:B1
Beam no
Given Data
Fcu =
300 kg/cm2
Fy =
4000 kg/cm2
Mu =
7.00 m.t.
Qu = 3.00 ton
Mt =
0.0 m.t.
0.1873
b=
20 cm
Design Procedueres
Fy =
4000 kg/cm2 From table get
Cmax/d =
μmax =
Mu
d min =
Rmax * Fcu/γc * b
take d ≈ 1.15 * d min =
40 cm
From d get c/d =
0.224
=
0.42
and
Rmax =
0.0129
and
μmin = 11/ Fy =
30.6 cm
11.99 m.t.
Mumax (moment Capacity for single reinforced section) = Rmax * Fcu/γc * b * d^2 =
Check cmin/d < c/d < cmax/d
Safe
Check Mu < Mumax
Safe
Mu
As =
Fy/γs * d * ( 1- 0.4*c/d)
=
5.52 cm2
0.83 cm2
As' = 0.15*As =
Main reinforcement
Choosen Φ main =
As actual =
16 mm
no. of Bars =
3
12 mm
no. of Bars =
2
6.03 cm2
μ = As/(b*d) =
0.008
Secondary reinforcement
Choosen Φ secondary =
As' actual =
μ' = As'/(b*d) =
2.263
0.003
Safe
Safe
Check μmin < μ < μmax
Check 0.1μ < μ' < 0.004
Check of Shear
qu =
qt =
dsi =
dti =
qcu =
qumax =
q tu =
qtumax =
Qu / (b*d) =
3*Mtorsion / (b² * t) =
((1+ (qt / qu)²)^0.5)^-1 =
((1+ (qt / qu)²)^0.5)^-1 =
0.75 * dsi *(Fcu/γc)^.5 =
2.2* dsi *(Fcu/γc)^.5 =
0.75 * dti *(Fcu/γc)^.5 =
2.2 * dti *(Fcu/γc)^.5 =
3.75 kg/cm2
0.00 kg/cm2
1.000
1.000
10.61 kg/cm2
30.00 kg/cm2
10.61 kg/cm2
31.11 kg/cm2
qu < qcu and qtu < qtcu
Use min stirrups 5Φ8 / m'
Closed stirrups for Shear & Torsion
Shear
qsteel = qu - 0.5*qcu =
0.00 kg/cm2
μ = n*Ast / (b * s) = qsteel / (Fystr/γs)
b < 40 cm take n = 2
2 Branches
Ast/Ss =
Torsion
0.00 kg/cm2
qsteel = qtu - 0.5*qtcu =
x1 , y1 are stirrup dimensions
Y1= 35 X1 = 15
1.43
α = 0.66 + 0.33* (y1 / x1) =
≤ 1.5
(qsteel * (b²*t)/3)
=
0.000 cm
Ast/St =
αt*x1*y1* (Fystr/γs)
0.000 cm
#DIV/0!
Choose Φstirrups
No. of stirrups = 100/St + 100/Ss =
Use Stirrups
#DIV/0! #DIV/0!
Longitudinal bars for torsion
As.l. =
2 * Astr * ( x1 + y1 ) * Fystr.
Choosen Φ secondary =
≤ 30 kg/cm2
St * Fy
12 mm
Shrinkage bar
t < 70 no need to shrinkage bar
#DIV/0!
2 Branches
=
0.00 cm2
no. of Bars =
0
Cmin/d =
0.0028
0.125
:B2
Beam no
Given Data
Fcu =
300 kg/cm2
Fy =
4000 kg/cm2
Mu =
10.00 m.t. Qu = 7.00 ton
Mt =
0.0 m.t.
0.1873
b=
20 cm
Design Procedueres
Fy =
4000 kg/cm2 From table get
Cmax/d =
μmax =
Mu
d min =
Rmax * Fcu/γc * b
take d ≈ 1.15 * d min =
45 cm
From d get c/d =
0.257
=
0.42
and
Rmax =
0.0129
and
μmin = 11/ Fy =
36.5 cm
15.17 m.t.
Mumax (moment Capacity for single reinforced section) = Rmax * Fcu/γc * b * d^2 =
Check cmin/d < c/d < cmax/d
Safe
Check Mu < Mumax
Safe
Mu
As =
Fy/γs * d * ( 1- 0.4*c/d)
=
7.12 cm2
1.07 cm2
As' = 0.15*As =
Main reinforcement
Choosen Φ main =
As actual =
16 mm
no. of Bars =
4
12 mm
no. of Bars =
2
8.05 cm2
μ = As/(b*d) =
0.009
Secondary reinforcement
Choosen Φ secondary =
As' actual =
μ' = As'/(b*d) =
2.263
0.003
Safe
Safe
Check μmin < μ < μmax
Check 0.1μ < μ' < 0.004
Check of Shear
qu =
qt =
dsi =
dti =
qcu =
qumax =
q tu =
qtumax =
Qu / (b*d) =
3*Mtorsion / (b² * t) =
((1+ (qt / qu)²)^0.5)^-1 =
((1+ (qt / qu)²)^0.5)^-1 =
0.75 * dsi *(Fcu/γc)^.5 =
2.2* dsi *(Fcu/γc)^.5 =
0.75 * dti *(Fcu/γc)^.5 =
2.2 * dti *(Fcu/γc)^.5 =
7.78 kg/cm2
0.00 kg/cm2
1.000
1.000
10.61 kg/cm2
30.00 kg/cm2
10.61 kg/cm2
31.11 kg/cm2
qu < qcu and qtu < qtcu
Use min stirrups 5Φ8 / m'
Closed stirrups for Shear & Torsion
Shear
qsteel = qu - 0.5*qcu =
0.00 kg/cm2
μ = n*Ast / (b * s) = qsteel / (Fystr/γs)
b < 40 cm take n = 2
2 Branches
Ast/Ss =
Torsion
qsteel = qtu - 0.5*qtcu =
x1 , y1 are stirrup dimensions
α = 0.66 + 0.33* (y1 / x1) =
(qsteel * (b²*t)/3)
Ast/St =
αt*x1*y1* (Fystr/γs)
0.000 cm
#DIV/0!
Choose Φstirrups
No. of stirrups = 100/St + 100/Ss =
Use Stirrups
#DIV/0! #DIV/0!
Longitudinal bars for torsion
As.l. =
2 * Astr * ( x1 + y1 ) * Fystr.
Choosen Φ secondary =
≤ 30 kg/cm2
St * Fy
12 mm
Shrinkage bar
t < 70 no need to shrinkage bar
#DIV/0!
2 Branches
=
0.00 cm2
no. of Bars =
0
0.00 kg/cm2
Y1= 40
1.5
≤ 1.5
=
X1 = 15
0.000 cm
Cmin/d =
0.0028
0.125
Column no
:C1
Given Data
Fcu = 350 kg/cm2 Fy = 3600 kg/cm2 PU =
90 ton
b=
20 cm Ø bars =
Ø 16 Ø stirrups =
Ø 8 Column location
Design Procedure
Pu = 0.35 Fcu Ac + 0.67 Fy As
Assume μ = 1.50%
1%
As = 0.01 Ac
Pu = Ac ( 0.35 Fcu + 0.0067 Fy )
Get Ac required =
567.18 cm2
b = 20 cm
t = 30 cm
t < 5b O.K.
As = 0.01 Ac =
8.51 cm2
Ø = 16 mm
n= 6
Bar spacing= 11.73
Ac actual =
600.00 cm2
As actual =
12.07 cm2
As min = Max of ( 0.006 Achoosen and 0.008 A required ) =
As max =
0.4%
Check As
24.00 cm2
600 =
*
As min < As < As max
S (stirrups spacing) = the Smallest value of 15Ø and 20 cm =
V (stirrups volume per meter height) = ( 100 / S ) * 2( X + Y ) * Asc =
V min = 0.0025 * 100 * b * t =
Check stirrups
4.80 cm2
Safe
150.0 cm3
20 cm
201.1 cm3
Safe
Interior
Column no
:C2
Given Data
Fcu = 350 kg/cm2 Fy = 3600 kg/cm2 PU =
120 ton
b=
20 cm Ø bars =
Ø 16 Ø stirrups =
Ø 8 Column location
Design Procedure
Pu = 0.35 Fcu Ac + 0.67 Fy As
Assume μ = 1.50%
1%
As = 0.01 Ac
Pu = Ac ( 0.35 Fcu + 0.0067 Fy )
Get Ac required =
756.24 cm2
b = 20 cm
t = 40 cm
t < 5b O.K.
As = 0.01 Ac =
11.34 cm2
Ø = 16 mm
n= 6
Bar spacing= 15.07
Ac actual =
800.00 cm2
As actual =
12.07 cm2
As min = Max of ( 0.006 Achoosen and 0.008 A required ) =
As max =
0.4%
Check As
32.00 cm2
800 =
*
As min < As < As max
S (stirrups spacing) = the Smallest value of 15Ø and 20 cm =
V (stirrups volume per meter height) = ( 100 / S ) * 2( X + Y ) * Asc =
V min = 0.0025 * 100 * b * t =
Check stirrups
6.40 cm2
Safe
200.0 cm3
20 cm
251.4 cm3
Safe
Interior
:F1
Footing no
Given Data
Fcu = 300 Kg/cm2
Fy = 3600 Kg/cm2
Pu = 90.0 t.
qna = 20 t/m2
b = 0.40 m
a = 0.20 m
t P.C. = 0.00 m
Design Procedueres
Pw = Pu /1.5 =
60.0 t
A p.c. = Pw/qna =
3.0 m2
L' =1.85 m
B' =1.65 m
Fn = Pu / L' *B' =
29.48 t/m2
P.C. thickness < 20 cm does not take into consideration
L = L'
B = B'
L =1.85 m
B =1.65 m
q = Pu / L *B =
29.48 t/m2
Depth for moment
Mu 1-1 = q * B * ( L - b )² /8 =
12.79 m.t.
Mu 2-2 = q * L * ( B - a )² /8 =
14.34 m.t.
d1-1 = 5 * ( Mu / Fcu * B ) ^0.5 =
25.4 cm
d2-2 = 5 * ( Mu / Fcu * L ) ^0.5 =
25.4 cm
d B.M. = 25.4 cm
Depth for shear
Q 1-1 = q * B * ( L - b - d )/2 =
35270.3 -2.43 d
Q 2-2 = q * L * ( B - a - d )/2 =
39545.5 -2.73 d
qcu = 0.49 (Fcu / 1.5)^0.5 =
6.93 Kg/cm2
1
q 1-1 = Q 1-1 / ( B * d1-1 )
25.4 cm
Depth for punching
Qp = q (L*B - (a+d)*(b+d)) =
87641
qp*2( a+b+2d )*d=
1697.1 d
2
-177 d -2.95d²
14.14 Kg/cm2
≤ (Fcu/1.5)^0.5
qpu = (0.5 + a/b) ( Fcu / 1.5 ) ^0.5 =
-176.9 d
59.5 d²
+ 1874 d
d punch = 25.7 cm
-2.95d² =
2
40 cm
+ 57 d²
qp * 2( a+b+2d ) = Qp
87641
20 cm
25.4 cm
d 2-2 = q ( B - a ) / (( qcu + q/2 ) /2) =
d shear = 25.4 cm
1.65 m
d 1-1 = q ( L - b ) / (( qcu + q/2 ) /2) =
1.65 m
q 2-2 = Q 2-2 / ( L * d2-2 )
1 1.85 m
1697.1 d + 57 d²
-87641.3 = 0
1.85 m
Take d = 25.7 cm
Reinforcement steel
Long direction
As req = Mu1-1 / Fy * J * d =
16.71 cm2
As min = μ min * B * d =
16.71 cm2
As =
6.37 cm2
Use 9 Ø16
As act. = 18.10 cm2
Short direction
6 Ø16 /m
As req = Mu2-2 / Fy * J * d =
18.73 cm2
As min = μ min * L * d =
18.73 cm2
As =
7.14 cm2
As (central zone of width B) =
Use 9
Use 1
As act. =
17.66 cm2
Ø16
As (end zones) =
Safe
5 Ø16 /m
Safe
5 Ø16 /m
Safe
1.07 cm2
Ø16
20.11 cm2
6 Ø16/m
0.5 Ø16
9 Ø16
50 cm
0 cm
1.65 m
1.85 m
1.85 m
d final =
43 cm
t final =
50 cm
0.5 Ø16
:F2
Footing no
Given Data
Fcu = 300 Kg/cm2
Fy = 3600 Kg/cm2
Pu = 120.0 t.
qna = 20 t/m2
b = 0.50 m
a = 0.20 m
t P.C. = 0.00 m
Design Procedueres
Pw = Pu /1.5 =
80.0 t
A p.c. = Pw/qna =
4.0 m2
L' =2.20 m
B' =1.90 m
Fn = Pu / L' *B' =
28.71 t/m2
P.C. thickness < 20 cm does not take into consideration
L = L'
B = B'
L =2.20 m
B =1.90 m
q = Pu / L *B =
28.71 t/m2
Depth for moment
Mu 1-1 = q * B * ( L - b )² /8 =
19.70 m.t.
Mu 2-2 = q * L * ( B - a )² /8 =
22.82 m.t.
d1-1 = 5 * ( Mu / Fcu * B ) ^0.5 =
29.4 cm
d2-2 = 5 * ( Mu / Fcu * L ) ^0.5 =
29.4 cm
d B.M. = 29.4 cm
Depth for shear
Q 1-1 = q * B * ( L - b - d )/2 =
46363.6 -2.73 d
Q 2-2 = q * L * ( B - a - d )/2 =
53684.2 -3.16 d
qcu = 0.49 (Fcu / 1.5)^0.5 =
6.93 Kg/cm2
1
q 1-1 = Q 1-1 / ( B * d1-1 )
29.2 cm
Depth for punching
Qp = q (L*B - (a+d)*(b+d)) =
117129
qpu = (0.5 + a/b) ( Fcu / 1.5 ) ^0.5 =
qp*2( a+b+2d )*d=
1781.9 d
2
-201 d -2.87d²
12.73 Kg/cm2
≤ (Fcu/1.5)^0.5
-201.0 d
-2.87d² =
2
50 cm
+ 51 d²
qp * 2( a+b+2d ) = Qp
117129
20 cm
29.2 cm
d 2-2 = q ( B - a ) / (( qcu + q/2 ) /2) =
d shear = 29.2 cm
1.90 m
d 1-1 = q ( L - b ) / (( qcu + q/2 ) /2) =
1.90 m
q 2-2 = Q 2-2 / ( L * d2-2 )
1 2.20 m
1781.9 d + 51 d²
53.8 d²
+ 1983 d -117129.2 = 0
d punch = 31.7 cm
2.20 m
Take d = 31.7 cm
Reinforcement steel
Long direction
As req = Mu1-1 / Fy * J * d =
20.88 cm2
As min = μ min * B * d =
20.88 cm2
As =
9.05 cm2
Use 11 Ø16
As act. = 22.13 cm2
Short direction
6 Ø16 /m
As req = Mu2-2 / Fy * J * d =
24.17 cm2
As min = μ min * L * d =
24.17 cm2
As =
10.47 cm2
As (central zone of width B) =
22.40 cm2
Use 12 Ø16
As (end zones) =
Use 1
As act. =
Safe
6 Ø16 /m
Safe
3 Ø16 /m
Use smaller Ø
1.77 cm2
Ø16
26.15 cm2
6 Ø16/m
0.5 Ø16
12 Ø16
50 cm
0 cm
1.90 m
2.20 m
2.20 m
d final =
43 cm
t final =
50 cm
0.5 Ø16
:S2
Slab no
Given Data
Fcu =
250 kg/cm2
3600 kg/cm2
Fy =
F.C. =
150 kg/cm2
L.L. =
200 kg/cm2
a=
4.0 m
Take ts =
12 cm
Design Procedure
Slab Dimensions
a (short span ) =
4.00 m
b ( Long span ) =
6.00 m
b/a=
1.5
Simply supported
ma =
1
Simply supported
mb =
1
Two way slab
ts min code specification
ts min for safe deflection
ts min
Assume ts the max of
ts min=
a / 35 =
12 cm
ts min=
r=
=
1.50
α *Wsu * a^2
Muα =
8
=
1.140
β *Wsu * b^2
Muβ =
8
=
0.665
ma * a
36 + 9 ( b / a )
10 cm
10 cm
0.95 t/m2
Wsu = 1.4 * ( γc * ts min + F.C. ) + 1.6 * L.L. =
mb * b
b (0.8 + Fy/1500)
α=
0.600
and
β=
0.156
Design section as rectangular section with width = 100 cm
Cmax/d= 0.44
μ max = 0.0126
Mu
=
5.9 cm
Rmax * Fcu/γc * 100
take d= max of d min and (ts min - 1.5) =
μ min =
0.0015
d min =
10.5 cm
get c/d =
0.125
Bottom reinforcement
As short =
Muα
Fy * d * ( 1-0.4*c/d )
As short =
As long =
=
3.6 cm2
8 Ø08
As =
Muβ
Fy * d * ( 1-0.4*c/d )
As long =
=
4.0 cm2
μ = As / (100*d ) =
2.5 cm2
5 Ø08
As =
0.004
μmin<μ<μmax
Use Minimum 2.514 cm² ( 5Φ8 )
2.5 cm2
μ = As / (100*d ) =
0.002
μmin<μ<μmax
Upper reinforcement
ts ≤ 16 cm no need for upper reinforcement
As short = 0.2 * As bottom short =
As short =
0
0
As long = 0.2 * As bottom long =
As long =
0
0
0.00 cm2
Minimum 2.514 cm² ( 5Φ8 )
0.00 cm2
Minimum 2.514 cm² ( 5Φ8 )
Check deflection
n=
15
221359.4 kg/cm2
Ec = 14000 * Fcu½ =
14400.000 cm4
Ig = B ts³ / 12 =
moment @ N.A. = 0
B C² / 2 + n As' ( C - d' ) - n As ( d - C ) = 0
C² + 1.20686 C -12.67 = 0
Get C = 3.007 cm
Icr = B C³ / 3 + n As' ( C- d' )² + n As ( d - C )² =
29.8 kg/cm2
Fctr = 0.75 * Fcu⅔ =
y=ts/2=
4294.267 cm2
6.0 cm
Mcr = Fctr * Ig / y =
0.7156 m.t.
1.30 m.t.
Ma = ( γc * ts + F.C. + L.L. )*L²/8 =
5980.17 cm4
Ie = ( Mcr / Ma )³ Ig + [ 1- ( Mcr / Ma )³ ] Icr =
Δ D.L. + L.L.
Δ D.L.
=
=
5 ww DL+LL L⁴
384 Ec Ie
5 ww DL L⁴
=
=
384 Ec Ie
α = 2- 1.2(As'/As) =
5E+11
5.1E+11
3.5E+11
5.1E+11
=
0.9820454
cm
=
0.6798776
cm
2
1.360 cm
Δ Creep = α * Δ D.L. =
Δ 1 = Δ D.L.+L.L. + Δ creep =
Δ 2 = Δ D.L. + Δ creep =
2.342 cm
2.040 cm
1.600 cm
Δ allowable (all loads) = Δ D.L. + Δ creep = L / 250 =
Δ allowable (D.L. + Creep)= Δ D.L. + Δ creep = min of (L / 350 and 2 cm ) =
1.143 cm
Unsafe Insrease ts
Unsafe Insrease ts
1.143 cm
Safe
14 cm
Take ts =
22866.667 cm4
Ig = B ts³ / 12 =
moment @ N.A. = 0
B C² / 2 + n As' ( C - d' ) - n As ( d - C ) = 0
C² + 1.20686 C -15.09 = 0
Get C = 3.327 cm
6305.029 cm2
Icr = B C³ / 3 + n As' ( C- d' )² + n As ( d - C )² =
1.1364 m.t.
Mcr = Fctr * Ig / y =
1.40 m.t.
Ma = ( γc * ts + F.C. + L.L. )*L²/8 =
15163.03 cm4
Ie = ( Mcr / Ma )³ Ig + [ 1- ( Mcr / Ma )³ ] Icr =
Δ D.L. + L.L.
Δ D.L.
=
=
α = 2- 1.2(As'/As) =
5 ww DL+LL L⁴
384 Ec Ie
5 ww DL L⁴
384 Ec Ie
=
5.4E+11
1.3E+12
3.8E+11
1.3E+12
=
0.4171036
=
0.2979312
2
Δ Creep = α * Δ D.L. =
0.596 cm
Δ 1 = Δ D.L.+L.L. + Δ creep =
Δ 2 = Δ D.L. + Δ creep =
=
1.013 cm
0.894 cm
Δ allowable (all loads) = Δ D.L. + Δ creep = L / 250 =
1.600 cm
Δ allowable (D.L. + Creep)= Δ D.L. + Δ creep = min of (L / 350 and 2 cm ) =
Safe
b=
6.0 m
:S3
Slab no
Given Data
Fcu =
250 kg/cm2
3600 kg/cm2
Fy =
F.C. =
150 kg/cm2
L.L. =
200 kg/cm2
a=
5.0 m
Take ts =
15 cm
Design Procedure
Slab Dimensions
a (short span ) =
5.00 m
b ( Long span ) =
5.50 m
b/a=
1.1
Simply supported
ma =
1
Simply supported
mb =
1
Two way slab
ts min code specification
ts min for safe deflection
ts min
Assume ts the max of
ts min=
a / 35 =
15 cm
ts min=
r=
=
1.10
α *Wsu * a^2
Muα =
8
=
1.319
β *Wsu * b^2
Muβ =
8
=
1.154
ma * a
36 + 9 ( b / a )
12 cm
10 cm
1.06 t/m2
Wsu = 1.4 * ( γc * ts min + F.C. ) + 1.6 * L.L. =
mb * b
b (0.8 + Fy/1500)
α=
0.400
and
β=
0.289
Design section as rectangular section with width = 100 cm
Cmax/d= 0.44
μ max = 0.0126
Mu
=
6.4 cm
Rmax * Fcu/γc * 100
take d= max of d min and (ts min - 1.5) =
μ min =
0.0015
d min =
13.5 cm
get c/d =
0.125
Bottom reinforcement
As short =
Muα
Fy * d * ( 1-0.4*c/d )
As short =
As long =
=
3.3 cm2
7 Ø08
As =
Muβ
Fy * d * ( 1-0.4*c/d )
As long =
=
3.5 cm2
μ = As / (100*d ) =
0.003
μmin<μ<μmax
μ = As / (100*d ) =
0.003
μmin<μ<μmax
3.2 cm2
7 Ø08
As =
3.5 cm2
Upper reinforcement
ts ≤ 16 cm no need for upper reinforcement
As short = 0.2 * As bottom short =
As short =
0
0
As long = 0.2 * As bottom long =
As long =
0
0
0.00 cm2
Minimum 2.514 cm² ( 5Φ8 )
0.00 cm2
Minimum 2.514 cm² ( 5Φ8 )
Check deflection
n=
15
221359.4 kg/cm2
Ec = 14000 * Fcu½ =
28125.000 cm4
Ig = B ts³ / 12 =
moment @ N.A. = 0
B C² / 2 + n As' ( C - d' ) - n As ( d - C ) = 0
C² + 1.05600 C -14.26 = 0
Get C = 3.284 cm
Icr = B C³ / 3 + n As' ( C- d' )² + n As ( d - C )² =
29.8 kg/cm2
Fctr = 0.75 * Fcu⅔ =
y=ts/2=
6691.120 cm2
7.5 cm
Mcr = Fctr * Ig / y =
1.1182 m.t.
2.27 m.t.
Ma = ( γc * ts + F.C. + L.L. )*L²/8 =
9267.99 cm4
Ie = ( Mcr / Ma )³ Ig + [ 1- ( Mcr / Ma )³ ] Icr =
Δ D.L. + L.L.
Δ D.L.
=
=
5 ww DL+LL L⁴
384 Ec Ie
5 ww DL L⁴
=
=
384 Ec Ie
α = 2- 1.2(As'/As) =
9.1E+11
7.9E+11
6.6E+11
7.9E+11
=
1.1503587
cm
=
0.8330184
cm
2
1.666 cm
Δ Creep = α * Δ D.L. =
Δ 1 = Δ D.L.+L.L. + Δ creep =
Δ 2 = Δ D.L. + Δ creep =
2.816 cm
2.499 cm
2.000 cm
Δ allowable (all loads) = Δ D.L. + Δ creep = L / 250 =
Δ allowable (D.L. + Creep)= Δ D.L. + Δ creep = min of (L / 350 and 2 cm ) =
1.429 cm
Unsafe Insrease ts
Unsafe Insrease ts
1.429 cm
Safe
17 cm
Take ts =
40941.667 cm4
Ig = B ts³ / 12 =
moment @ N.A. = 0
B C² / 2 + n As' ( C - d' ) - n As ( d - C ) = 0
C² + 1.05600 C -16.37 = 0
Get C = 3.552 cm
9031.266 cm2
Icr = B C³ / 3 + n As' ( C- d' )² + n As ( d - C )² =
1.6278 m.t.
Mcr = Fctr * Ig / y =
2.42 m.t.
Ma = ( γc * ts + F.C. + L.L. )*L²/8 =
18719.66 cm4
Ie = ( Mcr / Ma )³ Ig + [ 1- ( Mcr / Ma )³ ] Icr =
Δ D.L. + L.L.
Δ D.L.
=
=
α = 2- 1.2(As'/As) =
5 ww DL+LL L⁴
384 Ec Ie
5 ww DL L⁴
384 Ec Ie
=
9.7E+11
1.6E+12
7.2E+11
1.6E+12
=
0.6088139
=
0.4517007
2
Δ Creep = α * Δ D.L. =
0.903 cm
Δ 1 = Δ D.L.+L.L. + Δ creep =
Δ 2 = Δ D.L. + Δ creep =
=
1.512 cm
1.355 cm
Δ allowable (all loads) = Δ D.L. + Δ creep = L / 250 =
2.000 cm
Δ allowable (D.L. + Creep)= Δ D.L. + Δ creep = min of (L / 350 and 2 cm ) =
Safe
b=
5.5 m
:S4
Slab no
Given Data
Fcu =
250 kg/cm2
3600 kg/cm2
Fy =
F.C. =
150 kg/cm2
L.L. =
200 kg/cm2
a=
6.0 m
Take ts =
18 cm
Design Procedure
Slab Dimensions
a (short span ) =
6.00 m
b ( Long span ) =
8.00 m
b/a=
1.33333333
Simply supported
ma =
1
Simply supported
mb =
1
Two way slab
ts min code specification
ts min for safe deflection
ts min
Assume ts the max of
ts min=
a / 35 =
18 cm
ts min=
r=
=
1.34
α *Wsu * a^2
Muα =
8
=
2.714
β *Wsu * b^2
Muβ =
8
=
1.809
ma * a
36 + 9 ( b / a )
15 cm
10 cm
1.16 t/m2
Wsu = 1.4 * ( γc * ts min + F.C. ) + 1.6 * L.L. =
mb * b
b (0.8 + Fy/1500)
α=
0.520
and
β=
0.195
Design section as rectangular section with width = 100 cm
Cmax/d= 0.44
μ max = 0.0126
Mu
=
9.1 cm
Rmax * Fcu/γc * 100
take d= max of d min and (ts min - 1.5) =
μ min =
0.0015
d min =
16.5 cm
get c/d =
0.125
Bottom reinforcement
As short =
Muα
Fy * d * ( 1-0.4*c/d )
As short =
As long =
=
5.5 cm2
8 Ø10
As =
Muβ
Fy * d * ( 1-0.4*c/d )
As long =
=
6.3 cm2
μ = As / (100*d ) =
0.004
μmin<μ<μmax
μ = As / (100*d ) =
0.003
μmin<μ<μmax
4.1 cm2
6 Ø10
As =
4.7 cm2
Upper reinforcement
ts > 16 cm use upper reinforcement
As short = 0.2 * As bottom short =
As short =
5 Ø08
2.51 cm2
Minimum 2.514 cm² ( 5Φ8 )
As long = 0.2 * As bottom long =
As long =
5 Ø08
2.15 cm2
Minimum 2.514 cm² ( 5Φ8 )
Check deflection
n=
15
221359.4 kg/cm2
Ec = 14000 * Fcu½ =
48600.000 cm4
Ig = B ts³ / 12 =
moment @ N.A. = 0
B C² / 2 + n As' ( C - d' ) - n As ( d - C ) = 0
C² + 2.63991 C -32.25 = 0
Get C = 4.510 cm
Icr = B C³ / 3 + n As' ( C- d' )² + n As ( d - C )² =
29.8 kg/cm2
Fctr = 0.75 * Fcu⅔ =
y=ts/2=
16953.975 cm2
9.0 cm
1.6102 m.t.
Mcr = Fctr * Ig / y =
3.60 m.t.
Ma = ( γc * ts + F.C. + L.L. )*L²/8 =
19785.72 cm4
Ie = ( Mcr / Ma )³ Ig + [ 1- ( Mcr / Ma )³ ] Icr =
Δ D.L. + L.L.
Δ D.L.
5 ww DL+LL L⁴
384 Ec Ie
5 ww DL L⁴
=
=
=
=
384 Ec Ie
α = 2- 1.2(As'/As) =
2.7E+12
1.7E+12
2E+12
1.7E+12
=
1.6028288
cm
=
1.2021216
cm
1.52
1.827 cm
Δ Creep = α * Δ D.L. =
Δ 1 = Δ D.L.+L.L. + Δ creep =
Δ 2 = Δ D.L. + Δ creep =
3.430 cm
3.029 cm
2.400 cm
Δ allowable (all loads) = Δ D.L. + Δ creep = L / 250 =
Δ allowable (D.L. + Creep)= Δ D.L. + Δ creep = min of (L / 350 and 2 cm ) =
1.714 cm
Unsafe Insrease ts
Unsafe Insrease ts
1.714 cm
Safe
25 cm
Take ts =
130208.333 cm4
Ig = B ts³ / 12 =
moment @ N.A. = 0
B C² / 2 + n As' ( C - d' ) - n As ( d - C ) = 0
C² + 2.63991 C -45.45 = 0
Get C = 5.549 cm
36696.093 cm2
Icr = B C³ / 3 + n As' ( C- d' )² + n As ( d - C )² =
4.3140 m.t.
Mcr = Fctr * Ig / y =
4.39 m.t.
Ma = ( γc * ts + F.C. + L.L. )*L²/8 =
125589.11 cm4
Ie = ( Mcr / Ma )³ Ig + [ 1- ( Mcr / Ma )³ ] Icr =
Δ D.L. + L.L.
Δ D.L.
5 ww DL+LL L⁴
384 Ec Ie
=
5 ww DL L⁴
=
α = 2- 1.2(As'/As) =
384 Ec Ie
=
=
3.3E+12
1.1E+13
2.6E+12
1.1E+13
=
0.3077526
=
0.2446239
1.454
Δ Creep = α * Δ D.L. =
0.356 cm
Δ 1 = Δ D.L.+L.L. + Δ creep =
Δ 2 = Δ D.L. + Δ creep =
0.664 cm
0.600 cm
Δ allowable (all loads) = Δ D.L. + Δ creep = L / 250 =
2.400 cm
Δ allowable (D.L. + Creep)= Δ D.L. + Δ creep = min of (L / 350 and 2 cm ) =
Safe
b=
8.0 m
Column no
:C4
Given Data
Fcu = 250 kg/cm2 Fy = 3600 kg/cm2 PU =
160 ton
b=
20 cm Ø bars =
Ø 16 Ø stirrups =
Ø 8 Column location
Design Procedure
Pu = 0.35 Fcu Ac + 0.67 Fy As
Assume μ =1.50%
1%
As = 0.01 Ac
Pu = Ac ( 0.35 Fcu + 0.0067 Fy )
Get Ac required =
1293.66 cm2
b = 20 cm
t = 65 cm
t < 5b O.K.
As = 0.01 Ac =
19.40 cm2
Ø = 16 mm
n = 10
Bar spacing=
Ac actual =
1300.00 cm2
As actual =
20.11 cm2
As min = Max of ( 0.006 Achoosen and 0.008 A required ) =
As max =
0.4%
Check As
52.00 cm2
1300 =
*
As min < As < As max
S (stirrups spacing) = the Smallest value of 15Ø and 20 cm =
V (stirrups volume per meter height) = ( 100 / S ) * 2( X + Y ) * Asc =
V min = 0.0025 * 100 * b * t =
Check stirrups
10.40 cm2
Safe
325.0 cm3
20 cm
377.1 cm3
13.4
Safe
Interior
: F3
Footing no
Given Data
Fcu = 300 Kg/cm2
Fy = 3600 Kg/cm2
Pu = 160.0 t.
qna = 20 t/m2
b = 0.70 m
a = 0.20 m
t P.C. = 0.00 m
Design Procedueres
Pw = Pu /1.5 =
106.7 t
A p.c. = Pw/qna =
5.3 m2
L' =2.60 m
B' =2.10 m
Fn = Pu / L' *B' =
29.30 t/m2
P.C. thickness < 20 cm does not take into consideration
L = L'
B = B'
L =2.60 m
B =2.10 m
q = Pu / L *B =
29.30 t/m2
Depth for moment
Mu 1-1 = q * B * ( L - b )² /8 =
27.77 m.t.
Mu 2-2 = q * L * ( B - a )² /8 =
34.38 m.t.
d1-1 = 5 * ( Mu / Fcu * B ) ^0.5 =
33.2 cm
d2-2 = 5 * ( Mu / Fcu * L ) ^0.5 =
33.2 cm
d B.M. = 33.2 cm
Depth for shear
Q 1-1 = q * B * ( L - b - d )/2 =
58461.5 -3.08 d
Q 2-2 = q * L * ( B - a - d )/2 =
72381 -3.81 d
qcu = 0.49 (Fcu / 1.5)^0.5 =
6.93 Kg/cm2
1
q 1-1 = Q 1-1 / ( B * d1-1 )
33.2 cm
Depth for punching
Qp = q (L*B - (a+d)*(b+d)) =
155897
qpu = (0.5 + a/b) ( Fcu / 1.5 ) ^0.5 =
qp*2( a+b+2d )*d=
2000.1 d
2
-264 d -2.93d²
11.11 Kg/cm2
≤ (Fcu/1.5)^0.5
-263.7 d
-2.93d² =
2
70 cm
+ 44 d²
qp * 2( a+b+2d ) = Qp
155897
20 cm
33.2 cm
d 2-2 = q ( B - a ) / (( qcu + q/2 ) /2) =
d shear = 33.2 cm
2.10 m
d 1-1 = q ( L - b ) / (( qcu + q/2 ) /2) =
2.10 m
q 2-2 = Q 2-2 / ( L * d2-2 )
1 2.60 m
2000.1 d + 44 d²
47.4 d²
+ 2264 d -155897.4 = 0
d punch = 38.2 cm
2.50 m
Take d = 38.2 cm
Reinforcement steel
Long direction
As req = Mu1-1 / Fy * J * d =
24.42 cm2
As min = μ min * B * d =
24.42 cm2
As =
12.05 cm2
Use 13 Ø16
As act. = 26.15 cm2
Short direction
7 Ø16 /m
As req = Mu2-2 / Fy * J * d =
30.23 cm2
As min = μ min * L * d =
30.23 cm2
As =
14.92 cm2
As (central zone of width B) =
27.01 cm2
Use 14 Ø16
As (end zones) =
Use 2
As act. =
Safe
7 Ø16 /m
Safe
4 Ø16 /m
Use smaller Ø
3.22 cm2
Ø16
32.18 cm2
7 Ø16/m
1 Ø16
14 Ø16
60 cm
0 cm
2.10 m
2.60 m
2.60 m
d final =
53 cm
t final =
60 cm
1 Ø16