𝜌 = 0.5(0.0577) = 0.0289 𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

0.0289(248) = 0.2556 28

𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 − 0.59𝜔) 𝑅𝑛 = 28(0.2556)[1 − 0.59(0.2556)] 𝑅𝑛 = 6.0776 𝑀𝑃𝑎 𝑀𝑢 = 𝜑𝑀𝑛 = 𝜑𝑅𝑛 𝑏𝑑2

540 x 106 = 0.90(8.307)(300)𝑑 2

𝑑 = 𝟒𝟗𝟏 𝒎𝒎

PROBLEM 2.7 A concrete one-way slab has a total thickness of 120 mm. The slab will be reinforced with 12-mm-diameter bars with 𝑓𝑦 = 275 𝑀𝑃𝑎.Concrete strength𝑓′𝑐 = 21 𝑀𝑃𝑎. Determine the required spacing 12 mm main bar if the total factored moment acting on 1-m width of slab is 23 kN-m width of slab is 23 kN-m. Clear concrete cover is 20 mm. SOLUTION Note: Slabs are practically singly reinforced because of its small depths.

Effective depth, d= 120 -20-1/2(12) = 94 mm Width, b = 1000 mm 𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑 2 23 x 106 = 0.90 𝑅𝑛 (1000)(94)2 𝑅𝑛 = 2.892

𝜌=

𝜌=

0.85𝑓 ′ 𝑐 𝑓𝑦

(1 − √1 −

2𝑅𝑛 ) 0.85𝑓 ′ 𝑐

0.85(21) 2(2.982) (1 − √1 − ) 275 0.85(21)

𝜌𝑚𝑎𝑥 =

𝜌𝑚𝑖𝑛 =

0.75 𝑥 0.85𝑓′𝑐 𝛽1 600 = 0.0284 𝑓𝑦 (600 + 𝑓𝑦 )

1.4 = 0.00509 𝑓𝑦

𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.1154(1000)(94) 𝐴𝑠 = 1085 𝒎𝒎𝟐

Spacing of bars (for walls and slabs using unit width): 𝑏

𝑠=𝑁

𝑠=

𝑠=

1000 𝐴𝑠 𝐴𝑏

1000𝐴𝑏 𝐴𝑠

Eq. 2-17 1000𝐴𝑏 𝑠= 𝐴𝑠

𝜋 1000 𝑥 4 (12)2 𝑠= 1085 𝑠 = 𝟏𝟎𝟎 𝒎𝒎

PROBLEM 2.8 A 2.8 m square column fooring has a total thickness of 47 mm. The factored moment at critical section for moment is 640 kN-m. Assume 𝑓′𝑐 = 21 𝑀𝑃𝑎 and 𝑓𝑦 = 275 𝑀𝑃𝑎. Clear concrete cover is 75 mm. Determine the required number of 20 mm tension bars. SOLUTION Effective depth, d=470-75-1/2(20) = 385 mm Width, b =2800 mm Design strength, 𝑀𝑢 = 640 𝑘𝑁 − 𝑚

Maximum and minimum requirements: 𝜌𝑚𝑎𝑥 = 0.75 𝑥

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+𝑓𝑦 )

= 0.0284

𝑀𝑢 𝑚𝑎𝑥 = 2528𝑘𝑁 − 𝑚 (Procedure is not shown anymore see Problem 2.2) 𝐴𝑠 𝑚𝑖𝑛 =

1.4 𝑏𝑤 𝑑 𝑓𝑦

= 5488 𝑚𝑚2

Singly reinforced: 𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2

𝜌=

0.85𝑓 ′ 𝑐 2𝑅𝑛 (1 − √1 − ) 𝑓𝑦 0.85𝑓 ′ 𝑐

𝜌=

0.85(21) 2(1.713 (1 − √1 − ) 275 0.85(21)

640 𝑥 106 = 0.90𝑅𝑛 (2800)(385)2 𝑅𝑛 = 1.713 𝑀𝑃𝑎

𝜌 = 0.00656 𝐴𝑠 = 𝜌 𝑏 𝑑

𝐴𝑠 = 0.00656(2800)(385) 𝐴𝑠 = 7074 𝑚𝑚2 > 𝐴𝑠 𝑚𝑖𝑛

Number of 20 mm bars: 𝑁=

𝐴𝑠 𝐴𝑏

7074 𝑁=𝜋 2 4 (20) 𝑁 = 22.5 𝑠𝑎𝑦 𝟐𝟑 𝒃𝒂𝒓𝒔

PROBLEM 2.9 Design a rectangular beam reinforced for tension only to carry a dead load moment of 60 kN-m (including its own weight) and a live load moment of 48 kN- m. Use 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 and 𝑓𝑦 = 276 𝑀𝑃𝑎. SOLUTION Required strength: 𝑀𝑢 = 1.4 𝑀𝑏 + 1.7 𝑀𝐿

𝑀𝑢 = 1.4(60) + 1.7(48) 𝑀𝑢 = 165.6 𝑘𝑁 − 𝑚

(Note: this already includes the weight of beam)

𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑏 =

0.85(20.7)(0.85)(600) 276(600 + 276)

𝜌𝑏 = 0.0371 𝜌𝑚𝑖𝑛 =

1.4 = 0.00507 𝑓𝑦 Note: this is the author’s suggestion

Try 𝜌 = 60% 𝜌𝑏 𝜌 = 0.6(0.0371) = 0.02226

𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

0.02226(276) 20.7

𝜔 = 0.2968 𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 − 0.59𝜔)

𝑅𝑛 = 20.7(0.2968)[1 − 0.59(0.2968)] 𝑅𝑛 = 5.068

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2

165.6 x 106 = 0.90(5.068)𝑏𝑑2 𝑏𝑑 2 = 36.296 𝑥 106 𝑚𝑚3

Try d = 1.75 b

b=228 mm say 230 mm d=399 say 30 mm

𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.02226(230)2(400) 𝐴𝑠 = 2.049 𝑚𝑚2

Summary: b = 230 mm d = 400 mm 𝑨𝒔 = 𝟐, 𝟎𝟒𝟗 𝒎𝒎𝟐 PROBLEM 2.10 Design a singly reinforced rectangular beam for a 6-m simple span to support a superimposed dead load of 29 kN/m and a live load of 44 kN/m. 24𝑘𝑁 Assume normal weight concrete with = 𝑚3 . Use 𝜌𝑚𝑎𝑥, 𝑓′𝑐 = 34 𝑀𝑃𝑎, 𝑎𝑛𝑑 𝑓𝑦 = 345 𝑀𝑃𝑎.

SOLUTION Weight of beam: (this is the author’s assumption) Assuming a 300 mm x 600 mm, 𝑊_𝑏 = 24 𝑥 (0.3 0.6) = 4.32𝑘𝑁/𝑚 𝑊𝑢 = 1.4 (29 + 4.32) + 1.7(44) 𝑊𝑢 = 121.448 𝑘𝑁/𝑚

𝑊𝑏 = 1.4 𝑊𝐿 + 1.7 𝑊𝐿 . 𝑀𝑢 =

𝑊𝑢 𝐿2

𝑀𝑢 =

8

121.448(6)2 8

𝑀𝑢 = 546.516 𝑘𝑁 − 𝑚 𝛽1 = 0.85 − 𝜌𝑏 =

0.05 = 0.821 7(34 − 30)

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+ 𝑓𝑦 )

𝜌𝑏 =

0.85(34)(0.821)(600) 345(600+345)

𝜌𝑏 = 0.04369

𝜌 = 𝜌𝑚𝑎𝑥 = 0.75 (0.04369) 𝜌𝑚𝑖𝑛 = 𝜔=

𝜌 = 0.03277 > 𝜌𝑚𝑖𝑛

√𝑓′𝑐 = 0.00423 4𝑓𝑦

𝜌𝑓𝑦

𝜔=

𝑓′𝑐

𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 > 31.36 𝑀𝑃𝑎

0.03277(345) 34

𝜔 = 0.332

𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 − 0.56 𝜔)

Assume d = 1.75 b

𝑅𝑛 = 34(0.332)[1 − 0.59(0.332)] 𝑅𝑛 = 9.087 𝑀𝑃𝑎

(this is the author’s assumption)

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏 𝑑 2

546.516 x 106 = 0.90(9.087)(𝑏)(1. 75𝑏)2 𝑏 = 279.4 𝑚𝑚 & 𝑑 = 489 𝑚𝑚

Use b = 280 mm, d = 490 mm Minimum beam the thickness (Section 409.6.2.1) ℎ𝑚𝑖𝑛 =

𝑓𝑦 𝐿 (0.4 + ) 16 700

ℎ𝑚𝑖𝑛 =

6000 345 (0.4 + ) 16 700

ℎ𝑚𝑖𝑛 = 335 𝑚𝑚 𝑂𝐾 𝐴𝑠 = 𝜌 𝑏 𝑑

𝐴𝑠 = 0.03277(280)(490) 𝐴𝑠 = 4496 𝑚𝑚2

Using 32 mm bars (#100): 𝑁=

𝐴𝑠 𝐴𝑏

4496 𝑁=𝜋 2 4 (32) 𝑁 = 5.6 𝑠𝑎𝑦 6 𝑏𝑎𝑟𝑠

ℎ = 490 + (25) + 32 + 20 ℎ = 554.5 𝑚𝑚 > ℎ𝑚𝑖𝑛 Beam weight = 24 (0.28)(0.5545) Beam weight = 3.73 kN/m < 4.32(OK)

PROBLEM 2.11 A propped cantilever beam shown in Figure 2.6 is made of reinforced concrete having a width of 290 mm overall depth of 490 mm. The beam is loaded with uniform dead load of 35 kN/m (including its own weight), and a uniform live load of 55 kN/m. Given 𝑓′𝑐 = 24 𝑀𝑃𝑎, 𝑓𝑦 = 415 𝑀𝑃𝑎.Concrete cover is 60 mm from the centroid of the bars. Determine the required tension steel area for maximum positive moment. Assume EI=constant.

Figure 2.6 SOLUTION Given: 𝑓′𝑐 = 24 𝑀𝑃𝑎 𝑓𝑦 = 415 𝑀𝑃𝑎 𝑓𝑦ℎ = 275 𝑀𝑃𝑎 𝑏 = 290 𝑚𝑚 𝐻 = 490 𝑚𝑚 𝑑′ = 60 𝑚𝑚

𝑊𝐷 = 35 𝑘𝑁/𝑚 𝑊𝐿 = 55 𝑘𝑁/𝑚 𝑑 = 490 − 60 = 430 𝑚𝑚 𝑊𝑢 = 1.4𝑊𝐷 + 1.7 𝑊𝐿

𝑊𝑢 = 1.4 (35) + 1.7 (55) 𝑊𝑢 = 142.5 𝑘𝑁/𝑚

Solve for moment reactions using the three-moment equation: 𝑀𝐵 = −142.5 (2)(1) = −285 𝑘𝑁 − 𝑚 Mo Lo + 2𝑀𝐴 (𝐿𝑜 + 𝐿1 ) + 𝑀𝐵 𝐿1 +

6𝐴0 ̅̅̅̅ 𝑎0 𝐿0

0 + 2𝑀𝐴 (0 + 6 ) + (−285 )(6) + 0 +

+

6𝐴1 ̅̅̅ 𝑏0 𝐿1

142.5(6)3 4

=0

=0

𝑀𝐴 = −498.75𝑘𝑁 − 𝑚 𝑀𝐴 = 𝑀𝐴 𝑟𝑖𝑔ℎ𝑡

-489.75 = R(6)- 142.5(8)(4) R=676.875 kN

𝑅𝐴 = 𝑊𝑢 𝐿 − 𝑅

𝑅𝐴 = 142.5(8) − 676.875 𝑅𝐴 = 463.125 𝑘𝑁

Maximum positive moment: 𝑉𝐷 = 0

𝑀𝐷 = 𝑅𝑋 − 𝑊𝑢

(𝑥+2)2 2

𝑊𝑢 (2 + 𝑥) − 𝑅 = 0 142.5(2 + x) - 676.875 = 0 x = 2.75 m

𝑀𝐷 = 676.875(2.75) − 142.5

(2.75+2)2

𝑀𝐷 = 253.828 𝑘𝑁 − 𝑚 Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 : 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑏 =

0.85(24)(0.85)600 415(600 + 415)

𝜌𝑏 = 0.0247 𝜌𝑚𝑎𝑥 = 0.75 𝜌𝑏

𝜔𝑚𝑎𝑥 =

𝜌𝑚𝑎𝑥 𝑓𝑦 𝑓′𝑐

𝜌𝑚𝑎𝑥 = 0.75 (0.0247) 𝜌𝑚𝑎𝑥 = 0.01852 𝜔𝑚𝑎𝑥 =

0.01852(415) 24

𝜔𝑚𝑎𝑥 = 0.3203

2

𝑅𝑛 𝑚𝑎𝑥 = 𝑓 ′ 𝑐 𝜔(1 − 0.59 𝜔) 𝑀𝑛 𝑚𝑎𝑥 = 𝑅𝑛 𝑏 𝑑 2

𝑅𝑛 𝑚𝑎𝑥 = 415(0.3203)[1 − 0.59(0.3203)] 𝑀 𝑛 𝑚𝑎𝑥 = 6.235(290)(430)2 𝑀𝑛 𝑚𝑎𝑥 = 334.316 𝑘𝑁 − 𝑚

𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90(334.316) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 300.884 𝑘𝑁 − 𝑚 At a point of maximum positive moment: 𝑀𝑢 = 253.828 𝑘𝑁 − 𝑚 < 𝜑 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2

𝜌=

0.85𝑓′𝑐 𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.85𝑓′𝑐

𝜌=

0.85(24) 2(5.26) [1 − √1 − ] 415 0.85(24)

(Singly reinforced) 253.828 x 106 = .90 𝑅𝑛 (290)(430)2 𝑅𝑛 = 5.26 𝑀𝑃𝑎

𝜌 = 0.01495 𝐴𝑠 = 𝜌 𝑏 𝑑

𝐴𝑠 = 0.01495(290)(430) 𝐴𝑠 = 1,864 𝑚𝑚2

ANALYSIS OF RECTANGULAR BEAMS WHERE STEEL YIELDS (𝒇𝑺 = 𝒇𝒀 ) PROBLEM 2.12(CE MAY 1999) A reinforced concrete rectangular beam with b = 400 mm and d= 720 mm is reinforced for tension only with 6-25 mm diameter bars. If 𝑓 ′ 𝑐 = 21 𝑀𝑃𝑎 and 𝑓𝑦 = 400 𝑀𝑃𝑎, 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔: a) The coefficient of resistance 𝑅𝑛 of the beam. b) The ultimate moment capacity of the beam. SOLUTION 𝜌𝑏 =

0.85𝑓 ′ 𝑐 𝛽1

𝑓𝑦 (600 + 𝑓𝑦 ) 0.85(21)(0.85)(600) = 𝜌𝑏 400(600 + 400) 𝜌𝑏 = 0.02276 𝐴𝑠 = 6 𝑥 𝜌=

𝐴𝑠 𝑏𝑑

𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜋 (25)2 = 2945 𝑚𝑚2 4 𝜌=

𝜔=

2945 = 0.01023 < 𝜌𝑏 (𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠) 400(720)

0.01023(400) = 0.195 21

𝑅𝑛 = 𝑓′𝑐 𝜔 (1 − 0.56𝜔)

𝑅𝑛 = 21(0.195)[1 − 0.59(0.195)] 𝑅𝑛 = 𝟑. 𝟔𝟐 𝑴𝑷𝒂

𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2 𝑀𝑢 = 0.90(3.62)(400)(720)2 𝑀𝑢 = 𝟔𝟕𝟓. 𝟔𝟕 𝒌𝑵 − 𝒎 Answer

Answer

PROBLEM 2.13 A rectangular beam reinforced for tension only has b= 300 m, d = 490 mm. The tension steel area provided is 4,500 sq. mm. Determine the ultimate moment capcity of the beam in kN-m. Assume 𝑓′𝑐 = 27 𝑀𝑃𝑎, 𝑓𝑦 = 275 𝑀𝑃𝑎. SOLUTION 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+ 𝑓𝑦 )

𝜌𝑏 =

0.85(27)(0.85)(600) 275(600+275)

𝜌𝑏 = 0.02276 𝐴

4,500

𝜌 = 𝑏𝑑𝑠 𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜌 = 300(490) 𝜔=

0.0361(275) 27

𝜔 = 0.3118 𝑅𝑛 = 𝑓′𝑐 𝜔 (1 − 0.59 𝜔)

𝑅𝑛 = 27(0.3118)[1 − 0.59(0.3118)] 𝑅𝑛 = 6.87 𝑀𝑃𝑎

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2

𝑀𝑢 = 0.90(6.87)(300)(490)2 𝑀𝑢 = 𝟒𝟒𝟓. 𝟑 𝒌𝑵 − 𝒎

PROBLEM 2.14 A rectangular beam has b = 300 mm, d = 500 mm, 𝐴𝑠 = 3 − 25 𝑚𝑚, 𝑓′𝑐 = 34.2 𝑀𝑃𝑎, grade 60 reinforcement (𝑓𝑦 = 414 𝑀𝑃𝑎). Calculate the design moment 𝑀𝑢 .

SOLUTION 𝛽1 = 0.85 −

0.85𝑓′𝑐 𝛽1 600

𝜌𝑏 =

𝜌𝑏 =

𝑓𝑦 (600+ 𝑓𝑦 )

𝐴𝑠 = 𝜌=

0.05 (34.2 − 30) = 0.82 7 0.85(34.2)(0.82)(600) 414(600+414)

𝜌𝑏 = 0.03407

𝜋 (25)2 𝑥 3 = 1473 𝑚𝑚2 4

𝐴𝑠

𝜌=

𝑏𝑑

1473 300(500)

𝜌 = 0.00982 < 𝜌𝑏 𝑆𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠

Check if the beam satisfies the minimum requirement: √𝑓′𝑐 𝜌𝑚𝑖𝑛 = = 0.00353 𝑂𝐾 4𝑓𝑦 𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

0.00982(414) 34.2

𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 − 0.59𝜔)

𝑅𝑛 = 34.2(0.1188)[1 − 0.59(0.1188)] 𝑅𝑛 = 3.779 𝑀𝑃𝑎

𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2

𝑀𝑢 = 0.90(3.779)(300)(500)2 𝑀𝑢 = 𝟐𝟓𝟓. 𝟏𝟏 𝒌𝑵 − 𝒎

PROBLEM 2.15 A 130-mm-thick-one-way slab is reinforced with 12-mm-diameter tension bars spaced at 110 on centers. Concrete cover is 20 mm, concrete strength 𝑓′𝑐 = 21 MPa and steel yield strength 𝑓𝑦 = 275 𝑀𝑃𝑎. Unit weight of concrete is 23.5 kN/𝑚3 . a) What is the ultimate moment capacity of the slab? b) If the slab is simply supported over a span of 4 m, what safe uniform live load pressure can the slab carry?

SOLUTION a) Consider 1 m width of slab, b = 1000 mm Effective depth: d = h – cover- 1/2 𝑑𝑏 d = 130-20-1/2(12)=104 mm

𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝐴𝑠 = 𝐴𝑏 𝑥 𝑁 𝜋

𝐴𝑠 = 4 (12)2

𝜌𝑏 =

0.85(21)(0.85)(600) 275(600 + 275) 𝜌𝑏 = 0.0378

𝑏 𝑠 𝐴𝑠 = 1028 𝑚𝑚2 𝐴𝑠 = 𝐴𝑏 𝑥

1000 100

𝐴𝑠 1028 𝜌= 𝑏𝑑 1000(104) 𝜌 = 0.00989

𝜌=

Check if the beam satisfies the minimum steel requirement on flexures: 1.4 = 0.00509 𝑓𝑦

𝜌𝑚𝑖𝑛 =

𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

𝑂𝐾

0.00989(275) 21 𝜔 = 0.129

𝑅𝑛 = 𝑓′𝑐 𝜔 (1 − 0.59 𝜔)

𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2

b) 𝑀𝑢 =

𝑅𝑛 = 21(0.129)(1 − 0.59(0.129)] 𝑅𝑛 = 2.511 𝑀𝑃𝑎

𝑀𝑢 = 0.90(2.511)(1000)(104)2 𝑀𝑢 = 𝟐𝟒. 𝟒𝟒𝟑 𝒌𝑵 − 𝒎 𝑊𝑢 𝐿2 8

24.443 =

𝑊𝑢 (4)2 8

𝑊𝑢 = 12.222 𝑘𝑁/𝑚

Dead load pressure, 𝜌𝐷 = 𝛾𝑐 x thickness of concrete. Dead load pressure, 𝜌𝐷 = 23.5 𝑥 0.13 = 3.055𝑘𝑃𝑎 𝑊𝑢 = 1.4𝑊𝐿 + 1.7 𝑊𝐿

𝑊𝑢 = 1.4(𝜌𝐷 𝑏) + 1.7 (𝜌𝐿 𝑏) 12.222 = 1.4(3.055 𝑥 1) + 1.7(𝜌𝐿 𝑥 1) 𝜌𝐿 = 𝟒. 𝟔𝟕𝟑 𝒌𝑷𝒂

PROBLEM 2.16 A rectangular beam with b = 250 mm and d = 460 m is reinforced for tension only with 3-25 mm bars. The beam is simply supported over a span of 6 m and carries a uniform dead load of 680 N/m including its own weight. Calculate the uniform live load that the beam can carry. Assume 𝑓𝑦 = 276.5 𝑀𝑃𝑎 and 𝑓′𝑐 = 20.7 𝑀𝑃𝑎. SOLUTION 𝜋 𝐴𝑠 = 3 𝑥 (25)2 = 1479 𝑚𝑚2 4 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

0.85(20.7)(0.85)(600) 276.5(600 + 276.5) 𝜌𝑏 = 0.03703 𝜌𝑏 =

𝐴𝑠 1.473 𝜌= 𝑏𝑑 250(460) 𝜌 = 0.01281 < 𝜌𝑏 (𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠) 𝜌=

Check if the beam satisfies the minimum steel requirement on flexure: 𝜌𝑚𝑖𝑛 =

𝜔=

1.4 = 0.00506 𝑓𝑦

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

𝑂𝐾

0.01281(276.5) 20.7 𝜔 = 0.171

𝑅𝑛 = 𝑓′𝑐 𝜔(1 − 0.59𝜔)

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2

𝑀𝑢𝑚𝑎𝑥

𝑊𝑢 𝐿2 = 8

𝑅𝑛 = 27(0.171)[1 − 0.59(0.171)] 𝑅𝑛 = 3.183 𝑀𝑃𝑎

𝑀𝑢 = 0.90(3.183)(250)(460)2 𝑀𝑢 = 151.56 𝑘𝑁 − 𝑚 𝑊𝑢 (6)2 151.56 = 8 𝑊𝑢 = 33.68 𝑘𝑁/𝑚 33.68 = 1.4 (0.68) + 1.7 𝑊𝐿𝐿 𝑊𝐿𝐿 = 𝟏𝟗. 𝟐𝟓 𝒌𝑵 − 𝒎

𝑊𝑢 = 1.4 𝑊𝐷𝐿 + 1.7 𝑊𝐿𝐿

PROBLEM 2.17 (CE JANUARY 2008) A reinforced concrete rectangular beam has a width of 300 mm and an effective depth of 55 mm. The beam is reinforced with six 25-mmdiameter tension bars. Steel yield 𝑓𝑦 is 415 MPa and concrete strength 𝑓′𝑐 is 28 MPa. a) What is the balanced steel ratio? b) What is the maximum steel area for singly reinforced? c) What is the nominal moment capacity of the beam? SOLUTION a) Balanced steel ratio: 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝛽1 = 0.85

𝜌𝑏 =

0.85(28)(0.85)600 415(600 + 415) 𝜌𝑏 = 0.028816 𝜌𝑏 = 𝟐. 𝟖𝟖%

b) Maximum steel area 𝐴𝑠 𝑚𝑎𝑥 = 𝜌𝑚𝑎𝑥 𝑏𝑑

𝐴𝑠 𝑚𝑎𝑥 = (0.75 𝜌𝑏 ) 𝑏𝑑 𝐴𝑠 𝑚𝑎𝑥 = (0.75 𝑥 0.028816)(300)(5) 𝐴𝑠 𝑚𝑎𝑥 = 𝟑, 𝟐𝟒𝟐 𝒎𝒎𝟐

c) Nominal moment capacity Using 6-25 mm bars: 𝜋 𝐴𝑠 = (25)2 𝑥 6 = 2,945 𝑚𝑚2 4 𝜌=

𝐴𝑠 𝑏𝑑

𝜌=

2,945 = 𝜌 = 0.01963 < 𝜌𝑏 (𝑡𝑒𝑛𝑠𝑖𝑜 𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠) 300(500)

𝜌𝑓𝑦 0.01963(415) 𝜔= = 0.291 𝑓′𝑐 28 𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 + 0.59 𝜔) = 𝑅𝑛 = 28(0.291)(1 − 0.59 𝑥 0.291) 𝑅𝑛 = 6.7494 𝑀𝑃𝑎 𝜔=

𝑀𝑛 = 𝑅𝑛 𝑏𝑑 2 = 𝑀𝑛 = 6.7494(300)(500)2 𝑀𝑛 = 𝟓𝟎𝟔. 𝟐 𝒌𝑵 − 𝒎 PROBLEM 2.18 A 350 mm x 500 mm rectangular is reinforced for tension only with 5-28 mm bars. The beam has an effective depth of 446 mm. The beam carries a uniform dead load of 4.5 kN/m (including its own weight), a uniform live load of 3 kN/m, and concentrated dead load of P and 2P as shown in Figure 2.7. Assume 𝑓𝑦 = 414 𝑀𝑃𝑎, 𝑓′𝑐 = 34.5 𝑀𝑃𝑎. Calculate the following: a) The ultimate moment capacity of the section in kN-m, and b) The maximum value of P in kN. 2P

2m

P

2m

2m

SOLUTION 𝛽1 = 0.85 −

0.05 (34.5 − 30) = 0.818 7

𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝐴𝑠 =

𝜋 (28)2 𝑥 5 = 3079 𝑚𝑚2 4

𝐴

𝜌𝑏 =

0.85(34.5)(0.818)(600) = 0.03428 414(600 + 414)

3079

𝜌 = 𝑏𝑑𝑠

𝜌 = 300(446) = 𝜌 = 0.01972 < 𝜌𝑏 𝑆𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠

Check if the beam satisfies the minimum requirement: 𝜌𝑚𝑖𝑛 =

𝜔=

√𝑓′𝑐 = 0.00355 4𝑓𝑦

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

0.01972(414) = 0.2367 34.5

𝑅𝑛 = 𝑓′𝑐 𝜔(1 − 0.59𝜔) = 𝑅𝑛 = 34.5(0.2367)[1 − 0.59(0.2367)]=7.025 MPa 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2 = 𝑀𝑢 = 0.90(7.025)(300)(446)2 = 440.18 kN - m

1.4(2P)

1.4P

𝑊𝑢 = 1.4(4.5) + 1.7(3) = 11.4𝑘𝑁/𝑚

A

𝑅𝑎

B 2m

C 2m

D 2m

Figure 2.8 – Beam with factored loads For the given loads, the maximum moment can occur at B or C: 𝑀𝑐 = 1.4𝑃(2) + 11.4(2)(1) 440.18 = 1.4P(2) + 11.4(2)(1) 𝑃 = 149 𝑘𝑁

At point C: Set 𝑀𝑐 = 𝑀𝑢 At point B: (First solve for 𝑅𝐴 ) ∑ 𝑀𝑐 = 0

4 𝑅𝐴 + 1.4𝑃(2) = 2.8𝑃(2) + 11.4(6)(1) 𝑅𝐴 = 17.1 + 0.7 𝑃

∑ 𝑀𝐵 𝑙𝑒𝑓𝑡 Set 𝑀𝐵 = 𝑀𝑢

𝑀𝐵 = (17.1 + 0.7𝑃) − 11.4(2)(1) 440.18 = (17.1 + 0.7 𝑃)(2) − 11.4(2)(1) 𝑃 = 306.27 𝑘𝑁

Thus the maximum value of P such that 𝑀𝑢 will not exceed 440.18 kN-m is 149 kN.

ANALYSIS OF RECTANGULAR BEAMS WHERE STEEL DOES NOT YIELDS (𝒇𝑺 ≠ 𝒇𝒀 ) PROBLEM 2.19 A rectangular beam has b = 300 mm, d = 500 mm, 𝐴𝑠 = 6 − 32 𝑚𝑚, 𝑓′𝑐 = 27.6 𝑀𝑃𝑎, grade 60 reinforcement (𝑓𝑦 = 414 𝑀𝑃𝑎). Calculate the ultimate moment capacity of the beam. SOLUTION 𝜌𝑏 =

𝐴𝑠 = 𝜌=

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑏 =

0.85(27.6)(0.85)(600) 414(600 + 414) 𝜌𝑏 = 0.0285

𝜋 (32)2 𝑥 6 = 4825 𝑚𝑚2 4

𝐴𝑠

𝜌=

𝑏𝑑

4825 300(500)

= 𝜌 = 0.03217 > 𝜌𝑏 𝑆𝑡𝑒𝑒𝑙 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑦𝑖𝑒𝑙𝑑

0.85

b=.300

c=0.85

ab

d=500

a

500-a/2 From Eq. 2-18

=4825 T=

𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝑓𝑠 = 600

500 − 𝑐 𝑐

∑ 𝐹𝐻 = 0 𝑇=𝐶

𝐴𝑠 𝑓𝑠 = 0.85 𝑓′𝑐 𝑎 𝑏, (4825)600

500−𝑐 𝑐

𝑎 = 𝛽1 𝑐 = 0.85 𝑐

= 0.85(27.6)(0.85𝑐)(300) 𝑐 2 = 484𝑐 − 241,964 = 0 𝑐 = 306.2 𝑚𝑚

𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝑓𝑠 = 600

500−306 306

= 𝑓𝑠 = 379.65 𝑀𝑃𝑎

𝑎 = 𝛽1 𝑐 = 0.85(306.2) 𝑎 = 260.3 𝑚𝑚 𝑎 𝜑𝑀𝑛 = 𝜑𝐴𝑠 𝑓𝑠 (𝑑 − ) 2 𝜑𝑀𝑛 = 0.90(4825)(379.65)(500 −

260.3 ) 2

𝜑𝑀𝑛 = 𝟔𝟎𝟗. 𝟖 𝒌𝑵 − 𝒎 PROBLEM 2.20 A rectangular beam reinforced for tension only has b=300 mm, d = 490 mm. The tension steel area provided is 7-25 mm diameter bars with 𝑓𝑦 = 415 𝑀𝑃𝑎. 𝑓′𝑐 = 21𝑀𝑃𝑎. Calculate the ultimate moment capacity of the beam.

SOLUTION 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝐴𝑠 =

𝜋 (25)2 𝑥 7 = 3436 𝑚𝑚2 4

𝐴

𝜌𝑏 =

0.85(21)(0.85)(600) 415(600 + 415) 𝜌𝑏 = 0.02161

3436

𝜌 = 𝑏𝑑𝑠

𝜌 = 300(490)

𝜌 = 0.02337 > 𝜌𝑏 𝑆𝑡𝑒𝑒𝑙 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑦𝑖𝑒𝑙𝑑

b=300

0.85 c=0.85

ab

d=490

a

490-a/2 =3436 From Eq.2-18: T=

𝑓𝑠 = 600

𝑑−𝑐 𝑐

= 𝑓𝑠 = 600

490 − 𝑐 𝑐

∑ 𝐹𝐻 = 0 𝑇=𝐶 (3436)600

𝐴𝑠 𝑓𝑠 490−𝑐 𝑐

= 0.85 𝑓′𝑐 𝑎 𝑏, 𝑎 = 𝛽1 𝑐 = 0.85 𝑐 = 0.85(221)(0.85𝑐)(300) = 𝑐 = 296.24 𝑚𝑚

𝑓𝑠 = 600

𝑑−𝑐 𝑐

490 − 296.24 296.24 𝑓𝑠 = 392.43 𝑀𝑃𝑎 < 𝑓𝑦

𝑓𝑠 = 600

𝑎 = 𝛽1 𝑐 = 0.85(392.43) 𝑎 = 251.81 𝑚𝑚 𝑎

𝜑𝑀𝑛 = 𝜑𝑇 (𝑑 − 2)

𝑎

𝜑𝑀𝑛 = 𝜑𝐴𝑠 𝑓𝑠 (𝑑 − 2)

251.81

𝜑𝑀𝑛 = 0.90(3436)(392.43)(490 − 2 ) 𝜑𝑀𝑛 = 𝟒𝟒𝟏. 𝟖𝟔 𝒌𝑵 − 𝒎

ANALYSIS & DESIGN OF SINGLY REINFORCED NON-RECTANGULAR BEAMS PROBLEM 2.21 Compute the ultimate moment capacity of the beam shown in Figure 2.9. Assume 𝑓𝑦 = 345 𝑀𝑃𝑎 and 𝑓′𝑐 = 21 𝑀𝑃𝑎.

125 125

125

700mm

125

4-32mm 75 375mm Figure 2.9

SOLUTION Note: This is not a rectangular beam. Some formulas derived above (such as𝜌, 𝜌𝑏 , 𝑅𝑛 ) may not be applicable. The moment can be computed using the assumptions in the Code and the conditions of equilibrium.

𝜋 (32)2 𝑥 4 4 𝐴𝑠 = 3217 𝑚𝑚2 𝐴𝑠 =

Solve for the balanced 𝐴𝑠 to determine whether the given steel yield or not.

600𝑑 600 + 𝑓𝑦

=

600(625) = 396.825𝑚𝑚 600 + 345

a

625mm

𝑐𝑏 =

125 125 125

125

𝐶𝑏 =:

From Eq. 2-11

𝑎 = 𝛽1 𝑐 𝑎 = 0.85(396.825)= 337.3 𝑚𝑚 𝐴𝑐 = 337.3(375) − 125(125) = 110,863 𝑚𝑚2 𝑇=𝐶 𝐴𝑠𝑏 𝑓𝑦 = 0.8𝑓′𝑐 𝐴𝑐 𝐴𝑠𝑏 (345) = 0.85(21)(110,863) 𝐴𝑠𝑏 = 5,736 𝑚𝑚2

4-32mm 375mm

Since 𝐴𝑠 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 < 𝐴𝑠𝑏 , tension steel yields. 𝐶=𝑇 0.85𝑓 ′ 𝑐 (𝑎𝑏 − 1252 ) = 𝐴𝑠 𝑓𝑦 0.85(21)(𝑎 𝑥 375 − 1252 ) = 3,217(345) 𝑎 = 207.5 𝑚𝑚

𝑀𝑛 = 𝑀𝑛1 − 𝑀𝑛2

𝑎 125 = 𝑀𝑛 = 𝐶1 (𝑑 − ) − 𝐶2 (𝑑 − ) 2 2

𝑀𝑛 = 0.85(21)(207.5)(375)(625 −

207.5 ) 2

𝑀𝑛 = 567.03 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(567.03) 𝜑𝑀𝑛 = 𝟓𝟏𝟎. 𝟑𝟑 𝒌𝑵 − 𝒎 PROBLEM 2.22 Compute the ultimate moment capacity of the beam shown in Figure 2.10. Assume 𝑓𝑦 = 345𝑀𝑃𝑎 and 𝑓′𝑐 = 21 𝑀𝑃𝑎.

𝐶𝑏 =

600 𝑑 600 + 𝑓𝑦

𝐶𝑏 =

600(375) 600 + 345

𝐶𝑏 = 238 𝑚𝑚 𝑎𝑏 = 𝛽1 𝐶𝑏 𝑎𝑏 = 0.85(238) 𝑎𝑏 = 202.4 𝑚𝑚 𝑥 375 5 = 𝑥= 𝑎 𝑎 450 6

375

Solve for 𝐴𝑠 :

3-22mm 75 375m m

450mm

SOLUTION 𝜋 𝐴𝑠 = (22)2 𝑥 3 4 𝐴𝑠 = 1,140 𝑚𝑚2

Figure 2.10

𝑥 = 168.7𝑚𝑚

x

𝐴𝑐 = 17,066 𝑚𝑚2

𝑇 = 𝐶𝐶 𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓′𝑐 𝐴𝑐 𝐴𝑠𝑏 (345) = 0.85(21)(17,066) 𝐴𝑠𝑏 = 883 𝑚𝑚2 < 𝐴𝑠

322mm 375m m

7 5

Since𝐴𝑠 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 > 𝐴𝑠𝑏 , tension steel does not yield (𝑓𝑠 < 𝑓𝑦 ) solve for c: 𝐶𝐶 = 𝑇 𝑓𝑠 = 600 𝑎 = 𝛽1 𝑐

0.85𝑓′𝑐 𝐴𝑐 = 𝐴𝑠 𝑓𝑠 𝑑−𝑐 5 2 𝑑−𝑐 0.85(21) 𝑎 = 1140 𝑥 600 𝑐 12 𝑐 7.437(0. 85𝑐)2 = 684,00

375−𝑐 𝑐

𝑐 = 250.92 𝑚𝑚 𝑎 = 𝛽1 𝑐

𝑎 = 0.85(250.92) = 213.3 𝑚𝑚

2 2 𝑀𝑛 = 𝐶𝑐 𝑥 (𝑑 − 𝑎) = 0.85𝑓 ′ 𝑐 𝐴𝑐 (𝑑 − 𝑎) 3 3 5 2 𝑀𝑛 = 0.85(21) (213. 29)2 𝑥 [375 − (213.3)] 12 3 𝑀𝑛 = 78.77 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90 𝑥 78.77 = 𝟕𝟎. 𝟖𝟗 𝒌𝑵 − 𝒎

m

c

d375 450m m 450m

𝐴𝑐 = 1/2(𝑥)(𝑎) 1 5 5 𝐴𝑐 = 2 𝑥 6 𝑎 𝑥 𝑎 = 𝑎 𝑎2

a

d-(2/3)a

T

PROBLEM 2.23 A hallow beam is shown in Figure 2.11. Assume 𝑓′𝑐 = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 345 𝑀𝑃𝑎. a) Calculate the required tension steel area when 𝑀𝑢 = 800𝑘𝑁 − 𝑚. b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? e) Calculate the required tension steel area when 𝑀𝑢 = 1200𝑘𝑁 − 𝑚.

Figure 2.11Hallow beam

SOLUTION To guide us whether “a: will exceed 150 mm or not, let us solve the design moment when a=150 mm. d = 800 – 75 = 725 mm

𝑎

𝜑𝑀𝑛 = 𝜑𝐶𝐶 (𝑑 − 2)

𝜑𝑀𝑛 = 0.90 𝑥 0.85(28)(150) (725 −

150 2

)

𝜑𝑀𝑛 = 1044.225 𝑘𝑁 − 𝑚 a) 𝑀𝑢 = 800 𝑘𝑁 − 𝑚 Since the required 𝑀𝑢 = 800 𝑘𝑁 − 𝑚 < 1044.25 𝑘𝑁 − 𝑚, 𝑎 < 150 𝑚𝑚. Assuming tension steel yields: 𝑎 𝑀𝑢 = 𝜑𝐶𝑐 (𝑑 − ) 2 𝑎 𝑀𝑢 = 𝜑0.85𝑓′𝑐 𝑎 𝑏(𝑑 − ) 2 800 𝑥 106 = 0.90 𝑥 0.85(28)𝑎(500)(725 − 0.5𝑎) 𝑀𝑢 = 𝜑𝑀𝑛

𝑎 = 111.6𝑚𝑚 < 150 𝑚𝑚 Check is steel yields: 𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝑓𝑠 = 600

725 − 131.2 = 2,712 𝑀𝑃𝑎 > 𝑓𝑦 131.3

𝑤ℎ𝑒𝑟𝑒 𝑐 =

𝑇=𝐶

𝑎 = 131.3 𝑚𝑚 𝛽1 𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏 𝐴𝑠 (345) = 0.85(28)(111.6)(500) 𝐴𝑠 = 𝟑, 𝟖𝟓𝟎 𝒎𝒎𝟐

b) Balanced condition (See Figure 2.12) 𝐶𝑏 = 𝑎 = 𝛽1 𝐶𝑏

600𝑑 600 + 𝑓𝑦

𝐶𝑏 =

600(725) = 460.32𝑚𝑚 600 + 345

𝑎 = 0.85(460.32) = 391.3 𝑚𝑚

𝑧 = 𝑎 − 150 = 241.27 𝑚𝑚 𝐴1 = 500(150) = 75,000 𝑚𝑚2 1 𝑦1 = 725 − = 650𝑚𝑚 2(150) 𝐴1 = 125(241.27) = 30,159 𝑚𝑚2 1 𝑦2 = 725 − 150 − = 454.37 2(241.27) 𝑀𝑏𝑛 = 𝐶1 𝑦1 + 2𝐶2 𝑦2 𝑀𝑏𝑛 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 2𝐴2 𝑦2 ) 𝑀𝑏𝑛 = 0.85(28)[75,000 𝑥 650 + 2 𝑥 30,159 𝑥 454.37] 𝑀𝑏𝑛 = 1812.52𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90 𝑥 1812.52 = 𝟏𝟔𝟑𝟏. 𝟑 𝒌𝑵 − 𝒎 500m m

25 0

12 5 150

12 5

725

1 2

2

z

a

T

Figure 2.12

c) Maximum steel area, 𝐴𝑠 𝑚𝑎𝑥 𝑇 = 𝐶1 + 𝐶2 𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 2𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(28)(75,00 + 2 𝑥 30,159) 𝐴𝑠𝑏 = 9,335 𝑚𝑚2 𝐴𝑠 𝑚𝑎𝑥 = 0.75 𝐴𝑠𝑏

𝐴𝑠 𝑚𝑎𝑥 = 0.75(9,335) 𝐴𝑠 𝑚𝑎𝑥 = 𝟕, 𝟎𝟎𝟏 𝒎𝒎𝟐

d) Maximum moment , 𝑀𝑢 𝑚𝑎𝑥 : Refer to Figure 2.12:

𝐶1 + 𝐶2 = 𝑇 0.85(28)[75,000 + 2𝐴2 ] = 7,001(245) 𝐴2 = 13,244 𝑚𝑚2 𝐴2 = 125 𝑧

𝑦2 = 725 −

𝑀𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥

13,244 = 125 𝑧 𝑧 = 105.95 𝑚𝑚 1501 = 522.03 𝑚𝑚 2(105.95)

= 𝐶1 + 𝑦1 + 2𝐶2 𝑦2 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 2𝐴2 𝑦2 ) = 0.85(28)[75,00 𝑥 650 + 2 𝑥 13,244 𝑥 522.03] = 1489.34 𝑘𝑁 − 𝑚

𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90 𝑥 1189.34 = 𝟏𝟑𝟒𝟎. 𝟒 𝒌𝑵 − 𝒎 e) 𝑀𝑢 = 1200𝑘𝑁 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥 Refer to Figure 2.12 𝐴1 = 75,000 𝑚𝑚2

𝑦1 = 650 𝑚𝑚

(𝑆𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑)

𝐴2 = 125𝑧

𝑦2 = 575 − 0.5𝑧

𝑀𝑢 = 𝜑𝑀𝑛 1200 𝑥 106 = 0.90 𝑥 0.85 𝑓 ′ 𝑐 (𝐴1 𝑦1 + 2𝐴2 𝑦2 ) 1200 𝑥 106 = 0.90 𝑥 0.85(28)[75,000(650) + 2(125𝑧)(575 − 0.5𝑧)] 𝑧 = 53.04𝑚𝑚 𝐴𝑐 = 𝐴1 + 𝐴2

𝐴𝑐 = 75,000 + 2 𝑥 125(53.04) 𝐴𝑐 = 88,259.2 𝑚𝑚2

𝑇=𝐶 𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝐴𝑐 𝐴𝑠 (345) = 0.85(28)(88,259.2) 𝐴𝑠 = 𝟔, 𝟎𝟖𝟗 𝒎𝒎𝟐

BEAM DEFLECTION PROBLEM PROBLEM 2.24 A reinforced concrete beam is 350 mm wide and 600 mm deep. The beam is simply supported over a span of 8 m and carries a uniform dead load of 11 kN/m including its own weight and a uniform live load of 15 kN/m. The beam is reinforced tension bars of 530 mm. 𝑓′𝑐 = 20.7 𝑀𝑃𝑎, 𝑓𝑦 = 344.8 𝑀𝑃𝑎, 𝑓𝑟 = 2.832 𝑀𝑃𝐴. Modulus of elasticity of concrete 𝐸𝑐 = 21,650 𝑀𝑃𝑎 and 𝐸𝑠 = 200 𝐺𝑃𝑎. a) Calculate the maximum instantaneous deflection due to service loads. b) Calculate the deflection for the same loads after five years assuming that 40% of the live load is sustained. SOLUTION b = 350 mm

Figure 2.13

6 – 25 mm Ø

c h = 600 mm

d = 530 mm

b = 350 mm

N.A.

d-c

Effective moment of inertia, 𝐼𝑒 :

Eq. 2-19

𝑀𝑐𝑟 3 𝑀𝑐𝑟 3 𝐼𝑒 = ( ) 𝐼𝑔 + [1 − ( ) ] 𝐼𝑐𝑟 ≤ 𝐼𝑔 𝑀𝑎 𝑀𝑎 𝐼𝑔 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑔𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑏ℎ3 𝐼𝑔 = 12

350(600)3 = = 6300 𝑥 106 𝑚𝑚4 12

𝑀𝑐𝑟 =

𝑓𝑟 𝐼𝑔 𝑦𝑡

𝑀𝑐𝑟 =

2.832(600 𝑥 10)6 = 59.472 𝑘𝑁 − 𝑚 600/2

𝑤ℎ𝑒𝑟𝑒 𝑦𝑡 = 1/2(600) = 300 𝑚𝑚

𝑀𝑎 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 𝑖𝑛 𝑏𝑒𝑎𝑚 𝑤𝐿2 𝑀𝑎 = 8 𝑀𝑎 =

𝑤 = 𝑤𝐷 + 𝑤𝐿 = 11 + 15 = 26 𝑘𝑁/𝑚

26(8)2 = 208𝑘𝑁 − 𝑚 8

𝐼𝑐𝑟 = Moment of inertia of cracked section with steel transformed to concrete

From Figure 2.13: 𝐸

Modular ratio, 𝑛 = 𝐸𝑠 = 9.238 𝑐

𝑛 𝐴𝑠 = 9.328 𝑥 6 𝑥

𝜋 (25)2 = 27,208 𝑚𝑚2 4

Solve for c: Moment of area above N.A. = Moment of area below N.A. 350 x c x c/2 = 27,208(350-c) c = 219.7 mm

𝐼𝑐𝑟 = 𝐼𝑁𝐴 𝐼𝑐𝑟 =

𝑏𝑐 3 = + 𝑛 𝐴𝑠 (𝑑 − 𝑐)2 3

350(219.7)3 + 27,208(530 − 219.7)2 3

𝐼𝑐𝑟 = 3,857 𝑥 106 𝑚𝑚3 𝑀𝑐𝑟 3 𝑀𝑐𝑟 3 𝐼𝑒 = ( ) 𝐼𝑔 + [1 − ( ) ] 𝐼𝑐𝑟 𝑀𝑎 𝑀𝑎 59.472 3 59.472 3 6 𝐼𝑒 = ( ) 𝑥 600 𝑥 10 + [1 − ( ) ] 𝑥 3,857 𝑥 106 208 208 𝐼𝑒 = 3,914 𝑥 106 𝑚𝑚4 < 𝐼𝑔

(𝑂𝐾)

a) Instantaneous Deflection: 𝛿=

5𝑤𝐿4 384 𝐸𝑐 𝐼𝑒

𝛿=

2(26)(8000)4 384(21,650)(3,914 𝑥 106 )

𝛿 = 𝟏𝟔. 𝟑𝟔 𝒎𝒎 b) Long-term Deflection Since only 40% of the live load was sustained: w = 11 + 0.4(15) = 17 kN/m

5𝑤𝐿4

Instantaneous deflection 𝛿 = 384 𝐸

𝑐 𝐼𝑒

𝛿=

5(17)(8)4 (1000)4 384(21,650)(3,914 𝑥 106 ) 𝛿 = 10.7 𝑚𝑚

Note: Since deflections are directly proportional to the load, the instantaneous deflection due to sustained load can be found by ratio and proportion using the result in Part”a”. 𝛿1 16.36 = 17 26 𝛿1 = 10.7 𝑚 Long-term deflection = 𝛿 + 𝛿1 𝜆=

𝜉 1 + 50 𝜌′

𝜉=2 𝜌′ = 0 𝜆=

𝑓𝑜𝑟 5 𝑦𝑒𝑎𝑟𝑠 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡

2 =2 1 + 50(0)

Long-term deflection = 16.36 + 2(10.7) Long-term deflection = 37.76 mm

PROBLEM 2.25 (CE NOVEMBER 2002) The continuous reinforced concrete beam shown in Figure 2.14 is subjected to a uniform service dead load of 16 k/m and a service live load of 32 kN/m,resulting in the bending moment diagram shown. Twenty percent of the live load will be sustained in nature, while 80% will be applied only intermittently. The concrete strength 𝑓𝑐 = 17.2𝑀𝑃𝑎. The modulus of elasticity of concrete is given by the expression 𝐸𝑐 = 4700 𝑆𝑞𝑟𝑡(𝑓 ′ 𝑐 ) and the modulus of rapture is given by the expression 𝑓𝑟 = 0.7 𝑆𝑞𝑟𝑡(𝑓 ′ 𝑐 ). Determine the following: a) The effective moment of inertia at the supports (maximum negative moment). b) The effective moment of inertia for the continuous member. c) The additional deflection (in addition to the initial deflection) after 5 years, under the sustained loading if the instantaneous deflection due to the combined service dead and live load is 5 mm.

Figure 2.14

y 560 mm

Gross Section I=0.0715 y=310 mm

y

AT SUPPORTS

Cracked Section I=0.00573 y=159 mm

1900 mm y

620 mm

560 mm

y

n As

Gross Section I=0.0138 y=194 mm

AT MIDSPAN

Cracked Section I=0.00573 y=107 mm

7.6 m 5-32 mmø

3-32 mmø

5-32 mmø

145 kN-m

202 kN-m

202 kN-m

SOLUTION 𝐸𝑐 = 4700√𝑓′𝑐 = 4700√17.2 = 19,492 𝑀𝑃𝐴 𝑓𝑟 = 0.70√𝑓′𝑐 = 0.7 √17.2 = 2.903 𝑀𝑃𝑎 a) Effective moment of inertia at the supports Maximum moment, 𝑀𝑢 = 202𝑘𝑁 − 𝑚 Distance from NA of gross section to extreme tension fiber, 𝑌𝑡 = 310 𝑚𝑚 Moment of inertia of gross section, 𝐼𝑔 = 0.00715 𝑚4 Moment of inertia of cracked section, 𝐼𝑔 = 0.00573 𝑚4 𝑀𝑐𝑟 =

𝑓𝑟 𝐼𝑔 𝑦𝑡

𝑀𝑐𝑟 =

2.903(0.00715 𝑥 10004 ) 10

𝑀𝑐𝑟 = 66.959 𝑘𝑁 − 𝑚

𝑀𝑐𝑟 3 𝑀𝑐𝑟 3 𝐼𝑒 = ( ) 𝐼𝑔 + [1 − ( ) ] 𝐼𝑐𝑟 𝑀𝑎 𝑀𝑎 66.9593 66.959 3 𝐼𝑒 = ( ) 𝑥 0.00715 + [1 − ( ) ] 𝑥0.00573 202 202 𝐼𝑒 = 𝟎. 𝟎𝟎𝟓𝟕𝟖𝟏𝟕 𝒎𝟒 b) Effective moment of inertia for the continuous member 𝐼𝑒 =

(𝐼𝑒 )𝑚𝑎𝑥 𝑝𝑜𝑠 𝑚𝑒𝑛𝑡 + (𝐼𝑒 )𝑚𝑎𝑥 𝑛𝑒𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 3

(𝑆𝑒𝑐𝑡. 409.6.2.4)

At maximum negative moment (at support) 𝐼𝑒 = 0.0057817 𝑚4 Solving for 𝐼𝑒 at maximum positive moment (at midspan) 𝐼𝑔 = 0.0138 𝑚4 (𝑏𝑜𝑡𝑡𝑜𝑚 𝑓𝑖𝑏𝑒𝑟𝑠 𝑖𝑛 𝑡𝑒𝑛𝑠𝑖𝑜𝑛) 𝑌𝑡 = 620 − 194 = 246 𝑚𝑚 𝐼𝑐𝑟 = 0.00513 𝑚4 𝑀𝑐𝑟 =

𝑓𝑟 𝐼𝑔 𝑌𝑡

𝑀𝑐𝑟 =

2.903(0.00715 𝑥 10004 ) 310

𝑀𝑐𝑟 = 66.959 𝑘𝑁 − 𝑚 𝑀𝑐𝑟 3 𝑀𝑐𝑟 3 𝐼𝑒 = ( ) 𝐼𝑔 + [1 − ( ) ] 𝐼𝑐𝑟 𝑀𝑎 𝑀𝑎 (𝐼𝑒 )𝑚𝑎𝑥 𝑝𝑜𝑠 𝑚𝑒𝑛𝑡 + (𝐼𝑒 )𝑚𝑎𝑥 𝑛𝑒𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 2 0.0057817 + 0.007932 𝐼𝑒 = = 𝟎. 𝟎𝟎𝟔𝟖𝟓𝟕 𝒎𝟒 2 𝐼𝑒 =

c) Additional long term deflection= long term deflection x 𝜆 𝜆=

𝜉 1 + 50𝜌′

𝜌′ = 0(𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑚𝑖𝑑𝑠𝑝𝑎𝑛) 𝜉 = 2(𝑎𝑓𝑡𝑒𝑟 5 𝑦𝑒𝑎𝑟𝑠) 𝜆=

2 =2 1+0

Solving for the instantaneous deflection under sustained loading: Instantaneous deflection = 5mm (given) Instantaneous loading = 16 kN/m + 32 kN/m Instantaneous loading = 48 kN/m Sustained loading = 16 + 20%(32) Sustained loading = 22.4 kN/m Sine deflection is directly proportional to the load: 𝛿1 5 = 22.4 48 𝛿1 = 2.33 𝑚 Additional long term deflection = 2.333 x 𝜆 =2.333 x 2 Additional long term deflection = 4.67 mm

ONE WAY SLAB PROBLEMS Problem 2.36 Design a one-way slab having a simple span 3 m. The slab is to carry a uniform live load of 7,500 Pa. Assume 𝑓′𝑐 = 27.6 𝑀𝑃𝑎 and 𝑓𝑦 = 276 𝑀𝑃𝑎 for main and temperature bars. The slab is not exposed to earth or weather. Use unit weight of concrete 𝛾𝑐 = 23.5 𝑘𝑀/𝑚3. SOLUTION Consider 1 m strip of slab, b= 1000 m Uniform live load, 𝑤𝐿 = 7.5 𝐾𝑃𝑎 𝑥 1𝑚 = 7.5 𝑘𝑁/𝑚 Minimum slab thickness from Table 2.1: 𝐿

𝑓𝑦

ℎ𝑚𝑖𝑛 = 20 (0.4 + 700)

ℎ𝑚𝑖𝑛 =

3000 20

276

(0.4 + 700)

ℎ𝑚𝑖𝑛 = 119 𝑚𝑚 (𝑢𝑠𝑒 120 𝑚𝑚)

10 mm temp. bars

B = 1000 mm d h = 120 mm

102mm main bars

Cover +

/2

Effective depth: d = 120-20 mm (covering)-1/2 bar diameter (12mm) d=94 mm Weight of slab: 𝑊𝑠 = 𝛾𝑐𝑜𝑛𝑐 𝑥 𝑏 𝑥 ℎ

𝑊𝑠 = 23.5 (1)(0.12) 𝑊𝑠 = 2.82 𝑘𝑁/𝑚

Factored floor pressure load: 𝑊𝑢 = 1.4𝑤𝑠 + 1.7 𝑤𝐿

𝑀𝑢 =

𝑊𝑢 = 1.4(2.82) + 1.7(7.5) 𝑊𝑢 = 16.698 𝑘𝑁/𝑚

𝑊𝑢 𝐿2

16.698(3)2

𝑀𝑢 = 2 𝑀𝑢 = 18.785 𝑘𝑁 − 𝑚

8

𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2

𝜌=

𝜌=

0.85𝑓 ′ 𝑐 𝑓𝑦

[1 − √1 −

18.785 𝑥 106 = 0.90 𝑅𝑛 (1000)(94)2 𝑅𝑛 = 2.362 𝑀𝑃𝑎

𝑅𝑢 ] 0.85𝑓 ′ 𝑐

0.85(27.6) 2(2.362) [1 − √1 − ] 276 0.85(27.6)

𝜌 = 0.009039 Check for 𝜌𝑚𝑖𝑛 and 𝜌𝑚𝑎𝑥 : 𝜌𝑚𝑖𝑛 =

1.4 = 0.00507 𝑓𝑦

𝜌𝑚𝑎𝑥 =

0.75 0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝑂𝐾

𝜌𝑚𝑎𝑥 =

0.75 0.85(27.6)(0.85)600 276(600 + 276)

𝜌𝑚𝑎𝑥 = 0.037 > 0.009309 (𝑂𝐾)

𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.009039(1000)(94) 𝐴𝑠 = 850 𝑚𝑚2 per meter width of slab

Using 12-mm main bars: Spacing s =

𝐴𝑏𝑎𝑟 𝐴𝑠

𝑥 1000

𝑠=

𝜋 (12)2 4

850

𝑥 1000

𝑠 = 138 𝑚𝑚 𝑠𝑎𝑦 135 𝑚𝑚 Maximum spacing required by the Code: a) 3(ℎ) = 3(120) = 360 𝑚𝑚 b) 450 𝑚𝑚

𝑂𝐾

Thus, use 12 mm main bars at 135 mm o.c. Temperature bars: (Grade 275) 𝐴𝑡 = 0.002𝑏ℎ

Spacing =

𝐴𝑏𝑎𝑟 𝐴𝑠

𝐴𝑡 = 0.002(1000)(120) 𝐴𝑡 = 240 𝑚𝑚2 𝑥 1000

𝑠=

𝜋 (10)2 4

240

𝑥 1000

𝑠 = 327 𝑚𝑚 𝑠𝑎𝑦 325 𝑚𝑚 Maximum spacing required by the Code: a) 5ℎ = 5(120) = 600𝑚𝑚 b) 450 mm OK Thus, use 10 mm temperature bars at 325 mm o.c.

PROBLEM 2.27 Design a one-way slab to carry a service live load of 4000 Pa. The slab has a length of 4m with both ends continuous. Assume 𝑓′𝑐 = 21 𝑀𝑃𝑎 and 𝑓𝑦 = 415 𝑀𝑃𝑎 for main bars and 𝑓𝑦 = 276 𝑀𝑃𝑎 for temperature bars. Steel cover is 20 mm. Unit weight of concrete is 23.5 kN/𝑚3 .

SOLUTION Consider 1 m strip, b = 1000 mm Uniform live load, 𝑤𝐿 = 4 𝑘𝑝𝑎 𝑥 1𝑚 = 4 𝑘𝑁/𝑚 Minimum slab thickness from Table 2.1: ℎ𝑚𝑖𝑛 =

𝐿 28

ℎ𝑚𝑖𝑛 =

4000 28

ℎ𝑚𝑖𝑛 = 143 𝑚𝑚 (𝑢𝑠𝑒 150 𝑚𝑚) Weight of beam (DL): 𝑤𝐷 = 𝛾𝑐𝑜𝑛𝑐 𝑥 𝑏 𝑥 ℎ

𝑤𝐷 = 23.5(1)(0.15) 𝑤𝐷 = 3.525 𝑘𝑃𝑎

𝑤𝑢 = 1.4 𝑤𝐷 + 1.4 𝑤𝐿

𝑤𝑢 = 1.4(3.525) + 1.7(4) 𝑤𝑢 = 11.735 𝑘𝑁/𝑚

Maximum factored moment, Section 408.4 (See Page 29) LL < 3 DL Column

Column

Column

Spandrel Beam

Shear

Moment

Effective depth, d = 1.50 – 20 – 1/2 (12) Effective depth, d = 124 mm At midspan: 𝑤𝑢 𝐿𝑛2 𝑀𝑢 = 16 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2

11.735 (4)2 𝑀𝑢 = 16

𝑀𝑢 = 11.735 𝑘𝑁 − 𝑚

11.735 𝑥 106 = 0.90 𝑅𝑛 (1000)(124)2 𝑅𝑛 = 0.848 𝑀𝑃𝑎

𝜌=

0.85𝑓′𝑐 2𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.85𝑓′𝑐

𝜌=

0.85(21) 2(0.848) [1 − √1 − ] 415 0.85(21)

𝜌 = 0.0021

𝜌𝑚𝑖𝑛 =

1.4 = 0.00337 > 0.0021 𝑓𝑦

Use 𝜌 = 𝜌𝑚𝑖𝑛 = 0.00337 𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.00337(1000)(124) 𝐴𝑠 = 418 𝑚𝑚2

Spacing, s =

𝐴𝑠𝑏 𝐴𝑠

𝑥 1000 =

𝑠 =

𝜋 (12)2 4

418

𝑥1000

𝑠 = 271 𝑠𝑎𝑦 270 𝑚𝑚 Maximum spacing required by the Code: a) 3 ℎ = 3(150) = 450 𝑚𝑚 b) 450 mm Thus, use 12 mm bottom bars at 270 mm o.c. at midspan At support: 𝑀𝑢 =

𝑤𝑢 𝐿𝑛 2 10

𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2

𝑀𝑢 =

=

11.735(4)2 10 𝑀𝑢 = 18.776 𝑘𝑁 − 𝑚

18.776 𝑥 106 = 0.90𝑅𝑛 (1000)(124)2 𝑅𝑛 = 1.357 𝑀𝑃𝑎

𝜌=

0.85𝑓′𝑐 1 − 2𝑅𝑛 [1 − √ ] 𝑓𝑦 0.85𝑓′𝑐

𝜌=

0.85(21) 2(1.357) [1 − √1 − ] 415 0.85(21)

𝜌𝑚𝑎𝑥 = 0.0034 > 𝜌𝑚𝑖𝑛 𝜌𝑚𝑎𝑥 = 0.75

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑚𝑎𝑥 = 0.75

0.85(21)(0.85)600 415(600 + 415)

𝜌𝑚𝑎𝑥 = 0.0162 > 0.0034 Use 𝜌 = 0.034 𝐴𝑠 = 𝜌𝑏𝑑

Spacing, 𝑠 =

𝐴𝑠 = 0.0034(1000)(124) 𝐴𝑠 = 422 𝑚𝑚2 𝐴𝑠𝑏 𝐴𝑠

𝑥 1000

𝑠=

𝜋 (12)2 4

422

𝑥 1000

𝑆𝑝𝑎𝑐𝑖𝑛𝑔 = 268 𝑠𝑎𝑦 265 𝑚𝑚 Thus, use 12 mm top bars @ 265 mm o.c. at support Temperature bars (10 mm): (𝜌𝑡 = 0.002) 𝐴𝑡 = 0.002𝑏ℎ = 𝐴𝑡 = 0.002(1000)(150)

Spacing, s =

𝐴𝑠𝑏 𝐴𝑠

𝑥 1000 = 𝑠 =

𝜋 (10)2 4

300

𝐴𝑡 = 300 𝑚𝑚2

𝑥 1000

𝑠 = 261 𝑠𝑎𝑦 260 𝑚𝑚

Maximum spacing required by the Code: a) 5ℎ = 5(150) = 750 𝑚𝑚 b) 450 mm

Thus, use 10 mm temperature bars @ 260 mm o.c.

150 mm

10 mm temperature bars @ 260 mm o.c.

12 mm main bars @ 265 mm o.c. L/4

L/2

L/4

PROBLEM 2.28

A one-way slab having a simple span of 3 m is 160 mm thick. The slab is reinforced with 12 mm tension bars (𝑓𝑦 = 275 𝑀𝑃𝑎) spaced at 140 mm o.c. Steel covering is 20 mm. Calculate the uniform live load pressure that a slab can carry. Use 𝑓′𝑐 = 20.7 𝑀𝑃𝑎. Unit weight of concrete is 23.5 kN/𝑚3 . SOLUTION

Consider 1 m strip of slab, b = 1000 m

𝑤𝑑 = 𝛾𝑐 𝑏 ℎ 𝑤𝑑 = 23.5(1)(0.16) 𝑤𝑑 = 3.76 𝑘𝑁 − 𝑚

Dead load:

d = 160 – 20 – 1/2(12) d = 134 mm

Effective depth:

Steel area, 𝐴𝑠 =

1000 𝑠

𝜌=

𝐴𝑠 = 𝑏𝑑

𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝑥 𝐴𝑠

140

𝑥

𝜋 24

(12)2

807.8 1000(134) 𝜌 = 0.006028

𝜌=

=

𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2

= 𝜌𝑏 =

𝜔=

0.85(20.7)(0.85)(600) 275(600 + 275) 𝜌𝑏 = 0.037 > 𝜌 (𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠)

0.006028(275) 20.7 𝑅𝑛 = 20.7(0.0801)[1 − 0.59(0.0801)] 𝑅𝑛 = 1.58 𝑀𝑃𝑎

𝑀𝑢 = 20.7(0.0801)[1 − 0.59(0.0801)] 𝑀𝑢 = 25.5334 𝑘𝑁 − 𝑚

𝑤𝑢 𝐿2 𝑤𝑢 (3)2 = 25.5334 = 8 8

𝑤𝑢 = 1.4 𝑤𝐷𝐿 + 1.7 𝑤𝐿𝐿 𝑤𝐿𝐿

1000

𝐴𝑠 = 807.8 𝑚𝑚2

𝑅𝑛 = 𝑓′𝑐 𝜔(1 − 0.59𝜔) =

𝑀𝑢 =

= 𝐴𝑠 =

𝑤𝑢 = 22.696 kN/m

= 22.696 = 1.4(3.76) + 1.7 𝑤𝐿𝐿 𝑤𝐿𝐿 = 10.25 𝑘𝑁/𝑚 = 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑥 𝑏 10.25 = Uniform pressure x 1 Uniform live load pressure = 10.25 kPa

Solved Problems Using 2010 NSCP PROBLEM 2.29 A reinforced concrete beam has width of 310 mm and an effective depth of 490 mm. 𝑓′𝑐 = 30 𝑀𝑃𝑎, 𝑓𝑦 = 415 𝑀𝑃𝑎. Determine the following: a) The balanced steel area b) The maximum steel area for singly reinforced condition c) The maximum design strength if the beam is singly reinforced d) The required steel area if the beam is subjected to dead load moment of 120 kN-m and live load moment of 170 kN-m.

SOLUTION Since 𝑓′𝑐 > 28 𝑀𝑃𝑎; 𝛽1 = 0.85 −

0.05 ′ (𝑓 𝑐 − 28) 7

𝛽1 = 0.85 − 𝛽1 = 0.836

0.05 (30 − 28) 7

a) Balanced steel area: 𝜌𝑏 =

0.85𝑓 ′ 𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝐴𝑠𝑏 = 𝜌𝑏 𝑏 𝑑

0.85(30)(0.836)(600) 415(600 + 415) 𝜌𝑏 = 0.03036 𝜌𝑏 =

𝐴𝑠𝑏 = 0.03036(310)(490) 𝐴𝑠𝑏 = 𝟒, 𝟔𝟏𝟏 𝒎𝒎𝟐

b) Maximum steel area when beam is singly reinforced: From Eq. 2-24:

3 0.85𝑓 ′ 𝑐 𝛽1 𝑓𝑦 3 0.85(30)(0.836)

𝜌𝑚𝑎𝑥 = 7 𝜌𝑚𝑎𝑥 = 7

415(600+415)

𝜌𝑚𝑎𝑥 = 0.0221

𝐴𝑠 𝑚𝑎𝑥 = 𝜌𝑚𝑎𝑥 𝑏 𝑑

𝐴𝑠 𝑚𝑎𝑥 = 0.0221(310)(490) 𝐴𝑠 𝑚𝑎𝑥 = 𝟑, 𝟑𝟒𝟑 𝒎𝒎𝟐

c) Maximum design strength, 𝜑𝑀𝑛 𝑚𝑎𝑥 : 𝜀 = 0.004, 𝑓𝑠 = 800 𝑀𝑃𝑎 51

𝑀𝑛 𝑚𝑎𝑥 =

3

𝑀𝑛 𝑚𝑎𝑥 = 140 𝛽1 𝑓′𝑐 𝑏𝑑 2 (1 − 14 𝛽1 )

From Eq. 2-25 :

51 3 (0.836)(30)(310)(490)2 (1 − 𝑥 0.836) 140 14 𝑀𝑛 𝑚𝑎𝑥 = 558.05 𝑘𝑁 − 𝑚

From Eq. 2-26: 𝜑 = 0.65 + 0.25 𝜑 = 0.65 + 0.25

800−𝑓𝑦 1000−𝑓𝑦 800−415 1000−415

𝜑 = 0.8145 𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.8145(558.05) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 𝟒𝟓𝟒. 𝟓𝟓 𝒌𝑵 − 𝒎 d) 𝑀𝑢 = 1.2𝑀𝐷 + 1.6 𝑀𝐿

𝑀𝑢 = 1.2(120) + 1.6(170) 𝑀𝑢 = 451.45 𝑘𝑁 − 𝑚

Thus, the beam is singly reinforced. Determine if the beam is tension-controlled: 459

3

From Eq. 2-22: 𝜑𝑀𝑡𝑛 = 1600 𝛽1 𝑓′𝑐 𝑏𝑑 2 (1 − 16 𝛽1 ) 𝜑𝑀𝑡𝑛 = 451.45 𝑘𝑁 − 𝑚 Since the required 𝑀𝑢 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑀𝑡𝑛 , the section is tension controlled. 𝜑 = 0.90

𝑀𝑢 = 𝜑𝑀𝑛 𝑎 𝑀𝑛 = 𝜑 𝑥 0.85 𝑓′𝑐 𝑎 𝑏 (𝑑 − 2)

𝑎 416 𝑥 106 = 0.90 𝑥 0.85(30)(𝑎)(310)(490 − ) 2 𝑎 = 139.06 𝑚𝑚 Check if it is really tension-controlled: 𝑐=

𝑎 139.06 = = 166.4 𝑚𝑚 𝛽1 0.836

𝑓𝑠 = 600

𝑑−𝑐 490 − 166.4 = 600 = 1,167 𝑀𝑃𝑎 𝑐 166.4 > 1,000 𝑀𝑃𝑎 (𝑂𝐾)

PROBLEM 2.30 Given the following data for a rectangular beam: width 𝑏 = 320𝑚𝑚, effective depth 𝑑 = 520 𝑚𝑚, 𝑓′𝑐 = 27 𝑀𝑃𝑎, 𝑓𝑦 = 345 𝑀𝑃𝑎. Dead load moment 𝑀𝐷 = 180 𝑘𝑁 − 𝑚, Live load moment 𝑀𝐿 = 167 𝑘𝑁 − 𝑚. 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 𝑎𝑟𝑒𝑎. SOLUTION 𝛽1 = 0.85 𝑀𝑈 = 1.2 𝑀𝐷 + 1.6 𝑀𝐿

𝑀𝑢 = 1.2(180) + 1.6(167) 𝑀𝑢 = 483.2 𝑘𝑁 − 𝑚

Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 to determine if compression steel area is required.

𝑀𝑛 𝑚𝑎𝑥 =

51 3 𝛽1 𝑓′𝑐 𝑏𝑑 2 (1 − 𝛽) 140 14 1

𝑀𝑛 𝑚𝑎𝑥 =

51 3 (0.85)(27)(320)(520)2 (1 − 𝑥 0.85) 140 14

𝑀𝑛 𝑚𝑎𝑥 = 591.64 𝑘𝑁 − 𝑚 𝜑 = 0.65 + 0.25

800−𝑓𝑦 1000−𝑓𝑦

=0.8237

𝜑𝑀𝑛 𝑚𝑎𝑥 = 487.31 𝑘𝑁 − 𝑚 > 𝑀𝑢

(𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑)

Solve for 𝜑𝑀𝑡𝑛 to determine if the section is tension-controlled. 𝜑𝑀𝑛 =

459 3 𝛽1 𝑓′𝑐 𝑏𝑑2 (1 − 𝛽1 ) = 478.9 𝑘𝑁 − 𝑚 1600 16

Since 𝑀𝑢 > 𝜑𝑀𝑡𝑛 , the section is within “transition region’, i.e 0.65 < 𝜑 < 0.90 𝑀𝑢 = 𝜑𝑀𝑛 = 𝜑 𝑥 0.85 𝑓′𝑐 𝑎𝑏 (𝑑 − 𝑎/2) 520 − 𝑐 600 − 345 𝑓𝑠 − 𝑓𝑦 𝑐 𝜑 = 0.65 + 0.25 = 0.65 + 0.25 1000 − 𝑓𝑦 1000 − 345 𝜑=

119.084 + 0.2893 𝑐

𝜑 = 0.85𝑐 119.084 + 0.2893) 𝑥 0.85(27)(0.85𝑐)(320)(520 𝑐 − 1/2𝑥 0.85𝑐)

483.2 𝑥 106 = (

𝑐 = 208.8 𝑚𝑚 𝑎 = 𝛽1 𝑐 = 177.45 𝑚𝑚 𝑇=𝐶

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏 𝐴𝑠 (34.5) = 0.850(27)(177.45)(320) 𝐴𝑠 = 3,777 𝑚𝑚2

PROBLEM 2.31 Given the following properties of a rectangular concrete beam: b = 280 mm, d = 480 mm, 𝑓′𝑐 = 21 𝑀𝑃𝑎, 𝑓𝑦 = 415 𝑀𝑃𝑎. The beam is reinforced for tension only. Determine the design strength under the following conditions. a) When the beam is reinforced with three 25 mm diameter bars. b) When the beam is reinforced with four 25 mm diameter bars. c) When the beam is reinforced with seven 25 mm diameter bars.

SOLUTION 𝛽1 = 0.85 𝜌𝑏 = 𝐴𝑏 =

𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 28 𝑀𝑃𝑎

0.85𝑓 ′ 𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

= 𝜌𝑏 =

0.85(21)(0.85)(600) = 𝜌𝑏 = 0.0216 415(600 + 415)

𝜋 (25)2 = 490.87 𝑚𝑚2 4

a) 𝐴𝑠 = 3 𝑥 𝐴𝑏 = 1473 𝑚𝑚2 𝜌=

𝐴𝑠 𝑏𝑑

=

1473 280(480) 𝜌 = 0.01096 < 𝜌𝑏

𝜌=

(𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠)

𝐶=𝑇 0.85𝑓 ′ 𝑐 𝑎 𝑏 = 𝐴𝑠 𝑓𝑦 0.85(21)(𝑎)(280) = 1473(415) 𝑎 = 122.28 𝑚𝑚 𝑎 𝑐= = 143.86 𝑚𝑚 𝛽1 𝑓𝑠 = 600

𝑑−𝑐 480 − 143.86 = 600 = 1,402 𝑀𝑃𝑎 > 1,000 𝑀𝑃𝑎 𝑐 143.86

The section is tension-controlled, 𝜑 = 0.90 𝑀𝑛 = 𝐶𝑐 (𝑑 − 𝑎/2) = 𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏 (𝑑 − 𝑎/2) 𝑀𝑛 = 0.85(21)(122.28)(280)(480 − 122.28/2) 𝑀𝑛 = 255.87 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(255.87) 𝜑𝑀𝑛 = 𝟐𝟑𝟎. 𝟐𝟖 𝒌𝑵 − 𝒎 b) 𝐴𝑠 = 4 𝑥 𝐴𝑏 = 1963 𝑚𝑚2 𝐴𝑠 1963 = 𝑏𝑑 280(480) 𝜌 = 0.014961 < 𝜌𝑏 𝜌=

(𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠)

𝐶=𝑇 0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠 𝑓𝑦 0.85(21)(𝑎)(280) = 1963(415) 𝑎 = 163.04 𝑚𝑚 𝑐=

𝑎 = 191.81 𝑚𝑚 𝛽1

𝑓𝑠 = 600

𝑑−𝑐 480 − 191.81 = 600 = 901.5 𝑀𝑃𝑎 < 1,000 𝑀𝑃𝑎 𝑐 191.81

The section within” transition region”, i. e 0.65 < 𝜑 < 0.90 𝜑 = 0.65 + 0.25

𝑓𝑠 − 𝑓𝑦 1000 − 𝑓𝑦

= 𝜑 = 0.65 + 0.25

901.5 − 415 1000 − 415

𝜑 = 0.858 𝑀𝑛 = 𝐶𝑐 (𝑑 − 𝑎/2) = 𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏 (𝑑 − 𝑎/2) 𝑀𝑛 = 0.85(21)(163.04)(280)(480 − 163.04/2) 𝑀𝑛 = 324.504 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.858(324.504) 𝜑𝑀𝑛 = 𝟐𝟕𝟖. 𝟑𝟗𝟔 𝒌𝑵 − 𝒎 c) 𝐴𝑠 = 7 𝑥 𝐴𝑏 = 3436 𝑚𝑚2 𝜌=

𝐴𝑠 3436 = = 0.02557 > 𝜌𝑏 𝑏𝑑 280(480)

(𝑠𝑡𝑒𝑙 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑦𝑖𝑒𝑙𝑑)

The section is compression-controlled, 𝜑 = 0.65 𝑇=𝐶 𝐴𝑠 𝑓𝑠 = 0.85𝑓′𝑐 𝑎 𝑏 480 − 𝑐 3436 𝑥 600 = 0.85(21)(0.85𝑐)(280) 𝑐 𝑐 = 297.56 𝑚𝑚 𝑎 = 𝛽1 𝑐 = 252.92 𝑚𝑚 𝑀𝑛 = 𝐶𝑐 )𝑑 − 𝑎/2) = 𝑀𝑛 = 0.85𝑓 ′ 𝑐 𝑎 𝑏 (𝑑 − 𝑎/2) 𝑀𝑛 = 0.85(21)(252.92)(280)(480 − 252.92/2) 𝜑𝑀𝑛 = 0.65(446.91) 𝜑𝑀𝑛 = 𝟐𝟗𝟎. 𝟒𝟗 𝒌𝑵 − 𝒎

PROBLEM 2.32 A hallow beam is shown in Figure 2.16. Assume 𝑓′𝑐 = 28 𝑀𝑃𝑎 and 𝑓𝑦 = 345 𝑀𝑃𝑎. a) Calculate the required tension steel area when 𝑀𝑢 = 800 𝑘𝑛 − 𝑚 b) What is the balanced moment capacity of the beam? c) What is the maximum steel area under singly reinforced condition? d) What is the maximum design moment strength under singly reinforced condition? e) Calculate the required tension steel area when 𝑀𝑢 = 1200 𝑘𝑁 − 𝑚.

500 mm 250

125

500 150

800 mm

150

125

75 mm

Figure 2.16 - Hallow beam SOLUTION This problem is the same as Problem 2.23. 𝑑 = 800 − 75 = 725 𝑚𝑚

To guide us whether “a” will exceed 150 mm or not, let us solve the design moment when a =150 mm. 𝑐=

𝑎 = 176.47 𝑚𝑚 𝛽1

𝑓𝑠 = 600

𝑑−𝑐 = 1,865 𝑀𝑃𝑎 > 1000 𝑀𝑃𝑎 𝑐

𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠, 𝜑 = 0.90

𝜑𝑀𝑛 = 𝜑𝐶𝑐 (𝑑 − 𝑎/2) = 0.90 𝑥 0.85(28)(150)(500)(725 − 150/2) 𝜑𝑀𝑛 = 1044.225 𝑘𝑁 − 𝑚 a) 𝑀𝑢 = 800 𝑘𝑁 − 𝑚 Since the required 𝑀𝑢 = 800 𝑘𝑁 − 𝑚 < 1044.225 𝑘𝑁 − 𝑚, 𝑎 < 150 𝑚𝑚. 𝑀𝑢 = 𝜑𝑀𝑛 = 𝜑𝐶𝑐 (𝑑 − 𝑎/2) 𝑀𝑢 = 𝜑0.85𝑓′𝑐 𝑎 𝑏(𝑑 − 𝑎/2) 800 𝑥 106 = 0.90 𝑥 0.85(28)𝑎(500)(725 − 0.5𝑎) 𝑎 = 111.6 𝑚𝑚 < 150 𝑚𝑚 Stress in steel 𝑑−𝑐 𝑎 𝑤ℎ𝑒𝑟𝑒 𝑐 = = 131.3 𝑚𝑚 𝑐 𝛽1 725 − 131.2 𝑓𝑠 = 600 = 2,712 𝑀𝑃𝑎 > 𝑓𝑦 𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠 131.3 𝑓𝑠 = 600

𝑇 = 𝐶𝑐

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏 𝐴𝑠 (345) = 0.85(28)(111.6)(500) 𝐴𝑠 = 𝟑, 𝟖𝟓𝟎 𝒎𝒎𝟐

b) Balanced condition: 𝜑 = 0.65

𝐶𝑏 =

600𝑑 600 + 𝑓𝑦

= 𝐶𝑏 =

600(725) = 460.32 𝑚𝑚 600 + 345

𝑎 = 𝛽1 𝑐𝑏 = 𝑎 = 0.85(460.32) = 391.3 𝑚𝑚 500 mm 125

250 125 150

N

a 725

T

𝑧 = 𝑎 − 150 = 241.27𝑚𝑚 𝐴1 = 500(150) = 75,000 𝑚𝑚2 𝐴1 = 125(241.27) = 30,159 𝑚𝑚2 𝑀𝑏𝑛 𝑀𝑏𝑛 𝑀𝑏𝑛 𝑀𝑏𝑛

Figure 2.17 𝑦1 = 725 − 1/2(150) = 650 𝑚𝑚 𝑦2 = 725 − 150 − 1/2(241.27) = 454.37

= 𝐶1 𝑦1 + 2𝐶2 𝑦2 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 2𝐴2 𝑦2 ) = 0.85(28)[75,000 𝑥 650 + 2 𝑥 30,159 𝑥 454.37] = 1812.52 𝑘𝑁 − 𝑚

𝜑𝑀𝑏𝑛 = 0.65 𝑥 1812.52 𝜑𝑀𝑏𝑛 = 𝟏𝟏𝟕𝟖. 𝟏𝟒 𝒌𝑵 − 𝒎

c) Maximum steel area, 𝐴𝑠 𝑚𝑎𝑥 𝐶𝑚𝑎𝑥 =

3 𝑑 = 310.71 𝑚𝑚 7

𝑅𝑒𝑓𝑒𝑟 𝑡𝑜 𝐹𝑖𝑔𝑢𝑟𝑒 2.17 𝑎 = 𝛽1 𝑐𝑚𝑎𝑥 = 264.11 𝑚𝑚

𝑧 = 𝑎 − 150 = 114.11 𝑚𝑚 𝐴1 = 500(150) = 75,000 𝑚𝑚2 𝑦1 = 725 − 1/2(150) = 650 𝑚𝑚 2 𝐴2 = 125(114.11) = 14,263 𝑚𝑚 𝑦2 = 725 − 150 − 1/2(114.11)=517.95 𝑇 = 𝐶1 + 𝐶2 𝐴𝑠 𝑚𝑎𝑥 𝑓𝑦 = 0.85 𝑓 ′ 𝑐 (𝐴1 + 2𝐴2 ) 𝐴𝑠 𝑚𝑎𝑥 (345) = 0.85(28)[75,000 𝑥 650 + 2 𝑥 14,263] 𝐴𝑠 𝑚𝑎𝑥 = 𝟕, 𝟏𝟒𝟐 𝒎𝒎𝟐 d) Maximum moment, 𝑀𝑛 𝑚𝑎𝑥 : 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑛 𝑚𝑎𝑥

= 𝐶1 𝑦1 + 2𝐶2 𝑦2 = 0.85𝑓′𝑐 + (𝐴1 𝑦1 + 2𝐴2 𝑦2 ) = 0.85(28)[75,000 𝑥 650 + 2 𝑥 14,263 𝑥 517.95] = 1511.9 𝑘𝑁 − 𝑚

𝜑 = 0.65 + 0.25

800 − 𝑓𝑦 = 0.824 1000 − 𝑓𝑦

𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.824 𝑥 1511.9 𝜑𝑀𝑛 𝑚𝑎𝑥 = 𝟏𝟐𝟒𝟓. 𝟑 𝒌𝑵 − 𝒎 e) 𝑀𝑢 = 1200 𝑘𝑁 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥 Refer to Figure 2.17 𝑀𝑢 = 𝜑0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 2𝐴2 𝑦2 )

(𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑)

𝜑 = 0.65 + 0.25

𝜑 = 0.65 + 0.25

𝑓𝑠 + 𝑓𝑦 1000 − 𝑓𝑦

𝑓𝑠 = 600

𝑑−𝑐 𝑐

725 − 𝑐 − 345 166.03 𝑐 = + 0.2893 1000 − 345 𝑐

600

𝑧 = 𝑎 − 150 = 0.85𝑐 − 150 𝐴2 = 125𝑧 = 106.25𝑐 − 18,750 𝑦2 = 725 − 150 − 1/2𝑧=575-1/2(0.85c-150) 𝑦2 = 650 − 0.425𝑐 𝑀𝑢 = 𝜑0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 2𝐴2 𝑦2 ) 166.03 1200 𝑥 106 = ( + 0.2893) 0.85(28)[75,000](650) 𝑐 + 2(106.25𝑐 − 18,750)(650 − 0.425𝑐)] 𝑐 = 398.7 𝑚𝑚 𝜑=

166.03 + 0.2893 = 0.706 398.7

𝐴2 = 106.25(398.7) − 18,750 = 23,615 𝑚𝑚2 𝑇=𝐶

𝐴𝑠 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 2𝐴2 ) 𝐴𝑠 (345) = 0.85(28)(75,000 + 2 𝑥 23,615) 𝐴𝑠 = 8,432 𝑚𝑚2

PROBLEM 2.33 Design a singly reinforced rectangular beam to carry dead load moment of 110 kN-m (including self weight) and live load moment of180 kN-m.

Use steel ratio 𝜌 = 0.65𝜌𝑏 and take 𝑑 = 1.9𝑏. Assume 𝑓𝑦 = 276 𝑀𝑃𝑎 and 𝑓′𝑐 = 21 𝑀𝑃𝑎. SOLUTION 𝑀𝑢 = 1.2 𝑀𝐷 + 1.6 𝑀𝐿

𝜌𝑏 =

𝑀𝑢 = 1.2(110) + 1.6(180) 𝑀𝑢 = 420 𝑘𝑁 − 𝑚

0.85𝑓′𝑐 𝛽1 600 = 0.03765 𝑓𝑦 (600 + 𝑓𝑦 )

𝑁𝑜𝑡𝑒: 𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 < 28 𝑀𝑃𝑎

𝜌 = 0.65 𝜌𝑏 = 0.02447 𝜔=

𝜌𝑓𝑦 = 0.322 𝑓′𝑐

𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 − 0.59 𝜔)] = 5.473 𝑀𝑃𝑎

𝐶𝑏 =

600𝑑 600 + 𝑓𝑦

𝐶𝑏 = 0.685𝑑

Note: For singly reinforced rectangular beam, 𝜌 is directly proportional to c. Thus, 𝑐 = 0.65 𝑐𝑏 𝑐 = 0.445𝑑 𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝜑 = 0.65 + 0.25

𝑑 − 0.445𝑑 0.445𝑑 𝑓𝑠 = 747.7 𝑀𝑃𝑎 < 1000 𝑀𝑃𝑎 "𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛" 𝑓𝑠 = 600

𝑓𝑠 − 𝑓𝑦 1000 − 𝑓𝑦

𝜑 = 0.65 + 0.25 𝜑 = 0.813

747.7 − 276 1000 − 276

𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2 420 𝑥 106 = 0.813(5.473)(𝑏)(1.9𝑏)2 𝑏 = 𝟐𝟗𝟕 𝒎𝒎 𝑑 = 1.9𝑏 = 𝟓𝟔𝟒 𝒎𝒎 𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.02447(297)(564) 𝐴𝑠 = 𝟒, 𝟏𝟎𝟎𝟎 𝒎𝒎𝟐

PROBLEM 2.34 Repeat Problem 2.33 using a steel ratio 𝜌 = 0.5𝜌𝑏 SOLUTION 𝑀𝑢 = 420𝑘𝑁 − 𝑚 𝜌𝑏 = 0.03765 𝜌 = 0.5𝜌𝑏 = 0.01883 𝜔=

𝜌𝑓𝑦 (1 − 0.59𝜔) = 4.438 𝑀𝑃𝑎 𝑓 ′𝑐

600𝑑 𝐶𝑏 = 0.685𝑑 600 + 𝑓𝑦 𝑐 = 0.5𝑐𝑏 = 0.34247 𝑑 𝐶𝑏 =

𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝑑 − 0.34247𝑑 0.324247𝑑 𝑓𝑠 = 1152 𝑀𝑃𝑎 > 1000 𝑀𝑃𝑎, 𝜑 = 0.90 𝑓𝑠 = 600 𝑑 −

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏 𝑑2 = 420 𝑥 106 = 0.90(5.473)(𝑏)(1.9𝑏)2 𝑏 = 𝟑𝟎𝟖 𝒎𝒎 𝑑 = 1.9𝑏 = 𝟓𝟖𝟓 𝒎𝒎 𝐴𝑠 = 𝜌𝑏𝑑

𝐴𝑠 = 0.01883(308)(585) 𝐴𝑠 = 𝟑, 𝟑𝟗𝟎 𝒎𝒎𝟐

SUPPLEMENTARY PROBLEMS PROBLEM 2.35 A rectangular beam has 𝑏 = 250 𝑚𝑚, 𝑑 = 350 𝑚𝑚, 𝑓𝑦 = 414 𝑀𝑃𝑎, 𝑓′𝑐 = 20.7 𝑀𝑃𝑎. Determine (a) the maximum design moment if the beam is singly reinforced and (b) the required steel area if the beam is required to carry a dead load moment of 50 kN-m and a live load moment of 30 kN-m. Use the 2001 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝑎) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 148.3 𝑘𝑁𝑚 𝑏) 𝐴𝑠 = 1075 𝑚𝑚2 PROBLEM 2.36 Repeat Problem 2.35 using the 2010 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝑎) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 130.8 𝑘𝑁 − 𝑚 𝑏) 𝐴𝑠 = 1056 𝑚𝑚2 PROBLEM 2.37 Design a rectangular beam reinforced for tension only carry dead load moment of 85 kN-m (including its estimated weight) and a live load of 102 kN-m. Use 𝜌 = 0.6𝜌𝑏 and use d= 1.75b. Assume 𝑓𝑦 = 276 𝑀𝑃𝑎 and 𝑓′𝑐 = 28 𝑀𝑃𝑎. Use the 2001 NSCP 𝐴𝑛𝑠𝑤𝑒𝑟: 𝑏 = 250 𝑚𝑚, 𝑑 = 436 𝑚𝑚, 𝐴𝑠 = 3,273 𝑚𝑚2 PROBLEM 2.38 Repeat Problem 2.37 using the 2010 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟 ∶ 𝑏 = 246 𝑚𝑚, 𝑑 = 430 𝑚𝑚, 𝐴𝑠 = 3182 𝑚𝑚2

PROBLEM 2.39 A reinforced concrete beam has the following properties: Use 2001 NSCP) beam with, 𝑏 = 320𝑚𝑚 effective depth, 𝑑 = 640 𝑚𝑚 concrete strength, 𝑓′𝑐 = 25𝑀𝑃𝑎 reinforcing steel, 𝑓𝑦 = 400 𝑀𝑃𝑎 reinforcing steel modulus, 𝐸𝑠 = 200,000 𝑀𝑃𝑎 service dead load moment 350 = 𝑘𝑁 − 𝑚 a) If the beam is to be designed for a balanced condition, find the required area of steel area reinforcement, design balanced moment, and the corresponding service live load moment. b) Find the maximum steel area, the maximum design moment, and the corresponding service live load moment if the beam is to be designed as singly reinforced. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝑎) 𝐴𝑠𝑏 = 5,549 𝑚𝑚2 , 𝜑𝑀𝑛 = 952.44, 𝑀𝐿 = 272 𝑘𝑁 − 𝑚 𝑏) 𝐴𝑠 𝑚𝑎𝑥 = 4,162 𝑚𝑚2 , 𝜑 𝑀𝑛 𝑚𝑎𝑥 = 775.46, 𝑀𝐿 = 168 𝑘𝑁 − 𝑚 PROBLEM 2.40 Repeat Problem 2.39 using the 2010 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝑎) 5,549 𝑚𝑚2 , 𝜑𝑀𝑛 = 687.87 𝑘𝑁 − 𝑚, 𝑀𝐿 = 167.42 𝑘𝑁 − 𝑚 𝑏) 𝐴𝑠 𝑚𝑎𝑥 = 3,963 𝑚𝑚2 , 𝜑𝑀𝑛 𝑚𝑎𝑥 = 677,7𝑀𝐿 = 161 𝑘𝑁 − 𝑚 PROBLEM 2.41 Calculate the ultimate moment capacity of a rectangular beam with 𝑏 = 350 𝑚𝑚, 𝑑 = 450 𝑚𝑚, 𝐴𝑠 = 5 − 25 𝑚𝑚. Assume 𝑓′𝑐 = 24 𝑀𝑃𝑎. 𝑓𝑦 = 345 𝑀𝑃𝑎. Use 2001 NSCP 𝐴𝑛𝑠𝑤𝑒𝑟: 𝜑𝑀𝑛 = 366.2 𝑘𝑁 − 𝑚 PROBLEM 2.42 Repeat Problem 2.41 using the 2010 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝜑𝑀𝑛 = 366.2 𝑘𝑁 − 𝑚

PROBLEM 2.43 Calculate the ultimate moment capacity of a rectangular beam with 𝑏 = 350 𝑚𝑚, 𝑑 = 540 𝑚𝑚, 𝐴𝑠 = 7 − 28 𝑚𝑚. Assume 𝑓′𝑐 = 24 𝑀𝑃𝑎, 𝑓𝑦 = 345 𝑀𝑃𝑎. Use 2010 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝜑𝑀𝑛 = 582.9 𝑘𝑁 − 𝑚 PROBLEM 2.44 Repeat Problem 2.43 using the 2010 NSCP. 𝐴𝑛𝑠𝑤𝑒𝑟: 𝜑𝑀𝑛 = 514.3 𝑘𝑁 − 𝑚 PROBLEM 2.45 Calculate the ultimate moment capacity of a rectangular beam with 𝑏 = 300 𝑚𝑚, 𝑑 = 500 𝑚𝑚, 𝐴𝑠 = 9 − 28 𝑚𝑚2 . Assume 𝑓′𝑐 = 34 𝑀𝑃𝑎, 𝑓𝑦 = 414 𝑀𝑃𝑎. Use 2010 NSCP 𝐴𝑛𝑠𝑤𝑒𝑟: 𝜑𝑀𝑛 = 729.6 𝑘𝑁 − 𝑚 PROBLEM 2.46 Repeat Problem 2.45 using the 2010 NSCP. 𝐴𝑠𝑛𝑤𝑒𝑟: 𝜑𝑀𝑛 = 522.5 𝑘𝑁 − 𝑚

SOLVED PROBLEMS IN T-BEAMS USING 2001 NSCP PROBLEM 3.1 Determine the effective flange with for symmetrical T-beam with a span of 6 m. The beam width of web is 250 mm, the slab thickness is 120 mm, and the clear distance to adjacent beams is 3m. SOLUTION For symmetrical T-beam, the effective flange width is the smallest of: 1. 1/4 span = 6000/4 = 1500 mm 2. 16𝑡 + 𝑏𝑤 = 16(120) + 250 = 2170 𝑚𝑚 3. clear spacing of beams + 𝑏𝑤 = 3000 + 250 = 3250 𝑚𝑚 Therefore 𝑏𝑓 = 𝟏𝟓𝟎𝟎 𝒎𝒎 PROBLEM 3.2 Given the following elements of a T-beam: Flange width, 𝑏𝑓 = 1200 𝑚𝑚 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑓′𝑐 = 30 𝑀𝑃𝑎 Flange thickness, 𝑡 = 130 𝑚𝑚 𝑆𝑡𝑒𝑒𝑙 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝑓𝑦 = 345 𝑀𝑃𝑎 Width of web, 𝑏𝑤 = 290 𝑚𝑚 Effective depth, 𝑑 = 470 𝑚𝑚 If the beam is reinforced for tension only, determine the ultimate moment capacity when the depth of compression concrete flange equals the flange thickness or 𝑎 = 𝑡. SOLUTION 𝑀𝑛 = 0.8 𝑓′𝑐 𝑏𝑓 𝑎(𝑑 − 𝑎⁄2) 𝑊ℎ𝑒𝑛 𝑎 = 𝑡 Eq. 3-5

𝑀𝑛 = 0.85 𝑓 ′ 𝑐 𝑏𝑓 𝑡(𝑑 − 𝑡⁄2)

𝑀𝑓𝑛 = 0.85(30)(120)(130)(470 −

180 ) 2

𝑀𝑓𝑛 = 1611 𝑘𝑁 − 𝑚 𝜑 𝑀𝑛 = 0.90 𝑥 1611 = 𝟏𝟒𝟓𝟎 𝒌𝒏 − 𝒎

PROBLEM 3.3 Given the following elements of a T-beam: Flange width, 𝑏𝑓 = 900 𝑚𝑚 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 Flange thickness, 𝑡 = 110 𝑚 𝑆𝑡𝑒𝑒𝑙 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ, 𝑓𝑦 = 414 𝑀𝑃𝑎 Width of web, 𝑏𝑤 = 310 𝑚𝑚 Effective depth, 𝑑 = 460 𝑚𝑚 If the beam is reinforced for tension only, determine the following: a) The balanced steel area b) The nominal and ultimate balanced moment capacity c) The maximum steel area d) The nominal and ultimate maximum moment capacity SOLUTION 𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 30 𝑀𝑃𝑎 a) Balanced condition 𝐶𝑏 =

600𝑑 600 + 𝑓𝑦

600(460) 600 + 414 𝐶𝑏 = 272.2 𝑚𝑚

𝐶𝑏 =

d = 460 mm

𝑎 = 𝛽1 𝑐

𝑎 = 0.85(272.2) 𝑎 = 231.4 𝑚𝑚 > 𝑡 = 900mm

t=10 0 C a

z T

=250 mm Figure 3.3

𝑧 = 𝑎 − 𝑡 = 121.4 𝑚𝑚 𝐴1 = 𝑏𝑓 𝑥 𝑡 = 900(110) = 99,000 𝑚𝑚2 𝐴2 = 𝑏𝑤 𝑥 𝑧 = 310(121.4) = 37,622 𝑚𝑚2 𝐴𝑐𝑏 = 𝐴1 + 𝐴2 = 136,622 𝑚𝑚2 𝑇 = 𝐶1 + 𝐶2

𝐴𝑠𝑏 𝑓𝑦 = 0.85 𝑓 ′ 𝑐 ( 𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (414) = 0.85(20.7)99,000 + 37,622) 𝐴𝑠𝑏 = 𝟓, 𝟖𝟎𝟔 𝒎𝒎𝟐 → 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 𝑎𝑟𝑒𝑎

𝑦1 = 𝑑 − 𝑡⁄2 = 405 𝑚𝑚 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 289.3 𝑚𝑚 𝑀𝑏𝑛 𝑀𝑏𝑛 𝑀𝑏𝑛 𝑀𝑏𝑛

= 𝐶2 𝑦1 + 𝐶2 𝑦2 = 0.85 𝑓 ′ 𝑐 ( 𝐴1 𝑦1 + 𝐴2 𝑦2 ) = 0.85(20.7)[99,000(405) + 37,622(289.3)] = 𝟓𝟗𝟕 𝒌𝑵 − 𝒎 → 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑚𝑜𝑚𝑒𝑛𝑡

𝑀𝑏𝑛 = 0.90(897) 𝑀𝑏𝑛 = 𝟖𝟎𝟕. 𝟑 𝒌𝑵 − 𝒎 b)

→ 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑚𝑜𝑚𝑒𝑛𝑡

Maximum steel area and moment. Refer to Figure 3.3.

𝐴𝑠 𝑚𝑎𝑥 = 0.75 𝐴𝑠𝑏

𝐴𝑠 𝑚𝑎𝑥 = 0.75(5806) 𝐴𝑠 𝑚𝑎𝑥 = 𝟒, 𝟑𝟓𝟓 𝒎𝒎𝟐 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑒𝑒𝑙 𝑎𝑟𝑒𝑎

𝐴𝑐 𝑚𝑎𝑥 = 0.75 𝐴𝑐𝑏 𝐴𝑐 𝑚𝑎𝑥 𝐴𝑐 𝑚𝑎𝑥 = 𝐴1 + 𝐴2

𝐴𝑐 𝑚𝑎𝑥 = 0.75(136,622) = 102,466 𝑚𝑚2 > 𝐴1 , 𝑡ℎ𝑢𝑠 𝑎 > 𝑡 102,466=99,000 + 310(z) 𝑧 = 11.2 𝑚𝑚

𝐴2 = 102,466 − 99,000 = 3,466 𝑚𝑚2 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 344.41 𝑚𝑚 𝑀𝑛 𝑚𝑛𝑥 = 𝐶1 𝑦1 + 𝐶2 𝑦2

𝑀𝑛 𝑚𝑎𝑥 = 0.85 𝑓 ′ 𝑐 ( 𝐴1 𝑦1 + 𝐴2 𝑦2 )

𝑀𝑛 𝑚𝑎𝑥 = 0.85(20.7)[99,000(405) + 3,466(289.3)] 𝑀𝑛 𝑚𝑎𝑥 = 𝟕𝟐𝟔. 𝟓 𝒌𝑵 − 𝒎 → 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑚𝑎𝑥 𝑚𝑜𝑚𝑒𝑛𝑡 𝑀𝑛 𝑚𝑎𝑥 = 0.90(726.5) 𝑀𝑛 𝑚𝑎𝑥 = 𝟔𝟓𝟑. 𝟖 𝒌𝑵 − 𝒎 → 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 PROBLEM 3.4 A T-beam has the following properties: 𝑏𝑓 = 820 𝑚𝑚, 𝑏𝑤 = 250 𝑚𝑚, 𝑑 =

470 𝑚𝑚, 𝑡 = 100 𝑚𝑚. Concrete compressive strength 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 and steel area for the following load conditions: a) 𝑀𝐷 = 150𝑘𝑛 − 𝑚, 𝑀𝐿 = 120 𝑘𝑁 − 𝑚 b) 𝑀𝐷 = 175 𝑘𝑁 − 𝑚, 𝑀𝐿 = 190 𝑘𝑁 − 𝑚 SOLUTION

𝛽1 = 0.85 𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 ↑ 𝜑 𝑀𝑛 𝑤ℎ𝑒𝑛 𝑎 = 𝑡 𝑡 𝜑𝑀𝑓𝑛 = 0.85𝑓′𝑐 𝑏𝑓 𝑡 (𝑑 − ) = 545.375 𝑘𝑁 − 𝑚𝑚 2 𝑆𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝜑𝑀𝑛 𝑚𝑎𝑥 : Balanced condition: 600𝑑 𝑐𝑏 = = 278.11 𝑚𝑚 600 + 𝑓𝑦

d = 470 mm

𝑎 < 𝛽1 𝑐𝑏 = 236.39 𝑚𝑚 > 𝑡 = 820mm

t=1 00 C a

z T

FIGURE 3.4

=250 mm

𝑧 = 𝑎 − 𝑡 = 136.39 𝑚𝑚 𝐴1 = 𝑏𝑓 𝑡 = 82,000 𝑚𝑚2 𝐴2 = 𝑏𝑤 𝑧 = 34,098 𝑚𝑚2 𝐴𝑐𝑏 = 𝐴1 + 𝐴2 = 116,098 𝑚𝑚2 Maximum condition: 𝐴𝑐 𝑚𝑎𝑥 = 0.75 𝐴𝑐 𝑏 = 87,073 𝑚𝑚2 > 𝐴1 𝐴2 = 𝐴𝑐 𝑚𝑎𝑥 − 82,000 = 5,073 𝑚𝑚2 𝐴𝑠 𝑧= = 20.29 𝑚𝑚 𝑏𝑤 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 359.85 𝑚𝑚 𝜑𝑀𝑛 𝑚𝑎𝑥 = 𝑀𝑛 + 𝑀𝑛2 = 𝑀𝑓𝑛 + 0,85𝑓′𝑐 𝐴2 𝑦2 𝜑𝑀𝑛 𝑚𝑎𝑥 = 574.28 𝑘𝑁 − 𝑚 a) 𝑀𝐷 = 150 𝑘𝑁 − 𝑚, 𝐴𝑀𝐿 = 120 𝑘𝑁 − 𝑚

𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿 = 414 𝑘𝑁 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥 , 𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 Since 𝑀𝑢 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑀𝑓𝑛 , "a" is less than t.

d = 470 mm

t =100

=820 mm

C d -a/2 T

𝑀𝑢 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 (𝑑 − 𝑎⁄2)

414 𝑥 106 = 0.90(0.85)(20.7)𝑎(820)(470 − 𝑎⁄2)

𝑎 = 73.6 𝑚𝑚 𝑇=𝐶

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 𝐴𝑠 = 2,565 𝑚𝑚2

Minimum 𝐴𝑠 is the smaller of: √𝑓′𝑐 2𝑓𝑦

𝑏𝑤 𝑑 = 646 𝑚𝑚2

√𝑓′𝑐 4𝑓𝑦

𝑏𝑓 𝑑 = 1059 𝑚𝑚2

Thus, 𝐴𝑠 = 𝟐, 𝟓𝟔𝟓 𝒎𝒎𝟐 b) 𝑀𝐷 = 175 𝑘𝑁 − 𝑚, 𝑀𝐿 = 190 𝑘𝑁 − 𝑚 𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿 = 568 𝑘𝑁 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥 , 𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 Since 𝑀𝑢 is more than 𝑀𝑓𝑛, "𝑎"is more than t. = 820mm

d = 470 mm

t=100 C a

z

T

=250 mm 𝑀𝑢 = 𝜑𝑀𝑓𝑛 + 𝜑𝑀𝑛2 568 𝑥 106 = 545.375 + 0.90(0.85)(20.7)(250)𝑧(470 − 100𝑧⁄2) 𝑧 = 15.78 𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 3,946 𝑚𝑚2

𝑇 = 𝐶1 + 𝐶2

𝐴𝑠 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠 (414) = 0.85(20.7)(82,000 + 3946) 𝐴𝑠 = 𝟑, 𝟔𝟓𝟑 𝒎𝒎𝟐

PROBLEM 3.5 Design a T-beam for a floor system for which 𝑏𝑤 = 300 𝑚𝑚 and 𝑑 = 550 𝑚𝑚. The beams are 4.5 m long and spaced at 3 mo.c. The slab thickness is 100 mm. 𝑀𝐷 = 450 𝑘𝑁 − 𝑚(𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑖𝑡𝑠 𝑜𝑤𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 ), 𝑀𝐿 = 350 𝑘𝑁 − 𝑚. 𝑓′𝑐 = 27 𝑀𝑃𝑎, 𝑓𝑦 = 415 𝑀𝑃𝑎. SOLUTION 𝛽1 = 0.85 𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿

𝑀𝑢 = 1.4(450) + 1.7(350) 𝑀𝑢 = 1225 𝑘𝑁 − 𝑚

Solve for bf: 𝑏𝑓 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑜𝑓: 1. L/4 = 1.125 m 2. 16𝑡 + 𝑏𝑤 = 16(100) + 300 = 1,900 𝑚𝑚 3. center-to center spacing of beams = 3 m Thus, 𝑏𝑓 = 1,125 𝑚𝑚 Solve for 𝜑𝑀𝑛 𝑤ℎ𝑒𝑛 𝑎 = 𝑡 = 100𝑚𝑚, 𝜑 = 0.90 𝜑𝑀𝑓𝑛 = 𝜑0.85𝑓′𝑐 𝑡 𝑏𝑓 (𝑑 − 𝑡⁄2) 𝜑𝑀𝑛 = 1161.844 𝑘𝑁 − 𝑚 Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 to determine if compression steel is needed. 600𝑑

𝑐𝑏 = 600+𝑓 = 325.123 𝑚𝑚 𝑦

𝑎 = 𝛽1 𝑐𝑏 = 276.355 𝑚𝑚 > 𝑡

= 1125mm

d =550mm

t=1 00 C a

z 4 5 T =300 mm

𝑀𝑢 = 𝜑𝑀𝑓𝑛 + 𝜑𝑐2 𝑦2

𝑧 1225 𝑥 106 = 1161.844 𝑥 106 + 0.90 𝑥 0.85(27)(300𝑧)(450 − ) 2 z=23.25 mm 𝐴2 = 𝑏𝑤 𝑧 = 6975.02 𝑚𝑚2 𝑇 = 𝑐1 + 𝑐2

𝐴𝑠 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠 (415) = 0.85(27)(112,500 + 6,975.02) 𝐴𝑠 = 6,607 𝑚𝑚2

Minimum 𝐴𝑠 is the smaller value of: √𝑓′𝑐 2𝑓𝑦

𝑏𝑤 𝑑 = 1033 𝑚𝑚2

√𝑓′𝑐 4𝑓𝑦

𝑏𝑓 𝑑 = 1937 𝑚𝑚2

Thus, 𝐴𝑠 = 𝟔, 𝟔𝟎𝟕 𝒎𝒎𝟐 PROBLEM 3.6 Determine the ultimate moment capacity of reinforced concrete T-beam with the following properties: Flange width b = 1500 mm, web width 𝑏𝑤 = 250 𝑚𝑚, effective depth d = 600 mm, slab thickness t = 100 mm. Assume

𝑓′𝑐 = 20.7 𝑀𝑃𝑎 and 𝑓𝑦 = 345 𝑀𝑃𝑎. The beam is reinforced with six 28 mm bars. SOLUTION Solve for balanced 𝐴𝑠 : 600 𝑑 = 380.95 𝑚𝑚 600 + 𝑓𝑦 𝑎 = 𝛽1 𝑐𝑏 = 323.81 > 𝑡 𝑧 = 𝑎 − 𝑡 = 22381 𝑚𝑚 𝐴1 = 𝑏𝑓 𝑡 = 150,000 𝐴2 = 𝑏𝑤 𝑧 = 55,952 𝑚𝑚2 𝑐𝑏 =

= 1500mm

d =600mm

t=100 C

a

z

=250 mm 𝑇=𝐶

𝐴𝑠 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(20.7)(150,00 + 55,952) 𝐴𝑠𝑏 = 10,503 𝜋

Steel area provided, 𝐴𝑠 = 6 𝑥 4 (28)2 = 3,695 𝑚𝑚2 > 𝐴𝑠𝑏 steel yields Therefore, 𝑓𝑠 = 𝑓𝑦 𝐶=𝑇 0.85 𝑓′𝑐 𝐴𝑐 = 𝐴𝑠 𝑓𝑦 0.85(20.7) 𝐴𝑐 = 3,695(345) 𝐴𝑐 = 72,441 < 𝐴1 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 "a" is less than t

=1500 mm

d = 600 mm

t =100

C d -a/2

T 𝐴𝑐 = 𝑎 𝑏𝑓

72,441 = a (1500) 𝑎 = 48.29 𝑚𝑚2

𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 (𝑑 − 𝑎⁄2)

𝑀𝑛 = 0.85(20.7)(48.29)(1500)(600 − 48.29⁄2) 𝑀𝑛 = 733.99 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(733.99) 𝜑𝑀𝑛 = 𝟔𝟔𝟎. 𝟔 𝒌𝑵 − 𝒎 PROBLEM 3.7 Given the following properties of T-beam: Flange width, 𝑏𝑓 = 900 𝑚𝑚 𝑓′𝑐 = 21 𝑀𝑃𝑎 Flange thickness, t=1200 𝑓𝑦 = 345 𝑀𝑃𝑎 Width of web, 𝑏𝑤 = 400 𝑚𝑚 Effective depth, d = 580 mm Service deal load, 𝑀𝐷 = 410 𝑘𝑛 − 𝑚 Determine the safe service live load if the beam is reinforced for tension only with twelve (12) 28-mm-diameter bars. SOLUTION 𝛽1 = 0.85; 𝜑 = .90

= 900mm

d =580mm

t=120 C

a

z

𝜋 𝐴𝑠 = 12 𝑥 (28)2 = 7,389 𝑚𝑚2 4 𝐴1 = 𝑏𝑓 𝑡 = 108,000 𝑚𝑚2

=400 mm

Solve for balance 𝐴𝑠 : 600𝑑 𝑐𝑏 = = 368.25 𝑚𝑚 600 + 𝑓𝑦 𝑐 = 𝛽1 𝑐𝑏 = 313.02 𝑚𝑚 > 𝑡 𝑧 = 𝑎 = 𝑡 = 193.02 𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 77,206 𝑚𝑚2 𝑇=𝐶

𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(21)(108,000 + 77,206) 𝐴𝑠𝑏 = 9,582 𝑚𝑚2

Steel area provided is less than the balanced steel area. Steel yields. 𝐶=𝑇

0.85𝑓′𝑐 𝐴𝑐 = 𝐴𝑠 𝑓𝑦 0.85(21)𝐴𝑐 = 7,389(345) 𝐴𝑐 = 142,813 𝑚𝑚2 > 𝐴1

a>𝑡

= 900mm

d =580mm

t=120 C a

z 46 0 T

=400 mm 𝐴𝑐 = 𝐴1 + 𝐴2

142,813=108,000+𝐴2 𝐴2 = 34,813 𝑚𝑚2

𝐴2 = 𝑏𝑤 𝑧

34,813 = 400z 𝑧 = 87.03 𝑚𝑚

𝑦1 = 𝑑 − 𝑡⁄2 = 520 𝑚𝑚 𝑧 𝑦2 = 𝑑 − 𝑡 − = 416.48 𝑚𝑚 2

𝑀𝑛 = 𝐶1 𝑦1 + 𝐶2 𝑦2 𝑀𝑛 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴2 𝑦2 ) 𝑀𝑛 = 0.85(21)[108,000(520) + 34,813(416.48)] 𝑀𝑛 = 1,261.3 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(1,261.3) 𝜑𝑀𝑛 = 1135.138 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 𝑀𝑢

𝑀𝑢 = 1.4 𝑀𝐷 + 1.7𝑀𝐿 1,135.138 = 1.4(410) + 1.7𝑀𝐿 𝑀𝐿 = 𝟑𝟑𝟎. 𝟎 𝒌𝑵 − 𝒎

PROBLEM 3.8 The section of a reinforced concrete T-beam is shown in Figure 3.5. The beam is reinforced with 10 32-mm-diameter tension bars with 𝑓𝑦 = 415 𝑀𝑃𝑎. Concrete strength 𝑓′𝑐 = 32 𝑀𝑃𝑎. If the total service dead load moment on the beam is 330 kN-m, determine the safe service live load moment.

=500mm

d=530mm

t = 120mm 10-32 mm SOLUTION 𝜋 𝐴𝑠 = 10 𝑥 (32)2 4 𝐴𝑠 = 8,042 𝑚𝑚2 𝐴1 = 𝑏𝑓 𝑡 = 60,000 𝑚𝑚2

𝛽1 = 0.836

t = 120mm

0.05 (32 − 30) 7

Solve for balance 𝐴𝑠 : 600𝑑 600 + 𝑓𝑦 𝐶𝑏 = 313.3 𝑚𝑚 𝑎 = 𝛽1 𝐶𝑏 = 261.83 𝑚𝑚 > 𝑡 𝑧 = 𝑎 − 𝑡 = 141.83 𝑚𝑚

Figure 3.5 =500m m a

d=530mm

𝛽1 = 0.85 −

=320mm

z

𝐶𝑏 =

𝐴2 = 𝑏𝑤 𝑧 = 45,385.5 𝑚𝑚2

=320m m

𝑇=𝐶

𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(21)(60,00 + 45,385.5) 𝐴𝑠𝑏 = 6,907 𝑚𝑚2

Since 𝐴𝑠 > 𝐴𝑠𝑏 , tension steel does not yield

=500m m

t = 120mm

d=530m m

a z

T

𝐴1 = 60,000 𝑚𝑚2 𝐴2 = 𝑏𝑤 𝑧 = 𝑏𝑤 (𝑎 − 𝑡) 𝐴2 = 𝑏𝑤 (𝛽1 𝑐 − 𝑡) 𝑑−𝑐 𝑓𝑠 = 600 𝑐 𝑇 = 𝐶1 + 𝐶2

=320m m

𝐴𝑠 𝑓𝑠 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 530−𝑐

8,042 𝑥 600 𝑐 = 0.85(32)[60,000 + 320(0.836𝑐 − 120)] 𝑐 = 327.95 𝑚𝑚 𝑎 = 𝛽1 𝑐 = 261.83 𝑚𝑚 𝑧 = 𝑎 − 𝑡 = 141.83 𝑚𝑚

𝐴2 = 𝑏𝑤 𝑧 = 49,303 𝑚𝑚2 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 332.97mm 𝑦1 = 𝑑 − 𝑡⁄2 = 470mm

𝑀𝑛 = 𝐶1 𝑦1 + 𝐶2 𝑦2

𝑀𝑛 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴1 𝑦2 ) 𝑀𝑛 = 0.85(32)[60,000(470) + 49,303(332.97)] 𝑀𝑛 = 1,213.56 𝑘𝑁 − 𝑚

𝜑𝑀𝑛 = 0.90(1,213.56) 𝜑𝑀𝑛 = 1,092.2 𝑘𝑁 − 𝑚 𝑀𝑢 = 𝜑𝑀𝑛

𝑀𝑢 = 1.4𝑀𝐷 + 1.7 𝑀𝐿 1,092 = 1.4(330) + 1.7 𝑀𝐿 𝑀𝐿 = 𝟑𝟕𝟎. 𝟕 𝒌𝑵 − 𝒎

SOLVED PROBLEMS IN T-BEAMS USING 2010 NSCP PROBLEM 3.9 Repeat Problem 3.3 using the 2010 NSCP. SOLUTION Given: 𝑏𝑓 = 900 𝑚𝑚 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 𝑡 = 110 𝑚𝑚 𝑓𝑦 = 414 𝑀𝑃𝑎 𝑏𝑤 = 3210 𝑚𝑚 𝑑 = 460 𝑚𝑚 𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓 ′ 𝑐 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 28 𝑀𝑃𝑎 a) Balanced condition, 𝜑 = 0.65 𝑐𝑏 =

600 𝑑 600 + 𝑓𝑦

600(460) 600 + 414 𝑐𝑏 = 272.2 𝑚𝑚 𝑐𝑏 =

𝑎 = 𝛽1 𝑐

𝑎 = 0.85(272.2) 𝑎 = 231.4 𝑚𝑚 > 𝑡

= 900mm

d = 460 mm

t=110 C a

z

T =310 mm Figure 3.6

𝑧 = 𝑎 − 𝑡 = 121.4 𝑚𝑚 𝐴1 = 𝑏𝑓 𝑥 𝑡 = 900(110) = 99,000 𝑚𝑚2 𝐴2 = 𝑏𝑤 𝑥 𝑧 = 310(121.4) = 37,622 𝑚𝑚2 𝐴𝑐𝑏 = 𝐴1 + 𝐴2 = 136,622 𝑚𝑚2 𝑇 = 𝑐1 + 𝑐2

𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (414) = 0.85(20.7)(99,000 + 37,622) 𝐴𝑠𝑏 = 𝟓, 𝟖𝟎𝟔 𝒎𝒎𝟐 → 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 𝑎𝑟𝑒𝑎

𝑦1 = 𝑑 − 𝑡⁄2 = 405 𝑚𝑚 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 289.3 𝑚𝑚 𝑀𝑏𝑛 = 𝑐1 𝑦1 + 𝑐2 𝑦2 𝑀𝑏𝑛 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴2 𝑦2 ) 𝑀𝑏𝑛 = 0.85(20.7)[99,000(405)37,622(289.3)] 𝑀𝑏𝑛 = 𝟖𝟗𝟕 𝒌𝑵 − 𝒎 → 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝜑𝑀𝑏𝑛 = 0.65(897) 𝜑𝑀𝑏𝑛 = 𝟓𝟖𝟑 𝒌𝑵 − 𝒎 → 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 b) Maximum steel area and moment. Refer to Figure 3.6. 800 − 𝑓𝑦 3 𝑐 = 𝑑 = 197.14 𝑚𝑚; 𝜑 = 0.65 + 0.25 = 0.815 7 1000 − 𝑓𝑦 𝑎 = 𝛽1 𝑐

𝑎 = 0.85(197.14) 𝑎 = 167.6 𝑚𝑚

𝑧 = 𝑎 − 𝑡 = 57.571 𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 310(57.6) = 17,847 𝑚𝑚2 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 321.21 𝑚𝑚 𝑇 = 𝑐1 + 𝑐2 𝐴𝑠 𝑚𝑎𝑥

𝐴𝑠 𝑚𝑎𝑥 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠 𝑚𝑎𝑥 (414) = 0.85𝑓 ′ 𝑐 (99,000 + 17,847) = 𝟒𝟗𝟔𝟔 𝒎𝒎𝟐 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠𝑡𝑒𝑒𝑙 𝑎𝑟𝑒𝑎

𝑀𝑛 𝑚𝑎𝑥 = 𝑐1 𝑦1 + 𝑐2 𝑦2 𝑀𝑛 𝑚𝑎𝑥 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴2 𝑦2 ) 𝑀𝑛 𝑚𝑎𝑥 = 0.85(20.7)[99,000(415) + 17,847(321.2)] 𝑀𝑛 𝑚𝑎𝑥 = 𝟖𝟎𝟔. 𝟑𝟒 𝒌𝑵 − 𝒎 → 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑚𝑎𝑥 𝑚𝑜𝑚𝑒𝑛𝑡 𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.815(806.34) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 𝟔𝟓𝟔. 𝟗 𝒌𝑵 − 𝒎 → 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 PROBLEM 3.10 Repeat Problem 3.2 using the 2010 NSCP. SOLUTION Given: 𝑏𝑓 = 1200 𝑚𝑚 𝑡 = 130 𝑚𝑚 𝑑 = 470 𝑚𝑚

𝑏𝑤 = 290 𝑚𝑚 𝑓′𝑐 = 30 𝑀𝑃𝑎 𝑓𝑦 = 345𝑀𝑃𝑎

𝑀𝑓𝑛 = 0.85𝑓′𝑐 𝑡 𝑏𝑓(𝑑 − 𝑡⁄2)

𝑀𝑓𝑛 = 0.85(30)(1200)(130)(470 − 130⁄2) 𝑀𝑓𝑛 = 1611𝑘𝑁 − 𝑚 Solving for 𝜑: 𝑎 = 130 𝑚𝑚 0.05 ′ 𝛽1 = 0.85 − (𝑓 𝑐 − 28) = 0.836 7 𝑐 = 𝑎⁄𝛽 = 155.56 𝑚𝑚 1 𝑑−𝑐 𝑓𝑠 = 600 = 1213 𝑀𝑃𝑎 > 1000 𝑀𝑃 𝑐 𝜑𝑀𝑓𝑛 = 090(1611) 𝜑𝑀𝑓𝑛 = 𝟏𝟒𝟓𝟎 𝒌𝑵 − 𝒎

"𝑡𝑒𝑛𝑠𝑖𝑜𝑛 − 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠, 𝜑 = 0.90"

PROBLEM 3.11 Repeat Problem 3.4 using the 2010 NSCP. Additional questions: c) Find the required steel area if 𝑀𝐷 = 195 𝑘𝑁 − 𝑚 and 𝑀𝐿 = 210 𝑘𝑁 − 𝑚. d) Find the maximum design moment so that section is tensioncontrolled if it is reinforced for tension only.

SOLUTION Given: 𝑏𝑓 = 820 𝑚𝑚 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 𝑏𝑤 = 250 𝑚𝑚 𝑓𝑦 = 414 𝑀𝑃𝑎 𝑑 = 470 𝑚𝑚 𝑡 = 100 𝑚𝑚 𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 < 28 < 𝑀𝑃𝑎 Solve for 𝜑𝑀𝑛 when 𝑎 = 𝑡: 𝑀𝑓𝑛 = 0.85𝑓′𝑐 𝑏𝑓 𝑡(𝑑 − 𝑡⁄2) = 605.97 𝑘𝑁 − 𝑚𝑚 𝑐 = 𝑎⁄𝛽 = 117.65 𝑚𝑚 1

𝑓𝑠 = 600

𝑑−𝑐 = 1797𝑀𝑃𝑎 > 1000𝑀𝑃𝑎, 𝜑 = 0.90 𝑐

𝜑𝑀𝑓𝑛 = 𝟓𝟒𝟓. 𝟑𝟕𝟓 𝒌𝑵 − 𝒎 Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 : 3 𝐶𝑚𝑎𝑥 = 𝑑 = 201.43 𝑚𝑚 7 800 − 𝑓𝑦 = 0.815 1000 − 𝑓𝑦 = 171.21 𝑚𝑚 > 𝑡

𝜑 = 0.65 + 0.25 𝑎 = 𝛽1 𝑐𝑚𝑎𝑥

= 820mm

d = 470 mm

t=100 C

a

z

T 𝑧 = 𝑎 − 𝑡 = 71.21 𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 17,803.6 𝑚𝑚2 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 334.39𝑚𝑚

=250 mm

𝑀𝑛 𝑚𝑎𝑥 = 𝑀𝑓𝑛 + 0.85𝑓′𝑐 𝐴2 𝑦2 𝑀𝑛 𝑚𝑎𝑥 = 710.72 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 𝑚𝑎𝑥 = 579 𝑘𝑁 − 𝑚 a) 𝑀𝐷 = 150 𝑘𝑁 − 𝑚, 𝑀𝐿 = 120 𝑘𝑁 − 𝑚 𝑀𝑢 = 1.2𝑀𝐷 + 1.6 𝑀𝐿 = 372 𝑘𝑁 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥, 𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 Since 𝑀𝑢 is less than 𝜑𝑀𝑓𝑛 ,”a” is less than t.

d = 470 mm

t =100

=820 mm

C d -a/2 T

Assume 𝜑 = 0.90 𝑀𝑢 = 𝜑0.85𝑓′𝑐 𝑎 𝑏𝑓 (𝑑 − 𝑎⁄2) 372 𝑥 106 = 0.90(0.85)(20.7)𝑎(820)(470 − 𝑎⁄2) 𝑎 = 65.52 𝑚𝑚 𝑐 = 𝑎⁄𝛽 = 77.08𝑚𝑚 1 𝑑−𝑐 𝑓𝑠 = 600 = 3,058 𝑀𝑃𝑎 > 1000 𝑀𝑃𝑎, 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠, 𝜑 = 0.90 𝑐 𝑇=𝐶

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 𝐴𝑠 (345) = 0.85(20.7)(65.52)(820) 𝐴𝑠 = 2,283 𝑚𝑚2

Minimum 𝐴𝑠 is the smaller value of: 𝑓′𝑐 𝑏 𝑑 = 646 𝑚𝑚2 2𝑓𝑦 𝑤

√𝑓′𝑐 𝑏 𝑑 = 1059 𝑚𝑚2 4𝑓𝑦 𝑓

Thus, 𝐴𝑠 = 𝟐, 𝟐𝟖𝟑 𝒎𝒎𝟐 b) 𝑀𝐷 = 175 𝑘𝑁 − 𝑚, 𝑀𝐿 = 190 𝑘𝑁 − 𝑚 𝑀𝑢 = 1.2 𝑀𝐷 + 1.6 𝑀𝐿 = 514 𝑘𝑁 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥 , 𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 Since 𝑀𝑢 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝜑𝑀𝑓𝑛 , "𝑎"is less than t. Assume 𝜑 = 0.90 𝑀𝑢 = 𝜑0.85𝑓′𝑐 𝑎 𝑏𝑓 (𝑑 − 𝑎⁄2) 514 𝑥 106 = 0.90(0.85)(20.7)𝑎(820)(470 − 𝑎⁄2) 𝑎 = 93.53 𝑚𝑚 𝑐 = 𝑎⁄𝛽 = 110.03 𝑚𝑚 1

t =100

d = 470 mm

=820 mm

C d -a/2

T 𝑓𝑠 = 600 𝑇=𝐶

𝑑−𝑐 = 1,963 𝑀𝑃𝑎 > 1000𝑀𝑃𝑎, 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠, 𝜑 = 0.90 𝑐 𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 𝐴𝑠 (345) = 0.85(20.7)(93.53)(820) 𝐴𝑠 = 𝟑, 𝟐𝟓𝟗 𝒎𝒎𝟐

c) 𝑀𝐷 = 195 𝑘𝑁 − 𝑚, 𝑀𝐿 = 210 𝑘𝑁 − 𝑚 𝑀𝑢 = 1.2𝑀𝐷 + 1.6𝑀𝐿 = 570 𝑘𝑛 − 𝑚 < 𝜑𝑀𝑛 𝑚𝑎𝑥 , 𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 Since 𝑀𝑢 𝑖𝑠 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝜑𝑀𝑢 , "𝑎"is more than t.

= 820mm

d =470mm

t=10 0 C

a

z

T =250 mm

Assume 𝜑 = 0.90 𝑀𝑢 = 𝜑𝑀𝑓𝑛 + 𝜑𝑀𝑛2 570 𝑥 106 = 545.375 + 0.90(0.85)(20.7)(250)𝑧(470 − 100 − 𝑧⁄2) 𝑧 = 17.05 𝑚𝑚 𝑎 = 𝑡 + 𝑧 = 117.05 𝑚𝑚; 𝑐 = 𝑎⁄𝛽1 = 137.7𝑚𝑚 𝑑−𝑐 = 1448𝑀𝑃𝑎 > 1000𝑀𝑃𝑎, 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠, 𝜑 = 0.90 𝑐 𝐴2 = 𝑏𝑤 𝑧 = 3,908 𝑚𝑚2 𝑓𝑠 = 600

𝐴𝑠 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 )

𝑇 = 𝐶1 + 𝐶2 𝐴𝑠 = 3,666 𝑚𝑚2 3

d) 𝑐 = 𝑏 𝑑 = 176.25 𝑚𝑚, 𝜑 = 0.90 𝑎 = 𝛽1 𝑐 = 149.81 𝑚𝑚 > 𝑡

= 820mm

d = 470 mm

t=100 C

a

z

T =250 mm 𝑧 = 𝑎 − 𝑡 = 49.81𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 12,453 𝑚𝑚2 𝑧 𝑦2 = 𝑑 − 𝑡 − = 3450.9𝑚𝑚 2

𝑀𝑡𝑛 = 𝑀𝑓𝑛 + 0.85 𝑓′𝑐 𝐴2 𝑦2 𝑀𝑡𝑛 = 681.59 𝑘𝑁 − 𝑚 𝜑𝑀𝑡𝑛 = 𝟔𝟏𝟑. 𝟒 𝒌𝑵 − 𝒎 Note: If 𝑀𝑢 is less than or equal to 𝜑𝑀𝑛 , the beam is tension-controlled. PROBLEM 3.12 Repeat Problem 3.6 using the 2010 NSCP. SOLUTION Given: 𝑏𝑓 = 1500𝑚𝑚 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 𝑏𝑤 = 250 𝑚𝑚 𝑓𝑦 = 345 𝑀𝑃𝑎 𝑑 = 600 𝑚𝑚 𝛽1 = 0.85 𝐴𝑠 = 6 − 28 𝑚𝑚 = 3,694 𝑚𝑚2 Solve for balanced 𝐴𝑠 : 600𝑑

𝑐𝑏 = 600+𝑓 = 380.95 𝑚𝑚 𝑦

𝑎 = 𝛽1 𝑐𝑏 = 323.81 > 𝑡 𝐴1 = 𝑏𝑓 𝑡 = 150,000 𝐴2 = 𝑏𝑤 𝑧 = 55,952 𝑚𝑚2

= 1500mm

d =600mm

t=100 C

a

z

=400 mm

𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(20.7)(150,000 + 55,952) 𝐴𝑠𝑏 = 10,503

𝑇=𝐶

Steel area provided, 𝐴𝑠 = 6 𝑥

𝜋 4

(28)2 = 3,695 𝑚𝑚2 < 𝐴𝑠𝑏 "𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠"

therefore, 𝑓𝑠 = 𝑓𝑦 𝐶=𝑇

0.85𝑓′𝑐 𝐴𝑐 = 𝐴𝑠 𝑓𝑦 0.85(20.7)𝐴𝑐 = 3,695(345) 𝐴𝑐 = 72,441 < 𝐴1 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 "𝑎" is less than t

t =100

d = 600 mm

=1500 mm

C d -a/2 T

𝐴𝑐 = 𝑎 𝑏𝑓

72,441 = 𝑎(1500) 𝑎 = 48.29 𝑚𝑚2

Solve for 𝜑: 𝑐 = 𝑎⁄𝛽 = 56.82 𝑚𝑚 1 𝑓𝑠 = 600

𝑑−𝑐 = 5,736𝑀𝑃𝑎 > 1000𝑀𝑃𝑎 𝑐

therefore 𝜑 = 0.90

𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠

𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 (𝑑 − 𝑎⁄2) 𝑀𝑛 = 0.85(20.7)(48.29)(1500)(600 − 48.29⁄2) 𝑀𝑛 = 733.99 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(733.99) 𝜑𝑀𝑛 = 𝟔𝟔𝟎. 𝟔 𝒌𝑵 − 𝒎 PROBLEM 3.13 Repeat Problem 3.7 using 2010 NSCP. SOLUTION Given the following properties of a T-beam: Flange width, 𝑏𝑓 = 900 𝑚𝑚 𝑓′𝑐 = 21 𝑀𝑃𝑎 Flange thickness, 𝑡 = 120 𝑚𝑚 𝑓𝑦 = 345 𝑀𝑃𝑎 Width of web, 𝑏𝑤 = 400 𝑚𝑚 𝐴𝑠 = 7,389 𝑚𝑚2 Effective depth, 𝑑 = 580 𝑚𝑚 Service deal load, 𝑀𝐷 = 410 𝑘𝑁 − 𝑚 𝛽1 = 0.85; 𝜑 = 0.90 𝜋 𝐴𝑠 = 12 𝑥 4 (28)2 = 7,389 𝑚𝑚2 𝐴1 = 𝑏𝑓 𝑡 = 108,000 𝑚𝑚2

𝐶𝑏 =

600𝑑 = 368.25 𝑚𝑚 600 + 𝑓𝑦

d =580mm

Solve for balance 𝐴𝑠 :

= 900mm

t=120 C

a

z

𝑎 = 𝛽1 𝑐𝑏 = 313.02 𝑚𝑚 > 𝑡 𝑧 = 𝑎 − 𝑡 = 193.02 𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 77,206 𝑚𝑚2 𝑇=𝐶

𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(21)(108,000 + 77,206) 𝐴𝑠𝑏 = 9,582 𝑚𝑚2

=400 mm

Steel area provided is less than the balanced steel area. Steel yields. 𝐶=𝑇

0.85𝑓′𝑐 𝐴𝑐 = 𝐴𝑠 𝑓𝑦 0.85(21)𝐴𝑐 = 7,389(345) 𝐴𝑐 = 142,813 𝑚𝑚2 > 𝐴1 = 900mm

d = 580 mm

t=120 C

a

“a” >t

z 46 0 T =400 mm

𝐴𝑐 = 𝐴1 + 𝐴2 𝐴2 = 𝑏𝑤 𝑧

142,813 = 108,000 + 𝐴2 𝐴2 = 34,813 𝑚𝑚2 34,813 = 400𝑧 𝑧 = 87.03 𝑚𝑚

𝑦1 = 𝑑 − 𝑡⁄2 = 520 𝑚𝑚 𝑦2 = 𝑑 − 𝑡 − 𝑧⁄2 = 416.48 𝑚𝑚 𝑀𝑛 𝑀𝑛 𝑀𝑛 𝑀𝑛

= 𝐶1 𝑦1 + 𝐶2 𝑦2 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴2 𝑦2 ) = 0.85(21)[108,000(520) + 34,813(416.48)] = 1,261.6 𝑘𝑛 − 𝑚

Solve for 𝜑: 𝑎 = 𝑡 + 𝑧 = 203.03 𝑚𝑚

𝑐 = 𝑎⁄𝛽 = 243.57𝑚𝑚 1

𝑓𝑠 = 600

𝑑−𝑐 = 828.76 𝑀𝑃𝑎 < 1000 𝑀𝑃𝑎 𝑐

Since 𝑓𝑦 < 𝑓𝑠 < 1000 𝑀𝑃𝑎,Transition region 𝜑 = 0.65 + 0.25

𝑓𝑠 − 𝑓𝑦 = 0.8346 1000 − 𝑓𝑦

𝜑𝑀𝑛 = 0.8346(1,261.3) 𝜑𝑀𝑛 = 1,052.703 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 𝑀𝑢

𝑀𝑢 = 1.2 𝑀𝐷 + .6 𝑀𝐿 1,052.703 = 1.2(410) + 1.7 𝑀𝐿 𝑀𝐿 = 𝟑𝟓𝟎. 𝟒𝟒 𝒌𝑵 − 𝒎

PROBLEM 3.14 Repeat Problem 3.8 using 2010 NSCP.

t = 120mm 10-32 mm

=320mm

Figure 3.7

d=530mm

=500mm

t =120 mm

=500mm

d=530mm

a Z

SOLUTION 𝐴𝑠 = 10 𝑥

𝜋 (32)2 4

=320mm

𝐴𝑠 = 8,042 𝑚𝑚2 𝐴1 = 𝑏𝑓 𝑡 = 60,000 𝑚𝑚2 𝛽1 = 0.85 − 𝛽1 = 0.821

0.05 (32 − 28) 7

Solve for balanced 𝐴𝑠 : 600𝑑 600 + 𝑓𝑦 𝑎 = 313.3 𝑚𝑚 𝑎 = 𝛽1 𝑐𝑏 = 257.35 𝑚𝑚 > 𝑡 𝑧 = 𝑎 − 𝑡 = 137.35 𝑚𝑚 𝐴2 = 𝑏𝑤 𝑧 = 43,953 𝑚𝑚2 𝑐𝑏 =

𝑇=𝐶

𝐴𝑠𝑏 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠𝑏 (345) = 0.85(21)(60,000 + 43,953) 𝐴𝑠𝑏 = 6,813 𝑚𝑚2

Since 𝐴𝑠 > 𝐴𝑠𝑏 , 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑦𝑖𝑒𝑙𝑑

=500m m

t =120 mm

d=530mm

a Z

T

=320m m

𝜑 = 0.65 𝑐𝑝𝑚𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙𝑠

𝐴1 = 60,000 𝑚𝑚2 𝐴2 = 𝑏𝑤 𝑧 = 𝑏𝑤 (𝑎 − 𝑡) = 𝑏𝑤 (𝛽1 𝑐 − 𝑡) 𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝑇 = 𝐶1 + 𝐶2

𝐴𝑠 𝑓𝑠 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 )

530 − 𝑐 = 0.85(32)[60,000 + 320(0.821𝑐 − 1200)] 𝑐 𝑐 = 329.27 𝑚𝑚 8,042 𝑥 600

𝑎 = 𝛽1 𝑐 = 270.47 𝑚𝑚 𝑧 = 𝑎 − 𝑡 = 150.47 𝑚𝑚 𝑦1 = 𝑑 − 𝑡⁄2 = 470 𝑚𝑚 𝑀𝑛 = 𝐶1 𝑦1 + 𝑐2 𝑦2

𝐴2 = 𝑏𝑤 𝑧 = 48,151 𝑚𝑚2 𝑧 𝑦2 = 𝑑 − 𝑡 − 2 = 334.76 𝑚𝑚 𝑀𝑛 = 0.85𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴2 𝑦2 ) 𝑀𝑛 = 0.85(32)[60,000(470) + 48,151(334.76)] 𝑀𝑛 = 1,205.48 𝑘𝑁 − 𝑚

𝜑𝑀𝑛 = 0.65(1,205.48) 𝜑𝑀𝑛 = 783.56 𝑘𝑛 − 𝑚 𝑀𝑢 = 𝜑𝑀𝑛

𝑀𝑢 = 1.2 𝑀𝐷 + 1.6𝑀𝐿 78.56 = 1.2(330) + 1.6 𝑀𝐿 𝑀𝐿 = 𝟐𝟒𝟐. 𝟐𝟑 𝒌𝑵 − 𝒎

DESIGN PROBLEMS Doubly Reinforced Beams PROBLEM 3.15 a .305-mm wide rectangular beam has an overall depth of 560 mm. The beam is reinforced with four 25-mm-diameter compression bars. The centroid fiber. Assume 𝑓𝑦 = 415 𝑀𝑃𝑎 and 𝑓′𝑐 = 29 𝑀𝑃𝑎. Determine the following: a) The balanced tension steel area and the nominal and ultimate balanced moment. b) The maximum tension steel area and the nominal and ultimate maximum moment. SOLUTION 305 mm

d=490 mm d-d’420

70 mm

a

4-25 mm

+

=

70 mm 𝛽1 = 0.85 a) Balanced condition 𝑐𝑏 =

𝑎 = 𝛽1 𝑐𝑏

600𝑑 600 + 𝑓𝑦

600(490) 600 + 415 𝑐𝑏 = 289.66 𝑚𝑚

𝑐𝑏 =

𝑎 = 0.85(289.66) 𝑎 = 246.21 𝑚𝑚

𝑓𝑠𝑐 = 600

𝑑−𝑐

𝑓𝑠𝑐 = 600

𝑐

289.66−70 289.66

𝑓𝑠𝑐 = 455 𝑀𝑃𝑎 > 𝑓𝑦 yield 𝑓𝑠𝑐 = 𝑓𝑦 𝑇1 = 𝐶𝑐

𝑇2 = 𝐶′𝑠

𝐴𝑠1 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏 𝐴𝑠1 (415) = 0.85(29)(246.21)(305) 𝐴𝑠1 = 4,460 𝑚𝑚2 𝐴𝑠2 𝑓𝑦 = 𝐴′𝑠 𝑓𝑦 𝐴𝑠2 = 1,964 𝑚𝑚2

Balanced steel area, 𝐴𝑠𝑏 = 𝐴𝑠1 + 𝐴𝑠2 = 𝟔, 𝟐𝟒𝟐 𝒎𝒎𝟐 𝑀𝑛𝑏 = 𝐶𝑐 (𝑑 − 𝑎⁄2) + 𝐶 ′ 𝑠 (𝑑 − 𝑑 ′ ) 𝑎 𝑀𝑛𝑏 = 0.85𝑇𝑓′𝑐 𝑎 𝑏 (𝑑 − ) + 𝐴′ 𝑠 𝑓𝑦 (𝑑 − 𝑑 ′ ) 2 𝑀𝑛𝑏 = 0.85(29)(246.21)(305)(490 − 246.21⁄2) 𝑀𝑛𝑏 = 𝟏, 𝟎𝟐𝟏. 𝟒 𝒌𝑵 − 𝒎 𝜑𝑀𝑛𝑏 = 0.90(1,021.4) = 𝟗𝟏𝟗. 𝟐𝟒 𝒌𝑵 − 𝒎 b) Maximum tension steel area: According to Section 410.4.3, for members with compression reinforcement, the portion of 𝜌𝑏 equalized by compression reinforcement need not be reduced by the 0.75 factor. 𝐴𝑠1 = 0.75𝐴𝑠1 𝐴𝑠2 = 1,964 𝑚𝑚2 𝐴𝑠 𝑚𝑎𝑥 = 𝐴𝑠1 𝑚𝑎𝑥 + 𝐴𝑠2

𝐴𝑠1 𝑚𝑎𝑥 = 0.75(4,460) 𝐴𝑠1 𝑚𝑎𝑥 = 3,345 𝑚𝑚2

𝐴𝑠 𝑚𝑎𝑥 = 3,345 + 1,964 𝐴𝑠 𝑚𝑎𝑥 = 𝟓, 𝟑𝟎𝟗 𝒎𝒎𝟐

𝐶𝑐 = 𝑇1

0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠1𝑚𝑎𝑥 𝑓𝑦 0.85(29)(𝑎)(305) = 3,345(415) 𝑎 = 184.7 𝑚𝑚

𝑐 = 𝑎/𝛽1

𝑓′𝑠 = 600

𝑐 = 184.7/0.85 𝑐 = 217.2 𝑚𝑚 𝑐−𝑑′ 𝑐

217.2−70

𝑓′𝑠 = 600 217.2 𝑓′𝑠 = 406.7 𝑀𝑃𝑎 < 𝑓𝑦

(𝑤𝑖𝑙𝑙 𝑛𝑜𝑡 𝑦𝑖𝑒𝑙𝑑)

𝑀𝑛 𝑚𝑎𝑥 = 𝐶𝑐 (𝑑 − 𝑎⁄2) + 𝐶 ′ 𝑠 (𝑑 − 𝑑′ ) 𝑀𝑛 𝑚𝑎𝑥 = 0.85𝑓′𝑐 𝑎 𝑏(𝑑 − 𝑎⁄2) + 𝐴′ 𝑠 𝑓 ′ 𝑠 (𝑑 − 𝑑 ′ ) 𝑀𝑛 𝑚𝑎𝑥 = 0.85(29)(184.7)(305)(490 − 184.7⁄2) +1964(406.7)(490 − 70) 𝑀𝑛 𝑚𝑎𝑥 = 𝟖𝟖𝟕. 𝟒𝟓 𝒌𝑵 − 𝒎 𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90(887.45) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 𝟕𝟗𝟖. 𝟕 𝒌𝑵 − 𝒎 PROBLEM 3.16 (CE NOVEMBER 2009) A reinforced concrete beam has width of 300 mm and effective depth of 460 mm. The beam is reinforced with 2-28 mm compression bars placed 70 mm from extreme concrete. Concrete strength 𝑓′𝑐 = 35 𝑀𝑃𝑎 and steel strength 𝑓𝑦 = 345 𝑀𝑃𝑎. a) What is the balanced steel area considering the contribution of the compression steel? b) What is the maximum tension steel area allowed by the code?

SOLUTION 𝛽1 = 0.85 −

0.05 (35 − 30) = 0.814 7

𝐴′𝑠 =

𝜋 (28)2 𝑥 2 = 1,232 𝑚𝑚2 4

a) Balanced condition considering compression steel: 𝑐𝑏 =

600 𝑑 600 + 𝑓𝑦

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑐𝑏 =

600(460) 600 + 345

𝑓′𝑠 = 600

292 − 70 292

𝑓′𝑠 = 456 𝑀𝑃𝑎 > 𝑓𝑦 , 𝑡ℎ𝑢𝑠 𝑓′𝑠 = 𝑓𝑦 = 345 𝑀𝑃𝑎 𝐶𝑐 + 𝐶𝑠 = 𝑇 0.85 𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓′𝑠 = 𝐴𝑠 𝑓𝑦 0.85(35)(0.814 𝑥 292)(300) + 1232(345) = 𝐴𝑠 (345) 𝐴𝑠 = 𝟕, 𝟑𝟖𝟒 𝒎𝒎𝟐 b) Maximum steel area: For rectangular beams: 𝐶𝑚𝑎𝑥 = 0.75 𝑐𝑏 = 0.75(292) 𝐶𝑚𝑎𝑥 = 219.05 𝑚𝑚 𝑎 = 𝛽1 𝑐𝑚𝑎𝑥 = 178.37 𝑚𝑚 𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑓′𝑠 = 600

219.05 − 70 219.05

𝑓′𝑠 = 408 𝑀𝑃𝑎 > 𝑓𝑦 , 𝑡ℎ𝑢𝑠 𝑓′𝑠 = 𝑓𝑦 = 345 𝑀𝑃𝑎 𝐶𝑐 + 𝐶𝑠 = 𝑇

0.85𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓′𝑠 = 𝐴𝑠 𝑓𝑦 0.85(35)(178.37)(300) +1232(345) = 𝐴𝑠 (345) 𝐴𝑠 = 𝟓, 𝟖𝟒𝟔 𝒎𝒎𝟑

PROBLEM 3.17 A rectangular beam has b=300 mm and d= 490 mm. Concrete compressive strength 𝑓′𝑐 = 27.6 𝑀𝑃𝑎 and steel yield strength 𝑓𝑦 = 276 𝑀𝑃𝑎. Compressive steel if required shall have its centroid 60 mm from extreme concrete fiber. Calculate the required tension steel area if the factored moment 𝑀𝑢 is 620 kN-m. SOLUTION This is the same problem in Chapter 2. Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 : 𝑐𝑏 =

600𝑑 = 335.616 𝑚𝑚 600 + 𝑓𝑦

𝑎𝑏 = 𝛽1 𝑐𝑏 = 285.27 𝑚𝑚 𝑎 𝑀𝑛 𝑚𝑎𝑥 = 0.85𝑓′𝑐 𝑎 𝑏(𝑑 − ) 2 𝑀𝑛 𝑚𝑎𝑥 = 0.85(27.6)(213.96)(300)(490 − 213.96⁄2) 𝑀𝑛 𝑚𝑎𝑥 = 576.76 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90(576.76) = 519 𝑘𝑁 − 𝑚 Since 𝑀𝑢 = 620 𝑘𝑛 − 𝑚 > 𝜑𝑀𝑛 𝑚𝑎𝑥 , the beam must be doubly reinforced.

b d ’

ca

=

d – a/2

+

d – d’

𝑀𝑛1 = 𝑀𝑛 𝑚𝑎𝑥 = 576.76 𝑘𝑁 − 𝑚 𝑀𝑛2 =

𝑀𝑢 − 𝑀𝑛1 𝜑

620 0.90 − 576.76 = 112.13 𝑘𝑁 − 𝑚

𝑀𝑛2 = 𝑀𝑛2

𝑎 = 213.96 𝑚𝑚 𝐴𝑠1 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏

𝐴𝑠1 (276) = 0.85(27.6)(213.96)(300) 𝐴𝑠1 = 5456 𝑚𝑚2

Note: 𝐴𝑠1 = 𝐴𝑠 𝑚𝑎𝑥 Solve for 𝑓′𝑠 : 𝑐 = 𝑎⁄𝛽 = 251.71 𝑚𝑚 1

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

251.71 − 60 251.71 𝑓′𝑠 = 457 𝑀𝑃𝑎 > 𝑓𝑦 𝑓′𝑠 = 600

Compression steel yields

𝑈𝑠𝑒 𝑓′𝑠 = 𝑓𝑦 𝑀𝑛2 = 𝑇2 (𝑑 − 𝑑′ )

112.13 𝑥 106 = 𝐴𝑠2 (276)(490 − 60) 𝐴𝑠2 = 945 𝑚𝑚2

Tension steel area, 𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2 = 𝟔𝟒𝟎𝟏 𝒎𝒎𝟐 Compression steel: 𝑐′𝑠 = 𝑇2

𝐴′𝑠 𝑓𝑦 = 𝐴𝑠2 𝑓𝑦 𝐴′𝑠 = 𝐴𝑠2 𝐴′𝑠 = 𝟗𝟒𝟓 𝒎𝒎𝟐

PROBLEM 3.18 A rectangular beam has b=310 mm and d=460 mm. The beam will be designed to carry a service dead load of 230 kN-m and service live load of190 kn-m. Compression reinforcement if necessary will have its centroid 70 mm from extreme concrete fiber. Determine the required steel area. Use 𝑓′𝑐 = 30𝑀𝑃𝑎 and 𝑓𝑦 = 415 𝑀𝑃𝑎. SOLUTION 𝛽1 = 0.85 𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿

𝑀𝑢 = 1.4(230) + 1.7(190) 𝑀𝑢 = 645 𝑘𝑁 − 𝑚

Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 : Note: For rectangular beams, 𝑐𝑚𝑎𝑥 = 0.75 𝑐𝑏 𝑐𝑚𝑎𝑥 = 0.75

600𝑑 = 203.94 𝑚𝑚 600 + 𝑓𝑦

𝑎 = 𝛽1 𝑐𝑚𝑎𝑥 = 173.35 𝑚𝑚 𝑀𝑛 𝑚𝑎𝑥 = 0.85𝑓′𝑐 𝑎 𝑏 (𝑑 − 𝑎⁄2) 𝑀𝑛 𝑚𝑎𝑥 = 0.85(30)(173.35)(310)(460 − 173.35/2) 𝑀𝑛 𝑚𝑎𝑥 = 511.58 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90(511.58) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 460.42 𝑘𝑁 − 𝑚 Since 𝑀𝑢 = 645 𝑘𝑁 − 𝑚 > 𝜑𝑀𝑛 𝑚𝑎𝑥, 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 𝑖𝑠 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦

b

d ’

ca

=

d– a/2

+

d– d’

𝑀𝑛1 = 𝑀𝑛 𝑚𝑎𝑥 = 511.58 𝑘𝑁 − 𝑚 𝑀𝑛2 =

𝑀𝑢 − 𝑀𝑛1 𝜑

𝑀𝑛2 =

645 − 511.58 0.90

𝑀𝑛2 = 205.088 𝑘𝑁 − 𝑚 𝑐 = 𝑐𝑚𝑎𝑥 = 203.94 𝑚𝑚 𝑎 = 173.35 𝑚𝑚 Tension Steel: 𝑇1 = 𝐶𝑐

𝑀𝑛2 = 𝑇2 (𝑑 − 𝑑′ ) 𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2

𝐴𝑠1 𝑓𝑦 = 0.85 𝑓′𝑐 𝑎 𝑏 𝐴𝑠1 (415) = 0.85(30)(173.35)(310) 𝐴𝑠1 = 3,302 𝑚𝑚2 205.088 x 106 = 𝐴𝑠2 (415)(460 − 70) 𝐴𝑠2 = 1,267 𝑚𝑚2 𝐴𝑠 = 3,302 + 1,267 𝐴𝑠 = 𝟒, 𝟓𝟔𝟗 𝒎𝒎𝟐

Compression steel: 𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑓′𝑠 = 600

203.94 − 70 203.94

Compression steel does not yield, 𝑓′𝑠 = 394.06 𝑃𝑎 𝐶′𝑠 = 𝑇2

𝐴′𝑠 𝑓′𝑠 = 𝐴𝑠2 𝑓𝑦 𝐴′ 𝑠 (394.06) = 1,267(415) 𝑨′𝒔 = 𝟏𝟑𝟑𝟒 𝒎𝒎𝟐

PROBLEM 3.19 A floor system consists of a 100-mm concrete slab supported by continuous T beam with 9 m span, 1.2 m on centers as shown in Figure 3.10. Web dimensions, as determined by negative-moment requirements, are 𝑏𝑤 = 280 𝑚𝑚, and 𝑑 = 500𝑚𝑚. Concrete cover is 70 mm from the centroid of the bars. The beam is subjected to a maximum positive factored moment of 1080 kN-m. Use 𝑓′𝑐 = 21 𝑀𝑃𝑎, 𝑓𝑦 = 415 𝑀𝑃𝑎. Unit weight of concrete is 23.5 kN/𝑚3 . a) Calculate the required tension steel area at the point of maximum positive moment. b) Using the tributary area method, what is the uniform service dead load acting on the beam? c) Calculate the uniform service live load acting on the beam.

A

L=9m

B

L=9m

SOLUTION 𝑓′𝑐 = 21𝑀𝑃𝑎 𝑏𝑤 = 280 𝑚𝑚 𝑓𝑦 = 414 𝑀𝑃𝑎 𝑑 = 500𝑚𝑚 𝛽1 = 0.85 𝑑 ′ = 70 𝑚𝑚 Maximum factored moment, 𝑀𝑢 𝑚𝑎𝑥 = 1080 𝑘𝑁 − 𝑚

C

Effective flange width, 𝑏𝑓 : 1. L/4=9/4=2.25 m 2. 16t+𝑏𝑤 = 16(100) + 280 = 1.88 𝑚 3. 𝑆𝑜𝑐 = 1.2𝑚 Use 𝑏𝑓 = 1.2 𝑚 Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 to determine if compression steel is required. Balanced condition: 𝑐=

600𝑑 600 + 𝑓𝑦

𝑐=

𝑎 = 𝛽1 𝑐

𝑎 = 0.85(295.57) 𝑎 = 251.23 𝑚𝑚 = 1200mm

t=100

d =500mm

600(500) 600 + 415

C

a

z

=280 mm 𝑧 =𝑎−𝑡

𝑧 = 251.23100 𝑧 = 151.23 𝑚𝑚

𝐴1 = 1200 𝑥 100 = 120,000 𝑚𝑚2 𝐴2 = 280(151.23) = 42,345 𝑚𝑚2 𝐴𝑐𝑏 = 𝐴1 + 𝐴2 = 162,345 𝑚𝑚2

Maximum condition: 𝐴𝑐 𝑚𝑎𝑥 = 0.75 𝐴𝑐𝑏

𝐴𝑐 𝑚𝑎𝑥 = 0.75(162,345) 𝐴𝑐 𝑚𝑎𝑥 = 121,759 𝑚𝑚2 > 𝐴1 = 1200mm

d =500mm

t=100 C a

z T =280 mm

𝐴2 = 𝐴𝑐 𝑚𝑎𝑥 − 𝐴1

𝐴2 = 121,759 − 120,000 𝐴2 == 1,759 𝑚𝑚2

𝐴2 = 𝑏𝑤 𝑧

1,759 = 280 𝑧 𝑧 = 6.28 𝑚𝑚

𝑦1 = 𝑑 − 𝑡/2

𝑦1 = 500 − 100/2 𝑦1 = 450 𝑚𝑚

𝑦2 = 𝑑 − 𝑡 − 𝑧/2

𝑦2 = 500 − 100 − 6.28/2 𝑦2 = 396.86 𝑚𝑚

𝑀𝑛 𝑚𝑎𝑥 = 𝐶1 𝑦1 + 𝑐2 𝑦2

𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90(976.36) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 878.72 𝑘𝑁 − 𝑚

𝑀𝑛 𝑚𝑎𝑥 = 0.85 𝑓 ′ 𝑐 (𝐴1 𝑦1 + 𝐴2 𝑦2 ) 𝑀𝑛 𝑚𝑎𝑥 = 0.85(21)(120,000 𝑥 450) +1,759 𝑥 396.86 𝑀𝑛 𝑚𝑎𝑥 = 976.36 𝑘𝑁 − 𝑚

Since 𝑀𝑢 = 1,080 𝑘𝑁 − 𝑚 > 𝜑𝑀𝑛 𝑚𝑎𝑥 , the compression reinforcement must be provided.

=1200mm

d=500mm

430

500 mm

Z

d’=7 0

=820mm 𝑎 =𝑡+𝑧

𝑎 = 100 + 6.28 𝑎 = 106.28 𝑚𝑚

𝑐 = 𝑎/𝛽1

𝑐 = 106.28/0.85 𝑐 = 125.04

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑓′𝑠 = 600

125.04 − 70 125.04

𝑓′𝑠 = 264.1 𝑀𝑃𝑎 < 𝑓𝑦 𝑀𝑛1 = 𝑀𝑛 𝑚𝑎𝑥 = 976.36 𝑘𝑁 − 𝑚 𝐴𝑠1 = 𝐴𝑠 𝑚𝑎𝑥 𝑇1 = 𝐶1 + 𝐶2 𝐴𝑠1

𝐴𝑠1 𝑓𝑦 = 0.85𝑓 ′ 𝑐 (𝐴1 + 𝐴2 ) 𝐴𝑠1 (415) = 0.85(21)(120,000 + 1,759) = 5,237 𝑚𝑚2

d-d’

d’=70

a

𝑀𝑛2 = 𝑀𝑛 − 𝑀𝑛1

𝑀𝑛2 =

𝑀𝑢

− 𝑀𝑛1

𝜑 1080

𝑀𝑛2 = 0.90 − 976.36 𝑀𝑛2 = 223.64 𝑘𝑁 − 𝑚 𝑀𝑛2 = 𝑇2 (𝑑 − 𝑑′ )

𝑀𝑛2 = 𝐴𝑠2 𝑓𝑦 (𝑑 − 𝑑 ′ ) 223.64 𝑥 106 = 𝐴𝑠2 (415)(500 − 70) 𝐴𝑠2 = 1,253 𝑚𝑚2

Tension steel area, 𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2 = 𝟔, 𝟒𝟗𝟎 𝒎𝒎𝟐 Compression steel area: 𝐶′𝑠 = 𝑇2

𝐴′𝑠 𝑓′𝑠 = 𝐴𝑠2 𝑓𝑦 𝐴′ 𝑠 (264.1) = 1,253(415) 𝐴′𝑠 = 1,969 𝑚𝑚2

b) Dead load=weight of concrete: Area=1.2(0.1)+0.28(0.47)=0.2516 𝑚3 𝑤𝑐 = 𝛾𝑐 𝑥 𝐴𝑟𝑒𝑎 𝑤𝑐 = 23.5(0.2516) 𝑤𝑐 = 𝟓. 𝟗𝟏𝟐𝟔 𝒌𝑵⁄𝒎

→ 𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑

c) Uniform live load

5-32 mmøA

7.6 m 3-32

L=9 mmø m 145 kN-m

5-32 5-32 B mmø mmø

7.6 m 3-32

L=9 mmø m

145 kN-m 202 kNm

5-32

Cmmø

Maximum positive moment (at midspan) 𝑤𝑢 =

𝑤𝑢 𝐿2 24

𝑤𝑢 = 1.4𝑤𝐷 + 1.7𝑤𝐿

𝑤𝑢 (9)2 24 𝑤𝑢 = 320 𝑘𝑁/𝑚 1,080 =

320 = 1.4(5.9126) + 1.7𝑤𝐿 𝑤𝐿 = 𝟏𝟖𝟑. 𝟑𝟕 𝒌𝑵⁄𝒎 → 𝑙𝑖𝑣𝑒 𝑙𝑜𝑎𝑑

INVESTIGATION (ANALYSIS) PROBLEMS PROBLEM 3.20 The beam shown in Figure 3.11 is subjected to a maximum service dead load moment of 230 kN-m. Determine the service live load that the beam can carry. Use 𝑓′𝑐 = 20.7 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑓𝑦 = 345 𝑀𝑃𝑎.

350 mm

60 mm

600 mm

540 mm

2-28 mm

4-36 mm

Figure 3.11 SOLUTION 𝜋 𝐴𝑠 = 4 (36)2 𝑥 4 = 4,072 𝑚𝑚2 𝜋

𝐴′𝑠 = 4 (28)2 𝑥2 = 1,232 𝑚𝑚2

b

d’

ca

=

d – a/2+

d – d’

Assume all steel yield: 𝑓𝑠 = 𝑓′𝑠 = 𝑓𝑦 𝐴𝑠2 = 𝐴′𝑠 = 1,232 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 2,840 𝑚𝑚2 𝐶𝑐 = 𝑇1

𝑐=

𝑎 = 187.18 𝑚𝑚 𝛽1

0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠1 𝑓𝑦 0.85(20.7)𝑎(350) = 2,840(345) 𝑎 = 159.1 𝑚𝑚

𝑓𝑠 = 600

𝑑−𝑐 𝑐

600 − 187.18 187.18 𝑓𝑠 = 1.323 > 𝑓𝑦 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 𝑦𝑖𝑒𝑙𝑑𝑠

𝑓𝑠 = 600

𝑐 − 𝑑′ 𝑐

187.18 − 60 187.18 𝑓𝑠 = 407.7 > 𝑓𝑦 compression steel yields

𝑓𝑠 = 600

𝑓𝑠 = 600

Assumption is correct, all steel yield. 𝑀𝑛 = 𝑀𝑛1 + 𝑀𝑛2 𝑎 𝑀𝑛 = 𝑇1 (𝑑 − ) + 𝑇2 (𝑑 − 𝑑′ ) 2 𝑎 𝑀𝑛 = 𝐴𝑠1 𝑓𝑦 (𝑑 − ) + 𝐴𝑠2 𝑓𝑦 (𝑑 − 𝑑 ′ ) 2 159.1 𝑀𝑛 = 2,840(345) (600 − ) + 1,232(345)(600 − 60) 2 𝑀𝑛 = 739.4 𝑘𝑛 − 𝑚 𝜑𝑀𝑛 = 0.90(739.4) = 665.43 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 𝑀𝑢 = 1.4𝑀𝐷 + 1.7𝑀𝐿 665.43 = 1.4(230) + 1.7 𝑀𝐿 𝑀𝐿 = 𝟐𝟎𝟐. 𝟎𝟐 𝒌𝑵 − 𝒎

PROBLEM 3.21 A rectangular beam has the following properties: Width, b=400 mm 𝑓𝑦 = 415 𝑀𝑃𝑎 Effective depth, d=620 mm 𝑓′𝑐 = 22 𝑀𝑃𝑎 Tension bars, 3 pcs 25-mm-diameter d’=70 mm Determine the design strength of the beam and the safe service live load if the service dead load is 320 kN-m. SOLUTION 𝜋 𝐴𝑠 = 10 𝑥 (28)2 = 6,158 𝑚𝑚2 4 𝜋 𝐴′𝑠 = 3 𝑥 (25)2 = 1,473 𝑚𝑚2 4 Assume all steel yields: 𝐴𝑠2 = 𝐴′𝑠 = 1,473 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 4,685 𝑚𝑚2

b d’

ca d – a/2+

=

0.85𝑓 ′ 𝑐 𝑎 𝑏 = 𝐴𝑠1 𝑓𝑦 𝑐 = 𝑎⁄𝛽 = 305.8 𝑚𝑚 1

0.85(22)𝑎(400) = 4,685(415) 𝑎 = 260 𝑚𝑚

d – d’

𝑑−𝑐 = 616.5 𝑀𝑃𝑎 > 𝑓𝑦 𝑐 𝑐 − 𝑑′ 𝑓′𝑠 = 600 = 463 𝑀𝑃𝑎 > 𝑓𝑦 𝑐 𝑓𝑠 = 600

(𝑦𝑖𝑒𝑙𝑑) (𝑦𝑖𝑒𝑙𝑑)

All steel yields. Assumption is correct 𝑀𝑛 = 𝑀𝑛1 + 𝑀𝑛2 𝑀𝑛 = 𝑇1 (𝑑 − 𝑎⁄2) + 𝑇2 (𝑑 − 𝑑′ ) 𝑀𝑛 = 𝐴𝑠1 𝑓𝑦 (𝑑 − 𝑎⁄2) + 𝐴𝑠2 (𝑑 − 𝑑 ′ ) 𝑀𝑛 = 4,685(415)(620 − 260⁄2) + 1,473(415)(620 − 70) 𝑀𝑛 = 1288.9 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(1288.9) = 1,160 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 𝑀𝑢 = 1.4 𝑀𝐷 + 1.7𝑀𝐿 1160 = 1.4(320) + 1.7𝑀𝐿 𝑀𝐿 = 𝟒𝟏𝟗 𝒌𝑵 − 𝒎 PROBLEM 3.22 A 12-m long rectangular reinforced concrete beam is simply supported at its ends. The beam is provided with an addition support at the mid span. Width of beam is 300 mm and the overall depth is 450 mm. The beam is reinforced with 25-mm-diameter bars, four bars at the tension side and 2 bars at the compression side .Concrete protective coverings is 70 mm form the centroid of the bars. Concrete strength 𝑓′𝑐 = 30 𝑀𝑃𝑎 and steel yield 𝑓𝑦 = 415 𝑀𝑃𝑎. Use 0.75 𝜌𝑏 = 0.023. a) Determine the depth of the compression block. b) Determine the nominal moment capacity of the beam. c) Determine the factored uniform load, including its own weight, the beam can carry. SOLUTION 𝛽1 = 0.85 𝑓′𝑐 = 30𝑀𝑃𝑎 𝑓𝑦 = 415 𝑀𝑃𝑎

300 mm 70 mm

𝜋 (25)2 4 𝐴𝑠 = 1963 𝑚𝑚2 𝜋 𝐴′𝑠 = 2 𝑥 (25)2 4 𝐴′𝑠 = 982 𝑚𝑚2 𝐴𝑠 = 4 𝑥

Assuming all steel yields: 𝐴𝑠2 = 𝐴′𝑠 = 982 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 982 𝑚𝑚2

450 mm

310 mm

380 mm

2-25 mm

4-25 mm 70 mm

𝐶𝑐 = 𝑇1

0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠1 𝑓𝑦 0.85(30)𝑎(30) = 982(415) 𝑎 = 53.26 𝑚𝑚

𝑐 = 𝑎⁄𝛽 = 62.66 𝑚𝑚 < 70 𝑚𝑚 compression steel does not yield 1

Assuming tension steel yields and compression steel does not. 𝑇 = 𝐶𝑐 + 𝐶′𝑠

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓′𝑠 1963(415) = 0.85(30(0.85𝑐)(300) 𝑐−70 +982 𝑥 600 𝑐 𝑐 = 98.87 𝑚𝑚

𝑓′𝑠 = 600 𝑓𝑠 = 600

98.86 − 70 = 175.17 𝑀𝑃𝑎 < 𝑓𝑦 98.86

𝑑−𝑐 = 1,706 > 𝑓𝑦 (𝑦𝑖𝑒𝑙𝑑) 𝑐

𝑎 = 𝛽1 𝑐 = 𝟖𝟒. 𝟎𝟑 𝒎𝒎 → 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑛 𝑃𝑎𝑟𝑡 𝑎

𝑀𝑛 = 𝐶𝑐 (𝑑 − 𝑎⁄2) + 𝐶𝑠 (𝑑 − 𝑑 ′ ) 𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏 + 𝐴′ 𝑠 𝑓 ′ 𝑠 (𝑑 − 𝑑′ ) 𝑀𝑛 = 𝟐𝟕𝟎. 𝟓𝟖 𝒌𝑵 − 𝒎 → 𝑎𝑠𝑛𝑤𝑒𝑟 𝑖𝑛 𝑝𝑎𝑟𝑡 𝑏 𝜑𝑀𝑛 = 0.90𝑀𝑛 𝜑𝑀𝑛 = 243.53 𝑘𝑁 − 𝑚 c) Maximum factored uniform load:

Factored load, A

B

C

By there-moment equation: 𝑀𝐴 𝐿1 + 2𝑀𝐵 (𝐿1 + 𝐿2 ) + 𝑀𝑐 𝐿2 +

6𝐴1 𝑎̅1 6𝐴2 𝑎̅2 + =0 𝐿1 𝐿2

𝑀𝐴 = 𝑀𝐶 = 0 6𝐴1 𝑎̅1 𝑤𝑢 𝐿1 3 = 𝐿1 4 6𝐴2 𝑎̅2 𝑤𝑢 𝐿1 3 = 𝐿2 4 𝑤𝑢 (6)3 𝑤𝑢 (6)3 0 + 2𝑀𝐵 (+6 +) + 0 + + =0 4 4 𝑤𝑢 = 𝟓𝟒. 𝟏𝟐 𝒌𝑵/𝒎

PROBLEM 3.23 (CE NOVEMBER 2010) A 6 meter long simply supported reinforced concrete beam has a width of 350mm and an overall depth of 470 mm. The beam is reinforced with 228 mm compression bars on top and 4-28 tension bars at the bottom, each located 70 mm from the extreme concrete fiber. Concrete strength 𝑓′𝑐 = 20.7 𝑀𝑃𝑎, and steel yield strength 𝑓𝑦 = 415 𝑀𝑃𝑎.Determine the following: a) Depth of compression blocks assuming both tension and compression steel yields. b) What is the ultimate moment capacity of the beam in kN-m? c) Determine the additional concentrated live load that can be applied at midspan if the dead load including the weight of the beam is 20 kN/m.

SOLUTION Given : 𝐿 = 6𝑚 𝑏 = 350 𝑚𝑚 𝑑 = 400𝑚𝑚 𝑑 ′ = 70 𝑚𝑚

𝑓′𝑐 = 20.7 𝑀𝑃𝑎 𝑓𝑦 = 415 𝑀𝑃𝑎 𝑑𝑏 = 28 𝑚𝑚 𝜋

Tension steel area 𝐴𝑠 = 4 (28)2 𝑥 4 = 2463 𝑚𝑚2 𝜋

Compression steel area, 𝐴′𝑠 = 4 (28)2 𝑥 2 = 132 𝑚𝑚2 𝛽1 = 0.85 𝜑 = 0.90 Assuming tension & compression steel yields: 𝐴𝑠2 = 𝐴′𝑠 = 1232 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 1232 𝑚𝑚2 𝑐𝑐 = 𝑇1

0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠1 𝑓𝑦 0.85(20.7)𝑎(350) = 1232(415) 𝑎 = 𝟖𝟑 𝒎𝒎 → 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑛 𝑃𝑎𝑟𝑡 𝑎

𝑐 = 𝑎/𝛽1

𝑐 = 83/0.85 𝑐 = 97.64 𝑚𝑚

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

97.64 − 70 97.64 𝑓′𝑠 = 170 𝑀𝑃𝑎 < 𝑓𝑦 𝑓′𝑠 = 600

Thus, compression steel does not yield.

Since compression steel does not yield, 𝑓′𝑠 = 600

𝑐−𝑑′ 𝑐

Assuming tension steel yields: 𝐶𝑐 + 𝐶𝑠 = 𝑇𝑠 0.85𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓′𝑠 = 𝐴𝑠 𝑓𝑦 0.85(20.7)(0.85𝑐)(350) + 1232 𝑥 600 𝑐 = 130.08 𝑚𝑚 𝑎 = 𝛽1 𝑐 = 110.6 𝑚𝑚 𝑓𝑠 = 600

𝑑−𝑐 𝑐

𝑐 − 70 = 2463(415) 𝑐

𝑓𝑠 = 600

400 − 130.08 130.08

𝑓𝑠 = 1245 > 𝑓𝑦 (𝑦𝑖𝑒𝑙𝑑) 𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑓′𝑠 = 600

130.08 − 70 130.08

𝑓′𝑠 = 277.11 𝑀𝑃𝑎 < 𝑓𝑦 𝑎 𝑀𝑛 = 𝑐𝑐 (𝑑 − ) + 𝐶 ′ 𝑠 (𝑑 − 𝑑′ ) 2 𝑎 𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏 (𝑑 − ) + 𝐴′ 𝑠 𝑓 ′ 𝑠 (𝑑 − 𝑑′ ) 2 𝑀𝑛 = 0.85(20.7)(110.6)(350)(400 − 110.6⁄2) + 1232(277.11)(400 − 70) 𝑀𝑛 = 347.33 𝑘𝑁 − 𝑚

Ultimate moment capacity= 𝜑𝑀𝑛 = 0.90(347.33) Ultimate moment capacity= 𝜑𝑀𝑛 = 312.6 𝒌𝑵 − 𝒎 → 𝑎𝑛𝑠𝑤𝑒𝑟 𝑖𝑛 𝑃𝑎𝑟𝑡 𝑏

3 m

3 m

L=6 m 𝑀𝑢 = 𝜑𝑀𝑛 = 312.6 𝑘𝑁 − 𝑚 𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿

𝑀𝑢 = 1.4

𝑀𝐷 𝐿2 8

+ 1.7

20(6)2

𝑃𝐿 𝐿 4

312.6 = 1.4 8 + 1.7 𝑃𝐿 = 𝟕𝟑. 𝟏𝟕𝟓 𝒌𝑵

𝑃𝐿 (6)2 4

PROBLEM 3.24 A beam section is shown in Figure 3.12. The beam will be subjected to a maximum service dead load of 215 kN-m. What is the safe service live load moment for this beam? Use 𝑓′𝑐 = 21 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑓𝑦 = 415 𝑀𝑃𝑎.

525mm 8 - 25mm

30 mm

650 mm

Figure 3.12

25 mm

360 mm

30 mm

360 mm d’

𝜋

(25)2 4

𝑑′ = 30 +

𝜋 4

(25)2

d

1 = 42.5 𝑚𝑚 2(25)

8 - 25mm

Effective depth to extreme tension bar: 1 𝑑𝑡 = 650 − 30 − = 607.5 𝑚𝑚 2(25) Effective depth (to centroid of tension bar) 𝑑 = 650 − 30 − 25 − 1/2(25) 𝑑 = 582.5 𝑚𝑚 𝑀𝐷 − 215 𝑘𝑁 − 𝑚 𝑓′𝑐 = 21 𝑀𝑃𝑎 𝑓𝑦 = 415 𝑀𝑃𝑎 Assume all steel yields: 𝐴𝑠2 = 𝐴′𝑠 = 2,454 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 1,473 𝑚𝑚2 𝐶𝑐 = 𝑇1

0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠1 𝑓𝑦 0.85(21)𝑎(360) = 1.473(415) 𝑎 = 95.1 𝑚𝑚

𝑐 = 𝑎/𝛽1

𝑐 = 95.1/0.85 𝑐 = 111.9 𝑚𝑚

𝑓′𝑠 = 600

𝑐−𝑑′ 𝑐

𝑓′𝑠 = 600

111.9−42.5 111.9

30 mm

650 mm

Compression steel, 𝐴′𝑠 = 5 𝑥 𝐴′𝑠 = 2,454 𝑚𝑚2

5- 25mm

25 mm

SOLUTION 𝛽1 = 0.85 Tension steel, 𝐴𝑠 = 8 𝑥 𝐴𝑠 = 3,927 𝑚𝑚2

30 mm

𝑓′𝑠 = 372 𝑀𝑃𝑎 < 𝑓𝑦 Compression steel does not yield.

360 mm d’

25 mm

d

5-5-25mm 25mm

c

a d-a/2

d-d’

8 - 825mm 25mm

T

𝑵𝒐𝒕𝒆: 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑙𝑎𝑤𝑦𝑒𝑟𝑠 𝑜𝑓 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑏𝑎𝑟𝑠 𝑤ℎ𝑖𝑐ℎ 𝑜𝑏𝑣𝑖𝑜𝑢𝑠𝑙𝑦 𝑦𝑖𝑒𝑙. 𝑇ℎ𝑢𝑠, 𝑡ℎ𝑖𝑒𝑟 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑒𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑓𝑦 𝑎𝑛𝑑 𝑡ℎ𝑖𝑒𝑟 𝑐𝑔 𝑖𝑠 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒𝑖𝑟 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑. 𝑇 = 𝐶𝑐 + 𝐶′𝑠

𝐴𝑠 𝑓𝑦 = 0.85𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓′𝑠 𝑐−𝑑′

𝐴𝑠 𝑓𝑦 = 0.85 𝑓 ′ 𝑐 (𝛽1 𝑐)𝑏 + 𝐴′𝑠 𝑥 600 𝑐 3,927(415) = 0.85(21)(0.85𝑐)(360) 𝑐−42.5 +2,454 𝑥 600 𝑐 𝑐 = 122.38 𝑚𝑚 𝑓′𝑠 = 600

𝑎 = 𝛽1 𝑐

𝑐−𝑑′ 𝑐

122.38−42.5

𝑓′𝑠 = 600 122.38 𝑓𝑠 = 391.64 𝑀𝑃𝑎 < 𝑓𝑦 𝑎 = 0.85(122.38) 𝑎 = 104.03 𝑚𝑚

𝑎 𝑀𝑛 = 𝐶𝑐 (𝑑 − ) + 𝐶 ′ 𝑠 (𝑑 − 𝑑′ ) 2 𝑎 𝑀𝑛 = 0.85 𝑓′𝑐 𝑎 𝑏 (𝑑 − ) + 𝐴′𝑠 𝑓 ′ 𝑠 (𝑑 − 𝑑 ′ ) 2 104.03 𝑀𝑛 = 0.85(21)(104.03)(360)(582.5 − ) 2 +2,454(391.64)(582.5 − 42.5) 𝑀𝑛 = 873.68 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(873.68) 𝜑𝑀𝑛 = 786.31 𝑘𝑛 − 𝑚 𝑀𝑢 = 𝜑 𝑀𝑛

𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿 786.31 = 1.4(215) + 1.7(𝑀𝐿 ) 𝑀𝐿 = 𝟐𝟖𝟓. 𝟓 𝒌𝑵 − 𝒎

PROBLEM 3.25 A beam section is shown in Figure 3.13. The beam will be subjected to a maximum service dead load of 360 kN-m. What is the safe service live load moment for this beam? Use 𝑓′𝑐 = 21 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑓𝑦 = 415 𝑀𝑃𝑎.

Compression steel, 𝐴′𝑠 = 2 𝑥

𝐴′𝑠 = 982 𝑚𝑚2

𝜋 4

(25)2

𝑑′ = 30 + 1⁄2(25) = 42.5 𝑚𝑚

30 mm 2- 25mm

10 - 28mm

650 mm

(28) 4

2

28 mm

Tension steel, 𝐴𝑠 = 10 𝑥 𝐴𝑠 = 6,158 𝑚𝑚2

𝛽1 = 0.85

𝜋

Effective depth (to centroid of tension bars) 𝑑 = 650 − 30 − 28 − 1/2(28) 𝑑 = 578 𝑚𝑚 𝑀𝐷 = 360 𝑘𝑁 − 𝑚 𝑓′𝑐 = 21 𝑀𝑃𝑎 𝑓𝑦 = 415 𝑀𝑃𝑎

30 mm

Figure 3.13

320 mm 2 - 25mm

650 mm

d’

30 mm

28 mm

d

10 - 28mm

30 mm

Assume all steel yields: 𝐴𝑠2 = 𝐴′𝑠 = 982 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 5,176 𝑚𝑚2 𝐶𝑐 = 𝑇1

0.85𝑓′𝑐 𝑎 𝑏 = 𝐴𝑠1 𝑓𝑦 0.85(21)𝑎(320) = 5,176(415) 𝑎 = 376.04 𝑚𝑚

𝑐 = 𝑎/𝛽1

𝑐 = 376.04/0.85 𝑐 = 442.4 𝑚𝑚

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑓′𝑠 = 600

442.4 − 42.5 442.4

𝑓𝑠 = 222 𝑀𝑃𝑎 < 𝑓𝑦 Tension steel does not yield.

320 mm d’

10 - 28mm 825mm

𝑵𝒐𝒕𝒆: 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑡𝑤𝑜 𝑙𝑎𝑦𝑒𝑟𝑠 𝑜𝑓 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑏𝑎𝑟𝑠 𝑤ℎ𝑖𝑐ℎ ℎ𝑎𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑓𝑦 . 𝑇ℎ𝑢𝑠, 𝑡ℎ𝑖𝑒𝑟 𝑐𝑔 𝑖𝑠 𝑛𝑜𝑡 𝑙𝑜𝑐𝑎𝑡𝑒𝑑 𝑎𝑡 𝑡ℎ𝑖𝑒𝑟 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑. 𝑑1 = 650 − 30 − 14 = 606 𝑚𝑚 𝑑2 = 650 − 30 − 28 − 28 − 14 = 550 𝑚𝑚 𝐴𝑠𝑡1 = 𝐴𝑠𝑡2 = 5 𝑥

𝜋 (28)2 = 3,079 𝑚𝑚2 4

𝑇1 + 𝑇2 = 𝐶𝑐 + 𝐶′𝑠 𝐴𝑠𝑡1 𝑓𝑠1 + 𝐴𝑠𝑡2 𝑓𝑠2 = 0.85 𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓𝑦 𝑑1 − 𝑐 𝑑2 − 𝑐 𝐴𝑠𝑡1 600 + 𝐴𝑠𝑡2 600 = 0.85𝑓′𝑐 𝑎 𝑏 + 𝐴′𝑠 𝑓𝑦 𝑐 𝑐 606 − 𝑐 550 − 𝑐 + 3,079 𝑥 600 𝑐 𝑐 = 0.85(21)(0.85𝑐)(320) + 982(415) 3,079 𝑥 600

𝑐 = 363.9 𝑚𝑚 𝑎 = 𝛽1 𝑐 = 309.29 𝑚𝑚

d-d’

c a

d-a/2

28 mm

5- 25mm 525mm

d

𝑓𝑠1 = 600

𝑑1 − 𝑐 𝑐

𝑓𝑠2 = 600

𝑑2 − 𝑐 𝑐

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

𝑎 = 𝛽1 𝑐

𝑓𝑠1 = 600

606 − 363.9 363.9

𝑓𝑠1 = 399.25 𝑀𝑃𝑎 < 𝑓𝑦 550 − 363.9 𝑓𝑠2 = 600 363.9 𝑓𝑠2 = 306.9 𝑀𝑃𝑎 < 𝑓𝑦 363.9 − 42.5 𝑓′𝑠 = 600 363.9 𝑓′𝑠 = 530 𝑀𝑃𝑎 > 𝑓𝑦 𝑎 = 0.85(345.4) 𝑎 = 301.2 𝑚𝑚

Solve for d: 𝑇1 = 𝐴𝑠𝑡1 𝑓𝑠1 𝑇2 = 𝐴𝑠2 𝑓𝑠2

𝑇1 = 3,079(399.25) 𝑇1 = 1,229.2 𝑘𝑁 𝑇2 = 3,079 (306.9) 𝑇2 = 944.9 𝑘𝑁

𝑇 𝑥 𝑑 = 𝑇1 𝑥 𝑑1 + 𝑇2 𝑥 𝑑2 2,174.1 𝑑 = 1,229.2(606) + 944.9(550) 𝑑 = 581.66𝑚 𝑎 𝑀𝑛 = 𝑐𝑐 (𝑑 − ) + 𝐶 ′ 𝑠 (𝑑 − 𝑑′ ) 2 𝑎 𝑀𝑛 = 0.85 𝑓′𝑐 𝑎 𝑏 (𝑑 − ) + 𝐴′𝑠 𝑓𝑦 (𝑑 − 𝑑′ ) 2 𝑀𝑛 = 0.85(21)(309.29)(320)(578 − 309.29⁄2) + 982(415)(581.66 − 42.5) 𝑀𝑛 = 974.07 𝑘𝑛 − 𝑚 𝜑𝑀𝑛 = 0.90(947.07) = 876.65 𝑘𝑁 − 𝑚 𝑀𝑢 = 𝜑𝑀𝑛

𝑀𝑢 = 1.4 𝑀𝐷 + 1.7 𝑀𝐿 876.65 = 1.4(360) + 1.7(𝑀𝐿 ) 𝑀𝐿 = 𝟐𝟏𝟗. 𝟐𝟏 𝒌𝑵 − 𝒎

PROBLEM 3.26 Calculate the design flexural strength of the T-beam shown in Figure 3.14. Use 𝑓′𝑐 = 27 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑓𝑦 = 350𝑀𝑃𝑎.

=600mm 25mm t=100mm 3-23mm

Figure 3.14

10mm stirrup 10-25mm 25mm

20mm =300mm

.

SOLUTION 𝛽1 = 0.85

𝜋 (25)2 = 4,909 𝑚𝑚2 4 𝜋 𝐴′𝑠 = 3 𝑥 (22)2 = 1,140 𝑚𝑚2 4 Flange area, 𝐴𝑓 = 600(110) = 66,000 𝑚𝑚2 𝐴𝑠 = 10 𝑥

Assume all steel yields: 𝑑 = 110 + 390 − 20 − 10 − 25 − 1⁄2(25) = 432.5𝑚𝑚 𝑑′ = 25 + 10 + 1⁄2 (22) = 46𝑚𝑚 𝐴𝑠2 = 𝐴′𝑠 = 1,140 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 3,768 𝑚𝑚2

=390mm

Area of compression concrete: 𝐶𝑐 = 𝑇1

0.85𝑓′𝑐 𝐴𝑐 = 𝐴𝑠1 𝑓𝑦 0.85(27)𝐴𝑐 = 3,768(350) 𝐴𝑐 = 57,468 𝑚𝑚2 < 𝐴𝑓 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 < 𝑡 57,469 = 𝑎(600) 𝑎 = 95.8 𝑚𝑚 < 𝑡 𝑐 = 95.8/0.85 𝑐 = 112.7 𝑚𝑚

𝐴𝑐 = 𝑎 𝑏𝑓 𝑐 = 𝑎⁄𝛽1

𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

112.7 − 46 112.7 𝑓′𝑠 = 355 𝑀𝑃𝑎 > 𝑓𝑦

𝑓′𝑠 = 600

(𝑦𝑖𝑒𝑙𝑑)

=600mm t=110mm

25mm

𝑓𝑠 = 600

𝑑−𝑐 𝑐

10-25mm 1025mm =300mm

10mm stirrup 20mm

432.5 − 112.7 112.7 𝑓𝑠 = 1,703 > 𝑓𝑦 (𝑦𝑖𝑒𝑙𝑑)

𝑓𝑠 = 600

d-d’

d

d-a/2

3-22mm 323mm

=390mm

25mm

Verify if the upper layer of tension steel yields 𝑑2 = 𝑑 − 1⁄2(25) − 1⁄29(25) = 407.5 𝑚𝑚 𝑓𝑠2 = 600

𝑑2 − 𝑐 = 1,567 𝑀𝑃𝑎 > 𝑓𝑦 𝑐

(𝑦𝑖𝑒𝑙𝑑)

All steel yields, assumption is correct: 𝑎 𝑀𝑛 = 𝐶𝑐 (𝑑 − ) + 𝐶 ′ 𝑠 (𝑑 − 𝑑′ ) 2 𝑎 𝑀𝑛 = 0.85𝑓′𝑐 𝑎 𝑏𝑓 (𝑑 − ) + 𝐴′ 𝑠 𝑓𝑦 (𝑑 − 𝑑′ ) 2 𝑀𝑛 = 0.85(27)(95.8)(600)432.5 − 46) + 1,140(350)(432.5 − 46) 𝑀𝑛 = 661.5 𝑘𝑁 − 𝑚 𝜑𝑀𝑛 = 0.90(661.5) 𝜑𝑀𝑛 = 𝟓𝟗𝟓. 𝟒 𝒌𝑵 − 𝒎 PROBLEM 3.27 Calculate the design flexural strength of the T-beam shown in Figure 3.15. Use 𝑓′𝑐 = 25 𝑀𝑃𝑎 𝑎𝑛𝑑 𝑓𝑦 = 345 𝑀𝑃𝑎.

=600mm 25mm t=100mm 2-22mm

Figure 3.15

10mm stirrup 10-28mm

25mm

20mm =315mm

=390mm

SOLUTION 𝛽1 = 0.85

𝜋 (28)2 = 6,158 𝑚𝑚2 4 𝜋 𝐴′𝑠 = 2 𝑥 (22)2 = 760 𝑚𝑚2 4 Flange area, 𝐴𝑓 = 600(100) = 60,000 𝑚𝑚2 𝐴𝑠 = 10 𝑥

Assume all steel yields: 𝑑 = 100 + 390 − −20 − 10 − 28 − 1⁄2(25) = 419.5 𝑚𝑚 𝑑′ = 25 + 10 + 1⁄2(22) = 46 𝑚𝑚 𝐴𝑠2 = 𝐴′𝑠 = 760 𝑚𝑚2 𝐴𝑠1 = 𝐴𝑠 − 𝐴𝑠2 = 5,397 𝑚𝑚2 Area of compression concrete: 𝐶𝑐 = 𝑇1

𝐴𝑐 = 𝐴𝑓 + 𝐴𝑤 𝐴𝑤 = 𝑏𝑤 𝑧

0.85𝑓′𝑐 𝐴𝑐 = 𝐴𝑠1 𝑓𝑦 0.85(25)𝐴𝑐 = 5,397(345) 𝐴𝑐 = 87,626 𝑚𝑚2 > 𝐴𝑓 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎 > 𝑡 87,626 = 60,00 + 𝐴𝑤 𝐴𝑤 = 27,626 𝑚𝑚2 27,626 = 315 𝑧 𝑧 = 87.7 𝑚𝑚

𝑎 = 100 + 𝑧 = 187.7 𝑚𝑚 𝑐 = 𝑎⁄𝛽1 = 220.83 𝑚𝑚 𝑓′𝑠 = 600

𝑐 − 𝑑′ 𝑐

220.83 − 46 220.83 𝑓′𝑠 = 475 𝑀𝑃𝑎 > 𝑓𝑦

𝑓′𝑠 = 600

(𝑦𝑖𝑒𝑙𝑑)