- 1. 1
- 2. Analysis of the Section 1.1 Strength design method assumption The analysis of a cross section to determine its moments of resistance at ultimate limit state is based on the following assumptions: 1- A plane section (perpendicular to the axis of bending) before bending remains a plane after bending, so that the strains distribution in the concrete and the reinforcement (tension or compression) must be linear distribution of strains. 2- The tensile strength of concrete is zero and the reinforcement carries all the tensile forces (in fact the tensile strength of concrete is only about 10 per cent of the compressive strength) 3- The bond between the concrete and the reinforcement is perfect, so no slip occurs to ensure that the tensile forces to be transferred by bond to the reinforcement 4- A compressive concrete stress is approximately proportional to strain only up to moderate loading, with an increase in load, the compressive stress-strain diagram takes shape similar to the concrete compressive stress-strain curve shown in figure 1.1; Figure 1.1: Short term stress strain curve for normal weight concrete design curve of the BS8110 characteristic curve 0.67 fcu /γm fck fcu 0.0035 Stress Strain Notes: As assumed by the BS8110, the ultimate compressive strain is 0.0035, 0.67 is the size factor, γm equal to 1.5 (Table 2.2 of the BS8110) 2
- 3. 5- The maximum useable concrete compressive strain at the extreme fiber is assumed to be equal to 0.0035 6- For stress in steel reinforcement less than yield stress (fy ), the steel stress is proportional to strain (εs) and equal to Es εs. Stress-strain diagram of steel reinforcement is shown in Figure 1.2. Figure 1.2: Stress strain diagram of steel reinforcement 1.2 Section Analysis Based on the assumption previously stated in section 1.1, we can now examine the strains, stresses and forces that exist in reinforced concrete beam as the load increases from zero to the magnitude that would cause failure: A) At very small load before cracks occurring (at cracking load); both concrete and steel reinforcement resist the tensile forces at the bottom of the beam (tension side, below the natural exist). In this case, the stresses are vary linearly from the neutral axis (zero position) of the section and proportioned to strains as shown in figure 1.3. Characteristic curve Design curve E = 200 kN/mm2 fy 0.95 fy Stress Strain 3
- 4. 4 Figure 1.3: Flexural behaviour of RC beam at very small load (below cracking load) b) Section A - A c) Strains distribution d) Stresses distribution fcu (compression) fcu (tension) fs (tension) εc (compression) εS (tension) h d εc (tension) a) Reinforced concrete beam A B A A N.A.
- 5. B) At moderate load which starts from cracking load and up to before yielding load. When the tensile stresses at the bottom of the beam exceed the tensile strength of the concrete, the cracks in concrete (hair crack) occur. Since the concrete cannot transmit any tension forces across a crack, the steel reinforcement then resist the entire tension forces. In this case; concrete compression on stresses are still assumed to be proportional to strains in concrete as shown in Figure 1.4. 5 A A A B a) Reinforced concrete beam N.A. fcu (allowable stress) fs (allowable stress) εc (tension) εc (compression) εs (tension) b) Section A-A beam c) Strains distribution d) Stresses distribution Figure 1.4: Flexural behaviour of RC beam at moderate load
- 6. C) At yielding load, this is started from the yield point of the steel reinforcement up to the ultimate load (before failure). In this case the trend of stress strain distribution in concrete section of the beam (above the neutral axis) is similar to that of concrete stress-strain curve (shown in Figure 1.1). the stresses and strains distribution in the section of the beam are shown in Figure 1.5. 6 concrete crush A A A B a) Reinforced concrete beam N.A. fs = fy (yield stress) 0.67 fcu (ultimate stress) εs εs = 0.0035 (as a limit) b) Section A-A c) Strains distribution d) Stresses distribution Figure 1.5 : Flexural behaviour of RC beam at ultimate load
- 7. As shown in Figure 1.5 (c), the shape of stresses distribution is regtangular parabolic shape. Some problems are encountered with the rectangular parabolic shape; such as calculation of the area under the curve and determine the centroid location. 1.3: Equivalent rectangular stress block Equivalent rectangular stress block was adopted by BS8110 for purposes of simplification and practical application. With respect to equivalent stress distribution block shown in figure 1.6(c); the average stress intensity is taken as 0.45 fcu and is assumed to be acted on the upper edge of the RC beam cross section. The area is defined by the width b and a depth of s. 7 N.A. fy x fs = fy (yield stress) 0.67 fcu /γm 0.67 fcu / 1.5 = 0.45 fcu s =0.9 x d b a) Section b) Rectangular parabolic stress block c) Equivalent rectangular stress block Figure 1.6: Rectangular parabolic stress block and equivalent rectangular stress block.
- 8. The equivalent rectangular stress block does not extend to the neutral axis of the section but has a depth of s, which is equal to 0.9x. This will result the centroid of the stress block being s/2, which is equal to 0.45x from the top edge of the section. The area and the centroid of the two stresses block (rectangular parabolic and equivalent stress block) are approximately equal. Thus the moment of resistance of the section will be similar using calculation of either type of stress block. 1.4: Types of failure and beam section classification. Three failure types could be happened to RC beam loaded in flexure and that depends on the amount of steel provided in the section. The three types of failure are as following. A – Balanced reinforced beam: In this type of failure the concrete crush at the compression area (above the neutral axis) and the steel reinforcement yield simultaneously at ultimate load. The amount of steel reinforcement required can be calculated by equating the compression force in the concrete and the tension force in the steel (equilibrium principle). Example 2.1: For the section shown in the Figure 1.7, determine the depth of the neutral axis for the balanced design case. fy = 460 N/mm2 and Es = 200 KN/mm2 8
- 9. Solution: Section Strains xd y x c − ∑ = ∑ 9 b h d h d b N.A. εc = 0.0035 εy x d-x Figure 1.7
- 10. 0.615dexceedshouldnotxcrush;concrete beforestatelimintultimatetheatareatensioninentreinforcemsteeltheofyieldingensuretoHence, limitstateultimatetheat 1equation d.x x.x)(d. .c .y . y mmkNE sand.γ m E s γ mf y y .........x .......yx)(dc 6150 00219000350 1equationinngsubstitutithatso 00350 and 002190 000200 051460 200051 2 = ∗=−∗ ∑ = ∑ = ∑ = == ∑ = ∑ ∑ ∗=−∗ B – Under reinforced beam: if the amount of steel reinforcement provided is less than that of case balanced reinforced beam, the steel will reach at ultimate yielding stress before the concrete fails at the ultimate load. This type of failure is gradual and giving warning of failure. In this case the depth of the neutral axis must be less than 0.615d. The BS8110 was adopted for the ultimate state design, (where the moment redistribution not greater than 10%); x ≤ 0.5d C – Over reinforced beam: If the amount of the steel reinforcement provided is greater than that of case of balanced reinforcement beam; then the concrete fails before steel reaches yielding stress. This type of failure is sudden and without warning of failure. The depth of the neutral axis in this case is greater than 0.615d. Example 2 10
- 11. The beam shown in the Figure 1.8 is made of concrete with compressive strength (fcu) of 35N/mm2 and steel reinforcement with strength (fy) of 460N/mm2 . Determine the ultimate moment of resistance of the cross section. Solution: Fcc = Stress ∗ Area Fcc = 0.45fcu ∗S ∗ b S = 0.9∗x Fcc = 0.45 ∗35∗0.9∗300 Fcc = 4252.5∗x Assume the steel has been yielded; fst = 0.95fy 11 b = 300mm As = 1450mm2 d = 550 mm 0.45 fcu x s N.A. Fcc Fst z b = 300mm As = 1450mm2 d = 550 mm Figure 1.8 cross section stress block
- 12. Fst = 0.95 fy ∗ As Fst = 0.95 ∗ 460 ∗ 1450 Fst = 633650 N For equilibrium; Fcc = Fst 4252.5 ∗ x = 633650 4252.5 633650 x = x = 149 mm Since, x (=149) < 0.615d (=338.25), therefore the steel has been yielded and fst = 0.95fy as assumed. Moment of resistance of the section M = Fst ∗ Z = 0.95fy As (d -0.9x/2) = 0.95 fy As (d – 0.9x / 2) = 0.95∗460 ∗1450 (550 – (0.9 ∗149) /2) ∗ 10 -6 = 306.02 KN.m Example 3 12
- 13. As = 1450mm2 305 mm305 mm h =610mm d =590mm 0.9x 0.45fcu Fcc N.A. Fst Compression zone area = (0.9x)2 / 2 cross - section 305 mm 305 mm Determine the ultimate moment of resistance of the cross section shown in Figure 1.9, given that the characteristic strength of the steel reinforcement is 460N/mm2 and the characteristic strength of concrete is 30 N/mm2 . 13 Solution: Z 0.9x stresses block (2/3) (0.9x) Figure 1.9
- 14. Fcc = 0.45fcu ∗ area of compression zone area = 0.45 ∗ 30 ∗ (0.9x)2 / 2 = 5.467 x2 Assume the steel reinforcement has been yielded, Fst = 0.95fy Fst = 0.95 fy As = 0.95 ∗ 460 ∗1450 = 633650 N/mm2 From equilibrium Fcc = Fst 5.467∗ x2 = 63350 x (=340.44 mm) < 0.615d (=362.85), therefore, the steel reinforcement has been yield and Fst = 0.95fy as assumed. Moment of resistance of the section Z = d – 2/3 (0.9x) = 590 -2/3 (0.9 x 340.44) = 385.736 mm M = Fst x Z = 0.95 fy As + Z = 0.95 x 460 x1450 x 385.736 x 10-6 = 244.421 KN.m 14
- 15. 1.5 Analysis of flanged section In the analysis of flanged section (T-section or L- section) is important to determine the location of the neutral axis whether within the flange or within the web. Example 4 A flanged beam section (T- section) is shown in the Figure 1.10. Determine the maximum moment which can be applied to the section. The characteristic strength of the materials are fy = 460 N/mm2 and fcu = 35 N/mm2 . Solution Assume that the neutral axis lies within the flange and the steel reinforcement is yielded ( Fst = 0.95fy ) 15 bf = 850mm hf = 150mm d = 450mm As = 1450 mm2 Figure 1.10
- 16. Fcc = 0.45 fcu ∗bf ∗ 0.9x = 0.45 ∗ 35 ∗850 ∗0.9x = 12048.7 x Fst = 0.95fy As = 0.95 x 460 x 1450 = 633650 N From equilibrium Fcc = Fst 12048.7 x = 633650 x = 633650 / 12048.7 = 52.6 mm < hf (=150) The neutral axis lies within the flange as assumed. X ( = 52.6mm ) < 0.615d (= 276.75mm) The steel will have yielded as assumed thus, fs = 0.95fy The resistance moment of the section M = Fst ∗z 16 Fcc d = 450mm x 0.45fcu S=0.9x 0.9x /2 Z Fst N.A. bf = 850mm hf = 150 mm cross section stress block
- 17. = 633650 ∗ (d – 0.9x / 2) ∗10-6 = 270.144 KN.m Example 5 Determine the moment of resistance of the T –section shown in the Figure 1.11. fy = 460 N/mm2 and fcu = 20N/mm2 Solution Assume that the neutral axis lies within the flange and the steel reinforcement is yielded so that fs = 0.95fy 17 bf = 850mm 0.45fcu 0.9xx Fcc As= 2200 mm2 300mm d = 450mm bf = 850mm hf = 100mm Figure 1.11
- 18. Fcc = 0.45 fcu ∗bf ∗0.9x = 0.45 ∗20 ∗850 ∗0.9x = 6885 x Fst = 0.95fy As = 0.95 ∗ 460 ∗2200 = 96.1400 Fcc = Fst 6885 x = 961400 x = 961400 / 6885 = 139.6 mm > hf ( = 100mm) The neutral axis lies within the web 18 Fst N.A. bf = 850mm 0.567fcu Fst Z2 = d-100/2 Z1 = d-100-(0.9x-100)/2 Fcx Fcw 100 mmS=0.8x N.A Cross section Stress block d=450mm
- 19. Fcf = 0.45 fcu bf hf = 0.45 ∗ 20∗850 ∗100 = 765000N Fcw = 0.45 fcu bw (0.9x – 100) = 0.45 ∗20 ∗300 (0.9x – 100) = 2430 x – 270000 Fst = 0.95 fy As = 961400 Fst = Fcf + Fcw 961400 = 765000 + 2430x – 270000 x = 191.9 mm < 0.615d (= 276.75) Therefore, the steel will have yielded as assumed thus, Fs = 0.95fy Taking moment about the centroid of the steel reinforcement. M = Fcw ∗Z1 + Fcf ∗Z2 = [(2430 x – 27000) ∗ [ d – (0.9x – 100) / ] + 765000 ( d – (100/2)] ∗10-6 = [(2430 x 191.9 – 27000)x[450–100-(0.9x-100)/2 ]+765000 (450 – (100/2)] ∗ 10-6 = [ 196317 x 313.645 + 306000000] 10-6 = 367.57 KN.m 1.6 The ultimate moment of resistance of singly reinforced section and analysis of doubly reinforced section: 19 bw = 300mm
- 20. When the applied design moment exceeds the concrete capacity; compression steel reinforcement is required to supplement the load - carrying capacity of the concrete. As specified by BS 8110 that the upper limit of the lever arm (Z) is 95 d and the lower limit of Z is 0.755d, while the depth of the neutral axis (x) should not exceeds 0.5d to ensure that the steel reinforcement will have yielded before the concrete in compression area is failed, therefore the RC beam will give signs of warning of the failure. To determine the moment capacity of the single reinforced section shown in Figure 1.12; the depth of the neutral axis assumed to be 0.5d. a) section b) strain distribution c) stress block Figure 1.12: strain distribution and stress block of single reinforced section Fc = 0.45 fcu b ∗0.45 d = 0.202 b d fcu And Z = d – 0.9x /2 = d – 0.45d / 2 = 0.775d Taking moment about the centroid area of the steel reinforcement 20 h d b x =0.5d 0.0035 0.45fcu 0.9x = 0.45d Fcc Fst z N.A
- 21. M = Fc ∗ Z = 0.202 bd fcu ∗0.775d = 0.156 bd2 fcu (capacity of the singly reinforced section based on concrete strength) Taking moment about the centroid of the compression area M = Fs ∗ Z = 0.95 fy As Z = 0.95 fy As (d – 0.45d / 2) = 0.736 fy As d (capacity of the singly reinforced based on steel reinforcement strength) Therefore, the capacity of the singly reinforced section (the ultimate moment of resistance) is given by the lesser of M = 0.156 fcu bd2 fcu (based on concrete strength) or M = 0.736 fy As d (based on steel reinforcement strength) Example 6 Determine the maximum ultimate moment which can be carried by the section shown in Figure 1.13. The characteristic strengths are fy = 460 N/mm2 for steel reinforcement and fcu = 30 N/mm for concrete. 21 b = 300mm h = 450mm d=400mm
- 22. Solution Section capacity based on concrete strength M = 0.156 bd2 fcu = 0.156 ∗300∗4002 ∗ 300∗10-6 = 224.64 KN.m Section capacity based on steel strength M = 0.736 fy As d = 0.736 ∗460 ∗1200 ∗400 ∗ 10-6 = 162.5 KN.m The maximum ultimate moment which can be carried by the section is 162.5 kN.m In the analysis of doubly reinforced section, to ensure that the compression steel will have yielded; d'/x should be less than 0.37 (for fy = 460 N/mm2 ) as shown in the analysis performed on the doublly reinforced section shown in Figure 1.14. 22 As = 1200mm2 As′ x 0.0035 εsc d' x-d' Figure 1.13
- 23. Assume steel reinforcement in compression area are yielded εsc= 0.95 fy / Es = 0.95 ∗460 ∗200000 = 0.00219 0.0035 / x = 0.00219 /( x-d') (x – d') / x = 0.00219 / 0.0035 1 – (d' / x) = 0.62 d' / x = 0.37 for other grades of steel, with x = 0.5d d' / d = 0.185 If d'/ x > 0.37 (or d'/d > 0.185), then it is necessary to determine the value of εsc from strains distribution, then and then to compute fsc from fsc = Es xεsc where Es = 200 000 N/mm2 Example 7 23 As d-x N.A Est Figure 1.14
- 24. Compute the design strength (the ultimate moment of resistance) of the doubly reinforced cross-section as shown in Figure 1.15; fy = 460 N/mm2 and fcu = 30 N/mm2 Solution Form equilibrium. Fst = Fcc + Fsc (assume that the steel reinforcement in the tension and compression are yielded values ( fst = fsc = 0.95 fy) 0.95 fy As = 0.45 fcu b ∗ 0.9 x + 0.95 fy As' x = 0.95 fy (As – As') / 0.405 fcu b x = 0.95 ∗ 460 (2325 – 403) / 0.405 ∗30 ∗250 x = 276.51 mm < 0.615d ( = 319.8mm) Therefor, steel reinforcement in tension will have yielded as assumed 24 b=250mm As' = 403mm2 As = 1325mm2 d = 520mm d' =50mm As′ As 0.45fcu Fst N.A Z2=d-50 Z1=d-0.9x/2 Fsc Fcc0.9xx Figure 1.15
- 25. d' / x = 50 / 276.51 = 0.18 < 0.37 So the steel reinforcement in the compression will have yielded as assumed. Taking moment about the centroid of the steel in tension M = Fcc x Z1 + Fsc ∗Z2 M = 0.45 fcu b ∗0.9 x (d – 0.9x / 2) + 0.95 fy As' (d – 50) = [0.45 x 30 x 250 x 0.9 276.51 (520 – 0.9x276.51/ 2) + 0.95 x 460 x 403 (520-50) ∗10-6 = 415 KN. m 1.7 Derivation of Design Formulae for Singly and Doubly Rectangular Reinforced Concrete Beams: To derive the equations stated in clause 3.4.4.4 of the BS8110:1995; Figure 1.16 is referred: Force= stress ∗ area Fs = 0.95 fy As Fc = 0.45 fcu ∗b ∗0.9x = 0.405 fcubx b dh As Z=d-0.45x Fs Fc 0.45fcu N.A 25 Figure 1.16
- 26. For equilibrium Fc= Fs [0.405 fcu b x = 0.95 fy As] ∗ d 1 d x = 2.345 ∗ fcu fy ∗ bd As Note that d x increases with increasing of As and decreases with increasing of the section area (bd). Taking moment about the centroid of the steel area: M = Fc ∗z = (0.405 fcu b x) (d - 0.45 x) Rearrange the above formulae = (0.405 d x ) (1-0.45 x) fcu b d2 Note that moment increases with increasing the value of d x Assume; k= (0.405 d x ) (1-0.45 x) ∴ M=kfcubd2 ( the applied bending moment on the section) Where the simplified stress block is used; BS8110:1995 limits d x should not exceed 0.5 and x d′ (for doubly reinforced section) should not exceed 0.37 (for fy=460 N/mm2 ). This limit in d x and x d′ is to ensure that the steel reinforcement will have yielded (fs=0.95fy) and the design will be under 26
- 27. reinforced section. Therefore, the maximum moment capacity of a singly reinforced beam based on concrete strength is calculating form (limiting x=0.5d) Mu= 0.202 fcu bd2 z Where; z= d-0.45 x = d-0.45 (0.5 d) = 0.775 d (minimum value of z) ∴Mu = (0.202 fcu bd) (0.775d) = 0.156 fcu bd2 Where k′ = 0.156 ∴ Mu= k′ fcu bd2 (the maximum moment capacity of singly reinforced beam based on concrete stregth) Where the applied moment exceeds the maximum moment capacity of the section then the excess (M-Mu) to be resisted by using steel reinforcement (As′ ) in the concrete compression area to supplement the load-carrying capacity of the concrete. The neutral axis depth (x) to be maintained at the maximum permitted value i.e 0.5d. By referring to the Figure 1.17: b dh As Z=d-0.45x Fs Fc 0.45fcu N.A As′ d′ Fs′ z1 =d-d′ 27 Figure 1.17
- 28. F s′ = 0.95 fy As′ M-Mu=Fs′ ∗ z1 = 0.95 fy As′ ∗ (d-d′) )d(df0.95 MuM As y ′− − =′ Where Mu= k′ fcu bd2 ∴ )d(dfy0.95 d2bfcukM As ′− ′− =′ An area of steel in tension must then provided to balance the total compressive forces in concrete and the compression reinforcement: Fs=Fs′ + Fc 0.95 fy As = 0.95 fy As′ + 0.202 fcu bd ∴ As fy0.95 bdfcu0.202 As ′+=′ Where; Mu=0.202fcubd∗z ∴ As zfy0.95 Mu As ′+= Note: to prove that z= d {0.5+ 9.025.0 k− }; refer to Figure 1.16 M = Fc ∗ z = 0.45 fcu b∗ 0.9x ∗z Where; z = d - 2 0.9x and 0.9x = 2(d-z) ∴M= 0.45 fcu b∗ 2(d-z) ∗ z 28
- 29. = 0.9 fcu b (d-z) ∗ z Where; k = dbfcu M 2 ∗∗ d z d z 0.9 k 2 2 −= And (z/d)2 – (z/d) + k/0.9 = 0 Solve the above quadratic equation z= d {0.5+ 9.025.0 k− } note: the above equations are also applicable to flanged beams where the neutral axis lies within the flange. Example 8 The ultimate design moment to be resisted by the section in the Figure 1.18 is 150 kN.m. Determine the area of tension reinforcement (As) required. The characteristic strength of steel fy=460 N/mm2 and the characteristic strength of concrete fcu = 40 N/mm2 : 29 b=200mm d= 387mm h= 450 mm As Figure 1.18
- 30. Solution k = dbfcu M 2 ∗∗ = 40387200 10150 2 6 ∗∗ ∗ = 0.125 < 0.156 Therefore, the section is singly reinforced: z= d {0.5+ 9.025.0 k− } = d {0.5+ 9.0125.025.0 − } = 0.83d < 0.95 d As= zfy0.95 M = mm 3870.834600.95 10150 2 6 = ∗∗∗ ∗ Example 9 Design the steel reinforcement for the section of dimension; 200 mm wide and 300 mm effective depth, while the inset of compression steel is 40 mm. the section to resist an applied moment of 123.3 kN.m. Solution k = dbfcu M 2 ∗∗ = 300200 123.3 2 6 ∗∗ ∗ 30 10 = 0.22 > k′ (=0.156) Therefore, compression steel reinforcement is required 30
- 31. z= d {0.5+ 9.0156.025.0 − } 0.775 d (zmin) < 0.776 d < 0.95 d (Z max) ∴z = 0.775 d = 0.776 ∗300 = 232.8 mm x = 0.45 z)(d − = 45.0 )8.232300( − = 149.33 mm 37.0267.0 33.149 40 x d <== ′ ∴The compression steel will have yielded )d-d(f y0.95 2bdf cu)k-(k ′ ′ =′As = )40300(46095.0 300200300.156)-(0.22 2 −∗∗ ∗∗∗ = 304.17 mm2 As = sA 0.95f bdfk y 2 cu ′+ ′ z = 17.304 232.84600.95 300200300.156 2 + ∗∗ ∗∗∗ = 1132.215 mm2 1.8 Derivation of design formulae for flanged reinforced concrete beams: The design procedure of flanged beam depends on the location of the neutral axis. The neutral axis may lie in the flange or in the web as shown in the Figure 1.19 31
- 32. A) Neutral axis in the flange Z= d – (0.9/2)x Or 0.9x = 2(d-z) If 0.9 x ≤ hf xhf d bf N.A N.A x a) Neutral axis in the flange b) Neutral axis in the web xhf d bf N.A 0.9x Fc 0.45x Fs Z=d-0.45x 32 Figure 1.19
- 33. ∴the neutral axis does lie within the flange as assumed and As= zfy0.95 M B) The neutral axis in the wed Fcf = 0.45fcu ∗ hf ∗ (b-bw) Fcw = 0.45fcu ∗0.9x∗bw Where; d x = 0.5 x = 0.5∗d 0.9 x = 0.9∗0.5d Therefore, Fcw=0.202 fcu bw d Z1 = d – 0.5 hf Z2 = d – 0.5 ∗0.9x = d – 0.5 ∗0.45d = 0.775 d Fs = 0.95 fy As Taking moment about the centroid of the flange M = Fs ∗z1 - Fcw ∗(z1-z2) = 0.95fyAs (d - 0.5hf)-0.202 fcu bw d (d-0.5hf - 0.775d) x hf d bf N.A 0.9x Fcf 0.45fcu Fs Z1 Fcw Z2 bw 33
- 34. = 0.95 fy As (d – 0.5hf) – 0.202 fcu bw d (0.225 d – 0.5 hf) ∴As = ( ) ( )h0.5df0.95 h0.45ddbf0.1M f fw y cu − −+ As stated in the BS8110; the above equation to calculate As for singly reinforced flanged section only when hf< 0.45d. For section with tension reinforced only; the applied moment must not exceed the moment of resistance of the concrete which can be calculated as following: M = Fcf.z1+Fcw.z2 = 0.45 fcu hf (b-bw) (d-0.5hf)+ 0.155 fcu bw d2 Thus; b bw0.15 2d h1 b bw1 d h 0.45 dbfcu M ff 2 + − −= or M = βf fcubd2 Where βf can be obtained from Table 3.6 of the BS8110 The equation for the steel area As only applies when the ultimate moment to be resisted by the section is less than βf fcubd2 . If the applied moment is greater than the resistance moment of the concrete section then the excess (M-Mu) to be resisted by steel reinforcement in the compression area: x hf d bf N.A 0.9x Fcf 0.45fcu Fs dFcw bw Fs′ d′ 34
- 35. ∴ M – Mc = Fs′ ∗z Where z = (d-d′) and Fs′ = 0.95 fy As′ ∴ M – Mc = 0.95 fy As′ (d - d′) As′ = )d(df0.95 M y ′− − c M For equilibrium; Fs = Fcf + Fcw +Fs′ So that: As = ( ) As fy0.95 bbhff0.45dbf0.2 wfcuwcu ′+ −+ Take note that x d′ should be less than o.37 to ensure that the steel in compression area will have yielded. Example 10 A ‘T’ beam has effective flange width 900 mm, flange thickness 150 mm, web width 250 mm, and overall depth 500 mm. characteristic strength for concrete 30 N/mm2 and for steel 460 N/mm2 . Assume that the centroid of reinforcement is placed 50 mm from the top or bottom of the beam, design suitable reinforcement for ultimate design moments: a) 300 kN.m, b) 700 kN.m, c) 800 kN.m Solution a) Determine the location of the neutral axis: 35
- 36. the moment resistance of the flange (Mf): Mf= 0.45 fcu bf hf (d-hf/2) = [ ] 6 10)2/150(450(1509003045.0 − ∗−∗∗∗∗ = 683.437 kN.m > the applied moment (=300 kN.m) ∴the neutral axis lies within the flange cu 2 fbd M k = = 30450900 10300 2 6 ∗∗ ∗ = 0.054 < / k (=0.156) ∴ the section is singly reinforced section }k/0.9-0.25d{0.5z += = }0.054/0.9-0.25d{0.5 + = 0.935d < zmax (=0.95d) 4500.935z ∗= = 420.75 mm zf95.0 M As y = = 75.42046095.0 10300 6 ∗∗ ∗ = 1631.6 mm2 b) Mf (=683.437 kN.m) < 700 kN.m ∴the neutral axis lies within the web 2 wcufwffcu dbf155.0)0.45h-d)(b(bh0.45fMc +−= 36 1 2 2 x
- 37. = [ ] 62 1045025030155.0)15045.0450)(250900(1503045.0 − ∗∗∗∗+∗−−∗∗∗ = 738.87 > 700 kN.m ∴the section is singly reinforced section hf (=150 mm) < 0.5d (0.5∗450=225) )0.5h-d(f95.0 )h-d(0.45db0.1fM As fy fwcu+ = = )1505.0450(46095.0 )15045045.0(450250301.010700 6 ∗−∗∗ −∗∗∗∗∗+∗ = 4379 mm2 c) Mc (=738.87 kN.m) < 800 kN.m ∴the section is doubly reinforced section x = 0.5d = 0.5∗450 = 225 mm 37.0222.0 225 50 x d/ <== ∴the compression reinforcement will have yielded )d-(d0.95f Mc-M As / y / = = )50450(46095.0 10)87.738800( 6 −∗ ∗− = 349.7 mm2 / y wffcuwcu As f95.0 )b-b(h0.45fdb0.2f As + + = = 7.349 46095.0 )250900(1503045.0450250302.0 + ∗ −∗∗∗+∗∗∗ = 4906 mm2 37
- 38. 1.10 Minimum and maximum percentages of reinforcement in beams and minimum clear spacing between bars: As stated in cl. 3.12.5 of the BS 8110:Part 1:1997, the minimum percentages of tension and compression reinforcement of rectangular and flanged beams are given in Table 3.25 of the BS 8110. While the area of tension or compression reinforcement should not exceed 4% of the cross sectional are of the beam as stated in cl. 3.12.6.1 of the BS 8110. As stated in cl.3.12.11.1 of the BS 8110, the horizontal distance between bars should be not less than the maximum size of coarse aggregate (hagg.) + 5 mm, while the vertical distance between bars should be not less than 2hagg./3. Example 11 A simply supported beam of 7 m span carries a characteristic dead load of 10 kN/m and live load of 5 kN/m. The beam dimensions are breadth 250 mm, effective depth (d) 450 mm and overall height (h) 500mm. The concrete grade is 35 N/mm2 and the steel grade is 460 N/mm2 , while the nominal cover is 37.5 mm. The maximum aggregate size is 20 mm, design the beam to resist the design flexural moment Solution Design load = 1.4∗10+1.6∗5 = 22 kN/m 38
- 39. 134.75 kN.m Ultimate moment = 8 2wl = 8 722 2 ∗ = 134.75 kN.m cu 2 fbd M k = = 35450250 10134.75 2 6 ∗∗ ∗ = 0.076 0.156)(k =′< ∴singly reinforced section }k/0.9-0.25d{0.5z += = }0.076/0.9-0.25d{0.5 + = 0.906d < zmax (=0.95d) 450906.0z ∗= = 407.7 mm 0.45 z)-(d x = 39 7m 22 kN/m
- 40. = 0.45 407.7)-(450 = 94 mm < 0.5d (0.5 ∗450=225mm), ∴tension steel reinforcement will have yielded zf95.0 M As y = = 7.40746095.0 1075.134 6 ∗∗ ∗ = 756.32 mm2 100 50025013.0 Asmin ∗∗ = = 162.5 mm2 500250 100 4 Asmax ∗∗= = 5000 mm2 )5000(As756.32)As()5.162(As maxmin =<=<= use steel bar size 25 mm; Area of one steel bar = =∗µ2 5.12 490.87 mm2 No. of steel bars = 87.490 32.756 = 1.54 2≅ As provided = 2∗490.87 = 981.74 mm2 spacing between bars = 1 2522.372250 ∗−∗− = 125 > hagg max. +5 (=20+5=25mm) 40 2Y25