330932238 Analysis and Design of Beams Problems PDF

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Summary

ANALYSIS AND DESIGN OF BEAMS PROBLEMSPROBLEM 2.A reinforced concrete rectangular beam 300 mm wide has an effectivedepth of 460 mm and is reinforced for tension only. Assuming 𝑓′𝑐= 21 𝑀𝑃𝑎and 𝑓𝑦= 345𝑀𝑃𝑎 , determine the balance steel area in sq.SOLUTION𝜌𝑏= 85𝑓′ 𝑐𝛽1600𝑓𝑦(600+𝑓𝑦)𝛽1= 0. 85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐&l...

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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 )

𝐴𝑠𝑏 = 𝜌𝑏 𝑏𝑑

𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 < 30𝑀𝑃𝑎 𝜌𝑏 =

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

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 =

𝜔𝑚𝑎𝑥 = 0.371

27.6

𝑅𝑛 𝑚𝑎𝑥 = 𝑓 ′ 𝑐 𝜔(1 − 0.59𝜔)

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

𝑀𝑛 𝑚𝑎𝑥 = 8.001(300)(490)2 𝑀𝑛 𝑚𝑎𝑥 = 576.279𝑥 106 𝑁 − 𝑚𝑚 𝑀𝑛 𝑚𝑎𝑥 = 576.279𝑘𝑁 − 𝑚𝑚 𝑀𝑢 𝑚𝑎𝑥 = 0.90 𝑥 576.279 𝑀𝑢 𝑚𝑎𝑥 = 518.65 𝑘𝑁 − 𝑚

𝑀𝑛 𝑚𝑎𝑥 = 𝑅𝑛 𝑚𝑎𝑥 𝑏𝑑2 𝑀𝑢 𝑚𝑎𝑥 = 𝜑𝑀𝑛 𝑚𝑎𝑥

a) 𝑀𝑢 = 20𝑘𝑁 − 𝑚 < 𝑀𝑢 𝑚𝑎𝑥 = (𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑) 𝑀𝑢 = 𝜑𝑅𝑛 𝑏𝑑2 20 x 106 = 0.90𝑅𝑛 (300)(490)2 𝑅𝑛 = 0.309 𝑀𝑃𝑎 𝜌= 𝜌= 𝜌𝑚𝑖𝑛 = 𝐴𝑠 = 𝜌𝑏𝑑

0.85𝑓′𝑐 2𝑅𝑛 ] [1 − √1 − 0.085𝑓′𝑐 𝑓𝑦

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

1.4 √𝑓′𝑐 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 − 0.85𝑓′𝑐

𝑓𝑦 0.85(27.6) 276

2(2.16) ] [1 − √1 − 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 − 0.85𝑓′𝑐 𝑓𝑦

0.85(27.6) 2(7.48) ] [1 − √1 − 0.85(27.6) 276

𝐴𝑠 = 𝜌 𝑏 𝑑

𝜌 = 0.03384 > 𝜌𝑚𝑖𝑛

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:

𝑀𝑢 = 𝑢8 𝑀𝑢 = 8 𝑀𝑢 = 𝟏𝟔𝟕. 𝟗𝟑 𝒌𝑵 − 𝒎 𝑊 𝐿2

53.738(5)2

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

415(600+415) 𝜌𝑏 = 𝜌𝑏 = 0.02881 𝜌𝑚𝑎𝑥 = 0.021261

0.85(28)(0.85)(600)

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 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)

𝑅𝑛 = 𝜌= 𝜌=

𝑀𝑢

𝑅𝑛 = 𝑅𝑛 = 6.936𝑀𝑃𝑎

314 .805 𝑥 (10 0. 90(300) 4160)2

𝜑𝑏𝑑2

0.85𝑓′𝑐 𝑓𝑦

2𝑅𝑛 ] [1 − √1 − 0.85𝑓′𝑐

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

𝐴𝑠 = 0.02031(300)(410) 𝐴𝑠 = 2498𝑚𝑚 2 𝜋 𝜋 𝐴𝑠 = 4 𝑑𝑏2 𝑁 2498 = 4 (20)2 N 𝑁 = 7.95 𝑠𝑎𝑦 𝟖 𝒃𝒂𝒓𝒔 𝐴𝑠 = 𝜌 𝑏 𝑑

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 =

𝜌𝑚𝑎𝑥 = 0.15𝜌𝑏

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝛽1 = 0.85 𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 < 10 𝑀𝑃𝑎

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

𝜔=

𝜌𝑓𝑦 𝑓′

𝜔=

𝑐

30 0.02323(414) 𝜔 = 0.3209

𝑀𝑢 𝑚𝑎𝑥 = 𝜑𝑓′𝑐 𝜔 𝑏 𝑑 2 (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 Solve for 𝜌: 𝜌= 𝜌=

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

2𝑅𝑛 0.85𝑓′𝑐 (1 − √1 − ) 𝑓𝑦 0.85𝑓′𝑐

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 = 0.03199

𝜌 𝑚𝑎𝑥 = 0.75𝜌𝑏

𝜌𝑏 =

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

𝜌𝑚𝑎𝑥 = 0.75(0.03199) 𝜌𝑚𝑎𝑥 = 0.02399

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

𝐴𝑠 𝑚𝑎𝑥 = 𝜌 𝑚𝑎𝑥 𝑏𝑑

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

𝑊𝑢 𝐿2 8

Solve for 𝑀𝑢 𝑚𝑎𝑥

𝑊𝑢 = 1.4(25) + 1.7(40) 𝑊𝑢 = 103 𝑘𝑁/𝑚 103(6)2 8 𝑀𝑢 =463.5kN-m 𝑀𝑢 =

𝜔=

𝜌 𝑚𝑎𝑥 𝑓𝑦 𝑓′𝑐

0.0299 21(312) 𝜔 = 0.356

𝑀𝑢 𝑚𝑎𝑥 = 𝜑𝑓 ′ 𝑐 𝜔𝑏𝑑 2 (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 = 0.39 𝑅𝑛 (300)(550)2 463.5 𝑥 106 = 0.9 𝑅𝑛 (300)(550)2 𝑅𝑛 = 5.67 𝑀𝑃𝑎

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2

𝜌= 𝜌=

0.85𝑓′𝑐 2𝑅𝑛 (1 − √1 − ) 𝑓𝑦 0.85 𝑓,𝑐

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: 𝐴𝑠

Number of 25-mm bars=

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 𝑀𝑃𝑎method. and steel yield strength 𝑓𝑦 = 248 𝑀𝑃𝑎. Solve using the strength design 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 𝑓′𝑐 = 28 𝑀𝑃𝑎 𝑓𝑦 = 248 𝑀𝑃𝑎

Given: 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 )

b) Effective depth using 𝜌 = 0.5𝜌𝑏 𝜌 = 0.5(0.0577) = 0.0289 𝜔=

𝜌𝑓𝑦 𝑓′𝑐

𝜔=

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

𝜌𝑏 = 0.0577 = 𝟓. 𝟕𝟕%

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 𝑑2 23 x 106 = 0.90 𝑅𝑛 (1000)( 94)2 𝑅𝑛 = 2.892 𝜌= 𝜌=

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

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

𝜌𝑚𝑎𝑥 =

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

1.4 𝜌𝑚𝑖𝑛 = 𝑓 = 0.00509 𝑦

𝐴𝑠 = 0.1154(1000)(94) 𝐴𝑠 = 1085 𝒎𝒎𝟐

𝐴𝑠 = 𝜌𝑏𝑑

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

𝑏 𝑁

𝑠=

𝑠=

1000 𝐴𝑠 𝐴𝑏

1000𝐴𝑏 𝐴𝑠

Eq. 2-17

1000𝐴𝑏 𝑠= 𝐴𝑠

𝑠=

𝜋 1000 𝑥4 (12)2 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

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

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

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

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

(Note: this already includes the weight of beam) 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑚𝑖𝑛 =

𝜌𝑏 =

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

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

𝜌𝑏 = 0.0371

1.4 = 0.00507 𝑓𝑦

Try 𝜌 = 60% 𝜌𝑏

Note: this is the author’s suggestion

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

)𝑏𝑑2 165.6 x 106 = 0.906(5.068 2 3 𝑏𝑑 = 36.296 𝑥 10 𝑚𝑚

𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2 Try d = 1.75 b

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

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

𝐴𝑠 = 𝜌𝑏𝑑 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 121.448(6)2 𝑀𝑢 = 8

𝑊𝑏 = 1.4 𝑊𝐿 + 1.7 𝑊𝐿 . 𝑀𝑢 =

𝑊𝑢 𝐿2 8

𝛽1 = 0.85 − 𝜌𝑏 =

0.05 = 0.821 7(34 − 30)

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+ 𝑓𝑦 )

𝜌𝑏 =

𝑀𝑢 = 546.516 𝑘𝑁 − 𝑚

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

𝜌𝑏 = 0.04369

𝜌 = 𝜌𝑚𝑎𝑥 = 0.75 (0.04369)

√𝑓′𝑐 𝜌𝑚𝑖𝑛 = 4𝑓 = 0.00423 𝑦 𝜔=

𝜌𝑓𝑦

𝜔=

𝑓′𝑐

0.03277(345)

𝑅𝑛 = 𝑓 ′ 𝑐 𝜔(1 − 0.56 𝜔)

Assume d = 1.75 b 𝑀𝑢 = 𝜑 𝑅𝑛 𝑏 𝑑2

𝜌 = 0.03277 > 𝜌𝑚𝑖𝑛

34

𝜔 = 0.332

𝑠𝑖𝑛𝑐𝑒 𝑓′𝑐 > 31.36 𝑀𝑃𝑎

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

(this is the author’s assumption)

Use b = 280 mm, d = 490 mm

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

Minimum beam the thickness (Section 409.6.2.1) ℎ𝑚𝑖𝑛 =

𝐿 𝑓𝑦 ) (0.4 + 700 16

𝐴𝑠 = 𝜌 𝑏 𝑑

ℎ𝑚𝑖𝑛 =

6000 345 (0.4 + ) 16 700

ℎ𝑚𝑖𝑛 = 335 𝑚𝑚 𝑂𝐾

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 𝑎 0 𝐿0

0 + 2𝑀𝐴 (0 + 6 ) + (−285 )(6) + 0 +

+

6𝐴1 𝑏 0 𝐿1

142.5(6)3 4

=0

=0

𝑀𝐴 = −498.75𝑘𝑁 − 𝑚

𝑀𝐴 = 𝑀𝐴 𝑟𝑖𝑔ℎ𝑡

𝑅𝐴 = 𝑊𝑢 𝐿 − 𝑅 Maximum positive moment: 𝑉𝐷 = 0

𝑀𝐷 = 𝑅𝑋 − 𝑊𝑢

(𝑥+2)2 2

Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 : 𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑚𝑎𝑥 = 0.75 𝜌𝑏 𝜔𝑚𝑎𝑥 =

𝜌𝑚𝑎𝑥 𝑓𝑦 𝑓′𝑐

-489.75 = R(6)- 142.5(8)(4) R=676.875 kN 𝑅𝐴 = 142.5(8) − 676.875 𝑅𝐴 = 463.125 𝑘𝑁

𝑊𝑢 (2 + 𝑥) − 𝑅 = 0 142.5(2 + x) - 676.875 = 0 x = 2.75 m 𝑀𝐷 = 676.875(2.75) − 142.5 𝑀𝐷 = 253.828 𝑘𝑁 − 𝑚 𝜌𝑏 =

(2.75+2)2 2

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

𝜌𝑏 = 0.0247

𝜌𝑚𝑎𝑥 = 0.75 (0.0247) 𝜌𝑚𝑎𝑥 = 0.01852 𝜔𝑚𝑎𝑥 =

0.01852(415) 24

𝜔𝑚𝑎𝑥 = 0.3203

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 𝑎 0 𝐿0

0 + 2𝑀𝐴 (0 + 6 ) + (−285 )( 6) + 0 +

𝑀𝐴 = −498.75𝑘𝑁 − 𝑚

𝑀𝐴 = 𝑀𝐴 𝑟𝑖𝑔ℎ𝑡

𝑅𝐴 = 𝑊𝑢 𝐿 − 𝑅 Maximum positive moment: 𝑉𝐷 = 0

+

6𝐴1 𝑏 0 𝐿1

142.5(6)3 4

=0

=0

-489.75 = R(6)- 142.5(8)(4) R=676.875 kN 𝑅𝐴 = 142.5(8) − 676.875 𝑅𝐴 = 463.125 𝑘𝑁

𝑊𝑢 (2 + 𝑥 ) − 𝑅 = 0

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

(𝑥+2)2 2

Solve for 𝜑𝑀𝑛 𝑚𝑎𝑥 :

𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600 + 𝑓𝑦 )

𝜌𝑚𝑎𝑥 = 0.75 𝜌𝑏 𝜔𝑚𝑎𝑥 =

𝜌𝑚𝑎𝑥 𝑓𝑦 𝑓′𝑐

𝑅𝑛 𝑚𝑎𝑥 = 𝑓 ′ 𝑐 𝜔(1 − 0.59 𝜔) 𝑀 = 𝑅 𝑏 𝑑2

𝑀𝐷 = 676.875(2.75) − 142.5 𝑀𝐷 = 253.828 𝑘𝑁 − 𝑚 𝜌𝑏 =

(2.75+2)2 2

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

𝜌𝑏 = 0.0247

𝜌𝑚𝑎𝑥 = 0.75 (0.0247) 𝜌𝑚𝑎𝑥 = 0.01852 𝜔𝑚𝑎𝑥 =

0.01852(415) 24

𝜔𝑚𝑎𝑥 = 0.3203

𝑅𝑛 𝑚𝑎𝑥 = 415(0.3203)[1 − 0.59(0.3203)] 𝑀 = 6.235(290)(430)2

𝑛 𝑚𝑎𝑥

𝑛

𝜑𝑀𝑛 𝑚𝑎𝑥 = 0.90(334.316) 𝜑𝑀𝑛 𝑚𝑎𝑥 = 300.884 𝑘𝑁 − 𝑚

𝑛 𝑚𝑎𝑥

𝑀𝑛 𝑚𝑎𝑥 = 334.316 𝑘𝑁 − 𝑚

At a point of maximum positive moment: 𝑀𝑢 = 253.828 𝑘𝑁 − 𝑚 < 𝜑 𝑀𝑛 𝑚𝑎𝑥 𝑀𝑢 = 𝜑𝑅𝑛 𝑏 𝑑2

𝜌=

(Singly reinforced)

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

0.85𝑓′𝑐 𝑅𝑛 [1 − √1 − ] 𝑓𝑦 0.85𝑓′𝑐

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

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) 𝐴𝑠 = 6 𝑥

𝜌=

𝜔=

𝐴𝑠 𝑏𝑑

𝜌𝑓𝑦 𝑓′𝑐

𝜌𝑏 = 0.02276

𝜋 (25)2 = 2945 𝑚𝑚2 4

𝜔=

𝜌=

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 =

𝜔=

𝜌=

𝜌𝑓𝑦 𝑓′𝑐

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+ 𝑓𝑦 )

𝐴𝑠 𝑏𝑑

𝜌=

𝜌𝑏 = 4,500 300(490)

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

𝜌𝑏 = 0.02276

0.0361(275) 𝜔= 27

𝜔 = 0.3118

𝑅𝑛 = 𝑓′𝑐 𝜔 (1 − 0.59 𝜔) 𝑀𝑢 = 𝜑 𝑅𝑛 𝑏𝑑2

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

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 0.05 𝛽1 = 0.85 − 7 (34.2 − 30) = 0.82

𝜌𝑏 =

𝐴𝑠 =

𝜌=

𝜌𝑏 =

0.85𝑓′𝑐 𝛽1 600 𝑓𝑦 (600+ 𝑓𝑦 )

0.85(34.2)(0.82)(600) 414(600+414)

𝜋 (25)2 𝑥 3 = 1473 𝑚𝑚2 4 𝜌=

𝐴𝑠

𝑏𝑑

1473

300(500)

𝜌𝑏 = 0.03407