+ All Categories
Home > Documents > EC2 Continuous Beam Design

EC2 Continuous Beam Design

Date post: 09-Oct-2015
Category:
Upload: ikanyu79
View: 131 times
Download: 3 times
Share this document with a friend
Description:
Design Continuous concrete beam using EC 2 method. This is a calculation sheet.
Popular Tags:

of 14

Transcript
  • Project

    Job Ref.

    Section

    Sheet no./rev.

    1

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    CONTINUOUS BEAM EXAMPLE

    RC BEAM ANALYSIS & DESIGN (EN1992-1)

    In accordance with UK national annex

    TEDDS calculation version 2.1.15

    Load Envelope - Combination 1

    0.0

    64.830

    mm 8000

    1A

    6000

    2B C

    Load Envelope - Combination 2

    0.0

    64.830

    mm 8000

    1A

    6000

    2B C

    Load Envelope - Combination 3

    0.0

    64.830

    mm 8000

    1A

    6000

    2B C

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    2

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Support conditions

    Support A Vertically restrained

    Rotationally sprung (35136 kNm/rad)

    Support B Vertically restrained

    Rotationally sprung (35136 kNm/rad)

    Support C Vertically restrained

    Rotationally sprung (35136 kNm/rad)

    Applied loading

    Permanent full UDL 25.8 kN/m

    Variable full UDL 20 kN/m

    Load combinations

    Load combination 1 Support A Permanent 1.35

    Variable 1.50

    Span 1 Permanent 1.35

    Variable 1.50

    Support B Permanent 1.35

    Variable 1.50

    Span 2 Permanent 1.35

    Variable 1.50

    Support C Permanent 1.35

    Variable 1.50

    Load combination 2 Support A Permanent 1.35

    Variable 1.50

    Span 1 Permanent 1.35

    Variable 1.50

    Support B Permanent 1.35

    Variable 1.50

    Span 2 Permanent 1.00

    Variable 0.00

    Support C Permanent 1.00

    Variable 0.00

    Load combination 3 Support A Permanent 1.00

    Variable 0.00

    Span 1 Permanent 1.00

    Variable 0.00

    Support B Permanent 1.35

    Variable 1.50

    Span 2 Permanent 1.35

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    3

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Variable 1.50

    Support C Permanent 1.35

    Variable 1.50

    Analysis results

    Maximum moment support A; MA_max = -177 kNm; MA_red = -142 kNm; 20%

    Maximum moment span 1 at 3692 mm; Ms1_max = 265 kNm; Ms1_red = 315 kNm; 18.9%

    Maximum moment support B; MB_max = -385 kNm; MB_red = -270 kNm; 30%

    Maximum moment span 2 at 3375 mm; Ms2_max = 145 kNm; Ms2_red = 159 kNm; 9.2%

    Maximum moment support C; MC_max = -78 kNm; MC_red = -78 kNm;

    Maximum shear support A; VA_max = 239 kN; VA_red = 249 kN

    Maximum shear support A span 1 at 449 mm; VA_s1_max = 210 kN; VA_s1_red = 220 kN

    Maximum shear support B; VB_max = -287 kN; VB_red = -287 kN

    Maximum shear support B span 1 at 7556 mm; VB_s1_max = -258 kN; VB_s1_red = -258 kN

    Maximum shear support B span 2 at 445 mm; VB_s2_max = 212 kN; VB_s2_red = 212 kN

    Maximum shear support C; VC_max = -170 kN; VC_red = -189 kN

    Maximum shear support C span 2 at 5551 mm; VC_s2_max = -141 kN; VC_s2_red = -160 kN

    Maximum reaction at support A; RA = 239 kN

    Unfactored permanent load reaction at support A; RA_Permanent = 92 kN

    Unfactored variable load reaction at support A; RA_Variable = 72 kN

    Maximum reaction at support B; RB = 528 kN

    Unfactored permanent load reaction at support B; RB_Permanent = 210 kN

    Unfactored variable load reaction at support B; RB_Variable = 163 kN

    Maximum reaction at support C; RC = 170 kN

    Unfactored permanent load reaction at support C; RC_Permanent = 59 kN

    Unfactored variable load reaction at support C; RC_Variable = 46 kN

    Rectangular section details

    Section width; b = 300 mm

    Section depth; h = 500 mm

    300

    Concrete details (Table 3.1 - Strength and deformation characteristics for concrete)

    Concrete strength class; C40/50

    Characteristic compressive cylinder strength; fck = 40 N/mm2

    Characteristic compressive cube strength; fck,cube = 50 N/mm2

    Mean value of compressive cylinder strength; fcm = fck + 8 N/mm2 = 48 N/mm2

    Mean value of axial tensile strength; fctm = 0.3 N/mm2 (fck/ 1 N/mm2)2/3 = 3.5 N/mm2

    Secant modulus of elasticity of concrete; Ecm = 22 kN/mm2 [fcm/10 N/mm2]0.3 = 35220 N/mm2

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    4

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Partial factor for concrete (Table 2.1N); C = 1.50

    Compressive strength coefficient (cl.3.1.6(1)); cc = 0.85

    Design compressive concrete strength (exp.3.15); fcd = cc fck / C = 22.7 N/mm2

    Maximum aggregate size; hagg = 20 mm

    Reinforcement details

    Characteristic yield strength of reinforcement; fyk = 500 N/mm2

    Partial factor for reinforcing steel (Table 2.1N); S = 1.15

    Design yield strength of reinforcement; fyd = fyk / S = 435 N/mm2

    Nominal cover to reinforcement

    Nominal cover to top reinforcement; cnom_t = 35 mm

    Nominal cover to bottom reinforcement; cnom_b = 35 mm

    Nominal cover to side reinforcement; cnom_s = 35 mm

    Support A

    300

    2 x 8 shear legs at 200 c/c

    4 x 16 bars

    4 x 16 bars

    Rectangular section in flexure (Section 6.1)

    Minimum moment factor (cl.9.2.1.2(1)); 1 = 0.25

    Design bending moment; M = max(abs(MA_red), 1 abs(Ms1_red)) = 142 kNm

    Depth to tension reinforcement; d = h - cnom_t - v - top / 2 = 449 mm

    Percentage redistribution; mrA = 20 %

    Redistribution ratio; = min(1 - mrA, 1) = 0.800

    K = M / (b d2 fck) = 0.059

    K' = 0.598 - 0.181 2 - 0.21 = 0.153

    K' > K - No compression reinforcement is required

    Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 424 mm

    Depth of neutral axis; x = 2.5 (d - z) = 61 mm

    Area of tension reinforcement required; As,req = M / (fyd z) = 767 mm2

    Tension reinforcement provided; 4 16 bars

    Area of tension reinforcement provided; As,prov = 804 mm2

    Minimum area of reinforcement (exp.9.1N); As,min = max(0.26 fctm / fyk, 0.0013) b d = 246 mm2

    Maximum area of reinforcement (cl.9.2.1.1(3)); As,max = 0.04 b h = 6000 mm2

    PASS - Area of reinforcement provided is greater than area of reinforcement required

    Minimum bottom reinforcement at supports

    Minimum reinforcement factor (cl.9.2.1.4(1)); 2 = 0.25

    Area of reinforcement to adjacent span; As,span = 1963 mm2

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    5

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Minimum bottom reinforcement to support; As2,min = 2 As,span = 491 mm2

    Bottom reinforcement provided; 4 16 bars

    Area of bottom reinforcement provided; As2,prov = 804 mm2

    PASS - Area of reinforcement provided is greater than minimum area of reinforcement required

    Rectangular section in shear (Section 6.2)

    Design shear force at support A; VEd,max = abs(max(VA_max, VA_red)) = 249 kN

    Angle of comp. shear strut for maximum shear; max = 45 deg

    Maximum design shear force (exp.6.9); VRd,max = b z v1 fcd / (cot(max) + tan(max)) = 727 kN

    PASS - Design shear force at support is less than maximum design shear force

    Design shear force span 1 at 449 mm; VEd = max(VA_s1_max, VA_s1_red) = 220 kN

    Design shear stress; vEd = VEd / (b z) = 1.730 N/mm2

    Strength reduction factor (cl.6.2.3(3)); v1 = 0.6 [1 - fck / 250 N/mm2] = 0.504

    Compression chord coefficient (cl.6.2.3(3)); cw = 1.00

    Angle of concrete compression strut (cl.6.2.3);

    = min(max(0.5 Asin[min(2 vEd / (cw fcd v1),1)], 21.8 deg), 45deg) = 21.8 deg

    Area of shear reinforcement required (exp.6.13); Asv,req = vEd b / (fyd cot()) = 477 mm2/m

    Shear reinforcement provided; 2 8 legs at 200 c/c

    Area of shear reinforcement provided; Asv,prov = 503 mm2/m

    Minimum area of shear reinforcement (exp.9.5N); Asv,min = 0.08 N/mm2 b (fck / 1 N/mm2)0.5 / fyk = 304 mm2/m

    PASS - Area of shear reinforcement provided exceeds minimum required

    Maximum longitudinal spacing (exp.9.6N); svl,max = 0.75 d = 337 mm

    PASS - Longitudinal spacing of shear reinforcement provided is less than maximum

    Crack control (Section 7.3)

    Maximum crack width; wk = 0.3 mm

    Design value modulus of elasticity reinf (3.2.7(4)); Es = 200000 N/mm2

    Mean value of concrete tensile strength; fct,eff = fctm = 3.5 N/mm2

    Stress distribution coefficient; kc = 0.4

    Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm, 0.65), 1) = 1.00

    Actual tension bar spacing; sbar = (b - 2 (cnom_s + v) - top) / (Ntop - 1) = 66 mm

    Maximum stress permitted (Table 7.3N); s = 347 N/mm2

    Concrete to steel modulus of elast. ratio; cr = Es / Ecm = 5.68

    Distance of the Elastic NA from bottom of beam; y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov (cr - 1)) =

    245 mm

    Area of concrete in the tensile zone; Act = b y = 73539 mm2

    Minimum area of reinforcement required (exp.7.1); Asc,min = kc k fct,eff Act / s = 297 mm2

    PASS - Area of tension reinforcement provided exceeds minimum required for crack control

    Quasi-permanent value of variable action; 2 = 0.30

    Quasi-permanent limit state moment; MQP = abs(MA_c21) + 2 abs(MA_c22) = 81 kNm

    Permanent load ratio; RPL = MQP / M = 0.57

    Service stress in reinforcement; sr = fyd As,req / As,prov RPL = 237 N/mm2

    Maximum bar spacing (Tables 7.3N); sbar,max = 200 mm

    PASS - Maximum bar spacing exceeds actual bar spacing for crack control

    Minimum bar spacing

    Minimum bottom bar spacing; sbar,min = (b - 2 cnom_s - 2 v - bot) / (Nbot - 1) = 66 mm

    Minimum allowable bottom bar spacing; sbar_bot,min = max(bot, hagg + 5 mm, 20 mm) + bot = 41 mm

    Minimum top bar spacing; sbar,min = (b - 2 cnom_s - 2 v - top) / (Ntop - 1) = 66 mm

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    6

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Minimum allowable top bar spacing; sbar_top,min = max(top, hagg + 5 mm, 20 mm) + top = 41 mm

    PASS - Actual bar spacing exceeds minimum allowable

    Mid span 1

    300

    2 x 8 shear legs at 300 c/c

    4 x 25 bars

    2 x 16 bars

    Rectangular section in flexure (Section 6.1) - Positive midspan moment

    Design bending moment; M = abs(Ms1_red) = 315 kNm

    Depth to tension reinforcement; d = h - cnom_b - v - bot / 2 = 445 mm

    Percentage redistribution; mrs1 = Ms1_red / Ms1_max - 1 = 18.9 %

    Redistribution ratio; = min(1 - mrs1, 1) = 0.811

    K = M / (b d2 fck) = 0.133

    K' = 0.598 - 0.181 2 - 0.21 = 0.156

    K' > K - No compression reinforcement is required

    Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 384 mm

    Depth of neutral axis; x = 2.5 (d - z) = 151 mm

    Area of tension reinforcement required; As,req = M / (fyd z) = 1885 mm2

    Tension reinforcement provided; 4 25 bars

    Area of tension reinforcement provided; As,prov = 1963 mm2

    Minimum area of reinforcement (exp.9.1N); As,min = max(0.26 fctm / fyk, 0.0013) b d = 243 mm2

    Maximum area of reinforcement (cl.9.2.1.1(3)); As,max = 0.04 b h = 6000 mm2

    PASS - Area of reinforcement provided is greater than area of reinforcement required

    Rectangular section in shear (Section 6.2)

    Design shear force at span s1; VEd,max = max(Vs1_mid, 0 kN) = 131 kN

    Angle of comp. shear strut for maximum shear; max = 45 deg

    Maximum design shear force (exp.6.9); VRd,max = b z v1 fcd / (cot(max) + tan(max)) = 658 kN

    PASS - Design shear force at support is less than maximum design shear force

    Design shear force Span 1 - 2400 mm to 5600 mm; VEd = max(Vs1_mid, 0 kN) = 131 kN

    Design shear stress; vEd = VEd / (b z) = 1.139 N/mm2

    Strength reduction factor (cl.6.2.3(3)); v1 = 0.6 [1 - fck / 250 N/mm2] = 0.504

    Compression chord coefficient (cl.6.2.3(3)); cw = 1.00

    Angle of concrete compression strut (cl.6.2.3);

    = min(max(0.5 Asin[min(2 vEd / (cw fcd v1),1)], 21.8 deg), 45deg) = 21.8 deg

    Area of shear reinforcement required (exp.6.13); Asv,req = vEd b / (fyd cot()) = 314 mm2/m

    Shear reinforcement provided; 2 8 legs at 300 c/c

    Area of shear reinforcement provided; Asv,prov = 335 mm2/m

    Minimum area of shear reinforcement (exp.9.5N); Asv,min = 0.08 N/mm2 b (fck / 1 N/mm2)0.5 / fyk = 304 mm2/m

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    7

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    PASS - Area of shear reinforcement provided exceeds minimum required

    Maximum longitudinal spacing (exp.9.6N); svl,max = 0.75 d = 333 mm

    PASS - Longitudinal spacing of shear reinforcement provided is less than maximum

    Crack control (Section 7.3)

    Maximum crack width; wk = 0.3 mm

    Design value modulus of elasticity reinf (3.2.7(4)); Es = 200000 N/mm2

    Mean value of concrete tensile strength; fct,eff = fctm = 3.5 N/mm2

    Stress distribution coefficient; kc = 0.4

    Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm, 0.65), 1) = 1.00

    Actual tension bar spacing; sbar = (b - 2 (cnom_s + v) - bot) / (Nbot - 1) = 63 mm

    Maximum stress permitted (Table 7.3N); s = 350 N/mm2

    Concrete to steel modulus of elast. ratio; cr = Es / Ecm = 5.68

    Distance of the Elastic NA from bottom of beam; y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov (cr - 1)) =

    239 mm

    Area of concrete in the tensile zone; Act = b y = 71633 mm2

    Minimum area of reinforcement required (exp.7.1); Asc,min = kc k fct,eff Act / s = 288 mm2

    PASS - Area of tension reinforcement provided exceeds minimum required for crack control

    Quasi-permanent value of variable action; 2 = 0.30

    Quasi-permanent limit state moment; MQP = abs(Ms1_c21) + 2 abs(Ms1_c22) = 122 kNm

    Permanent load ratio; RPL = MQP / M = 0.39

    Service stress in reinforcement; sr = fyd As,req / As,prov RPL = 162 N/mm2

    Maximum bar spacing (Tables 7.3N); sbar,max = 250 mm

    PASS - Maximum bar spacing exceeds actual bar spacing for crack control

    Minimum bar spacing

    Minimum bottom bar spacing; sbar,min = (b - 2 cnom_s - 2 v - bot) / (Nbot - 1) = 63 mm

    Minimum allowable bottom bar spacing; sbar_bot,min = max(bot, hagg + 5 mm, 20 mm) + bot = 50 mm

    Minimum top bar spacing; sbar,min = (b - 2 cnom_s - 2 v - top) / (Ntop - 1) = 198 mm

    Minimum allowable top bar spacing; sbar_top,min = max(top, hagg + 5 mm, 20 mm) + top = 41 mm

    PASS - Actual bar spacing exceeds minimum allowable

    Deflection control (Section 7.4)

    Reference reinforcement ratio; m0 = (fck / 1 N/mm2)0.5 / 1000 = 0.006

    Required tension reinforcement ratio; m = As,req / (b d) = 0.014

    Required compression reinforcement ratio; 'm = As2,req / (b d) = 0.000

    Structural system factor (Table 7.4N); Kb = 1.3

    Basic allowable span to depth ratio (7.16b); span_to_depthbasic = Kb [11 + 1.5 (fck / 1 N/mm2)0.5 m0 / (m - 'm)

    + (fck / 1 N/mm2)0.5 ('m / m0)0.5 / 12] = 19.817

    Reinforcement factor (exp.7.17); Ks = min(As,prov / As,req 500 N/mm2 / fyk, 1.5) = 1.042

    Flange width factor; F1 = 1.000

    Long span supporting brittle partition factor; F2 = 1.000

    Allowable span to depth ratio; span_to_depthallow = min(span_to_depthbasic Ks F1 F2, 40 Kb) =

    20.640

    Actual span to depth ratio; span_to_depthactual = Ls1 / d = 17.998

    PASS - Actual span to depth ratio is within the allowable limit

    Support B

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    8

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    300

    2 x 8 shear legs at 150 c/c

    2 x 25 bars

    4 x 25 bars

    Rectangular section in flexure (Section 6.1)

    Design bending moment; M = abs(MB_red) = 270 kNm

    Depth to tension reinforcement; d = h - cnom_t - v - top / 2 = 445 mm

    Percentage redistribution; mrB = 30 %

    Redistribution ratio; = min(1 - mrB, 1) = 0.700

    K = M / (b d2 fck) = 0.114

    K' = 0.598 - 0.181 2 - 0.21 = 0.120

    K' > K - No compression reinforcement is required

    Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 394 mm

    Depth of neutral axis; x = 2.5 (d - z) = 126 mm

    Area of tension reinforcement required; As,req = M / (fyd z) = 1573 mm2

    Tension reinforcement provided; 4 25 bars

    Area of tension reinforcement provided; As,prov = 1963 mm2

    Minimum area of reinforcement (exp.9.1N); As,min = max(0.26 fctm / fyk, 0.0013) b d = 243 mm2

    Maximum area of reinforcement (cl.9.2.1.1(3)); As,max = 0.04 b h = 6000 mm2

    PASS - Area of reinforcement provided is greater than area of reinforcement required

    Rectangular section in shear (Section 6.2)

    Design shear force at support B; VEd,max = abs(max(VB_max, VB_red)) = 287 kN

    Angle of comp. shear strut for maximum shear; max = 45 deg

    Maximum design shear force (exp.6.9); VRd,max = b z v1 fcd / (cot(max) + tan(max)) = 676 kN

    PASS - Design shear force at support is less than maximum design shear force

    Design shear force span 1 at 7556 mm; VEd = abs(min(VB_s1_max, VB_s1_red)) = 258 kN

    Design shear stress; vEd = VEd / (b z) = 2.182 N/mm2

    Strength reduction factor (cl.6.2.3(3)); v1 = 0.6 [1 - fck / 250 N/mm2] = 0.504

    Compression chord coefficient (cl.6.2.3(3)); cw = 1.00

    Angle of concrete compression strut (cl.6.2.3);

    = min(max(0.5 Asin[min(2 vEd / (cw fcd v1),1)], 21.8 deg), 45deg) = 21.8 deg

    Area of shear reinforcement required (exp.6.13); Asv,req = vEd b / (fyd cot()) = 602 mm2/m

    Shear reinforcement provided; 2 8 legs at 150 c/c

    Area of shear reinforcement provided; Asv,prov = 670 mm2/m

    Minimum area of shear reinforcement (exp.9.5N); Asv,min = 0.08 N/mm2 b (fck / 1 N/mm2)0.5 / fyk = 304 mm2/m

    PASS - Area of shear reinforcement provided exceeds minimum required

    Maximum longitudinal spacing (exp.9.6N); svl,max = 0.75 d = 333 mm

    PASS - Longitudinal spacing of shear reinforcement provided is less than maximum

    Design shear force span 2 at 445 mm; VEd = max(VB_s2_max, VB_s2_red) = 212 kN

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    9

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Design shear stress; vEd = VEd / (b z) = 1.793 N/mm2

    Strength reduction factor (cl.6.2.3(3)); v1 = 0.6 [1 - fck / 250 N/mm2] = 0.504

    Compression chord coefficient (cl.6.2.3(3)); cw = 1.00

    Angle of concrete compression strut (cl.6.2.3);

    = min(max(0.5 Asin[min(2 vEd / (cw fcd v1),1)], 21.8 deg), 45deg) = 21.8 deg

    Area of shear reinforcement required (exp.6.13); Asv,req = vEd b / (fyd cot()) = 495 mm2/m

    Shear reinforcement provided; 2 8 legs at 150 c/c

    Area of shear reinforcement provided; Asv,prov = 670 mm2/m

    Minimum area of shear reinforcement (exp.9.5N); Asv,min = 0.08 N/mm2 b (fck / 1 N/mm2)0.5 / fyk = 304 mm2/m

    PASS - Area of shear reinforcement provided exceeds minimum required

    Maximum longitudinal spacing (exp.9.6N); svl,max = 0.75 d = 333 mm

    PASS - Longitudinal spacing of shear reinforcement provided is less than maximum

    Crack control (Section 7.3)

    Maximum crack width; wk = 0.3 mm

    Design value modulus of elasticity reinf (3.2.7(4)); Es = 200000 N/mm2

    Mean value of concrete tensile strength; fct,eff = fctm = 3.5 N/mm2

    Stress distribution coefficient; kc = 0.4

    Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm, 0.65), 1) = 1.00

    Actual tension bar spacing; sbar = (b - 2 (cnom_s + v) - top) / (Ntop - 1) = 63 mm

    Maximum stress permitted (Table 7.3N); s = 350 N/mm2

    Concrete to steel modulus of elast. ratio; cr = Es / Ecm = 5.68

    Distance of the Elastic NA from bottom of beam; y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov (cr - 1)) =

    239 mm

    Area of concrete in the tensile zone; Act = b y = 71633 mm2

    Minimum area of reinforcement required (exp.7.1); Asc,min = kc k fct,eff Act / s = 288 mm2

    PASS - Area of tension reinforcement provided exceeds minimum required for crack control

    Quasi-permanent value of variable action; 2 = 0.30

    Quasi-permanent limit state moment; MQP = abs(MB_c21) + 2 abs(MB_c22) = 189 kNm

    Permanent load ratio; RPL = MQP / M = 0.70

    Service stress in reinforcement; sr = fyd As,req / As,prov RPL = 244 N/mm2

    Maximum bar spacing (Tables 7.3N); sbar,max = 150 mm

    PASS - Maximum bar spacing exceeds actual bar spacing for crack control

    Minimum bar spacing

    Minimum bottom bar spacing; sbar,min = (b - 2 cnom_s - 2 v - bot) / (Nbot - 1) = 189 mm

    Minimum allowable bottom bar spacing; sbar_bot,min = max(bot, hagg + 5 mm, 20 mm) + bot = 50 mm

    Minimum top bar spacing; sbar,min = (b - 2 cnom_s - 2 v - top) / (Ntop - 1) = 63 mm

    Minimum allowable top bar spacing; sbar_top,min = max(top, hagg + 5 mm, 20 mm) + top = 50 mm

    PASS - Actual bar spacing exceeds minimum allowable

    Mid span 2

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    10

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    300

    2 x 8 shear legs at 300 c/c

    2 x 25 bars

    2 x 16 bars

    Rectangular section in flexure (Section 6.1) - Positive midspan moment

    Design bending moment; M = abs(Ms2_red) = 159 kNm

    Depth to tension reinforcement; d = h - cnom_b - v - bot / 2 = 445 mm

    Percentage redistribution; mrs2 = Ms2_red / Ms2_max - 1 = 9.2 %

    Redistribution ratio; = min(1 - mrs2, 1) = 0.908

    K = M / (b d2 fck) = 0.067

    K' = 0.598 - 0.181 2 - 0.21 = 0.184

    K' > K - No compression reinforcement is required

    Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 416 mm

    Depth of neutral axis; x = 2.5 (d - z) = 70 mm

    Area of tension reinforcement required; As,req = M / (fyd z) = 877 mm2

    Tension reinforcement provided; 2 25 bars

    Area of tension reinforcement provided; As,prov = 982 mm2

    Minimum area of reinforcement (exp.9.1N); As,min = max(0.26 fctm / fyk, 0.0013) b d = 243 mm2

    Maximum area of reinforcement (cl.9.2.1.1(3)); As,max = 0.04 b h = 6000 mm2

    PASS - Area of reinforcement provided is greater than area of reinforcement required

    Rectangular section in shear (Section 6.2)

    Shear reinforcement provided; 2 8 legs at 300 c/c

    Area of shear reinforcement provided; Asv,prov = 335 mm2/m

    Minimum area of shear reinforcement (exp.9.5N); Asv,min = 0.08 N/mm2 b (fck / 1 N/mm2)0.5 / fyk = 304 mm2/m

    PASS - Area of shear reinforcement provided exceeds minimum required

    Maximum longitudinal spacing (exp.9.6N); svl,max = 0.75 d = 333 mm

    PASS - Longitudinal spacing of shear reinforcement provided is less than maximum

    Design shear resistance (assuming cot() is 2.5); Vprov = 2.5 Asv,prov z fyd = 151.7 kN

    Shear links provided valid between 1400 mm and 5700 mm with tension reinforcement of 982 mm2

    Crack control (Section 7.3)

    Maximum crack width; wk = 0.3 mm

    Design value modulus of elasticity reinf (3.2.7(4)); Es = 200000 N/mm2

    Mean value of concrete tensile strength; fct,eff = fctm = 3.5 N/mm2

    Stress distribution coefficient; kc = 0.4

    Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm, 0.65), 1) = 1.00

    Actual tension bar spacing; sbar = (b - 2 (cnom_s + v) - bot) / (Nbot - 1) = 189 mm

    Maximum stress permitted (Table 7.3N); s = 249 N/mm2

    Concrete to steel modulus of elast. ratio; cr = Es / Ecm = 5.68

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    11

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Distance of the Elastic NA from bottom of beam; y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov (cr - 1)) =

    244 mm

    Area of concrete in the tensile zone; Act = b y = 73266 mm2

    Minimum area of reinforcement required (exp.7.1); Asc,min = kc k fct,eff Act / s = 413 mm2

    PASS - Area of tension reinforcement provided exceeds minimum required for crack control

    Quasi-permanent value of variable action; 2 = 0.30

    Quasi-permanent limit state moment; MQP = abs(Ms2_c21) + 2 abs(Ms2_c22) = 56 kNm

    Permanent load ratio; RPL = MQP / M = 0.35

    Service stress in reinforcement; sr = fyd As,req / As,prov RPL = 138 N/mm2

    Maximum bar spacing (Tables 7.3N); sbar,max = 300 mm

    PASS - Maximum bar spacing exceeds actual bar spacing for crack control

    Minimum bar spacing

    Minimum bottom bar spacing; sbar,min = (b - 2 cnom_s - 2 v - bot) / (Nbot - 1) = 189 mm

    Minimum allowable bottom bar spacing; sbar_bot,min = max(bot, hagg + 5 mm, 20 mm) + bot = 50 mm

    Minimum top bar spacing; sbar,min = (b - 2 cnom_s - 2 v - top) / (Ntop - 1) = 198 mm

    Minimum allowable top bar spacing; sbar_top,min = max(top, hagg + 5 mm, 20 mm) + top = 41 mm

    PASS - Actual bar spacing exceeds minimum allowable

    Deflection control (Section 7.4)

    Reference reinforcement ratio; m0 = (fck / 1 N/mm2)0.5 / 1000 = 0.006

    Required tension reinforcement ratio; m = As,req / (b d) = 0.007

    Required compression reinforcement ratio; 'm = As2,req / (b d) = 0.000

    Structural system factor (Table 7.4N); Kb = 1.3

    Basic allowable span to depth ratio (7.16b); span_to_depthbasic = Kb [11 + 1.5 (fck / 1 N/mm2)0.5 m0 / (m - 'm)

    + (fck / 1 N/mm2)0.5 ('m / m0)0.5 / 12] = 26.165

    Reinforcement factor (exp.7.17); Ks = min(As,prov / As,req 500 N/mm2 / fyk, 1.5) = 1.120

    Flange width factor; F1 = 1.000

    Long span supporting brittle partition factor; F2 = 1.000

    Allowable span to depth ratio; span_to_depthallow = min(span_to_depthbasic Ks F1 F2, 40 Kb) =

    29.301

    Actual span to depth ratio; span_to_depthactual = Ls2 / d = 13.498

    PASS - Actual span to depth ratio is within the allowable limit

    Support C

    300

    2 x 8 shear legs at 300 c/c

    4 x 16 bars

    4 x 16 bars

    Rectangular section in flexure (Section 6.1) - Positive midspan moment

    Minimum moment factor (cl.9.2.1.2(1)); 1 = 0.25

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    12

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Design bending moment; M = abs(MC_red) = 78 kNm

    Depth to tension reinforcement; d = h - cnom_t - v - top / 2 = 449 mm

    Percentage redistribution; mrC = 0 %

    Redistribution ratio; = min(1 - mrC, 1) = 1.000

    K = M / (b d2 fck) = 0.032

    K' = 0.598 - 0.181 2 - 0.21 = 0.207

    K' > K - No compression reinforcement is required

    Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 427 mm

    Depth of neutral axis; x = 2.5 (d - z) = 56 mm

    Area of tension reinforcement required; As,req = M / (fyd z) = 421 mm2

    Tension reinforcement provided; 4 16 bars

    Area of tension reinforcement provided; As,prov = 804 mm2

    Minimum area of reinforcement (exp.9.1N); As,min = max(0.26 fctm / fyk, 0.0013) b d = 246 mm2

    Maximum area of reinforcement (cl.9.2.1.1(3)); As,max = 0.04 b h = 6000 mm2

    PASS - Area of reinforcement provided is greater than area of reinforcement required

    Minimum bottom reinforcement at supports

    Minimum reinforcement factor (cl.9.2.1.4(1)); 2 = 0.25

    Area of reinforcement to adjacent span; As,span = 982 mm2

    Minimum bottom reinforcement to support; As2,min = 2 As,span = 245 mm2

    Bottom reinforcement provided; 4 16 bars

    Area of bottom reinforcement provided; As2,prov = 804 mm2

    PASS - Area of reinforcement provided is greater than minimum area of reinforcement required

    Rectangular section in shear (Section 6.2)

    Design shear force at support C; VEd,max = abs(max(VC_max, VC_red)) = 170 kN

    Angle of comp. shear strut for maximum shear; max = 45 deg

    Maximum design shear force (exp.6.9); VRd,max = b z v1 fcd / (cot(max) + tan(max)) = 731 kN

    PASS - Design shear force at support is less than maximum design shear force

    Design shear force span 2 at 5551 mm; VEd = abs(min(VC_s2_max, VC_s2_red)) = 160 kN

    Design shear stress; vEd = VEd / (b z) = 1.253 N/mm2

    Strength reduction factor (cl.6.2.3(3)); v1 = 0.6 [1 - fck / 250 N/mm2] = 0.504

    Compression chord coefficient (cl.6.2.3(3)); cw = 1.00

    Angle of concrete compression strut (cl.6.2.3);

    = min(max(0.5 Asin[min(2 vEd / (cw fcd v1),1)], 21.8 deg), 45deg) = 21.8 deg

    Area of shear reinforcement required (exp.6.13); Asv,req = vEd b / (fyd cot()) = 346 mm2/m

    Shear reinforcement provided; 2 8 legs at 300 c/c

    Area of shear reinforcement provided; Asv,prov = 335 mm2/m

    Minimum area of shear reinforcement (exp.9.5N); Asv,min = 0.08 N/mm2 b (fck / 1 N/mm2)0.5 / fyk = 304 mm2/m

    FAIL - Area of shear reinforcement provided is less than the minimum required

    Maximum longitudinal spacing (exp.9.6N); svl,max = 0.75 d = 337 mm

    PASS - Longitudinal spacing of shear reinforcement provided is less than maximum

    Crack control (Section 7.3)

    Maximum crack width; wk = 0.3 mm

    Design value modulus of elasticity reinf (3.2.7(4)); Es = 200000 N/mm2

    Mean value of concrete tensile strength; fct,eff = fctm = 3.5 N/mm2

    Stress distribution coefficient; kc = 0.4

    Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm, 0.65), 1) = 1.00

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    13

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Actual tension bar spacing; sbar = (b - 2 (cnom_s + v) - top) / (Ntop - 1) = 66 mm

    Maximum stress permitted (Table 7.3N); s = 347 N/mm2

    Concrete to steel modulus of elast. ratio; cr = Es / Ecm = 5.68

    Distance of the Elastic NA from bottom of beam; y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov (cr - 1)) =

    245 mm

    Area of concrete in the tensile zone; Act = b y = 73539 mm2

    Minimum area of reinforcement required (exp.7.1); Asc,min = kc k fct,eff Act / s = 297 mm2

    PASS - Area of tension reinforcement provided exceeds minimum required for crack control

    Quasi-permanent value of variable action; 2 = 0.30

    Quasi-permanent limit state moment; MQP = abs(MC_c21) + 2 abs(MC_c22) = 27 kNm

    Permanent load ratio; RPL = MQP / M = 0.34

    Service stress in reinforcement; sr = fyd As,req / As,prov RPL = 78 N/mm2

    Maximum bar spacing (Tables 7.3N); sbar,max = 300 mm

    PASS - Maximum bar spacing exceeds actual bar spacing for crack control

    Minimum bar spacing

    Minimum bottom bar spacing; sbar,min = (b - 2 cnom_s - 2 v - bot) / (Nbot - 1) = 66 mm

    Minimum allowable bottom bar spacing; sbar_bot,min = max(bot, hagg + 5 mm, 20 mm) + bot = 41 mm

    Minimum top bar spacing; sbar,min = (b - 2 cnom_s - 2 v - top) / (Ntop - 1) = 66 mm

    Minimum allowable top bar spacing; sbar_top,min = max(top, hagg + 5 mm, 20 mm) + top = 41 mm

    PASS - Actual bar spacing exceeds minimum allowable

    Rectangular section in flexure (Section 6.1) - Negative span moment

    Minimum moment factor (cl.9.2.1.2(1)); 1 = 0.25

    Design bending moment; M = abs(MC_neg) = 2 kNm

    Depth to tension reinforcement; d = h - cnom_b - v - bot / 2 = 449 mm

    Percentage redistribution; mrC = 0 %

    Redistribution ratio; = min(1 - mrC, 1) = 1.000

    K = M / (b d2 fck) = 0.001

    K' = 0.598 - 0.181 2 - 0.21 = 0.207

    K' > K - No compression reinforcement is required

    Lever arm; z = min((d / 2) [1 + (1 - 3.53 K)0.5], 0.95 d) = 427 mm

    Depth of neutral axis; x = 2.5 (d - z) = 56 mm

    Area of tension reinforcement required; As,req = M / (fyd z) = 11 mm2

    Tension reinforcement provided; 4 16 bars

    Area of tension reinforcement provided; As,prov = 804 mm2

    Minimum area of reinforcement (exp.9.1N); As,min = max(0.26 fctm / fyk, 0.0013) b d = 246 mm2

    Maximum area of reinforcement (cl.9.2.1.1(3)); As,max = 0.04 b h = 6000 mm2

    PASS - Area of reinforcement provided is greater than area of reinforcement required

    Crack control (Section 7.3)

    Maximum crack width; wk = 0.3 mm

    Design value modulus of elasticity reinf (3.2.7(4)); Es = 200000 N/mm2

    Mean value of concrete tensile strength; fct,eff = fctm = 3.5 N/mm2

    Stress distribution coefficient; kc = 0.4

    Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm, 0.65), 1) = 1.00

    Actual tension bar spacing; sbar = (b - 2 (cnom_s + v) - bot) / (Nbot - 1) = 66 mm

    Maximum stress permitted (Table 7.3N); s = 347 N/mm2

    Concrete to steel modulus of elast. ratio; cr = Es / Ecm = 5.68

  • Project

    Job Ref.

    Section

    Sheet no./rev.

    14

    Calc. by

    P

    Date

    8/10/2014

    Chk'd by

    Date

    App'd by

    Date

    Distance of the Elastic NA from bottom of beam; y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov (cr - 1)) =

    245 mm

    Area of concrete in the tensile zone; Act = b y = 73539 mm2

    Minimum area of reinforcement required (exp.7.1); Asc,min = kc k fct,eff Act / s = 297 mm2

    PASS - Area of tension reinforcement provided exceeds minimum required for crack control

    Quasi-permanent value of variable action; 2 = 0.30

    Quasi-permanent limit state moment; MQP = abs(MC_c21) + 2 abs(MC_c22) = 27 kNm

    Permanent load ratio; RPL = MQP / M = 13.19

    Service stress in reinforcement; sr = fyd As,req / As,prov RPL = 78 N/mm2

    Maximum bar spacing (Tables 7.3N); sbar,max = 300 mm

    PASS - Maximum bar spacing exceeds actual bar spacing for crack control

    Minimum bar spacing

    Minimum bottom bar spacing; sbar,min = (b - 2 cnom_s - 2 v - bot) / (Nbot - 1) = 66 mm

    Minimum allowable bottom bar spacing; sbar_bot,min = max(bot, hagg + 5 mm, 20 mm) + bot = 41 mm

    Minimum top bar spacing; sbar,min = (b - 2 cnom_s - 2 v - top) / (Ntop - 1) = 66 mm

    Minimum allowable top bar spacing; sbar_top,min = max(top, hagg + 5 mm, 20 mm) + top = 41 mm

    PASS - Actual bar spacing exceeds minimum allowable


Recommended