Pilar Jembatan
April 7, 2017 | Author: MAWAR08 | Category: N/A
Short Description
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Description
Bendung ….
PIER :
DIMENSIONS OF PIER Abutment Type Super Structure Type
Case I
B2
B5
AB 120A T-beam Road Bridge
Case II
B2
W2
W5
W2 Hrh
Hrb
+ 79.00
H5 H6
HT H1
Ho
B3
H2
W3
+ 78.00
hw B5
B4
W4
B4
B6
B6
ho
W5
W4
W6
W6
+ 69.00 + 68.20
H3 H4
+ 67.00
WT
BT
1 Input 1) Dimensions (unit: m) HT 12.00
Ho 9.80
H1 10.80
H2 10.20
H3 0.30
H4 0.90
B2 0.20
B3 0.20
B4 1.75
B5 1.25
B6 0.40
BT 4.75
W2 0.25
W3 0.25
W4 2.25
W5 2.75
W6 0.40
WT 7.25
2) Design parameters Unit Weight Soil Soil saturated Concrete water
1,800 2,000 2,400 1,000
Internal friction angle (degree) f Friction Coefficient =Tan f b =
Number of barrel
Reaction Normal Seismic
kg/m3 kg/m3 kg/m3 kg/m3
Type of bearing
Surcharge Load Load per m
21.87 tf/m2 (normal) 32.80 tf/m2 (seismic))
Horizontal seismic coefficient kh : for earth for structure
Young Modulus (reinforcement bar) Young Modulus ratio
H6 0.40
30 degree 0.5
Allowale bearing capacity Qa
Concrete Design Strength Creep strain coefficient (concrete) Reinforcement concrete Allowable stress Concrete Re-bar Shering Yielding Point of reinforcement Bar
H5 0.60
( max. ( max.
ho 0.80
hw 9.00
Hrh 0.75
Hrb 1.50
1
Vn=Rd+Rl 275.92 ton Ve=Rd 142.54 ton He= 25.66 ton left Fixed right Movable (Input Fixed or Movable) 0 kg/m2 0 ton/m
(normal, left + right) (seismic, left + right) (He= 2 kh Rd, for fixed + fixed) (He= kh Rd, for fixed + movable or movable + movable)
30.0 t/m2 for soil foundation) 45.0 t/m2 for soil foundation)
0.18 0.18 Normal condition 175 0.0035
Seismic condition 175 0.0035
sc =
kgf / m2
scs = ssa = ta =
kgf / m2 kgf / m2 kgf / m2
60 1850 5.5
90 2775 8.25
ssy =
kgf / m2
3000 2,100,000 24
3000 2,100,000 16
n
1/37
88347268.xls.ms_office, Input
2
Check 2.1 Stability analysis
Overturning
Sliding
Case I, II Normal condition -
Case I (parallel to bridge axis) Seismic condition e= 1.527 m BT/3= 1.583 m e < (BT/3) OK Fs=Hu/H 3.25 Fs > 1.25 OK Qmax = 30.15 t/m2 Qmax < Qa OK Qa = 32.80 t/m2
-
Sinking
Qmax = Qmax < Qa Qa =
11.95 OK 21.87
t/m2 t/m2
Case II (perpendicular to bridge axis) e= BT/3= e < (BT/3) Fs=Hu/H Fs > Qmax = Qmax < Qa Qa =
1.596 2.417 OK 3.25 1.25 OK 19.24 OK 32.80
m m
t/m2 t/m2
2.2 Structural analysis (1) Body Section A - A Case I (parallel to bridge axis) Normal Seismic Bar arrangement Tensile bar
Case II (perpendicular to bridge axis) Normal Seismic
f (mm) spacing (mm) As1 (cm2) f (mm) (vertical) spacing (mm) As (cm2) f (mm) (horizontal) interval (mm) Max interval (mm)
25 200 30.68 25 200 30.68
25 200 30.68 25 200 30.68 16 200 2410
25 200 60.87
25 200 60.87 25 200 60.87 16 200 487
Effective width (whole width) (cm) Concrete cover (cm) Effective height (cm) Design load Mf (t m) Nd (t) S (t) Checking of minimum reinforcement bar 1.7 Mf < Mc ? If no, check Mu Mu > Mc ? Required bar (cm2) Checking of allowable stress sc Compressive stress kgf/cm2 ss Bending stress kgf/cm2 ss' kgf/cm2 tm Mean shearing stress kgf/cm2
125.0 7.0 241.0 0.0 360.8 0.0
125.0 7.0 241.0 172.3 227.4 34.7
248.0 7.0 118.0 0.0 360.8 0.0
248.0 7.0 118.0 191.5 227.4 34.7
ok
ok
ok
0
-10.71
0
check Mu ok 38.1557
Compressive bar
Hoop bar
(vertical)
25 200
Design dimensions
0.0 ok 0.0 ok 0.0 ok
21.7 160.8 332.9 1.2
ok ok ok ok
0.0 ok 0.0 ok 0.0 ok
48.0 1,215.3 650.2 1.3
ok ok ok ok
Section B - B Case I (parallel to bridge axis) Normal Seismic Bar arrangement Tensile bar
Case II (perpendicular to bridge axis) Normal Seismic
f (mm) spacing (mm) As1 (cm2) f (mm) (vertical) spacing (mm) As (cm2) f (mm) (horizontal) interval (mm) Max interval (mm)
25 100 61.36 25 100 61.36
25 100 61.36 25 100 61.36 16 200 2410
25 100 121.74
25 100 121.74 25 100 121.74 16 200 1180
Effective width (whole width) (cm) Concrete cover (cm) Effective height (cm) Design load Mf (t m) Nd (t) S (t) Checking of minimum reinforcement bar 1.7 Mf < Mc ? If no, check Mu Mu > Mc ? Required bar (cm2) Checking of allowable stress sc Compressive stress kgf/cm2 ss Bending stress kgf/cm2 ss' kgf/cm2 tm Mean shearing stress kgf/cm2
125.0 7.0 241.0 0.0 360.8 0.0
125.0 7.0 241.0 373.3 227.4 42.8
248.0 7.0 118.0 0.0 360.8 0.0
248.0 7.0 118.0 392.6 227.4 42.8
ok
check Mu ok 146.7
Compressive bar
Hoop bar
(vertical)
25 100
Design dimensions
ok 0.0 0.0 ok 0.0 ok 0.0 ok
2/37
check Mu ok 35.5 45.1 1,171.6 665.8 1.5
0.0 ok ok ok ok
0.0 ok 0.0 ok 0.0 ok
72.7 2,157.8 966.0 1.6
ok ok ok ok
88347268.xls.ms_office, Input
(2) Footing Case I (parallel to bridge axis) Normal Seismic Lower Upper Lower Upper Bar arrangement Upper face
Lower face
f (mm) spacing (mm) As5 (cm2, >As7/3) f (mm) (additional bar) spacing (mm) As6 (cm2, >As5/2 and As8/3) f (mm) (tensile bar) spacing (mm) As7 (cm2) f (mm) (additional bar) spacing (mm) As8 (cm2, >As7/2) (tensile bar)
22 300 ok 60.19 >=40.1 22 300 ok 60.19 >=40.1 22 150 120.38 22 150 ok 120.38 >=60.2
Case II (perpendicular to bridge axis) Normal Seismic Lower Upper Lower Upper
22 300 ok 60.19 >=40.1 22 300 ok 60.19 >=40.1
22 22 300 ok 300 ok 91.87 >=61.2 91.87 >=61.2 22 22 300 ok 300 ok 91.87 >=61.2 91.87 >=61.2 22 150 183.73 22 ok 150 ok >=91.9 183.73 >=91.9
22 150 120.38 22 150 ok 120.38 >=60.2
22 150 183.73 22 150 183.73
475 6 114 99.0 0.0 111.0
475 10 110 319.3 0.0 305.3
475 6 114 99.0 0.0 111.0
725 10 110 202.0 0.0 246.2
ok
ok
ok
ok
ok
ok
ok
54.94
118.06
35.30
116.22
79.36
125.49
50.99
Design dimensions Effective width (whole width) (cm) 475 Concrete cover (cm) 10 Effective height (cm) 110 Design load Mf 146.8 Nd 0.0 S 168.5 Checking of minimum reinforcement bar 1.7 Mf < Mc ? If no, check Mu ok Mu > Mc ? Required bar (cm2) 84.46 Checking of allowable stress sc Compressive stress kgf/cm2 20.0 ok ss Bending stress kgf/cm2 1223.6 ok tm Mean shearing stress kgf/cm2 3.56 ok
16.7 ok 1548.5 ok 2.12 ok
50.9 ok 2618.6 ok 6.35 ok
19.8 ok 1529.8 ok 2.17 ok
18.0 ok 1103.1 ok 3.41 ok
725 6 114 142.9 0.0 125.5
15.8 ok 1465.5 ok 1.57 ok
725 10 110 339.4 0.0 277.6
35.4 ok 1823.7 ok 3.78 ok
725 6 114 142.9 0.0 125.5
18.8 ok 1447.8 ok 1.61 ok
(3) Beam Upper
Bar arrangement Upper face
Lower face
f (mm) spacing (mm) As (cm2) f (mm) spacing (mm) As (cm2)
16 300 11.06 16 300 11.06
Side
16 300 6.70 16 300 6.70
Design dimensions Effective width(cm) 165 Concrete cover (cm) 6 Effective height (cm) 94 Design load Mf 2.9 Nd 0.0 S 4.0 Checking of minimum reinforcement bar 1.7 Mf < Mc ? If no, check Mu ok Mu > Mc ? Required bar (cm2) 1.99 Checking of allowable stress sc Compressive stress kgf/cm2 2.5 ok ss Bending stress kgf/cm2 300.6 ok tm Mean shearing stress kgf/cm2 0.3 ok Stirrup If shearing stress is not sufficient, put stirrup or increase thickness of concrete. f (mm) 12 spacing (mm) 200 As (cm2) 5.65
3/37
100 6 159 0.5 0.0 0.7 ok 0.21 21.854 0.3 ok 52.1 ok 0.0 ok
88347268.xls.ms_office, Input
1. WEIGHT OF ABUTMENT AND MOMENT 1) Dimensions (unit: m) HT 12.00
Ho 9.80
H1 10.80
H2 10.20
H3 0.30
H4 0.90
B2 0.20
B3 0.20
B4 1.75
B5 1.25
B6 0.40
BT 4.75
W2 0.25
W3 0.25
W4 2.25
W5 2.75
W6 0.40
WT 7.25
2) Design parameters Unit Weight Soil Soil under water Concrete water Internal friction angle (degree) f Friction Coefficient =Tan f b = Allowale bearing capacity Qa
1,800 2,000 2,400 1,000 30.00 0.50 21.87 32.80
Horizontal seismic coefficient kh : for earth for structure
H5 0.60
H6 0.40
ho 0.80
hw 9.00
Hrh 0.75
Reaction Normal Seismic
kg/m3 kg/m3 kg/m3 kg/m3 degree
Vn=Rd+Rl 275.92 ton (normal) Ve=Rd 142.54 ton (seismic)) He= 25.66 ton (Superstructure x kh) Surcharge Load 0.00 kg/m2 Load per m 0.00 ton/m
tf/m2 tf/m2
( max. ( max.
30.00 t/m2 for soil foundation) 45.00 t/m2 for soil foundation)
0.18 0.18 1.65
0.20
(unit: m) Hrb 1.50
3.25 0.25
1.25
2.75
0.25
0.20 0.75 0.60
1 2
0.40 12.00
0.20 10.80
A 3
9.00
1.75
1.25
0.40
6 7
0.30
1.75
2.25
0.40
0.80
4 2.05 5
0.90
2.75
0.40
0.40
No. m3 1 2 3 4 5 Total Soil
6 7 Total
G. Total
3.22 1.75 30.40 5.76 30.99 72.12 25.07 4.57
29.64 101.76
unit weight t/m3 2.40 2.40 2.40 2.40 2.40 2.00 2.00
7.25
Vertical Distance Load Y t m 7.7 10.500 4.2 10.015 73.0 4.900 13.8 74.4 173.1 50.1 9.1 59.3 232.4
(2) Inertia force H Section B - B H = (Total weight of 1 to 3) x kh = 84.88 x 0.18 = Center y y = 480.58 \
1.20
B
3.55
(1) Weigh and center Section B-B Volum
2.25
y=o
4.75
Body
4.5
A
10.20
B
0.25 9.80
/
84.88 =
+
5.66 =
Moment WY t.m/m 81.1 42.0 357.5
No. Body
1 2 3 4 5
480.6
Total Soil
0.0 480.6
15.28 tf
5.66 m
4/37
3.22 1.75 13.96
Vertical Distance Load Y t m 2.40 7.7 6.000 2.40 4.2 5.515 2.40 33.5 2.650
Moment WY t.m/m 46.3 23.1 88.8
18.92
45.4
158.2
18.92
45.4
158.2
6 7 Total
G. Total
Section A - A H = (Total weight of 1 to 3) x kh = 45.42 x 0.18 =
8.18 tf
Center y y = 158.23
3.48 m
\
6.86
Section A-A Volum unit weight m3 t/m3
5.70
/
45.42 =
+
3.48 =
9.18
88347268.xls.ms_office, Stability
(3) Dynamic water pressure Section B - B b/h= 2.75 / 4.50 = 0.61111 < 2.00 Dynamic water pressure P P= 3 / 8 x Kh x Wo x b x h2 ; (b/h> 2.0), 3/4 kh Wo b2 h (1-b/4h) = 0.75 x 0.18 x 1 x 1.56 h/2 = 2.25 m y= 2.25 + 6.50 = 8.75 m Section B - B b/h= 2.75 / 9.80 = 0.28061 < 2.00 Dynamic water pressure P P= 3 / 8 x Kh x Wo x b x h2 ; (b/h> 2.0), 3/4 kh Wo b2 h (1-b/4h) = 0.75 x 0.18 x 1 x 1.56 h/2 = 4.5 m y= 4.5 + 2.00 = 6.50 m
; (b/h < 2.0) x 4.50
x
0.931
=
0.88
tf
; (b/h < 2.0) x 9.00
x
0.965
=
1.83
tf
(4) Buoyancy u Volume under water level Vo (body) Vo = 68.88 + 27.92
=
96.79 m3
LOAD AND MOMENT Case I-2
Case II-2
R
R H
0.75
H
y
y
X
X
STABILITY ANALYSIS Case I
Normal Condition 1 Moment and Acting Point Description Body Soil Reaction (bridge) Buoyancy u Total
V Load V (t) 173.1 59.3 275.9 -96.8 (100%) 411.5
1.1 Sinking (Bearing capacity) Allowale bearing capacity Qa Reaction from the Foundation
Q
max(min)
=
21.87 tf/m2
V WT × BT
Fondation Reaction Q Q = 11.95 tf/m2 Case I-2
Q is smaller than Qa? OK
Seismic Condition 1 Moment and Acting Point Description Body Soil Reaction (bridge) Dynamic water pressure Buoyancy (100%) S
V Load HLoad V (t) H (t) 173.08 15.28 59.28 142.54 25.66 1.83 -96.79 278.1 42.8
Distance (m) X Y 2.375 6.862 2.375 2.375 12.000 6.500 2.375
5/37
Moment (t.m) Mx My 411.1 104.9 140.8 338.5 307.9 11.9 -229.9 660.5 424.6
Combined Acting Point Xo=(SMx-SMy)/SV 0.848 m Eccentric Distance e=(BT/2-Xo) 1.527 m Bending Moment M =SV x e 424.64 t.m
88347268.xls.ms_office, Stability
2 Stability Analysis 2.1 Over Turnng e BT/6 ?
in case eBT/6
V 6M Q max(min) = ± WT × BT WT × BT2
X=3(BT/2-e) =
Maximiun Fondation Reaction Q max Q max = 23.652 tf/m2
Foundation Reaction Qe Q=2V/(WT.X)= 30.153 tf/m2
2.544 m
Minimum Fondation Reaction Q max Q min = -7.500 tf/m2
e V
H
Q
X Qmax is smaller than Qa? OK
Qe is smaller than Qa? OK
BT
Therefore Qmax = 30.153 tf/m2 Qmin = 0.000 tf/m2 X= 2.544 m Case II-2 Seismic Condition 1 Moment and Acting Point Description Body Soil Reaction (bridge) Dynamic water pressure Buoyancy (100%) S
2 Stability Analysis 2.1 Over Turnng e WT/6 ?
in case eWT/6
V 6M Qmax(min) = ± BT × WT BT × WT2
X=3(WT/2-e) =
Maximiun Fondation Reaction Q max Q max = 18.743 tf/m2
Foundation Reaction Qe Q=2V/(BT.X)= 19.238 tf/m2
6.087 m
Minimum Fondation Reaction Q max Q min = -2.591 tf/m2
e V
H
Q X
Qmax is smaller than Qa? OK
Qe is smaller than Qa? OK Therefore Qmax = 19.238 tf/m2 Qmin = 0.000 tf/m2 X= 6.087 m
6/37
WT
88347268.xls.ms_office, Stability
Bearing Capacity of soil (1) Design Data fB B
= =
30.00 o 4.75 m
cB z
= =
t/m2 m
0.00 1.20
gs ' L
= =
1.00 7.25
t/m3 m
(=gsat-gw)
(2) Ultimate Bearing Capacity of soil, (qu) Calculation of ultimate bearing capacity will be obtained by appliying the following Terzaghi's formula : qu
=
a c Nc + gs' z Nq + b gs' B Ng Shape factor (Table 2.5 of KP-06) a
=
b
1.23
Shape of footing
:
=
0.40
rectangular, B x L a 1.00 1.30 1.23 (= 1.09 + 0.21 B/L) (= 1.09 + 0.21 L/B) 1.30
Shape of footing 1 strip 2 square 3 rectangular, B x L (B < L) (B > L) 4 circular, diameter = B
b 0.50 0.40 0.40
0.30
Bearing capacity factor (Figure 2.3 of KP-06, by Capper) Nc
=
36.0
Nq
f
Nc 5.7 7.0 9.0 12.0 17.0 24.0 36.0 57.0 70.0 82.0
0 5 10 15 20 25 30 35 37 39 > a c Nc gs' z Nq b gs' B Ng qu
=
= = =
=
23.0 Nq 0.0 1.4 2.7 4.5 7.5 13.0 23.0 44.0 50.0 50.0
Ng
=
20.0
Ng 0.0 0.0 0.2 2.3 4.7 9.5 20.0 41.0 55.0 73.0
0.000 27.600 38.000
2 65.600 t/m
(3) Allowable Bearing Capacity of soil, (qa) qa
=
qu / 3
=
2 21.867 t/m
(safety factor =
3 , normal condition)
qae
=
qu / 2
=
2 32.800 t/m
(safety factor =
2 , seismic condition)
7/37
88347268.xls.ms_office, Stability
STRUCTURAL CALCULATION, PEIR BODY Section A - A
1 Load and Bending Moment of Peir Body Case 1 (parallel to bridge axis)
Case II (perpendicular to bridge axis)
1.65
3.25
1.25
0.25
0.20
2.75
0.25
0.20 0.75
+ 79.00
+ 79.00 + 78.40 + 78.00
0.60
1 2
0.40 5.50
0.25
0.20
4.50
+ 78.00
4.90 3
4.50 1.25
A
A
A
A
+ 73.50 + 68.20 + 67.00
Dimensions B2 0.20
B3 0.20
Seismic Coefficient
B5 1.25
soil
H1 10.80
kh=
H2 10.20
0.18
H5 0.60
unit:m Ho 9.80
H6 0.40
stryctur
kh=
0.18
1) Normal condition (Case I-1 and II-1) N= Rd + Rl + Dead load (1, 2, 3) = 275.924 + 84.877 = 360.801 t H= 0.000 t M= 0.000 t m 2) Seismic Case I-2 N= = = H= M= Case II-2 N= H= M=
Rd + 142.544 + 227.421 t 34.718 t 172.301 t m
Dead load (1, 2, 3) 84.877
Item He Dead load x Kh Dynamic water pressure P Total
227.421 t 34.718 t 191.545 t m
Horizontal Load t 25.658 8.180 0.880 34.718
Case I-2 Distance Y m 5.500 3.484 3.050
Moment My t.m 141.119 28.499 2.684 172.301
Case II-2 Distance Y m 6.250 3.484 3.050
Moment My t.m 160.362 28.499 2.684 191.545
2 Calculation of Required Reinforcement Bar 1) Effective section 2.48
Design
p d2 / 4
+
Calculation Therefore
bh
=
Case I
b= h=
1.25 m 2.48 m
Case II
b= h=
2.48 m 1.25 m
1.25
0.625
bh =
0.625 1.5 2.75
3 Summary of Intersectional Force Description Case I Case II
Normal Condition Moment Load Shearing M (tfm) N (tf) S (tf) 0.00 360.80 0.00 0.00 360.80 0.00
Seismic Condition Moment Load Shearing M (tfm) N (tf) S (tf) 172.30 227.42 34.72 191.54 227.42 34.72
4 3.1022 m 4 3.1022 m
4 Calculation of Required Reinforcement Bar as Rectangular Beam, Normal Condition 1) Cracking Moment Mc=
Case I
Zc*(s'ck + N/Ac)
where, Mc Zc
Mc= =
Cracking Moment Section Modulus Zc=b*h1^2/6 b=125 cm Tensile strength of Concrete (bending) s'ck = 0.5*sck^(2/3) s ck= 175 kgf/cm2 Axial force (kg) Area of Concrete = b*h1 thickness of section
s'ck
N Ac h1 b
Case II
34957421 kgf.cm/m 350 tf.m/m kgf.cm 1281333 cm3
17619668 kgf.cm/m 176 tf.m/m kgf.cm b=248 cm
645833 cm3
16 kgf/cm2 360801 31000 248 125
16 kgf/cm2
kgf cm2 cm cm
360801 31000 125 248
kgf cm2 cm cm
2) Checking of Cracking Moment and Design Bending Moment Design Bending Moment Mf Check Mf & Mc
0.000 tf.m/m
0.000 tf.m/m
1.7*Mf>Mc?, if yes check ultimate bending moment 1.7*Mf = 0.000 tf.m Mc= 349.574 tf.m 1.7*Mf>Mc? No, no need to check ultimate bending moment
0.000 tf.m 176.197 tf.m No, no need to check ultimate bending moment
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} where, Mu As s sy d
s'ck b
Mu= =
0 kgf.cm 0.000 tf.m
Mu= =
0 kgf.cm 0.000 tf.m
Ultimate Bending Moment Area of Tensile Bar Yielding point of Tensile Bar Effective height = h1-cover cover d1= 7 cm h1= 248.0 cm, 125.0 cm Design Compressive Strength of Concrete Effective Width
tf.m cm2 3000 kgf/cm2 241 cm
175 kgf/cm2 125 cm
175 kgf/cm2 248 cm
As=Mf/(s sa*j*d) s sa= Allowable Stress Rbar j= 1 -k/3 (=8/9 ) or k = n/{n+s sa/s ca) n= Young's modulus ratio s ca Allowable Stress Concrete
0.000 cm2 1850 kgf/cm2 0.854
0.000 cm2 1850 kgf/cm2 0.854
24 60 kgf/cm2
24 60 kgf/cm2
Check Mu & Mc
Mu = Mc = Mu>Mc?
tf.m cm2 3000 kgf/cm2 118 cm
(Spec >295 N/mm2)
0.000 tf.m 349.574 tf.m not applicable
Mu = 0.000 tf.m Mc = 176.197 tf.m not applicable
4) Bar Arrangement Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf? where, M1 Cs s m ssa sca n
M1= 81422113.06 kgf.cm = 814.221 tf.m
Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca
38727010 kgf.cm 387.270 tf.m
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24
Check M1 > Mf?
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24
M1= 814.221 tf.m Mf= 0.000 tf.m M1>Mf: Design Tensile Bar Only
387.270 tf.m 0.000 tf.m M1>Mf: Design Tensile Bar Only
(a) Tensile Bar Max Bar Area As max = Min Bar Area As min = Required Bar Area Apply f = Required Bar Nos Bar Area Mrs=ssa*As2(d-d2)
0.02*b*d = b*4.5%= As req= 25 @ Nos=b/pitch = As =
30.680 cm2
602.50 5.63 0 200 6.250
cm2 cm2 cm2 mm nos ok
Mrs= 13677553 kgf.cm = 136.776 tf.m where, Mrs Resistance Mooment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} As2= 30.677 ssa= 1850 d
25
@
585.28 11.16 0 200 12.400
60.868 cm2
cm2 cm2 cm2 mm nos ok
Mrs= 13287109 kgf.cm = 132.871 tf.m
As2= ssa=
60.866 1850
(Spec >295 N/mm2)
d
h
d1
d= d2=
241 0
M'= As'= M1= Mf=
0.000 0.000 814.221 0.000
d= d2= ssa=
241 0 1850 0 @ 30.680
d= d2=
118 0
(b) Compressive Bar, in case M1Acmin?
Case II
2026.929 cm2
2026.929 cm2
2775 kgf/m2 90 kgf/m2 227.421 tf
b*c=
31000
>
2775 kgf/m2 90 kgf/m2 227.421 tf Acmin
ok
31000
>
2) Cracking Moment Mc=
Zc*(s'ck + N/Ac)
where, Mc Zc
s'ck
Cracking Moment Section Modulus Zc=b*h2^2/6 b= Tensile strength of Concrete (bending) s'ck = 0.5*sck^(2/3) s ck=
= =
29444381 kgf.cm 294.444 tf.m kgf.cm 1281333.333 cm3 125 cm 15.643 kgf/cm2 175 kgf/cm2
14840918 kgf.cm 148.409 tf.m kgf.cm 645833.33 cm3 248 cm 15.643 kgf/cm2 175 kgf/cm2
Acmin
ok
N Ac h
Axial force Area of Concrete = b*h2 thickness of section
227.421 tf 31000 cm2 248 cm
227.421 tf 31000 cm2 125 cm
Minimum Reifircement Bar (a) As a beam (b) As a column
As min = b(m)*4.5 cm2 As min=0.008*Ac min
As min= As min=
5.63 cm2 16.22 cm2
As min= As min=
11.16 cm2 16.22 cm2
Maximum Reinforcement Bar (a) As a beam (b) As a column
As max = 0.002*b*d As max = 0.006*Ac
As max= As max=
602.50 cm2 1860.00 cm2
As max= As max=
585.28 cm2 1860.00 cm2
3) Checking of Cracking Moment and Design Bending Moment Design Bending Moment Mf Check Mf & Mc
172.301 tf.m
4) Checking of Required Reinforcement Bar
ssa Ms e M N n c h b d1 d s
325.626 tf.m 148.409 tf.m yes, check ultimate bending moment
Compute under eccentric load condition
As={[sc*(s/2)-N/(b*d)]/ssa}*b*d where, sc
191.545 tf.m
1.7*Mf>Mc?, if yes check ultimate bending moment 1.7*Mf = 292.912 tf.m Mc= 294.444 tf.m 1.7*Mf>Mc? No, no need to check ultimate bending moment
As=
-10.707 cm2
25.431 cm2
sc= Stress of concrete 54.725 kg.cm2 solve the equation Eq1 below Eq1 = sc^3 + [3*ssa/(2*n)-3*Ms/(b*d^2)]*sc^2 - 6*Ms/(n*b*d^2)*ssa*sc - 3*Ms/(n^2*b*d^2)*ssa^2 = 0 ssa = Allowable stress of Reinforcement Bar 2775 kg.cm2 Eccentric Moment, Ms=N(e+c) Ms= 43838447.42 kgf.cm Essentric Distance e=M/N e= 75.763 cm Bending Moment M= 172.301 tf.m Axial Force N= 227.421 tf Young's Modulus Ratio n= 16 c=h/2 - d1 c= 117.0 cm Height of Section h= 248.0 cm Width of section b= 125.0 cm Concrete Cover d1= 7.0 cm Effective Width of section d=h-d1 d= 241.0 cm s=n*sc/(n*sc+ssa) s= 0.240 [3*ssa/(2*n)-3*Ms/(b*d^2)]= 6*Ms/(n*b*d^2)*ssa= 3*Ms/(n^2*b*d^2)*ssa^2=
70.481 kg.cm2
2775 31776376 84.225 191.545 227.421 16 55.5 125.0 248.0 7.0 118.0 0.289
242.04 6283.55 544902
kg.cm2 kgf.cm cm tf.m tf cm cm cm cm cm
232.55 9575.97 830416
sc (trial)= 54.725187 Eq1 (trial)= 0.0006335 ok cross check 0.0006335 ok
sc (trial)= 70.481349 Eq1 (trial)= -2.11E-07 ok cross check -2.11E-07 ok
5) Ultimate Bending Moment Mu=c*(h/2-0.4X)+Ts'(h/2-d2)+Ts(h/2-d1) Mu=min(Mu1,Mu2)
where, Mu c sck b X Ts' As' As Es ecu h d1 d2
Mu= =
0 kgf.cm 0.000 tf.m in case X>0 in case X295 N/mm2) 227421 kgf a = 29512 b-Ts-N = -210255 - b*d2 = -654222 9.466 -2.342 233.078 tf.m 148.409 tf.m ok
6) Bar Arrangement d
d2
d1
Max Bar Area As max = Min Bar Area As min =
602.50 cm2 16.22 cm2
(a) Tensile Bar Required Bar Area Apply f = Column width Bar Area
As req= 25
@
b= As =
(b) Compressive Bar Required Bar Area Apply f = Column width Bar Area
As' req= 25
@
b= As' =
(c ) Hoop Bar f' min = Minimum Diameter Apply f' = Bar Interval shall satisfy the following conditions: Mc?, if yes check ultimate bending moment 1.7*Mf = 0.000 tf.m Mc= 349.574 tf.m 1.7*Mf>Mc? No, no need to check ultimate bending moment
0.000 tf.m 176.197 tf.m No, no need to check ultimate bending moment
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} where, Mu As s sy d
s'ck b
Mu= =
0 kgf.cm 0.000 tf.m
Mu= =
0 kgf.cm 0.000 tf.m
Ultimate Bending Moment Area of Tensile Bar Yielding point of Tensile Bar Effective height = h1-cover cover d1= 7 cm h1= 248.0 cm, 125.0 cm Design Compressive Strength of Concrete Effective Width
tf.m cm2 3000 kgf/cm2 241 cm
175 kgf/cm2 125 cm
175 kgf/cm2 248 cm
As=Mf/(s sa*j*d) s sa= Allowable Stress Rbar j= 1 -k/3 (=8/9 ) or k = n/{n+s sa/s ca) n= Young's modulus ratio s ca Allowable Stress Concrete
0.000 cm2 1850 kgf/cm2 0.854
0.000 cm2 1850 kgf/cm2 0.854
24 60 kgf/cm2
24 60 kgf/cm2
Check Mu & Mc
Mu = Mc = Mu>Mc?
tf.m cm2 3000 kgf/cm2 118 cm
(Spec >295 N/mm2)
0.000 tf.m 349.574 tf.m not applicable
(Spec >295 N/mm2)
Mu = 0.000 tf.m Mc = 176.197 tf.m not applicable
4) Bar Arrangement Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf? where, M1 Cs s m ssa sca n
M1= 81422113.06 kgf.cm = 814.221 tf.m
Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca
38727010 kgf.cm 387.270 tf.m
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24
Check M1 > Mf?
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24
M1= 814.221 tf.m Mf= 0.000 tf.m M1>Mf: Design Tensile Bar Only
387.270 tf.m 0.000 tf.m M1>Mf: Design Tensile Bar Only
(a) Tensile Bar Max Bar Area As max = Min Bar Area As min = Required Bar Area Apply f = Required Bar Nos Bar Area
0.02*b*d = b*4.5%= As req= 25 @ Nos=b/pitch = As =
61.359 cm2
Mrs=ssa*As2(d-d2)
h
602.50 5.63 0 100 12.500
cm2 cm2 cm2 mm nos ok
Mrs= 27356060 kgf.cm = 273.561 tf.m where, Mrs Resistance Mooment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} As2= 61.357 ssa= 1850 d d= 241 d1 d2= 0
15/37
25
585.28 11.16 0 100 24.800
@ 121.737 cm2
cm2 cm2 cm2 mm nos ok
Mrs= 26574672 kgf.cm = 265.747 tf.m
As2= ssa= d= d2=
121.735 1850 118 0
88347268.xls.ms_office, Body
(b) Compressive Bar, in case M1Acmin?
Case II
2026.929 cm2
2026.929 cm2
2775 kgf/m2 90 kgf/m2 227.421 tf
b*c=
31000
>
2775 kgf/m2 90 kgf/m2 227.421 tf Acmin
ok
31000
>
Acmin
ok
2) Cracking Moment Mc=
Zc*(s'ck + N/Ac)
where, Mc Zc
s'ck
N Ac h
= =
Cracking Moment Section Modulus Zc=b*h2^2/6 b= Tensile strength of Concrete (bending) s'ck = 0.5*sck^(2/3) s ck= Axial force Area of Concrete = b*h2 thickness of section
29444381 kgf.cm 294.444 tf.m kgf.cm 1281333.333 cm3 125 cm 15.643 175 227.421 31000 248
14840918 kgf.cm 148.409 tf.m kgf.cm 645833.3 cm3 248 cm
kgf/cm2 kgf/cm2 tf cm2 cm
15.643 175 227.421 31000 125
kgf/cm2 kgf/cm2 tf cm2 cm
Minimum Reifircement Bar (a) As a beam (b) As a column
As min = b(m)*4.5 cm2 As min=0.008*Ac min
As min= As min=
5.63 cm2 16.22 cm2
As min= As min=
11.16 cm2 16.22 cm2
Maximum Reinforcement Bar (a) As a beam (b) As a column
As max = 0.002*b*d As max = 0.006*Ac
As max= As max=
602.50 cm2 1860.00 cm2
As max= As max=
585.28 cm2 1860.00 cm2
16/37
88347268.xls.ms_office, Body
3) Checking of Cracking Moment and Design Bending Moment Design Bending Moment Mf Check Mf & Mc
373.321 tf.m
4) Checking of Required Reinforcement Bar
ssa Ms e M N n c h b d1 d s
667.360 tf.m 148.409 tf.m yes, check ultimate bending moment
Compute under eccentric load condition
As={[sc*(s/2)-N/(b*d)]/ssa}*b*d where, sc
392.565 tf.m
1.7*Mf>Mc?, if yes check ultimate bending moment 1.7*Mf = 634.646 tf.m Mc= 294.444 tf.m 1.7*Mf>Mc? yes, check ultimate bending moment
As=
23.632 cm2
97.798 cm2
Stress of concrete sc= 68.619 kg.cm2 solve the equation Eq1 below Eq1 = sc^3 + [3*ssa/(2*n)-3*Ms/(b*d^2)]*sc^2 - 6*Ms/(n*b*d^2)*ssa*sc - 3*Ms/(n^2*b*d^2)*ssa^2 = 0 Allowable stress of Reinforcement Bar ssa = 2775 kg.cm2 Eccentric Moment, Ms=N(e+c) Ms= 63940416.79 kgf.cm Essentric Distance e=M/N e= 164.154 cm Bending Moment M= 373.321 tf.m Axial Force N= 227.421 tf Young's Modulus Ratio n= 16 c=h/2 - d1 c= 117.0 cm Height of Section h= 248.0 cm Width of section b= 125.0 cm Concrete Cover d1= 7.0 cm Effective Width of section d=h-d1 d= 241.0 cm s=n*sc/(n*sc+ssa) s= 0.283 [3*ssa/(2*n)-3*Ms/(b*d^2)]= 6*Ms/(n*b*d^2)*ssa= 3*Ms/(n^2*b*d^2)*ssa^2=
95.805 kg.cm2
2775 51878346 172.616 392.565 227.421 16 55.5 125.0 248.0 7.0 118.0 0.356
233.74 9164.86 794765
kg.cm2 kgf.cm cm tf.m tf cm cm cm cm cm
215.09 15633.80 1355744
sc (trial)= 68.61883 Eq1 (trial)= -2.21E-09 ok cross check -2.21E-09 ok
sc (trial)= 95.80498 Eq1 (trial)= -0.000585 ok cross check -0.000585 ok
5) Ultimate Bending Moment Mu=c*(h/2-0.4X)+Ts'(h/2-d2)+Ts(h/2-d1) Mu=min(Mu1,Mu2)
where, Mu c sck b X Ts' As' As Es ecu h d1 d2
Mu= =
45176618 kgf.cm 451.766 tf.m in case X>0 in case X0 in case X295 N/mm2) 227421 kgf a = 29512 b-Ts-N = -161408 - b*d2 = -2515858
16.668 -2.452
12.364 -6.895
451.766 tf.m 294.444 tf.m ok
487.492 tf.m 148.409 tf.m ok
6) Bar Arrangement d
d2
d1 Max Bar Area As max = Min Bar Area As min = (a) Tensile Bar Required Bar Area Apply f = Column width Bar Area
602.50 cm2 16.22 cm2
As req= 25
@
b= As =
17/37
23.632 100 125 61.359
cm2 mm cm2 cm2
585.28 cm2 16.22 cm2
25 ok
@
97.798 100 248 121.737
cm2 mm cm2 cm2
ok
88347268.xls.ms_office, Body
(b) Compressive Bar Required Bar Area Apply f = Column width Bar Area
As' req= 25
11.816 100 125 61.359
@
b= As' =
(c ) Hoop Bar Minimum Diameter f' min = Apply f' = Bar Interval shall satisfy the following conditions: Mc?
84.459
54.937
116.218
79.355
1850 0.854
1850 0.854
1850 0.854
1850 0.854
24 60
24 60
24 60
24 60
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} where, Mu As s sy d
s'ck b
j= 1 -k/3 or
Check Mu & Mc
274.172 705.904 not applicable
185.961 705.904 not applicable
377.885 1644.503 not applicable
268.768 1644.503 not applicable
64458063 644.581
69231156 692.312
98383359 983.834
1.06E+08 1056.686
12.8436 0.4377 30.8333 1850 60 24
12.8436 0.4377 30.8333 1850 60 24
12.8436 0.4377 30.8333 1850 60 24
12.8436 0.4377 30.8333 1850 60 24
983.834 0.000 Yes
1056.686 0.000 Yes
4) Bar Arrangement Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf? where, M1 Cs s m ssa sca n Check M1 & Mf
M1= = Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca
kgf.cm/m tf.m/m
kgf/cm2 kgf/cm2
M1>Mf?, if "Yes", design tensile bar only, if "No", design tensile + compressive bars. M1= Mf=
tf.m tf.m
644.581 146.799 Yes
M1 > Mf?
22/37
692.312 98.958 Yes
88347268.xls.ms_office, Footing
(a) Tensile Bar Max Bar Area As max = Min Bar Area As min = Required Bar Area Apply f =
0.02*b*d = cm2 b*4.5%= cm2 As req= cm2
1045.0 21.4 84.459 22 150 31.667 120.375 ok
@ mm Nos=b/pitch = nos As = cm2
Required Bar Nos Bar Area Mrs=ssa*As2(d-d2)
Mrs= kgf.cm 24495631 = tf.m 244.956 where, Mrs Resistance Moment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} As2= 120.372 ssa= 1850 d d= 110 d1 d2= 0
h
1083.0 21.4 54.937 22 300 15.833 60.188 ok
1595.0 32.6 116.218 22 150 48.333 183.731 ok
1653.0 32.6 79.355 22 300 24.167 91.865 ok
12692430 126.924
37388068 373.881
19374414 193.744
60.182 1850 114 0
183.725 1850 110 0
91.865 1850 114 0
0.000 0.000 692.312 98.958
0.000 0.000 983.834 0.000
0.000 0.000 1056.686 0.000
(b) Compressive Bar, in case M1Mc?, if "Yes", check ultimate bending moment, if "No", no need to check ultimate bending moment. tf.m 542.860 168.228 577.061 tf.m 588.254 588.254 1370.419 1.7*Mf>Mc? No No No
243.004 1370.419 No
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]}
kgf.cm tf.m
38071026 380.710
11993607 119.936
40755988 407.560
17330798 173.308
3000 110 10 120 175 475 118.057 2775 0.886
3000 114 6 120 175 475 35.301 2775 0.886
3000 110 10 120 175 725 125.494 2775 0.886
3000 114 6 120 175 725 50.992 2775 0.886
16 90
16 90
16 90
16 90
where, Mu As s sy d
Ultimate Bending Moment tf.m Area of Tensile Bar cm2 Yielding point of Tensile Bar kgf/cm2 Effective height = h1-cover cm cover d1= cm h1= cm s'ck Design Compressive Strength of Concrete kgf/cm2 b Effective Width cm As=Mf/(s sa*j*d) cm2 s sa= Allowable Stress Rbar kgf/cm2 j= 1 -k/3 (=8/9 ) or k = n/{n+s sa/s ca) n= Young's modulus ratio s ca Allowable Stress Concrete kgf/cm2 Check Mu & Mc Mu = tf.m Mc = tf.m Mu>Mc? 4) Bar Arrangement
380.710 588.254 not applicable
119.936 588.254 not applicable
407.560 1370.419 not applicable
173.308 1370.419 not applicable
78297777 782.978
84095695 840.957
1.2E+08 1195.071
1.28E+08 1283.566
14.2724 0.3416 30.8333 2775 90 16
14.2724 0.3416 30.8333 2775 90 16
14.2724 0.3416 30.8333 2775 90 16
14.2724 0.3416 30.8333 2775 90 16
1195.071 339.448 Yes
1283.566 142.943 Yes
Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf?
M1= kgf.cm/m tf.m/m Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca kgf/cm2 kgf/cm2
where, M1 Cs s m ssa sca n Check M1 & Mf
M1>Mf?, if "Yes", design tensile bar only, if "No", design tensile + compressive bars. M1= tf.m 782.978 840.957 Mf= tf.m 319.329 98.958 M1 > Mf? Yes Yes
(a) Tensile Bar Max Bar Area As max = Min Bar Area As min = Required Bar Area Apply f = Required Bar Nos Bar Area
0.02*b*d = cm2 b*4.5%= cm2 As req= cm2
1045.0 21.4 118.057 22 150 31.667 120.375 ok
@ mm Nos=b/pitch = nos As = cm2
Mrs=ssa*As2(d-d2)
h
d
Mrs= kgf.cm 36743695 tf.m 367.437 where, Mrs Resistance Moment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} 120.372 ssa= 2775 d= 110 d2= 0
1083.0 21.4 35.301 22 300 15.833 60.188 ok
1595.0 32.6 125.494 22 150 48.333 183.731 ok
1653.0 32.6 50.992 22 300 24.167 91.865 ok
19039423 190.394
56082481 560.825
29060172 290.602
60.185 2775 114 0
183.726 2775 110 0
91.861 2775 114 0
d1
24/37
88347268.xls.ms_office, Footing
(b) Compressive Bar, in case M1Mc?, if yes check ultimate bending moment 1.7*Mf = 168.228 tf.m Mc= 588.254 tf.m 1.7*Mf>Mc?
25/37
No, no need to check ultimate bending moment
88347268.xls.ms_office, Footing
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} where, Mu As s sy d
Mu= 18653068 kgf.cm
Ultimate Bending Moment Area of Tensile Bar Yielding point of Tensile Bar Effective height = h1-cover cover d1= 10 cm h1= 120 cm Design Compressive Strength of Concrete Effective Width
s'ck b
tf.m cm2 3000 kgf/cm2 110 cm
Mu = Mc =
186.531 tf.m
(Spec >295 N/mm2)
175 kgf/cm2 725 cm
As=Mf/(s sa*j*d) s sa= Allowable Stress Rbar j= 1 -k/3 (=8/9 ) or k = n/{n+s sa/s ca) n= Young's modulus ratio s ca Allowable Stress Concrete Check Mu & Mc
=
186.531 tf.m 588.254 tf.m
56.934 cm2 1850 kgf/cm2 0.854 24 60 kgf/cm2
Mu>Mc?
not applicable
4) Bar Arrangement Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf? where, M1 Cs s m ssa sca n
M1= 98383359 kgf.cm/m
Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca
Check M1 > Mf?
M1= Mf=
=
983.834 tf.m/m
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24
983.834 tf.m 98.958 tf.m
M1>Mf: Design Tensile Bar Only
0.02*b*d = 1595.0 b*4.5%= 21.4 As req= 56.934 cm2 @ 150 mm Nos=b/pitch = 31.667 As = 120.375 cm2
cm2 cm2
(a) Tensile Bar Max Bar Area As max = Min Bar Area As min = Required Bar Area Apply f = 22 Required Bar Nos Bar Area
d
h
d1 nos ok
Pitch shall be same as that of toe
Mrs=ssa*As2(d-d2)
Mrs= 24495234 kgf.cm
where, Mrs Resistance Moment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} ssa= 1850 kgf/cm2 d= d2= 0 cm
As2=
=
244.952 tf.m
120.370 cm2
110 cm
(b) Compressive Bar, in case M1Mc ?, if yes, check ultimate bending moment 1.7*Mf = 5.014 tf.m Mc= 70.982 tf.m 1.7*Mf>Mc ?
29/37
No, no need to check ultimate bending moment
88347268.xls.ms_office, Beam
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} where, Mu As s sy d
s'ck b
Check Mu & Mc
Mu=
559252.6 kgf.cm
=
Ultimate Bending Moment Area of Tensile Bar Yielding point of Tensile Bar Effective height = h1-cover cover d1= 6 cm h1= 100.0 cm Design Compressive Strength of Concrete Effective Width
tf.m cm2 3000 kgf/cm2 94 cm
As=Mf/(s sa*j*d) s sa= Allowable Stress Rbar j= 1 -k/3 (=8/9 ) or k = n/{n+s sa/s ca) n= Young's modulus ratio s ca Allowable Stress Concrete
1.986 cm2 1850 kgf/cm2 0.854
Mu = Mc =
5.593 tf.m
(Spec >295 N/mm2)
175 kgf/cm2 165 cm
5.593 tf.m 70.982 tf.m
24 60 kgf/cm2
Mu>Mc ?
not applicable
4) Bar Arrangement Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf? where, M1 Cs s m ssa sca n
M1=
Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca
Check M1 > Mf ?
M1= Mf=
16350759 kgf.cm/m =
163.508 tf.m/m
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24
163.508 tf.m 2.949 tf.m
Compressive Tensile
M1>Mf: Design Tensile Bar Only
(a) Tensile Bar Max Bar Area As max = 0.02*b*d = 310.20 Min Bar Area As min = b*4.5%= 7.43 Required Bar Area As req= 1.986 cm2 Apply f = 16 @ 300 mm Required Bar Nos Nos=b/pitch = 5.5 Bar Area As = 11.058 cm2
cm2 cm2
d
h
d1 nos ok
Mrs=ssa*As2(d-d2) where, Mrs Resistance Moment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} ssa= 1850 kgf/cm2 d= d2= 0 cm
Mrs=
1922865 kgf.cm
As2=
11.057 cm2
=
19.229 tf.m
94 cm
(b) Compressive Bar, in case M1Mc ?, if yes, check ultimate bending moment 1.7*Mf = 0.903 tf.m Mc= 26.072 tf.m 1.7*Mf>Mc ?
No, no need to check ultimate bending moment
3) Ultimate Bending Moment Mu=As*s sy{d-(1/2)*[As*s sy]/[0.85*s ck*b]} where, Mu As s sy d
s'ck b
Check Mu & Mc
Mu=
100782.1 kgf.cm
Ultimate Bending Moment Area of Tensile Bar Yielding point of Tensile Bar Effective height = h1-cover cover d1= 6 cm h1= 165.0 cm Design Compressive Strength of Concrete Effective Width
tf.m cm2 3000 kgf/cm2 159 cm
As=Mf/(s sa*j*d) s sa= Allowable Stress Rbar j= 1 -k/3 (=8/9 ) or k = n/{n+s sa/s ca) n= Young's modulus ratio s ca Allowable Stress Concrete
0.211 cm2 1850 kgf/cm2 0.854
Mu = Mc =
1.008 tf.m 26.072 tf.m
=
1.008 tf.m
(Spec >295 N/mm2)
175 kgf/cm2 100 cm
Mu>Mc ?
24 60 kgf/cm2
not applicable
4) Bar Arrangement Checking of Single or Double bar arrangement M1= (d/Cs)^2*ssa*b >Mf? where, M1 Cs s m ssa sca n Check M1 > Mf ?
M1=
Resistance moment ={2m/[s*(1-s/3)]}^(1/2) (n*sca)/(n*sca+ssa) ssa/sca
M1= Mf=
283.526 tf.m 0.531 tf.m
28352576 kgf.cm/m =
283.526 tf.m/m
12.8436 0.4377 30.8333 1850 kgf/cm2 60 kgf/cm2 24 M1>Mf: Design Tensile Bar Only
31/37
88347268.xls.ms_office, Beam
(a) Tensile Bar Max Bar Area As max = 0.02*b*d = Min Bar Area As min = b*4.5%= Required Bar Area As req= 0.211 Apply f = 16 @ 300 Required Bar Nos Nos=b/pitch = Bar Area As = 6.702
318.00 cm2 4.50 cm2
d1 Pitch shall be same as that of toe
Mrs=ssa*As2(d-d2) where, Mrs Resistance Moment by tensile bar As2 As2= As-As1=As - M1/{ssa*(1-s/3)*d} ssa= 1850 kgf/cm2 d= d2= 0 cm
d
h
cm2 mm 3.3333333 nos cm2 ok
Mrs=
1971080 kgf.cm
As2=
6.701 cm2
=
19.711 tf.m
159 cm
(b) Compressive Bar, in case M1 Mc go to step 3
Step 3
Calculation of Mu
Step 4
Comparison between Mu and Mc Mu > Mc
OK
If No, effective height must be increased.
Revised 26-Nov-02
Discription of Case I-1, Case I-2, Case II-1, Case II-2 are as follows Direction Condition normal Case I-1 Case II-1 seismic Case I-2 Case II-2 The expression mantioned above are revised.
30-Nov-02
4-Dec-02
All sheets Some of inputting cells and calculation result cells have been rearranged. Safety factors against sliding have been set at the same figures as retaining wall. revised figure previous figure normal condition 1.5 2.0 seismic condition 1.2 1.25 Reinforcement bar arrangements for additional bars and distribution bars have been added. Modification of stress and additional bar arrangement
12-Dec-02
Minor change: sheets "Stability", "Body" and "Footing"
28-Dec-02
Minor correction: sheets "Body" and "Beam" (calculation results are same as revision 4)
2-Feb-03
Sheet "Input" Input Cell "L48"was modified to calculation cell. Input Cells "L49" and "L50" were added. Cells "I49", "K49", "K50", "L51", "N46", "N47", "N48", "N49" and "N50" were added.
N49" and "N50" were added.
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