Design Analysis
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STRUCTURAL ANALYSIS AND DESIGN
DATA AND SPECIFICATIONS : Owner Name of Project Location of Building Unit Wt. Of concrete Exterior Walls Slab Thickness ( t ) Live Loads ( LL ) Live Loads on stairs Super imposed DL height of CHB Stair
: MS. LEA MAYLA B. JACKSON : Proposed ONE STOREY RESL. BLDG : DUNKIN DRIVE ST., CARIG-ALIMANNAO, TUGUEGARAO CITY : 24.00 KN / m3. : 2.11 KN / m2. : 0.125 m. : 4.80 KN / m2. : 2.40 KN / m2. : 1.45 KN / m2. : 2.80 m. : 4.80 KN / m2.
Design Specifications / Criteria : fc' = fs = fy =
۷allow. = = μallow. =
20.70 Mpa 124.00 Mpa 280.00 Mpa
0.17(fc') 0.33(fc') -
for beam stress for punching shear
10.12(fc')/D
for Bond Shear
n= E= Sp= Fy=
10.00 17,300.00 Mpa 170.00 Mpa 0.70 Mpa
Seismic load provisions - Allowable 10 - 20 % of DL and LL
Prepared by :
RICARDO V. ALLAM Civil Engineer PRC No. : 53492 PRTR No. : 264358 D-Issued: : April 1, 2008 P-Issued: : Tuguegarao City
DESIGN OF BEAM :
B1
Size of BEAM
length(L) 6.00
-
DEAD LOAD DL : Wt. OF Beam = Wt. OF Slab = CHB WALL = S.I. Dead load = dead load ( B ) =
d (m,) 0.50
b x d x (24) = t x L x 24 x (2/3) = 2.103 x h = 1.45 x L x 2/3
= LIVE LOAD LL : Total Load
3.60 9.00 5.90 5.80
========= 24.30 14.40 ========= 38.70
b (m,) 0.30
KN/m KN/m KN/m KN/m
KN/m KN/m KN/m
USING " USD " : wu = (1.4DL+1.7LL)
58.49552 KN/m
Add seismic load provision : wu = 1.15 x wu Use Mu = wu x L^2 12
=
67.269848 KN/m
for most critical seciton
= 201.8095 KN. M CHECK FOR Mu =
d: Ǿ bd fy [ 1 - 0.59 x fy x
]
ρ min = 0.18
=
try
=
2.02E+08 d =
= =
=
fc' fy bd
x
2.9362788
201.809544 2.9362788 473.64 513.6428529
add mm.
THEREFORE ADOPT BEAM SECTION : CHECK FOR RSB :
0.013
40 mm steel cover 500.00 mm
500.00
mm
SAFE !!!
X
300.00 mm
As =
bd : use d =
=
1,794.00
Using
n
460.00 mm
16.00
=
8.922603132
mm^2 mm Ǿ RSB
6 pcs. 16 mm Ǿ RSB
say
@ SUPPORT
@ MIDSPAN
n.a.
d =
b=
500 mm.
300 mm
V
=
1.15 w x L
=
232.0809756 KN 2
vu
=
Φ bd =
V/
vu allow. = 0.17 fc'
=
1.978524941 Mpa
0.773453295 Mpa
Since vu allow. < vu actual therefore provide stirrups of 10 mm Ǿ RSB spaced 5 @ 0.05m.;4 at 0.10 m. : 3 at 0.15 m and rest at 0.20 m. on center. CHECKED FOR BENDING STRESS :
therefore
μallow. =
10.12 D
μ act. = =
V / (o x j x d)
μ actual
fc' =
2.877701231 Mpa
1.945186009 Mpa <
μ allow.
SAFE
!!!
Adopt 250 x 400 mm Beam Section w/ 6 - 16mmØ RSB & 2 - 12mmØ RSB provided stirrups of 10mmØ RSB spaced 5 @ 50mm, 4 @ 100mm, 3 @ 150mm and rest at 200mm O.C.
DESIGN OF COLUMNS :
C 1:
Design Design Load
P =
235.85
KN
Add 15 % for Seismic load provision : Pu=
271.22
Using
KN
=
Ag =
0.013
Pu / [ø (0.80) { 0.85 x fc' ( 1 - ρg ) + (fy x ρg) }]
=
23056.2542
mm^2 ?
CHECK ACTUAL COLUMN SIZE : Ag =
62500 Therefore
?
USE
250
mm^2
X
250
23056.2542 mm^2 Required
Very S A F E ! ! !
CHECK FOR STEEL REQUIREMENT : As =
ρg x Ag
Using
=
812.5 mm2
16
mm
ø RSB : n =
4.041034
4 pcs.
Check Area of Steel : Ast =
812.5 Therefore
mm^2
812.5 mm^2 = As req'd
Very S A F E ! ! !
therefore adopt : 250 x 250 column with 4 RSB with 10 mm ø RSB as stirrups spaced at 7 @ 0.05 m.; 3 @ 0.10 m.; 3 @ 0.15 m. and rest at 0.20 m. on center. check Spacing of stirrups : 1 16 ø = 2. 48 x tie ø = 3. Least dimension =
256 mm. 480 mm. 250 mm.
CHECK FOR Pu allow. : Pu = =
Φ (0.80) { 0.85 x Fc' x ( Ag - Ast ) + 607.82519
KN therefore S A F E
fy x Ast } 235.85 !!!
KN
16 mm^2
ø
DESIGN OF FOOTING: Use
Pu =
235.85
KN
Using Sp =
170.00
A req'd = Pu / Sp Using
:
APPROX. USE
=
1.387327059 m2
1.177848487 :
Sp net =
Mpa
1.2
Pu / A
1.2 X
=
1.2 163.7816667 KN/ m2
CHECK FOR BENDING SHEAR : use
ρmin =
=
w
=
fy / fc'
1.4 / fy
=
=
0.005
0.06763285
X =
0.475 0
1.2 Mu
0.25
= Sp x Ab x X/2 =
22.17194313
KN.m. 1.2 m.
DESIGN DEPTH d OF FOOTING :
Mu
= Φ fc' . Bd^2 x (ω) ( 1 - 0.59ω)
Mu
=
1451.666087
d^2
d
=
123.5857827
mm. + 100 mm steel cover
therefore
d
=
300
mm.
STEEL REQUIREMENT : As
=
min. L . d
min =
=
939.0602872
=
16
Ao =
201.0624
TRY
mm^2 mm ø RSB mm^2
0.0035
=
223.5858
N
=
As / Ao =
4.870491784
SAY
THEREFORE USE : BOTHWAYS FOR FOOTING .
16 mm ø RSB
5 PCS. -
5
PCS. 1.2
P=
X
16 mm ø RSB MATTING 1.2 X
235.85 Kn. P
300 mm.
L= w=
1.20 m. 1.20 m. 1.20
1.20
0.25
0.25
DESIGN OF FLOOR SLAB
DCE
fc' = 20.70 Mpa. fs = 124.00 Mpa. Consider typical Slab :
300
6.00 DCE
CE m = CE 4.00
S / L
=
1.5
CASE NO. 3 - TWO EDGE DISCONTTINUES :
SOLVE FOR THE SLAB THICKNESS : t=
perimeter / 180 =
111.1111111 mm.
total depth = t + steel cover(25mm) = say St =
136.1111111
DEAD LOAD : Slab = 24 x d = Ceiling = Floor fin.= super imposed dead load= total = Live load
say
111.1111111 mm.
136.111111 mm.
mm.
3.266666667 0.24 0.72 1.45 5.676666667
=
KN/m2 KN/m2 KN/m2 KN/m2 KN/m2
2.39 KN/m2
** Consider 1.0 m. Strip : Total w = 1.4 DL + 1.7 LL
=
12.0103333 KN/ m.
FROM TABLE ( Case 3 ) ::: Along short span / long span + M @ midspan = cw s^2 = - M @ DCE = - M @ CE =
14.124152 KN.m. 7.2062 KN.m. 10.665176 KN.m.
m =
1
Consider Max Moment to check for thickness of SLAB t =? : *** M= 14.124152 KN.m. d=
b = n = 10 ;
M / (R x b) =
R = 1/2 x fc' x k x j = d= 61.46631052
3.73842 Mpa mm. - add
d=
mm.
86.46631052
Therefore adopth Slab Thickness St = use effective d= @ Midspan use M=
1000 ( 1 m. strip ) k = 0.42 ; j = 0.86
25 mm. Steel cover d =
136.11111 mm.
136.1111111 mm. 111.1111111 mm. 14.124152 Mpa.
As = M / (fs x j x d) =
1192.02333 mm2
OK ! ! !
Using Spacing
12
mm ø RSB
S = Ast x 1000 / As =
Ast =
113.0976 mm2
94.8786799 mm.
CHECK FOR MAXIMUM SLAB SPACING S = : S = 3xt = 408.3333333 mm. THEREFORE ADOPT ;
S =
* * *@ DCE max
Adopt * * *@ CE max
94.87867987 mm.
M=
7.2062 Mpa
As = M / (fs x j x d) =
608.175169 mm2
S = Ast*1000/ As =
185.962213 mm.
S =
185.9622125 mm.
M=
S = Ast*1000/ As =
SUMMARY :
S = L/4
O K !!!
10.665176 Mpa
As = M / (fs x j x d) =
Adopt
408.33333 mm.
L/2
900.09925 mm^2 125.650144 mm.
125.6501436 mm. L/4
408.3333333 mm.
185.9622 mm. on center DCE 12 mm ø RSB @ 10 mm ø RSB @ 125mm
95 mm.
65.28
32.64 5.04 2.16 5.90 4.80 2.11 117.92
DATA AND SPECIFICATIONS : Owner Name of Project Location of Building Unit Wt. Of concrete Exterior Walls Slab Thickness ( t ) Live Loads ( LL ) Live Loads on stairs Super imposed DL height of CHB Stair
: MS. LEA MAYLA B. JACKSON : Proposed ONE STOREY RESL. BLDG : DUNKIN DRIVE ST., CARIG-ALIMANNAO, TUGUEGARAO CITY : 24.00 KN / m3. : 2.11 KN / m2. : 0.125 m. : 4.80 KN / m2. : 2.40 KN / m2. : 1.45 KN / m2. : 2.80 m. : 4.80 KN / m2.
Design Specifications / Criteria : fc' = fs = fy =
۷allow. = = μallow. =
20.70 Mpa 124.00 Mpa 280.00 Mpa
0.17(fc') 0.33(fc') -
for beam stress for punching shear
10.12(fc')/D
for Bond Shear
n= E= Sp= Fy=
10.00 17,300.00 Mpa 170.00 Mpa 0.70 Mpa
Seismic load provisions - Allowable 10 - 20 % of DL and LL
Prepared by :
RICARDO V. ALLAM Civil Engineer PRC No. : 53492 PRTR No. : 264358 D-Issued: : April 1, 2008 P-Issued: : Tuguegarao City
DESIGN OF BEAM :
B1
Size of BEAM
length(L) 6.00
-
DEAD LOAD DL : Wt. OF Beam = Wt. OF Slab = CHB WALL = S.I. Dead load = dead load ( B ) =
d (m,) 0.50
b x d x (24) = t x L x 24 x (2/3) = 2.103 x h = 1.45 x L x 2/3
= LIVE LOAD LL : Total Load
3.60 9.00 5.90 5.80
========= 24.30 14.40 ========= 38.70
b (m,) 0.30
KN/m KN/m KN/m KN/m
KN/m KN/m KN/m
USING " USD " : wu = (1.4DL+1.7LL)
58.49552 KN/m
Add seismic load provision : wu = 1.15 x wu Use Mu = wu x L^2 12
=
67.269848 KN/m
for most critical seciton
= 201.8095 KN. M CHECK FOR Mu =
d: Ǿ bd fy [ 1 - 0.59 x fy x
]
ρ min = 0.18
=
try
=
2.02E+08 d =
= =
=
fc' fy bd
x
2.9362788
201.809544 2.9362788 473.64 513.6428529
add mm.
THEREFORE ADOPT BEAM SECTION : CHECK FOR RSB :
0.013
40 mm steel cover 500.00 mm
500.00
mm
SAFE !!!
X
300.00 mm
As =
bd : use d =
=
1,794.00
Using
n
460.00 mm
16.00
=
8.922603132
mm^2 mm Ǿ RSB
6 pcs. 16 mm Ǿ RSB
say
@ SUPPORT
@ MIDSPAN
n.a.
d =
b=
500 mm.
300 mm
V
=
1.15 w x L
=
232.0809756 KN 2
vu
=
Φ bd =
V/
vu allow. = 0.17 fc'
=
1.978524941 Mpa
0.773453295 Mpa
Since vu allow. < vu actual therefore provide stirrups of 10 mm Ǿ RSB spaced 5 @ 0.05m.;4 at 0.10 m. : 3 at 0.15 m and rest at 0.20 m. on center. CHECKED FOR BENDING STRESS :
therefore
μallow. =
10.12 D
μ act. = =
V / (o x j x d)
μ actual
fc' =
2.877701231 Mpa
1.945186009 Mpa <
μ allow.
SAFE
!!!
Adopt 250 x 400 mm Beam Section w/ 6 - 16mmØ RSB & 2 - 12mmØ RSB provided stirrups of 10mmØ RSB spaced 5 @ 50mm, 4 @ 100mm, 3 @ 150mm and rest at 200mm O.C.
DESIGN OF COLUMNS :
C 1:
Design Design Load
P =
235.85
KN
Add 15 % for Seismic load provision : Pu=
271.22
Using
KN
=
Ag =
0.013
Pu / [ø (0.80) { 0.85 x fc' ( 1 - ρg ) + (fy x ρg) }]
=
23056.2542
mm^2 ?
CHECK ACTUAL COLUMN SIZE : Ag =
62500 Therefore
?
USE
250
mm^2
X
250
23056.2542 mm^2 Required
Very S A F E ! ! !
CHECK FOR STEEL REQUIREMENT : As =
ρg x Ag
Using
=
812.5 mm2
16
mm
ø RSB : n =
4.041034
4 pcs.
Check Area of Steel : Ast =
812.5 Therefore
mm^2
812.5 mm^2 = As req'd
Very S A F E ! ! !
therefore adopt : 250 x 250 column with 4 RSB with 10 mm ø RSB as stirrups spaced at 7 @ 0.05 m.; 3 @ 0.10 m.; 3 @ 0.15 m. and rest at 0.20 m. on center. check Spacing of stirrups : 1 16 ø = 2. 48 x tie ø = 3. Least dimension =
256 mm. 480 mm. 250 mm.
CHECK FOR Pu allow. : Pu = =
Φ (0.80) { 0.85 x Fc' x ( Ag - Ast ) + 607.82519
KN therefore S A F E
fy x Ast } 235.85 !!!
KN
16 mm^2
ø
DESIGN OF FOOTING: Use
Pu =
235.85
KN
Using Sp =
170.00
A req'd = Pu / Sp Using
:
APPROX. USE
=
1.387327059 m2
1.177848487 :
Sp net =
Mpa
1.2
Pu / A
1.2 X
=
1.2 163.7816667 KN/ m2
CHECK FOR BENDING SHEAR : use
ρmin =
=
w
=
fy / fc'
1.4 / fy
=
=
0.005
0.06763285
X =
0.475 0
1.2 Mu
0.25
= Sp x Ab x X/2 =
22.17194313
KN.m. 1.2 m.
DESIGN DEPTH d OF FOOTING :
Mu
= Φ fc' . Bd^2 x (ω) ( 1 - 0.59ω)
Mu
=
1451.666087
d^2
d
=
123.5857827
mm. + 100 mm steel cover
therefore
d
=
300
mm.
STEEL REQUIREMENT : As
=
min. L . d
min =
=
939.0602872
=
16
Ao =
201.0624
TRY
mm^2 mm ø RSB mm^2
0.0035
=
223.5858
N
=
As / Ao =
4.870491784
SAY
THEREFORE USE : BOTHWAYS FOR FOOTING .
16 mm ø RSB
5 PCS. -
5
PCS. 1.2
P=
X
16 mm ø RSB MATTING 1.2 X
235.85 Kn. P
300 mm.
L= w=
1.20 m. 1.20 m. 1.20
1.20
0.25
0.25
DESIGN OF FLOOR SLAB
DCE
fc' = 20.70 Mpa. fs = 124.00 Mpa. Consider typical Slab :
300
6.00 DCE
CE m = CE 4.00
S / L
=
1.5
CASE NO. 3 - TWO EDGE DISCONTTINUES :
SOLVE FOR THE SLAB THICKNESS : t=
perimeter / 180 =
111.1111111 mm.
total depth = t + steel cover(25mm) = say St =
136.1111111
DEAD LOAD : Slab = 24 x d = Ceiling = Floor fin.= super imposed dead load= total = Live load
say
111.1111111 mm.
136.111111 mm.
mm.
3.266666667 0.24 0.72 1.45 5.676666667
=
KN/m2 KN/m2 KN/m2 KN/m2 KN/m2
2.39 KN/m2
** Consider 1.0 m. Strip : Total w = 1.4 DL + 1.7 LL
=
12.0103333 KN/ m.
FROM TABLE ( Case 3 ) ::: Along short span / long span + M @ midspan = cw s^2 = - M @ DCE = - M @ CE =
14.124152 KN.m. 7.2062 KN.m. 10.665176 KN.m.
m =
1
Consider Max Moment to check for thickness of SLAB t =? : *** M= 14.124152 KN.m. d=
b = n = 10 ;
M / (R x b) =
R = 1/2 x fc' x k x j = d= 61.46631052
3.73842 Mpa mm. - add
d=
mm.
86.46631052
Therefore adopth Slab Thickness St = use effective d= @ Midspan use M=
1000 ( 1 m. strip ) k = 0.42 ; j = 0.86
25 mm. Steel cover d =
136.11111 mm.
136.1111111 mm. 111.1111111 mm. 14.124152 Mpa.
As = M / (fs x j x d) =
1192.02333 mm2
OK ! ! !
Using Spacing
12
mm ø RSB
S = Ast x 1000 / As =
Ast =
113.0976 mm2
94.8786799 mm.
CHECK FOR MAXIMUM SLAB SPACING S = : S = 3xt = 408.3333333 mm. THEREFORE ADOPT ;
S =
* * *@ DCE max
Adopt * * *@ CE max
94.87867987 mm.
M=
7.2062 Mpa
As = M / (fs x j x d) =
608.175169 mm2
S = Ast*1000/ As =
185.962213 mm.
S =
185.9622125 mm.
M=
S = Ast*1000/ As =
SUMMARY :
S = L/4
O K !!!
10.665176 Mpa
As = M / (fs x j x d) =
Adopt
408.33333 mm.
L/2
900.09925 mm^2 125.650144 mm.
125.6501436 mm. L/4
408.3333333 mm.
185.9622 mm. on center DCE 12 mm ø RSB @ 10 mm ø RSB @ 125mm
95 mm.
65.28
32.64 5.04 2.16 5.90 4.80 2.11 117.92
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