Reinforced Concrete Element Design of Rectangular Beams
LB3 Civil Engineering ITS Fakultas Teknik Sipil dan Perencanaan Institut Teknologi Sepuluh Nopember Surabaya
Last Edited 2014
Design of Rectangular Beams and One-Way Slabs In the previous chapter, the behaviour of reinforced concrete beams (and one-way slabs) was explained, and procedures given for the analysis of sections. Unlike the analysis problem, the design problem does not have a unique solution because the flexural strength of a section is dependent on its width and effective depth, and on the area of reinforcement; and there are several combinations of these which would give the required strength. Different designers may come up with different solutions, all of which may meet the desired requirements. A complete design of a beam involves considerations of safety under the ultimate limit states in flexure, shear, torsion and bond, as well as considerations of the serviceability limit states of deflection, crack-width, durability etc.
Required strength Required strength U shall be at least equal to the effects of factored loads in Eq. (9-1) through (9-7). The effect of one or more loads not acting simultaneously shall be investigated.
( 9 − 1) U = 1.4D U = 1.2D + 1.6L + 0.5(Lr or S or R ) ( 9 − 2) ( 9 − 3) U = 1.2D + 1.6(Lr or S or R ) + ( 1.0L or 0.5W ) U = 1.2D + 1.0W + 1.0L + 0.5(Lr or S or R) ( 9 − 4) U = 1.2D + 1.0E + 1.0L + 0.2S ( 9 − 5) U = 0.9D + 1.0W ( 9 − 6) U = 0.9D + 1.0E ( 9 − 7) Where: D = dead load, L = live load, Lr = roof live load, S = snow load, R = rain load, W = wind load, E = seismic load
Example:4.1 The compression gravity axial loads for a building column have been estimated with the following results: D = 150 k; live load from roof, Lr = 60 k; and live loads from floors, L = 300 k. Compression wind, W = 70 k; tensile wind, W = 60 k; seismic compression load = 50 k; and tensile seismic load = 40 k. Determine the critical design load using the SNI load combinations. Solution:
(9 − 1)U = 1.4D = (1.4)(150k ) = 210k (9 − 2)U = 1.2D + 1.6L + 0.5(Lr or S or R ) = (1.2)( 150k ) + (1.6)( 300k ) + (0.5)( 60k ) = 690k (9 − 3)( a)U = 1.2D + 1.6(Lr or S or R ) + (L or 0.5W ) = (1.2)(150k ) + (1.6)( 60k ) + (300k ) = 576k (b)U = 1.2D + 1.6(Lr or S or R ) + (L or 0.5W ) = (1.2)(150k ) + (1.6)( 60k ) + (0.5)( 70k ) = 311 (c)U = 1.2D + 1.6(Lr or S or R ) + (L or 0.5W ) = (1.2)(150k ) + (1.6)( 60k ) + (0.5)( −60k ) = 24 (9 − 4)( a)U = 1.2D + 1.0W + L + 0.5(Lr or S or R ) = (1.2)( 150k ) + (1.0)( 70k ) + (300k ) + 0.5(60k ) = (b)U = 1.2D + 1.0W + L + 0.5(Lr or S or R ) = (1.2)( 150k ) + (1.0)( −60k ) + (300k ) + 0.5(60k ) (9 − 5)( a)U = 1.2D + 1.0E + L + 0.2S = (1.2)( 150k ) + (1.0)( 50k ) + ( 300k ) + (0.2)( 0k ) = 530k (b)U = 1.2D + 1.0E + L + 0.2S = (1.2)( 150k ) + (1.0)( −40k ) + (300k ) + (0.2)(0k ) = 440k (9 − 6)( a)U = 0.9D + 1.0W = (0.9)( 150k ) + (1.0)( 70k ) = 205k (b)U = 0.9D + 1.0W = (0.9)( 150k ) + (1.0)( −60k ) = 75k (9 − 7)( a)U = 0.9D + 1.0E = (0.9)( 150) + (1.0)( 50k ) = 185k (b)U = 0.9D + 1.0E = (0.9)( 150) + (1.0)( −40k ) = 95k
Design of Rectangular Beams Before the design of an actual beam is attempted, several miscellaneous topics need to be discussed : (1) Beam proportions, ratio d to b, in 1 12 to 2 (2) Deflection, from best practice h ≥ l/10 to l/14 this value is higher than SNI. (3) Estimated beam weight. The weight of the beam to be selected must be included in the calculation of the bending moment to be resisted. Reinforced concrete weigth is 240 kN/m3 . (4) Selection of bars. After the required reinforcing area is calculated. (5) Cover. Use SNI 7.7 for requirements.
Design Procedure: 1. Set Mu = φ Mn = φRn bd2 2. Assume ρ = (0.4 − 0.6) ρb
f c ρb = 0.85 β 1 f y
600 600 + f y
3. Find the flexural resistance factor Rn Rn = ρ f y
ρm 1− 2
m=
4. Determine the required dimensions b, d Mn Mu = bd = φ Rn Rn 2
f y 0.85 f c
5. Determine the required steel area for the chosen b, d As = ρbd 1 ρ = m
1−
2mRn 1− f y
,
Mn Rn = 2 , bd
m=
f y 0.85 f c
6. Check for minimum steel reinforcement area As,min
f c 1.4 = 0.25 bd ≥ bd f y f y
4 if As, provided ≥ As,required not need to use As,min 3 7. Check for strain (ε t ≥ 0.005) tension-controlled section. 8. Check for steel bars arrangement in section.
Values of ρb, ρmin , ρ at ε t for various strength f c β 1 f y = 240 ρ balanced ρ at ε = 0.004 ρ at ε = 0.005 ρ min f y = 320 ρ balanced ρ at ε = 0.004 ρ at ε = 0.005 ρ min f y = 400 ρ balanced ρ at ε = 0.004 ρ at ε = 0.005 ρ min
20 0.85 0.0430 0.0258 0.0226 0.0058 0.0294 0.0194 0.0169 0.0044 0.0217 0.0155 0.0135 0.0035
25 0.85 0.0538 0.0323 0.0282 0.0058 0.0368 0.0242 0.0212 0.0044 0.0271 0.0194 0.0169 0.0035
30 0.84 0.0634 0.0381 0.0333 0.0058 0.0434 0.0285 0.0250 0.0044 0.0320 0.0228 0.0200 0.0035
35 0.80 0.0708 0.0425 0.0372 0.0062 0.0485 0.0319 0.0279 0.0046 0.0357 0.0255 0.0223 0.0037
40 0.76 0.0773 0.0464 0.0406 0.0066 0.0530 0.0348 0.0305 0.0049 0.0390 0.0278 0.0244 0.0040
Reinforcement Area Table
Minimum beam width
Example:4.2 Calculate the area of steel reinforcement required for the beam 300x650mm. Mu = 360 kN-m, with f c = 30 MPa and f y = 400 MPa. Assume D25 with one layer arrangement, and ∅10 for stirrup. Solution: 25 d = h − decking − ∅stirrup − D25 = 650 40 10 − − − 2 2 = 587.5 mm take φ = 0.9 for tension controlled. Mn Mu 360 × 106 Rn = 2 = = = 3.86 MPa bd φbd2 0.9 × 300 × 587.52 f y 400 = = 15.69 m= 0.85 f c 0.85 × 30 1 2mRn 1 2· 15.69· 3.86 ρ = = = 1− 1− 1− 1− m f y 15.69 400 0.0105 ρmin = 0.0035 From table. ⇒ ρreq > ρmin ok As = ρbd = 0.0105 × 300 × 587.5 = 1850.625 mm2
Use 4D25, with As = 1963.50 mm2 (from steel area table) Check for strain: As f y 1963.50 × 400 a= = = 102.66 mm 0.85 f c b 0.85 × 30 × 300 f c − 28 β 1 = 0.85 − 0.005 = 0.836 7 102.66 a c= = = 122.8 mm β 1 0.836 d−c 587.5 − 122.8 ε t = 0.003 = 0.003 = 0.01135 > 0.005 Ok. c 122.8
Check for bar placement: 300 − 40 × 2 − 10 × 2 − 4 × 25 = 33.33 > 25 mm. Ok. Sb = 3
Example:4.3 Select an economical rectangular beam sizes and select bars . The beam is a simply supported span of a 12 m and it is to carry a live load of 20 kN/m and a dead load of 25 kN/m including beam weight. Select f c = 28 MPa and f y = 400 MPa. Assumed d ≈ 2b Solution:
W u = 1.2DL + 1.6LL = 1.2 · 25 + 1.6 · 20 = 62 kN/m W u l2 62 × 122 Mu = Mmax = = = 1116 kN-m 8 8 Take φ = 0.9 for tension controlled section. take ρ = 0.4 ρb , ρ b = 0.032 (from table), ρ = 0.4 × 0.032 = 0.012 f y 400 m= = = 16.807 0.85 f c 0.85 × 28 ρm 0.012 × 16.807 Rn = ρ f y 1 − = 0.012 × 400 1 − = 4.36 MPa 2 2 6 6 M · · 1116 10 1116 10 3 u bd2 = = = 4b3 so that b = = 414.28 mm φ Rn 0.9 · 4.36 4 · 0.9 · 4.36
Take b = 400 mm, and d = 2 · b = 800 mm Mu 1116 · 106 Rn = = = 4.84 MPa. 2 2 φbd 0.9 · 400 · 800 1 2mRn 1 2· 16.807· 4.84 ρ = = 1− 1− 1− 1− 16.807 400 m f y 0.01367 ρmin = 0.0035 (from table) As = ρbd = 0.01367 · 400 · 800 = 4374.54 mm2 Use 4D28 + 4D25 two layers, with As = 2463.01 + 1963.5 = 4426.25 mm2
Check for strain. As f y 4426.25 · 400 a= = = 185.99 mm 0.85 f c 0.85 · 28 a 185.99 c= = = 218.81 mm β 1 0.85 28 dt = d + S2 + d2b = 800 + 25 + 2 2 = 826.5 mm dt − c 826.5 − 218.81 ε t = 0.003 = 0.003 = 0.00833 > 0.005 Ok. 218.81 c
=
Home Works
Solve Problem 4.34 and 4.35 of reference book.
