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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
EXAMPLES
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Sample No. 1 Determine the maximum positive moment and the negative moment at the fixed support.
58
7= 1.0879 6(
58
7) = 6.5278
Cantilever M= 8.4482 x (1.0879)2/2
MB= 5 kNm
FEMs= 1
12 (8.4482)(6.5278)2 = 30.000
Mc = 30 +25
2= 42.5
Point of zero shear:
=21.829
8.4482= 2.5839
=1
2(21.829)(2.5839) 5 = .
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Sample No. 2 Determine the moment at point c.
Columns steel pipe, outer = 300mm, thickness=10mm, G=200 GPa
Rafters aluminum, I section, Flanges, 250 x 20mm, web 450 x 10 mm, G=83 GPa
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
=
64(3004 2804) = 95.8891064
=1
12(2504903 2404503) = 628.521064
=. 75 (
200 95.8896
)
. 75 (20095.889
6 ) +83628.52
45
= 0.29130
= 0.7087
=1
12(10)(45)
2= 66.6667
= 66.667 +0.7087
2(66.667 0) = .
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
WSD SINGLY
Example 1. Determine the moment capacity of beam if the area of reinforcing bars is 1400
mm2. If fs=120MPa and fc=20MPa.
n = Es4.7fc
= 200
4.720= 9.5152 10
n = Asbd n = 1400
(200)(400)(10) = 0.17500
k = n + (n2) + 2(n)
k = 0.175 + (0.1752) + 2(0.175)
k = 0.44195
c = k d = 0.44195 x 400 = 176.78
Itr = bc3
3+ nAs(d c)
2 = (200)(176.78)3
3+ (10)(1400)(400 176.78)2
= .
Due to Concrete: Mcap = fcItr
c
Mcap = (0.45)(20)(1066.073)
(176.78)= .
Due to Steel: Mcap = fsItr
(d c)(n)
Mcap = (120)(1066.073)
(400176.78)(10)= .
= .
As
200 mm
40
0 m
m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
80
2 w
1.7 w
BOTTOM BARS TOP BARS
70
4-32
2-20
70
4-32
SECTION B
A B C
SECTION AB
Example 2. Determine the maximum safe load w based on the following stresses, fc = 25 MPa
and fs = 138 MPa.
n = Es
4.7fc=
200
4.725= 8.5106 9
SECTION B:
d = 4 x 322x 80
4 x 322 + 2 x 202+ 375 = 441.93 mm
n = Asbd
n =
4 (4 x 32
2 + 2 x 202)
(400)(441.93)(9) = 0.195776
c = [n + (n2) + 2(n) ] d = 203.23 mm
Itr = bc3
3+ nAs(d c)
2 = (400)(203.233)3
3+ (9) (
4) (4 x 322 + 2 x 202)(441.93 203.23)2
= .
5 m 4 m
37
5
400
45
5
400
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Due to Concrete: Mcap = fcItr
c
Mneg = (0.45)(20)(3091.057)
(203.23)= .
Due to Steel: Mcap = fsItr
(d c)(n)
Mneg = (138)(3091.057)
(441.93203.23)(9)= .
= .
SECTION B:
d = 455 mm
n = Asbd
n =
4 (4 x 32
2)
(400)(455)(9) = 0.159082
c = [n + (n2) + 2(n) ] d = 194.2768 m
Itr = bc3
3+ nAs(d c)
2 = (400)(194.2768)3
3+ (9) (
4) (4 x 322)(455 194.2768)2
= .
Due to Concrete: Mcap = fcItr
c
Mpos = (0.45)(20)(2945.811)
(194.2768)= .
Due to Steel: Mcap = fsItr
(d c)(n)
M = (138)(2945.811)
(455194.2768)(9)= .
= .
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
= 1
5
15 +
0.754
= 0.51613
= 0.48387
FEMs:
= 2
12 (52) = 2.08334
= 1.7
12 (42) 1.5 = 3.4
FINAL Ms:
= 2.08334 (0.48387) + 3.4 (0.51613) = 2.76307
= 2.08334 + 1
2(2.08334 w 3.4 w)(0.51613) = 1.74355 w
= 0
=
2+
= 5.2039
2() = 5.2039
= 2.60195
Mpos = (5.2039 2.60195
2) (2.76307) = 4.00708 w
4.00708 = 136.467
= .
Mneg = 171.1086
2.76307 = 171.1086
= .
= .
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
3-36mm
3-25mm
WSD DOUBLY
In the figure shown, determine the deflection at free end. d = 75mm, d = 595mm and b =
360mm. Use fc = 35 MPa.
=
4.7=
200
4.735= 7.2 7
Solve for As and As,
=
4(36)2 3 = 3053.632
= 7 3053.63 = 21375 2
=
4(25)2 3 = 1472.6 2
(2 1) = [(27) 1] 1472.6 = 1914.4 2
Let A = b/2, = 360
2= 180
= (2 1) + = 1914.4 + 21375 = 40519
= (2 1) + = (19144 75) + (21375 595) = 14153925
Substitute A, B and C on the equation for c, we obtain
c = 40519+ 405192+4(180+14153925)
2 180= 189.61
220 kN-m
40 kN-m
3.20 m
360
52
0
75
7
5
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Calculate Itr,
Itr = 3
3+ ( )
2 + ((2 1))( )2
= 360+189.612
3+ 21375(370.39)2 + 1914.4(114.61) = 4582.31064
Ec = 4700 = 470035 = 27806
= 3
3+
4
8=
220(3.2)3(10)12
3 27.806 4.5823102+
40(3.2)4(10)12
8 27.806 4.5823 1012
= .
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
WSD IRREGULAR
Example 1. Given L = 10m., find max w that the beam can carry safely
kNmMcapTherefore
mkNwmkN
L
MwwLMMcap
MPafs
kNmMcap
MPafc
mm
I
mmc
mmC
mm
therefore
or
mmcAssume
mmnA
sayn
tr
s
667.86
/9333.6m-kN 122.59 667.86
10
667.8688
8
110
)10)(56.173550(
)101.2364(138
138
steelon Based
59.1221056.173
)101.2364(9
92045.
stresses allowable Concreteon Based
101.2364
)56.173550(125663
)56.173(150)56.173)(100)(200()100)(200(
12
1
56.173
7911300)550(125662
)200(100
3256612566)200(100B
75mm.150/2A
sectionirregular:
5654700175000
)100550(1256650100 x 350
.100
125664)20(4
10
1052.9207.4
200
22
26
6
6
6
46
2
3
23
3
2
2
22
55
0
75 75 200
400
100
100
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
USD SINGLY
USD DOUBLY
USD IRREGULAR
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
ONE WAY SLAB
1. SAMPLE PROBLEM a. Determine the required thickness of a prismatic one way slab shown below (round
up to nearest cm). b. Determine the required spacing of the bars in the first interior support. Use grade 40, 10mm diameter bars and fc=25 MPa.
a.
24=
3100
24= 129.17mm.
28=
2900
28= 103.57 .
Reqd thickness= 129.17[0.4 +276
700]
T =102.60 say 110mm
b. wu =1.4[5+(0.11 24)] + 1.7(3) = 15.796 kPa
Mu = 1
10(15.796)(3.1)2 = 15.180 (governs)
Mu =1
12(15.796)(2.9)2 = 11.070 d=110-20-
10
2= 85
X=15.18
0.91000(0.0852)276 = 0.03363
m =276
0.8525= 12.988
=112
= 0.0089822
=1.4
4= 0.00507 0.004289
= 0.00507 <
Reqd s= (10)2
4(0.0089822)(85)= 102.87 ( 100)
Smax = 3Ats or 450= 330 (governs) > Reqd s Adopt smallest S:
S= 100 mm
3.4 m 3.2 m 3.2 m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
STRINGERS
Sample Problem No.1 Compute for the moment of the prismatic stringer shown below. Use superimposed dead load = 2.1 kPa and a live load of 2.4 kPa and a slab thickness = 110 mm. Assume beam width = 250mm. Also, assume that the dimension of the stringer is 200 x 500mm.
1. Spans are more than two. 2. Check all the span lengths if it is okay to use the ACI Moment Coefficient:
8
7= 1.14286 ()
7.2
7= 1.02857 ()
8.2
7.2= 1.0 ()
3. Convert Floor Load to Uniform Load:
Total Dead Load:
Superimposed Dead Load = 2.1 kPa
Weight of Slab = 24 * 0.11
= 2.64 kPa 4.74 kPa
stringers
8 @ 3.5 m = 28 m
stringers
Weight of Stringer = 24 * 0.2 * 0.5
= 2.4 kN/m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Loading for 8m span: m = 3.5/8=0.4375
WD = ( 4.74* 3.5)/8 ((3-(0.4375)^2)/2)*2 = 5.83661 kN/m + 2.4 kN/m = 8.23661 kN/m
WL = ( 2.4* 3.5)/8 ((3-(0.4375)^2)/2)*2
= 2.94902 kN/m Wu = 1.2 (8.23661) + 1.6 (2.94902) = 14.60236 kN/m Loading for 7m span: m = 3.5/7=0.5
WD = ( 4.74 * 3.5)/7 ((3-(0.5)^2)/2)*2
= 6.5175 kN/m + 2.4 kN/m = 8.9175 kN/m
WL = ( 2.4* 3.5)/7 ((3-(0.5)^2)/2)*2
= 3.3 kN/m Wu = 1.2 (8.9175) + 1.6 (3.3) = 15.981 kN/m Loading for 7.2m span: m = 3.5/7.2=0.48611
WD = ( 4.74* 3.5)/7.2 ((3-(0.48611)^2)/2)*2
= 6.368 kN/m + 2.4 kN/m = 8.768 kN/m
WL = ( 2.4* 3.5)/7.2 ((3-(0.48611)^2)/2)*2 = 3.22431 kN/m Wu = 1.2 (8.768) + 1.6 (3.22431) = 15.680 kN/m Loading for 8.2m span: m = 3.5/8.2=0.42683
WD = ( 4.74 * 3.5)/8.2 ((3-(0.42683)^2)/2)*2 = 5.70092 kN/m + 2.4 kN/m = 8.10092 kN/m
WL = ( 2.4* 3.5)/8.2 ((3-(0.42683)^2)/2)*2
= 2.88654 kN/m Wu = 1.2 (8.10092) + 1.6 (2.88654) = 14.33957 kN/m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
For 8m span: M = 1/10 Wuln2 = 1/10 (14.60236) (7.75)2 = 87.705 kN-m M = 1/14 Wuln2 = 1/14 (14.60236) (7.75)2 = 62.647 kN-m M = 1/24 Wuln2 = 1/24 (14.60236) (7.75)2 = 36.544 kN-m For 7m span: M = 1/11 Wuln2 = 1/11 (15.981) (6.75)2 = 66.194 kN-m M = 1/16 Wuln2 = 1/16 (15.981) (6.75)2 = 45.508 kN-m For 7.2m span: M = 1/11 Wuln2 = 1/11 (15.680) (6.95)2 = 68.853 kN-m M = 1/16 Wuln2 = 1/16 (15.680) (6.95)2 = 47.336 kN-m For 8.2m span: M = 1/10 Wuln2 = 1/10 (14.33957) (7.95)2 = 90.63 kN-m M = 1/14 Wuln2 = 1/14 (14.33957) (7.95)2 = 64.74 kN-m M = 1/24 Wuln2 = 1/24 (14.33957) (7.95)2 = 37.76 kN-m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
TWO WAY SLAB
SAMPLE PROBLEM 1. Determine the required thickness of the slab shown in the figure. Assume
all beam widths are 300mm x 600mm and all dimensions shown are center-to-center of beams.
Use Gr.40 bars and fc' = 25MPa.
Since the biggest panel is Panel C having clear spans of 4.9m x 4.2m, this is to be
considered:
= (0.8 +
1400)
36 + 9=
4900(0.8 +276
1400)
36 + 9 (49004200)
= 105.08 ~110
= . ~
5.0 m
5.2 m
5.0 m
4.2 m 4.5 m 4.2 m
A B
C D
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 2.
1.) Determine the reqd thickness of the floor using Grade 60 bars. 2.) Determine the design negative moment at the continuous edge in the long direction of
the middle strip of slab A if total DL is 6.3 kPa and LL is 2.4 kPa. 3.) Determine the design negative moment at the discontinuous edge of slab B in the column
strip in the short direction using the loads in problem 2. All beam widths = 300 mm
mm 110.07 h dq'
mm 991.1*90 mm 1500414 0.8 44001500
fy 0.80 lnh 1.)
:SOLUTIONS
Re
07.11010.1*
5.3
4.4*936
936
strip 1mm/ -kN
5020.62.4*4.11*032333.0Mb
032333.0005.0*2*33333.0029.0Cb
83333.04.2
3.5m
4 CASE
kPa 4.11 W
84.34.2*6.1W6.1
56.73.6*2.1W2.1
.)2
2
neg
neg
U
L
D
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
m/m-kN 1.14023
2*
3
5.1311Ma,
Strip)(Column Edge ousDiscontinu
1311.51339.29972.2Ma,
1339.23.5*3.84*0.045364LL Ma,
0.0453640.004*2*0.4545-0.049LL Ca,
9972.23.5*7.56*0.032364DL Ma,
0.0323640.004*2*0.4545-0.036DL Ca,
8 CASE
79545.04.4
3.5m
.)3
neg
pos
2
pos
pos
2
pos
pos
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 3.
Design the exterior panel of the two-way slab using direct design method.
Given Data:
fy= 276 MPa fc= 33 MPa
g= 0.002
max= 0.04173
min= 0.0052 m= 9.8396 Bar size= 12mm Assumed beam width= 0.2m Assumed Depth= 0.4m Dead Load = 1.83kPa Live Load = 1.9kPa Concrete Cover = 20mm
Span Length: Along z-direction=3m Along x-direction=2.15m Clear Span Length: Along z-direction=3-0.2= 2.8m Along x-direction=2.15-0.2= 1.95m
Therefore: ln = 2.8m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Slab Thickness: Assume conventional slab with stiff beams ( > 2.0)
= 2.8/1.95 = 1.4359
= (0.8 +
1400
)
36 + 9=
2.8(0.8 +276
1400)
36 + 9(1.4359)= 57.07~60
therefore: h=100mm since 100 is the minimum slab thickness
Load Calculation: Total Factored Load= 1.2(1.83+24*.1)+1.6(1.9)= 8.12kPa At Z-direction End Span, Slabs with beams between all support
Exterior Side Interior Side ln 2.8m 2.8m l1 3m 3m l2 1.075m 2.15m
lB=bh^3/12 1.61094 2.131094 lS=bh^3/12 9x1074 1.791084
Mo=2
2
8 3.28kN-m 13.13kN-m
17.8605 11.9070 X1 = assumed beam width = 200mm Y1 = assumed depth = 400mm X2 = h = 100mm Y2 = smallest between 4h and Y1-h = 300mm
= (1 0.63
)
3
3
= (1 0.63200
400) (
2003400
3) + (1 0.63
100
300) (
1003300
3) = 8.11084
=
2= 1.6193
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Based from the NSCP 2010 Code Sec. 413.7.3.3; Since the kind of slab has beams between all supports, the Interior Neg. Moment is 0.7, the Positive Moment is 0.63 and the Exterior Neg. Moment is 0.16.
Column strip for Exterior Negative Moment
Exterior Side Interior Side Column strip 0.5375m 1.075m
2 1 0.3583 0.7167 2 1 6.4 8.5333
2 1 = 0 75 75 2 1 > 1 94.25% 83.5% Col. Strip 94.25% 83.5%
Middle strip 5.75% 16.5% Moment 0.03kN-m 0.347kN-m
Overall width 0.5375m 1.075m Moment 0.056 kN-m/m 0.322 kN-m/m
Width of side 0.80625 m 0.5375 m Total Exterior Negative Moment = 0.1627 Column strip for Exterior Positive Moment
Exterior Side Interior Side Column strip 0.5375m 1.075m
2 1 0.3583 0.7167 2 1 6.4 8.5333
2 1 = 0 60 60 2 1 > 1 94.25% 83.5% Col. Strip 94.25% 83.5%
Middle strip 5.75% 16.5% Moment 0.108kN-m 0.108kN-m
Overall width 0.5375m 1.075m Moment 0.2 kN-m/m 0.1 kN-m/m
Width of side 0.80625 m 0.5375 m Total Exterior Positive Moment = 0.16013 At X-Direction End Span, Slabs with beams between all supports
Left Side Right Side ln 2.8m 2.8m l1 2.15m 2.15m l2 2.75m 2.75m
lB=bh^3/12 2.11094 21094 lS=bh^3/12 2.3x1084 21084
Mo=2
2
8 21.48kN-m 21.48kN-m
9.3091 9.3091
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
X1 = 200mm Y1 = 400mm X2 = 100mm Y2 = 300mm
= (1 0.63
)
3
3
= (1 0.63200
400) (
2003400
3) + (1 0.63
100
300) (
1003300
3) = 8.11084
=
2= 2.2595
Based from the NSCP 2010 Code Sec. 413.7.3.3; Since the kind of slab has beams between all supports, the Interior Neg. Moment is 0.7, the Positive Moment is 0.63 and the Exterior Neg. Moment is 0.16. Column strip for Exterior Negative Moment
Left Side Right Side Column strip 1.075m 1.075m
2 1 0.3583 0.7167 2 1 6.4 8.5333
2 1 = 0 75 75 2 1 > 1 94.25% 83.5% Col. Strip 94.25% 83.5%
Middle strip 5.75% 16.5% Moment 0.198kN-m 0.567kN-m
Overall width 0.5375m 1.075m Moment 0.368 kN-m/m 0.528 kN-m/m
Width of side 0.5375 m 0.5375 m Total Exterior Negative Moment = 0.4476 Column strip for Exterior Positive Moment
Left Side Right Side Column strip 1.075m 1.075m
2 1 0.3583 0.7167
2 1 6.4 8.5333 2 1 = 0 60 60
2 1 > 1 94.25% 83.5% Col. Strip 94.25% 83.5%
Middle strip 5.75% 16.5% Moment 0.108kN-m 0.108kN-m
Overall width 0.5375m 1.075m Moment 0.2 kN-m/m 0.1 kN-m/m
Total Exterior Positive Moment = 0.15012
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
DESIGN ALONG Z-DIRECTION For Bottom Bars Ext. Neg. Moment = 0.1627 d = h cc- .5db = 100 20 - 1.5(12) = 62mm
=
2=
(0.1627 106)
0.91000622276= 0.0002
= 1 1 2
=
1 1 2(9.8396)(0.0002)
9.8396= 0.0002
min= 0.0052
Since min is greater than , therefore use min
=
= 122
4
0.0052(62)= 350.39~350
Smax = 450 or 3h, whichever is lesser. Since 3h = 3x100, Therefore; Smax = 200mm Smax Since s is greater than Smax, therefore use Smax Spacing of the bottom bars is 200mm
For Top Bars (z-direction) For the design of top bars, same procedure with the bottom bars. The only difference is the maximum moment. Ext. Neg. Moment = 0.16 d = h cc- .5db = 100 20 - 1.5(12) = 62mm
=
2=
(0.16 106)
0.91000622276= 0.0002
= 1 1 2
=
1 1 2(9.8396)(0.0002)
9.8396= 0.0002
min= 0.0052
Since min is greater than , therefore use min
=
= 122
4
0.0052(62)= 350.39~350
Smax = 450 or 3h, whichever is lesser. Since 3h = 3x100, Therefore; Smax = 200mm Since s is greater than Smax, therefore use Smax Spacing of the top bars is 200mm
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
DESIGN ALONG X-DIRECTION For Bottom Bars Ext. Neg. Moment = 0.4476 d = h cc- .5db = 100 20 - 1.5(12) = 62mm
=
2=
(0.4476 106)
0.91000622276= 0.0005
= 1 1 2
=
1 1 2(9.8396)(0.0005)
9.8396= 0.0002
min= 0.0052
Since min is greater than , therefore use min
=
= 122
4
0.0052(62)= 350.39~350
Smax = 450 or 3h, whichever is lesser. Since 3h = 3x100, Therefore; Smax = 200mm Since s is greater than Smax, therefore use Smax Spacing of the bottom bars is 200mm For Top Bars (x-direction) For the design of top bars, same procedure with the bottom bars. The only difference is the maximum moment. Ext. Neg. Moment = 0.15 d = h cc- .5db = 100 20 - 1.5(12) = 62mm
=
2=
(0.15 106)
0.91000622276= 0.0002
= 1 1 2
=
1 1 2(9.8396)(0.0002)
9.8396= 0.0002
min= 0.0052
Since min is greater than , therefore use min
=
= 122
4
0.0052(62)= 350.39~350
Smax = 450 or 3h, whichever is lesser. Since 3h = 3x100, Therefore; Smax = 200mm Since s is greater than Smax, therefore use Smax Spacing of the top bars is 200mm
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SHEAR
SAMPLE PROBLEM 1. Design the shear reinforcements of the beam shown using 10 mm dia., two legged stirrups. The beam width is 300 mm and effective depth is 630 mm. Use fc = 24 MPa & Gr. 60
stirrups.
kN 14.402V
7
500600
7
4250
2
770V
kN 86.337V
7
600500
7
3250
2
770V
BA
BA
AB
AB
mm 248.44242
433.54s
mm 600
mm 300say (governs) mm 3152
mm 630
mm 54.433300
27608.1573s
kN 40.571Vc
mm 08.5714
102A 1063000324)1(17.0
bdcf'17.0Vc
min
max
2
2
V
3
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
m .780770
207.49-402.14
W
V-V x
kN 0 V
kN 86.708 630276157.08
s
dfA V
entreinforcem shear minimum For
kN 358.040.6370-402.14
dW-V V
:side rightof Design
U
UBA
1
U
yV
S
UBCmax
2
49.207708.8640.15785.
300
mm 100 @rest mm, 50 @ 1 mm, 10 use
mm 100say mm 103.53
157.40-0.85
358.04
10630276157.08
VV
dfAs
:support from d"" distance @
3-
CU
yV
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 1. Determine the 2 legged stirrups, 10 mm stirrups, 2h from fixed
support.
WDL = 50
2 PLL = 150
2
WLL = 30
2 fc = 27
PDL = 150
2 Grade 40 rebars, cc = 40 mm.
Main bars = 20 mm.
P
W
B = 300
4m h = 600
400 x 400
4m 5m
Solution:
IAB = 44 = 256
IBC = 3 63 = 648
DFBA = 0.75
256
40.75 256
4+
648
9
= 0.4
DFBC = 0.60
WU = (1.4 50) + (1.7 30) = 121
PU= = (1.4 150) + (1.7 80) = 345
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Fixed End Moments
MBC = 121 92
12 +
346 4 52
92 = 1243.91 kNm
MB = 121 92
12 +
346 42 5
92 = 1158.48 kNm
Final Moments
MB = 1243.91 0.4 = 497.56 kNm
MC = 1158.48 + 0.6
2 (1243.91) = 1531.65 kNm
Shears
VCB = 121 9
2
497.56 1531.65
9 +
4 ( 346 )
9 = 813.18 kNm
@ 2h = 2 60 = 1.2m from fixed support
VC = 1
627 300 540 10-3 = 140.29 kN
d = 600 40 10 20
2 = 540 mm
AV = 2 102
4 = 157.08 mm2
Reqd S = 157.08 276 540
667.98
0.85 140.3
10-3 = 36.265 mm
Min S = 3 157.08 276
2 27 300 = 41.717 mm > 36.265
Reqd S not allowed. Increase web width!
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
TORSION
SAMPLE PROBLEM 1.
a. Determine Vu and Tu at 0.75h from the fixed support b. Determine the torsion can be neglected. c. Determine Tn @ 0.75h from the support. d. Assuming torsion to be considered, determine if the section is adequate for shear and torsion
Span Length = 7m fc = 20 MPa Gr. 40 bars Concrete Cover = 50 mm 4-20 mm main bars 10 mm stirrups
Load: WDL = 10 kPa WLL = 6 kPa
Solution:
a. Determine Vu and Tu at 0.75h from the fixed support
Wu = 1.20 {(0.4*0.2 + 0.1*0.8) 24+10*0.10} + 1.60 (6*1.0)
= 26.208 kN/m
0.75*400 = 300 mm
Vu @ 0.75h = 26.208 (7/2 0.30) = 83.866 kN
Tu = 1.20 (0.1*24+10) +(1.60*6)*1.0*0.40 = 9.792 kN/m
Tu @ 0.75h = 9.792 (7/2 0.30) = 31.334 kN-m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
b. Determine if torsion can be neglected
Formula:
Tu
12 (
2
)
0.75 1.0 20
12 {
(200 400 + 100 300).2
200 + 400 + 500 + 100 + 300 + 300)
= 1878.918 kN/mm or 1.879 kN/m < 31.334
Torsion must be considered
c. Determine Tn @ 0.75h from the support
= 2
= 45
Ag = 0.85 Ao h = 0.85*90*290= 22185 mm
=0.75 2 221 5
3.14164 10
2 276
150
= 4809.04 N-mm or 4.809 kN-m
100 mm
300 mm 400 mm
200 mm
400-2*(50-10) = 290 mm
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
a. Assuming torsion to be considered, determine if the section is adequate for shear and torsion.
=
2
=31.334
0.75 2 22.185 276
= 3.41165
d = 400-50-10-20/2 =330 mm
= 0.17
= 0.17 1.0 20 200 300
= 501.770
=
=
83.8660.75
50.171
276 330
= 0.67682 s
2At +Av =3.4116s +0.67682s = 7.500s
Required s:
=4
3.14.16
410
7.50
= 41.888
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 2.
1. Determine the maximum design Vu and Tu.
2. Determine if torsion needs to be considered.
Stirrup size= 10 mm PL= 40 kN
Main bar size= 25 mm fc= 26 MPa WD= 5 kPa (including slab and beam weight) fy= 276 Mpa
WL= 3 kPa
Concrete Unit Weight= 24 kN/ m3
1.) WD = 5*1.3+(0.6*0.3+1*0.12)*24= 13.7 kN/m
WU= 1.2*13.7+1.6*3*1.3= 22.68 kN/m
PU= 1.28*100+1.6*40= 184 kN
VA= 22.688
2+ 184 [
4.5
8+
3.54.5
83 (4.5 3.5)]
VA= 199.88 kN
At d distance from A1
d= 600 (40 + 10 +25
2) = 537.5 mm
VU= 199.88 22.68 0.5375 = 187.69 kN
TU= PU e = 184 1.15 = 211.6 kN m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
TALA
GJ=
TBLBGJ
TA =LBLA
TB
TA+ TB = TU
TB =TU
LBLA
+ 1=
LATULA+LB
= TLAL
=TAL
TA= TU b
L+
tU L
2
tU=w*e=[1.2(5*1.3+1.3*0.12*24)+1.6*3*1.3]*0.65= 12.046 kN-m/m
TA= 211.6*4.5
8+
12.0468
2=167.21 kN-m
At d distance from A1,
TU=167.21-0.5875*12.046
TU=160.74 kN-m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SHORT COLUMNS
SAMPLE PROBLEM 1. Determine the forces Pn and Mn for c=200mm using the section shown below.
fc=24MPa and Gr. 60 bars.
.30.393
10*2
1702001734)70200(*)76.76571.725(0
.95.1693
76.76571.72517340
71.25710*25*4
*4*24)*0.85-390()'85.0(
414
390600*200
70200600*
765.7610*25*4
*4*390
414
390600*200
200330600*
173410*500*170*24*85.0)('85.0C
170200*85.0
3
32
32
3
C
1
mkNMn
MnM
mkNPn
PnF
kNAcffC
MPa
MPac
dcf
kNAfT
MPa
MPac
cdf
kNabcf
mmca
sscs
sc
ssts
st
200
tsc st
0.85 fc
Cs
Ts
Cc
200 130 70
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 2. Determine the values of Pno and Pn for the column section shown below using the following data:
fc = 28MPa, 10mm ties and concrete cover of 40mm. Gr. 60 bars, fy = 414 MPa Axial factored compressive load Pu = 600 kN Pnx = 2250 kN and Pny = 1300kN
Pno = 0.85fcAc + Astfy
Pno = 0.85(28) [4002
4252(8)] +
4252(8)(414)
= . .
spiralcolumn w/ 0.75
tiescolumn w/ 0.65;
P
1
P
1
P
1
P
1
nonynxn
)!(SAFE! kN P kN u 60081.713
104009.11
312.5340
1
1300
1
2250
11
3
n
n
n
P
xP
P
kN 2027.3471
)31182.5340(65.0
)'85.0(65.0
no
no
scno
P
P
fyAcAfP
400.00
400.008-25
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
mm
mm
25
75
SAMPLE PROBLEM 3. Determine the required pitch of the 10mm spiral for the column section with a
diameter of 600mm. Use concrete cover of 50mm and main bar diameter of 28mm. fc=30MPa Gr. 60
bars.
sch
sp
d
As
4 = mm791.43
)550(01438.0
])10(4
[4 2
791.3310791.43
Reqd. s=43.791mm
014348.0414
30*)144.1(*45.0
44.154.196349
34.282743
54.196349
)500(44
34.282743
)600(44
'*145.0
2
22
2
22
2
s
c
g
c
cc
g
g
ch
g
s
A
A
mmA
dA
mmA
DA
fyt
cf
A
A
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
LONG COLUMNS
SAMPLE PROBLEM 1. Determine the design forces Pu and Mu. Use K=0.90 (effective length factor)
SERVICE LOADS P Mtop (kN-m) Mbottom (kN-m)
Dead Loads 750 kN 40 20
Live Loads 450 kN 68 34 Solution r = 0.3(375) = 112.5 mm
6.495.112
6200*9.0
r
klu kNx
PC 6.3340)6200*9.0(
100539.1*2
102
5.02
1 M
M 138.1
6.3340*75.016201
4.0
75.01
C
u
m
PP
C
columnslongr
klu 40)5.0(1234 mkNMu 47.177686.1402.1138.1
4.0)5.0(9.06.0 mC mkNMkNP uu 47.177;1620
55556.01620
)750(2.1
1620)450(6.1)750(2.1
d
u
B
kNP
GPaEc 870.24287.4
210
4
100539.155556.01
870.24*345*12
1*4.0mmkNxEI
mm375
mm375
tieslateralmm
bars
12
326 topM
ml 2.6 u
bottomM
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 2. Determine if columns A (300x300) and B (400x400) are long or short under gravity
and earthquake load combination.
Beams are 300x600 mm
Column A
Lu = 5.0 0.3 = 4.7m
R = 0.3 x 300 = 90 mm
K = ?
= 0
=
34
5+
33
3362
8
= 0.533
From nomograph with side sway
=
1.08 4700
90= 56.4 > 22
Column B
Lu = 9-0.3 = 7.7 m
R = 0.3 x 400 = 120 mm
K = ?
= 0
=
44
8 +44
3362
7
= 1.267
From nomograph with side sway
=
1.2 7700
1200= 77 > 22
B A
7 7 8
3
3
5
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
SAMPLE PROBLEM 3.
A. Determine the design forces in the column shown fc= 20mPa Gr. 40
B. Determine the tie spacing requirements.
DFbc=
25503
425503
4+
3504
10
= 0.83887
DFba= 0.16113 (Solving for Distribution factor to use in solving the Final Moments) FEMs:
Mbc= 42[180
20+
120
30] =
Mcb= 42[180
30+
120
20] =
(Solving the FEMs of the given system, will be multiply to DF to get the Final Moments) Final Moments: Mb= 208*0.16113=35.515 kN-m (Mu2)
Mc= 192+ 0.83887
2 208=279.24 kN-m
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Ma= 35.515
2=16.758 kN-m (Mu1)
Pu= Vbc= 4
6(2 180 + 120) +
33.515279.24
4= . (Solving for axial load)
lu= 10,000 500
2= (lu must be decreased by the half of the beam thickness)
a= 0 (because it is fix)
b= 354
10
25503/4= . (using the equation in the 1st example)
From chart of non-sway frames, k= 0.54 r= 0.3*350mm=105mm (for rectangular columns 0.3h)
klu/r= 0.54 9750
105= . (using the formula for non-sway frames)
34 12[2/1] = 34 12 (1
2) = < .
Therefore: LONG COLUMN
Cm= 0.6 + 0.4 (1
2) = .
EI= 470020
1
123504
1+0.7 0.4 = .
PC= 2
()2= 2
6.18461013
(0.549750)2 103 =
=Cm
1 Pu
0.75Pc
= . < 1.0
Therefore: = . Therefore the design forces are: Pu= 258.57 kN Mu1= 16.758 kN-m Mu2= 33.515 kN-m
B. Required spacing of 10mmties 16db=16 20 = () 48dbt=48 10 = 480 Least column dimension= 350mm (rounding it up to the nearest 50)
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
DEVELOPMENT LENGTH
SAMPLE PROBLEM 1. A cantilever beam is reinforced with four 28-mm diameter bar with fy = 276 MPa.
Use fc = 27.6 MPa and assume side, top, and bottom cover to be greater than 65mm. Determine the
required development length if a 90 hook is used.
Using the Eq.1.8-1 from the development of standard hooks in tension and the modification factor no. 1
we will solve for the development length of standard hooks. Then check for ldh limits,
Using a 90 hook:
=
.24
ldh = (0.24*1*276/1 27.6)*28
ldh = 353.04mm
which ldh shall not be less than 8db which is 224mm or less than 150mm.
Modification factor for 90 hook (NSCP 2010 Sec. 412.6.3)
m = 0.70
Required development length, ldh = 353.04*0.70
Required development length, ldh = 247.13, say ldh = 250mm
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
Wu = 117 kN/m
3-28 mm 2-28 mm
SAMPLE PROBLEM 2. A rectangular beam having a width of 400 mm has an effective depth of 535 mm.
It is reinforced with 3-28 mm bars extend 100 mm past the centerline of supports.
= 1.0 (bottom bars), = 1.0 (uncoated reinforced), = 1.0 (normal weight concrete). The beam
carries a factored uniform load of Vu = 117 kN/m. The beam has a span between supports equal 6m.
Using fc = 20.7 MPa, fy = 414.7 MPa.
a.) Compute the required development length. b.) Compute the nominal moment capacity of the beam. c.) Compute the furnished development length and indicate if it is properly anchored at the
support or not.
a.) We can compute for the development length by using the Formula (see NSCP 412.3.2) given below and the factor that you can found in NSCP 412.3.4.Since the diameter of the bar is larger than 25 mm, we will use the formula in the table in that can see in NSCP. = 1.0 = 1.0 = 1.0 Ld = 3fy = 9 (414.7) (1) (1) (1) db 5 fc 10 20.7
3-28 mm
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REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
The db is the diameter of the bar which is 28 mm. Substitute it And we can get the development length. Ld = 54.69 db Ld = 54.69 (28) Then, the value of development length will be: Ld = 1531.29 mm b.) Nominal moment capacity: At first we must compute for the area in order to compute the moment capacity.
As = (28)2 (3) = 1847 mm2 4 T=Asfy And the, we can now compute for the value of a. T = 0.85fcab Asfy = 0.85fcab 1847 (414.7) = 0.85(20.7) (a) (400) a = 108.83 mm By getting the unknown value, we can now substitute it to compute for the moment capacity of the reinforced concrete. Mn = Asfy (d a/2) Mn = 1847(414.7)(535 108.83/2) Mn = 368 x 106 kN.m. (nominal moment strength).
-
REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE
EXAMPLES Copyright 2013
c.) Furnished development length: In getting the furnished development, we must got first the shear value to substitute in the formula. Vu = WL/2 =117(6) = 3541 kN 2 We use 1.3 because we must increase 30 percent in the value Mn/Vu when the ends of reinforcement are confined by a compressive reaction. It must not greater than the development that we computed for us to say it was properly anchored. Ld < 1.3Mn + La Vu Ld < 1.3(368) + 0.100 351 Ld < 1.463 mm 1531.29 mm > 1463 mm (not ok!) The furnished development length is less than the required development length and because of this we can say that the beam is not properly anchored at the supports.
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