2 Way Slab Design-roof Slab 611n54
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 3i3n4
Overview 26281t
& View 2 Way Slab Design-roof Slab as PDF for free.
More details 6y5l6z
- Words: 6,850
- Pages: 120
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
7214 4775 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
118 mm 4475 mm 4593 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 118 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
153 mm 150 mm 115 105 4590 7019 1.529
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.529 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.076 0.047 0.057 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 4475 mm
19.946 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.076 0.057
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 15.01 11.26
My (kNm/m) 9.28 6.91
573.75 mm 877.375 mm
Bar dia (mm) 10 10
Reinforcement Provided Spacing (mm) 220 250
Bar dia (mm) 10 10
Reinforcement Provided Spacing (mm) 250 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 314.16 mm2 Extension of reinforcement over = lx/5 =
918 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
235.62 mm2 8 mm 4.6875 5 229.5 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
117.81 mm2 6 mm 4.16666667 5 229.5 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
4.775 m 7.214 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 7.214
A
lx = 4.775
Beam 2-3 2.3875 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 357.00 OK 314.16 OK
Reinforcement Provided Area Remarks (mm2) 314.16 OK 314.16 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
1.135 0.851
0.28 0.20
316.98 234.16
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.842 0.627
0.20 0.15
211.37 155.68
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
4.775 (l)
max intensity of D DL =
22.38
Max. Moment =
42.53 4.25 26.72 2.67
Max. Shear or Rxtn =
max intensity of D DL =
22.38
2.439 (y)
2.3875
max Intensity of UDL =
22.38
Max. Moment =
124.34 12.43 54.02 5.40
(x) Max. Shear or Rxtn =
for fst
190 lower
pt kt =
0.2 0
for pt = kt =
0.27 1.323
for fst = kt =
216.16 1.43
SHORT SPAN pt provided 0.31 0.27
pt provided 0.30 0.30
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.181 0.404 0.181
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.181 0.377 0.181
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 216.16
kt only for mid span 1.67
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
190 higher 0.3 1.89 0.27 1.323
lower
240 higher 0.2 0.3 1.68 1.47 0.27 1.533
216.16 1.43
Check for Deflection Control prov < l/d l/d max provided max only for mid span 43.42 39.91304 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
4775 3607 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
88 mm 3307 mm 3395 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 88 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
123 mm 150 mm 115 105 3422 4580 1.338
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.338 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.067 0.047 0.050 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3307 mm
14.471 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.067 0.050
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 7.36 5.49
My (kNm/m) 5.16 3.84
427.75 mm 572.5 mm
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
684.4 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 171.1 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 171.1 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.607 m 4.775 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 4.775
A
lx = 3.607
Beam 2-3 1.8035 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.556 0.415
0.13 0.10
150.71 111.69
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.468 0.349
0.11 0.08
115.28 85.35
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.607 (l)
max intensity of D DL =
16.91
Max. Moment =
18.33 1.83 15.25 1.52
Max. Shear or Rxtn =
max intensity of D DL =
16.91
1.168 (y)
1.8035 (x)
max Intensity of UDL =
16.91
Max. Moment =
39.02 3.90 25.12 2.51
Max. Shear or Rxtn =
for fst
190 lower
pt kt =
0.2 0
for pt = kt =
0.21 0.189
for fst = kt =
225 1.22
SHORT SPAN pt provided 0.17 0.17
pt provided 0.19 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.132 0.404 0.132
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.132 0.377 0.132
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.22
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
190 higher 0.3 1.89 0.21 0.189
lower
240 higher 0.2 0.3 1.68 1.47 0.21 1.659
225 1.22
Check for Deflection Control prov < l/d l/d max provided max only for mid span 31.72 29.75652 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
7214 4216 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
104 mm 3916 mm 4020 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 104 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
139 mm 150 mm 115 105 4031 7019 1.741
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.741 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.064 0.037 0.048 0.028
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3916 mm
17.325 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.064 0.048
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.037 0.028
Reinforcement extension over s short span = long span =
Mx (kNm/m) 9.75 7.31
My (kNm/m) 5.64 4.27
503.875 mm 877.375 mm
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 220 250
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
0.125
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
806.2 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 4 268.733333 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 4 268.733333 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
4.216 m 7.214 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 7.214
A
lx = 4.216
Beam 2-3 2.108 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 228.48 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.737 0.553
0.18 0.13
201.59 149.79
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.511 0.387
0.12 0.09
126.20 94.92
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
4.216 (l)
max intensity of D DL =
19.76
Max. Moment =
29.27 2.93 20.83 2.08
Max. Shear or Rxtn =
max intensity of D DL =
19.76
2.998 (y)
2.108 (x)
max Intensity of UDL =
19.76
Max. Moment =
113.92 11.39 50.45 5.05
Max. Shear or Rxtn =
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.20 0.17
pt provided 0.19 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.158 0.404 0.158
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.158 0.377 0.158
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.5
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 39 35.05217 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
4216 3607 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
88 mm 3307 mm 3395 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 88 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
123 mm 150 mm 115 105 3422 4021 1.175
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.175 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.047 0.037 0.035 0.028
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3307 mm
14.471 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.037 0.028
Reinforcement extension over s short span = long span =
Mx (kNm/m) 5.16 3.84
My (kNm/m) 4.06 3.07
427.75 mm 502.625 mm
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 250 250
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
0.125
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
684.4 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 171.1 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 171.1 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.607 m 4.216 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 4.216
A
lx = 3.607
Beam 2-3
1.8035 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 314.16 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.390 0.291
0.09 0.07
104.86 77.71
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.368 0.279
0.09 0.06
90.32 68.05
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.607 (l)
max intensity of D DL =
16.91
Max. Moment =
18.33 1.83 15.25 1.52
Max. Shear or Rxtn =
max intensity of D DL = max Intensity of UDL =
16.91 16.91
0.609 (y)
1.8035
Max. Moment = Max. Shear or Rxtn =
28.40 2.84 20.40 2.04
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.27 0.17
pt provided 0.19 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.132 0.404 0.132
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.132 0.377 0.132
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.53
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl /12)
kN/m kN/m
(w1) (w2)
2
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
kNm t-m kN t
(w1x2/3 + w2y/2*(x+y/4)) (0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 39.78 29.75652 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
7214 3886 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
95 mm 3586 mm 3681 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 95 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
130 mm 150 mm 115 105 3701 7019 1.897
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.897 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.088 0.047 0.067 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3586 mm
15.779 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.088 0.067
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 11.30 8.60
My (kNm/m) 6.04 4.49
462.625 mm 877.375 mm
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
740.2 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 185.05 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 185.05 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.886 m 7.214 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 7.214
A
lx = 3.886
Beam 2-3 1.943 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.854 0.651
0.20 0.15
235.08 177.12
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.547 0.408
0.13 0.10
135.38 100.12
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.886 (l)
max intensity of D DL =
18.22
Max. Moment =
22.92 2.29 17.70 1.77
Max. Shear or Rxtn =
max intensity of D DL =
18.22
3.328 (y)
1.943 (x)
max Intensity of UDL =
18.22
Max. Moment =
107.04 10.70 48.01 4.80
Max. Shear or Rxtn =
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.23 0.17
pt provided 0.25 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.143 0.404 0.143
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.143 0.377 0.143
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.48
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 38.48 32.18261 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
3886 3607 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
88 mm 3307 mm 3395 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 88 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
123 mm 150 mm 115 105 3422 3691 1.079
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.079 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.052 0.047 0.039 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3307 mm
14.471 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.052 0.039
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 5.71 4.28
My (kNm/m) 5.16 3.84
427.75 mm 461.375 mm
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
684.4 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 171.1 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 171.1 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.607 m 3.886 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 3.886
A
lx = 3.607
Beam 2-3 1.8035
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.432 0.324
0.10 0.08
116.25 86.73
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.468 0.349
0.11 0.08
115.28 85.35
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.607 (l)
max intensity of D DL =
16.91
Max. Moment =
18.33 1.83 15.25 1.52
Max. Shear or Rxtn =
max intensity of D DL =
16.91
0.279 (x)
max Intensity of UDL =
16.91
Max. Moment =
22.75 2.27 17.61 1.76
(y) Max. Shear or Rxtn =
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.23 0.17
pt provided 0.25 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.132 0.404 0.132
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.132 0.377 0.132
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.53
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 39.78 29.75652 ok
DATA Span of the beam = Assume width of beam (b) = Assume depth of beam (D) = Assume clear cover = Assume main reinforcement dia (ϕ) = Assume width of columns = Concrete density = Concrete grade = Characteristic Compr. Strength (Fck) = Permissible stress in concrete (Fc) = Steel grade = Characteristic Tensile Strength (Fy) = Permissible stress in steel (Fs) = xu,max/d =
#REF!
Maximum moment acting on the beam =
#REF!
0.15 0.18 25 12 0.6 2.5 M 25 2548.42 1132.631 Fe 500 50968.400 29546.898 0.46
Torsion acting on the beam (if any) =
m m m mm mm m t/m3
(taken from below)
t/m2 t/m2 t/m2 t/m2
t-m 0.000 t-m
Equivalent Moment (Mu,lim) =
SIMPLE BEAM
(M) (T)
#REF!
t-m
a) From deflection criteria (L/d) =
#REF!
m
b) From Limiting Moment =
#REF!
m
max accepted =
#REF!
m
Provided =
0.180 m #REF! 1.20 (generally 1.5-2)
Adequacy of beam section Effective depth calculation
d/b =
cond factor 26
(REQD)
Steel Requirement Area of Main steel required =
256.966 mm2
Provide +
12 ϕ bars @ 3 nos. 0 ϕ bars @ 0 nos. = Pt =
339.292 mm2 1.26
ADEQUATE
Nos. in one layer = Spacing =
3 0.044
m 0.05 m
Say Side face Reinforcement =
NOT REQUIRED
OK
mm2
(on each side)
Provide 10 ϕ bars @ 2 nos. 12 ϕ bars @ 1 nos. =
+
270.177 mm2
CHECK BAR DIA AND NOS.
Development Length (Cl. 26.2.1, IS 456)= f = 12 ss = 230 tbd = 1.4
mm N/mm2 N/mm2
Ld =
M 25
0.49 m
Shear Stress in Concrete Pt = 100 As/b d =
1.257
Max shear acting (Vu) = shear stress acting =
M 25
#REF! #REF!
Design shear resistance (Vuc) =
t t/m2 65.24 t/m2
Design Shear (Vus = Vu - Vuc) =
#REF! #REF!
Assume dia of shear reinf. = legged = Area of steel (Asv) = spacing (sv) =
t/m2 t
4.01 t
Crack Widths (Service Loads) - Limiting value 0.2 mm 0.15 339.292 #REF! 20387359.84 2548419.98 1274209.99
(calculate from table)
8 mm 2 100.53 mm2 200 mm c/c
shear resistance provided =
b= Ast = M= Es = Ec = Ece =
(at d from ) #REF!
m mm2 t-m t/m2 t/m2 t/m2
#REF!
m= Cmin = h=D= d= dia of bar = bar spacing = effective cover =
16.00 25.00 0.18 0.18 12.00 0.05 31.00
t/m2 mm m m mm m mm
N.A at working loads
x^2 + 0.072x - 0.013 = 0
0.072 0.013
x=
0.084
m
Ic =
0.0001
m4
Determination of crack width at a point directly under a bar on tension face x1 = ε1 = a= εm =
0.096 m #REF! 0.18 m #REF!
acr = Wcr =
Cmin = 25.00 mm #REF!
mm
#REF!
Determination of crack width at a point midway between bars on tension face x1 = ε1 = a= εm =
0.096 m #REF! 0.18 m #REF!
acr = Wcr =
33.82 mm #REF!
mm
#REF!
Determination of crack width at a point bottom corner of beam on tension face x1 = ε1 =
0.096 m #REF!
a= εm =
0.18 m #REF!
acr = Wcr =
37.84 mm #REF!
mm
#REF!
Determination of crack width at a point 2/3(d-x) from N.A. x1 = ε1 = a= εm =
0.064 m #REF! 0.15 m #REF!
acr = Wcr =
38.66 mm #REF!
mm
#REF!
Cost of Conc Section rate of steel = total weight = cost of reinf steel =
72274 rs/mt #REF! mt #REF! rs
rate of M25 concrete = total vol = cost of conc =
6159.15 rs/m3 #REF! m3 #REF! rs
total cost =
#REF!
rs
SIMPLE BEAM (taken from below)
cond factor
ADEQUATE
(on each side)
CHECK BAR DIA AND NOS. (Cl. 26.5.1.3, IS 456)
(at d from )
(calculate from table)
DATA Span of the beam = Assume width of beam (b) = Assume depth of beam (D) = Assume clear cover = Assume main reinforcement dia (ϕ) = Assume width of columns = Concrete density = Concrete grade = Characteristic Compr. Strength (Fck) = Permissible stress in concrete (Fc) = Steel grade = Characteristic Tensile Strength (Fy) = Permissible stress in steel (Fs) = xu,max/d =
#REF!
Maximum moment acting on the beam =
#REF!
0.2 0.25 25 16 0.6 2.5 M 25 2548.42 1132.631 Fe 500 50968.400 29546.898 0.46
Torsion acting on the beam (if any) =
m m m mm mm m t/m3
(taken from below)
t/m2 t/m2 t/m2 t/m2
t-m 0.000 t-m
Equivalent Moment (Mu,lim) =
SIMPLE BEAM
(M) (T)
#REF!
t-m
a) From deflection criteria (L/d) =
#REF!
m
b) From Limiting Moment =
#REF!
m
max accepted =
#REF!
m
Provided =
0.250 m #REF! 1.25 (generally 1.5-2)
Adequacy of beam section Effective depth calculation
d/b =
cond factor 26
(REQD)
Steel Requirement Area of Main steel required =
475.862 mm2
Provide +
16 ϕ bars @ 3 nos. 0 ϕ bars @ 0 nos. = Pt =
603.186 mm2 1.21
ADEQUATE
Nos. in one layer = Spacing =
4 0.045 Say
Side face Reinforcement =
m 0.05 m
NOT REQUIRED
OK
mm2
(on each side)
Provide 10 ϕ bars @ 2 nos. 12 ϕ bars @ 1 nos. =
+
270.177 mm2
CHECK BAR DIA AND NOS.
Development Length (Cl. 26.2.1, IS 456)= f = 16 ss = 230 tbd = 1.4 Ld =
mm N/mm2 N/mm2
M 25
0.66 m
Shear Stress in Concrete Pt = 100 As/b d =
1.206
Max shear acting (Vu) = shear stress acting =
1.220
Design shear resistance (Vuc) =
M 25
t 24.400 t/m2 70.1 t/m2
Design Shear (Vus = Vu - Vuc) =
8 mm 2 100.53 mm2 200 mm c/c
shear resistance provided =
5.57 t
Crack Widths (Service Loads) - Limiting value 0.2 mm 0.2 603.186 #REF! 20387359.84 2548419.98 1274209.99
(calculate from table)
-45.700 t/m2 -2.285 t
Assume dia of shear reinf. = legged = Area of steel (Asv) = spacing (sv) =
b= Ast = M= Es = Ec = Ece =
(at d from ) OK
m mm2 t-m t/m2 t/m2 t/m2
OK
m= Cmin = h=D= d= dia of bar = bar spacing = effective cover =
16.00 25.00 0.25 0.25 16.00 0.05 33.00
t/m2 mm m m mm m mm
N.A at working loads
x^2 + 0.096x - 0.024 = 0
0.097 0.024
x=
0.114
m
Ic =
0.0003
m4
Determination of crack width at a point directly under a bar on tension face x1 = ε1 = a= εm =
0.136 m #REF! 0.25 m #REF!
acr = Wcr =
Cmin = 25.00 mm #REF!
mm
#REF!
Determination of crack width at a point midway between bars on tension face x1 = ε1 = a= εm =
0.136 m #REF! 0.25 m #REF!
acr = Wcr =
33.4 mm #REF!
mm
#REF!
Determination of crack width at a point bottom corner of beam on tension face x1 = ε1 =
0.136 m #REF!
a= εm =
0.25 m #REF!
acr = Wcr =
38.67 mm #REF!
mm
#REF!
Determination of crack width at a point 2/3(d-x) from N.A. x1 = ε1 = a= εm =
0.091 m #REF! 0.20 m #REF!
acr = Wcr =
48.12 mm #REF!
mm
#REF!
Cost of Conc Section rate of steel = total weight = cost of reinf steel =
72274 rs/mt #REF! mt #REF! rs
rate of M25 concrete = total vol = cost of conc =
6159.15 rs/m3 #REF! m3 #REF! rs
total cost =
#REF!
rs
SIMPLE BEAM (taken from below)
cond factor
ADEQUATE
(on each side)
CHECK BAR DIA AND NOS. (Cl. 26.5.1.3, IS 456)
(at d from )
(calculate from table)
Slab design (Simply ed) Data Input Dimension of longer span (ly) = Dimension of shorter span (lx) = Nominal cover for steel
4.5 m 3.16 m 30 mm
Imposed load = Surface finish =
2.5 kN/m2 1 kN/m2
Fck = Fy = Unit density of concrete
(IS 875_2)
20 N/mm2 500 N/mm2 25 kN/m3
Depth Calculation (Serviceablity) ly/lx =
1.42 < 2, two way
lx/D =
Min. Required Overall depth D = Provide D =
28 (as per conditions) (use avg of S/S & cont. where applicable for cont & discont. edge) 112.86 mm 120 mm
(min. 150mm)
Design Moments & Shear Dead load of slab = Ultimate Load (w) = Maximum Shear in either direction =
3 kN/m2 9.75 kN/m2 Vu =
w (lx/2) 15.41
kN/m
Mulim = Rlim bd^2 Fe 250 415 500 Value of Rlim to be used = Min. value of shear strength = multiplying factor (k) =
2.68
Rlim (N/mm2) 3.00 2.78 2.68 N/mm2
0.28 N/mm2 1.3
(Table 19, IS 456) (Cl. 40.2.1.1, IS 456)
For
ax
Mx (kNm/m)
Bar dia (mm)
Simply ed on Four Sides
0.099
9.64
8
Reinforcement Provided Spacing (mm) 200
For
ay
My (kNm/m)
Bar dia (mm)
Simply ed on Four Sides
0.051
4.97
8
Reinforcement Provided Spacing (mm) 200
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.16 m 4.5 m 9.75 kN/m2
The loads of triangular segment A will be transferred
Beam 1-2 ly =
A
lx =
Beam 2-3
nt. where applicable for cont & discont. edge)
Reinforcement Provided Area Remarks (mm2) 251.327
OK
SHORT SPAN Effective depth (d) Reqd Provided Remarks (mm) (mm) 59.97
86
OK
Strength (tc) N/mm2 0.364
Shear Adequacy Stress (tv) N/mm2 0.179
Reinforcement Provided Area Remarks (mm2) 251.327
OK
LONG SPAN Effective depth (d) Reqd Provided Remarks (mm) (mm) 43.04
86
OK
Strength (tc) N/mm2 0.364
Shear Adequacy Stress (tv) N/mm2 0.179
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.16 (l)
1.58 (x)
1.34 (y)
max intensity of D DL =
15.41
Max. Moment =
12.82 1.28 12.17 1.22
Max. Shear or Rxtn =
1.58 (x)
max intensity of D DL = max Intensity of UDL =
15.41 13.07
Max. Moment =
29.58 2.96 20.92 2.09
Max. Shear or Rxtn =
Shear Adequacy Remarks OK
Reinforcement Requirement (% of Area Min. Steel steel), pt (mm2) reqd 0.25
215
144
Shear Adequacy Remarks
Reinforcement Requirement (% of Area Min. Steel steel), pt (mm2) reqd
OK
0.25
215
egment B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m kN/m
(w1) (w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
(wl/4)
2
(0.5*(w1x+w2y)
144
7214 4775 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
118 mm 4475 mm 4593 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 118 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
153 mm 150 mm 115 105 4590 7019 1.529
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.529 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.076 0.047 0.057 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 4475 mm
19.946 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.076 0.057
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 15.01 11.26
My (kNm/m) 9.28 6.91
573.75 mm 877.375 mm
Bar dia (mm) 10 10
Reinforcement Provided Spacing (mm) 220 250
Bar dia (mm) 10 10
Reinforcement Provided Spacing (mm) 250 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 314.16 mm2 Extension of reinforcement over = lx/5 =
918 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
235.62 mm2 8 mm 4.6875 5 229.5 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
117.81 mm2 6 mm 4.16666667 5 229.5 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
4.775 m 7.214 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 7.214
A
lx = 4.775
Beam 2-3 2.3875 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 357.00 OK 314.16 OK
Reinforcement Provided Area Remarks (mm2) 314.16 OK 314.16 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
1.135 0.851
0.28 0.20
316.98 234.16
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.842 0.627
0.20 0.15
211.37 155.68
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
4.775 (l)
max intensity of D DL =
22.38
Max. Moment =
42.53 4.25 26.72 2.67
Max. Shear or Rxtn =
max intensity of D DL =
22.38
2.439 (y)
2.3875
max Intensity of UDL =
22.38
Max. Moment =
124.34 12.43 54.02 5.40
(x) Max. Shear or Rxtn =
for fst
190 lower
pt kt =
0.2 0
for pt = kt =
0.27 1.323
for fst = kt =
216.16 1.43
SHORT SPAN pt provided 0.31 0.27
pt provided 0.30 0.30
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.181 0.404 0.181
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.181 0.377 0.181
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 216.16
kt only for mid span 1.67
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
190 higher 0.3 1.89 0.27 1.323
lower
240 higher 0.2 0.3 1.68 1.47 0.27 1.533
216.16 1.43
Check for Deflection Control prov < l/d l/d max provided max only for mid span 43.42 39.91304 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
4775 3607 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
88 mm 3307 mm 3395 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 88 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
123 mm 150 mm 115 105 3422 4580 1.338
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.338 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.067 0.047 0.050 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3307 mm
14.471 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.067 0.050
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 7.36 5.49
My (kNm/m) 5.16 3.84
427.75 mm 572.5 mm
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
684.4 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 171.1 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 171.1 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.607 m 4.775 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 4.775
A
lx = 3.607
Beam 2-3 1.8035 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.556 0.415
0.13 0.10
150.71 111.69
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.468 0.349
0.11 0.08
115.28 85.35
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.607 (l)
max intensity of D DL =
16.91
Max. Moment =
18.33 1.83 15.25 1.52
Max. Shear or Rxtn =
max intensity of D DL =
16.91
1.168 (y)
1.8035 (x)
max Intensity of UDL =
16.91
Max. Moment =
39.02 3.90 25.12 2.51
Max. Shear or Rxtn =
for fst
190 lower
pt kt =
0.2 0
for pt = kt =
0.21 0.189
for fst = kt =
225 1.22
SHORT SPAN pt provided 0.17 0.17
pt provided 0.19 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.132 0.404 0.132
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.132 0.377 0.132
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.22
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
190 higher 0.3 1.89 0.21 0.189
lower
240 higher 0.2 0.3 1.68 1.47 0.21 1.659
225 1.22
Check for Deflection Control prov < l/d l/d max provided max only for mid span 31.72 29.75652 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
7214 4216 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
104 mm 3916 mm 4020 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 104 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
139 mm 150 mm 115 105 4031 7019 1.741
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.741 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.064 0.037 0.048 0.028
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3916 mm
17.325 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.064 0.048
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.037 0.028
Reinforcement extension over s short span = long span =
Mx (kNm/m) 9.75 7.31
My (kNm/m) 5.64 4.27
503.875 mm 877.375 mm
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 220 250
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
0.125
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
806.2 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 4 268.733333 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 4 268.733333 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
4.216 m 7.214 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 7.214
A
lx = 4.216
Beam 2-3 2.108 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 228.48 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.737 0.553
0.18 0.13
201.59 149.79
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.511 0.387
0.12 0.09
126.20 94.92
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
4.216 (l)
max intensity of D DL =
19.76
Max. Moment =
29.27 2.93 20.83 2.08
Max. Shear or Rxtn =
max intensity of D DL =
19.76
2.998 (y)
2.108 (x)
max Intensity of UDL =
19.76
Max. Moment =
113.92 11.39 50.45 5.05
Max. Shear or Rxtn =
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.20 0.17
pt provided 0.19 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.158 0.404 0.158
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.158 0.377 0.158
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.5
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 39 35.05217 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
4216 3607 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
88 mm 3307 mm 3395 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 88 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
123 mm 150 mm 115 105 3422 4021 1.175
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.175 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.047 0.037 0.035 0.028
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3307 mm
14.471 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.037 0.028
Reinforcement extension over s short span = long span =
Mx (kNm/m) 5.16 3.84
My (kNm/m) 4.06 3.07
427.75 mm 502.625 mm
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 250 250
Bar dia (mm) 8 8
Reinforcement Provided Spacing (mm) 250 250
0.125
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
684.4 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 171.1 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 171.1 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.607 m 4.216 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 4.216
A
lx = 3.607
Beam 2-3
1.8035 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 314.16 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 201.06 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.390 0.291
0.09 0.07
104.86 77.71
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.368 0.279
0.09 0.06
90.32 68.05
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.607 (l)
max intensity of D DL =
16.91
Max. Moment =
18.33 1.83 15.25 1.52
Max. Shear or Rxtn =
max intensity of D DL = max Intensity of UDL =
16.91 16.91
0.609 (y)
1.8035
Max. Moment = Max. Shear or Rxtn =
28.40 2.84 20.40 2.04
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.27 0.17
pt provided 0.19 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.132 0.404 0.132
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.132 0.377 0.132
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.53
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl /12)
kN/m kN/m
(w1) (w2)
2
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
kNm t-m kN t
(w1x2/3 + w2y/2*(x+y/4)) (0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 39.78 29.75652 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
7214 3886 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
95 mm 3586 mm 3681 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 95 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
130 mm 150 mm 115 105 3701 7019 1.897
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.897 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.088 0.047 0.067 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3586 mm
15.779 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.088 0.067
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 11.30 8.60
My (kNm/m) 6.04 4.49
462.625 mm 877.375 mm
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
740.2 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 185.05 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 185.05 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.886 m 7.214 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 7.214
A
lx = 3.886
Beam 2-3 1.943 (x)
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.854 0.651
0.20 0.15
235.08 177.12
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.547 0.408
0.13 0.10
135.38 100.12
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.886 (l)
max intensity of D DL =
18.22
Max. Moment =
22.92 2.29 17.70 1.77
Max. Shear or Rxtn =
max intensity of D DL =
18.22
3.328 (y)
1.943 (x)
max Intensity of UDL =
18.22
Max. Moment =
107.04 10.70 48.01 4.80
Max. Shear or Rxtn =
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.23 0.17
pt provided 0.25 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.143 0.404 0.143
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.143 0.377 0.143
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.48
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 38.48 32.18261 ok
2 way Slab design Data Input C/C distance of longer span (Ly) = C/C distance of shorter span (Lx) = Assume widths = Nominal cover = Bar dia = unit density of concrete = Grade of Concrete (Fck) Grade of Steel (Fy) =
3886 3607 300 30 10 25 25 500
mm mm mm mm mm kN/m3 Mpa Mpa
Depth Calculation Assume effective depth (dx) = Clear short span = effective short span =
88 mm 3307 mm 3395 mm
By deflection criteria of beams, span/effective depth ratio = 26 modification factor = 1.5 effective depth = 88 mm Overall depth D = Provide D = Effective depth (dx) = Effective depth (dy) = Effective short span (lx) = Effective long span (ly) = Aspect ratio (r) = ly/lx =
(goal seek for diff = 0)
diff =
123 mm 150 mm 115 105 3422 3691 1.079
mm mm mm mm
Loading Self weight of slab = Imposed Load = Finishes =
3.75 1.5 1 6.25 factored load (x1.5) =
Design Moments & Shear Moment Coefficients
kN/m2 kN/m2 kN/m2 kN/m2
9.375 kN/m2
(IS 875_2)
ly/lx = 1.079 Negative moment at cont. edge Positive moment at mid span
Short span Long Span (αx) (αy) 0.052 0.047 0.039 0.035
Shear avg effective depth (d) = clear short span (lxn) = Design shear = =
110 mm 3307 mm
14.471 kN/m
multiplying factor (k) =
1.3
(Cl. 40.2.1.1, IS 456)
Detailing 1) Main Reinforcement Maximum spacing of bars minimum of 3d or 300 mm 3dx = 3dy = Maximum spacing of bars =
345 mm 315 mm 300
For
ax
Negative moments at continuous edge Positive moment at Mid span
0.052 0.039
For
ay
Negative moments at continuous edge Positive moment at Mid span
0.047 0.035
Reinforcement extension over s short span = long span =
Mx (kNm/m) 5.71 4.28
My (kNm/m) 5.16 3.84
427.75 mm 461.375 mm
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
Bar dia (mm) 10 8
Reinforcement Provided Spacing (mm) 300 250
0.125l
2) Torsional reinforcement area of reinforcement for maximum midspan moment A= 201.06 mm2 Extension of reinforcement over = lx/5 =
684.4 mm
For extreme corners = 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
150.80 mm2 8 mm 3 5 171.1 mm
For 1 edge continuous = 0.5 * 3/4 A = Assume bar dia = Nos. = Povide = Spacing =
75.40 mm2 6 mm 2.66666667 5 171.1 mm
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.607 m 3.886 m 9.375 kN/m2
The loads of triangular segment A will be transfer
Beam 1-2 ly = 3.886
A
lx = 3.607
Beam 2-3 1.8035
(goal seek for diff = 0)
0
(Cl. 40.2.1.1, IS 456)
SHORT SPAN Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Provided Area Remarks (mm2) 261.80 OK 201.06 OK
Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.432 0.324
0.10 0.08
116.25 86.73
180 180
LONG SPAN Reinforcement Requirement R = M/bd^2
(% of steel), pt
Area (mm2)
Min. Steel reqd
0.468 0.349
0.11 0.08
115.28 85.35
180 180
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.607 (l)
max intensity of D DL =
16.91
Max. Moment =
18.33 1.83 15.25 1.52
Max. Shear or Rxtn =
max intensity of D DL =
16.91
0.279 (x)
max Intensity of UDL =
16.91
Max. Moment =
22.75 2.27 17.61 1.76
(y) Max. Shear or Rxtn =
for fst
240 lower
pt kt =
0.2 1.68
for pt = kt =
0.2 1.68
for fst = kt =
248 1.64
SHORT SPAN pt provided 0.23 0.17
pt provided 0.25 0.19
Strength (tc) N/mm2 0.36 0.311
Shear Adequacy Stress (tv) k*tc N/mm2 0.468 0.132 0.404 0.132
Strength (tc) N/mm2 0.304 0.29
Shear Adequacy Stress (tv) k*tc N/mm2 0.395 0.132 0.377 0.132
Check for Deflection Contro Remarks OK OK
Remarks OK OK
fst N/mm2 259.62
kt only for mid span 1.53
nt B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m
(w1)
(wl/4)
effects on beam will be due to slab on both sides, so (x2)
effects on beam will be due to slab on both sides, so (x2)
kN/m
(w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
2
(0.5*(w1x+w2y)
240 higher 0.3 1.47 0.2 1.68
lower
290 higher 0.2 0.3 1.4 1.23 0.2 1.4
248 1.64
Check for Deflection Control prov < l/d l/d max provided max only for mid span 39.78 29.75652 ok
DATA Span of the beam = Assume width of beam (b) = Assume depth of beam (D) = Assume clear cover = Assume main reinforcement dia (ϕ) = Assume width of columns = Concrete density = Concrete grade = Characteristic Compr. Strength (Fck) = Permissible stress in concrete (Fc) = Steel grade = Characteristic Tensile Strength (Fy) = Permissible stress in steel (Fs) = xu,max/d =
#REF!
Maximum moment acting on the beam =
#REF!
0.15 0.18 25 12 0.6 2.5 M 25 2548.42 1132.631 Fe 500 50968.400 29546.898 0.46
Torsion acting on the beam (if any) =
m m m mm mm m t/m3
(taken from below)
t/m2 t/m2 t/m2 t/m2
t-m 0.000 t-m
Equivalent Moment (Mu,lim) =
SIMPLE BEAM
(M) (T)
#REF!
t-m
a) From deflection criteria (L/d) =
#REF!
m
b) From Limiting Moment =
#REF!
m
max accepted =
#REF!
m
Provided =
0.180 m #REF! 1.20 (generally 1.5-2)
Adequacy of beam section Effective depth calculation
d/b =
cond factor 26
(REQD)
Steel Requirement Area of Main steel required =
256.966 mm2
Provide +
12 ϕ bars @ 3 nos. 0 ϕ bars @ 0 nos. = Pt =
339.292 mm2 1.26
ADEQUATE
Nos. in one layer = Spacing =
3 0.044
m 0.05 m
Say Side face Reinforcement =
NOT REQUIRED
OK
mm2
(on each side)
Provide 10 ϕ bars @ 2 nos. 12 ϕ bars @ 1 nos. =
+
270.177 mm2
CHECK BAR DIA AND NOS.
Development Length (Cl. 26.2.1, IS 456)= f = 12 ss = 230 tbd = 1.4
mm N/mm2 N/mm2
Ld =
M 25
0.49 m
Shear Stress in Concrete Pt = 100 As/b d =
1.257
Max shear acting (Vu) = shear stress acting =
M 25
#REF! #REF!
Design shear resistance (Vuc) =
t t/m2 65.24 t/m2
Design Shear (Vus = Vu - Vuc) =
#REF! #REF!
Assume dia of shear reinf. = legged = Area of steel (Asv) = spacing (sv) =
t/m2 t
4.01 t
Crack Widths (Service Loads) - Limiting value 0.2 mm 0.15 339.292 #REF! 20387359.84 2548419.98 1274209.99
(calculate from table)
8 mm 2 100.53 mm2 200 mm c/c
shear resistance provided =
b= Ast = M= Es = Ec = Ece =
(at d from ) #REF!
m mm2 t-m t/m2 t/m2 t/m2
#REF!
m= Cmin = h=D= d= dia of bar = bar spacing = effective cover =
16.00 25.00 0.18 0.18 12.00 0.05 31.00
t/m2 mm m m mm m mm
N.A at working loads
x^2 + 0.072x - 0.013 = 0
0.072 0.013
x=
0.084
m
Ic =
0.0001
m4
Determination of crack width at a point directly under a bar on tension face x1 = ε1 = a= εm =
0.096 m #REF! 0.18 m #REF!
acr = Wcr =
Cmin = 25.00 mm #REF!
mm
#REF!
Determination of crack width at a point midway between bars on tension face x1 = ε1 = a= εm =
0.096 m #REF! 0.18 m #REF!
acr = Wcr =
33.82 mm #REF!
mm
#REF!
Determination of crack width at a point bottom corner of beam on tension face x1 = ε1 =
0.096 m #REF!
a= εm =
0.18 m #REF!
acr = Wcr =
37.84 mm #REF!
mm
#REF!
Determination of crack width at a point 2/3(d-x) from N.A. x1 = ε1 = a= εm =
0.064 m #REF! 0.15 m #REF!
acr = Wcr =
38.66 mm #REF!
mm
#REF!
Cost of Conc Section rate of steel = total weight = cost of reinf steel =
72274 rs/mt #REF! mt #REF! rs
rate of M25 concrete = total vol = cost of conc =
6159.15 rs/m3 #REF! m3 #REF! rs
total cost =
#REF!
rs
SIMPLE BEAM (taken from below)
cond factor
ADEQUATE
(on each side)
CHECK BAR DIA AND NOS. (Cl. 26.5.1.3, IS 456)
(at d from )
(calculate from table)
DATA Span of the beam = Assume width of beam (b) = Assume depth of beam (D) = Assume clear cover = Assume main reinforcement dia (ϕ) = Assume width of columns = Concrete density = Concrete grade = Characteristic Compr. Strength (Fck) = Permissible stress in concrete (Fc) = Steel grade = Characteristic Tensile Strength (Fy) = Permissible stress in steel (Fs) = xu,max/d =
#REF!
Maximum moment acting on the beam =
#REF!
0.2 0.25 25 16 0.6 2.5 M 25 2548.42 1132.631 Fe 500 50968.400 29546.898 0.46
Torsion acting on the beam (if any) =
m m m mm mm m t/m3
(taken from below)
t/m2 t/m2 t/m2 t/m2
t-m 0.000 t-m
Equivalent Moment (Mu,lim) =
SIMPLE BEAM
(M) (T)
#REF!
t-m
a) From deflection criteria (L/d) =
#REF!
m
b) From Limiting Moment =
#REF!
m
max accepted =
#REF!
m
Provided =
0.250 m #REF! 1.25 (generally 1.5-2)
Adequacy of beam section Effective depth calculation
d/b =
cond factor 26
(REQD)
Steel Requirement Area of Main steel required =
475.862 mm2
Provide +
16 ϕ bars @ 3 nos. 0 ϕ bars @ 0 nos. = Pt =
603.186 mm2 1.21
ADEQUATE
Nos. in one layer = Spacing =
4 0.045 Say
Side face Reinforcement =
m 0.05 m
NOT REQUIRED
OK
mm2
(on each side)
Provide 10 ϕ bars @ 2 nos. 12 ϕ bars @ 1 nos. =
+
270.177 mm2
CHECK BAR DIA AND NOS.
Development Length (Cl. 26.2.1, IS 456)= f = 16 ss = 230 tbd = 1.4 Ld =
mm N/mm2 N/mm2
M 25
0.66 m
Shear Stress in Concrete Pt = 100 As/b d =
1.206
Max shear acting (Vu) = shear stress acting =
1.220
Design shear resistance (Vuc) =
M 25
t 24.400 t/m2 70.1 t/m2
Design Shear (Vus = Vu - Vuc) =
8 mm 2 100.53 mm2 200 mm c/c
shear resistance provided =
5.57 t
Crack Widths (Service Loads) - Limiting value 0.2 mm 0.2 603.186 #REF! 20387359.84 2548419.98 1274209.99
(calculate from table)
-45.700 t/m2 -2.285 t
Assume dia of shear reinf. = legged = Area of steel (Asv) = spacing (sv) =
b= Ast = M= Es = Ec = Ece =
(at d from ) OK
m mm2 t-m t/m2 t/m2 t/m2
OK
m= Cmin = h=D= d= dia of bar = bar spacing = effective cover =
16.00 25.00 0.25 0.25 16.00 0.05 33.00
t/m2 mm m m mm m mm
N.A at working loads
x^2 + 0.096x - 0.024 = 0
0.097 0.024
x=
0.114
m
Ic =
0.0003
m4
Determination of crack width at a point directly under a bar on tension face x1 = ε1 = a= εm =
0.136 m #REF! 0.25 m #REF!
acr = Wcr =
Cmin = 25.00 mm #REF!
mm
#REF!
Determination of crack width at a point midway between bars on tension face x1 = ε1 = a= εm =
0.136 m #REF! 0.25 m #REF!
acr = Wcr =
33.4 mm #REF!
mm
#REF!
Determination of crack width at a point bottom corner of beam on tension face x1 = ε1 =
0.136 m #REF!
a= εm =
0.25 m #REF!
acr = Wcr =
38.67 mm #REF!
mm
#REF!
Determination of crack width at a point 2/3(d-x) from N.A. x1 = ε1 = a= εm =
0.091 m #REF! 0.20 m #REF!
acr = Wcr =
48.12 mm #REF!
mm
#REF!
Cost of Conc Section rate of steel = total weight = cost of reinf steel =
72274 rs/mt #REF! mt #REF! rs
rate of M25 concrete = total vol = cost of conc =
6159.15 rs/m3 #REF! m3 #REF! rs
total cost =
#REF!
rs
SIMPLE BEAM (taken from below)
cond factor
ADEQUATE
(on each side)
CHECK BAR DIA AND NOS. (Cl. 26.5.1.3, IS 456)
(at d from )
(calculate from table)
Slab design (Simply ed) Data Input Dimension of longer span (ly) = Dimension of shorter span (lx) = Nominal cover for steel
4.5 m 3.16 m 30 mm
Imposed load = Surface finish =
2.5 kN/m2 1 kN/m2
Fck = Fy = Unit density of concrete
(IS 875_2)
20 N/mm2 500 N/mm2 25 kN/m3
Depth Calculation (Serviceablity) ly/lx =
1.42 < 2, two way
lx/D =
Min. Required Overall depth D = Provide D =
28 (as per conditions) (use avg of S/S & cont. where applicable for cont & discont. edge) 112.86 mm 120 mm
(min. 150mm)
Design Moments & Shear Dead load of slab = Ultimate Load (w) = Maximum Shear in either direction =
3 kN/m2 9.75 kN/m2 Vu =
w (lx/2) 15.41
kN/m
Mulim = Rlim bd^2 Fe 250 415 500 Value of Rlim to be used = Min. value of shear strength = multiplying factor (k) =
2.68
Rlim (N/mm2) 3.00 2.78 2.68 N/mm2
0.28 N/mm2 1.3
(Table 19, IS 456) (Cl. 40.2.1.1, IS 456)
For
ax
Mx (kNm/m)
Bar dia (mm)
Simply ed on Four Sides
0.099
9.64
8
Reinforcement Provided Spacing (mm) 200
For
ay
My (kNm/m)
Bar dia (mm)
Simply ed on Four Sides
0.051
4.97
8
Reinforcement Provided Spacing (mm) 200
NOTES FOR TABLE 1) Bending moment coefficients taken from Table 26, (Annex D), IS 456 2) Moment values are calculated per unit width 3) ONLY INPUT VALUES IN THE TABLE 4) % steel (pt) calculated from Table 19, IS 456 pt = 100 As/bd 5) Minimum area of steel = 0.12% of bD 6) Minimum bar dia to be used = 8mm
Loads for ing Beams lx = ly = UDL on slab =
3.16 m 4.5 m 9.75 kN/m2
The loads of triangular segment A will be transferred
Beam 1-2 ly =
A
lx =
Beam 2-3
nt. where applicable for cont & discont. edge)
Reinforcement Provided Area Remarks (mm2) 251.327
OK
SHORT SPAN Effective depth (d) Reqd Provided Remarks (mm) (mm) 59.97
86
OK
Strength (tc) N/mm2 0.364
Shear Adequacy Stress (tv) N/mm2 0.179
Reinforcement Provided Area Remarks (mm2) 251.327
OK
LONG SPAN Effective depth (d) Reqd Provided Remarks (mm) (mm) 43.04
86
OK
Strength (tc) N/mm2 0.364
Shear Adequacy Stress (tv) N/mm2 0.179
e loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be transferred to beam 2-3
3.16 (l)
1.58 (x)
1.34 (y)
max intensity of D DL =
15.41
Max. Moment =
12.82 1.28 12.17 1.22
Max. Shear or Rxtn =
1.58 (x)
max intensity of D DL = max Intensity of UDL =
15.41 13.07
Max. Moment =
29.58 2.96 20.92 2.09
Max. Shear or Rxtn =
Shear Adequacy Remarks OK
Reinforcement Requirement (% of Area Min. Steel steel), pt (mm2) reqd 0.25
215
144
Shear Adequacy Remarks
Reinforcement Requirement (% of Area Min. Steel steel), pt (mm2) reqd
OK
0.25
215
egment B will be transferred to beam 2-3
kN/m
(w)
kNm t-m kN t
(wl2/12)
kN/m kN/m
(w1) (w2)
kNm t-m kN t
(w1x /3 + w2y/2*(x+y/4))
(wl/4)
2
(0.5*(w1x+w2y)
144
Related Documents 3h463d
2 Way Slab Design-roof Slab 611n54
November 2019 53
Two - Way Slab Design - 2 704f
October 2020 0
Two Way Slab Deflection n1e4h
December 2019 77
Two Way Slab Design.pdf 1s3h5v
October 2019 91
One Way Solid Slab 2o6012
December 2019 47
One Way Continuous Slab 6t3n3j
December 2019 59More Documents from "Sambhav Poddar" 14b5n
2 Way Slab Design-roof Slab 611n54
November 2019 53
Barrage Design 4i1y11
October 2020 0
Courier Services x372s
January 2021 0
Watch Keeping Contents Of Bridge Procedures Guide 11697
November 2021 0
Pwm n613g
December 2019 66