Post on 02-Apr-2018
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consulte o site:
http://www.clubedoconcreto.com.br/
http://www.clubedoconcreto.com.br/http://www.clubedoconcreto.com.br/7/27/2019 Muro L e Gravidade
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CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOG
BSCE-5 2nd Excel Program 18-Aug-09
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program. =)
****====DESIGN OF GRAVITY RETAINING WALL====****
1. Soil or gravel without fine particles,
highly permeable.
2. Sand or gravel with silt mixture, low
permeability
3. Silty sand, sand and gravel with high
clay content
4. Medium or stiff clay
5. Soft clay, silt
Table 1: Unit weights w, effective angles of internal friction , and coefficients of friction with concrete f.
0.5 m
qs = 20 kPa
3.50 m
1.10 m 0.60 m
Solution: (Use class 2 of the table given above)Composite Section
3.50 m
1.10 m
Soil pressure coefficient, Rankine equation for horizontal soil surface
= 30 Passive soil pressure coefficient
w = 18.85 kN/mqs = 20.00 kPa = 3.00
h' = 1.061 m
Distances computation
Active soil pressure coefficient c1 = B/2 = 2.0000 m
c2 = e/2 = 0.1250 m
= 0.3333 c3 = e + a/2 = 0.5000 m
c4 = (B - 2e - a)/3 + e + a = 1.7500 m
c5 = B - e + e/2 = 3.8750 m
c6 = (B - 2e - a)2/3 + e + a = 2.7500 m
c7 = (B - e - a)/2 + e + a = 2.3750 m
0.2
5
0.2
5
15.70 - 18.85 25 - 35 0.2 - 0.4
14.10 - 17.25 20 -25 0.2 - 0.3
17.25 - 18.85 23 - 30 0.3 - 0.4
18.85 - 20.40 25 - 35 0.4 - 0.5
w, kN/m , degrees f, coefficient
17.25 - 18.85 33 -40 0.5 -0.6
7W7c
a
3W 5c
1W
2W
4W
5W
6W
1c
2c
3c4c
6c
'h
h
e
B
d
sin1
sin1
ahC
sin1
sin1
phC
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Given retaining wall dimensions:
a = 0.50 m Passive soil pressure:
b = 1.10 m h = b= 1.10 m
c = 3.50 m
d = 0.60 m = 34.213 kN
e = 0.25 m
Active soil pressure: = 0.3667 m
h = b + c = 4.60 m Tentative wall base dimension:
= 97.144 kN B = 4.00 m
= 1.7754 m
Check retaining wall stability wc = 23.60 kN/m
Friction coeff., f = 0.50
component weights Wi ci RM=Wici
W1 = Bdwc = 56.640 2.0000 113.280
W2 = e(b - d)ws = 2.356 0.1250 0.295
W3 = a(b + c - d)wc = 47.200 0.5000 23.600
W4 = (B - 2e - a)(b + c - d)wc/2 = 141.600 1.7500 247.800
W5 = (B - 2e - a)(b + c - d)ws/2 = 113.100 3.8750 438.263
W6 = e(b + c - d)ws = 18.850 2.7500 51.838
W7 = qs(B - e - a) = 65.000 2.3750 154.375
Wi = 444.746 Wici = 1029.450
Overturning moment: OM Factor of safety against overturning:
OM = Pahyah = 172.47 kN-m
Location of resultant with respect to toe: = 5.969008 > 2.00, ok!
= 1.9269 m Factor of safety against sliding:
= 0.0731 m = 2.641285 > 1.50, ok!
B/3 = 1.33 m
the middle third of the base. No tension will occur on the foundation.
qmax = 121.61 kPa
qmin = 100.76 kPa
qa = 143 kPa
qmax < qa, the wall is safe against soil bearing.
0.5 m
Retaining Wall Details qs = 20 kPa
3.50 m
1.10 m 0.60 m
4.00 m
0.2
5 0.2
5
'22
1hhwhCP ahah
'23'3
2
hh
hhhyah
2
2
1whCP phph
3hyph
iv
WR
OMRMx
xB
e 2
2minmax
61
B
e
B
Rq v
ahah
ii
goverturninyPOM
cWRMFS
ah
phiv
slid ingP
PWffRFS
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Given retaining wall dimensions:
a = 0.50 m Passive soil pressure:
b = 1.10 m h = b= 1.10
c = 3.50 m
d = 0.60 m = 34.2128
e = 0.25 m
Active soil pressure: = 0.36667
h = b + c = 4.60 m Tentative wall base dimension:
= 97.1443 kN B = 4.00 m
= 1.77536 m
Check retaining wall stability: wc = 23.60 kN/m
Friction coeff., f = 0.50
component weights Wi ci RM=Wici
W1 = Bdwc = 56.640 2.0000 113.280
W2 = e(b - d)ws = 2.356 0.1250 0.295
W3 = a(b + c - d)wc = 47.200 0.5000 23.600
W4 = (B - 2e - a)(b + c - d)wc/2 = 141.600 1.7500 247.800
W5 = (B - 2e - a)(b + c - d)ws/2 = 113.100 3.8750 438.263
W6 = e(b + c - d)ws = 18.850 2.7500 51.838
W7 = qs(B - e - a) = 65.000 2.3750 154.375
Wi = 444.746 Wici = 1029.450
Overturning moment: OM Factor of safety against overturning:
OM = Pahyah = 172.47 kN-m
Location of resultant with respect to toe: = 5.969008 > 2.00, ok!
= 1.9269 m Factor of safety against sliding:
= 0.0731 m = 2.641285 > 1.50, ok!
B/3 = 1.33 m
the middle third of the base. No tension will occur on the foundation.
qmax = 121.6103 kPa
qmin = 100.7628 kPa
qa = 143 kPa
0.5 m
Retaining Wall Details qs = 20.00
3.50 m 0.2
5
0.2
5
1.10 m 0.60 m
4.00 m
'22
1hhwhCP ahah
'23'3
2
hh
hhhyah
iv WR
OMRMx
xB
e 2
2minmax
61
B
e
B
Rq v
ahah
ii
goverturninyPOM
cWRMFS
ah
phiv
sl idi ngP
PWffRFS
2
21 whCP phph
3
hy
ph
2minmax
61
B
e
B
Rq v
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CARMEL B. SABADO CE-162 PROF. GERONID
BSCE-5 2nd Excel Program
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)
****====Design of Cantilever Retaining Wall====****
Right Side Loading
Given:
fc' = 20.70 Mpa Retaining wall dimensions:
fy = 414.00 Mpa a = 0.30 m
s = 18.82 kN/m3 c = 4.50 m
= 40 o
= 0.50 Tentative dimentions:
c = 23.60 kN/m3 B = 3.20 m
qa = 143.50 kPa b = 0.40 m
qs = 19.30 kPa d = 0.50 m
3.65 m
b(main) 25 mm Use Wu = 1.4DL + 1.7LL + 1.7H
b(temp) 16 mm
a
Es = 200000 Mpa
shear = 0.85 qs = 19.30 kPa
flexure
= 0.9
smax = [ 3t , 450 ]min c
h' = 1.026 m
Cantilever Retaining Wall Figure:
d
b
Property Line
a
h'
c h
d
b
B
backfill height =
W1
C1
W2
C2
W3
W4
W5
C3
C4
C5
W6
C6
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Soil pressure coefficient, Rankine equation for horizontal surface:
Active soil pressure coeffecient: Cah = 0.217443
Active soil pressure: h = 5.00 m
Pah = 1 Cahh(h+2h') =
2 72.137 kN
yah = h2
+ 3hh' =
3(h + 2h') 1.9091 m
Check the retaining wall stability:
components weights Wi ci RM = Wici
W1 = 37.760 1.6000 60.4160
W2 = 31.860 0.1500 4.7790
W3 = 5.310 0.3333 1.7700
W4 = 237.132 1.8000 426.8376
W5 = 4.235 0.3333 1.4115
W6 = 55.97 1.75 97.9475
Wi = 372.267 RM = 593.1616
Overturning moment: OM = 137.7138 kN-m
Factor of safety against overturning:
FSoverturning = RM = 4.307 > 2.00 safe!!!!
OM
Factor of safety against sliding:
FSsliding = (f Wi) = 2.580 > 1.50 safe!!!!
Pah
Check for bearing pressure: qa = 143.50 kPa
Location of resultant with respect to toe:
x = RM - OM = 1.2234 m
Wi
e = B - x = 0.376554 m
2
B / 3 = 1.07 m
The middle third of the base where No tension will occur on the foundation.
q = Wi 1 + 6e qmax = 142.001 kPa
B B2
qmin = 90.666 kPa
Since qma < qa, wall is safe againts soil bearing.
Design of stem:
Ve
p1=qs
P1 P2
p2 = Cahwsy
y/2
yM
V
d Vmax Mmax
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Design of Base:
19.30
5.00
0.50
Note: The expected worst condition of loading, the passive earth pressure of soil is generally
neglected due to erosion of top soil, and the soil bearing at the toe is neglected due to
the empending action to overturn.
Use: 1.4 for DL
1.7 for LL and service load bearing pressure
qmax x 1.7 = 241.401 kPa V = (-Ws-Wc-qs)L = -470.11
qmin x 1.7 = 154.132 kPa M = (-Ws-Wc-qs)L2
/2 = -658.15Ws = 1.4sc = 118.566 kPa
qs x 1.7 = 32.810 kPa
Wc = 1.4cd = 16.520 kPa
e = 0.377 m
L= B - b) = 2.800 m
Try d = 400 mm
b = 1000 mm
Ru = Mu = 4.570502 As,flexure = bd = 5216.20
fbd2 s = 1000Ao = 94.11
As
= .85fc' 1 - 1 - 2Ru = 0.013041 As,temp = bd =0.002bd = 800.00
fy .85fc' stemp = 1000 Atemp = 251.33
As
Use: = 0.01304
Check for shear: Vuc = fc' bd = 257.8178 kN/m > V, safe!!!!
6
qmin
qmax
qs =
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Retaining Wall Details:
Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.
0.30
25 mm @
16 mm temp. @ 16
260 oc bw mm temp @
5.00 25 mm @
m
25 mm @
0.50
16 mm temp. bars
250 oc bw
0.40 meters
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CARMEL B. SABADO CE-162 PROF. GERONID
BSCE-5 2nd Excel Program
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)
****====Design of Cantilever Retaining Wall===****
Left Side Loading
Given:
fc' = 27.50 Mpa Retaining wall dimensions: Tentative dimentions:
fy = 275.00 Mpa a = 0.30 m B = 3.00
c = 23.60 kN/m3 H = 3.20 m b = 0.40
s = 17.25 kN/m3 h = 1.00 m D = 0.50
= 35 o
= 0.45 Surcharge load: Es = 200000
qa = 120.00 kPa qs2 = 19.50 kPa shear = 0.85b(main) = 25 mm qs1 = 19.20 kPa flexure = 0.9b(temp) = 16 mm
smax = [ 3t , 450 ]min
Minimun factor safety requirements:
Overturning = 2.00
Sliding = 1.50
Cantilever Retaining Wall Figure:
qs2 = 19.50 kPa
a
h' = 1.11 m
H' = 1.13 m
H
qs1 = 19.20
b
B
Property Line
a
H' qs2
H
qs1
h'
h
D
x
b
B
W3
W4
W5
W2
W1
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Soil pressure coefficient, Rankine equation for horizontal surface:
Active soil pressure coeffecient: Passive soil pressure coefficient:
Cah = 0.271 Cph = 3.690
h = 3.20 m h = 1.61
Active soil pressure:
Pah = 51.549 kN Passive soil pressure:
yah = 1.467 m Pph = 177.889
yph = 1.079
Check the retaining wall stability:
components weights Wi xi Mi B-xi Mx
W1 = 22.656 0.1500 3.3984 2.850 64.5696
W2 = 3.776 0.3333 1.2587 2.667 10.069333
W3 = 35.400 1.5000 53.1000 1.500 53.1
W4 = 144.690 1.7000 245.9730 1.300 188.097
W5 = 18.225 0.3780 6.8890 2.622 47.785952
Total = 202.091 310.6191 RM = 363.62189
Overturning moment: OM = 75.6346 kN-m
= 0.031239833 x = 0.066
Factor of safety against overturning:
FSoverturning = RM = 4.808 > 2.00 safe!!!
OMFactor of safety against sliding:
FSsliding = (f Wi)+Pph = 5.215 > 1.50 safe!!!
Pah
Check for bearing pressure:
B / 3 = 1.000 m
x = RM - OM = 1.4250 m within 1/3 o
Wiq = 2WT , when x < 1/3 B = 94.543 kPa
3x
qmax = [4B - 6x]WT/B
2
= 77.46 kPa safe!!!qmin = [6x - 2B]WT/B
2= 57.26 kPa safe!!!
Design of stem:
Vu = 1.7 [qsH + 0.5CahwsH2
= 146.768 kN/m
Mu = 1.7 Mmax = 220.696 kN-m/m
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M = qsy2/2 + 1/3Cahwsy
3
y M Amain = Dmain2
= 490.874
0.50 2.693 kN-m 4
1.00 11.797 kN-m Atemp = Dtemp2
= 201.0621.50 27.196 kN-m 4
2.00 51.466 kN-m min = 0.5 [1.4/fy, fc'/4fy] = 0.00238372.50 85.284 kN-m max = .75 .85fc'b1 600 = 0.03715713.00 129.821 kN-m fy 600+fy
Vuc = fc' = 0.74296
Depth as required by shear: b = 1000 mm
d = Vu = 197.56 mm
Vucb
Design for flexure:
Try h = = 400 mm
d = h - (100+s/2) = 287.5 mmRu = Mu = 2.966719
fbd2
= .85fc' 1 - 1 - 2Ru = 0.011576fy .85fc'
Use: = 0.011576376
As = bd = 3328.208 mm2/m
spacing, S = [1000Ao/As, 3t,450]min = 147 mm oc
Atemp = 0.0018bd = 517.50 mm2/m
spacing, S = [1000Atemp/As, 5t,450]min = 388 mm oc
Design of Toe:
Vu = 1.7b [(qmax + qmin)/2 - (wcd + ws (h+h'))]= 97.4797 kN/m
Mu = 1.7 [(B-b)2/2][qmin +2/3(B-b)(qmax - qmin) - [wc(d)-ws(h+h')]]
= 253.0727332 kN-m/m
min = 0.5 [1.4/fy, fc'/4fy] = 0.002383656max = .75 .85fc'b1 600 = = 0.037157143
fy 600+fy
Vuc = fc' = 0.7866 kN/m6
Depth as required by shear: b = 1000 mmd = Vu = 123.9243 mm
Vucb
Design for flexure:
Try h = 400 mm
d = h - (100+s/2) = 287.5 mmRu = Mu = 3.401944
fbd2
= .85fc' 1 - 1 - 2Ru = 0.013432 Use: =fy .85fc'
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As = bd = 3861.697 mm2/m
spacing, S = [1000Ao/As, 3t,450]min = 127.00 mm oc
3t = 1200.00
450 = 450
Compare: 1000Ao/AS AND 3t, compare 3t and 450 whichever is smallest.
SINCE 127.00 mm < 450.00 mm
Use: 127.00 mm
Atemp = 0.0018bd = 517.50 mm2/m
spacing, S = [1000Atemp/As, 5t,450]min = 489 mm oc
Retaining Wall Details:
0.30
25
137
16
3.70 338
25
117.00
0.50
3.00
16 mmtemp @0.40 338 oc
Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.
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CARMEL B. SABADO CE-162 PROF. GERONID
BSCE-5 2nd Excel Program
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)
****====DESIGN OF A CANTILEVER RETAINING WALL===****
(with shear key)
fc' = 20.7 Mpa Figure: x=1.5m
fy = 414 Mpa
ws = 18.82 kN/m qa = 19.2 kPa
= 35
f= 0.5
wc = 23.6 kN/m e=3.65m
qa = 143.5 kPa
qs = 19.2 kPa
backfill height = 3.65 m
Use Wu = 1.4DL + 1.7LL + 1.7H c=1.0m
Use 25mm for main rebars, 16mm for
temperature bars.
Solution:
Composite section and location of forces
e
c
d
f
Given retaining wall dimensions:
a = 0.20 m Distances:
c = 1.00 m x1 = B/2 = 1.5000
e = 3.65 m x2 =xh + a/2 = 1.6000
xh = 1.50 m x3 = xh + a + (b - a)/3 = 1.7667
x4 = xh/2 = 0.7500
Tentative dimensions: x5 = xh + b + (B - xh - b)/2 = 2.4500
B = 3.00 m x6 = xh + a + 2(b - a)/3 = 1.8333
b = 0.40 m x7 = xh + a + (B - xh - a)/2 = 2.3500
d = 0.50 m x8 = xh + g/2 = 1.7000
f = 0.40 m H = d + c + e = 5.1500
g = 0.40 m h' = qs/ws = 1.0202
h = 0.20 m
Active soil pressure coefficient Passive soil pressure coefficient
= 0.27099 = 3.69
Active soil pressure: h = 5.15 m Passive soil pressure: h = 1.50
h
H
ax
e
b
B
sin1
sin1
ahC
sin1
sin1
phC
7W
6W
5W
4W 3W
2W
1W
8W
ahx7
x
6x
5x
2x
4x3x
1x
8x
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= 94.42831 kN = 78.13017
= 1.960 m = 0.500
Check retaining wall stability:
Component weights Wi xi RM=Wixi
W1 = Bdwc = 35.4 1.5000 53.1
W2 = a(c + e)wc = 21.948 1.6000 35.1168
W3 = 0.5(b - a)(c + e)wc = 10.974 1.7667 19.3874
W4 = c(x)ws = 28.23 0.7500 21.1725
W5 = (B - x - b)(c + e)ws = 96.2643 2.4500 235.847535
W6 = 0.5(b - a)(c + e)ws = 8.7513 1.8333 16.04405
W7 = qs(B - x - a) = 24.96 2.3500 58.656
W8 = 0.5f(g +h)qs = 2.304 1.7000 3.9168
Wi = 228.8316 Wixi = 443.241085
Overturning moment: OM Factor of safety against overturning:
OM = Pahyah = 185.101 kN-m
Location of resultant with respect to toe: = 2.39458534 > 2.00, ok!
= 1.12808 m Factor of safety against sliding:
= 0.37192 m = 2.03907049 > 1.50, ok!
B/3 = 1.00 m > e, Rv will fall within the middle third of the base.
No tension will occur on the foundation.
qmax = 95.19002 kPa
qmin = 57.36438 kPa
qa = 143.5 kPa
qmax < qa, the wall is safe against soil bearing
Design of stem:
p1 = qs
P1
M y y/2 P2
V y/3
p2 = Cahwsx
Ve
d
Vmax Mmax
STEM P V M
Soil pressure at level y: Shear equation at level y:
p1 = qs = 19.2 kPa Vy = P1 + P2 = qsy + 0.5Cahwsy
p2 = Cahwsx = 5.10003 y kPa P1 = qsy = 19.2
P2 = 0.5Cahwsy = 2.55
'22
1hhwhCP ahah
2
2
1whCP phph
'23'3
2
hh
hhhyah
3
hyph
iv WR
OMRMx
x
B
e 2
ahah
ii
goverturninyPOM
cWRMFS
ah
phiv
sl id ing P
PWffR
FS
2minmax
61
B
e
B
Rq v
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Moment equation at level y:
My = P1y1 +P2y2 = qsy/2 + 0.5Cahwsy/3
M1 = P1y1 = qsy/2 = 9.6 y
M2 = P2y2 = 0.5Cahwsy/3 = 0.85000547 y
Given:
Level, y Vy Vu=1.7Vy My Mu=1.7My
0.00 0.000 0.000 0.000 0.000 Es = 200
0.50 10.238 17.404 2.506 4.261 fy = 414
1.00 21.750 36.975 10.450 17.765 fc' = 20.7
1.50 34.538 58.714 24.469 41.597 fshear= 0.85
2.00 48.600 82.620 45.200 76.840 fflexure = 0.90
2.50 63.938 108.694 73.281 124.578 Db = 25
3.00 80.550 136.935 109.350 185.895 Dtemp = 16
3.50 98.438 167.344 154.044 261.875 Smax = [3t, 450]min
4.00 117.600 199.920 208.000 353.6014.50 138.038 234.664 271.857 462.156
4.65 144.418 245.510 293.039 498.167
Compute:
490.874 mm 201.062 mm
= 0.0033816
= 0.01603
try d = 400 mm
= 3.4595
= 0.00939 ok!
As,flexure = bd = 3757.84 mm/m
= 130.627 mm oc
As,temp = tempbd = 0.002bd = 800 mm/m
= 251.327 mm oc
Check for shear:
= 257.818 kN/m
At d distance from bottom of stem:
y = 4.25 m
Vud = 1.7(19.2y + 3.13667y) = 235.03537 kN/m
At 3.00 m
Try d = 300 mm
= 2.29500
= 0.00596 ok!
2
4bo DA
2
4temptemp
DA
yf
4.1min
ys
s
y
c
fE
E
f
f
003.
003.'85.75. 1max
2
/
bd
MR uu
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
s
o
A
As
1000
temps
temp
tempA
As
,
1000
bdf
Vc
vcuc6
'
2
/
bd
MR uu
'85.211'85.2111
c
u
y
c
y
u
fR
ff
fR
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As,flexure = bd = 1788.48 mm/m
= 274.464 mm oc
As,temp = tempbd = 0.002bd = 600 mm/m
= 335.103 mm oc
Design of heel and toe:
ws2 Qs
subject to erosion ws1 a b
heel toe
a b
L1 L2
Use load factor:
1.4 for DL qmax qmin
1.7 for ll and service load bearing pressures q1 q2
qmax x 1.7 = 161.823 kPa e = 0.371923
qmin x 1.7 = 97.51944 kPa At a, x =B/2 - xh = 0
ws1 = 1.4(ws)c = 26.348 kPa q1 = 129.6712
ws2 = 1.4(ws)(c + e) = 122.5182 kPa At b, x = [B/2 - (xh + b)] = -0.4
qs x 1.7 = 32.64 kPa q2 = 103.9498
Wc = 1.4(wc)d = 16.52 kPa L1 = xh = 1.50
L2 = B - (xh + b) = 1.10
Va = .5(qmax - q1)L1 + q1L1 - (ws1=0)L1 - WcL1 = 193.840711 kN
Ma = (qmax - q1)L1/3 + (q1 - (ws1 = 0) - Wc)L1/2 = 151.408996 kN-m
Vb = (.5(q2 - qmin)L2 = 0) + [(qmin = 0) - ws2 - wc - qs]L2 = -188.84602 kN
Mb = [(q2 - qmin) = 0]L2/6 + [(qmin = 0) - ws2 - wc - qa]L2/2 = -103.86531 kN-m
ws1 is taken equal to zero due to erosion of top soil at the heel, the expected worst condition of loading.
At the toe, the worst condition of loading is when the wall is at empending action to overturn, soil bearing
at the heel is assumed to be zero.
At Heel:
try d = 400 mm
= 1.0514514
= 0.0026205 not ok!-use pmin
As,flexure = bd = 1352.657 mm/m
= 362.896026 mm oc
As,temp = tempbd = 0.002bd = 800 mm/m
= 251.327412 mm oc
s
o
A
As
1000
temps
temp
tempA
As
,
1000
2/bdMR uu
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
s
o
A
As
1000
temps
temp
tempA
A
s,
1000
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Check for shear:
= 257.81777 kN/m > Va, safe
At Toe:try d = 400 mm
= 0.72129
= 0.0017795 not ok!-use pmin
As,flexure = bd = 1352.657 mm/m
= 362.896026 mm oc
As,temp= tempbd = 0.002bd = 800 mm/m
= 251.327412 mm oc
Check for shear:
= 257.81777 kN/m > Va, safe
Design of Key:
pph1 pah1
f
pph2 h pah2
Cah = 0.27099 Cph = 3.69
At pah1: At pph1:
yah = h' + c + e + d = 6.170 m yph = c = 1.00
pah1 = Cahwsyah1 = 31.468178 kPa pph1 = Cphwsyph1 = 69.44904
At pah2: At pph2:
yah = h' + c + e + d + f = 6.5702 m yph = c + f = 1.40
pah2 = Cahwsyah2 = 33.508191 kPa pph2 = Cphwsyph2 = 97.22866
Vah = 1.7[pah1f + (pah2 - pah1)f/2] Vph = 1.7[pah1f + (pah2 - pah1)f/2]
Vah = 22.091966 kN Vph = 56.67042
Mah = 1.7[pah1f/2 + (pah2 - pah1)f
/3] Mph = 1.7[pah1f
/2 + (pah2 - pah1)f
/3]Mah = 4.465 kN-m Mph = 11.964
Net Shear:
Vu = Vph - Vah = 34.5784538 kN
Net Moment:
Mu = Mph - Mah = 7.499 kN-m
try d = 300 mm
= 0.09258
= 0.00022 not ok!-use pmin
bdf
Vc
vcuc6
'
2
/
bd
MR uu
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
s
o
A
As
1000
temps
temp
tempA
As
,
1000
bdf
Vc
vcuc6
'
'hhwCp ahah whCp phph
2
/
bd
MR u
u
'85.
2
11
'85.2
11
1
c
u
y
c
y
u
f
R
f
f
f
R
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As,flexure = bd = 1014.49275 mm/m
= 483.861369 mm oc
As,temp = tempbd = 0.002bd = 600 mm/m
= 335.103216 mm oc
Check for shear:
= 193.36332 kN/m > Va, safe
Details:
Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.
1.00 m
1.50 m
0
mmtemp @ 250 oc bw
3.650
mmtemp @ 250 oc bw
0.40
m
0.40
m
0.00 m 0.50 -0.10 m
0 mmtemp @
0.00 m 250 oc bw
s
o
A
As
1000
temps
temp
tempA
As
,
1000
bdf
Vc
vcuc6
'
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ES P. ANCOG
18-Aug-09
m
m
m
m
m
m
m
- 0.4
- 0.3
- 0.4
- 0.5
fficient
-0.6
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m
kN
m
!
!
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ES P. ANCOG
18-Aug-09
Pah
yah
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/3
y2
y3
0.003382
0.016032
safe!!!!
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kN
kN-m
mm2/m
mm cc
mm2/m
mm cc
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240 oc
250 oc
250 oc
360 oc
meters
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ES P. ANCOG
18-Aug-09
m
m
m
Mpa
kPa
D
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m
kN
m
base
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mm2
mm
2
kN/m
0.013431989
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mm @oc
mmtemp @oc
mm @oc
m
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kN
m
y
y
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GPa
MPa
MPa
mm
mm
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m
m
kPa
m
m
m
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m
kPa
m
kPa
kN
kN-m
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mm @
25
240 oc
16
250 oc
mmtemp @
16
120 oc16
360 oc
mm @
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CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOG
BSCE-5 2nd Excel Program 18-Aug-09
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)
****====Design of Cantilever Retaining Wall===****
Right Side Loading
Given:
fc' = 20.70 Mpa Retaining wall dimensions:
fy = 414.00 Mpa a = 0.30 m
s = 18.82 kN/m3 c = 4.50 m
= 40 o
= 0.50 Tentative dimentions:
c = 23.60 kN/m3 B = 3.20 m
qa = 143.50 kPa b = 0.40 m
qs = 19.30 kPa d = 0.50 m3.65 m
b(main) 25 mm Use Wu = 1.4DL + 1.7LL + 1.7H
b(temp) 16 mm
a
Es = 200000 Mpa
shear = 0.85 qs = 19.30 kPa
flexure = 0.9
smax = [ 3t , 450 ]min c
h' = 1.026 m
Cantilever Retaining Wall Figure:
d
b
Property Line
a
h'
c h
Pah
yah
db
B
Soil pressure coefficient, Rankine equation for horizontal surface:
Active soil pressure coeffecient: Cah = 0.217443
Active soil pressure: h = 5.00 m
Pah = 1 Cahh(h+2h') =
2 72.137 kN
yah =h
2+ 3hh' =
3(h + 2h') 1.9091 m
backfill height =
W1
C1
W2C2
W3
W4
W5
C3
C4
C5
W6
C6
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Check the retaining wall stability:
components weights Wi ci RM = Wici
W1 = 37.760 1.6000 60.4160
W2 = 31.860 0.1500 4.7790
W3 = 5.310 0.3333 1.7700
W4 = 237.132 1.8000 426.8376
W5 = 4.235 0.3333 1.4115
W6 = 55.97 1.75 97.9475
Wi = 372.267 RM = 593.1616
Overturning moment: OM = 137.7138 kN-m
Factor of safety against overturning:
FSoverturning = RM = 4.307 > 2.00 safe!!!!
OM
Factor of safety against sliding:
FSsliding = (f Wi) = 2.580 > 1.50 safe!!!!
Pah
Check for bearing pressure: qa = 143.50 kPa
Location of resultant with respect to toe:
x = RM - OM = 1.2234 m
Wi
e = B - x = 0.376554 m
2
B / 3 = 1.07 m
The middle third of the base where No tension will occur on the foundation.
q = Wi 1 + 6e qmax = 142.001 kPa
B B2
qmin = 90.666 kPa
Since qma < qa, wall is safe againts soil bearing.
Design of stem:
Soil pressure at y: Moment equation at level y:
p1 = qs = 19.30 kPa My = p1y1 + p2y2 = qsy2/2 + 0.5cahwsy
3/3
p2 = cahwsy = 4.09227 y kPa M1 = qsy2/2 = 9.65 y
2
M2 = 0.5cahwsy3/3 = 0.682 y
3
Shear equation at level y:
Vy= p1 + p2 = qsy + 0.5cahwsy2
At level y = 4.50 m
p1 = qsy = 19.30 y Vu = 218.1 kN
2.04614 y2
Mu = 437.9 kN-m
Db2
= 490.874 mm2
min = 1.4 / fy = 0.003382
p2 = 0.5cahwsy2
=
Ve
p1=qs
P1 P2
p2 = Cahwsy
y/2yM
V
dVmax Mmax
Stem P MV
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4 max = .75 .85fc'b1 600 = 0.016032
Atemp = Dtemp2
= 201.062 mm2
fy 600+fy
4
Try d = 400 mm ; b = 1000 mm
Ru = Mu = 3.040685
fbd2
= .85fc' 1 - 1 - 2Ru = 0.00812
fy .85fc'
Use: = 0.00812
As,flexure = bd = 3248.172 mm2/m
s = 1000Ao = 151.1231 mm cc
As
As,temp = bd =0.002bd = 800 mm2/m
stemp = 1000 Atemp = 251.3274 mm cc
As
Check for shear: Vuc = fc' bd = 257.8178 kN/m
6
At d distance from the bottom of stem:
yd = 4.10 m
Vud = 192.9935 kN/m < Vuc, safe!!!!
At y = 3.00 m
Try d = 300 mm Ru = Mu = 2.209270
fbd2
= .85fc' 1 - 1 - 2Ru = 0.005722
fy .85fc'
Use: = 0.005722
As,flexure = bd = 1716.459 mm2/m
s = 1000Ao = 285.9805 mm cc
As
As,temp = bd =0.002bd = 600 mm2/m
stemp = 1000 Atemp = 335.1032 mm cc
As
Design of Base:
19.30
5.00
0.50
Note: The expected worst condition of loading, the passive earth pressure of soil is generally
neglected due to erosion of top soil, and the soil bearing at the toe is neglected due to
the empending action to overturn.
qmin
qmax
qs =
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Use: 1.4 for DL
1.7 for LL and service load bearing pressure
qmax x 1.7 = 241.401 kPa V = (-Ws-Wc-qs)L = -470.11 kN
qmin x 1.7 = 154.132 kPa M = (-Ws-Wc-qs)L2/2 = -658.15 kN-m
Ws = 1.4sc = 118.566 kPaqs x 1.7 = 32.810 kPa
Wc = 1.4cd = 16.520 kPa
e = 0.377 m
L= B - b) = 2.800 m
Try d = 400 mm
b = 1000 mm
Ru = Mu = 4.570502 As,flexure = bd = 5216.20 mm2/m
fbd2
s = 1000Ao = 94.11 mm cc
As
= .85fc' 1 - 1 - 2Ru = 0.013041 As,temp = bd =0.002bd = 800.00 mm2/m
fy .85fc' stemp = 1000 Atemp = 251.33 mm cc
As
Use: = 0.01304
Check for shear: Vuc = fc' bd = 257.8178 kN/m > V, safe!!!!
6
Retaining Wall Details:
Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.
0.30
25 mm @ 240 oc
16 mm temp. @ 16
260 oc bw mm temp @ 250 oc
5.00 25 250 oc
m
25 mm @ 360 oc
0.50 meters
16 mm temp. bars
250 oc bw
0.40 meters
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CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOG
BSCE-5 2nd Excel Program 18-Aug-09
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)
****====Design of Cantilever Retaining Wall===****
Left Side Loading
Given:
fc' = 27.50 Mpa Retaining wall dimensions: Tentative dimentions:
fy = 275.00 Mpa a = 0.30 m B = 3.00 m
c = 23.60 kN/m3 H = 3.20 m b = 0.40 m
s = 17.25 kN/m3 h = 1.00 m D = 0.50 m
= 35 o
= 0.45 Surcharge load: Es = 200000 Mpaqa = 120.00 kPa qs2 = 19.50 kPa shear = 0.85
b(main) = 25 mm qs1
= 19.20 kPa flexure
= 0.9
b(temp) = 16 mm
smax = [ 3t , 450 ]min
Minimun factor safety requirements:
Overturning = 2.00
Sliding = 1.50
Cantilever Retaining Wall Figure:
qs2 = 19.50 kPa
a
h' = 1.11 m
H' = 1.13 m
H
qs1 = 19.20 kPa
b
B
Property Line
a
H' qs2
H
qs1
h'
h
D
x
bB
D
W3
W4
W5
W2
W1
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Soil pressure coefficient, Rankine equation for horizontal surface:
Active soil pressure coeffecient: Passive soil pressure coefficient:
Cah = 0.271 Cph = 3.690
h = 3.20 m h = 1.61 m
Active soil pressure:
Pah = 51.549 kN Passive soil pressure:
yah = 1.467 m Pph = 177.889 kN
yph = 1.079 m
Check the retaining wall stability:
components weights Wi xi Mi B-xi Mx
W1 = 22.656 0.1500 3.3984 2.850 64.5696
W2 = 3.776 0.3333 1.2587 2.667 10.06933
W3 = 35.400 1.5000 53.1000 1.500 53.1
W4 = 144.690 1.7000 245.9730 1.300 188.097
W5 = 18.225 0.3780 6.8890 2.622 47.78595Total = 202.091 310.6191 RM = 363.6219
Overturning moment: OM = 75.6346 kN-m
= 0.03124 x = 0.066
Factor of safety against overturning:
FSoverturning = RM = 4.808 > 2.00 safe!!!
OM
Factor of safety against sliding:
FSsliding = (f Wi)+Pph = 5.215 > 1.50 safe!!!Pah
Check for bearing pressure:
B / 3 = 1.000 m
x = RM - OM = 1.4250 m within 1/3 of base
Wiq = 2WT , when x < 1/3 B = 94.543 kPa
3x
qmax = [4B - 6x]WT/B2
= 77.46 kPa safe!!!
qmin = [6x - 2B]WT/B2
= 57.26 kPa safe!!!
Design of stem:
Vu = 1.7 [qsH + 0.5CahwsH2
= 146.768 kN/m
Mu = 1.7 Mmax = 220.696 kN-m/m
M = qsy
2
/2 + 1/3Cahwsy
3
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y M Amain = Dmain2
= 490.874 mm2
0.50 2.693 kN-m 4
1.00 11.797 kN-m Atemp = Dtemp2
= 201.062 mm2
1.50 27.196 kN-m 4
2.00 51.466 kN-m min = 0.5 [1.4/fy, fc'/4fy] = 0.002383662.50 85.284 kN-m max = .75 .85fc'b1 600 = 0.037157143.00 129.821 kN-m fy 600+fy
Vuc = fc' = 0.7429 kN/m6
Depth as required by shear: b = 1000 mm
d = Vu = 197.56 mm
Vucb
Design for flexure:
Try h = = 400 mmd = h - (100+s/2) = 287.5 mmRu = Mu = 2.966719
fbd2
= .85fc' 1 - 1 - 2Ru = 0.011576fy .85fc'
Use: = 0.011576
As = bd = 3328.208 mm2/m
spacing, S = [1000Ao/As, 3t,450]min = 147 mm oc
Atemp = 0.0018bd = 517.50 mm2/m
spacing, S = [1000Atemp/As, 5t,450]min = 388 mm oc
Design of Toe:
Vu = 1.7b [(qmax + qmin)/2 - (wcd + ws (h+h'))]
= 97.4797 kN/m
Mu = 1.7 [(B-b)2/2][qmin +2/3(B-b)(qmax - qmin) - [wc(d)-ws(h+h')]]
= 253.0727 kN-m/m
min = 0.5 [1.4/fy, fc'/4fy] = 0.002384max = .75 .85fc'b1 600 = 0.037157
fy 600+fy
Vuc = fc' = 0.7866 kN/m6
Depth as required by shear: b = 1000 mm
d = Vu = 123.9243 mm
VucbDesign for flexure:
Try h = 400 mm
d = h - (100+s/2) = 287.5 mmRu = Mu = 3.401944
fbd2
= .85fc' 1 - 1 - 2Ru = 0.013432 Use: = 0.01343199fy .85fc'
As = bd = 3861.697 mm2/m
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spacing, S = [1000Ao/As, 3t,450]min = 127.00 mm oc
3t = 1200.00
450 = 450
Compare: 1000Ao/AS AND 3t, compare 3t and 450 whichever is smallest.
SINCE 127.00 mm < 450.00 mm
Use: 127.00 mm
Atemp = 0.0018bd = 517.50 mm2/m
spacing, S = [1000Atemp/As, 5t,450]min = 489 mm oc
Retaining Wall Details:
0.30
25 mm @137 oc
16 mmtemp @3.70 338 oc
25 mm @117.00 oc
0.50 m
3.00
16 mmtemp @0.40 338 oc
Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.
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CARMEL B. SABADO CE-162 PROF. GERONIDES P. ANCOG
BSCE-5 2nd Excel Program 18-Aug-09
*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)
****====DESIGN OF A CANTILEVER RETAINING WALL===****
(with shear key)
fc' = 20.7 Mpa Figure: x=1.5m
fy = 414 Mpa
ws = 18.82 kN/m qa = 19.2 kPa
= 35
f= 0.5
wc = 23.6 kN/m e=3.65m
qa = 143.5 kPa
qs = 19.2 kPa
backfill height = 3.65 m
Use Wu = 1.4DL + 1.7LL + 1.7H c=1.0m
Use 25mm for main rebars, 16mm for
temperature bars.
Solution:
Composite section and location of forces
e
c
d
f
Given retaining wall dimensions:
a = 0.20 m Distances:
c = 1.00 m x1 = B/2 = 1.5000 m
e = 3.65 m x2 =xh + a/2 = 1.6000 m
xh = 1.50 m x3 = xh + a + (b - a)/3 = 1.7667 m
x4 = xh/2 = 0.7500 m
Tentative dimensions: x5 = xh + b + (B - xh - b)/2 = 2.4500 m
B = 3.00 m x6 = xh + a + 2(b - a)/3 = 1.8333 m
b = 0.40 m x7 = xh + a + (B - xh - a)/2 = 2.3500 m
d = 0.50 m x8 = xh + g/2 = 1.7000 m
f = 0.40 m H = d + c + e = 5.1500 m
g = 0.40 m h' = qs/ws = 1.0202 m
h = 0.20 m
Active soil pressure coefficient Passive soil pressure coefficient
= 0.27099 = 3.69
Active soil pressure: h = 5.15 m Passive soil pressure: h = 1.50 m
h
H
ax
e
b
B
sin1
sin1
ahC
sin1
sin1
phC
7W
6W
5W
4W 3W
2W
1W
8W
ah
x7
x
6x
5x
2x
4x
3x
1x
8x
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= 94.4283 kN = 78.1302 kN
= 1.960 m = 0.500 m
Check retaining wall stability:
Component weights Wi xi RM=Wixi
W1 = Bdwc = 35.4 1.5000 53.1
W2 = a(c + e)wc = 21.948 1.6000 35.1168
W3 = 0.5(b - a)(c + e)wc = 10.974 1.7667 19.3874
W4 = c(x)ws = 28.23 0.7500 21.1725
W5 = (B - x - b)(c + e)ws = 96.2643 2.4500 235.848
W6 = 0.5(b - a)(c + e)ws = 8.7513 1.8333 16.0441
W7 = qs(B - x - a) = 24.96 2.3500 58.656
W8 = 0.5f(g +h)qs = 2.304 1.7000 3.9168
Wi = 228.832 Wixi = 443.241
Overturning moment: OM Factor of safety against overturning:
OM = Pahyah = 185.101 kN-m
Location of resultant with respect to toe: = 2.39459 > 2.00, ok!
= 1.12808 m Factor of safety against sliding:
= 0.37192 m = 2.03907 > 1.50, ok!
B/3 = 1.00 m > e, Rv will fall within the middle third of the base.
No tension will occur on the foundation.
qmax = 95.19 kPa
qmin = 57.3644 kPa
qa = 143.5 kPa
qmax < qa, the wall is safe against soil bearing
Design of stem:
p1 = qs
P1
M y y/2 P2
V y/3
p2 = Cahwsx
Ved
Vmax Mmax
STEM P V M
Soil pressure at level y: Shear equation at level y:
p1 = qs = 19.2 kPa Vy = P1 + P2 = qsy + 0.5Cahwsy
p2 = Cahwsx = 5.10003 y kPa P1 = qsy = 19.2 y
P2 = 0.5Cahwsy = 2.55 y
Moment equation at level y:
My = P1y1 +P2y2 = qsy/2 + 0.5Cahwsy/3
M1 = P1y1 = qsy/2 = 9.6 y
'22
hhwhCP ahah 2
2whCP phph
'23'3
2
hh
hhhyah
3
hyph
iv WR
OMRMx
xB
e 2
ahah
ii
goverturninyPOM
cWRMFS
ah
phiv
slidingP
PWffRFS
2minmax
61
B
e
B
Rq v
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M2 = P2y2 = 0.5Cahwsy/3 = 0.85001 y
Given:
Level, y Vy Vu=1.7Vy My u=1.7My
0.00 0.000 0.000 0.000 0.000 Es = 200 GPa
0.50 10.238 17.404 2.506 4.261 fy = 414 MPa
1.00 21.750 36.975 10.450 17.765 fc
' = 20.7 MPa
1.50 34.538 58.714 24.469 41.597 fshear= 0.85
2.00 48.600 82.620 45.200 76.840 fflexure = 0.90
2.50 63.938 108.694 73.281 124.578 Db = 25 mm
3.00 80.550 136.935 109.350 185.895 Dtemp = 16 mm
3.50 98.438 167.344 154.044 261.875 Smax = [3t, 450]min
4.00 117.600 199.920 208.000 353.601
4.50 138.038 234.664 271.857 462.156
4.65 144.418 245.510 293.039 498.167
Compute:
490.874 mm 201.062 mm
= 0.00338
= 0.01603
try d = 400 mm
= 3.4595
= 0.00939 ok!
As,flexure = bd = 3757.84 mm/m
= 130.627 mm oc
As,temp = tempbd = 0.002bd = 800 mm/m
= 251.327 mm oc
Check for shear:
= 257.818 kN/m
At d distance from bottom of stem:
y = 4.25 m
Vud = 1.7(19.2y + 3.13667y) = 235.035 kN/m
At 3.00 m
Try d = 300 mm
= 2.29500
= 0.00596 ok!
As,flexure = bd = 1788.48 mm/m
= 274.464 mm oc
2
4bo
DA
2
4temptemp
DA
yf
4.1min
ys
s
y
c
fE
E
f
f
003.
003.'85.75. 1
max
2
/
bd
MR u
u
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
s
o
A
As
1000
temps
temp
tempA
As
,
1000
bdf
Vc
vcuc6
'
2
/
bd
MR u
u
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
oAs 1000
7/27/2019 Muro L e Gravidade
53/55
7/27/2019 Muro L e Gravidade
54/55
try d = 400 mm
= 0.72129
= 0.00178 not ok!-use pmin
As,flexure = bd = 1352.66 mm/m
= 362.896 mm oc
As,temp= tempbd = 0.002bd = 800 mm/m
= 251.327 mm oc
Check for shear:
= 257.818 kN/m > Va, safe
Design of Key:
pph1 pah1
f
pph2 h pah2
Cah = 0.27099 Cph = 3.69
At pah1: At pph1:
yah = h' + c + e + d = 6.170 m yph = c = 1.00 m
pah1 = Cahwsyah1 = 31.4682 kPa pph1 = Cphwsyph1 = 69.449 kPa
At pah2: At pph2:
yah = h' + c + e + d + f = 6.5702 m yph = c + f = 1.40 m
pah2 = Cahwsyah2 = 33.5082 kPa pph2 = Cphwsyph2 = 97.2287 kPaVah = 1.7[pah1f + (pah2 - pah1)f/2] Vph = 1.7[pah1f + (pah2 - pah1)f/2]
Vah = 22.092 kN Vph = 56.6704 kN
Mah = 1.7[pah1f/2 + (pah2 - pah1)f/3] Mph = 1.7[pah1f/2 + (pah2 - pah1)f/3]
Mah = 4.465 kN-m Mph = 11.964 kN-m
Net Shear:
Vu = Vph - Vah = 34.5785 kN
Net Moment:
Mu = Mph - Mah = 7.499 kN-m
try d = 300 mm
= 0.09258
= 0.00022 not ok!-use pmin
As,flexure = bd = 1014.49 mm/m
= 483.861 mm oc
As,temp = tempbd = 0.002bd = 600 mm/m
2
/
bd
MR uu
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
s
o
A
As
1000
temps
temp
tempA
As
,
1000
bdf
Vc
vcuc6
'
'hhwCp ahah whCp phph
2/
bdMR uu
'85.
211
'85.211
1
c
u
y
c
y
u
f
R
f
f
f
R
s
o
A
As
1000
7/27/2019 Muro L e Gravidade
55/55
= 335.103 mm oc
Check for shear:
= 193.363 kN/m > Va, safe
Details:
Note: the spacing of the reinforcement is upon the desire of the designer based on the computed bar spacing above.
0.20 m
3.65 m 25 mm @
240 oc
16 mmtemp @
16 250 ocmmtemp @ 250 oc bw
25 mm @
1.00 120 oc
16 25
mmtemp @ 250 oc bw mm @ 360 oc
0.50
m
0.40
m
1.50 m 0.40 1.10 m
16 mmtemp @
0.20 m 250 oc bw
temps
temp
tempA
As
,
1000
bdf
Vc
vcuc6
'