Steps In Deg A Rectangular Beam 4u3y6t

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STEPS IN DEG A RECTANGULAR BEAM 1. Compute the factored loads: Wu = 1.4 D. L. + 1.7 L.L. 2. Compute the ultimate moment Mu Mu = Wu L2/8 (for uniformly distributed load, simply ed) 3. Compute for ρmax ρmax = 0.75 ρb (rho balance) ρb = 0.85fc’ β 600 fy (600 + fy) Take note of the value of β depends on fc’ β = 0.85 for fc’ ≤30 MPa β = 0.85 – 0.05 (fc’ -30) (1/7) for each 7MPa in excess of 30 MPa but not to be taken < 0.65 4. 5. 6. 7.

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Compute ρminimum ; ρmin = 1.4/fy Assume values of ρ ranging from ρmin to .50ρmax Compute the value of w = ρfy/fc’ (tension reinforcement index) Compute for “ d “ using the relation Mu = Ø b d2 fc’ w ( 1 – 0.59 w ) Ø = 0.90 for bending Use the relation b = d/2 for approximate size Compute for steel area requirements using the relation As = ρ b d Assume bar diameter, and then solve for the number of bars, N Check for actual ρ ρ = As/(bd) ρmin < ρ < ρmax ρmax = 0.75 ρb Check for its moment capacity T=C As fy = 0.85 fc’ ab A = (As fy)/(0.85 fc’ b ) Mu = Ø As fy ( d- a/2) > actual Mu ok, safe! Ø = 0.85 for shear Vu = max. shear at a distance “ d “ from the

Vc = (1/6) (√fc’) (bd) for allowable shear 13. Compute for the spacing of 10mm Ø stirrups if Vu > Ø Vc/2 S = Av fy d Vs Vs = Vu – Vc Ø Min Av = b s 3fy Av = (∏/4) (102) (2)

2 14. Compute the development length Ld = 0.02 Ab fy ( √fc’ ) Ab = area of one bar Min. Ld = 0.06 db fy or 300 mm db = diameter of one bar ; h = d + 10 + ½ diameter of bar + 40 _________________________________________________________________________________ EXAMPLE 1 A rectangular beam reinforced for tension only is to carry a dead load moment of 150 kN-m and a live load of 140 kN-m. Use 65 mm steel covering with fc’ = 20.70 MPa and fy = 414. Use ρ = .50 ρmax 1. Determine the dimension of the beam. 2. Determine the number of 25 mmØ bars that could be used. 3. Determine the ultimate moment capacity of the beam being used. 4. Check the strain condition if steel will yield first Solution: 1. Dimension of the beam Mu = 1.4MD + 1.7ML Mu = 1.4(150) + 1.7(140) Mu = 448 kN-m ρb = [0.85(20.7)(.85)(600)]/[414(600+414)] ρb = 0.0214 ρmax = 0.75(0.0214) = 0.016 ρ = 0.50 ρmax = 0.05(0.016) = 0.008 W = (ρ fy)/fc’ = [0.008 (414]/20.7 = 0.16 Mu = Ø fc’ bd2 w ( 1 – 059 w ) R = fc’ w ( 1- 0.59 w ) R = 20.7(0.16)[ 1- (0.59)(0.16)] R=3 Mu = Øbd2 R 448 x 106 = 0.90 bd2 (3) bd2 = 166 x 106 Try b = 350 mm

d2 = 166 x 106 / 350 d = 689 say 690 mm Total depth = 690 + 65 = 755 mm Trial section: 350 mm x 755 mm

3 2. COMPUTE FOR THE NUMBER OF BARS FOR REINFORCEMENT Mu = Øbd2 R’ 448 x 106 = 0.90 (350) (690)2 R’ R’ = 2.99 (required) Compute the for the real value of ρ 0.008/R = ρ/R’ R = [0.008 (2.99)]/3 R = 0.00797 As = 0.0079 (350) (690) As = 1926 mm2 Using 25mm Ø bars (∏/4)(25)2 N = 192 6 N = 3.92, say 4 bars 3. MOMENT CAPACITY OF THE BEAM USED As = (∏/4)(25)2 (4) As = 1963 mm2 ρ = As /(bd) ρ = 1963/[(350)(690)] ρ = 0.0081 ρmax = 0.016 ρmin = 1.4/414 = 0.0034 ρmin < ρ < ρmax , Steel yields. The design is ok. C=T 0.85 fc’ ab = As fy 0.85 (20.7) a (350) = 1963(4)(414) a = 131.97 Mu = Ø T ( d - 0.5a ) Mu = 0.90 (As fy) [ d- a/2] Mu 0.90 (1963)(414)[690 – 0.5(131.97)] Mu = 456 x 106 N-mm > 448 x 106 N-mm (safe)

Check by strain condition a = βc 131.97 = 0.85c C = 155.26 Ɛs/534.74 = 0.003/155.26 Ɛs = 0.0103 Ɛy = 414/200 000 Ɛy = 0.00207 < Ɛs (Steel yields) Assumption is correct

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EXAMPLE 2 Design a reinforced rectangular beam reinforced for tension only for a beam having three continuous span of 5.4 m ( clear span ) . If the beam carries a dead load of 9 kN/m (including its own weight) and a live load of 18 kN/m. Use 1983 ACI specifications and ultimate strength method with fc’ = 28 MPa, fy = 345 MPa.

Solution: LL/DL = 18/19 =2 < 3 OK

Try ρ = 0.50 ρmax ρ = 0.014

Wu = 1.4 DL + 1.7 LL

w = ρ fy/fc’

= 1.4 (9) + 1.7 (18)

w = 0.014(345)/28

= 43.2 kN/m

w = 0.1725

Mu = Wu L2/10

Mu = Ø fc’ bd2 w ( 1- 0.59 w)

= 43.2 (5.4)2/10

R = fc’ (w) (1 – 0.59w) ; R = 4.34 Mu = 0.90bd2 (4.34)

Mu = 125.97 kN-m

Mu = 3.906 bd2 (substitute value of Mu = 125.97x106)

ρb = 0.85fc’β 600 fy( 600 + fy ) ρb = 0.85(2.8)(.85)(600) 345(600 + 345)

125.97 x 106 = 3.906 bd2 bd2 = 32.25 x 106 Try b = 200 mm

ρb = 0.037

d = 402 mm , say 400mm

ρmax = 0.75 (0.037)

Mu = ØRbd2

ρmax = 0.028

125.97 x 106 = 0.90(200)(4002) R’

R’ = 4.37 New ρ 0.014/4.34 = ρ/4.37 Ρ = 0.0141

5 At interior , M = wL2 /10 As = ρbd = 0.0141 (200) (400) As = 1128 mm2 Using 25 mm Ø (∏/4)(252) N = 1128 N = 2.3 , say 3 bars As = 3 (∏/4)(252) As = 1472 mm2 AT CENTER SPAN: M = wL2 /16 By proportion AS/10 = 1128/16 As = 705 mm2 Using 20mm Ø (∏/4) (20)2 N = 705 N = 2.24, say 3 bars At mid-span, outer span: M = (1/14) WL2/14 As/10 = 1128/14 As = 806 mm2 Using 20 mm Ø (∏/4) (20)2 N = 806 mm2 N = 2.56 say 3 bars Check for Web reinforcement

WEB REINFORCEMENT The design of bending for shear is based on the assumption that concrete resist part of the shear and any excess over and above what the concrete could carry should be resisted by shear reinforcement which may take in several forms a) vertical stirrups b) inclined stirrups and c) the main reinforcement bend at ends to act as inclined stirrups. When beams are subjected to external loads they produced not only bending moments but also internal shear force. The shear does not produce failure directly on the vertical plane on which they act but it’s major effect is it includes tensile stresses on diagonal plane which is called diagonal tension. Since concrete has a relatively low tensile strength, it would produce cracks making an angle of 45° with the plane on which the shear acts. When this diagonal tension stresses in combination with bending stresses created by moment exceed the tensile strength of concrete, diagonal cracking will occur that can split the beam without warning.

6 Shear produces two types of diagonal cracks in reinforced concrete beams. 1. Web-shear cracks 2. Flexure-shear cracks DRAWING

Shear force that the concrete alone could carry, Vc Vc = (1/6)√𝑓𝑐 ′ bd b = width of rectangular beam or width of web for a T – beam d = effective depth of beam Ø = strength reduction factor = 0.85 ØVc = Ø(1/6)√𝑓𝑐 ′ bd Vu = design shear force which results from the application of factored loads. Vu = (WuL/2) - Wud (for simply ed beam) a) When Vu < ØVc/2 , no web reinforcement is required. However, the code requires that a minimum area of shear reinforcement be provided in all reinforced concrete flexural except as follows: 1. In slabs and footings 2. In beams with a total depth less than 250 mm, 2.5 times the flange thickness or one half the width of web, whichever is greater. 3. In concrete joist construction 4. Where Vu < (ØVc)/2 b) Minimum area of web reinforcement: Min. Av = Vs/(3fy) Av = total cross-sectional area of web reinforcement in mm2 Av = 2 As where As is the cross-sectional area of stirrups in mm2 b = width of rectangular beam or web width for T-beams in mm S = spacing of stirrups center to center in mm Fy = yield strength of web reinforcement in MPa C) Spacing of stirrups: 1. S = (Av fy d)/Vs Vs = Vn – Vc Vn = Vu/Ø Vc = (1/6)√𝑓𝑐 ′ bd Vs = nominal shear strength provided by shear reinforcement. Shear Stress = Vu/(Øbd) Shear Stress = Ø(Vc+Vs)/(Øbd) Shear Stress = Vc/(bd) + Vs/(bd) Vs = (Vu/Ø) – Vc When Vu > Øvc/2 , stirrups are needed

7 3. For 45° stirrups S = (1.414 Av fy d)/Vs Vu < ØVc + ØVs Vs = (Vu – ØVc)/Ø Vs = (Vu/Ø) – Vc Vs should not exceed (2/3)√𝑓𝑐 ′ bd irrespective of amount of web reinforcement d. Maximum spacing of stirrups 1. Max. S = d/2 or 600 mm, whichever is smaller when Vs<[(1/3)√𝑓𝑐 ′ bd] 2. Max S = d/4 or 300 mm when Vs > [(1/3)√𝑓𝑐 ′ bd] 3. Desirable spacing of stirrups = 100 mm NOTE: Stirrups should be required if Vu > (ØVc)/2

(CONTINUATION OF EXAMPLE) At interior s Vu = 1.15WuL/2 Vu = 1.15 (43.2)(5.4)/2 Vu = 134.136 kN At a distance “d” from the Vu = 134.136 – 43.2 (0.40) Vu = 116.856 kN = 116856 N

ØVc = Ø(1/6)√𝑓𝑐′ bd ØVc = Ø1/6√𝑓𝑐 ′ (200)(400) 6 ØVc = 59970 ØVc/2 = 2998 < Vu , stirrups are needed Vs = (Vu/Ø) – Vc Vc = 59970/.85 Vc = 70553 Vs = (116856/0.85) – 70553 Vs = 66925

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Spacing of 10mm Ø stirrups S = (Av fy d) / Vs Av = (∏/4) (102) (2) = 157 mm2 S = 157(345) (400)/ 66925 S = 324 mm O.C. Max. S = d/2 when Vs < 1/3 √𝑓𝑐′ bd 1/3√28 (200)(400) = 141107 Vs = 66925 N Vs < 141107 ok Max S = 400/2 = 200mm o.c. Min. stirrup area to be provided Av = bS /(3fy) Av = 200(200)/[(3)(345)] Av = 38.65 < 157 mm2 ok DETAIL