CONSULTING Engineering Calculation Sheet jxxx 1 Effects From Structural Analysis Design axial force, F (tension -ve and compression +ve) (ensure < 0.1f cu b w h 0 kn OK Design shear force, V d 4223 kn Design bending moment, M 11963 knm Design torsion moment, T 1500 knm Material Properties Characteristic strength of concrete, f cu / f c ' (f cu 75 or 105N/mm 2 40 32 N/mm 2 OK Yield strength of longitudinal steel, f y 460 N/mm 2 Yield strength of shear and torsion link steel, f yv 460 N/mm 2 Section Dimensions Span (effective width, deflection, long'l shear and deep beam calcs) 3.750 m Available beam spacing 4.000 m (effective width calcs; usual spacing for interior beams; half for edge beams) Beam section type and support condition (section type for bending, defl'n calcs; support condition for LTB restraint, eff width, defl'n, long'l shear calcs) Ratio b b (1.0 for no moment redistribution, 0.70, 1.30) 1.00 OK Section Type and Support Condition Option Selection Downstand Beam Upstand Beam Support Effect Slab Type Defl'n Support Effect Slab Type Defl'n S/S Sag Precast Rect-s/s Yes S/S Sag Precast Rect-s/s Yes S/S Sag Insitu T/L-s/s Yes S/S Sag Insitu Rect-s/s Yes Cont. Sag Precast Rect-cont. Yes Cont. Sag Precast Rect-cont. Yes Cont. Sag Insitu T/L-cont. Yes Cont. Sag Insitu Rect-cont. Yes Cont. Hog Precast Rect-cont. Cont. Hog Precast Rect-cont. Cont. Hog Insitu Rect-cont. Cont. Hog Insitu T/L-cont. Cant. Hog Precast Rect-cant. Yes Cant. Hog Precast Rect-cant. Yes Cant. Hog Insitu Rect-cant. Yes Cant. Hog Insitu T/L-cant. Yes Overall depth, h (includes insitu slab thks; excludes precast slab thks; span/ 2800 mm OK Depth of flange, h f (bending flanged beam, longitudinal shear calcs) 300 mm Width (rectangular) or web width (flanged), b w 1500 mm Cover to all reinforcement, cover (usually 35 (C35) or 30 (C40) internal; 40 Add cover to tension steel (due to transverse steel layer(s)), cover add,t Add cover to compression steel (due to transverse steel layer(s)), cover add,c Effective depth to tension steel, d = h - cover - MAX(f link, f link,t, cover add,t ) - [ Effective depth to compression steel, d' = cover + MAX(f link, f link,t, cover add,c ) 35 mm 0 mm 0 mm 2415 mm 198 mm Longitudinal Reinforcement Details Tension steel reinforcement diameter, f t 25 mm Tension steel reinforcement number, n t 72 Tension steel area provided, A s,prov = n t.p.f 2 t /4 35343 mm 2 Compression steel reinforcement diameter, f c (where applicable) 25 mm Compression steel reinforcement number, n c 72 Compression steel area provided, A s,prov ' = n c.p.f 2 c /4 35343 mm 2 Number of layers of tension steel, n layers,tens 6 layer(s) Spacer for tension steel, s r,tens ( MAX (f t, 25mm)) 100 mm OK Number of layers of compression steel, n layers,comp Spacer for compression steel, s r,comp = MAX (f c, 25mm) 6 layer(s) 25 mm
CONSULTING Engineering Calculation Sheet jxxx 2 Member Design - RC Beam Shear Reinforcement Details XX 23/10/2017 Shear link diameter, f link 25 mm Number of shear links in a cross section, i.e. number of legs, n leg 5 Area provided by all shear links in a cross-section, A sv,prov = p.f 2 link /4.n leg 2454 mm 2 Pitch of shear links, S 200 mm Torsion Reinforcement Details Torsion link diameter, f link,t 25 mm Number of torsion links in a cross section, i.e. number of legs, n leg,t = 2 2 Area provided by all torsion links in a cross-section, A sv,prov,t = p.f 2 link,t /4.n leg,t 982 mm 2 Pitch of torsion links, S t 200 mm Note that further longitudinal steel A s,t must be provid 4640 18203 18203 mm 2 Utilisation Summary Item UT Remark Rectangular beam tension steel 34% 31% 34% OK ) Rectangular beam compression s Rectangular beam % min tension 15% 31% 31% OK Rectangular beam % min compression reinforcement Rectangular beam % max tension and compression rei 21% OK Flanged beam tension steel Flanged beam compression steel Flanged beam % min tension reinforcement Flanged beam % min compression reinforcement Flanged beam % max tension and compression reinfor Rectangular beam shear ultimate 23% 29% 29% OK Rectangular beam shear design c 20% 27% 27% OK Rectangular beam torsion ultima 12% 16% 16% OK Rectangular beam torsion design 23% 27% 27% OK Rectangular beam shear and tors 35% 34% 35% OK Rectangular or flanged beam deflection requirements 3% OK Total utilisation rectangular beam 35% OK Total utilisation rectangular beam (excluding def 35% OK Total utilisation flanged beam Detailing requirements OK % Tension reinforcement (rectangular) 0.84 % % Compression reinforcement (rectangular) % % Tension reinforcement (flanged) % % Compression reinforcement (flanged) % Estimated steel reinforcement quantity (125-160kg/m 3 ) 285 kg/m 3 7850. [(A s,prov +A s,prov ') / (b w or b).(h or h f ) + (A sv,prov.(h+anc.)/s+a sv,prov,t.(h+b w +anc.)/s t ) / Estimated steel reinforcement quantity (125-160kg/m 3 ) 399 kg/m 3 IStructE 11000. [(A s,prov +A s,prov ') / (b w or b).(h or h f ) + (A sv,prov.(h+anc.)/s+a sv,prov,t.(h+b w +anc.)/s t ) [Note that steel quantity in kg/m 3 can be obtained from 110.0 x % rebar]; Material cost: concrete, c 315 units/m 3 steel, s 4600 units/tonne Reinforced concrete material cost = [c+(est. rebar quant).s].(b w o 9028 units/m Ductility of failure mechanism Section Under Reinforced Rectangular beam crack width 50% OK Max LTB stability (compression flange) restraints spacing, L LTB 90.0 m Note s/s / cont L LTB = MIN (60(b w or b), 250(b w or b) 2 /d) and cant L LTB = MIN (25b w, 100b w 2 / OK
jxxx 3 Member Design - RC Beam Material Stresses XX 23/10/2017 Axial stress, F/b w h 0.00 N/mm 2 Shear stress, V d /b w d 1.17 N/mm 2 Bending stress, 6M/(b w or b).d 2 8.20 N/mm 2 Torsion stress, 2T/((h 2 min )(h max -h min /3)) 0.58 N/mm 2 Material Stress-Strain Curves Detailing Instructions Not to Scale h f = h = 2800 mm Comp Steel = d = 2415 mm d' = 198 mm Shear Links = 5 legs of T25@200mm pitch Torsion Links = 2 legs of T25@200mm pitch Cover = 35 mm Concrete = 40 MPa Rebars = 460 MPa Links = 460 MPa b w = 1500 mm Tension Steel = 72 T25@123mm centres Diagram Produced for Sagging Design; Flip if Hogging Design; / b w.h + A sv,prov,h.(h - n layers,tens.s r,tens - n layers,comp.s r,comp - 2.cover)/S h / b w.h]; No curtailment; No laps; / b w.h + A sv,prov,h.(h - n layers,tens.s r,tens - n layers,comp.s r,comp - 2.cover)/S h / b w.h]; No curtailment; No laps; /d);
CONSULTING Engineering Calculation Sheet jxxx 4 Additional Longitudinal Shear Rectangular or Flanged Beam Utilisation Summary Longitudinal shear between web and flange Longitudinal shear within web Length under consideration, Dx (span/2 s/s, span/4 cont, span cant) 938 mm Applicability of longitudinal shear design Not Applicable Longitudinal Shear Between Web and Flange (EC2) Longitudinal shear stress limit to prevent crushing Longitudinal shear stress limit for no transverse reinforcement Required design transverse reinforcement per unit length Longitudinal Shear Between Web and Flange (BS5400-4) Longitudinal shear force limit per unit length Required nominal transverse reinforcement per unit length Longitudinal Shear Between Web and Flange Mandatory Cr Longitudinal Shear Within Web (EC2) Longitudinal shear stress limit Longitudinal Shear Within Web () Longitudinal shear stress limit for no nominal / design vertical rein Required nominal vertical reinforcement per unit length Required design vertical reinforcement per unit length Longitudinal Shear Within Web (BS5400-4) Longitudinal shear force limit per unit length Required nominal vertical reinforcement per unit length Longitudinal Shear Within Web Mandatory Criteria Additional Deep Beam Rectangular Beam Utilisation Summary Deep beam design Span to depth ratio, span / h 1.34 Applicability of deep beam design Applicable Concrete type Horizontal shear link diameter, f link,h mm Number of horizontal shear links in a horizontal section, i.e. number of legs, 4 Pitch of horizontal shear links, S h 225 mm Clear distance from edge of load to face of support, a 1 or x e 1750 mm Detailing requirements OK OK Reynolds Tension steel (deep beam) 46% OK Tension steel zone depth, T zone (sag s/s, sag cont, hog cant) 513 mm Tension steel zone depth, T zone (hog cont) Minimum breadth for deep beam, b w 567 mm OK Shear ultimate force (deep beam) 48% OK Shear design capacity (deep beam) 31% OK CIRIA Guide 2 Bending ultimate moment (deep beam) 21% OK Tension steel (deep beam) 49% OK Tension steel zone depth, T zone (sag s/s, sag cont, hog cant) 560 mm Tension steel zone depth, T zone (hog cont) upper band 17% 560 mm Tension steel zone depth, T zone (hog cont) lower band 83% 1680 mm Shear ultimate force (deep beam) 43% OK Shear design capacity (deep beam) 29% OK
Neutral axis 0.90 > b b > 1.10 0.90 b b 1.10 Neutral axis, x 0.90 > > 1.10 CONSULTING Engineering Calculation Sheet jxxx 5 Member Design - RC Beam Bending Rectangular Beam (Singly or Doubly Reinforced) () XX 23/10/2017 M u = K'f cu bd 2 Note b here is b w ; 54590 knm Ratio, K' 0.156 b b 1.10 b b K' = K' = 0.156 mm BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 cl.3.4.4.4 cl.3.4.4.4 Note the expression for K' ensures that the section remains ductile i.e. ensuring that failure occurs with the gradual yielding of the tension steel and not by a sudden catastrophic compression failure of the concrete; VALID Note b here is b w ; 0.034 To scheme beam, choose d such that K = M/(b w d 2 f cu ) < K' (0.156 for b b =1.0 and NSC) for no compression steel, i.e. d > 1131 and h > 1516 mm z <=0.95d 2294 mm limit, x limit 0.90 Depth of neutral axis, x x limit 268 1208 mm OK 268 mm BC2 x = cl.3.4.4.4 cl.3.4.4.4 1208 mm BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 cl.3.4.4.4 cl.3.4.4.4 Tension steel 11932 mm 2 Back-analysis of x in (0.67f cu /1.5).(s.b w )=(f y /1.05).(A s,prov ) 642 mm OK 0.90x mm BC2 s = cl.3.4.4.4 cl.3.4.4.4 Note for an under reinforced section, require x x limit for yielding of tension steel;
CONSULTING Engineering Calculation Sheet jxxx 6 0.90 > b b > 1.10 0.90 b b 1.10 Neutral axis, x To scheme beam, with compression steel, choose d such that K = M/(b w d 2 f cu ) < 10/f cu for non-excessive comp steel, i.e. d > 893 and h > 1278 mm Depth of neutral axis, x x limit Neutral axis limit, x limit x = BC2 cl.3.4.4.4 cl.3.4.4.4 BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 cl.3.4.4.4 cl.3.4.4.4 K' Compression steel 2 Tension steel z 2 Back-analysis of x in (0.67f cu /1.5).(s.b w )+(f y /1.05).(A s,prov ')=(f y /1 BC2 s = cl.3.4.4.4 cl.3.4.4.4 Note for an under reinforced section, require e st = e cc (d-x)/x e y for yielding of tension steel and e sc = e cc (x-d')/x e y for yielding of compression steel, where e cc =0.0035 for f cu 60N/mm and e cc =0.0035 - (f cu - 60)/50000 for f cu > 60N/mm 2 and e y =(f y /1.05)/E s ; Tension steel area provided 35343 mm 2 Tension steel area provided utilisation 34% OK Compression steel area provided 2 Compression steel area provided utilisation % Min tension reinforcement 0.84 % TR49 % Min tension reinforcement (>= 0.0024b w h G250; >= MAX (0.0013, 0.0013(f cu /40) 2/3 ).b w h G4 cl.3.1.7 % Min tension reinforcement utilisation 15% OK % Min compression reinforcement (>= 0.002b w h) % % Min compression reinforcement utilisation % Max tension reinforcement (<= 0.04b w h) 0.84 % % Max tension reinforcement utilisation 21% OK % Max compression reinforcement (<= 0.04b w h) % % Max compression reinforcement utilisation % Max tension or compression reinforcement utilisation 21% OK
CONSULTING Engineering Calculation Sheet jxxx 7 Bending Rectangular Beam (Singly or Doubly Reinforced) (ACI318) ACI318 Singly Reinforced Rectangular Section Doubly Reinforced Rectangular Section Required nominal flexural strength coefficient of resistance, R n = M* / fb w d 2 1.36 N/mm 2 Strength reduction factor for tension-controlled sections, f = 0.90 0.90 Ratio of A s to b w d, r w = 0.85f c '/f y.[1- (1-2R n /(0.85f c '))] 0.30 % Tension steel, A s = MAX(r w b w d, A s,min ) 11079 mm 2 A s,min = MAX(0.25b w d f c '/f y, 1.4b w d/f y ) 11079 mm 2 Factor, b 1 = MAX(MIN(0.85-(0.05/7).(f c '-28),0.85),0.65) 0.82 Note b 1 is the factor relating depth of equivalent rectangular compressive stress block to neutral axis depth; Depth of equivalent rectangular stress block, a = A s.f y /(0.85f c 'b w ) 126 mm Distance, c = a/b 1 153 mm Note c is the distance from extreme compression fiber to neutral axis (for singly reinforced section); Distance, d t = h - cover - MAX(f link, f link,t, cover add,t ) - f t /2 2728 mm Note d t is the distance from extreme compression fiber to centroid of extreme layer of longitudinal tension st e t = 0.003(d t /c-1) 0.050 Note e t is the net tensile strain in extreme layer of longitudinal tension steel at nominal strength; no compression steel VALID Tension steel, A s 11079 mm 2 m 2 compression steel required Ratio of A s to b w d, r w = 0.319b 1 f c '/f y. d t /d % R nt = r w f y (1-0.59r w f y /f c ') N/mm 2 Note R nt is the required nominal flexural strength coefficient of resistance for rectangular beam corresponding to ε t =0.005; M nt = R nt b w d 2 knm Note M nt is the required nominal flexural strength corresponding to ε t =0.005; M n ' = M*/f - M nt knm Note M n ' is the required nominal flexural strength that need to be resisted by the compression reinforcement; Distance, c t = 0.375d t Note c t is the distance from extreme compression fiber to neutral axis (for doubly reinforced section); Stress in compression reinforcement, f s ' = 0.003E s (1-d'/c t ) f y N/mm 2 Modulus of elasticity of reinforcement, E s = 200,000 N/mm 2 Compression steel, A s ' = M n ' / [f s '(d-d')] 2 Tension steel, A s = r w b w d + M n ' / [f y (d-d')] 2 no valid solution
CONSULTING Engineering Calculation Sheet jxxx 8 ACI318 Tension steel area provided 35343 mm 2 Tension steel area provided utilisation 31% OK Compression steel area provided 2 Compression steel area provided utilisation % Min tension reinforcement 0.84 % cl.9.6.1.2 % Min tension reinforcement (>= MAX (0.25b w d f c '/f y, 1.4b w d/f y )) % Min tension reinforcement utilisation 31% OK teel;
0.90 > b b > 1.10 0.90 b b 1.10 Neutral axis, x 0.90 > b b > 1.10 0.90 b b 1.10 CONSULTING Engineering Calculation Sheet jxxx 9 Member Design - RC Beam Bending Flanged Beam (Singly or Doubly Reinforced) XX 23/10/2017 For cantilevers, width shown is applicable for downstand beams as rect- sections. For upstand beams, T- or L- sections will apply with insitu Span m Depth of flange, h f Effective width, b = MIN(b w + function (span, section, structure), beam spac (bending flanged beam, deflection calcs flanged beam) K = M/bd 2 f cu z = d. [0.5 + (0.25-K/0.9) 0.5 ] <= 0.95d Depth of compression stress block, s = 2.(d-z) (applicable for all f cu ) Compression Stress Block in Flange (s <= h f ) M u = K'f cu bd 2 Ratio, K' K' = K' = knm BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 cl.3.4.4.4 cl.3.4.4.4 Note the expression for K' ensures that the section remains ductile i.e. ensuring that failure occurs with the gradual yielding of the tension steel and not by a sudden catastrophic compression failure of the concrete; To scheme beam, choose d such that K = M/(bd 2 f cu ) < K' (0.156 for b b =1.0 and NSC) for no compression steel, i.e. d > and h > z <=0.95d Depth of neutral axis, x x limit Neutral axis limit, x limit x = BC2 cl.3.4.4.4 cl.3.4.4.4 BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 cl.3.4.4.4 cl.3.4.4.4 Tension steel 2 Back-analysis of x in (0.67f cu /1.5).(s.b)=(f y /1.05).(A s,prov ) BC2 s = cl.3.4.4.4 cl.3.4.4.4 Note for an under reinforced section, require x x limit for yielding of tension steel;
0.90 > b b > 1.10 0.90 b b 1.10 Neutral axis, x CONSULTING Engineering Calculation Sheet jxxx 10 Member Design - RC Beam XX 23/10/2017 To scheme beam, with compression steel, choose d such that K = M/(b d 2 f cu ) < 10/f cu for non-excessive comp steel, i.e. d > and h > Depth of neutral axis, x x limit Neutral axis limit, x limit x = BC2 cl.3.4.4.4 cl.3.4.4.4 BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 cl.3.4.4.4 cl.3.4.4.4 K' Compression steel 2 Tension steel z mm 2 Back-analysis of x in (0.67f cu /1.5).(s.b)+(f y /1.05).(A s,prov ')=(f y /1.0 BC2 s = cl.3.4.4.4 cl.3.4.4.4 Note for an under reinforced section, require e st = e cc (d-x)/x e y for yielding of tension steel and e sc = e cc (x-d')/x e y for yielding of compression steel, where e cc =0.0035 for f cu 60N/mm and e cc =0.0035 - (f cu - 60)/50000 for f cu > 60N/mm 2 and e y =(f y /1.05)/E s ;
0.90 b b 1.10 CONSULTING Engineering Calculation Sheet jxxx 11 Member Design - RC Beam XX 23/10/2017 Compression Stress Block in Web (s > h f AND h f <={0.45,0.36,0.30} Note simplified method as equations within this section only valid for b b 0.90 and 1.10 as x = 0.5d; BC2 cl.3.4.4.5 cl.3.4.4.5 cl.3.4.4.5 knm Ratio, K' K' = BC2 cl.3.4.4.4 [ b b =1.0] cl.3.4.4.4 If K f < K' no compression steel k 1 k 2 2 BC2 cl.3.4.4.5 cl.3.4.4.5 cl.3.4.4.5 m 2 Back-analysis of x in (0.67f cu /1.5).[b.h f +(s-h f ).b w ]=(f y /1.05).(A s,p BC2 s = x limit =0.5d cl.3.4.4.4 cl.3.4.4.4 Note for an under reinforced section, require x x limit for yielding of tension steel; If K f > K' compression steel required Compression steel, A s ' = [M-(K'f cu b w d 2 +M uf )]/[0.95f y (d-d')] 2 Tension steel, A s = [{0.20,0.18,0.16}f cu b w d+0.45f cu h f (b-b w )]/(0.9 2 Note the coefficient 0.20, 0.18 or 0.16 is used for f cu 60, 75 or 105N/mm 2 respectively; x=0.5d Back-analysis of x in (0.67f cu /1.5).[b.h f +(s-h f ).b w ]+(f y /1.05).(A s,p BC2 s = cl.3.4.4.4 cl.3.4.4.4 Note for an under reinforced section, require e st = e cc (d-x)/x e y for yielding of tension steel and e sc = e cc (x-d')/x e y for yielding of compression steel, where e cc =0.0035 for f cu 60N/mm and e cc =0.0035 - (f cu - 60)/50000 for f cu > 60N/mm 2 and e y =(f y /1.05)/E s ;
jxxx 12 Compression Stress Block in Web (s > h f AND h f > {0.45,0.36,0.30}d Note complex method as equations within this section valid for b b 0.70 and 1.30; BC2 cl.3.4.4.5 cl.3.4.4.5 cl.3.4.4.5 Tension steel area provided 2 Tension steel area provided utilisation Compression steel area provided 2 Compression steel area provided utilisation % Min tension reinforcement % TR49 % Min tension reinforcement (>= 0.0032b w h G250; >= MAX (0.0018, 0.0018(f cu /40) 2/3 ).b w h G4 cl.3.1.7 % Min tension reinforcement utilisation % Min compression reinforcement (>= 0.004bh f flange; >= 0.002b w h web) % % Min compression reinforcement utilisation % Max tension reinforcement (<= 0.04b w h) % % Max tension reinforcement utilisation % Max compression reinforcement (<= 0.04bh f flange; <= 0.04b w h web) % % Max compression reinforcement utilisation % Max tension or compression reinforcement utilisation m 2
jxxx 13 Member Design - RC Beam Shear Rectangular Beam () Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Note that d is the effective depth to the tension reinforcement irrespective of whether the section is sagging at midspans or hogging at supports. It follows then, for midspans, tension steel is the bottom steel whilst for supports, tension steel is the top steel; XX 23/10/2017 Ultimate shear stress, v ult = V d /b w d (< 0.8f cu 0.5 & {5.0,7.0}N/mm 2 ) 1.17 N/mm 2 BC2 Note the ultimate shear stress limit of 5.0 or 7.0N/mm 2 is used for f cu 60 or 105N/mm 2 respeccl.3.4.5.2 Ultimate shear stress utilisation 23% OK Design shear stress, v d = V d /b w d 1.17 N/mm 2 (Conservatively, shear capacity enhancement by either calculating v d at d from support and comparing against unenhanced v c as clause 3.4.5.10 or calculating v d at support and comparing against enhanced v c within 2d of the support as clause 3.4.5.8 ignored;) Area of tension steel reinforcement provided, A s,prov 35343 mm 2 r w = 100A s,prov /b w d 0.98 % v c = (0.79/1.25)(r w f cu /25) 1/3 (400/d) 1/4 ; r w <3; f cu <80; (400/d) 1/4 >(0.67 or 1. 0.73 N/mm 2 BC2 cl.3.4.5.4 Minimum shear strength, v r = MAX (0.4, 0.4 (MIN (80, f cu )/40) 2/3 ) 0.40 N/mm 2 BC2 cl.3.4.5.3 Check v d < 0.5v c for no links (minor elements) INVALID Concrete shear capacity v c.(b w d) 2656 kn Check 0.0 < v d < v r + v c for nominal links Provide nominal links A sv /S > v r.b w /(0.95f yv ) i.e. A sv /S > Concrete and nominal links shear capacity (v r + v c ).(b w d) 1.37 mm 2 /mm 4105 kn Check v d > v r + v c for design links Provide shear links A sv /S > b w (v d -v c )/(0.95f yv ) i.e. A sv /S > Concrete and design links shear capacity (A sv,prov /S+A sv,prov,t /S t ).(0.95f yv ).d + v c.(b w d) VALID 1.48 mm 2 /mm 20787 kn Area provided by all shear and torsion links in a cross-section, A sv,prov + A sv,pr 3436 mm 2 Tried A sv,prov /S + A sv,prov,t /S t value 17.18 mm 2 /mm Design shear resistance utilisation 20% OK Note while minimum links should be provided in all beams of structural importance, it may be satisfactory to omit them in members of minor structural importance such as lintels or where the maximum design shear stress is less than half v c ;
jxxx 14 Shear Rectangular Beam (ACI318) Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Note that d is the effective depth to the tension reinforcement irrespective of whether the section is sagging at midspans or hogging at supports. It follows then, for midspans, tension steel is the bottom steel whilst for supports, tension steel is the top steel; ACI318 Ultimate shear stress, v ult = V d */b w d (< v c + f0.66 f c ') 1.05 N/mm 2 Ultimate shear stress utilisation 29% OK Design shear stress, v d = V d */b w d 1.05 N/mm 2 (Conservatively, shear capacity enhancement by calculating v d at d from support as clause 9.4.3.2 ACI318 ignored;) Strength reduction factor for shear, f = 0.75 0.75 Area of tension steel reinforcement provided, A s,prov 35343 mm 2 r w = 100A s,prov /b w d 0.98 % v c = fv c /b w d 0.78 N/mm 2 where l = {1.00 NWC, 0.75 LWC} [Simplified] f V c = Note for reference only; 2599 kn [Detailed MIN(a-c)] f V c = 2830 kn [Detailed (a)] f V c = 2830 kn [Detailed (b)] f V c = 2897 kn [Detailed (c)] f V c = 4434 kn where l = {1.00 NWC, 0.75 LWC} Minimum shear strength, v r = MAX (0.35, 0.062 (f c ') 1/2 ) 0.35 N/mm 2 Check v d < 0.5v c for no links (minor elements) Concrete shear capacity v c.(b w d) INVALID 2830 kn Check 0.0 < v d < v c for nominal links Provide nominal links A sv /S > v r.b w /MIN(f yv,420mpa) i.e. A sv /S > Concrete and nominal links shear capacity (fv r + v c ).(b w d) 1.25 mm 2 /mm 3781 kn Check v d > v c for design links Provide shear links A sv /S > b w (v d -v c )/(fmin(f yv,420mpa)) i.e. A sv /S Concrete and design links shear capacity f(a sv,prov /S+A sv,prov,t /S t ).MIN(f yv,420mpa).d + v c.(b w d) VALID 1.26 mm 2 /mm 15900 kn Area provided by all shear and torsion links in a cross-section, A sv,prov + A sv,pr 3436 mm 2 Tried A sv,prov /S + A sv,prov,t /S t value 17.18 mm 2 /mm Design shear resistance utilisation 27% OK Note while minimum links should be provided in all beams of structural importance, it may be satisfactory to omit them in members of minor structural importance such as lintels or where the maximum design shear stress is less than half v c ;
jxxx 15 Torsion Rectangular Beam () Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Larger dimension of rectangular section, h max = MAX (h, b w ) 2800 mm Smaller dimension of rectangular section, h min = MIN (h, b w ) 1500 mm Design torsion stress, v t = 2T/((h 2 min )(h max -h min /3)) (< 0.8f 0.5 cu & {5.0,7.0}N 0.58 N/mm 2 Note the ultimate torsion stress limit of 5.0 or 7.0N/mm 2 is used for f cu 60 or 105N/mm 2 respectively; Design ultimate torsion stress utilisation 12% OK Check v t < v t.min for no torsion links BC2 Concrete torsion resistance, v t.min = MIN (0.6, 0.067 f cu ) 0.42 N/mm 2 cl.2.3 Check v t > v t.min for design torsion links VALID Horizontal dimension of closed link, x 1 = b w - 2 x cover - f link,t 1405 mm Vertical dimension of closed link, y 1 = h - 2 x cover - f link,t 2705 mm Provide closed torsion links A sv,t /S t > T / (0.8x 1 y 1 (0.95f yv )) i.e. A sv 1.13 mm 2 /mm Provide longitudinal bars around inside perimeter of links, A s,t 4640 mm 2 Note require A s,t > (A sv,t /S t )(f yv /f y )(x 1 +y 1 ); Area provided by all torsion links in a cross-section, A sv,prov,t 982 mm 2 Tried A sv,prov,t /S t value 4.91 mm 2 /mm Design torsion resistance utilisation 23% OK Combined Shear and Torsion Rectangular Beam () Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Ultimate shear and torsion stresses v ult + v t (< 0.8f cu 0.5 & {5.0,7.0}N/mm 2 ) 1.75 N/mm 2 BC2 Note the ultimate shear and torsion stress limit of 5.0 or 7.0N/mm 2 is used for f cu 60 or 105N/mcl.3.4.5.2 Ultimate shear and torsion stresses utilisation 35% N/mm 2 OK Check design shear and torsion links Provide single torsion link leg > A sv /S/(n leg +n leg,t ) + A sv,t /S t /n leg,t 0.78 mm 2 /mm Area provided by single torsion link leg, A sv,prov,t /n leg,t 491 mm 2 Tried A sv,prov,t /n leg,t /S t value 2.45 mm 2 /mm Design shear and torsion resistance utilisation 32% OK
jxxx 16 Torsion Rectangular Beam (ACI318) Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; ACI318 Larger dimension of rectangular section, h max = MAX (h, b w ) 2800 mm Smaller dimension of rectangular section, h min = MIN (h, b w ) 1500 mm Design torsion moment, T* (< f0.66 f c '.1.7(x 1 y 1 ) 2 /2/(x 1 +y 1 )) 1345 knm Strength reduction factor for torsion, f = 0.75 0.75 Design ultimate torsion stress utilisation 16% OK Check T < ft th for no torsion links Concrete torsion resistance, ft th = f0.083l f c '.(h min.h max ) 2 /[2.(h mi where l = {1.00 NWC, 0.75 LWC} Check T > ft th for design torsion links 719 knm VALID Horizontal dimension of closed link, x 1 = b w - 2 x cover - f link,t 1405 mm Vertical dimension of closed link, y 1 = h - 2 x cover - f link,t 2705 mm Provide closed torsion links A sv,t /S t > T* / (f0.85x 1 y 1 MIN(f yv,420m 1.32 mm 2 /mm Note Provide longitudinal bars around inside perimeter of lin 18203 18203 mm 2 Note require A s,t > (A sv,t /S t )(f yv /f y )(x 1 +y 1 ) > A l,min where both f yv and f y are limited to 420M Area provided by all torsion links in a cross-section, A sv,prov,t 982 mm 2 Tried A sv,prov,t /S t value 4.91 mm 2 /mm Design torsion resistance utilisation 27% OK Combined Shear and Torsion Rectangular Beam (ACI318) Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Ultimate shear and torsion stresses [v ult 2 + v t 2 ] 1/2 (< v c + f0.66 f c ') 1.14 N/mm 2 Design torsion stress, v t = T*.2.(x 1 +y 1 )/1.7/(x 1 y 1 ) 2 0.45 N/mm 2 Ultimate shear and torsion stresses utilisation 32% N/mm 2 OK Check design shear and torsion links Provide single torsion link leg > A sv /S/(n leg +n leg,t ) + A sv,t /S t /n leg,t Provide single torsion link leg > MAX(0.031 f c ',0.175).b w /MIN(f yv, 0.84 mm 2 /mm 0.63 mm 2 /mm Area provided by single torsion link leg, A sv,prov,t /n leg,t 491 mm 2 Tried A sv,prov,t /n leg,t /S t value 2.45 mm 2 /mm Design shear and torsion resistance utilisation 34% OK
jxxx 17 Detailing Requirements Rectangular or Flanged Beam All detailing requirements met? OK Min tension steel reinforcement diameter, f t (>=12mm) 25 mm OK Min tension steel reinforcement pitch (b w -2.cover-2.MAX(f link,f link,t )-f t )/(n t /n l 123 mm OK Max tension steel reinforcement pitch (b w -2.cover-2.MAX(f link,f link,t )-f t )/(n t /n 123 mm OK Note that max pitch assumes no moment redistribution in beam, if less than 20% moment redistribution then pitch to be less than 175mm for G250 and less than 150mm for G460; Min compression steel reinforcement diameter, f c (>=12mm) 25 mm OK Min compression steel reinforcement pitch (b w or b-2.cover-2.max(f link,f link,t ) Max compression steel reinforcement pitch (b w or b-2.cover-2.max(f link,f link,t Min shear link diameter, f link (>=6mm, >=f c /4) 25 mm OK Min torsion link diameter, f link,t (>=6mm, >=f c /4) 25 mm OK Shear link pitch, S (<=0.75d, <=12f c, <=300mm, >=MAX(100mm, 50+12. 200 mm OK Torsion link pitch, S t (<=0.75d, <=12f c, <=300mm, >=MAX(100mm, 50+1 200 mm OK A sv,prov / (b w.s) + A sv,prov,t / (b w.s t ) (>0.10% G460; >0.17% G250) 1.15 % OK A sv,prov / (b w.s) + A sv,prov,t / (b w.s t ) (<4.00%) 1.15 % OK Note require an overall enclosing link; Note require additional restraining links for each alternate longitudinal Note lacer bars of 16mm are required at the sides of beams more than 750mm deep at 250mm pitch;
jxxx 18 Deflection Criteria Rectangular or Flanged Beam bar;
jxxx 19 Span 3.750 m Span / effective depth ratio 1.6 Basic span / effective depth ratio criteria 26.0 Multiplier C 1,rect or flanged 1.00 cl.3.4.6.3 Multiplier C 1,span more or less than 10m 1.00 cl.3.4.6.4 Modification factor for tension C 2 M/(b w or b).d 2 1.37 N/mm 2 cl.3.4.6.2 104 N/mm 2 Modification 1.92 T.3.10 Modification factor for compression C 3 100A s,prov '/(b w or b).d cl.3.4.6.2 Modification 1.00 T.3.11 Modified span / effective depth ratio criteria 50.0 Deflection utilisation 3% OK
jxxx 20 Crack Width Estimation Rectangular Beam Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Maximum crack width criteria, w allow 0.30 mm Load factor, LF = 1.4 conservatively 1.4 SLS bending moment, M sls = M / LF 8545 knm SLS axial force (tension +ve and compression -ve), F sls = -F / LF 0 kn Uncracked elastic modulus, E uncracked = 20+0.2f cu 28 GPa Cracked elastic modulus, E cracked = E uncracked / 2 14 GPa Steel elastic modulus, E s 200 GPa Modulus ratio, a e = E s / E cracked 14.3 Reinforcement ratio, r w = A s,prov /b w d 0.01 Factor, a e.r w 0.14 Depth of neutral axis, x = a e.r w.[(1+2/(a e r w )) 1/2-1].d 982 mm Lever arm, z = d-x/3 2088 mm Steel tensile service stress (flexural), f s1 = M sls /z / A s,prov 116 N/mm 2 Steel tensile (tensile +ve and compressive -ve) service stress (axial), f s2 0 N/mm 2 Note if F sls > 0, f s2 = F sls /(A s,prov +A s,prov '), else f s2 = a e.(a s,prov +A s,prov ')/[ a e.(a s,prov +A s,prov ')+(b w.h-a s,prov -A s,prov ')].F sls /(A s,prov +A s,prov '); Steel tensile service stress utilisation, (f s1 +f s2 ) / (0.8f y ) < 1.00 31% OK Concrete compressive service stress (flexural and axial), f c 6 N/mm 2 Note if F sls > 0, f c = 2(M sls /z)/(b w x), else f c = 2(M sls /z)/(b w x) + (b w.h-a s,prov -A s,prov ')/[ a e.(a s,prov +A s,prov ')+(b w.h-a s,prov -A s,prov ')].ABS(F sls )/(b w.h-a s,prov +A s,prov '); Concrete compressive service stress utilisation, f c / (0.45f cu ) < 1.00 31% OK Strain at tension face, e 1 = (h-x)/(d-x).(f s1 +f s2 )/E s 0.73 x10-3 Strain stiffening, e 2 = b w (h-x) 2 /[3E s A s,prov (d-x)] 0.16 x10-3 Mean strain, e m = e 1 - e 2 0.57 x10-3 Distance to face of extreme rebar, c f = cover + MAX(f link,f link,t ) Distance to centroid of extreme rebar, c = c f + f t /2 Distance, s = (b w -2.cover-2.MAX(f link,f link,t )-f t )/(n t /n layers,tens -1) Distance, a c1 = [c²+c²] 1/2 - f t /2 Distance, a c2 = [c²+(s/2)²] 1/2 - f t /2 Distance, a cr = MAX(a c1, a c2 ) Note a c1 is not applicable for continuous width sections, i.e. slabs or walls; 60 mm 73 mm 123 mm 90 mm 83 mm 90 mm a c1 a c2 s c Maximum crack width, w max = 3a cr e m /[1+2.(a cr -c f )/(h-x)] 0.15 mm Maximum crack width utilisation, w max / w allow < 1.00 50% OK Note for a particular section and force / moment, crack widths can be reduced by increasing steel area, reducing spacing between rebars and reducing concrete cover (limited to durability requirements). The employment of smaller diameter bars at closer centres is thus preferable to larger diameter bars at further centres. There should be a provision for longitudinal steels at the side faces of beams of moderate depths; Crack width utilisation 50% OK
jxxx 21 Member Design - RC Beam Longitudinal Shear Between Web and Flange Rectangular or Flanged Beam (EC2) Note that this check is performed for both rectangular and flanged section designs, although theoretically only applicable in the latter case; XX 23/10/2017 EC2 Longitudinal shear stress, K S. v Ed N/mm 2 Longitudinal shear stress, N/mm 2 cl.6.2.4 Change of normal force in flange half over Dx, DF d = K B.(M*-0)/z kn Note conservatively factor, K B = 0.5(b eff - b w )/b eff employed even if neutral axis within web; Lever arm, z Note if neutral axis within web, for simplicity, z = d - h f /2; Thickness of the flange at the junctions, h f Length under consideration, Dx m Note the maximum value that may be assumed for D x is half the distance between the section where the moment is 0 and the section where the moment is maximum. However, since D F d is also calculated over D x based on a variation of moment of ~ M/2-0 say, it is deemed acceptable to use for D x the full distance between the section where the moment is 0 and the section where the moment is maximum based on a variation of moment of M - 0 and factored by K S. Shear stress distribution factor, K S For UDLs, K S may be taken as 2.00 for simply supported beams, 1.33 for continuous beams and 2.00 for cantilever beams; Effective width, b eff = MIN(b w + function (span, section, structure Note for rectangular sections, b eff equivalent to that of T-sections assumed; Width (rectangular) or web width (flanged), b w cl.6.2.4 Longitudinal shear stress limit to prevent crushing, N/mm 2 cl.6.2.4 Design compressive strength, f cd N/mm 2 Strength reduction factor for concrete cracked in shear, n with a cc =1.0, g C =1.5 cl.3.1.6 cl.6.2.2 Longitudinal shear stress limit to prevent crushing utilisation, (K S.v Ed )/(nf cd sin Longitudinal shear stress limit for no transverse reinforcement, 0.4f ctd N/mm 2 cl.6.2.4 Design tensile strength, f ctd N/mm 2 with a ct =1.0, g C =1.5 cl.3.1.6 N/mm 2 T.3.1 N/mm 2 T.3.1 N/mm 2 T.3.1 Characteristic cylinder strength of concrete, f ck N/mm 2 T.3.1 Characteristic cube strength of concrete, f cu N/mm 2 T.3.1 Longitudinal shear stress limit for no transverse reinforcement utilisation, (K S Required design transverse reinforcement per unit length, A sf /s f > 2 /m Note area of transverse steel to be provided should be the greater of 1.0A sf /s f and cl.6.2.4 0.5A sf /s f + area required for slab bending; Note K S factored onto v Ed herein; Design yield strength of reinforcement, f yd = f y / g S, g S =1.15 N/mm 2 cl.2.4.2.4 Thickness of the flange at the junctions, h f Angle, q f 45.0 degrees cl.6.2.4 Provided transverse reinforcement per unit length, A e 3927 mm 2 /m Required design transverse reinforcement per unit length utilisation, (A sf /s f )/
jxxx 22 Member Design - RC Beam BS5400-4 Longitudinal Shear Between Web and Flange Rectangular or Flanged Beam (BS5400-4) Note that this check is performed for both rectangular and flanged section designs, although theoretically only applicable in the latter case; XX 23/10/2017 Longitudinal shear force per unit length, V 1 = K S. DF d / Dx Change of normal force in flange half over Dx, DF d = K B.(M*-0)/z kn/m kn Note conservatively factor, K B = 0.5(b eff - b w )/b eff employed even if neutral axis within web; Lever arm, z Note if neutral axis within web, for simplicity, z = d - h f /2; Thickness of the flange at the junctions, h f Length under consideration, Dx m Note D x is the beam length between the point of maximum design moment and the point of zero moment; Shear stress distribution factor, K S The longitudinal shear should be calculated per unit length. For UDLs, K S may be taken as 2.00 for simply supported beams, 1.33 for continuous beams and 2.00 for cantilever beams; Effective width, b eff = MIN(b w + function (span, section, structure Note for rectangular sections, b eff equivalent to that of T-sections assumed; Width (rectangular) or web width (flanged), b w cl.7.4.2.3 Longitudinal shear force limit per unit length, V 1,limit kn/m (a) kn/m cl.7.4.2.3 (b) kn/m cl.7.4.2.3 Concrete bond constant, k 1 T.31 Ultimate longitudinal shear stress limit, n 1 N/mm 2 T.31 Surface type T.31 Length of shear plane, L s = h f Provided transverse reinforcement per unit length, A e 3927 mm 2 /m Note reinforcement provided for coexistent bending effects and shear reinforcement crossing the shear plane, provided to resist vertical shear, may be included provided they are fully anchored; Characteristic strength of reinforcement, f y N/mm 2 cl.7.4.2.3 Longitudinal shear force limit per unit length utilisation, V 1 /V 1,limit Required nominal transverse reinforcement per unit length, 0.15%L s 2 /m cl.7.4.2.3 Required nominal transverse reinforcement per unit length utilisation, 0.15%
Longitudinal shear stress limit, v Rdi N/mm 2 cl.6.2.5 CONSULTING Engineering Calculation Sheet jxxx 23 Longitudinal Shear Within Web Rectangular or Flanged Beam (EC2) EC2 Longitudinal shear stress, N/mm 2 cl.6.2.5 Ratio, b = 1.0 cl.6.2.5 Transverse shear force, V Ed = V d * kn cl.6.2.5 Lever arm, z m cl.6.2.5 Note if neutral axis within web, for simplicity, z = d - h f /2; Width of the interface, b i = b w cl.6.2.5 Note c.f ctd = 0.00 if s n is negative (tension); cl.6.2.5 Roughness coefficient, c cl.6.2.5 Roughness coefficient, m cl.6.2.5 Design tensile strength, f ctd N/mm 2 with a ct =1.0, g C =1.5 cl.3.1.6 N/mm 2 N/mm 2 N/mm 2 T.3.1 T.3.1 T.3.1 Characteristic cylinder strength of concrete, f ck N/mm 2 T.3.1 Characteristic cube strength of concrete, f cu N/mm 2 T.3.1 Normal stress across longitudinal shear interface, s n = 0 N/mm 2 Reinforcement ratio, r = A s / A i cl.6.2.5 Area of reinforcement, A s = A sv,prov /S + A sv,prov,t /S t 2 /m Note that the area of reinforcement crossing the shear interface may include ordinary shear reinforcement with adequate anchorage at both cl.6.2.5 sides of the interface; Area of the joint, A i = 1000.b i 2 /m Design yield strength of reinforcement, f yd = f yv / g S, g S =1.15 N/mm 2 cl.2.4.2.4 Angle of reinforcement, a = 90.0 degrees cl.6.2.5 Design compressive strength, f cd N/mm 2 Strength reduction factor for concrete cracked in shear, n with a cc =1.0, g C =1.5 cl.3.1.6 cl.6.2.2 Longitudinal shear stress limit utilisation, v Edi /v Rdi
jxxx 24 Longitudinal Shear Within Web Rectangular or Flanged Beam () Longitudinal shear stress, n h = K S. DF c / (b w.dx) N/mm 2 cl.5.4.7.2 Change of total compression force over Dx, DF c = (M-0)/z kn cl.5.4.7.1 Lever arm, z Note if neutral axis within web, for simplicity, z = d - h f /2; Length under consideration, Dx m Note D x is the beam length between the point of maximum design moment and the point of zero moment; Shear stress distribution factor, K S The average design shear stress should then be distributed in proportion to the vertical design shear force diagram to give the horizontal shear stress at any point along the length of the member. For UDLs, K S maybe taken as 2.00 for simply supported beams, 1.33 for continuous beams and 2.00 for cantilever beams; Width (rectangular) or web width (flanged), b w cl.5.4.7.2 cl.5.4.7.2 Longitudinal shear stress limit for no nominal / design vertical reinforcement N/mm 2 Surface type T.5.5 Longitudinal shear stress limit for no nominal / design vertical reinforcement Required nominal vertical reinforcement per unit length, 0.15%b w 2 /m cl.5.4.7.3 Provided vertical reinforcement per unit length, A e 2 /m Note A e = A sv,prov / S + A sv,prov,t / S t ; Required nominal vertical reinforcement per unit length utilisation, 0.15%b w / Note UT set to 0% if longitudinal shear stress limit for no nominal vertical reinforcement UT <= 100%; Required design vertical reinforcement per unit length, A h 2 /m cl.5.4.7.4 Required design vertical reinforcement per unit length utilisation, A h /A e Note UT set to 0% if longitudinal shear stress limit for no design vertical reinforcement UT <= 100%;
jxxx 25 Longitudinal Shear Within Web Rectangular or Flanged Beam (BS5400-4) BS5400-4 Longitudinal shear force per unit length, V 1 = K S. DF c / Dx Change of total compression force over Dx, DF c = (M-0)/z Lever arm, z Note if neutral axis within web, for simplicity, z = d - h f /2; Length under consideration, Dx kn/m kn m Note D x is the beam length between the point of maximum design moment and the point of zero moment; Shear stress distribution factor, K S The longitudinal shear should be calculated per unit length. For UDLs, K S may be taken as 2.00 for simply supported beams, 1.33 for continuous beams and 2.00 for cantilever beams; Width (rectangular) or web width (flanged), b w cl.7.4.2.3 Longitudinal shear force limit per unit length, V 1,limit kn/m (a) kn/m cl.7.4.2.3 (b) kn/m cl.7.4.2.3 Concrete bond constant, k 1 T.31 Ultimate longitudinal shear stress limit, n 1 N/mm 2 T.31 Surface type T.31 Length of shear plane, L s = b w Provided vertical reinforcement per unit length, A e 2 /m Note A e = A sv,prov / S + A sv,prov,t / S t ; Note reinforcement provided for coexistent bending effects and shear reinforcement crossing the shear plane, provided to resist vertical shear, may be included provided they are fully anchored; Characteristic strength of reinforcement, f yv N/mm 2 cl.7.4.2.3 Longitudinal shear force limit per unit length utilisation, V 1 /V 1,limit Required nominal vertical reinforcement per unit length, 0.15%L s 2 /m cl.7.4.2.3 Required nominal vertical reinforcement per unit length utilisation, 0.15%L s / Note UT set to 0% if longitudinal shear force limit per unit length for no nominal vertical reinforcement UT <= 100%;
jxxx 26 Deep Beam Rectangular Beam Note that this design check is performed for both rectangular and flanged sections, adopting rectangular section equations in the case of flanged sections; Note this section assumes a top loaded beam with no web openings ; Note this section assumes that the deep beam is laterally and rotationally restrained at the top and bottom of the beam overall depth, h (ref. CIRIA Guide 2 cl.2.2.3); Span to depth ratio, span / h 1.34 Applicability of deep beam design Applicable Note deep beam design is applicable for {1.0 span / h 2.0 s/s; 1.0 span / h 2.5 cont; 0.5 span / h 1.0 cant} (Reynolds cl.21.4.1 and CIRIA Guide 2 cl.1.3); OK Concrete type Horizontal shear link diameter, f link,h 25 mm Number of horizontal shear links in a horizontal section, i.e. number of legs, 4 Area provided by all horizontal shear links in a horizontal section, A sv,prov,h = 1963 mm 2 Pitch of horizontal shear links, S h 225 mm 2.p.(f link or f link,t ) 2 /4 / (b w.(s or S t )) (>0.20% G460; >0.25% G250) 0.33 % OK 2.p.f link,h 2 /4 / (b w.s h ) (>0.20% G460; >0.25% G250) 0.29 % OK Tension steel area provided (deep beam) 35343 mm 2 CIRIA % Min tension reinforcement (deep beam) 0.84 % Guide 2 % Min tension reinforcement (deep beam) (>= 0.52 f cu /0.95f y ) cl.2.6.2 % Min tension reinforcement (deep beam) utilisation 89% OK All detailing requirements met? OK
jxxx 27 Deep Beam Bending Reynolds Design bending moment, M 11963 knm Tension steel (deep beam), A s,db = 1.75M / [f y.h] 16254 mm 2 T.148 Note that the factor 1.75 is obtained from 1 / (lever arm factor 0.6 x material factor 0.95); cl.21.4.1 Tension steel zone depth, T zone (sag s/s, sag cont, hog cant) 513 mm cl.21.4.1 Tension steel zone depth, T zone (hog cont) Note A s,db to be distributed over depth of T zone from tension face; cl.21.4.1 Note T zone = (5h - span lim ) / 20 s/s sag and cont sag, (5h - 2 x span lim ) / 20 cant hog; cl.21.4.1 / SE Tension steel area provided (deep beam) 35343 mm 2 Tension steel area provided (deep beam) utilisation 46% OK Deep Beam Shear Reynolds Design shear force, V d 4223 kn Note the ultimate shear force limit is b w.h.f c '/10 g m where f c ' is the cylinder comp strength and g cl.21.4.1 Ultimate shear force utilisation 48% OK Area of tension steel reinforcement provided, A s,prov 35343 mm 2 Clear distance from edge of load to face of support, a 1 1750 mm T.148 Note for UDLs, concentrate total UDL at {span/4 s/s and cont, span/2 cant} from the support(s); T.148 Ratio a 1 /h 0.63 OK Note ensure a 1 /h is not greatly outside range of 0.23 to 0.70; T.148 Angle between horizontal bar and critical diagonal crack, q = tan -1 (h/a 1 ) 58.0 degrees T.148 Empirical coefficient, k 1 = {0.70 NWC, 0.50 LWC} 0.70 T.148 Empirical coefficient, k 2 = {100 plain round bars, 225 deformed bars} 225 N/mm 2 T.148 Cylinder splitting tensile strength, f t = 0.5(f cu ) 0.5 3.16 N/mm 2 T.148 Minimum breadth for deep beam, b w MAX {0, 0.65V d /[k 1.(h-0.35a 1 ).f t ]} 567 mm OK Number of rows of horizontal shear links in a vertical cross-section, n 10 Note the no. of rows of horizontal shear links reduced to account for T zone, i.e. (h - T zone )/S h ; Design horizontal links shear capacity, V r = k 2 SA sv,prov,h.a 2.sin 2 q/h 1404 kn T.148 Note the summation of the depths at which the horizontal shear links intersect the diagonal crack, S a 2 is calculated as S h.n.(n+1)/2 where n is the number of rows of horizontal shear links; Check V d < 1.0V 1 for no horizontal links (minor elements) VALID T.148 Concrete shear capacity, V 1 12195 kn T.148 Note V 1 = MAX[0,k 1.(h-0.35a 1 ).f t.b w ]+k 2.A s,prov.d.sin 2 q /h; T.148 Check V d > 0.0 for design horizontal links VALID T.148 Concrete and design horizontal links shear capacity V r + V 1 13600 kn T.148 Design shear resistance (deep beam) utilisation 31% OK
jxxx 28 Member Design - RC Beam Deep Beam Bending Design bending moment, M Note the ultimate bending moment limit is 0.12f cu b w h 2 ; XX 11963 knm CIRIA Guide 2 cl.2.4.1 Ultimate bending moment utilisation 21% OK 23/10/2017 Tension steel (deep beam), A s,db = M / [0.95f y.z] 17217 mm 2 cl.2.4.1 Lever arm at which the tension steel (deep beam) acts, z 1590 mm cl.2.4.1 Simply supported, z = 0.2 x span lim + 0.4h 1870 mm cl.2.4.1 ELF Continuous, z = 0.2 x span lim + 0.3h 1590 mm cl.2.4.1 Cantilever, z = 0.4 x span lim + 0.4h 2240 mm SELF Tension steel zone depth, T zone (sag s/s, sag cont, hog cant) 560 mm cl.2.4.1 Tension steel zone depth, T zone (hog cont) upper band 17% 560 mm cl.2.4.1 Tension steel zone depth, T zone (hog cont) lower band 83% 1680 mm cl.2.4.1 Note A s,db to be distributed over depth of T zone from tension face; cl.2.4.1 Note T zone = 0.2h s/s sag, cont sag and cant hog; cl.2.4.1 Note T zone = 0.2h cont hog upper band 0.5(span lim /h-1)a s,db, 0.2h-0.8h cont hog lower band remainder; Tension steel area provided (deep beam) 35343 mm 2 Tension steel area provided (deep beam) utilisation 49% OK Deep Beam Shear Design shear force, V d 4223 kn Note the ultimate shear force limit is min{b w.h. n u,2b w.h 2 n c k s /x e } where n u is the ultimate concrete shear strength from CP 110 T.6 and T.26 replaced by min{0.8f cu 0.5,{5.0,7.0}N/mm 2 } and n c is the design concrete shear strength from CP 110 T.5 and T.25 replaced by v c ; CIRIA Guide 2 cl.2.4.2 Factor, k s = 1.0 for h/b w < 4, else 0.6 1.0 cl.2.4.2 Ultimate shear force utilisation 43% OK Area of tension steel reinforcement provided, A s,prov 35343 mm 2 Clear distance from edge of load to face of support, x e 1750 mm cl.2.4.2 Note for UDLs, concentrate total UDL at {span/4 s/s and cont, span/2 cant} from the support(s); cl.2.4.2 Ratio x e /h 0.63 OK Note ensure x e /h is not greatly outside range of 0.23 to 0.70; cl.3.4.2 Angle between horizontal bar and critical diagonal crack, q = tan -1 (h/x e ) 58.0 degrees cl.2.4.2 Empirical coefficient, l 1 = {0.44 NWC, 0.32 LWC} 0.44 cl.3.4.2 Empirical coefficient, l 2 = {0.85 plain round bars, 1.95 deformed bars} 1.95 N/mm 2 cl.3.4.2 Number of rows of horizontal shear links in a vertical cross-section, n 9 Note the no. of rows of horizontal shear links reduced to account for T zone, i.e. (h - T zone )/S h ; Design horizontal links shear capacity, V r = 100l 2 SA sv,prov,h.y r.sin 2 q/h 996 kn cl.3.4.2 Note the summation of the depths at which the horizontal shear links intersect the diagonal crack, S y r is calculated as S h.n.(n+1)/2 where n is the number of rows of horizontal shear links; Check V d < 1.0V for no horizontal links (minor elements) VALID cl.3.4.2 Concrete shear capacity, V ( 1.3l 1 f cu.b w.h) 13406 kn cl.3.4.2 Note V = MAX[0, l 1.(h-0.35x e ). f cu.b w ]+100 l 2.A s,prov.d.sin 2 q /h; cl.3.4.2 Note require [100 l 2.A s,prov.d.sin 2 q /h] / V 0.20; 0.32 OK Check V d > 0.0 for design horizontal links VALID cl.3.4.2 Concrete and design horizontal links shear capacity V r + V ( l 1.n 14401 kn cl.3.4.2 Note n max = 1.3 f cu ; T.5 Note require [V r + 100 l 2.A s,prov.d.sin 2 q /h] / [V r + V] 0.20; 0.37 OK Design shear resistance (deep beam) utilisation 29% OK
jxxx 29 Detailing Instructions (Deep Beam)
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jxxx 31 Scheme Design
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jxxx 39 Note optional method of limiting allowable steel stress to full crack width calculation method;
jxxx 40 Typical Initial Span / Effective Depth Ratios
jxxx 41 Notes on Application to Upstand Beams Rect - s/s Rect - continuous Hog in continuous beam with precast slab Deflections irrelevant Rect - cantilever Hog in cantilever beam with precast slab Deflections relevant T - s/s T - continuous Hog in continuous interior beam with insitu slab Deflections irrelevant T - cantilever Hog in cantilever interior beam with insitu slab Deflections relevant L - s/s L - continuous Hog in continuous edge beam with insitu slab Deflections irrelevant L - cantilever Hog in cantilever edge beam with insitu slab Deflections relevant Rect - s/s Sag in s/s beam with precast or insitu slab Deflections relevant Rect - continuous Sag in continuous beam with precast or insitu slab Deflections relevant