Reference Description Results 5.5.1 Design for Critical BM
ULS Design Inner face of wall
Design ULS Bending (BM ULS) 38kNm/m
Width of the section (b) 1000mm
Section Height 400mm
Concrete Cover 50mm
Main R.F Bar Diameter 12mm
spacing 150mm
Shear/ Secondary R.F Bar Diameter 10mm
Concrete Strength ( fcu) 30N/mm2
R.F Strength (fy) 460N/mm2
cl .5.8.4.1, Check for Minimm area of R.F
Part 4 , BS Effective Depth (d) 344 mm
5400 Provided R.F (As) 754 mm2
Min. R.F percentage 0.15%
Min. Area of R.F 516 mm2
As> Min. R.F, Hence Satisfactory eq.5, Part Lever Arm (z) calculation
4, BS 5400 The lever Arm of mm
the Section (z) =
cl.5.3.2.2 z = 331.28 mm
0.95 d = 326.8 mm
cl.5.3.2.2 value z is greater than 0.95d
Hence z = 326.80 mm
Ultimate Resistant (Mu ) calculation
eq. 1 Mu = ( 0.87fy) Asz
= 98.61 kNm
eq. 2 Mu= 0.15fcubd2
= 532.51 kNm
cl.5.3.2.3 Mu is taken as the lesser of eqn 1 and eqn2
Hence Mu = 98.61 kNm
Check for ULS bending Moment
Mu > BM ULS , Hence provided R.F adequate SLS DESIGN Servicible Moment = 22 kNm Moment ,Permanent loads, Mg = 19 kNm Moment , Live loads , Mq = 3kNm
{
1-1.1 fy As}
d fcu bdSection Height = 400 mm
Concrete Cover = 50 mm
Dist com. face to crack
point ,a' = 300 mm
Dia of Main Bar = 12 mm
E steel = 200 kNmm2
E concrete,short term = 28 kNmm2
(Table 3, part 4)
A steel,provided = 754 mm2
Max Bar Spacing = 150 mm
f y = 460 Nmm2
f cu = 30 Nmm2
Limitations
Allowable crack width = 0.25 mm Stress Limitation Factor, Concrete = 0.38 Stress Limitation Factor, Steel = 0.75 Effective depth d = 344 mm
cl.4.3.2.1.b, Econcrete,long term = 14 kNmm2
part 4 Modular ratio (α) = 14.29
Taking first moment of area about neutral axis :
x = 75.98 mm x =depth to N/A z = 318.67 mm z=lever arm
cl.4.1.1.3. Max steel stress ,fs = Ms/(As*z)
& Table 2 = 91.56 N/mm2
0.75 fy = 345 N/mm2
fs<Limiting stress of steel ,Hence Satisfactory
cl.4.1.1.3. Max Con. Stress, fcb = 1.82 N/mm2
& Table 2 0.38 fcu = 11.4 N/mm2
fcb<Limiting stress of concrete,Hence Satisfactory Єs=strain at R/F level
Єs = 0.000457813
Cl 5.8.8.2 Є1=strain at notional surface
Є1 = 0.000493684
Allow for stiffening effect of concrete Eqn 25
Hence T 16 @
provided as
Cl 5.8.8.2
a
cr = 87.60 mm main RF of theinner face of
Eqn 24 Design crack width wing wall
W
= 0.0971335
Design crack width < Allowable crack width, Hence Satisfactory Inner face of wall secondary r/f
Design ULS Bending (BM ULS) 12kNm/m
Width of the section (b) 1000mm
Section Height 400mm
Concrete Cover 50mm
Main R.F Bar Diameter 12mm
spacing 150mm
Shear/ Secondary R.F Bar Diameter 10mm
Concrete Strength ( fcu) 30N/mm2
R.F Strength (fy) 460N/mm2
cl .5.8.4.1, Check for Minimm area of R.F
Part 4 , BS Effective Depth (d) 333 mm
5400 Provided R.F (As) 754 mm2
Min. R.F percentage 0.15%
Min. Area of R.F 499.5 mm2
As> Min. R.F, Hence Satisfactory eq.5, Part Lever Arm (z) calculation
4, BS 5400 The lever Arm of mm
the Section (z) =
cl.5.3.2.2 z = 320.28 mm
0.95 d = 316.35 mm
cl.5.3.2.2 value z is greater than 0.95d
Hence z = 316.35 mm
Ultimate Resistant (Mu ) calculation
eq. 1 Mu = ( 0.87fy) Asz
= 95.46 kNm
eq. 2 Mu= 0.15fcubd2
= 499.00 kNm
cl.5.3.2.3 Mu is taken as the lesser of eqn 1 and eqn2
Hence Mu = 95.46 kNm
Check for ULS bending Moment
Mu > BM ULS , Hence provided R.F adequate SLS DESIGN
{
1-1.1 fy As}
d fcu bdServicible Moment = 8kNm Moment ,Permanent loads, Mg = 7kNm Moment , Live loads , Mq = 1kNm Section Height = 400 mm Concrete Cover = 50 mm
Dist com. face to crack
point ,a' = 300 mm
Dia of Main Bar = 12 mm
E steel = 200 kNmm2
E concrete,short term = 28 kNmm2
(Table 3, part 4)
A steel,provided = 754 mm2
Max Bar Spacing = 150 mm
f y = 460 Nmm2
f cu = 30 Nmm2
Limitations
Allowable crack width = 0.25 mm Stress Limitation Factor, Concrete = 0.38 Stress Limitation Factor, Steel = 0.75 Effective depth d = 333 mm
cl.4.3.2.1.b, Econcrete,long term = 14 kNmm2
part 4 Modular ratio (α) = 14.29
Taking first moment of area about neutral axis :
x = 74.61 mm x =depth to N/A z = 308.13 mm z=lever arm
cl.4.1.1.3. Max steel stress ,fs = Ms/(As*z)
& Table 2 = 34.43 N/mm2
0.75 fy = 345 N/mm2
fs<Limiting stress of steel ,Hence Satisfactory
cl.4.1.1.3. Max Con. Stress, fcb = 0.70 N/mm2
& Table 2 0.38 fcu = 11.4 N/mm2
fcb<Limiting stress of concrete,Hence Satisfactory Єs=strain at R/F level
Єs = 0.000172173
Є1 = 0.000193495 Allow for stiffening effect of concrete
Eqn 25 Hence T 12 @ Єm = 0.000193495 150 mm c/c provided as Cl 5.8.8.2
a
cr = 87.60 mm main RF of the inner face ofEqn 24 Design crack width wing wall
W
= 0.0381291 Hence T 10 @
Design crack width < Allowable crack width, Hence Satisfactory 150 mm c/c provided as distribution rf of the wall
Hence T 10 @
Cl 5.8.4.2 Min area of main reinforcement in compression face = 0.12% of Area 150mm c/c
Minimum reinforcement area = 412.8 mm2/m provided as
Distribution reinforcement = 10 mm bars @ main RF of the
Cl 7.5.9 150 c/c ( 471.2 mm2) outer face of
wing wall
Cl 5.8.4.2 Minimum area of secondary reinforcement = 0.12% of Area Minimum reinforcement area = 412.8 mm2/m
Cl 7.5.9 Distribution reinforcement = 10 mm bars @ Hence T 10 @
150 c/c ( 471.2 mm2) 150 mm c/c provided as distribution rf of the wall
5.2.3 Check for ULS Shear
ULS Shear Force = 54 KN/m Methodology
cl 5.4.1 The shear forces in a slab may be obtained from general elastic Cl
5.3.3.1,Part Shear Stress is checked against the maximum allowable stress of 0.75√fcu or 4, BS 54000 analysis 4.75N/mm2 , whichever is lesser
cl.5.4..4.1 If the Shear stress is less than ξsvc as per cl.5.3.3, then it is considered as no addditonal r.f is required
cl.5.3.3.3 shear enhancement is allowed for section within distance av<2d
ULS Shear Force = 54 KN/m
ULS Shear Stress = 0.135N/mm2
Section Height = 400 mm
Concrete Strength
(f cu ) = 30 N/mm2
Main Bar Dia. = 12 mm
R.F Area (As ) = 754 mm2
Disribution Bar Dia. ( if
out side of Main Bar) = 10 mm
Shear Enhancement = No
av = 2.5 mm
Effective Depth (d) = 344
Shear Stress (v) = V/bd
= 0.157 N/mm2
cl. 5.3.3.1 Limiting Shear stress = 4.75 N/mm2 v< Limiting shear stress, Hence Satisfactory
Table 9 Depth Facotor (ξs) = 1.10
Table 8 vc = 0.405 N/mm2
ξsvc = 0.444 N/mm2
shear enhancement Checking
(2d/av ) = 1 against Shear
Resisisting S.S = 0.444 N/mm2 Satisfied,
ResisistingS.S< LimitingS.S. Yes Section is O.K
Hence resisting S.S 0.444 N/mm2
Is resisiting S.S> v Yes Satisfactory Against Shear
