example problems

REINFORCED CONCRETE DESIGN Engr. ALBERTO C. CAETE

EXAMPLES Copyright 2013

EXAMPLES



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 = .



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



=

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) = .



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



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




c

Mneg = (0.45)(20)(3091.057)

(203.23)= .


(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

= .


c

Mpos = (0.45)(20)(2945.811)

(194.2768)= .


(d c)(n)

M = (138)(2945.811)

(455194.2768)(9)= .

= .



= 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

= .

= .



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



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

= .



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



USD SINGLY

USD DOUBLY

USD IRREGULAR



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



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



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



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



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



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



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



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



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



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%



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



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%



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%



Total Exterior Positive Moment = 0.15012



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


=

= 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



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


=

= 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


=

= 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



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



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



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



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!



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



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



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



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



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



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



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



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



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



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



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



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)



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



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



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).



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.

example problems

Documents

caete examples copyright

gpa reinforced concrete

n n2

asbd n

fcitrc mcap

section b

fsitrd cn mcap

b c section ab example