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1
BS8110 REF. CALCULATION OUTPUT
I/ Design BeamGB9A effective span= 7.7 m
1 -Loading Slab Panel 7.7 x 2.3 m and 0.1 mTHK (Slab on one side of the beam)
Dead Load
Slab 24 kN/m³ x 2.08 m³ / 7.7 = 6.48 kN/m
Beam 25 kN/m³ x 1.05 m³ / 7.7 = 3.4 kN/m
Other 1 kN/m² x 2.25 m = 2.25 kN/m
Wall 10cm 1.2 kN/m² x 2.25 m = 2.7 kN/m
Wall 20cm 0 kN/m² x 0 m = 0 kN/m
Floor Finis 1.5 kN/m² x 2.25 m = 3.38 kN/m
Total Dead Lo ### kN/m (Trapezoidal Dead load per m lenght
Live Load or Imposed load on one side of the beam)
Floor LL 4 kN/m² x 2.25 m³ / m = 9 kN/m
Total live Load 9 kN/m (Trapezoidal Live load per m lenght
Design Load on one side of the beam)
Design Load1.6 x ### )+( 1.4 x 9.00 ) = 42 kN/m (Trapezoidal load= 41.73 kN/m)
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed ### x 0.73 ) = 30 kN/m
Load on the other side of th 30 - ( ### x 30 / 100 )= 21 kN/m (Reduce 30% for the load on
the other side)
Total Load on both side of t 30 + ### = ### kN/m
2 -Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = ### kN/m
Msup
Mmid
2
3 -Beam Section
h=L
toL
=7700
to7700
» h= 550 mm » h= 400 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5400 » b= 248 mm » b= 200 mm
Cover: C= 25 mm 6 mm ; 20 mm
d= = 400 - 25 - 3 - 10 = 362 mm » d'= 35 mm
ii -Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 255.89 kNm
k=M
=256 x
0.325 > 0.156200 x 131044 x 30
z =d.[0.5+(0.25-(k'/0.9))^(1/ 2)]=362.[0.5+(0.25-(0.132/ 0.9)^(1/ 2))]= 297 mm
that z not exceed 0.95d= 344 mm so z = 344 mm
» [(k'. fcu.b.d^2)/(0.87.fy.z)]+ = [(0.13x30x200x362^2)x10^6) / (0.87x390x343.
3087.5 » Assume 11 T 20 » ###
» P=100(As,req / bd) 4.32 > 0.13 < 4 » OK
b -Define Steel Area at Mid span Mmax= 160 kNm
k=M
=160 x
0.203 > 0.156200 x 131044 x 30
z =d.[0.5+(0.25-(k'/0.9))^(1/ 2)]=.[0.5+(0.25-(0.132/ 0.9)^(1/ 2))]= 297 mm
that z not exceed 0.95d= 344 mm so z = 297 mm
» [(k'. fcu.b.d^2)/(0.87.fy.z)]+ = [(0.13x30x200x362^2) x10^6)/ (0.87x3
2285.9 » Assume 9 T 20 » ###
;Assume Steel Bar ØDR= DT=
h - C - DR/2 - DT/2
106 (Doubly Reinforced
Section)bd2fcu
As =
As = mm2 As= mm2
106 (Doubly Reinforced
Section)bd2fcu
As =
As = mm2 As= mm2
3
» P=100(As,req / bd) 3.53 > 0.13 < 4 » OK
c -Check Deflection
=160 x
= 6.10200 x 1E+05
fs =5
fy =5
x 390 x###
= 197 (Service Stress)8 8 ###
modification factor = 0.55 +477 - fs
= 1120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/depth= 26 x 1 = 25
Actual span/depth = lx / dx = ### / 362 = 21
» »Allowable span/depth= 25 >Actual span/depth= 21 » OK
c -Link
Max Shear at Support
= 168 kN
= / bd = 168x(10^3) / 200x362= 2.32 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.63 x [((100x2826)/(200x362))^(1/3)]x(400/362)^(1/4)x(30/25)^(1/3)
= 1.08
= 2.32 N/mm² > +0.4=1.48 N/mm² (Case 3 in table3.7 in Code BS8110)
Assume 6 mm » Area of link Asv= 55
» Link Spacin = 48 mm » 80 mm
Hence R6@80
Shear at L/4 from the support
= 150 kN
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
Vmax
Vstress Vmax
Vcapacity
Vstress Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Vl/4
4
= 2.07 N/mm²
= 1.08 N/mm² Assume 6 mm » Area of link Asv= 55
» Link Spacin = 60 mm » 180 mm
Hence R6@180
Vstress
Vcap. mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
5
RE-Design Section of Ground GB9A
BS8110 REF. CALCULATION OUTPUT
I/ Design BeamGB9A effective span= 5 m
1 -Loading Slab Panel 5 x 2.5 m and 5 x 2.5 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.12 x 2.5 = 7.2 kN/m 24 kN/m³ x 0.12 x 2.5 = 7.2 kN/m
Beam 24 kN/m³ x 0.35 x 0.2 = 1.68 kN/m
Other 0 kN/m²x 2.5 = 0 kN/m 0 kN/m²x 2.5 = 0 kN/m
Wall 10cm 1.2 kN/m²x 3.3 = 3.96 kN/m 1.2 kN/m²x 3.3 = 3.96 kN/m
Wall 20cm 0 kN/m²x 0 = 0 kN/m 0 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 3.3 = 4.95 kN/m 1.5 kN/m²x 3.3 = 4.95 kN/m
Total Dead Load ### kN/m ### kN/m
Live Load or Imposed load
Floor LL 1.5 kN/m² x 2.5 m = 3.75 kN/m 1.5 kN/m² x 2.5 m = 3.75 kN/m
Total live Load 3.8 kN/m 3.8 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 17.79) + (1.6 x 3.75) = ### kN/m (1.4 x 16.11) + (1.6 x 3. = ### kN/m
Convert Trapezoidal load to uniformly distributed load for bending moment
Uniformly distributed 30.9 x 0.7 = 21 kN/m 28.6 x 0.7 = 19 kN/m
Total uniformly distributed load
20.6 + 19 = 39 kN/m
6
BS8110 REF. CALCULATION OUTPUT
2 -Bending Moment
Max moment at suppo = - ql²/12 = -82.14 kN/m
Max moment at mid s = ql²/24 = 41.07 kN/m
3 -Beam Section
h=L
toL
=5000
to5000
» h= 357 mm » h= 350 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5350 » b= 161 mm » b= 200 mm
Cover: C= 25 mm 6 mm ; 16 mm
d= = 350 - 25 - 3 - 8 = 314 mm d'= 33 mm
4 -Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 82.14 kNm
k=M
=82 x
0.139 < 0.156200 x 98596 x 30
z = d.[0.5+(0.25-(k/0.9))^(1/ 2)]=314.[0.5+(0.25-(0.139/ 0.9)^(1/ 2))]= 254 mm
that z not exceed 0.95d= 298 mm so z = 298 mm
x = ( 314 - 298 )/ 0.45 = 34.89
» Compression Steel Area
» No Steel Require
» Tension Steel Area
» M /0.87 fy.z = [(82x10^6) / (0.87x390x298.3)]
811.59 » Assume 5 T 16 » ###
» P=100(As,req / bd) 1.44 > 0.13 < 4 » OK
Msup
Mmid
;Assume Steel Bar ØDR= DT=
h - C - DR/2 - DT/2
106 (Singly Rienforced
Section)bd2fcu
As =
As = mm2 As= mm2
7
BS8110 REF. CALCULATION OUTPUT
b -Define Steel Area at Mid sp Mmax= 41.07 kNm
k=M
=41 x
0.069 < 0.156200 x 98596 x 30
z = d.[0.5+(0.25-(k/0.9))^(1/ 2)]=.[0.5+(0.25-(0.069/ 0.9)^(1/ 2))]= 288 mm
that z not exceed 0.95d= 298 mm so z = 298 mm
x = ( 0 - 298 )/ 0.45 = -663
» Compression Steel Area
» No Steel Require
» Tension Steel Area
» M /0.87 fy.z = [(41x10^6) / (0.87x390x298.3)]
405.8 » Assume 3 T 16 603
» P=100(As,req / bd) 0.96 > 0.13 < 4 » OK
c -Check Deflection
=41.1 x
= 2.08200 x 98596
fs =5
fy =5
x 390 x406
= 164 (Service Stress)8 8 603
modification factor = 0.55 +477 - fs
= 1.94120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/depth= 26 x 1.94 = 50
Actual span/depth = lx / dx = ### / 314 = 16
» »Allowable span/depth= 50 >Actual span/depth= 16 » OK
106 (Singly Rienforced
Section)bd2fcu
As =
As = mm2 As= mm2
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
8
BS8110 REF. CALCULATION OUTPUT
d -Link
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 30.9 x 0.5 = 15 kN/m 28.6 x 0.5 = 14 kN/m
Total uniformly distributed load
15.5 + 14 = 30 kN/m Total load per m lenght for
bending moment 29.7 kN/m
Max Shear at Support
= ql/2 = 74 kN
= / bd = 1.18 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.83 N/mm²
= 1.18 N/mm² > +0.4=1.23 N/mm²(Case 3 in table3.7 in Code BS8110)
Assume 6 mm » Area of link Asv= 55
» Link Spacin = 170 mm » 120 mm
Hence R6@120
Shear at L/4 from the support
= q(l-0.25l)/2 = 55.7 kN
= 0.89 N/mm²
= 0.83 N/mm² Assume 6 mm » Area of link Asv= 55
» Link Spacin = ### mm » 150 mm
Hence R6@150
Vmax
Vstress Vmax
Vcapacity
Vstress Vcapacity
mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Vl/4
Vstress
Vcap. mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
9
Check Existing Section of Ground Beam GB7 (Along Grid 4 from Grid E-G)
BS8110 REF. CALCULATION OUTPUT
I/ Check ExistiGB7 effective span= 6.6 m
A Loading Slab Panel 6.6 x 3.3 m and 6.6 x 1.1 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.1 x 3.3 = 7.92 kN/m 24 kN/m³ x 0.1 x 1.1 = 2.64 kN/m
Beam 24 kN/m³ x 0.2 x 0.4 = 1.92 kN/m
Roof Load 0 kN/m²x 1.7 = 0 kN/m 0 kN/m²x 0.83 = 0 kN/m
Wall 10cm 1.2 kN/m²x 3.3 = 3.96 kN/m 1.2 kN/m²x 3.3 = 3.96 kN/m
Wall 20cm 0 kN/m²x 0 = 0 kN/m 0 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 3.3 = 4.95 kN/m 1.5 kN/m²x 1.1 = 1.65 kN/m
Total Dead Load ### kN/m 8.3 kN/m
Live Load or Imposed load
Floor LL 4 kN/m² x 3.3 m = 13.2 kN/m 4 kN/m² x 1.1 m = 4.4 kN/m
Roof LL 0 kN/m² x 1.7 m = 0 kN/m 0 kN/m² x 0.83 m = 0 kN/m
Total live Load 13 kN/m 4.4 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 18.75) + (1.6 x 13.2) = ### kN/m (1.4 x 8.25) + (1.6 x 4.4) = ### kN/m
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed 47.4 x 1 = 47 kN/m 18.6 x 1 = 19 kN/m
Total uniformly distributed load
47.4 + 18.6 = 65.96 kN/m
10
BS8110 REF. CALCULATION OUTPUT
B Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = ### kN/m
C -Beam Section
h=L
toL
=6600
to6600
» h= 471.43 mm » h= 400 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5)400 » b= 212.14 mm » b= 200 mm
Cover: C= 25 mm 8 mm ; 20 mm
d= = 400 - 25 - 4 - 10 = 361 mm d'= 35 mm
D-Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 239.43 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 4 T 20 » ### 0 T 25 » 0
Total Steel Area 1884 mm²
» P=100(As,req / bd) 2.36 > 0.13 < 4 » OK
b -Define Steel Area at Mid spanMmax= 119.72 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 4 T 20 » ###
Total Steel Area 1884 mm²
» P=100(As,req / bd) 2.36 > 0.13 < 4 » OK
Msup
Mmid
;Assume Steel Bar Ø DR= DT=
h - C - DR/2 - DT/2
As'= mm2
As= mm2 As'= mm2
As'= mm2
As= mm2
11
BS8110 REF. CALCULATION OUTPUT
c -Link or Stirrup
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 47.4 x 1 = 47 kN/m 18.6 x 1 = 19 kN/m
Total uniformly distributed load
47.4 + 18.6 = 65.96 kN/m
Max Shear at Support
= ql/2 = 218 kN
= / bd = 3.01 N/mm²
Max Shear Stress = 4.38 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.95 N/mm²
+0.4= 1.35 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
6 mm » Area of link Asv= 57
» Link Spacin = 30 mm » 30 mm
Hence R6@30
» Existing Link Ø 6 mm @ 150 mm not enough
Vmax
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
12
BS8110 REF. CALCULATION OUTPUT
Shear at L/4
= = 163 kN
= / bd = 2.26 N/mm²
Max Shear Stress = 1.6 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.95 N/mm²
+0.4= 1.35 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
6 mm » Area of link Asv= 57
» Link Spacin = 47 mm » 50 mm
Hence R6@50
» Existing Link Ø 6 mm @ 150 mm not enough
d -Check Deflection
=120 x
= 4.59200 x 130321
fs =5
fy =2
x 390 x1884
= 260 (Service Stress)8 3 1884
modification factor = 0.55 +477 - fs
= 1120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/dep = 26 x 0.987 = 25.674
Actual span/depth = l /### = 6600 / 250 = 26.4 mm
Vmax qll/4 / 2
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume Ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
13
» »Allowable span/depth= 25.67 <Actual span/depth= 26 » NO
BS8110 REF. CALCULATION OUTPUT
e -Check Cracking
Refer 3.12.11.2.3
1/ CLear distance between horizontal tension bars
Allowable clear distance between bars
= 155 (table 3.28)
=
= 47 mm < 155 OK
Clear Distance between the face of the beam and nearest longitudinal bar in tension
=
y =
= 43 mm
Allowable distance
= 155x0.5 = 77.5 mm
» = 50.81 mm < 77.5 bar should be provided in side faces of beam to control cracking
Distance between bars
S1 (b-2(cover)-2(Ølinks)-2(Øbar))/2
S2 ((y2+y2)^(1/2))-(Øbar/2)
cover+Ølink+(Øbar/2)
S2
14
Sb ≤ 250 mm
Minimum size bar
Ø >
> 11.32 mm Provide DB12 @ 250
((Sbb/fy)^(1/2))
15
Check Existing Section of Ground Beam GB9 (Along Grid E from Grid 7'-9')
BS8110 REF. CALCULATION OUTPUT
I/ Check ExistiGB9 effective span= 7.7 m
A Loading Slab Panel 7.7 x 2.2 m and 7.7 x 1.7 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.1 x 2.2 = 5.28 kN/m 24 kN/m³ x 0.1 x 1.65 = 3.96 kN/m
Beam 24 kN/m³ x 0.2 x 0.4 = 1.92 kN/m
Roof Load 0 kN/m²x 1.7 = 0 kN/m 0 kN/m²x 0.83 = 0 kN/m
Wall 10cm 1.2 kN/m²x 3.3 = 3.96 kN/m 1.2 kN/m²x 3.3 = 3.96 kN/m
Wall 20cm 0 kN/m²x 0 = 0 kN/m 0 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 2.2 = 3.3 kN/m 1.5 kN/m²x 1.65 = 2.47 kN/m
Total Dead Load ### kN/m ### kN/m
Live Load or Imposed load
Floor LL 4 kN/m² x 2.2 m = 8.8 kN/m 4 kN/m² x 1.65 m = 6.6 kN/m
Roof LL 0 kN/m² x 1.7 m = 0 kN/m 0 kN/m² x 0.83 m = 0 kN/m
Total live Load 8.8 kN/m 6.6 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 14.46) + (1.6 x 8.8) = ### kN/m (1.4 x 10.4) + (1.6 x 6.6) = ### kN/m
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed 34.3 x 1 = 34 kN/m 25.1 x 1 = 25 kN/m
Total uniformly distributed load
34.3 + 25.1 = 59.44 kN/m
16
BS8110 REF. CALCULATION OUTPUT
B Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = ### kN/m
C -Beam Section
h=L
toL
=7700
to7700
» h= 550 mm » h= 400 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5)400 » b= 247.5 mm » b= 200 mm
Cover: C= 25 mm 8 mm ; 20 mm
d= = 400 - 25 - 4 - 10 = 361 mm d'= 35 mm
D-Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 293.67 kNm
» Compression Steel Area
Existing Steel Area 2 T 25 » 981
» Tension Steel Area
Existing Steel Area 2 T 20 » 628 3 T 25 » ###
Total Steel Area 3081 mm²
» P=100(As,req / bd) 3.85 > 0.13 < 4 » OK
b -Define Steel Area at Mid spanMmax= 146.83 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 6 T 25 » ###
Total Steel Area 3572 mm²
Msup
Mmid
;Assume Steel Bar Ø DR= DT=
h - C - DR/2 - DT/2
As'= mm2
As= mm2 As'= mm2
As'= mm2
As= mm2
17
» P=100(As,req / bd) 4.46 > 0.13 < 4 » OK
BS8110 REF. CALCULATION OUTPUT
c -Link or Stirrup
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 34.3 x 1 = 34 kN/m 25.1 x 1 = 25 kN/m
Total uniformly distributed load
34.3 + 25.1 = 59.44 kN/m
Max Shear at Support
= ql/2 = 229 kN
= / bd = 3.17 N/mm²
Max Shear Stress = 4.38 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 1.11 N/mm²
+0.4= 1.51 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
6 mm » Area of link Asv= 57
» Link Spacin = 30 mm » 30 mm
Hence R6@30
» Existing Link Ø 6 mm @ 150 mm not enough
Vmax
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
18
BS8110 REF. CALCULATION OUTPUT
Shear at L/4
= = 172 kN
= / bd = 2.38 N/mm²
Max Shear Stress = 2 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 1.11 N/mm²
+0.4= 1.51 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
6 mm » Area of link Asv= 57
» Link Spacin = 49 mm » 50 mm
Hence R6@50
» Existing Link Ø 6 mm @ 150 mm not enough
d -Check Deflection
=147 x
= 5.63200 x 130321
fs =5
fy =2
x 390 x3572
= 260 (Service Stress)8 3 3571.8
modification factor = 0.55 +477 - fs
= 0.91120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Vmax qll/4 / 2
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume Ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
19
Allowable span/dep = 26 x 0.907 = 23.573
Actual span/depth = l /### = 7700 / 250 = 30.8 mm
» »Allowable span/depth= 23.57 <Actual span/depth= 31 » NO
BS8110 REF. CALCULATION OUTPUT
e -Check Cracking
Refer 3.12.11.2.3
1/ CLear distance between horizontal tension bars
Allowable clear distance between bars
= 155 (table 3.28)
=
= 42 mm < 155 OK
Clear Distance between the face of the beam and nearest longitudinal bar in tension
=
y =
= 45.5 mm
Allowable distance
= 155x0.5 = 77.5 mm
» = 51.85 mm < 77.5 bar should be provided in side faces of beam to control cracking
Distance between bars
S1 (b-2(cover)-2(Ølinks)-2(Øbar))/2
S2 ((y2+y2)^(1/2))-(Øbar/2)
cover+Ølink+(Øbar/2)
S2
20
Sb ≤ 250 mm
Minimum size bar Ø >
> 11.323 mm
Provide DB12 @ 250
((Sbb/fy)^(1/2))
21
Check Existing Section of First Floor Beam B2a (Along Grid E from Grid 7'-9')
BS8110 REF. CALCULATION OUTPUT
I/ Check ExistiB2a effective span= 7.7 m
A Loading Slab Panel 7.7 x 2.2 m and 7.7 x 1.7 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.12 x 2.2 = 6.34 kN/m 24 kN/m³ x 0.12 x 1.65 = 4.75 kN/m
Beam 24 kN/m³ x 0.25 x 0.6 = 3.6 kN/m
Roof Load 0 kN/m²x 1.7 = 0 kN/m 0 kN/m²x 0.83 = 0 kN/m
Wall 10cm 1.2 kN/m²x 3.3 = 3.96 kN/m 1.2 kN/m²x 3.3 = 3.96 kN/m
Wall 20cm 0 kN/m²x 0 = 0 kN/m 0 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 2.2 = 3.3 kN/m 1.5 kN/m²x 1.65 = 2.47 kN/m
Total Dead Load ### kN/m ### kN/m
Live Load or Imposed load
Floor LL 4 kN/m² x 2.2 m = 8.8 kN/m 4 kN/m² x 1.65 m = 6.6 kN/m
Roof LL 0 kN/m² x 1.7 m = 0 kN/m 0 kN/m² x 0.83 m = 0 kN/m
Total live Load 8.8 kN/m 6.6 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 17.2) + (1.6 x 8.8) = ### kN/m (1.4 x 11.19) + (1.6 x 6.6) = ### kN/m
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed 38.2 x 1 = 38 kN/m 26.2 x 1 = 26 kN/m
Total uniformly distributed load
38.2 + 26.2 = 64.38 kN/m
22
BS8110 REF. CALCULATION OUTPUT
B Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = ### kN/m
C -Beam Section
h=L
toL
=7700
to7700
» h= 550 mm » h= 600 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5)600 » b= 247.5 mm » b= 250 mm
Cover: C= 25 mm 8 mm ; 20 mm
d= = 600 - 25 - 4 - 10 = 561 mm d'= 35 mm
D-Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 318.07 kNm
» Compression Steel Area
Existing Steel Area 2 T 25 » 981
» Tension Steel Area
Existing Steel Area 2 T 20 » 628 3 T 25 » ###
Total Steel Area 3081 mm²
» P=100(As,req / bd) 2.05 > 0.13 < 4 » OK
b -Define Steel Area at Mid spanMmax= 159.04 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 6 T 25 » ###
Total Steel Area 3572 mm²
Msup
Mmid
;Assume Steel Bar Ø DR= DT=
h - C - DR/2 - DT/2
As'= mm2
As= mm2 As'= mm2
As'= mm2
As= mm2
23
» P=100(As,req / bd) 2.38 > 0.13 < 4 » OK
BS8110 REF. CALCULATION OUTPUT
c -Link or Stirrup
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 38.2 x 1 = 38 kN/m 26.2 x 1 = 26 kN/m
Total uniformly distributed load
38.2 + 26.2 = 64.38 kN/m
Max Shear at Support
= ql/2 = 248 kN
= / bd = 1.77 N/mm²
Max Shear Stress = 4.38 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.80 N/mm²
+0.4= 1.20 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 90 mm » 90 mm
Hence R8@90
» Existing Link Ø 8 mm @ 125 mm not enough
Vmax
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
24
Shear at L/4
= = 186 kN
= / bd = 1.33 N/mm²
Max Shear Stress = 2 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.80 N/mm²
+0.4= 1.20 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 166 mm » 160 mm
Hence R8@160
» Existing Link Ø 8 mm @ 125 mm safe enough
d -Check Deflection
=159 x
= 2.02250 x 314721
fs =5
fy =2
x 390 x3572
= 260 (Service Stress)8 3 3571.8
modification factor = 0.55 +477 - fs
= 1.54120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/dep = 26 x 1.544 = 40.145
Actual span/depth = l /### = 7700 / 250 = 30.8 mm
» »Allowable span/depth= 40.15 >Actual span/depth= 31 » OK
Vmax qll/4 / 2
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume Ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
25
e -Check Cracking
Refer 3.12.11.2.3
1/ CLear distance between horizontal tension bars
Allowable clear distance between bars
= 155 (table 3.28)
=
= 67 mm < 155 OK
Clear Distance between the face of the beam and nearest longitudinal bar in tension
=
y =
= 45.5 mm
Allowable distance
= 155x0.5 = 77.5 mm
» = 51.85 mm < 77.5 bar should be provided in side faces of beam to control cracking
Distance between bars
Sb ≤ 250 mm
S1 (b-2(cover)-2(Ølinks)-2(Øbar))/2
S2 ((y2+y2)^(1/2))-(Øbar/2)
cover+Ølink+(Øbar/2)
S2
26
Minimum size bar
Ø >
> 12.66 mm
Provide DB12 @ 250
((Sbb/fy)^(1/2))
27
Check Existing Section of Roof Floor Beam RB1A (Along Grid H from Grid 2-4)
BS8110 REF. CALCULATION OUTPUT
I/ Check ExistiRB1A effective span= 6.6 m
A Loading Slab Panel 6.6 x 1.65 m and 6.6 x 1.7 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.12 x 1.65 = 4.75 kN/m 24 kN/m³ x 0.12 x 1.65 = 4.75 kN/m
Beam 24 kN/m³ x 0.2 x 0.4 = 1.92 kN/m
Roof Load 1.5 kN/m²x 1.7 = 2.55 kN/m 1.5 kN/m²x 0.83 = 1.24 kN/m
Wall 10cm 0 kN/m²x 3.3 = 0 kN/m 0 kN/m²x 3.3 = 0 kN/m
Wall 20cm 0 kN/m²x 0 = 0 kN/m 0 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 1.65 = 2.47 kN/m 1.5 kN/m²x 1.65 = 2.47 kN/m
Total Dead Load ### kN/m 8.5 kN/m
Live Load or Imposed load
Floor LL 1.5 kN/m² x 1.65 m = 2.47 kN/m 1.5 kN/m² x 1.65 m = 2.47 kN/m
Roof LL 1.5 kN/m² x 1.7 m = 2.55 kN/m 1.5 kN/m² x 0.83 m = 1.24 kN/m
Total live Load 5 kN/m 3.7 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 11.7) + (1.6 x 5.03) = ### kN/m (1.4 x 8.47) + (1.6 x 3.72) = ### kN/m
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed 24.4 x 1 = 24 kN/m 17.8 x 1 = 18 kN/m
Total uniformly distributed load
24.4 + 17.8 = 42.23 kN/m
28
BS8110 REF. CALCULATION OUTPUT
B Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = 76.64 kN/m
C -Beam Section
h=L
toL
=6600
to6600
» h= 471.43 mm » h= 400 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5)400 » b= 212.14 mm » b= 200 mm
Cover: C= 25 mm 8 mm ; 20 mm
d= = 400 - 25 - 4 - 10 = 361 mm d'= 35 mm
D-Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 153.29 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 2 T 20 » 628 2 T 16 » 402
Total Steel Area 1658 mm²
» P=100(As,req / bd) 2.07 > 0.13 < 4 » OK
b -Define Steel Area at Mid spanMmax= 76.64 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 4 T 20 » ###
Total Steel Area 1884 mm²
Msup
Mmid
;Assume Steel Bar Ø DR= DT=
h - C - DR/2 - DT/2
As'= mm2
As= mm2 As'= mm2
As'= mm2
As= mm2
29
» P=100(As,req / bd) 2.36 > 0.13 < 4 » OK
BS8110 REF. CALCULATION OUTPUT
c -Link or Stirrup
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 24.4 x 1 = 24 kN/m 17.8 x 1 = 18 kN/m
Total uniformly distributed load
24.4 + 17.8 = 42.23 kN/m
Max Shear at Support
= ql/2 = 139 kN
= / bd = 1.93 N/mm²
Max Shear Stress = 4.38 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.91 N/mm²
+0.4= 1.31 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 107 mm » 100 mm
Hence R8@100
» Existing Link Ø 8 mm @ 125 mm not enough
Vmax
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
30
Shear at L/4
= = 105 kN
= / bd = 1.45 N/mm²
Max Shear Stress = 1.6 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.91 N/mm²
+0.4= 1.31 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 202 mm » 200 mm
Hence R8@200
» Existing Link Ø 8 mm @ 125 mm safe enough
d -Check Deflection
=76.6 x
= 2.94200 x 130321
fs =5
fy =2
x 390 x1884
= 260 (Service Stress)8 3 1884
modification factor = 0.55 +477 - fs
= 1.23120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/dep = 26 x 1.233 = 32.065
Actual span/depth = l /### = 6600 / 250 = 26.4 mm
» »Allowable span/depth= 32.07 >Actual span/depth= 26 » OK
Vmax qll/4 / 2
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume Ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
31
e -Check Cracking
Refer 3.12.11.2.3 BS8110
1/ CLear distance between horizontal tension bars
Allowable clear distance between bars
= 155 (table 3.28)
=
= 47 mm < 155 OK
Clear Distance between the face of the beam and nearest longitudinal bar in tension
=
y =
= 43 mm
Allowable distance
= 155x0.5 = 77.5 mm
» = 50.81 mm < 77.5 bar should be provided in side faces of beam to control cracking
Distance between bars
Sb ≤ 250 mm
S1 (b-2(cover)-2(Ølinks)-2(Øbar))/2
S2 ((y2+y2)^(1/2))-(Øbar/2)
cover+Ølink+(Øbar/2)
S2
32
Minimum size bar
Ø >
> 11.32 mm
Provide DB12 @ 250
((Sbb/fy)^(1/2))
33
Check Existing Section of Roof Floor Beam RB15B (Along Grid E from Grid 1-5)
BS8110 REF. CALCULATION OUTPUT
I/ Check ExistiRB15B effective span= 9.9 m
A Loading Slab Panel 9.9 x 2.2 m and 9.9 x 1.7 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.12 x 2.2 = 6.34 kN/m 24 kN/m³ x 0.12 x 1.65 = 4.75 kN/m
Beam 24 kN/m³ x 0.25 x 0.6 = 3.6 kN/m
Roof Load 1.5 kN/m²x 1.7 = 2.55 kN/m 1.5 kN/m²x 0.83 = 1.24 kN/m
Wall 10cm 0 kN/m²x 3.3 = 0 kN/m 0 kN/m²x 3.3 = 0 kN/m
Wall 20cm 0 kN/m²x 0 = 0 kN/m 0 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 2.2 = 3.3 kN/m 1.5 kN/m²x 1.65 = 2.47 kN/m
Total Dead Load ### kN/m 8.5 kN/m
Live Load or Imposed load
Floor LL 1.5 kN/m² x 2.2 m = 3.3 kN/m 1.5 kN/m² x 1.65 m = 2.47 kN/m
Roof LL 1.5 kN/m² x 1.7 m = 2.55 kN/m 1.5 kN/m² x 0.83 m = 1.24 kN/m
Total live Load 5.9 kN/m 3.7 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 15.79) + (1.6 x 5.85) = ### kN/m (1.4 x 8.47) + (1.6 x 3.72) = ### kN/m
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed 31.5 x 1 = 31 kN/m 17.8 x 1 = 18 kN/m
Total uniformly distributed load
31.5 + 17.8 = 49.27 kN/m
34
BS8110 REF. CALCULATION OUTPUT
B Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = ### kN/m
C -Beam Section
h=L
toL
=9900
to9900
» h= 707.14 mm » h= 600 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5)600 » b= 318.21 mm » b= 250 mm
Cover: C= 25 mm 8 mm ; 20 mm
d= = 600 - 25 - 4 - 10 = 561 mm d'= 35 mm
D-Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 402.44 kNm
» Compression Steel Area
Existing Steel Area 2 T 25 » 981
» Tension Steel Area
Existing Steel Area 2 T 20 » 628 3 T 25 » ###
Total Steel Area 3081 mm²
» P=100(As,req / bd) 2.05 > 0.13 < 4 » OK
b -Define Steel Area at Mid spanMmax= 201.22 kNm
» Compression Steel Area
Existing Steel Area 2 T 20 » 628
» Tension Steel Area
Existing Steel Area 6 T 25 » ###
Total Steel Area 3572 mm²
Msup
Mmid
;Assume Steel Bar Ø DR= DT=
h - C - DR/2 - DT/2
As'= mm2
As= mm2 As'= mm2
As'= mm2
As= mm2
35
» P=100(As,req / bd) 2.38 > 0.13 < 4 » OK
BS8110 REF. CALCULATION OUTPUT
c -Link or Stirrup
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 31.5 x 1 = 31 kN/m 17.8 x 1 = 18 kN/m
Total uniformly distributed load
31.5 + 17.8 = 49.27 kN/m
Max Shear at Support
= ql/2 = 244 kN
= / bd = 1.74 N/mm²
Max Shear Stress = 4.38 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.80 N/mm²
+0.4= 1.20 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 93 mm » 90 mm
Hence R8@90
» Existing Link Ø 8 mm @ 125 mm not enough
Vmax
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
36
Shear at L/4
= = 183 kN
= / bd = 1.3 N/mm²
Max Shear Stress = 2 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 0.80 N/mm²
+0.4= 1.20 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 173 mm » 170 mm
Hence R8@170
» Existing Link Ø 8 mm @ 125 mm safe enough
d -Check Deflection
=201 x
= 2.56250 x 314721
fs =5
fy =2
x 390 x3572
= 260 (Service Stress)8 3 3571.8
modification factor = 0.55 +477 - fs
= 1.34120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/dep = 26 x 1.336 = 34.727
Actual span/depth = l /### = 9900 / 250 = 39.6 mm
» »Allowable span/depth= 34.73 <Actual span/depth= 40 » NO
Vmax qll/4 / 2
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume Ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
37
e -Check Cracking
Refer 3.12.11.2.3
1/ CLear distance between horizontal tension bars
Allowable clear distance between bars
= 155 (table 3.28)
=
= 67 mm < 155 OK
Clear Distance between the face of the beam and nearest longitudinal bar in tension
=
y =
= 45.5 mm
Allowable distance
= 155x0.5 = 77.5 mm
» = 51.85 mm < 77.5 bar should be provided in side faces of beam to control cracking
Distance between bars
Sb ≤ 250 mm
S1 (b-2(cover)-2(Ølinks)-2(Øbar))/2
S2 ((y2+y2)^(1/2))-(Øbar/2)
cover+Ølink+(Øbar/2)
S2
38
Minimum size bar
Ø >
> 12.66 mm
Provide DB12 @ 250
((Sbb/fy)^(1/2))
39
Check Existing Section of First Floor Beam B1 (Along Grid G from Grid 13-15)
BS8110 REF. CALCULATION OUTPUT
I/ Check ExistiB1 effective span= 6.6 m
A Loading Slab Panel 6.6 x 2.2 m and 6.6 x 0 m and 0.1 mTHK
Dead Load
Left Side of the beam Right Side of the beam
Slab 24 kN/m³ x 0.12 x 2.2 = 6.34 kN/m 24 kN/m³ x 0.12 x 0 = 0 kN/m
Beam 24 kN/m³ x 0.2 x 0.3 = 1.44 kN/m
Roof Load 0 kN/m²x 1.7 = 0 kN/m 0 kN/m²x 0.83 = 0 kN/m
Wall 10cm 0 kN/m²x 3.3 = 0 kN/m 0 kN/m²x 3.3 = 0 kN/m
Wall 20cm 2.3 kN/m²x 4.3 = 9.89 kN/m 2.3 kN/m²x 0 = 0 kN/m
Floor Finish 1.5 kN/m²x 2.2 = 3.3 kN/m 1.5 kN/m²x 0 = 0 kN/m
Total Dead Load ### kN/m 0.0 kN/m
Live Load or Imposed load
Floor LL 4 kN/m² x 2.2 m = 8.8 kN/m 4 kN/m² x 0 m = 0 kN/m
Roof LL 0 kN/m² x 1.7 m = 0 kN/m 0 kN/m² x 0.83 m = 0 kN/m
Total live Load 8.8 kN/m 0 kN/m
Triangle or Trapezoidal Design Load
(1.4 x 20.97) + (1.6 x 8.8) = ### kN/m (1.4 x 0) + (1.6 x 0) = 0.0 kN/m
Convert Trapezoidal load to uniformly distributed load
Uniformly distributed 43.4 x 1 = 43 kN/m 0.0 x 1 = 0 kN/m
Total uniformly distributed load
43.4 + 0 = 43.43 kN/m
40
BS8110 REF. CALCULATION OUTPUT
B Bending Moment
Max moment at suppo = - ql²/12 = ### kN/m
Max moment at mid s = ql²/24 = 78.83 kN/m
C -Beam Section
h=L
toL
=6600
to6600
» h= 471.43 mm » h= 300 mm10 15 10 15
b=(0.4 to 0.5)h= (0.4 to 0.5)300 » b= 212.14 mm » b= 200 mm
Cover: C= 25 mm 8 mm ; 20 mm
d= = 300 - 25 - 4 - 10 = 261 mm d'= 35 mm
D-Define Steel Area
Assume fcu= 30 N/mm² ; fy v= 250 N/mm² fy = 390 N/mm²
a -Define Steel Area at Support Mmax= 157.66 kNm
» Compression Steel Area
Existing Steel Area 2 T 16 » 402
» Tension Steel Area
Existing Steel Area 4 T 20 » ### 0 T 25 » 0
Total Steel Area 1658 mm²
» P=100(As,req / bd) 2.76 > 0.13 < 4 » OK
b -Define Steel Area at Mid spanMmax= 78.83 kNm
» Compression Steel Area
Existing Steel Area 2 T 16 » 628
» Tension Steel Area
Existing Steel Area 3 T 16 » 603
Total Steel Area 1231 mm²
Msup
Mmid
;Assume Steel Bar Ø DR= DT=
h - C - DR/2 - DT/2
As'= mm2
As= mm2 As'= mm2
As'= mm2
As= mm2
41
» P=100(As,req / bd) 2.05 > 0.13 < 4 » OK
BS8110 REF. CALCULATION OUTPUT
c -Link or Stirrup
Convert Trapezoidal load to uniformly distributed load for bending moment
Left Side of the beam Right Side of the beam
Uniformly distributed 43.4 x 1 = 43 kN/m 0.0 x 1 = 0 kN/m
Total uniformly distributed load
43.4 + 0 = 43.43 kN/m
Max Shear at Support
= ql/2 = 143 kN
= / bd = 2.75 N/mm²
Max Shear Stress = 4.38 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 1.09 N/mm²
+0.4= 1.49 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 66 mm » 70 mm
Hence R8@70
» Existing Link Ø 8 mm @ 125 mm not enough
Vmax
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
42
Shear at L/4
= = 107 kN
= / bd = 2.06 N/mm²
Max Shear Stress = 1.39 N/mm²
= (0.79/1.25) [(100As/bd)^(1/3)] [(400/d)^(1/4)] [(fcu/25)^(1/3)]
= 1.09 N/mm²
+0.4= 1.49 N/mm²
» Vc+0.4 Vs < 0.8(fcu^(1/2))
8 mm » Area of link Asv= 101
» Link Spacin = 113 mm » 100 mm
Hence R8@100
» Existing Link Ø 8 mm @ 125 mm not enough
d -Check Deflection
=78.8 x
= 5.79200 x 68121
fs =5
fy =2
x 390 x1231
= 260 (Service Stress)8 3 1230.9
modification factor = 0.55 +477 - fs
= 0.9120
Bassic span/depth = 26 ( table3.9 in Code BS8110 Part 1 1985)
Allowable span/dep = 26 x 0.897 = 23.329
Actual span/depth = l /### = 6600 / 250 = 26.4 mm
» »Allowable span/depth= 23.33 <Actual span/depth= 26 » NO
Vmax qll/4 / 2
Vstress Vmax
0.8(fcu^(1/2))
Vcapacity
Vcapacity
Asv≥(bv Sv(vs-vc))/0.87fy Assume Ø mm2
Sv≤(Asv x 0.87fy)/(bv(vs-vc))
Mmax 106
bd2
As,req
As,pro
(0.9+(M / bd2))
43
e -Check Cracking
Refer 3.12.11.2.3
1/ CLear distance between horizontal tension bars
Allowable clear distance between bars
= 155 (table 3.28)
=
= 51 mm < 155 OK
Clear Distance between the face of the beam and nearest longitudinal bar in tension
=
y =
= 41 mm
Allowable distance
= 155x0.5 = 77.5 mm
» = 50 mm < 77.5 bar should be provided in side faces of beam to control cracking
Distance between bars
Sb ≤ 250 mm
S1 (b-2(cover)-2(Ølinks)-2(Øbar))/2
S2 ((y2+y2)^(1/2))-(Øbar/2)
cover+Ølink+(Øbar/2)
S2
44
Minimum size bar
Ø >
> 11.32 mm
Provide DB12 @ 250
((Sbb/fy)^(1/2))