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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 πβ²π = 21 πππ and ππ¦ = 345πππ, determine the balance steel area in sq.mm. SOLUTION ππ =
0.85πβ²π π½1 600 ππ¦ (600+ππ¦ )
π½1 = 0.85 π ππππ πβ²π < 30πππ ππ =
0.85(21)(0.85)(600) 345(600 + 345) ππ = 0.02792
π΄π π = ππ ππ
PROBLEM 2.2 A rectangular beam has b = 300 mm and d =490 mm. Concrete compressive strength πβ²π = 27.6πππ and steel yield strength ππ¦ = 276 πππ. Calculate the required tension steel area if the factored moment ππ’ is (a) 20 kN-m, (b) 140 kN-m, (c) 485 kN-m, and (d)620 kN-m. SOLUTION Solve for ππππ₯ πππ ππ’ πππ₯ : ππ =
0.85πβ²π π½1 600 ππ¦ (600+ππ¦ )
ππππ₯ = 0.75 ππ ππππ₯ =
ππππ₯ ππ¦ πβ² π
ππ =
0.85(27.6)0.85(600) 276(600+276)
ππ = 0.0495 ππππ₯ = 0.75(0.0495) ππππ₯ = 0.0371 0.03711(276) ππππ₯ = 27.6 ππππ₯ = 0.371
π
π πππ₯ = π β² π π(1 β 0.59π) π
π πππ₯ ππ πππ₯ = π
π πππ₯ ππ2
π
π πππ₯ = 27.6(0.371)[1 β 0.59(0.37)] = 8.001πππ
ππ πππ₯ = 8.001(300)(490)2 ππ πππ₯ = 576.279π₯106 π β ππ ππ πππ₯ = 576.279ππ β ππ ππ’ πππ₯ = 0.90 π₯ 576.279 ππ’ πππ₯ = 518.65 ππ β π
ππ’ πππ₯ = πππ πππ₯
a) ππ’ = 20ππ β π < ππ’ πππ₯ = (π πππππ¦ ππππππππππ) ππ’ = ππ
π ππ 2 20 x 106 = 0.90π
π (300)(490)2 π
π = 0.309 πππ
π=
π=
0.85πβ²π 2π
π [1 β β1 β ] ππ¦ 0.085πβ²π
0.85(27.6) 2(0.309 [1 β β1 β ] 276 0.85(27.6) π = 0.00113 < ππππ
ππππ =
π΄π = πππ
βπβ²π 1.4 ππ πβ²π > 31.36πππ, ππ‘βπππ€ππ π ππππ = 4ππ¦ ππ¦ 1.4 ππππ = = 0.005072 ππ¦ π΄π = 0.00572(300)(490) π¨π = ππππππ
b) ππ’ = 140ππ β π < ππ’ πππ₯ (singly reinforced) ππ’ = ππ
π ππ2 140 x 106 = 0.90 π
π (300)(490)2 π
π = 2.16 πππ
π=
π=
π΄π = πππ
0.85πβ²π 2π
π [1 β β1 β ] ππ¦ 0.85πβ²π
0.85(27.6) 2(2.16) [1 β β1 β ] 276 0.85(27.6)
π = 0.00822 > ππππ π΄π = 0.00822(300)(490) π΄π = π, ππππππ
c) ππ’ = 485 ππ β π < ππ’πππ₯ (singly reinforced) ππ’ = ππ
π ππ2 485 x 102 = 0.90π
π (300)(490)2 π
π = 7.48 πππ
π=
π=
0.85πβ²π 2π
π [1 β β1 β ] ππ¦ 0.85πβ²π
0.85(27.6) 2(7.48) [1 β β1 β ] 276 0.85(27.6) π = 0.03384 > ππππ
π΄π = π π π
π΄π = 0.03384(300)(490) π΄π = π, ππππππ
d) ππ’ = 600 ππ β π > ππ’πππ₯ 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 ππ¦ = 415 MPa and concreteπβ²π = 28 πππ. Concrete cover is 70 mm from the centroid of the steel area. Unit weight concrete is 23.5kN/π3 .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 πβ²π = 300 πππ d=480-70=410 mm π½1 = 0.85 1.4 ππ¦ = 415 πππ ππππ = π = 0.00337 π¦
Bar diameter , ππ = 20 π ππ Weight of beam, π€π = πΎπ π΄π = 23.5(0.3 π₯ 0.48 ) = 3.384 π a) Maximum factored moment on the beam. Factored load, ππ’ = 1.4(3.384 + 18) + .7 (14) Factored load, ππ’ = 53.738 ππ/π Maximum factored moment: π πΏ2
53.738(5)2
ππ’ = π’8 ππ’ = 8 ππ’ = πππ. ππ ππ΅ β π b) ππ’ = 280 ππ β π
Solve for ππ’πππ₯ to determine whether compression steel is needed ππ =
0.85πβ²π π½1 600 ππ¦ (600+ ππ¦ )
ππππ₯ = 0.75 ππ ππππ₯ =
ππππ₯ ππ¦ πβ²π
ππ =
0.85(28)(0.85)(600) 415(600+415)
ππ = 0.02881 ππππ₯ = 0.021261 ππππ₯ = 0.03203
π
π πππ₯ = π β² π ππππ₯ (1 β 0.59 ππππ₯ ) = 7.274 ππ’ πππ₯ = π π
ππππ₯ ππ2 = 330.14 ππ β π Required ππ’ = 280 ππ β π ππππ 415 0.85(28)
π΄π = π π π π΄π = 0.01755(300)(410) π΄π = 2159ππ2 π π π΄π = ππ2 2159 = (20)2 π 4 4 N=6.9 say 7 bars
1. ππ’ = 240ππ ππ‘ ππππ πππ π_π = 3.384 ππ/π (weight of beam) ππ’ =
ππ’ πΏ (1.4ππ )πΏ2 + = 314.805 ππ β π < ππ’ πππ₯ 4 8
(π πππππ¦)
314.805 π₯ 106
π
π
π = ππππ’ 2
π
π = 0.90(300)(410)2 π
π = 6.936πππ
π=
0.85πβ²π 2π
π [1 β β1 β ] ππ¦ 0.85πβ²π
π=
0.85(28) 2π
π [1 β β1 β ] = 002031 > ππππ 415 0.85πβ²π
π΄π = π π π
π΄π = 0.02031(300)(410) π΄π = 2498ππ2 π π π΄π = 4 ππ2 π 2498 = 4 (20)2 N π = 7.95 π ππ¦ π ππππ
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 πβ²π = 30 πππ and ππ¦ = 414 πππ, determine the tension steel area if the beam is to resist an ultimate moment of 650 kN-m. SOLUTION Solve for ππππ₯ and ππ’πππ₯ : ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
0.85(30)(0.85)(600) 414(600 + 414) ππ = 0.031 ππππ₯ = 0.75(0.031) ππππ₯ = 0.0232 ππ =
ππππ₯ = 0.15ππ
π½1 = 0.85 π ππππ πβ²π < 10 πππ
π=
πππ¦ πβ²π
π=
0.02323(414) 30 π = 0.3209
ππ’ πππ₯ = ππβ²π π π π2 (1 β 0.59π) ππ’ πππ₯ = 0.90(30)(0.3209)(300)(600)2 [1-0.59(0.309) ππ’ πππ₯ = 758.1 ππ β π > ππ’ Sinceππ’ < ππ’ πππ₯ , the beam may be designed as singly reinforced. π
π = 6.687 πππ
650 x 106 = 0.90π
π (300)(600)2 π
π = 6.687πππ
Solve for π:
π=
0.85πβ²π 2π
π (1 β β1 β ) ππ¦ 0.85πβ²π
π=
0.85(30) 2(6.687) [1 β β1 β ] = 0.0191 > ππππ 414 0.85(30)
ππππ =
1.4 = 0.00338 ππ¦
π΄π = πππ
π΄π = 0.0191(300)(600) π΄π = 3442 ππ2
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 πβ²π = 21 πππ and ππ¦ = 312 πππ. 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: ππ =
0.85 π β² π π½1 600
ππ¦ (600 + ππ¦ ) π½1 = 0.85 π ππππ ππ ππ πππ π π‘βππ 30 πππ ππ =
0.85(21)(0.85)(600) 312(312 + 600)
ππ = 0.03199 π πππ₯ = 0.75ππ
ππππ₯ = 0.75(0.03199) ππππ₯ = 0.02399
π΄π πππ₯ = π πππ₯ ππ
π΄π πππ₯ = 0.02399(300)(550) π΄π πππ₯ = π, πππ πππ
b) Required tension steel area: Factored load: ππ’ = 1.4 π· + 1.7 πΏ
ππ’ = 1.4(25) + 1.7(40) ππ’ = 103 ππ/π
Required strength: ππ’ =
ππ’ πΏ2 8
Solve for ππ’ πππ₯
103(6)2 8 ππ’ =463.5kN-m
ππ’ =
π=
π πππ₯ ππ¦ πβ²π
0.0299(312) 21 π = 0.356
π=
ππ’ πππ₯ = ππ β² π πππ2 (1 β 0.59π) ππ’ πππ₯ = 0.90(30)(0.356)(300)(550)2 [1 β 0.59(0.356)] ππ’ πππ₯ = 536.5 ππ β π > ππ’ π πππππ¦ ππππππππππ ππ’ = π π
π ππ2
ππ’ = 0.39 π
π (300)(550)2 463.5 π₯ 106 = 0.9 π
π (300)(550)2 π
π = 5.67 πππ
π=
0.85πβ²π 2π
π (1 β β1 β ) ππ¦ 0.85 π,π
π=
0.85(21) 2(5.67) [1 β β1 β ] 312 0.85(21)
π = 0.02269 π΄π = πππ
π΄π = 0.002269(300)(550) π΄π = 3743 ππ2
c) Number of 25 mm bars: Number of 25-mm bars=
π΄π π΄π 25 3.743
Number of 25-mm bars=π 4
(25)2
= 7.63 π ππ¦ 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 πβ²π = 28 πππ and steel yield strength ππ¦ = 248 πππ. 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 π 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: πβ²π = 28 πππ ππ¦ = 248 πππ
b=300 mm h=600 mm ππ’ = 540 ππ β π π½1 = 0.85 a) Balanced steel ratio: ππ =
0.85π β² π π½1 600
ππ =
ππ¦ (600 + ππ¦ )
0.85(28)(0.85)600 248(600 + 248)
ππ = 0.0577 = π. ππ% b) Effective depth using π = 0.5ππ π = 0.5(0.0577) = 0.0289 π=
πππ¦ πβ²π
π=
0.0289(248) = 0.2556 28
π
π = π β² π π(1 β 0.59π) π
π = 28(0.2556)[1 β 0.59(0.2556)] π
π = 6.0776 πππ ππ’ = πππ = ππ
π ππ2
540 x 106 = 0.90(8.307)(300)π 2
π = πππ ππ
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 ππ¦ = 275 πππ.Concrete strengthπβ²π = 21 πππ. 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 ππ’ = ππ
π π π 2 23 x 106 = 0.90 π
π (1000)(94)2 π
π = 2.892
π=
π=
0.85π β² π ππ¦
(1 β β1 β
2π
π ) 0.85π β² π
0.85(21) 2(2.982) (1 β β1 β ) 275 0.85(21)
ππππ₯ =
ππππ =
0.75 π₯ 0.85πβ²π π½1 600 = 0.0284 ππ¦ (600 + ππ¦ )
1.4 = 0.00509 ππ¦
π΄π = πππ
π΄π = 0.1154(1000)(94) π΄π = 1085 πππ
Spacing of bars (for walls and slabs using unit width): π
π =π
π =
π =
1000 π΄π π΄π
1000π΄π π΄π
Eq. 2-17 1000π΄π π = π΄π
π 1000 π₯ 4 (12)2 π = 1085 π = πππ ππ
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 πβ²π = 21 πππ and ππ¦ = 275 πππ. 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, ππ’ = 640 ππ β π
Maximum and minimum requirements: ππππ₯ = 0.75 π₯
0.85πβ²π π½1 600 ππ¦ (600+ππ¦ )
= 0.0284
ππ’ πππ₯ = 2528ππ β π (Procedure is not shown anymore see Problem 2.2) π΄π πππ =
1.4 ππ€ π ππ¦
= 5488 ππ2
Singly reinforced: ππ’ = ππ
π π π2
π=
0.85π β² π 2π
π (1 β β1 β ) ππ¦ 0.85π β² π
π=
0.85(21) 2(1.713 (1 β β1 β ) 275 0.85(21)
640 π₯ 106 = 0.90π
π (2800)(385)2 π
π = 1.713 πππ
π = 0.00656 π΄π = π π π
π΄π = 0.00656(2800)(385) π΄π = 7074 ππ2 > π΄π πππ
Number of 20 mm bars: π=
π΄π π΄π
7074 π=π 2 4 (20) π = 22.5 π ππ¦ ππ ππππ
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 πβ²π = 20.7 πππ and ππ¦ = 276 πππ. SOLUTION Required strength: ππ’ = 1.4 ππ + 1.7 ππΏ
ππ’ = 1.4(60) + 1.7(48) ππ’ = 165.6 ππ β π
(Note: this already includes the weight of beam)
ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
ππ =
0.85(20.7)(0.85)(600) 276(600 + 276)
ππ = 0.0371 ππππ =
1.4 = 0.00507 ππ¦ Note: this is the authorβs suggestion
Try π = 60% ππ π = 0.6(0.0371) = 0.02226
π=
πππ¦ πβ²π
π=
0.02226(276) 20.7
π = 0.2968 π
π = π β² π π(1 β 0.59π)
π
π = 20.7(0.2968)[1 β 0.59(0.2968)] π
π = 5.068
ππ’ = π π
π ππ2
165.6 x 106 = 0.90(5.068)ππ2 ππ 2 = 36.296 π₯ 106 ππ3
Try d = 1.75 b
b=228 mm say 230 mm d=399 say 30 mm
π΄π = πππ
π΄π = 0.02226(230)2(400) π΄π = 2.049 ππ2
Summary: b = 230 mm d = 400 mm π¨π = π, πππ πππ 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. 24ππ Assume normal weight concrete with = π3 . Use ππππ₯, πβ²π = 34 πππ, πππ ππ¦ = 345 πππ.
SOLUTION Weight of beam: (this is the authorβs assumption) Assuming a 300 mm x 600 mm, π_π = 24 π₯ (0.3 0.6) = 4.32ππ/π ππ’ = 1.4 (29 + 4.32) + 1.7(44) ππ’ = 121.448 ππ/π
ππ = 1.4 ππΏ + 1.7 ππΏ . ππ’ =
ππ’ πΏ2
ππ’ =
8
121.448(6)2 8
ππ’ = 546.516 ππ β π π½1 = 0.85 β ππ =
0.05 = 0.821 7(34 β 30)
0.85πβ²π π½1 600 ππ¦ (600+ ππ¦ )
ππ =
0.85(34)(0.821)(600) 345(600+345)
ππ = 0.04369
π = ππππ₯ = 0.75 (0.04369) ππππ = π=
π = 0.03277 > ππππ
βπβ²π = 0.00423 4ππ¦
πππ¦
π=
πβ²π
π ππππ πβ²π > 31.36 πππ
0.03277(345) 34
π = 0.332
π
π = π β² π π(1 β 0.56 π)
Assume d = 1.75 b
π
π = 34(0.332)[1 β 0.59(0.332)] π
π = 9.087 πππ
(this is the authorβs assumption)
ππ’ = π π
π π π 2
546.516 x 106 = 0.90(9.087)(π)(1. 75π)2 π = 279.4 ππ & π = 489 ππ
Use b = 280 mm, d = 490 mm Minimum beam the thickness (Section 409.6.2.1) βπππ =
ππ¦ πΏ (0.4 + ) 16 700
βπππ =
6000 345 (0.4 + ) 16 700
βπππ = 335 ππ ππΎ π΄π = π π π
π΄π = 0.03277(280)(490) π΄π = 4496 ππ2
Using 32 mm bars (#100): π=
π΄π π΄π
4496 π=π 2 4 (32) π = 5.6 π ππ¦ 6 ππππ
β = 490 + (25) + 32 + 20 β = 554.5 ππ > βπππ 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 πβ²π = 24 πππ, ππ¦ = 415 πππ.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: πβ²π = 24 πππ ππ¦ = 415 πππ ππ¦β = 275 πππ π = 290 ππ π» = 490 ππ πβ² = 60 ππ
ππ· = 35 ππ/π ππΏ = 55 ππ/π π = 490 β 60 = 430 ππ ππ’ = 1.4ππ· + 1.7 ππΏ
ππ’ = 1.4 (35) + 1.7 (55) ππ’ = 142.5 ππ/π
Solve for moment reactions using the three-moment equation: ππ΅ = β142.5 (2)(1) = β285 ππ β π Mo Lo + 2ππ΄ (πΏπ + πΏ1 ) + ππ΅ πΏ1 +
6π΄0 Μ
Μ
Μ
Μ
π0 πΏ0
0 + 2ππ΄ (0 + 6 ) + (β285 )(6) + 0 +
+
6π΄1 Μ
Μ
Μ
π0 πΏ1
142.5(6)3 4
=0
=0
ππ΄ = β498.75ππ β π ππ΄ = ππ΄ πππβπ‘
-489.75 = R(6)- 142.5(8)(4) R=676.875 kN
π
π΄ = ππ’ πΏ β π
π
π΄ = 142.5(8) β 676.875 π
π΄ = 463.125 ππ
Maximum positive moment: ππ· = 0
ππ· = π
π β ππ’
(π₯+2)2 2
ππ’ (2 + π₯) β π
= 0 142.5(2 + x) - 676.875 = 0 x = 2.75 m
ππ· = 676.875(2.75) β 142.5
(2.75+2)2
ππ· = 253.828 ππ β π Solve for πππ πππ₯ : ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
ππ =
0.85(24)(0.85)600 415(600 + 415)
ππ = 0.0247 ππππ₯ = 0.75 ππ
ππππ₯ =
ππππ₯ ππ¦ πβ²π
ππππ₯ = 0.75 (0.0247) ππππ₯ = 0.01852 ππππ₯ =
0.01852(415) 24
ππππ₯ = 0.3203
2
π
π πππ₯ = π β² π π(1 β 0.59 π) ππ πππ₯ = π
π π π 2
π
π πππ₯ = 415(0.3203)[1 β 0.59(0.3203)] π π πππ₯ = 6.235(290)(430)2 ππ πππ₯ = 334.316 ππ β π
πππ πππ₯ = 0.90(334.316) πππ πππ₯ = 300.884 ππ β π At a point of maximum positive moment: ππ’ = 253.828 ππ β π < π ππ πππ₯ ππ’ = ππ
π π π2
π=
0.85πβ²π π
π [1 β β1 β ] ππ¦ 0.85πβ²π
π=
0.85(24) 2(5.26) [1 β β1 β ] 415 0.85(24)
(Singly reinforced) 253.828 x 106 = .90 π
π (290)(430)2 π
π = 5.26 πππ
π = 0.01495 π΄π = π π π
π΄π = 0.01495(290)(430) π΄π = 1,864 ππ2
ANALYSIS OF RECTANGULAR BEAMS WHERE STEEL YIELDS (ππΊ = ππ ) 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 π β² π = 21 πππ and ππ¦ = 400 πππ, πππ‘ππππππ π‘βπ ππππππ€πππ: a) The coefficient of resistance π
π of the beam. b) The ultimate moment capacity of the beam. SOLUTION ππ =
0.85π β² π π½1
ππ¦ (600 + ππ¦ ) 0.85(21)(0.85)(600) = ππ 400(600 + 400) ππ = 0.02276 π΄π = 6 π₯ π=
π΄π ππ
π=
πππ¦ πβ²π
π (25)2 = 2945 ππ2 4 π=
π=
2945 = 0.01023 < ππ (π π‘πππ π¦πππππ ) 400(720)
0.01023(400) = 0.195 21
π
π = πβ²π π (1 β 0.56π)
π
π = 21(0.195)[1 β 0.59(0.195)] π
π = π. ππ π΄π·π
ππ’ = ππ
π π π2 ππ’ = 0.90(3.62)(400)(720)2 ππ’ = πππ. ππ ππ΅ β π 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 πβ²π = 27 πππ, ππ¦ = 275 πππ. SOLUTION ππ =
0.85πβ²π π½1 600 ππ¦ (600+ ππ¦ )
ππ =
0.85(27)(0.85)(600) 275(600+275)
ππ = 0.02276 π΄
4,500
π = πππ π=
πππ¦ πβ²π
π = 300(490) π=
0.0361(275) 27
π = 0.3118 π
π = πβ²π π (1 β 0.59 π)
π
π = 27(0.3118)[1 β 0.59(0.3118)] π
π = 6.87 πππ
ππ’ = π π
π ππ2
ππ’ = 0.90(6.87)(300)(490)2 ππ’ = πππ. π ππ΅ β π
PROBLEM 2.14 A rectangular beam has b = 300 mm, d = 500 mm, π΄π = 3 β 25 ππ, πβ²π = 34.2 πππ, grade 60 reinforcement (ππ¦ = 414 πππ). Calculate the design moment ππ’ .
SOLUTION π½1 = 0.85 β
0.85πβ²π π½1 600
ππ =
ππ =
ππ¦ (600+ ππ¦ )
π΄π = π=
0.05 (34.2 β 30) = 0.82 7 0.85(34.2)(0.82)(600) 414(600+414)
ππ = 0.03407
π (25)2 π₯ 3 = 1473 ππ2 4
π΄π
π=
ππ
1473 300(500)
π = 0.00982 < ππ ππ‘πππ π¦πππππ
Check if the beam satisfies the minimum requirement: βπβ²π ππππ = = 0.00353 ππΎ 4ππ¦ π=
πππ¦ πβ²π
π=
0.00982(414) 34.2
π
π = π β² π π(1 β 0.59π)
π
π = 34.2(0.1188)[1 β 0.59(0.1188)] π
π = 3.779 πππ
ππ’ = ππ
π ππ2
ππ’ = 0.90(3.779)(300)(500)2 ππ’ = πππ. ππ ππ΅ β π
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 πβ²π = 21 MPa and steel yield strength ππ¦ = 275 πππ. Unit weight of concrete is 23.5 kN/π3 . 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 β cover- 1/2 ππ d = 130-20-1/2(12)=104 mm
ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
π΄π = π΄π π₯ π π
π΄π = 4 (12)2
ππ =
0.85(21)(0.85)(600) 275(600 + 275) ππ = 0.0378
π π π΄π = 1028 ππ2 π΄π = π΄π π₯
1000 100
π΄π 1028 π= ππ 1000(104) π = 0.00989
π=
Check if the beam satisfies the minimum steel requirement on flexures: 1.4 = 0.00509 ππ¦
ππππ =
π=
πππ¦ πβ²π
π=
ππΎ
0.00989(275) 21 π = 0.129
π
π = πβ²π π (1 β 0.59 π)
ππ’ = ππ
π ππ2
b) ππ’ =
π
π = 21(0.129)(1 β 0.59(0.129)] π
π = 2.511 πππ
ππ’ = 0.90(2.511)(1000)(104)2 ππ’ = ππ. πππ ππ΅ β π ππ’ πΏ2 8
24.443 =
ππ’ (4)2 8
ππ’ = 12.222 ππ/π
Dead load pressure, ππ· = πΎπ x thickness of concrete. Dead load pressure, ππ· = 23.5 π₯ 0.13 = 3.055πππ ππ’ = 1.4ππΏ + 1.7 ππΏ
ππ’ = 1.4(ππ· π) + 1.7 (ππΏ π) 12.222 = 1.4(3.055 π₯ 1) + 1.7(ππΏ π₯ 1) ππΏ = π. πππ ππ·π
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 ππ¦ = 276.5 πππ and πβ²π = 20.7 πππ. SOLUTION π π΄π = 3 π₯ (25)2 = 1479 ππ2 4 ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
0.85(20.7)(0.85)(600) 276.5(600 + 276.5) ππ = 0.03703 ππ =
π΄π 1.473 π= ππ 250(460) π = 0.01281 < ππ (π π‘πππ π¦πππππ ) π=
Check if the beam satisfies the minimum steel requirement on flexure: ππππ =
π=
1.4 = 0.00506 ππ¦
πππ¦ πβ²π
π=
ππΎ
0.01281(276.5) 20.7 π = 0.171
π
π = πβ²π π(1 β 0.59π)
ππ’ = π π
π ππ2
ππ’πππ₯
ππ’ πΏ2 = 8
π
π = 27(0.171)[1 β 0.59(0.171)] π
π = 3.183 πππ
ππ’ = 0.90(3.183)(250)(460)2 ππ’ = 151.56 ππ β π ππ’ (6)2 151.56 = 8 ππ’ = 33.68 ππ/π 33.68 = 1.4 (0.68) + 1.7 ππΏπΏ ππΏπΏ = ππ. ππ ππ΅ β π
ππ’ = 1.4 ππ·πΏ + 1.7 ππΏπΏ
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 ππ¦ is 415 MPa and concrete strength πβ²π 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: ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
π½1 = 0.85
ππ =
0.85(28)(0.85)600 415(600 + 415) ππ = 0.028816 ππ = π. ππ%
b) Maximum steel area π΄π πππ₯ = ππππ₯ ππ
π΄π πππ₯ = (0.75 ππ ) ππ π΄π πππ₯ = (0.75 π₯ 0.028816)(300)(5) π΄π πππ₯ = π, πππ πππ
c) Nominal moment capacity Using 6-25 mm bars: π π΄π = (25)2 π₯ 6 = 2,945 ππ2 4 π=
π΄π ππ
π=
2,945 = π = 0.01963 < ππ (π‘πππ ππ π π‘πππ π¦πππππ ) 300(500)
πππ¦ 0.01963(415) π= = 0.291 πβ²π 28 π
π = π β² π π(1 + 0.59 π) = π
π = 28(0.291)(1 β 0.59 π₯ 0.291) π
π = 6.7494 πππ π=
ππ = π
π ππ 2 = ππ = 6.7494(300)(500)2 ππ = πππ. π ππ΅ β π 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 ππ¦ = 414 πππ, πβ²π = 34.5 πππ. 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 π½1 = 0.85 β
0.05 (34.5 β 30) = 0.818 7
ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
π΄π =
π (28)2 π₯ 5 = 3079 ππ2 4
π΄
ππ =
0.85(34.5)(0.818)(600) = 0.03428 414(600 + 414)
3079
π = πππ
π = 300(446) = π = 0.01972 < ππ ππ‘πππ π¦πππππ
Check if the beam satisfies the minimum requirement: ππππ =
π=
βπβ²π = 0.00355 4ππ¦
πππ¦ πβ²π
π=
0.01972(414) = 0.2367 34.5
π
π = πβ²π π(1 β 0.59π) = π
π = 34.5(0.2367)[1 β 0.59(0.2367)]=7.025 MPa ππ’ = ππ
π ππ2 = ππ’ = 0.90(7.025)(300)(446)2 = 440.18 kN - m
1.4(2P)
1.4P
ππ’ = 1.4(4.5) + 1.7(3) = 11.4ππ/π
A
π
π
B 2m
C 2m
D 2m
Figure 2.8 β Beam with factored loads For the given loads, the maximum moment can occur at B or C: ππ = 1.4π(2) + 11.4(2)(1) 440.18 = 1.4P(2) + 11.4(2)(1) π = 149 ππ
At point C: Set ππ = ππ’ At point B: (First solve for π
π΄ ) β ππ = 0
4 π
π΄ + 1.4π(2) = 2.8π(2) + 11.4(6)(1) π
π΄ = 17.1 + 0.7 π
β ππ΅ ππππ‘ Set ππ΅ = ππ’
ππ΅ = (17.1 + 0.7π) β 11.4(2)(1) 440.18 = (17.1 + 0.7 π)(2) β 11.4(2)(1) π = 306.27 ππ
Thus the maximum value of P such that ππ’ will not exceed 440.18 kN-m is 149 kN.
ANALYSIS OF RECTANGULAR BEAMS WHERE STEEL DOES NOT YIELDS (ππΊ β ππ ) PROBLEM 2.19 A rectangular beam has b = 300 mm, d = 500 mm, π΄π = 6 β 32 ππ, πβ²π = 27.6 πππ, grade 60 reinforcement (ππ¦ = 414 πππ). Calculate the ultimate moment capacity of the beam. SOLUTION ππ =
π΄π = π=
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
ππ =
0.85(27.6)(0.85)(600) 414(600 + 414) ππ = 0.0285
π (32)2 π₯ 6 = 4825 ππ2 4
π΄π
π=
ππ
4825 300(500)
= π = 0.03217 > ππ ππ‘πππ ππππ πππ‘ π¦ππππ
0.85
b=.300
c=0.85
ab
d=500
a
500-a/2 From Eq. 2-18
=4825 T=
ππ = 600
πβπ π
ππ = 600
500 β π π
β πΉπ» = 0 π=πΆ
π΄π ππ = 0.85 πβ²π π π, (4825)600
500βπ π
π = π½1 π = 0.85 π
= 0.85(27.6)(0.85π)(300) π 2 = 484π β 241,964 = 0 π = 306.2 ππ
ππ = 600
πβπ π
ππ = 600
500β306 306
= ππ = 379.65 πππ
π = π½1 π = 0.85(306.2) π = 260.3 ππ π πππ = ππ΄π ππ (π β ) 2 πππ = 0.90(4825)(379.65)(500 β
260.3 ) 2
πππ = πππ. π ππ΅ β π 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 ππ¦ = 415 πππ. πβ²π = 21πππ. Calculate the ultimate moment capacity of the beam.
SOLUTION ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
π΄π =
π (25)2 π₯ 7 = 3436 ππ2 4
π΄
ππ =
0.85(21)(0.85)(600) 415(600 + 415) ππ = 0.02161
3436
π = πππ
π = 300(490)
π = 0.02337 > ππ ππ‘πππ ππππ πππ‘ π¦ππππ
b=300
0.85 c=0.85
ab
d=490
a
490-a/2 =3436 From Eq.2-18: T=
ππ = 600
πβπ π
= ππ = 600
490 β π π
β πΉπ» = 0 π=πΆ (3436)600
π΄π ππ 490βπ π
= 0.85 πβ²π π π, π = π½1 π = 0.85 π = 0.85(221)(0.85π)(300) = π = 296.24 ππ
ππ = 600
πβπ π
490 β 296.24 296.24 ππ = 392.43 πππ < ππ¦
ππ = 600
π = π½1 π = 0.85(392.43) π = 251.81 ππ π
πππ = ππ (π β 2)
π
πππ = ππ΄π ππ (π β 2)
251.81
πππ = 0.90(3436)(392.43)(490 β 2 ) πππ = πππ. ππ ππ΅ β π
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 ππ¦ = 345 πππ and πβ²π = 21 πππ.
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 asπ, ππ , π
π ) may not be applicable. The moment can be computed using the assumptions in the Code and the conditions of equilibrium.
π (32)2 π₯ 4 4 π΄π = 3217 ππ2 π΄π =
Solve for the balanced π΄π to determine whether the given steel yield or not.
600π 600 + ππ¦
=
600(625) = 396.825ππ 600 + 345
a
625mm
ππ =
125 125 125
125
πΆπ =:
From Eq. 2-11
π = π½1 π π = 0.85(396.825)= 337.3 ππ π΄π = 337.3(375) β 125(125) = 110,863 ππ2 π=πΆ π΄π π ππ¦ = 0.8πβ²π π΄π π΄π π (345) = 0.85(21)(110,863) π΄π π = 5,736 ππ2
4-32mm 375mm
Since π΄π ππππ£ππππ < π΄π π , tension steel yields. πΆ=π 0.85π β² π (ππ β 1252 ) = π΄π ππ¦ 0.85(21)(π π₯ 375 β 1252 ) = 3,217(345) π = 207.5 ππ
ππ = ππ1 β ππ2
π 125 = ππ = πΆ1 (π β ) β πΆ2 (π β ) 2 2
ππ = 0.85(21)(207.5)(375)(625 β
207.5 ) 2
ππ = 567.03 ππ β π πππ = 0.90(567.03) πππ = πππ. ππ ππ΅ β π PROBLEM 2.22 Compute the ultimate moment capacity of the beam shown in Figure 2.10. Assume ππ¦ = 345πππ and πβ²π = 21 πππ.
πΆπ =
600 π 600 + ππ¦
πΆπ =
600(375) 600 + 345
πΆπ = 238 ππ ππ = π½1 πΆπ ππ = 0.85(238) ππ = 202.4 ππ π₯ 375 5 = π₯= π π 450 6
375
Solve for π΄π :
3-22mm 75 375m m
450mm
SOLUTION π π΄π = (22)2 π₯ 3 4 π΄π = 1,140 ππ2
Figure 2.10
π₯ = 168.7ππ
x
π΄π = 17,066 ππ2
π = πΆπΆ π΄π π ππ¦ = 0.85πβ²π π΄π π΄π π (345) = 0.85(21)(17,066) π΄π π = 883 ππ2 < π΄π
322mm 375m m
7 5
Sinceπ΄π ππππ£ππππ > π΄π π , tension steel does not yield (ππ < ππ¦ ) solve for c: πΆπΆ = π ππ = 600 π = π½1 π
0.85πβ²π π΄π = π΄π ππ πβπ 5 2 πβπ 0.85(21) π = 1140 π₯ 600 π 12 π 7.437(0. 85π)2 = 684,00
375βπ π
π = 250.92 ππ π = π½1 π
π = 0.85(250.92) = 213.3 ππ
2 2 ππ = πΆπ π₯ (π β π) = 0.85π β² π π΄π (π β π) 3 3 5 2 ππ = 0.85(21) (213. 29)2 π₯ [375 β (213.3)] 12 3 ππ = 78.77 ππ β π πππ = 0.90 π₯ 78.77 = ππ. ππ ππ΅ β π
m
c
d375 450m m 450m
π΄π = 1/2(π₯)(π) 1 5 5 π΄π = 2 π₯ 6 π π₯ π = π π2
a
d-(2/3)a
T
PROBLEM 2.23 A hallow beam is shown in Figure 2.11. Assume πβ²π = 28 πππ and ππ¦ = 345 πππ. a) Calculate the required tension steel area when ππ’ = 800ππ β π. 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 ππ’ = 1200ππ β π.
Figure 2.11Hallow beam
SOLUTION To guide us whether βa: will exceed 150 mm or not, let us solve the design moment when a=150 mm. d = 800 β 75 = 725 mm
π
πππ = ππΆπΆ (π β 2)
πππ = 0.90 π₯ 0.85(28)(150) (725 β
150 2
)
πππ = 1044.225 ππ β π a) ππ’ = 800 ππ β π Since the required ππ’ = 800 ππ β π < 1044.25 ππ β π, π < 150 ππ. Assuming tension steel yields: π ππ’ = ππΆπ (π β ) 2 π ππ’ = π0.85πβ²π π π(π β ) 2 800 π₯ 106 = 0.90 π₯ 0.85(28)π(500)(725 β 0.5π) ππ’ = πππ
π = 111.6ππ < 150 ππ Check is steel yields: ππ = 600
πβπ π
ππ = 600
725 β 131.2 = 2,712 πππ > ππ¦ 131.3
π€βπππ π =
π=πΆ
π = 131.3 ππ π½1 π π‘πππ π¦πππππ
π΄π ππ¦ = 0.85πβ²π π π π΄π (345) = 0.85(28)(111.6)(500) π΄π = π, πππ πππ
b) Balanced condition (See Figure 2.12) πΆπ = π = π½1 πΆπ
600π 600 + ππ¦
πΆπ =
600(725) = 460.32ππ 600 + 345
π = 0.85(460.32) = 391.3 ππ
π§ = π β 150 = 241.27 ππ π΄1 = 500(150) = 75,000 ππ2 1 π¦1 = 725 β = 650ππ 2(150) π΄1 = 125(241.27) = 30,159 ππ2 1 π¦2 = 725 β 150 β = 454.37 2(241.27) πππ = πΆ1 π¦1 + 2πΆ2 π¦2 πππ = 0.85π β² π (π΄1 π¦1 + 2π΄2 π¦2 ) πππ = 0.85(28)[75,000 π₯ 650 + 2 π₯ 30,159 π₯ 454.37] πππ = 1812.52ππ β π πππ = 0.90 π₯ 1812.52 = ππππ. π ππ΅ β π 500m m
25 0
12 5 150
12 5
725
1 2
2
z
a
T
Figure 2.12
c) Maximum steel area, π΄π πππ₯ π = πΆ1 + πΆ2 π΄π π ππ¦ = 0.85π β² π (π΄1 + 2π΄2 ) π΄π π (345) = 0.85(28)(75,00 + 2 π₯ 30,159) π΄π π = 9,335 ππ2 π΄π πππ₯ = 0.75 π΄π π
π΄π πππ₯ = 0.75(9,335) π΄π πππ₯ = π, πππ πππ
d) Maximum moment , ππ’ πππ₯ : Refer to Figure 2.12:
πΆ1 + πΆ2 = π 0.85(28)[75,000 + 2π΄2 ] = 7,001(245) π΄2 = 13,244 ππ2 π΄2 = 125 π§
π¦2 = 725 β
ππ πππ₯ ππ πππ₯ ππ πππ₯ ππ πππ₯
13,244 = 125 π§ π§ = 105.95 ππ 1501 = 522.03 ππ 2(105.95)
= πΆ1 + π¦1 + 2πΆ2 π¦2 = 0.85π β² π (π΄1 π¦1 + 2π΄2 π¦2 ) = 0.85(28)[75,00 π₯ 650 + 2 π₯ 13,244 π₯ 522.03] = 1489.34 ππ β π
πππ πππ₯ = 0.90 π₯ 1189.34 = ππππ. π ππ΅ β π e) ππ’ = 1200ππ β π < πππ πππ₯ Refer to Figure 2.12 π΄1 = 75,000 ππ2
π¦1 = 650 ππ
(ππππππ¦ ππππππππππ)
π΄2 = 125π§
π¦2 = 575 β 0.5π§
ππ’ = πππ 1200 π₯ 106 = 0.90 π₯ 0.85 π β² π (π΄1 π¦1 + 2π΄2 π¦2 ) 1200 π₯ 106 = 0.90 π₯ 0.85(28)[75,000(650) + 2(125π§)(575 β 0.5π§)] π§ = 53.04ππ π΄π = π΄1 + π΄2
π΄π = 75,000 + 2 π₯ 125(53.04) π΄π = 88,259.2 ππ2
π=πΆ π΄π ππ¦ = 0.85πβ²π π΄π π΄π (345) = 0.85(28)(88,259.2) π΄π = π, πππ πππ
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. πβ²π = 20.7 πππ, ππ¦ = 344.8 πππ, ππ = 2.832 πππ΄. Modulus of elasticity of concrete πΈπ = 21,650 πππ and πΈπ = 200 πΊππ. 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 β 25 mm Γ
c h = 600 mm
d = 530 mm
b = 350 mm
N.A.
d-c
Effective moment of inertia, πΌπ :
Eq. 2-19
πππ 3 πππ 3 πΌπ = ( ) πΌπ + [1 β ( ) ] πΌππ β€ πΌπ ππ ππ πΌπ = ππππππ‘ ππ πππππ‘ππ ππ ππππ π π πππ‘πππ πβ3 πΌπ = 12
350(600)3 = = 6300 π₯ 106 ππ4 12
πππ =
ππ πΌπ π¦π‘
πππ =
2.832(600 π₯ 10)6 = 59.472 ππ β π 600/2
π€βπππ π¦π‘ = 1/2(600) = 300 ππ
ππ = πππ₯πππ’π ππππππ‘ ππ ππππ π€πΏ2 ππ = 8 ππ =
π€ = π€π· + π€πΏ = 11 + 15 = 26 ππ/π
26(8)2 = 208ππ β π 8
πΌππ = Moment of inertia of cracked section with steel transformed to concrete
From Figure 2.13: πΈ
Modular ratio, π = πΈπ = 9.238 π
π π΄π = 9.328 π₯ 6 π₯
π (25)2 = 27,208 ππ2 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
πΌππ = πΌππ΄ πΌππ =
ππ 3 = + π π΄π (π β π)2 3
350(219.7)3 + 27,208(530 β 219.7)2 3
πΌππ = 3,857 π₯ 106 ππ3 πππ 3 πππ 3 πΌπ = ( ) πΌπ + [1 β ( ) ] πΌππ ππ ππ 59.472 3 59.472 3 6 πΌπ = ( ) π₯ 600 π₯ 10 + [1 β ( ) ] π₯ 3,857 π₯ 106 208 208 πΌπ = 3,914 π₯ 106 ππ4 < πΌπ
(ππΎ)
a) Instantaneous Deflection: πΏ=
5π€πΏ4 384 πΈπ πΌπ
πΏ=
2(26)(8000)4 384(21,650)(3,914 π₯ 106 )
πΏ = ππ. ππ ππ b) Long-term Deflection Since only 40% of the live load was sustained: w = 11 + 0.4(15) = 17 kN/m
5π€πΏ4
Instantaneous deflection πΏ = 384 πΈ
π πΌπ
πΏ=
5(17)(8)4 (1000)4 384(21,650)(3,914 π₯ 106 ) πΏ = 10.7 ππ
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 Partβaβ. πΏ1 16.36 = 17 26 πΏ1 = 10.7 π Long-term deflection = πΏ + πΏ1 π=
π 1 + 50 πβ²
π=2 πβ² = 0 π=
πππ 5 π¦ππππ ππ ππππ π ππππ π‘βπππ ππ ππ πππππππ π πππ πππππππππππππ‘
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 ππ = 17.2πππ. The modulus of elasticity of concrete is given by the expression πΈπ = 4700 ππππ‘(π β² π ) and the modulus of rapture is given by the expression ππ = 0.7 ππππ‘(π β² π ). 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 mmΓΈ
3-32 mmΓΈ
5-32 mmΓΈ
145 kN-m
202 kN-m
202 kN-m
SOLUTION πΈπ = 4700βπβ²π = 4700β17.2 = 19,492 πππ΄ ππ = 0.70βπβ²π = 0.7 β17.2 = 2.903 πππ a) Effective moment of inertia at the supports Maximum moment, ππ’ = 202ππ β π Distance from NA of gross section to extreme tension fiber, ππ‘ = 310 ππ Moment of inertia of gross section, πΌπ = 0.00715 π4 Moment of inertia of cracked section, πΌπ = 0.00573 π4 πππ =
ππ πΌπ π¦π‘
πππ =
2.903(0.00715 π₯ 10004 ) 10
πππ = 66.959 ππ β π
πππ 3 πππ 3 πΌπ = ( ) πΌπ + [1 β ( ) ] πΌππ ππ ππ 66.9593 66.959 3 πΌπ = ( ) π₯ 0.00715 + [1 β ( ) ] π₯0.00573 202 202 πΌπ = π. πππππππ ππ b) Effective moment of inertia for the continuous member πΌπ =
(πΌπ )πππ₯ πππ ππππ‘ + (πΌπ )πππ₯ πππ ππππππ‘ 3
(ππππ‘. 409.6.2.4)
At maximum negative moment (at support) πΌπ = 0.0057817 π4 Solving for πΌπ at maximum positive moment (at midspan) πΌπ = 0.0138 π4 (πππ‘π‘ππ ππππππ ππ π‘πππ πππ) ππ‘ = 620 β 194 = 246 ππ πΌππ = 0.00513 π4 πππ =
ππ πΌπ ππ‘
πππ =
2.903(0.00715 π₯ 10004 ) 310
πππ = 66.959 ππ β π πππ 3 πππ 3 πΌπ = ( ) πΌπ + [1 β ( ) ] πΌππ ππ ππ (πΌπ )πππ₯ πππ ππππ‘ + (πΌπ )πππ₯ πππ ππππππ‘ 2 0.0057817 + 0.007932 πΌπ = = π. ππππππ ππ 2 πΌπ =
c) Additional long term deflection= long term deflection x π π=
π 1 + 50πβ²
πβ² = 0(π ππππ π‘βπππ ππ ππ πππππππ π πππ πππππππππππππ‘ ππ‘ ππππ πππ) π = 2(πππ‘ππ 5 π¦ππππ ) π=
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: πΏ1 5 = 22.4 48 πΏ1 = 2.33 π Additional long term deflection = 2.333 x π =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 πβ²π = 27.6 πππ and ππ¦ = 276 πππ for main and temperature bars. The slab is not exposed to earth or weather. Use unit weight of concrete πΎπ = 23.5 ππ/π3. SOLUTION Consider 1 m strip of slab, b= 1000 m Uniform live load, π€πΏ = 7.5 πΎππ π₯ 1π = 7.5 ππ/π Minimum slab thickness from Table 2.1: πΏ
ππ¦
βπππ = 20 (0.4 + 700)
βπππ =
3000 20
276
(0.4 + 700)
βπππ = 119 ππ (π’π π 120 ππ)
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: ππ = πΎππππ π₯ π π₯ β
ππ = 23.5 (1)(0.12) ππ = 2.82 ππ/π
Factored floor pressure load: ππ’ = 1.4π€π + 1.7 π€πΏ
ππ’ =
ππ’ = 1.4(2.82) + 1.7(7.5) ππ’ = 16.698 ππ/π
ππ’ πΏ2
16.698(3)2
ππ’ = 2 ππ’ = 18.785 ππ β π
8
ππ’ = ππ
π π π2
π=
π=
0.85π β² π ππ¦
[1 β β1 β
18.785 π₯ 106 = 0.90 π
π (1000)(94)2 π
π = 2.362 πππ
π
π’ ] 0.85π β² π
0.85(27.6) 2(2.362) [1 β β1 β ] 276 0.85(27.6)
π = 0.009039 Check for ππππ and ππππ₯ : ππππ =
1.4 = 0.00507 ππ¦
ππππ₯ =
0.75 0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
ππΎ
ππππ₯ =
0.75 0.85(27.6)(0.85)600 276(600 + 276)
ππππ₯ = 0.037 > 0.009309 (ππΎ)
π΄π = πππ
π΄π = 0.009039(1000)(94) π΄π = 850 ππ2 per meter width of slab
Using 12-mm main bars: Spacing s =
π΄πππ π΄π
π₯ 1000
π =
π (12)2 4
850
π₯ 1000
π = 138 ππ π ππ¦ 135 ππ Maximum spacing required by the Code: a) 3(β) = 3(120) = 360 ππ b) 450 ππ
ππΎ
Thus, use 12 mm main bars at 135 mm o.c. Temperature bars: (Grade 275) π΄π‘ = 0.002πβ
Spacing =
π΄πππ π΄π
π΄π‘ = 0.002(1000)(120) π΄π‘ = 240 ππ2 π₯ 1000
π =
π (10)2 4
240
π₯ 1000
π = 327 ππ π ππ¦ 325 ππ Maximum spacing required by the Code: a) 5β = 5(120) = 600ππ 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 πβ²π = 21 πππ and ππ¦ = 415 πππ for main bars and ππ¦ = 276 πππ for temperature bars. Steel cover is 20 mm. Unit weight of concrete is 23.5 kN/π3 .
SOLUTION Consider 1 m strip, b = 1000 mm Uniform live load, π€πΏ = 4 πππ π₯ 1π = 4 ππ/π Minimum slab thickness from Table 2.1: βπππ =
πΏ 28
βπππ =
4000 28
βπππ = 143 ππ (π’π π 150 ππ) Weight of beam (DL): π€π· = πΎππππ π₯ π π₯ β
π€π· = 23.5(1)(0.15) π€π· = 3.525 πππ
π€π’ = 1.4 π€π· + 1.4 π€πΏ
π€π’ = 1.4(3.525) + 1.7(4) π€π’ = 11.735 ππ/π
Maximum factored moment, Section 408.4 (See Page 29) LL < 3 DL Column
Column
Column
Spandrel Beam
Shear
Moment
Effective depth, d = 1.50 β 20 β 1/2 (12) Effective depth, d = 124 mm At midspan: π€π’ πΏπ2 ππ’ = 16 ππ’ = ππ
π ππ2
11.735 (4)2 ππ’ = 16
ππ’ = 11.735 ππ β π
11.735 π₯ 106 = 0.90 π
π (1000)(124)2 π
π = 0.848 πππ
π=
0.85πβ²π 2π
π [1 β β1 β ] ππ¦ 0.85πβ²π
π=
0.85(21) 2(0.848) [1 β β1 β ] 415 0.85(21)
π = 0.0021
ππππ =
1.4 = 0.00337 > 0.0021 ππ¦
Use π = ππππ = 0.00337 π΄π = πππ
π΄π = 0.00337(1000)(124) π΄π = 418 ππ2
Spacing, s =
π΄π π π΄π
π₯ 1000 =
π =
π (12)2 4
418
π₯1000
π = 271 π ππ¦ 270 ππ Maximum spacing required by the Code: a) 3 β = 3(150) = 450 ππ b) 450 mm Thus, use 12 mm bottom bars at 270 mm o.c. at midspan At support: ππ’ =
π€π’ πΏπ 2 10
ππ’ = ππ
π ππ2
ππ’ =
=
11.735(4)2 10 ππ’ = 18.776 ππ β π
18.776 π₯ 106 = 0.90π
π (1000)(124)2 π
π = 1.357 πππ
π=
0.85πβ²π 1 β 2π
π [1 β β ] ππ¦ 0.85πβ²π
π=
0.85(21) 2(1.357) [1 β β1 β ] 415 0.85(21)
ππππ₯ = 0.0034 > ππππ ππππ₯ = 0.75
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
ππππ₯ = 0.75
0.85(21)(0.85)600 415(600 + 415)
ππππ₯ = 0.0162 > 0.0034 Use π = 0.034 π΄π = πππ
Spacing, π =
π΄π = 0.0034(1000)(124) π΄π = 422 ππ2 π΄π π π΄π
π₯ 1000
π =
π (12)2 4
422
π₯ 1000
πππππππ = 268 π ππ¦ 265 ππ Thus, use 12 mm top bars @ 265 mm o.c. at support Temperature bars (10 mm): (ππ‘ = 0.002) π΄π‘ = 0.002πβ = π΄π‘ = 0.002(1000)(150)
Spacing, s =
π΄π π π΄π
π₯ 1000 = π =
π (10)2 4
300
π΄π‘ = 300 ππ2
π₯ 1000
π = 261 π ππ¦ 260 ππ
Maximum spacing required by the Code: a) 5β = 5(150) = 750 ππ 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 (ππ¦ = 275 πππ) spaced at 140 mm o.c. Steel covering is 20 mm. Calculate the uniform live load pressure that a slab can carry. Use πβ²π = 20.7 πππ. Unit weight of concrete is 23.5 kN/π3 . SOLUTION
Consider 1 m strip of slab, b = 1000 m
π€π = πΎπ π β π€π = 23.5(1)(0.16) π€π = 3.76 ππ β π
Dead load:
d = 160 β 20 β 1/2(12) d = 134 mm
Effective depth:
Steel area, π΄π =
1000 π
π=
π΄π = ππ
ππ =
0.85πβ²π π½1 600 ππ¦ (600 + ππ¦ )
π=
πππ¦ πβ²π
π₯ π΄π
140
π₯
π 24
(12)2
807.8 1000(134) π = 0.006028
π=
=
ππ’ = ππ
π π π2
= ππ =
π=
0.85(20.7)(0.85)(600) 275(600 + 275) ππ = 0.037 > π (π π‘πππ π¦πππππ )
0.006028(275) 20.7 π
π = 20.7(0.0801)[1 β 0.59(0.0801)] π
π = 1.58 πππ
ππ’ = 20.7(0.0801)[1 β 0.59(0.0801)] ππ’ = 25.5334 ππ β π
π€π’ πΏ2 π€π’ (3)2 = 25.5334 = 8 8
π€π’ = 1.4 π€π·πΏ + 1.7 π€πΏπΏ π€πΏπΏ
1000
π΄π = 807.8 ππ2
π
π = πβ²π π(1 β 0.59π) =
ππ’ =
= π΄π =
π€π’ = 22.696 kN/m
= 22.696 = 1.4(3.76) + 1.7 π€πΏπΏ π€πΏπΏ = 10.25 ππ/π = πππππππ ππππ π π’ππ π₯ π 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. πβ²π = 30 πππ, ππ¦ = 415 πππ. 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 πβ²π > 28 πππ; π½1 = 0.85 β
0.05 β² (π π β 28) 7
π½1 = 0.85 β π½1 = 0.836
0.05 (30 β 28) 7
a) Balanced steel area: ππ =
0.85π β² π π½1 600 ππ¦ (600 + ππ¦ )
π΄π π = ππ π π
0.85(30)(0.836)(600) 415(600 + 415) ππ = 0.03036 ππ =
π΄π π = 0.03036(310)(490) π΄π π = π, πππ πππ
b) Maximum steel area when beam is singly reinforced: From Eq. 2-24:
3 0.85π β² π π½1 ππ¦ 3 0.85(30)(0.836)
ππππ₯ = 7 ππππ₯ = 7
415(600+415)
ππππ₯ = 0.0221
π΄π πππ₯ = ππππ₯ π π
π΄π πππ₯ = 0.0221(310)(490) π΄π πππ₯ = π, πππ πππ
c) Maximum design strength, πππ πππ₯ : π = 0.004, ππ = 800 πππ 51
ππ πππ₯ =
3
ππ πππ₯ = 140 π½1 πβ²π ππ 2 (1 β 14 π½1 )
From Eq. 2-25 :
51 3 (0.836)(30)(310)(490)2 (1 β π₯ 0.836) 140 14 ππ πππ₯ = 558.05 ππ β π
From Eq. 2-26: π = 0.65 + 0.25 π = 0.65 + 0.25
800βππ¦ 1000βππ¦ 800β415 1000β415
π = 0.8145 πππ πππ₯ = 0.8145(558.05) πππ πππ₯ = πππ. ππ ππ΅ β π d) ππ’ = 1.2ππ· + 1.6 ππΏ
ππ’ = 1.2(120) + 1.6(170) ππ’ = 451.45 ππ β π
Thus, the beam is singly reinforced. Determine if the beam is tension-controlled: 459
3
From Eq. 2-22: πππ‘π = 1600 π½1 πβ²π ππ 2 (1 β 16 π½1 ) πππ‘π = 451.45 ππ β π Since the required ππ’ ππ πππ π π‘βππ ππ‘π , the section is tension controlled. π = 0.90
ππ’ = πππ π ππ = π π₯ 0.85 πβ²π π π (π β 2)
π 416 π₯ 106 = 0.90 π₯ 0.85(30)(π)(310)(490 β ) 2 π = 139.06 ππ Check if it is really tension-controlled: π=
π 139.06 = = 166.4 ππ π½1 0.836
ππ = 600
πβπ 490 β 166.4 = 600 = 1,167 πππ π 166.4 > 1,000 πππ (ππΎ)
PROBLEM 2.30 Given the following data for a rectangular beam: width π = 320ππ, effective depth π = 520 ππ, πβ²π = 27 πππ, ππ¦ = 345 πππ. Dead load moment ππ· = 180 ππ β π, Live load moment ππΏ = 167 ππ β π. π·ππ‘ππππππ π‘βπ ππππ’ππππ π‘πππ πππ π π‘πππ ππππ. SOLUTION π½1 = 0.85 ππ = 1.2 ππ· + 1.6 ππΏ
ππ’ = 1.2(180) + 1.6(167) ππ’ = 483.2 ππ β π
Solve for πππ πππ₯ to determine if compression steel area is required.
ππ πππ₯ =
51 3 π½1 πβ²π ππ 2 (1 β π½) 140 14 1
ππ πππ₯ =
51 3 (0.85)(27)(320)(520)2 (1 β π₯ 0.85) 140 14
ππ πππ₯ = 591.64 ππ β π π = 0.65 + 0.25
800βππ¦ 1000βππ¦
=0.8237
πππ πππ₯ = 487.31 ππ β π > ππ’
(π πππππ¦ ππππππππππ)
Solve for πππ‘π to determine if the section is tension-controlled. πππ =
459 3 π½1 πβ²π ππ2 (1 β π½1 ) = 478.9 ππ β π 1600 16
Since ππ’ > πππ‘π , the section is within βtransition regionβ, i.e 0.65 < π < 0.90 ππ’ = πππ = π π₯ 0.85 πβ²π ππ (π β π/2) 520 β π 600 β 345 ππ β ππ¦ π π = 0.65 + 0.25 = 0.65 + 0.25 1000 β ππ¦ 1000 β 345 π=
119.084 + 0.2893 π
π = 0.85π 119.084 + 0.2893) π₯ 0.85(27)(0.85π)(320)(520 π β 1/2π₯ 0.85π)
483.2 π₯ 106 = (
π = 208.8 ππ π = π½1 π = 177.45 ππ π=πΆ
π΄π ππ¦ = 0.85πβ²π π π π΄π (34.5) = 0.850(27)(177.45)(320) π΄π = 3,777 ππ2
PROBLEM 2.31 Given the following properties of a rectangular concrete beam: b = 280 mm, d = 480 mm, πβ²π = 21 πππ, ππ¦ = 415 πππ. 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 π½1 = 0.85 ππ = π΄π =
π ππππ πβ²π ππ πππ π π‘βππ 28 πππ
0.85π β² π π½1 600 ππ¦ (600 + ππ¦ )
= ππ =
0.85(21)(0.85)(600) = ππ = 0.0216 415(600 + 415)
π (25)2 = 490.87 ππ2 4
a) π΄π = 3 π₯ π΄π = 1473 ππ2 π=
π΄π ππ
=
1473 280(480) π = 0.01096 < ππ
π=
(π π‘πππ π¦πππππ )
πΆ=π 0.85π β² π π π = π΄π ππ¦ 0.85(21)(π)(280) = 1473(415) π = 122.28 ππ π π= = 143.86 ππ π½1 ππ = 600
πβπ 480 β 143.86 = 600 = 1,402 πππ > 1,000 πππ π 143.86
The section is tension-controlled, π = 0.90 ππ = πΆπ (π β π/2) = ππ = 0.85πβ²π π π (π β π/2) ππ = 0.85(21)(122.28)(280)(480 β 122.28/2) ππ = 255.87 ππ β π πππ = 0.90(255.87) πππ = πππ. ππ ππ΅ β π b) π΄π = 4 π₯ π΄π = 1963 ππ2 π΄π 1963 = ππ 280(480) π = 0.014961 < ππ π=
(π π‘πππ π¦πππππ )
πΆ=π 0.85πβ²π π π = π΄π ππ¦ 0.85(21)(π)(280) = 1963(415) π = 163.04 ππ π=
π = 191.81 ππ π½1
ππ = 600
πβπ 480 β 191.81 = 600 = 901.5 πππ < 1,000 πππ π 191.81
The section withinβ transition regionβ, i. e 0.65 < π < 0.90 π = 0.65 + 0.25
ππ β ππ¦ 1000 β ππ¦
= π = 0.65 + 0.25
901.5 β 415 1000 β 415
π = 0.858 ππ = πΆπ (π β π/2) = ππ = 0.85πβ²π π π (π β π/2) ππ = 0.85(21)(163.04)(280)(480 β 163.04/2) ππ = 324.504 ππ β π πππ = 0.858(324.504) πππ = πππ. πππ ππ΅ β π c) π΄π = 7 π₯ π΄π = 3436 ππ2 π=
π΄π 3436 = = 0.02557 > ππ ππ 280(480)
(π π‘ππ ππππ πππ‘ π¦ππππ)
The section is compression-controlled, π = 0.65 π=πΆ π΄π ππ = 0.85πβ²π π π 480 β π 3436 π₯ 600 = 0.85(21)(0.85π)(280) π π = 297.56 ππ π = π½1 π = 252.92 ππ ππ = πΆπ )π β π/2) = ππ = 0.85π β² π π π (π β π/2) ππ = 0.85(21)(252.92)(280)(480 β 252.92/2) πππ = 0.65(446.91) πππ = πππ. ππ ππ΅ β π
PROBLEM 2.32 A hallow beam is shown in Figure 2.16. Assume πβ²π = 28 πππ and ππ¦ = 345 πππ. a) Calculate the required tension steel area when ππ’ = 800 ππ β π 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 ππ’ = 1200 ππ β π.
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. π = 800 β 75 = 725 ππ
To guide us whether βaβ will exceed 150 mm or not, let us solve the design moment when a =150 mm. π=
π = 176.47 ππ π½1
ππ = 600
πβπ = 1,865 πππ > 1000 πππ π
ππππ πππ ππππ‘ππππ , π = 0.90
πππ = ππΆπ (π β π/2) = 0.90 π₯ 0.85(28)(150)(500)(725 β 150/2) πππ = 1044.225 ππ β π a) ππ’ = 800 ππ β π Since the required ππ’ = 800 ππ β π < 1044.225 ππ β π, π < 150 ππ. ππ’ = πππ = ππΆπ (π β π/2) ππ’ = π0.85πβ²π π π(π β π/2) 800 π₯ 106 = 0.90 π₯ 0.85(28)π(500)(725 β 0.5π) π = 111.6 ππ < 150 ππ Stress in steel πβπ π π€βπππ π = = 131.3 ππ π π½1 725 β 131.2 ππ = 600 = 2,712 πππ > ππ¦ π π‘πππ π¦πππππ 131.3 ππ = 600
π = πΆπ
π΄π ππ¦ = 0.85πβ²π π π π΄π (345) = 0.85(28)(111.6)(500) π΄π = π, πππ πππ
b) Balanced condition: π = 0.65
πΆπ =
600π 600 + ππ¦
= πΆπ =
600(725) = 460.32 ππ 600 + 345
π = π½1 ππ = π = 0.85(460.32) = 391.3 ππ 500 mm 125
250 125 150
N
a 725
T
π§ = π β 150 = 241.27ππ π΄1 = 500(150) = 75,000 ππ2 π΄1 = 125(241.27) = 30,159 ππ2 πππ πππ πππ πππ
Figure 2.17 π¦1 = 725 β 1/2(150) = 650 ππ π¦2 = 725 β 150 β 1/2(241.27) = 454.37
= πΆ1 π¦1 + 2πΆ2 π¦2 = 0.85π β² π (π΄1 π¦1 + 2π΄2 π¦2 ) = 0.85(28)[75,000 π₯ 650 + 2 π₯ 30,159 π₯ 454.37] = 1812.52 ππ β π
ππππ = 0.65 π₯ 1812.52 ππππ = ππππ. ππ ππ΅ β π
c) Maximum steel area, π΄π πππ₯ πΆπππ₯ =
3 π = 310.71 ππ 7
π
ππππ π‘π πΉπππ’ππ 2.17 π = π½1 ππππ₯ = 264.11 ππ
π§ = π β 150 = 114.11 ππ π΄1 = 500(150) = 75,000 ππ2 π¦1 = 725 β 1/2(150) = 650 ππ 2 π΄2 = 125(114.11) = 14,263 ππ π¦2 = 725 β 150 β 1/2(114.11)=517.95 π = πΆ1 + πΆ2 π΄π πππ₯ ππ¦ = 0.85 π β² π (π΄1 + 2π΄2 ) π΄π πππ₯ (345) = 0.85(28)[75,000 π₯ 650 + 2 π₯ 14,263] π΄π πππ₯ = π, πππ πππ d) Maximum moment, ππ πππ₯ : ππ πππ₯ ππ πππ₯ ππ πππ₯ ππ πππ₯
= πΆ1 π¦1 + 2πΆ2 π¦2 = 0.85πβ²π + (π΄1 π¦1 + 2π΄2 π¦2 ) = 0.85(28)[75,000 π₯ 650 + 2 π₯ 14,263 π₯ 517.95] = 1511.9 ππ β π
π = 0.65 + 0.25
800 β ππ¦ = 0.824 1000 β ππ¦
πππ πππ₯ = 0.824 π₯ 1511.9 πππ πππ₯ = ππππ. π ππ΅ β π e) ππ’ = 1200 ππ β π < πππ πππ₯ Refer to Figure 2.17 ππ’ = π0.85π β² π (π΄1 π¦1 + 2π΄2 π¦2 )
(π πππππ¦ ππππππππππ)
π = 0.65 + 0.25
π = 0.65 + 0.25
ππ + ππ¦ 1000 β ππ¦
ππ = 600
πβπ π
725 β π β 345 166.03 π = + 0.2893 1000 β 345 π
600
π§ = π β 150 = 0.85π β 150 π΄2 = 125π§ = 106.25π β 18,750 π¦2 = 725 β 150 β 1/2π§=575-1/2(0.85c-150) π¦2 = 650 β 0.425π ππ’ = π0.85π β² π (π΄1 π¦1 + 2π΄2 π¦2 ) 166.03 1200 π₯ 106 = ( + 0.2893) 0.85(28)[75,000](650) π + 2(106.25π β 18,750)(650 β 0.425π)] π = 398.7 ππ π=
166.03 + 0.2893 = 0.706 398.7
π΄2 = 106.25(398.7) β 18,750 = 23,615 ππ2 π=πΆ
π΄π ππ¦ = 0.85π β² π (π΄1 + 2π΄2 ) π΄π (345) = 0.85(28)(75,000 + 2 π₯ 23,615) π΄π = 8,432 ππ2
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 π = 0.65ππ and take π = 1.9π. Assume ππ¦ = 276 πππ and πβ²π = 21 πππ. SOLUTION ππ’ = 1.2 ππ· + 1.6 ππΏ
ππ =
ππ’ = 1.2(110) + 1.6(180) ππ’ = 420 ππ β π
0.85πβ²π π½1 600 = 0.03765 ππ¦ (600 + ππ¦ )
πππ‘π: π½1 = 0.85 π ππππ πβ²π < 28 πππ
π = 0.65 ππ = 0.02447 π=
πππ¦ = 0.322 πβ²π
π
π = π β² π π(1 β 0.59 π)] = 5.473 πππ
πΆπ =
600π 600 + ππ¦
πΆπ = 0.685π
Note: For singly reinforced rectangular beam, π is directly proportional to c. Thus, π = 0.65 ππ π = 0.445π ππ = 600
πβπ π
π = 0.65 + 0.25
π β 0.445π 0.445π ππ = 747.7 πππ < 1000 πππ "π‘ππππ ππ‘πππ" ππ = 600
ππ β ππ¦ 1000 β ππ¦
π = 0.65 + 0.25 π = 0.813
747.7 β 276 1000 β 276
ππ’ = ππ
π π π2 420 π₯ 106 = 0.813(5.473)(π)(1.9π)2 π = πππ ππ π = 1.9π = πππ ππ π΄π = πππ
π΄π = 0.02447(297)(564) π΄π = π, ππππ πππ
PROBLEM 2.34 Repeat Problem 2.33 using a steel ratio π = 0.5ππ SOLUTION ππ’ = 420ππ β π ππ = 0.03765 π = 0.5ππ = 0.01883 π=
πππ¦ (1 β 0.59π) = 4.438 πππ π β²π
600π πΆπ = 0.685π 600 + ππ¦ π = 0.5ππ = 0.34247 π πΆπ =
ππ = 600
πβπ π
π β 0.34247π 0.324247π ππ = 1152 πππ > 1000 πππ, π = 0.90 ππ = 600 π β
ππ’ = π π
π π π2 = 420 π₯ 106 = 0.90(5.473)(π)(1.9π)2 π = πππ ππ π = 1.9π = πππ ππ π΄π = πππ
π΄π = 0.01883(308)(585) π΄π = π, πππ πππ
SUPPLEMENTARY PROBLEMS PROBLEM 2.35 A rectangular beam has π = 250 ππ, π = 350 ππ, ππ¦ = 414 πππ, πβ²π = 20.7 πππ. 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. π΄ππ π€ππ: π) πππ πππ₯ = 148.3 πππ π) π΄π = 1075 ππ2 PROBLEM 2.36 Repeat Problem 2.35 using the 2010 NSCP. π΄ππ π€ππ: π) πππ πππ₯ = 130.8 ππ β π π) π΄π = 1056 ππ2 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 π = 0.6ππ and use d= 1.75b. Assume ππ¦ = 276 πππ and πβ²π = 28 πππ. Use the 2001 NSCP π΄ππ π€ππ: π = 250 ππ, π = 436 ππ, π΄π = 3,273 ππ2 PROBLEM 2.38 Repeat Problem 2.37 using the 2010 NSCP. π΄ππ π€ππ βΆ π = 246 ππ, π = 430 ππ, π΄π = 3182 ππ2
PROBLEM 2.39 A reinforced concrete beam has the following properties: Use 2001 NSCP) beam with, π = 320ππ effective depth, π = 640 ππ concrete strength, πβ²π = 25πππ reinforcing steel, ππ¦ = 400 πππ reinforcing steel modulus, πΈπ = 200,000 πππ service dead load moment 350 = ππ β π 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. π΄ππ π€ππ: π) π΄π π = 5,549 ππ2 , πππ = 952.44, ππΏ = 272 ππ β π π) π΄π πππ₯ = 4,162 ππ2 , π ππ πππ₯ = 775.46, ππΏ = 168 ππ β π PROBLEM 2.40 Repeat Problem 2.39 using the 2010 NSCP. π΄ππ π€ππ: π) 5,549 ππ2 , πππ = 687.87 ππ β π, ππΏ = 167.42 ππ β π π) π΄π πππ₯ = 3,963 ππ2 , πππ πππ₯ = 677,7ππΏ = 161 ππ β π PROBLEM 2.41 Calculate the ultimate moment capacity of a rectangular beam with π = 350 ππ, π = 450 ππ, π΄π = 5 β 25 ππ. Assume πβ²π = 24 πππ. ππ¦ = 345 πππ. Use 2001 NSCP π΄ππ π€ππ: πππ = 366.2 ππ β π PROBLEM 2.42 Repeat Problem 2.41 using the 2010 NSCP. π΄ππ π€ππ: πππ = 366.2 ππ β π
PROBLEM 2.43 Calculate the ultimate moment capacity of a rectangular beam with π = 350 ππ, π = 540 ππ, π΄π = 7 β 28 ππ. Assume πβ²π = 24 πππ, ππ¦ = 345 πππ. Use 2010 NSCP. π΄ππ π€ππ: πππ = 582.9 ππ β π PROBLEM 2.44 Repeat Problem 2.43 using the 2010 NSCP. π΄ππ π€ππ: πππ = 514.3 ππ β π PROBLEM 2.45 Calculate the ultimate moment capacity of a rectangular beam with π = 300 ππ, π = 500 ππ, π΄π = 9 β 28 ππ2 . Assume πβ²π = 34 πππ, ππ¦ = 414 πππ. Use 2010 NSCP π΄ππ π€ππ: πππ = 729.6 ππ β π PROBLEM 2.46 Repeat Problem 2.45 using the 2010 NSCP. π΄π ππ€ππ: πππ = 522.5 ππ β π
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. 16π‘ + ππ€ = 16(120) + 250 = 2170 ππ 3. clear spacing of beams + ππ€ = 3000 + 250 = 3250 ππ Therefore ππ = ππππ ππ PROBLEM 3.2 Given the following elements of a T-beam: Flange width, ππ = 1200 ππ πΆππππππ‘π π π‘πππππ‘β πβ²π = 30 πππ Flange thickness, π‘ = 130 ππ ππ‘πππ π π‘πππππ‘β, ππ¦ = 345 πππ Width of web, ππ€ = 290 ππ Effective depth, π = 470 ππ 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 π = π‘. SOLUTION ππ = 0.8 πβ²π ππ π(π β πβ2) πβππ π = π‘ Eq. 3-5
ππ = 0.85 π β² π ππ π‘(π β π‘β2)
πππ = 0.85(30)(120)(130)(470 β
180 ) 2
πππ = 1611 ππ β π π ππ = 0.90 π₯ 1611 = ππππ ππ β π
PROBLEM 3.3 Given the following elements of a T-beam: Flange width, ππ = 900 ππ πΆππππππ‘π π π‘πππππ‘β πβ²π = 20.7 πππ Flange thickness, π‘ = 110 π ππ‘πππ π π‘πππππ‘β, ππ¦ = 414 πππ Width of web, ππ€ = 310 ππ Effective depth, π = 460 ππ 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 π½1 = 0.85 π ππππ πβ²π ππ πππ π π‘βππ 30 πππ a) Balanced condition πΆπ =
600π 600 + ππ¦
600(460) 600 + 414 πΆπ = 272.2 ππ
πΆπ =
d = 460 mm
π = π½1 π
π = 0.85(272.2) π = 231.4 ππ > π‘ = 900mm
t=10 0 C a
z T
=250 mm Figure 3.3
π§ = π β π‘ = 121.4 ππ π΄1 = ππ π₯ π‘ = 900(110) = 99,000 ππ2 π΄2 = ππ€ π₯ π§ = 310(121.4) = 37,622 ππ2 π΄ππ = π΄1 + π΄2 = 136,622 ππ2 π = πΆ1 + πΆ2
π΄π π ππ¦ = 0.85 π β² π ( π΄1 + π΄2 ) π΄π π (414) = 0.85(20.7)99,000 + 37,622) π΄π π = π, πππ πππ β ππππππππ π π‘πππ ππππ
π¦1 = π β π‘β2 = 405 ππ π¦2 = π β π‘ β π§β2 = 289.3 ππ πππ πππ πππ πππ
= πΆ2 π¦1 + πΆ2 π¦2 = 0.85 π β² π ( π΄1 π¦1 + π΄2 π¦2 ) = 0.85(20.7)[99,000(405) + 37,622(289.3)] = πππ ππ΅ β π β πππππππ ππππππππ ππππππ‘
πππ = 0.90(897) πππ = πππ. π ππ΅ β π b)
β π’ππ‘ππππ‘π ππππππππ ππππππ‘
Maximum steel area and moment. Refer to Figure 3.3.
π΄π πππ₯ = 0.75 π΄π π
π΄π πππ₯ = 0.75(5806) π΄π πππ₯ = π, πππ πππ β πππ₯πππ’π π π‘πππ ππππ
π΄π πππ₯ = 0.75 π΄ππ π΄π πππ₯ π΄π πππ₯ = π΄1 + π΄2
π΄π πππ₯ = 0.75(136,622) = 102,466 ππ2 > π΄1 , π‘βπ’π π > π‘ 102,466=99,000 + 310(z) π§ = 11.2 ππ
π΄2 = 102,466 β 99,000 = 3,466 ππ2 π¦2 = π β π‘ β π§β2 = 344.41 ππ ππ πππ₯ = πΆ1 π¦1 + πΆ2 π¦2
ππ πππ₯ = 0.85 π β² π ( π΄1 π¦1 + π΄2 π¦2 )
ππ πππ₯ = 0.85(20.7)[99,000(405) + 3,466(289.3)] ππ πππ₯ = πππ. π ππ΅ β π β πππππππ πππ₯ ππππππ‘ ππ πππ₯ = 0.90(726.5) ππ πππ₯ = πππ. π ππ΅ β π β π’ππ‘ππππ‘π πππ₯πππ’π ππππππ‘ PROBLEM 3.4 A T-beam has the following properties: ππ = 820 ππ, ππ€ = 250 ππ, π =
470 ππ, π‘ = 100 ππ. Concrete compressive strength πβ²π = 20.7 πππ and steel area for the following load conditions: a) ππ· = 150ππ β π, ππΏ = 120 ππ β π b) ππ· = 175 ππ β π, ππΏ = 190 ππ β π SOLUTION
π½1 = 0.85 ππππ£π πππ β π ππ π€βππ π = π‘ π‘ ππππ = 0.85πβ²π ππ π‘ (π β ) = 545.375 ππ β ππ 2 ππππ£π πππ πππ πππ₯ : Balanced condition: 600π ππ = = 278.11 ππ 600 + ππ¦
d = 470 mm
π < π½1 ππ = 236.39 ππ > π‘ = 820mm
t=1 00 C a
z T
FIGURE 3.4
=250 mm
π§ = π β π‘ = 136.39 ππ π΄1 = ππ π‘ = 82,000 ππ2 π΄2 = ππ€ π§ = 34,098 ππ2 π΄ππ = π΄1 + π΄2 = 116,098 ππ2 Maximum condition: π΄π πππ₯ = 0.75 π΄π π = 87,073 ππ2 > π΄1 π΄2 = π΄π πππ₯ β 82,000 = 5,073 ππ2 π΄π π§= = 20.29 ππ ππ€ π¦2 = π β π‘ β π§β2 = 359.85 ππ πππ πππ₯ = ππ + ππ2 = πππ + 0,85πβ²π π΄2 π¦2 πππ πππ₯ = 574.28 ππ β π a) ππ· = 150 ππ β π, π΄ππΏ = 120 ππ β π
ππ’ = 1.4 ππ· + 1.7 ππΏ = 414 ππ β π < πππ πππ₯ , π πππππ¦ ππππππππππ Since ππ’ ππ πππ π π‘βππ πππ , "a" is less than t.
d = 470 mm
t =100
=820 mm
C d -a/2 T
ππ’ = 0.85πβ²π π ππ (π β πβ2)
414 π₯ 106 = 0.90(0.85)(20.7)π(820)(470 β πβ2)
π = 73.6 ππ π=πΆ
π΄π ππ¦ = 0.85πβ²π π ππ π΄π = 2,565 ππ2
Minimum π΄π is the smaller of: βπβ²π 2ππ¦
ππ€ π = 646 ππ2
βπβ²π 4ππ¦
ππ π = 1059 ππ2
Thus, π΄π = π, πππ πππ b) ππ· = 175 ππ β π, ππΏ = 190 ππ β π ππ’ = 1.4 ππ· + 1.7 ππΏ = 568 ππ β π < πππ πππ₯ , π πππππ¦ ππππππππππ Since ππ’ is more than πππ, "π"is more than t. = 820mm
d = 470 mm
t=100 C a
z
T
=250 mm ππ’ = ππππ + πππ2 568 π₯ 106 = 545.375 + 0.90(0.85)(20.7)(250)π§(470 β 100π§β2) π§ = 15.78 ππ π΄2 = ππ€ π§ = 3,946 ππ2
π = πΆ1 + πΆ2
π΄π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π (414) = 0.85(20.7)(82,000 + 3946) π΄π = π, πππ πππ
PROBLEM 3.5 Design a T-beam for a floor system for which ππ€ = 300 ππ and π = 550 ππ. The beams are 4.5 m long and spaced at 3 mo.c. The slab thickness is 100 mm. ππ· = 450 ππ β π(πππππ’ππππ ππ‘π ππ€π π€πππβπ‘ ), ππΏ = 350 ππ β π. πβ²π = 27 πππ, ππ¦ = 415 πππ. SOLUTION π½1 = 0.85 ππ’ = 1.4 ππ· + 1.7 ππΏ
ππ’ = 1.4(450) + 1.7(350) ππ’ = 1225 ππ β π
Solve for bf: ππ ππ π‘βπ π ππππππ π‘ ππ: 1. L/4 = 1.125 m 2. 16π‘ + ππ€ = 16(100) + 300 = 1,900 ππ 3. center-to center spacing of beams = 3 m Thus, ππ = 1,125 ππ Solve for πππ π€βππ π = π‘ = 100ππ, π = 0.90 ππππ = π0.85πβ²π π‘ ππ (π β π‘β2) πππ = 1161.844 ππ β π Solve for πππ πππ₯ to determine if compression steel is needed. 600π
ππ = 600+π = 325.123 ππ π¦
π = π½1 ππ = 276.355 ππ > π‘
= 1125mm
d =550mm
t=1 00 C a
z 4 5 T =300 mm
ππ’ = ππππ + ππ2 π¦2
π§ 1225 π₯ 106 = 1161.844 π₯ 106 + 0.90 π₯ 0.85(27)(300π§)(450 β ) 2 z=23.25 mm π΄2 = ππ€ π§ = 6975.02 ππ2 π = π1 + π2
π΄π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π (415) = 0.85(27)(112,500 + 6,975.02) π΄π = 6,607 ππ2
Minimum π΄π is the smaller value of: βπβ²π 2ππ¦
ππ€ π = 1033 ππ2
βπβ²π 4ππ¦
ππ π = 1937 ππ2
Thus, π΄π = π, πππ πππ PROBLEM 3.6 Determine the ultimate moment capacity of reinforced concrete T-beam with the following properties: Flange width b = 1500 mm, web width ππ€ = 250 ππ, effective depth d = 600 mm, slab thickness t = 100 mm. Assume
πβ²π = 20.7 πππ and ππ¦ = 345 πππ. The beam is reinforced with six 28 mm bars. SOLUTION Solve for balanced π΄π : 600 π = 380.95 ππ 600 + ππ¦ π = π½1 ππ = 323.81 > π‘ π§ = π β π‘ = 22381 ππ π΄1 = ππ π‘ = 150,000 π΄2 = ππ€ π§ = 55,952 ππ2 ππ =
= 1500mm
d =600mm
t=100 C
a
z
=250 mm π=πΆ
π΄π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (345) = 0.85(20.7)(150,00 + 55,952) π΄π π = 10,503 π
Steel area provided, π΄π = 6 π₯ 4 (28)2 = 3,695 ππ2 > π΄π π steel yields Therefore, ππ = ππ¦ πΆ=π 0.85 πβ²π π΄π = π΄π ππ¦ 0.85(20.7) π΄π = 3,695(345) π΄π = 72,441 < π΄1 π‘βπππππππ "a" is less than t
=1500 mm
d = 600 mm
t =100
C d -a/2
T π΄π = π ππ
72,441 = a (1500) π = 48.29 ππ2
ππ = 0.85πβ²π π ππ (π β πβ2)
ππ = 0.85(20.7)(48.29)(1500)(600 β 48.29β2) ππ = 733.99 ππ β π πππ = 0.90(733.99) πππ = πππ. π ππ΅ β π PROBLEM 3.7 Given the following properties of T-beam: Flange width, ππ = 900 ππ πβ²π = 21 πππ Flange thickness, t=1200 ππ¦ = 345 πππ Width of web, ππ€ = 400 ππ Effective depth, d = 580 mm Service deal load, ππ· = 410 ππ β π Determine the safe service live load if the beam is reinforced for tension only with twelve (12) 28-mm-diameter bars. SOLUTION π½1 = 0.85; π = .90
= 900mm
d =580mm
t=120 C
a
z
π π΄π = 12 π₯ (28)2 = 7,389 ππ2 4 π΄1 = ππ π‘ = 108,000 ππ2
=400 mm
Solve for balance π΄π : 600π ππ = = 368.25 ππ 600 + ππ¦ π = π½1 ππ = 313.02 ππ > π‘ π§ = π = π‘ = 193.02 ππ π΄2 = ππ€ π§ = 77,206 ππ2 π=πΆ
π΄π π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (345) = 0.85(21)(108,000 + 77,206) π΄π π = 9,582 ππ2
Steel area provided is less than the balanced steel area. Steel yields. πΆ=π
0.85πβ²π π΄π = π΄π ππ¦ 0.85(21)π΄π = 7,389(345) π΄π = 142,813 ππ2 > π΄1
a>π‘
= 900mm
d =580mm
t=120 C a
z 46 0 T
=400 mm π΄π = π΄1 + π΄2
142,813=108,000+π΄2 π΄2 = 34,813 ππ2
π΄2 = ππ€ π§
34,813 = 400z π§ = 87.03 ππ
π¦1 = π β π‘β2 = 520 ππ π§ π¦2 = π β π‘ β = 416.48 ππ 2
ππ = πΆ1 π¦1 + πΆ2 π¦2 ππ = 0.85π β² π (π΄1 π¦1 + π΄2 π¦2 ) ππ = 0.85(21)[108,000(520) + 34,813(416.48)] ππ = 1,261.3 ππ β π πππ = 0.90(1,261.3) πππ = 1135.138 ππ β π πππ = ππ’
ππ’ = 1.4 ππ· + 1.7ππΏ 1,135.138 = 1.4(410) + 1.7ππΏ ππΏ = πππ. π ππ΅ β π
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 ππ¦ = 415 πππ. Concrete strength πβ²π = 32 πππ. 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 π΄π = 10 π₯ (32)2 4 π΄π = 8,042 ππ2 π΄1 = ππ π‘ = 60,000 ππ2
π½1 = 0.836
t = 120mm
0.05 (32 β 30) 7
Solve for balance π΄π : 600π 600 + ππ¦ πΆπ = 313.3 ππ π = π½1 πΆπ = 261.83 ππ > π‘ π§ = π β π‘ = 141.83 ππ
Figure 3.5 =500m m a
d=530mm
π½1 = 0.85 β
=320mm
z
πΆπ =
π΄2 = ππ€ π§ = 45,385.5 ππ2
=320m m
π=πΆ
π΄π π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (345) = 0.85(21)(60,00 + 45,385.5) π΄π π = 6,907 ππ2
Since π΄π > π΄π π , tension steel does not yield
=500m m
t = 120mm
d=530m m
a z
T
π΄1 = 60,000 ππ2 π΄2 = ππ€ π§ = ππ€ (π β π‘) π΄2 = ππ€ (π½1 π β π‘) πβπ ππ = 600 π π = πΆ1 + πΆ2
=320m m
π΄π ππ = 0.85π β² π (π΄1 + π΄2 ) 530βπ
8,042 π₯ 600 π = 0.85(32)[60,000 + 320(0.836π β 120)] π = 327.95 ππ π = π½1 π = 261.83 ππ π§ = π β π‘ = 141.83 ππ
π΄2 = ππ€ π§ = 49,303 ππ2 π¦2 = π β π‘ β π§β2 = 332.97mm π¦1 = π β π‘β2 = 470mm
ππ = πΆ1 π¦1 + πΆ2 π¦2
ππ = 0.85π β² π (π΄1 π¦1 + π΄1 π¦2 ) ππ = 0.85(32)[60,000(470) + 49,303(332.97)] ππ = 1,213.56 ππ β π
πππ = 0.90(1,213.56) πππ = 1,092.2 ππ β π ππ’ = πππ
ππ’ = 1.4ππ· + 1.7 ππΏ 1,092 = 1.4(330) + 1.7 ππΏ ππΏ = πππ. π ππ΅ β π
SOLVED PROBLEMS IN T-BEAMS USING 2010 NSCP PROBLEM 3.9 Repeat Problem 3.3 using the 2010 NSCP. SOLUTION Given: ππ = 900 ππ πβ²π = 20.7 πππ π‘ = 110 ππ ππ¦ = 414 πππ ππ€ = 3210 ππ π = 460 ππ π½1 = 0.85 π ππππ π β² π ππ πππ π π‘βππ 28 πππ a) Balanced condition, π = 0.65 ππ =
600 π 600 + ππ¦
600(460) 600 + 414 ππ = 272.2 ππ ππ =
π = π½1 π
π = 0.85(272.2) π = 231.4 ππ > π‘
= 900mm
d = 460 mm
t=110 C a
z
T =310 mm Figure 3.6
π§ = π β π‘ = 121.4 ππ π΄1 = ππ π₯ π‘ = 900(110) = 99,000 ππ2 π΄2 = ππ€ π₯ π§ = 310(121.4) = 37,622 ππ2 π΄ππ = π΄1 + π΄2 = 136,622 ππ2 π = π1 + π2
π΄π π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (414) = 0.85(20.7)(99,000 + 37,622) π΄π π = π, πππ πππ β ππππππππ π π‘πππ ππππ
π¦1 = π β π‘β2 = 405 ππ π¦2 = π β π‘ β π§β2 = 289.3 ππ πππ = π1 π¦1 + π2 π¦2 πππ = 0.85π β² π (π΄1 π¦1 + π΄2 π¦2 ) πππ = 0.85(20.7)[99,000(405)37,622(289.3)] πππ = πππ ππ΅ β π β πππππππ ππππππππ ππππππ‘ ππππ = 0.65(897) ππππ = πππ ππ΅ β π β π’ππ‘ππππ‘π ππππππππ ππππππ‘ b) Maximum steel area and moment. Refer to Figure 3.6. 800 β ππ¦ 3 π = π = 197.14 ππ; π = 0.65 + 0.25 = 0.815 7 1000 β ππ¦ π = π½1 π
π = 0.85(197.14) π = 167.6 ππ
π§ = π β π‘ = 57.571 ππ π΄2 = ππ€ π§ = 310(57.6) = 17,847 ππ2 π¦2 = π β π‘ β π§β2 = 321.21 ππ π = π1 + π2 π΄π πππ₯
π΄π πππ₯ ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π πππ₯ (414) = 0.85π β² π (99,000 + 17,847) = ππππ πππ β πππ₯πππ’π π π‘πππ ππππ
ππ πππ₯ = π1 π¦1 + π2 π¦2 ππ πππ₯ = 0.85π β² π (π΄1 π¦1 + π΄2 π¦2 ) ππ πππ₯ = 0.85(20.7)[99,000(415) + 17,847(321.2)] ππ πππ₯ = πππ. ππ ππ΅ β π β πππππππ πππ₯ ππππππ‘ πππ πππ₯ = 0.815(806.34) πππ πππ₯ = πππ. π ππ΅ β π β π’ππ‘ππππ‘π πππ₯πππ’π ππππππ‘ PROBLEM 3.10 Repeat Problem 3.2 using the 2010 NSCP. SOLUTION Given: ππ = 1200 ππ π‘ = 130 ππ π = 470 ππ
ππ€ = 290 ππ πβ²π = 30 πππ ππ¦ = 345πππ
πππ = 0.85πβ²π π‘ ππ(π β π‘β2)
πππ = 0.85(30)(1200)(130)(470 β 130β2) πππ = 1611ππ β π Solving for π: π = 130 ππ 0.05 β² π½1 = 0.85 β (π π β 28) = 0.836 7 π = πβπ½ = 155.56 ππ 1 πβπ ππ = 600 = 1213 πππ > 1000 ππ π ππππ = 090(1611) ππππ = ππππ ππ΅ β π
"π‘πππ πππ β ππππ‘ππππ , π = 0.90"
PROBLEM 3.11 Repeat Problem 3.4 using the 2010 NSCP. Additional questions: c) Find the required steel area if ππ· = 195 ππ β π and ππΏ = 210 ππ β π. d) Find the maximum design moment so that section is tensioncontrolled if it is reinforced for tension only.
SOLUTION Given: ππ = 820 ππ πβ²π = 20.7 πππ ππ€ = 250 ππ ππ¦ = 414 πππ π = 470 ππ π‘ = 100 ππ π½1 = 0.85 π ππππ πβ²π < 28 < πππ Solve for πππ when π = π‘: πππ = 0.85πβ²π ππ π‘(π β π‘β2) = 605.97 ππ β ππ π = πβπ½ = 117.65 ππ 1
ππ = 600
πβπ = 1797πππ > 1000πππ, π = 0.90 π
ππππ = πππ. πππ ππ΅ β π Solve for πππ πππ₯ : 3 πΆπππ₯ = π = 201.43 ππ 7 800 β ππ¦ = 0.815 1000 β ππ¦ = 171.21 ππ > π‘
π = 0.65 + 0.25 π = π½1 ππππ₯
= 820mm
d = 470 mm
t=100 C
a
z
T π§ = π β π‘ = 71.21 ππ π΄2 = ππ€ π§ = 17,803.6 ππ2 π¦2 = π β π‘ β π§β2 = 334.39ππ
=250 mm
ππ πππ₯ = πππ + 0.85πβ²π π΄2 π¦2 ππ πππ₯ = 710.72 ππ β π πππ πππ₯ = 579 ππ β π a) ππ· = 150 ππ β π, ππΏ = 120 ππ β π ππ’ = 1.2ππ· + 1.6 ππΏ = 372 ππ β π < πππ πππ₯, π πππππ¦ ππππππππππ Since ππ’ is less than ππππ ,βaβ is less than t.
d = 470 mm
t =100
=820 mm
C d -a/2 T
Assume π = 0.90 ππ’ = π0.85πβ²π π ππ (π β πβ2) 372 π₯ 106 = 0.90(0.85)(20.7)π(820)(470 β πβ2) π = 65.52 ππ π = πβπ½ = 77.08ππ 1 πβπ ππ = 600 = 3,058 πππ > 1000 πππ, π‘πππ πππ ππππ‘ππππ , π = 0.90 π π=πΆ
π΄π ππ¦ = 0.85πβ²π π ππ π΄π (345) = 0.85(20.7)(65.52)(820) π΄π = 2,283 ππ2
Minimum π΄π is the smaller value of: πβ²π π π = 646 ππ2 2ππ¦ π€
βπβ²π π π = 1059 ππ2 4ππ¦ π
Thus, π΄π = π, πππ πππ b) ππ· = 175 ππ β π, ππΏ = 190 ππ β π ππ’ = 1.2 ππ· + 1.6 ππΏ = 514 ππ β π < πππ πππ₯ , π πππππ¦ ππππππππππ Since ππ’ ππ πππ π π‘βππ ππππ , "π"is less than t. Assume π = 0.90 ππ’ = π0.85πβ²π π ππ (π β πβ2) 514 π₯ 106 = 0.90(0.85)(20.7)π(820)(470 β πβ2) π = 93.53 ππ π = πβπ½ = 110.03 ππ 1
t =100
d = 470 mm
=820 mm
C d -a/2
T ππ = 600 π=πΆ
πβπ = 1,963 πππ > 1000πππ, π‘πππ πππ ππππ‘ππππ , π = 0.90 π π΄π ππ¦ = 0.85πβ²π π ππ π΄π (345) = 0.85(20.7)(93.53)(820) π΄π = π, πππ πππ
c) ππ· = 195 ππ β π, ππΏ = 210 ππ β π ππ’ = 1.2ππ· + 1.6ππΏ = 570 ππ β π < πππ πππ₯ , π πππππ¦ ππππππππππ Since ππ’ ππ ππππ π‘βππ πππ’ , "π"is more than t.
= 820mm
d =470mm
t=10 0 C
a
z
T =250 mm
Assume π = 0.90 ππ’ = ππππ + πππ2 570 π₯ 106 = 545.375 + 0.90(0.85)(20.7)(250)π§(470 β 100 β π§β2) π§ = 17.05 ππ π = π‘ + π§ = 117.05 ππ; π = πβπ½1 = 137.7ππ πβπ = 1448πππ > 1000πππ, π‘πππ πππ ππππ‘ππππ , π = 0.90 π π΄2 = ππ€ π§ = 3,908 ππ2 ππ = 600
π΄π ππ¦ = 0.85π β² π (π΄1 + π΄2 )
π = πΆ1 + πΆ2 π΄π = 3,666 ππ2 3
d) π = π π = 176.25 ππ, π = 0.90 π = π½1 π = 149.81 ππ > π‘
= 820mm
d = 470 mm
t=100 C
a
z
T =250 mm π§ = π β π‘ = 49.81ππ π΄2 = ππ€ π§ = 12,453 ππ2 π§ π¦2 = π β π‘ β = 3450.9ππ 2
ππ‘π = πππ + 0.85 πβ²π π΄2 π¦2 ππ‘π = 681.59 ππ β π πππ‘π = πππ. π ππ΅ β π Note: If ππ’ is less than or equal to πππ , the beam is tension-controlled. PROBLEM 3.12 Repeat Problem 3.6 using the 2010 NSCP. SOLUTION Given: ππ = 1500ππ πβ²π = 20.7 πππ ππ€ = 250 ππ ππ¦ = 345 πππ π = 600 ππ π½1 = 0.85 π΄π = 6 β 28 ππ = 3,694 ππ2 Solve for balanced π΄π : 600π
ππ = 600+π = 380.95 ππ π¦
π = π½1 ππ = 323.81 > π‘ π΄1 = ππ π‘ = 150,000 π΄2 = ππ€ π§ = 55,952 ππ2
= 1500mm
d =600mm
t=100 C
a
z
=400 mm
π΄π π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (345) = 0.85(20.7)(150,000 + 55,952) π΄π π = 10,503
π=πΆ
Steel area provided, π΄π = 6 π₯
π 4
(28)2 = 3,695 ππ2 < π΄π π "π π‘πππ π¦πππππ "
therefore, ππ = ππ¦ πΆ=π
0.85πβ²π π΄π = π΄π ππ¦ 0.85(20.7)π΄π = 3,695(345) π΄π = 72,441 < π΄1 π‘βπππππππ "π" is less than t
t =100
d = 600 mm
=1500 mm
C d -a/2 T
π΄π = π ππ
72,441 = π(1500) π = 48.29 ππ2
Solve for π: π = πβπ½ = 56.82 ππ 1 ππ = 600
πβπ = 5,736πππ > 1000πππ π
therefore π = 0.90
π‘πππ πππ ππππ‘ππππ
ππ = 0.85πβ²π π ππ (π β πβ2) ππ = 0.85(20.7)(48.29)(1500)(600 β 48.29β2) ππ = 733.99 ππ β π πππ = 0.90(733.99) πππ = πππ. π ππ΅ β π PROBLEM 3.13 Repeat Problem 3.7 using 2010 NSCP. SOLUTION Given the following properties of a T-beam: Flange width, ππ = 900 ππ πβ²π = 21 πππ Flange thickness, π‘ = 120 ππ ππ¦ = 345 πππ Width of web, ππ€ = 400 ππ π΄π = 7,389 ππ2 Effective depth, π = 580 ππ Service deal load, ππ· = 410 ππ β π π½1 = 0.85; π = 0.90 π π΄π = 12 π₯ 4 (28)2 = 7,389 ππ2 π΄1 = ππ π‘ = 108,000 ππ2
πΆπ =
600π = 368.25 ππ 600 + ππ¦
d =580mm
Solve for balance π΄π :
= 900mm
t=120 C
a
z
π = π½1 ππ = 313.02 ππ > π‘ π§ = π β π‘ = 193.02 ππ π΄2 = ππ€ π§ = 77,206 ππ2 π=πΆ
π΄π π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (345) = 0.85(21)(108,000 + 77,206) π΄π π = 9,582 ππ2
=400 mm
Steel area provided is less than the balanced steel area. Steel yields. πΆ=π
0.85πβ²π π΄π = π΄π ππ¦ 0.85(21)π΄π = 7,389(345) π΄π = 142,813 ππ2 > π΄1 = 900mm
d = 580 mm
t=120 C
a
βaβ >t
z 46 0 T =400 mm
π΄π = π΄1 + π΄2 π΄2 = ππ€ π§
142,813 = 108,000 + π΄2 π΄2 = 34,813 ππ2 34,813 = 400π§ π§ = 87.03 ππ
π¦1 = π β π‘β2 = 520 ππ π¦2 = π β π‘ β π§β2 = 416.48 ππ ππ ππ ππ ππ
= πΆ1 π¦1 + πΆ2 π¦2 = 0.85π β² π (π΄1 π¦1 + π΄2 π¦2 ) = 0.85(21)[108,000(520) + 34,813(416.48)] = 1,261.6 ππ β π
Solve for π: π = π‘ + π§ = 203.03 ππ
π = πβπ½ = 243.57ππ 1
ππ = 600
πβπ = 828.76 πππ < 1000 πππ π
Since ππ¦ < ππ < 1000 πππ,Transition region π = 0.65 + 0.25
ππ β ππ¦ = 0.8346 1000 β ππ¦
πππ = 0.8346(1,261.3) πππ = 1,052.703 ππ β π πππ = ππ’
ππ’ = 1.2 ππ· + .6 ππΏ 1,052.703 = 1.2(410) + 1.7 ππΏ ππΏ = πππ. ππ ππ΅ β π
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 π΄π = 10 π₯
π (32)2 4
=320mm
π΄π = 8,042 ππ2 π΄1 = ππ π‘ = 60,000 ππ2 π½1 = 0.85 β π½1 = 0.821
0.05 (32 β 28) 7
Solve for balanced π΄π : 600π 600 + ππ¦ π = 313.3 ππ π = π½1 ππ = 257.35 ππ > π‘ π§ = π β π‘ = 137.35 ππ π΄2 = ππ€ π§ = 43,953 ππ2 ππ =
π=πΆ
π΄π π ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π π (345) = 0.85(21)(60,000 + 43,953) π΄π π = 6,813 ππ2
Since π΄π > π΄π π , π‘πππ πππ π π‘πππ ππππ πππ‘ π¦ππππ
=500m m
t =120 mm
d=530mm
a Z
T
=320m m
π = 0.65 ππππππ π πππ ππππ‘ππππ
π΄1 = 60,000 ππ2 π΄2 = ππ€ π§ = ππ€ (π β π‘) = ππ€ (π½1 π β π‘) ππ = 600
πβπ π
π = πΆ1 + πΆ2
π΄π ππ = 0.85π β² π (π΄1 + π΄2 )
530 β π = 0.85(32)[60,000 + 320(0.821π β 1200)] π π = 329.27 ππ 8,042 π₯ 600
π = π½1 π = 270.47 ππ π§ = π β π‘ = 150.47 ππ π¦1 = π β π‘β2 = 470 ππ ππ = πΆ1 π¦1 + π2 π¦2
π΄2 = ππ€ π§ = 48,151 ππ2 π§ π¦2 = π β π‘ β 2 = 334.76 ππ ππ = 0.85π β² π (π΄1 π¦1 + π΄2 π¦2 ) ππ = 0.85(32)[60,000(470) + 48,151(334.76)] ππ = 1,205.48 ππ β π
πππ = 0.65(1,205.48) πππ = 783.56 ππ β π ππ’ = πππ
ππ’ = 1.2 ππ· + 1.6ππΏ 78.56 = 1.2(330) + 1.6 ππΏ ππΏ = πππ. ππ ππ΅ β π
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 ππ¦ = 415 πππ and πβ²π = 29 πππ. 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-dβ420
70 mm
a
4-25 mm
+
=
70 mm π½1 = 0.85 a) Balanced condition ππ =
π = π½1 ππ
600π 600 + ππ¦
600(490) 600 + 415 ππ = 289.66 ππ
ππ =
π = 0.85(289.66) π = 246.21 ππ
ππ π = 600
πβπ
ππ π = 600
π
289.66β70 289.66
ππ π = 455 πππ > ππ¦ yield ππ π = ππ¦ π1 = πΆπ
π2 = πΆβ²π
π΄π 1 ππ¦ = 0.85πβ²π π π π΄π 1 (415) = 0.85(29)(246.21)(305) π΄π 1 = 4,460 ππ2 π΄π 2 ππ¦ = π΄β²π ππ¦ π΄π 2 = 1,964 ππ2
Balanced steel area, π΄π π = π΄π 1 + π΄π 2 = π, πππ πππ πππ = πΆπ (π β πβ2) + πΆ β² π (π β π β² ) π πππ = 0.85ππβ²π π π (π β ) + π΄β² π ππ¦ (π β π β² ) 2 πππ = 0.85(29)(246.21)(305)(490 β 246.21β2) πππ = π, πππ. π ππ΅ β π ππππ = 0.90(1,021.4) = πππ. ππ ππ΅ β π b) Maximum tension steel area: According to Section 410.4.3, for members with compression reinforcement, the portion of ππ equalized by compression reinforcement need not be reduced by the 0.75 factor. π΄π 1 = 0.75π΄π 1 π΄π 2 = 1,964 ππ2 π΄π πππ₯ = π΄π 1 πππ₯ + π΄π 2
π΄π 1 πππ₯ = 0.75(4,460) π΄π 1 πππ₯ = 3,345 ππ2
π΄π πππ₯ = 3,345 + 1,964 π΄π πππ₯ = π, πππ πππ
πΆπ = π1
0.85πβ²π π π = π΄π 1πππ₯ ππ¦ 0.85(29)(π)(305) = 3,345(415) π = 184.7 ππ
π = π/π½1
πβ²π = 600
π = 184.7/0.85 π = 217.2 ππ πβπβ² π
217.2β70
πβ²π = 600 217.2 πβ²π = 406.7 πππ < ππ¦
(π€πππ πππ‘ π¦ππππ)
ππ πππ₯ = πΆπ (π β πβ2) + πΆ β² π (π β πβ² ) ππ πππ₯ = 0.85πβ²π π π(π β πβ2) + π΄β² π π β² π (π β π β² ) ππ πππ₯ = 0.85(29)(184.7)(305)(490 β 184.7β2) +1964(406.7)(490 β 70) ππ πππ₯ = πππ. ππ ππ΅ β π πππ πππ₯ = 0.90(887.45) πππ πππ₯ = πππ. π ππ΅ β π 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 πβ²π = 35 πππ and steel strength ππ¦ = 345 πππ. 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 π½1 = 0.85 β
0.05 (35 β 30) = 0.814 7
π΄β²π =
π (28)2 π₯ 2 = 1,232 ππ2 4
a) Balanced condition considering compression steel: ππ =
600 π 600 + ππ¦
πβ²π = 600
π β πβ² π
ππ =
600(460) 600 + 345
πβ²π = 600
292 β 70 292
πβ²π = 456 πππ > ππ¦ , π‘βπ’π πβ²π = ππ¦ = 345 πππ πΆπ + πΆπ = π 0.85 πβ²π π π + π΄β²π πβ²π = π΄π ππ¦ 0.85(35)(0.814 π₯ 292)(300) + 1232(345) = π΄π (345) π΄π = π, πππ πππ b) Maximum steel area: For rectangular beams: πΆπππ₯ = 0.75 ππ = 0.75(292) πΆπππ₯ = 219.05 ππ π = π½1 ππππ₯ = 178.37 ππ πβ²π = 600
π β πβ² π
πβ²π = 600
219.05 β 70 219.05
πβ²π = 408 πππ > ππ¦ , π‘βπ’π πβ²π = ππ¦ = 345 πππ πΆπ + πΆπ = π
0.85πβ²π π π + π΄β²π πβ²π = π΄π ππ¦ 0.85(35)(178.37)(300) +1232(345) = π΄π (345) π΄π = π, πππ πππ
PROBLEM 3.17 A rectangular beam has b=300 mm and d= 490 mm. Concrete compressive strength πβ²π = 27.6 πππ and steel yield strength ππ¦ = 276 πππ. Compressive steel if required shall have its centroid 60 mm from extreme concrete fiber. Calculate the required tension steel area if the factored moment ππ’ is 620 kN-m. SOLUTION This is the same problem in Chapter 2. Solve for πππ πππ₯ : ππ =
600π = 335.616 ππ 600 + ππ¦
ππ = π½1 ππ = 285.27 ππ π ππ πππ₯ = 0.85πβ²π π π(π β ) 2 ππ πππ₯ = 0.85(27.6)(213.96)(300)(490 β 213.96β2) ππ πππ₯ = 576.76 ππ β π πππ πππ₯ = 0.90(576.76) = 519 ππ β π Since ππ’ = 620 ππ β π > πππ πππ₯ , the beam must be doubly reinforced.
b d β
ca
=
d β a/2
+
d β dβ
ππ1 = ππ πππ₯ = 576.76 ππ β π ππ2 =
ππ’ β ππ1 π
620 0.90 β 576.76 = 112.13 ππ β π
ππ2 = ππ2
π = 213.96 ππ π΄π 1 ππ¦ = 0.85πβ²π π π
π΄π 1 (276) = 0.85(27.6)(213.96)(300) π΄π 1 = 5456 ππ2
Note: π΄π 1 = π΄π πππ₯ Solve for πβ²π : π = πβπ½ = 251.71 ππ 1
πβ²π = 600
π β πβ² π
251.71 β 60 251.71 πβ²π = 457 πππ > ππ¦ πβ²π = 600
Compression steel yields
ππ π πβ²π = ππ¦ ππ2 = π2 (π β πβ² )
112.13 π₯ 106 = π΄π 2 (276)(490 β 60) π΄π 2 = 945 ππ2
Tension steel area, π΄π = π΄π 1 + π΄π 2 = ππππ πππ Compression steel: πβ²π = π2
π΄β²π ππ¦ = π΄π 2 ππ¦ π΄β²π = π΄π 2 π΄β²π = πππ πππ
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 πβ²π = 30πππ and ππ¦ = 415 πππ. SOLUTION π½1 = 0.85 ππ’ = 1.4 ππ· + 1.7 ππΏ
ππ’ = 1.4(230) + 1.7(190) ππ’ = 645 ππ β π
Solve for πππ πππ₯ : Note: For rectangular beams, ππππ₯ = 0.75 ππ ππππ₯ = 0.75
600π = 203.94 ππ 600 + ππ¦
π = π½1 ππππ₯ = 173.35 ππ ππ πππ₯ = 0.85πβ²π π π (π β πβ2) ππ πππ₯ = 0.85(30)(173.35)(310)(460 β 173.35/2) ππ πππ₯ = 511.58 ππ β π πππ πππ₯ = 0.90(511.58) πππ πππ₯ = 460.42 ππ β π Since ππ’ = 645 ππ β π > πππ πππ₯, πππππππ π πππ π π‘πππ ππ πππππ π πππ¦
b
d β
ca
=
dβ a/2
+
dβ dβ
ππ1 = ππ πππ₯ = 511.58 ππ β π ππ2 =
ππ’ β ππ1 π
ππ2 =
645 β 511.58 0.90
ππ2 = 205.088 ππ β π π = ππππ₯ = 203.94 ππ π = 173.35 ππ Tension Steel: π1 = πΆπ
ππ2 = π2 (π β πβ² ) π΄π = π΄π 1 + π΄π 2
π΄π 1 ππ¦ = 0.85 πβ²π π π π΄π 1 (415) = 0.85(30)(173.35)(310) π΄π 1 = 3,302 ππ2 205.088 x 106 = π΄π 2 (415)(460 β 70) π΄π 2 = 1,267 ππ2 π΄π = 3,302 + 1,267 π΄π = π, πππ πππ
Compression steel: πβ²π = 600
π β πβ² π
πβ²π = 600
203.94 β 70 203.94
Compression steel does not yield, πβ²π = 394.06 ππ πΆβ²π = π2
π΄β²π πβ²π = π΄π 2 ππ¦ π΄β² π (394.06) = 1,267(415) π¨β²π = ππππ πππ
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 ππ€ = 280 ππ, and π = 500ππ. 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 πβ²π = 21 πππ, ππ¦ = 415 πππ. Unit weight of concrete is 23.5 kN/π3 . 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 πβ²π = 21πππ ππ€ = 280 ππ ππ¦ = 414 πππ π = 500ππ π½1 = 0.85 π β² = 70 ππ Maximum factored moment, ππ’ πππ₯ = 1080 ππ β π
C
Effective flange width, ππ : 1. L/4=9/4=2.25 m 2. 16t+ππ€ = 16(100) + 280 = 1.88 π 3. πππ = 1.2π Use ππ = 1.2 π Solve for πππ πππ₯ to determine if compression steel is required. Balanced condition: π=
600π 600 + ππ¦
π=
π = π½1 π
π = 0.85(295.57) π = 251.23 ππ = 1200mm
t=100
d =500mm
600(500) 600 + 415
C
a
z
=280 mm π§ =πβπ‘
π§ = 251.23100 π§ = 151.23 ππ
π΄1 = 1200 π₯ 100 = 120,000 ππ2 π΄2 = 280(151.23) = 42,345 ππ2 π΄ππ = π΄1 + π΄2 = 162,345 ππ2
Maximum condition: π΄π πππ₯ = 0.75 π΄ππ
π΄π πππ₯ = 0.75(162,345) π΄π πππ₯ = 121,759 ππ2 > π΄1 = 1200mm
d =500mm
t=100 C a
z T =280 mm
π΄2 = π΄π πππ₯ β π΄1
π΄2 = 121,759 β 120,000 π΄2 == 1,759 ππ2
π΄2 = ππ€ π§
1,759 = 280 π§ π§ = 6.28 ππ
π¦1 = π β π‘/2
π¦1 = 500 β 100/2 π¦1 = 450 ππ
π¦2 = π β π‘ β π§/2
π¦2 = 500 β 100 β 6.28/2 π¦2 = 396.86 ππ
ππ πππ₯ = πΆ1 π¦1 + π2 π¦2
πππ πππ₯ = 0.90(976.36) πππ πππ₯ = 878.72 ππ β π
ππ πππ₯ = 0.85 π β² π (π΄1 π¦1 + π΄2 π¦2 ) ππ πππ₯ = 0.85(21)(120,000 π₯ 450) +1,759 π₯ 396.86 ππ πππ₯ = 976.36 ππ β π
Since ππ’ = 1,080 ππ β π > πππ πππ₯ , the compression reinforcement must be provided.
=1200mm
d=500mm
430
500 mm
Z
dβ=7 0
=820mm π =π‘+π§
π = 100 + 6.28 π = 106.28 ππ
π = π/π½1
π = 106.28/0.85 π = 125.04
πβ²π = 600
π β πβ² π
πβ²π = 600
125.04 β 70 125.04
πβ²π = 264.1 πππ < ππ¦ ππ1 = ππ πππ₯ = 976.36 ππ β π π΄π 1 = π΄π πππ₯ π1 = πΆ1 + πΆ2 π΄π 1
π΄π 1 ππ¦ = 0.85π β² π (π΄1 + π΄2 ) π΄π 1 (415) = 0.85(21)(120,000 + 1,759) = 5,237 ππ2
d-dβ
dβ=70
a
ππ2 = ππ β ππ1
ππ2 =
ππ’
β ππ1
π 1080
ππ2 = 0.90 β 976.36 ππ2 = 223.64 ππ β π ππ2 = π2 (π β πβ² )
ππ2 = π΄π 2 ππ¦ (π β π β² ) 223.64 π₯ 106 = π΄π 2 (415)(500 β 70) π΄π 2 = 1,253 ππ2
Tension steel area, π΄π = π΄π 1 + π΄π 2 = π, πππ πππ Compression steel area: πΆβ²π = π2
π΄β²π πβ²π = π΄π 2 ππ¦ π΄β² π (264.1) = 1,253(415) π΄β²π = 1,969 ππ2
b) Dead load=weight of concrete: Area=1.2(0.1)+0.28(0.47)=0.2516 π3 π€π = πΎπ π₯ π΄πππ π€π = 23.5(0.2516) π€π = π. ππππ ππ΅βπ
β ππππ ππππ
c) Uniform live load
5-32 mmΓΈA
7.6 m 3-32
L=9 mmΓΈ m 145 kN-m
5-32 5-32 B mmΓΈ mmΓΈ
7.6 m 3-32
L=9 mmΓΈ m
145 kN-m 202 kNm
5-32
CmmΓΈ
Maximum positive moment (at midspan) π€π’ =
π€π’ πΏ2 24
π€π’ = 1.4π€π· + 1.7π€πΏ
π€π’ (9)2 24 π€π’ = 320 ππ/π 1,080 =
320 = 1.4(5.9126) + 1.7π€πΏ π€πΏ = πππ. ππ ππ΅βπ β πππ£π ππππ
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 πβ²π = 20.7 πππ πππ ππ¦ = 345 πππ.
350 mm
60 mm
600 mm
540 mm
2-28 mm
4-36 mm
Figure 3.11 SOLUTION π π΄π = 4 (36)2 π₯ 4 = 4,072 ππ2 π
π΄β²π = 4 (28)2 π₯2 = 1,232 ππ2
b
dβ
ca
=
d β a/2+
d β dβ
Assume all steel yield: ππ = πβ²π = ππ¦ π΄π 2 = π΄β²π = 1,232 ππ2 π΄π 1 = π΄π β π΄π 2 = 2,840 ππ2 πΆπ = π1
π=
π = 187.18 ππ π½1
0.85πβ²π π π = π΄π 1 ππ¦ 0.85(20.7)π(350) = 2,840(345) π = 159.1 ππ
ππ = 600
πβπ π
600 β 187.18 187.18 ππ = 1.323 > ππ¦ π‘πππ πππ π π‘πππ π¦πππππ
ππ = 600
π β πβ² π
187.18 β 60 187.18 ππ = 407.7 > ππ¦ compression steel yields
ππ = 600
ππ = 600
Assumption is correct, all steel yield. ππ = ππ1 + ππ2 π ππ = π1 (π β ) + π2 (π β πβ² ) 2 π ππ = π΄π 1 ππ¦ (π β ) + π΄π 2 ππ¦ (π β π β² ) 2 159.1 ππ = 2,840(345) (600 β ) + 1,232(345)(600 β 60) 2 ππ = 739.4 ππ β π πππ = 0.90(739.4) = 665.43 ππ β π πππ = ππ’ = 1.4ππ· + 1.7ππΏ 665.43 = 1.4(230) + 1.7 ππΏ ππΏ = πππ. ππ ππ΅ β π
PROBLEM 3.21 A rectangular beam has the following properties: Width, b=400 mm ππ¦ = 415 πππ Effective depth, d=620 mm πβ²π = 22 πππ Tension bars, 3 pcs 25-mm-diameter dβ=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 π΄π = 10 π₯ (28)2 = 6,158 ππ2 4 π π΄β²π = 3 π₯ (25)2 = 1,473 ππ2 4 Assume all steel yields: π΄π 2 = π΄β²π = 1,473 ππ2 π΄π 1 = π΄π β π΄π 2 = 4,685 ππ2
b dβ
ca d β a/2+
=
0.85π β² π π π = π΄π 1 ππ¦ π = πβπ½ = 305.8 ππ 1
0.85(22)π(400) = 4,685(415) π = 260 ππ
d β dβ
πβπ = 616.5 πππ > ππ¦ π π β πβ² πβ²π = 600 = 463 πππ > ππ¦ π ππ = 600
(π¦ππππ) (π¦ππππ)
All steel yields. Assumption is correct ππ = ππ1 + ππ2 ππ = π1 (π β πβ2) + π2 (π β πβ² ) ππ = π΄π 1 ππ¦ (π β πβ2) + π΄π 2 (π β π β² ) ππ = 4,685(415)(620 β 260β2) + 1,473(415)(620 β 70) ππ = 1288.9 ππ β π πππ = 0.90(1288.9) = 1,160 ππ β π πππ = ππ’ = 1.4 ππ· + 1.7ππΏ 1160 = 1.4(320) + 1.7ππΏ ππΏ = πππ ππ΅ β π 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 πβ²π = 30 πππ and steel yield ππ¦ = 415 πππ. Use 0.75 ππ = 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 π½1 = 0.85 πβ²π = 30πππ ππ¦ = 415 πππ
300 mm 70 mm
π (25)2 4 π΄π = 1963 ππ2 π π΄β²π = 2 π₯ (25)2 4 π΄β²π = 982 ππ2 π΄π = 4 π₯
Assuming all steel yields: π΄π 2 = π΄β²π = 982 ππ2 π΄π 1 = π΄π β π΄π 2 = 982 ππ2
450 mm
310 mm
380 mm
2-25 mm
4-25 mm 70 mm
πΆπ = π1
0.85πβ²π π π = π΄π 1 ππ¦ 0.85(30)π(30) = 982(415) π = 53.26 ππ
π = πβπ½ = 62.66 ππ < 70 ππ compression steel does not yield 1
Assuming tension steel yields and compression steel does not. π = πΆπ + πΆβ²π
π΄π ππ¦ = 0.85πβ²π π π + π΄β²π πβ²π 1963(415) = 0.85(30(0.85π)(300) πβ70 +982 π₯ 600 π π = 98.87 ππ
πβ²π = 600 ππ = 600
98.86 β 70 = 175.17 πππ < ππ¦ 98.86
πβπ = 1,706 > ππ¦ (π¦ππππ) π
π = π½1 π = ππ. ππ ππ β πππ π€ππ ππ ππππ‘ π
ππ = πΆπ (π β πβ2) + πΆπ (π β π β² ) ππ = 0.85πβ²π π π + π΄β² π π β² π (π β πβ² ) ππ = πππ. ππ ππ΅ β π β ππ ππ€ππ ππ ππππ‘ π πππ = 0.90ππ πππ = 243.53 ππ β π c) Maximum factored uniform load:
Factored load, A
B
C
By there-moment equation: ππ΄ πΏ1 + 2ππ΅ (πΏ1 + πΏ2 ) + ππ πΏ2 +
6π΄1 πΜ
1 6π΄2 πΜ
2 + =0 πΏ1 πΏ2
ππ΄ = ππΆ = 0 6π΄1 πΜ
1 π€π’ πΏ1 3 = πΏ1 4 6π΄2 πΜ
2 π€π’ πΏ1 3 = πΏ2 4 π€π’ (6)3 π€π’ (6)3 0 + 2ππ΅ (+6 +) + 0 + + =0 4 4 π€π’ = ππ. ππ ππ΅/π
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 πβ²π = 20.7 πππ, and steel yield strength ππ¦ = 415 πππ.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 : πΏ = 6π π = 350 ππ π = 400ππ π β² = 70 ππ
πβ²π = 20.7 πππ ππ¦ = 415 πππ ππ = 28 ππ π
Tension steel area π΄π = 4 (28)2 π₯ 4 = 2463 ππ2 π
Compression steel area, π΄β²π = 4 (28)2 π₯ 2 = 132 ππ2 π½1 = 0.85 π = 0.90 Assuming tension & compression steel yields: π΄π 2 = π΄β²π = 1232 ππ2 π΄π 1 = π΄π β π΄π 2 = 1232 ππ2 ππ = π1
0.85πβ²π π π = π΄π 1 ππ¦ 0.85(20.7)π(350) = 1232(415) π = ππ ππ β πππ π€ππ ππ ππππ‘ π
π = π/π½1
π = 83/0.85 π = 97.64 ππ
πβ²π = 600
π β πβ² π
97.64 β 70 97.64 πβ²π = 170 πππ < ππ¦ πβ²π = 600
Thus, compression steel does not yield.
Since compression steel does not yield, πβ²π = 600
πβπβ² π
Assuming tension steel yields: πΆπ + πΆπ = ππ 0.85πβ²π π π + π΄β²π πβ²π = π΄π ππ¦ 0.85(20.7)(0.85π)(350) + 1232 π₯ 600 π = 130.08 ππ π = π½1 π = 110.6 ππ ππ = 600
πβπ π
π β 70 = 2463(415) π
ππ = 600
400 β 130.08 130.08
ππ = 1245 > ππ¦ (π¦ππππ) πβ²π = 600
π β πβ² π
πβ²π = 600
130.08 β 70 130.08
πβ²π = 277.11 πππ < ππ¦ π ππ = ππ (π β ) + πΆ β² π (π β πβ² ) 2 π ππ = 0.85πβ²π π π (π β ) + π΄β² π π β² π (π β πβ² ) 2 ππ = 0.85(20.7)(110.6)(350)(400 β 110.6β2) + 1232(277.11)(400 β 70) ππ = 347.33 ππ β π
Ultimate moment capacity= πππ = 0.90(347.33) Ultimate moment capacity= πππ = 312.6 ππ΅ β π β πππ π€ππ ππ ππππ‘ π
3 m
3 m
L=6 m ππ’ = πππ = 312.6 ππ β π ππ’ = 1.4 ππ· + 1.7 ππΏ
ππ’ = 1.4
ππ· πΏ2 8
+ 1.7
20(6)2
ππΏ πΏ 4
312.6 = 1.4 8 + 1.7 ππΏ = ππ. πππ ππ΅
ππΏ (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 πβ²π = 21 πππ πππ ππ¦ = 415 πππ.
525mm 8 - 25mm
30 mm
650 mm
Figure 3.12
25 mm
360 mm
30 mm
360 mm dβ
π
(25)2 4
πβ² = 30 +
π 4
(25)2
d
1 = 42.5 ππ 2(25)
8 - 25mm
Effective depth to extreme tension bar: 1 ππ‘ = 650 β 30 β = 607.5 ππ 2(25) Effective depth (to centroid of tension bar) π = 650 β 30 β 25 β 1/2(25) π = 582.5 ππ ππ· β 215 ππ β π πβ²π = 21 πππ ππ¦ = 415 πππ Assume all steel yields: π΄π 2 = π΄β²π = 2,454 ππ2 π΄π 1 = π΄π β π΄π 2 = 1,473 ππ2 πΆπ = π1
0.85πβ²π π π = π΄π 1 ππ¦ 0.85(21)π(360) = 1.473(415) π = 95.1 ππ
π = π/π½1
π = 95.1/0.85 π = 111.9 ππ
πβ²π = 600
πβπβ² π
πβ²π = 600
111.9β42.5 111.9
30 mm
650 mm
Compression steel, π΄β²π = 5 π₯ π΄β²π = 2,454 ππ2
5- 25mm
25 mm
SOLUTION π½1 = 0.85 Tension steel, π΄π = 8 π₯ π΄π = 3,927 ππ2
30 mm
πβ²π = 372 πππ < ππ¦ Compression steel does not yield.
360 mm dβ
25 mm
d
5-5-25mm 25mm
c
a d-a/2
d-dβ
8 - 825mm 25mm
T
π΅πππ: πβπππ πππ π‘π€π πππ€π¦πππ ππ π‘πππ πππ ππππ π€βππβ πππ£πππ’π ππ¦ π¦πππ. πβπ’π , π‘βπππ π π‘πππ π ππ πππ πππ‘β π ππ‘ πππ’ππ π‘π ππ¦ πππ π‘βπππ ππ ππ πππππ‘ππ ππ‘ π‘βπππ ππππππ‘πππ ππππ‘ππππ. π = πΆπ + πΆβ²π
π΄π ππ¦ = 0.85πβ²π π π + π΄β²π πβ²π πβπβ²
π΄π ππ¦ = 0.85 π β² π (π½1 π)π + π΄β²π π₯ 600 π 3,927(415) = 0.85(21)(0.85π)(360) πβ42.5 +2,454 π₯ 600 π π = 122.38 ππ πβ²π = 600
π = π½1 π
πβπβ² π
122.38β42.5
πβ²π = 600 122.38 ππ = 391.64 πππ < ππ¦ π = 0.85(122.38) π = 104.03 ππ
π ππ = πΆπ (π β ) + πΆ β² π (π β πβ² ) 2 π ππ = 0.85 πβ²π π π (π β ) + π΄β²π π β² π (π β π β² ) 2 104.03 ππ = 0.85(21)(104.03)(360)(582.5 β ) 2 +2,454(391.64)(582.5 β 42.5) ππ = 873.68 ππ β π πππ = 0.90(873.68) πππ = 786.31 ππ β π ππ’ = π ππ
ππ’ = 1.4 ππ· + 1.7 ππΏ 786.31 = 1.4(215) + 1.7(ππΏ ) ππΏ = πππ. π ππ΅ β π
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 πβ²π = 21 πππ πππ ππ¦ = 415 πππ.
Compression steel, π΄β²π = 2 π₯
π΄β²π = 982 ππ2
π 4
(25)2
πβ² = 30 + 1β2(25) = 42.5 ππ
30 mm 2- 25mm
10 - 28mm
650 mm
(28) 4
2
28 mm
Tension steel, π΄π = 10 π₯ π΄π = 6,158 ππ2
π½1 = 0.85
π
Effective depth (to centroid of tension bars) π = 650 β 30 β 28 β 1/2(28) π = 578 ππ ππ· = 360 ππ β π πβ²π = 21 πππ ππ¦ = 415 πππ
30 mm
Figure 3.13
320 mm 2 - 25mm
650 mm
dβ
30 mm
28 mm
d
10 - 28mm
30 mm
Assume all steel yields: π΄π 2 = π΄β²π = 982 ππ2 π΄π 1 = π΄π β π΄π 2 = 5,176 ππ2 πΆπ = π1
0.85πβ²π π π = π΄π 1 ππ¦ 0.85(21)π(320) = 5,176(415) π = 376.04 ππ
π = π/π½1
π = 376.04/0.85 π = 442.4 ππ
πβ²π = 600
π β πβ² π
πβ²π = 600
442.4 β 42.5 442.4
ππ = 222 πππ < ππ¦ Tension steel does not yield.
320 mm dβ
10 - 28mm 825mm
π΅πππ: πβπππ πππ π‘π€π πππ¦πππ ππ π‘πππ πππ ππππ π€βππβ βππ£π πππππππππ‘ π π‘πππ π ππ πππ π π‘βππ ππ¦ . πβπ’π , π‘βπππ ππ ππ πππ‘ πππππ‘ππ ππ‘ π‘βπππ ππππππ‘πππ ππππ‘ππππ. π1 = 650 β 30 β 14 = 606 ππ π2 = 650 β 30 β 28 β 28 β 14 = 550 ππ π΄π π‘1 = π΄π π‘2 = 5 π₯
π (28)2 = 3,079 ππ2 4
π1 + π2 = πΆπ + πΆβ²π π΄π π‘1 ππ 1 + π΄π π‘2 ππ 2 = 0.85 πβ²π π π + π΄β²π ππ¦ π1 β π π2 β π π΄π π‘1 600 + π΄π π‘2 600 = 0.85πβ²π π π + π΄β²π ππ¦ π π 606 β π 550 β π + 3,079 π₯ 600 π π = 0.85(21)(0.85π)(320) + 982(415) 3,079 π₯ 600
π = 363.9 ππ π = π½1 π = 309.29 ππ
d-dβ
c a
d-a/2
28 mm
5- 25mm 525mm
d
ππ 1 = 600
π1 β π π
ππ 2 = 600
π2 β π π
πβ²π = 600
π β πβ² π
π = π½1 π
ππ 1 = 600
606 β 363.9 363.9
ππ 1 = 399.25 πππ < ππ¦ 550 β 363.9 ππ 2 = 600 363.9 ππ 2 = 306.9 πππ < ππ¦ 363.9 β 42.5 πβ²π = 600 363.9 πβ²π = 530 πππ > ππ¦ π = 0.85(345.4) π = 301.2 ππ
Solve for d: π1 = π΄π π‘1 ππ 1 π2 = π΄π 2 ππ 2
π1 = 3,079(399.25) π1 = 1,229.2 ππ π2 = 3,079 (306.9) π2 = 944.9 ππ
π π₯ π = π1 π₯ π1 + π2 π₯ π2 2,174.1 π = 1,229.2(606) + 944.9(550) π = 581.66π π ππ = ππ (π β ) + πΆ β² π (π β πβ² ) 2 π ππ = 0.85 πβ²π π π (π β ) + π΄β²π ππ¦ (π β πβ² ) 2 ππ = 0.85(21)(309.29)(320)(578 β 309.29β2) + 982(415)(581.66 β 42.5) ππ = 974.07 ππ β π πππ = 0.90(947.07) = 876.65 ππ β π ππ’ = πππ
ππ’ = 1.4 ππ· + 1.7 ππΏ 876.65 = 1.4(360) + 1.7(ππΏ ) ππΏ = πππ. ππ ππ΅ β π
PROBLEM 3.26 Calculate the design flexural strength of the T-beam shown in Figure 3.14. Use πβ²π = 27 πππ πππ ππ¦ = 350πππ.
=600mm 25mm t=100mm 3-23mm
Figure 3.14
10mm stirrup 10-25mm 25mm
20mm =300mm
.
SOLUTION π½1 = 0.85
π (25)2 = 4,909 ππ2 4 π π΄β²π = 3 π₯ (22)2 = 1,140 ππ2 4 Flange area, π΄π = 600(110) = 66,000 ππ2 π΄π = 10 π₯
Assume all steel yields: π = 110 + 390 β 20 β 10 β 25 β 1β2(25) = 432.5ππ πβ² = 25 + 10 + 1β2 (22) = 46ππ π΄π 2 = π΄β²π = 1,140 ππ2 π΄π 1 = π΄π β π΄π 2 = 3,768 ππ2
=390mm
Area of compression concrete: πΆπ = π1
0.85πβ²π π΄π = π΄π 1 ππ¦ 0.85(27)π΄π = 3,768(350) π΄π = 57,468 ππ2 < π΄π π‘βπππππππ π < π‘ 57,469 = π(600) π = 95.8 ππ < π‘ π = 95.8/0.85 π = 112.7 ππ
π΄π = π ππ π = πβπ½1
πβ²π = 600
π β πβ² π
112.7 β 46 112.7 πβ²π = 355 πππ > ππ¦
πβ²π = 600
(π¦ππππ)
=600mm t=110mm
25mm
ππ = 600
πβπ π
10-25mm 1025mm =300mm
10mm stirrup 20mm
432.5 β 112.7 112.7 ππ = 1,703 > ππ¦ (π¦ππππ)
ππ = 600
d-dβ
d
d-a/2
3-22mm 323mm
=390mm
25mm
Verify if the upper layer of tension steel yields π2 = π β 1β2(25) β 1β29(25) = 407.5 ππ ππ 2 = 600
π2 β π = 1,567 πππ > ππ¦ π
(π¦ππππ)
All steel yields, assumption is correct: π ππ = πΆπ (π β ) + πΆ β² π (π β πβ² ) 2 π ππ = 0.85πβ²π π ππ (π β ) + π΄β² π ππ¦ (π β πβ² ) 2 ππ = 0.85(27)(95.8)(600)432.5 β 46) + 1,140(350)(432.5 β 46) ππ = 661.5 ππ β π πππ = 0.90(661.5) πππ = πππ. π ππ΅ β π PROBLEM 3.27 Calculate the design flexural strength of the T-beam shown in Figure 3.15. Use πβ²π = 25 πππ πππ ππ¦ = 345 πππ.
=600mm 25mm t=100mm 2-22mm
Figure 3.15
10mm stirrup 10-28mm
25mm
20mm =315mm
=390mm
SOLUTION π½1 = 0.85
π (28)2 = 6,158 ππ2 4 π π΄β²π = 2 π₯ (22)2 = 760 ππ2 4 Flange area, π΄π = 600(100) = 60,000 ππ2 π΄π = 10 π₯
Assume all steel yields: π = 100 + 390 β β20 β 10 β 28 β 1β2(25) = 419.5 ππ πβ² = 25 + 10 + 1β2(22) = 46 ππ π΄π 2 = π΄β²π = 760 ππ2 π΄π 1 = π΄π β π΄π 2 = 5,397 ππ2 Area of compression concrete: πΆπ = π1
π΄π = π΄π + π΄π€ π΄π€ = ππ€ π§
0.85πβ²π π΄π = π΄π 1 ππ¦ 0.85(25)π΄π = 5,397(345) π΄π = 87,626 ππ2 > π΄π π‘βπππππππ π > π‘ 87,626 = 60,00 + π΄π€ π΄π€ = 27,626 ππ2 27,626 = 315 π§ π§ = 87.7 ππ
π = 100 + π§ = 187.7 ππ π = πβπ½1 = 220.83 ππ πβ²π = 600
π β πβ² π
220.83 β 46 220.83 πβ²π = 475 πππ > ππ¦
πβ²π = 600
(π¦ππππ)
