Example Problems

(1)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

(2)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

Sample No. 1 Determine the maximum positive moment and the negative moment at the fixed

support

. √58 7 = 1.0879𝑚 6( √58 7 ) = 6.5278 Cantilever M= 8.4482 x (1.0879)2_/2 MB= 5 kNm FEM’s= 1 12 (8.4482)(6.5278) 2_{= 30.000 𝑘𝑁𝑚} Mc = 30 +25 2 = 42.5 𝑘𝑁𝑚 Point of zero shear: 𝑋 =21.829

8.4482= 2.5839𝑚 𝑀𝑎𝑥 𝑀𝑝𝑜𝑠 =1

(3)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

Sample No. 2 Determine the moment at point c.

Columns – steel pipe, outer ø= 300mm, thickness=10mm, G=200 GPa

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REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES Copyright © 2013 𝐼𝑐𝑜𝑙 = 𝜋 64(300 4_{− 280}4_{) = 95.889𝑥10}6_𝑚𝑚4 𝐼𝑟𝑎𝑓𝑡 = 1 12(250𝑥490 3_{− 240𝑥450}3_{) = 628.52𝑥10}6_𝑚𝑚4 𝐷𝐹𝐵𝐴= . 75 (200 𝑥 95.889₆ ) . 75 (200𝑥95.889₆ ) +83𝑥628.52 4√5 = 0.29130 𝐷𝐹𝐵𝐶 = 0.7087 𝐹𝐸𝑀′𝑠 = 1 12(10)(4√5) 2 = 66.6667 𝑘𝑁𝑚 𝑀𝑐 = 66.667 +0.7087 2 (66.667 − 0) = 𝟗𝟎. 𝟐𝟗𝟎 𝒌𝑵𝒎

(5)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

WSD SINGLY

Example 1. Determine the moment capacity of beam if the area of reinforcing bars is 1400

mm

2

_{. If fs=120MPa and f’c=20MPa.}

n =

Es 4.7√f′c

=

200 4.7√20

= 9.5152 ≈ 10

ρ

n

=

As bd

n =

1400 (200)(400)

(10) = 0.17500

k = −ρ

_n

+ √(ρ

_n2

_{) + 2(ρ}

n

)

k = −0.175 + √(0.175

2

_{) + 2(0.175)}

k = 0.44195

c = k ∗ d = 0.44195 x 400 = 176.78

I

tr

=

bc

3

3 + nA

s

(d − c)

2

₌

(200)(176.78)

3

3 + (10)(1400)(400 − 176.78)

2

𝑰

𝒕𝒓

= 𝟏𝟎𝟔𝟔. 𝟎𝟕𝟑 𝒙𝟏𝟎

𝟔

𝒎𝒎

𝟒

Due to Concrete: M

cap

=

f

_c

I

_tr

c

M

_cap

=

(0.45)(20)(1066.073)

(176.78)

= 𝟓𝟒. 𝟐𝟕𝟓 𝐤𝐍 − 𝐦

Due to Steel: M

cap

=

f

_s

I

_tr

(d − c)(n)

M

cap

=

(120)(1066.073) (400−176.78)(10)

= 𝟓𝟕. 𝟑𝟏𝟏 𝐤𝐍 − 𝐦

∴ 𝑴

_𝒄𝒂𝒑

= 𝟓𝟒. 𝟐𝟕𝟓 𝒌𝑵 − 𝒎

As 200 mm 400 mm

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REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

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, f’c = 25 MPa

and fs = 138 MPa.

n =

Es

4.7√f′c

=

200 4.7√25

= 8.5106 ≈ 9

SECTION B:

d =

4 x 32

2

_{x 80}

4 x 32

2

_{+ 2 x 20}

2

+ 375 = 441.93 mm

ρ

_n

=

A

s

bd

n =

𝜋

4 (4 x 32

2

+ 2 x 20

2

)

(400)(441.93)

(9) = 0.195776

c = [−ρ

n

+ √(ρ

n2

) + 2(ρ

n

) ] d = 203.23 mm

Itr = bc3 3 + nAs(d − c) 2₌(400)(203.233) 3 3 + (9) ( 𝜋 4) (4 x 32 2_{+ 2 x 20}2_{)(441.93 − 203.23)}2

𝑰

_𝒕𝒓

= 𝟑𝟎𝟗𝟏. 𝟎𝟓𝟕 𝒙𝟏𝟎

𝟔

_𝒎𝒎

𝟒 5 m 4 m 375 400 455 400

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REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

Due to Concrete: M

cap

=

f

_c

I

_tr

c

M

_neg

=

(0.45)(20)(3091.057)

(203.23)

= 𝟏𝟕𝟏. 𝟏𝟎𝟖𝟔 𝐤𝐍 − 𝐦

Due to Steel: M

cap

=

f

_s

I

_tr

(d − c)(n)

M

neg

=

(138)(3091.057) (441.93−203.23)(9)

= 𝟏𝟗𝟖. 𝟓𝟓𝟗𝟕 𝐤𝐍 − 𝐦

∴ 𝑴

_𝒏𝒆𝒈

= 𝟏𝟕𝟏. 𝟏𝟎𝟖𝟔 𝒌𝑵 − 𝒎

SECTION B:

d = 455 mm

ρ

_n

=

A

s

bd

n =

𝜋

4 (4 x 32

2

)

(400)(455)

(9) = 0.159082

c = [−ρ

_n

+ √(ρ

_n2

_{) + 2(ρ}

n

) ] d = 194.2768 m

Itr = bc3 3 + nAs(d − c) 2₌(400)(194.2768) 3 3 + (9) ( 𝜋 4) (4 x 32 2_{)(455 − 194.2768)}2

𝑰

_𝒕𝒓

= 𝟐𝟗𝟒𝟓. 𝟖𝟏𝟏 𝒙𝟏𝟎

𝟔

𝒎𝒎

𝟒

Due to Concrete: M

_cap

=

f

c

I

tr

c

M

pos

=

(0.45)(20)(2945.811)_(194.2768)

= 𝟏𝟑𝟔. 𝟒𝟔𝟕 𝐤𝐍 − 𝐦

Due to Steel: M

_cap

=

f

s

I

tr

(d − c)(n)

M

_𝑝𝑜𝑠

=

(138)(2945.811)

(455−194.2768)(9)

= 𝟏𝟕𝟑. 𝟐𝟒𝟓 𝐤𝐍 − 𝐦

(8)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

𝐷𝐹

_𝐵𝐴

=

1

5 ⁄

1

5 ⁄ + 0.75 4

⁄

= 0.51613

𝐷𝐹

𝐵𝐶

= 0.48387

FEM’s:

𝑀

_𝐴𝐵

=

2 𝑤

12 (5

2

_{) = 2.08334 𝑤}

𝑀

𝐵𝐶

=

1.7 𝑤

12 (4

2

_{) 𝑥 1.5 = 3.4 𝑤}

FINAL M’s:

𝑀

_𝐵

= 2.08334 𝑤 (0.48387) + 3.4 𝑤 (0.51613) = 2.76307 𝑤

𝑀

𝐴

= 2.08334 𝑤 +

1 2

(2.08334 w – 3.4 w)(0.51613) = 1.74355 w

𝑀

_𝐶

= 0

𝑉

𝐵𝐶

=

𝑤𝐿

2 +

𝑀

_𝐵

− 𝑀

_𝐴

𝐿

= 5.2039 𝑤

2𝑤(𝑥) = 5.2039 𝑤

𝑥 = 2.60195

M

_pos

= (

5.2039 𝑥 2.60195

2 ) (2.76307) = 4.00708 w

4.00708 𝑤 = 136.467

𝒘 = 𝟑𝟒. 𝟎𝟓𝟔𝟓 𝒌𝑵 𝒎

⁄

M

neg

= 171.1086

2.76307 𝑤 = 171.1086

𝒘 = 𝟔𝟏. 𝟗𝟐𝟕 𝒌𝑵 𝒎

⁄

∴ 𝒘 = 𝟑𝟒. 𝟎𝟓𝟔𝟓 𝒌𝑵 𝒎

⁄

(9)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

WSD DOUBLY

In the figure shown, determine the deflection at free end. d’ = 75mm, d = 595mm and b =

360mm. Use f’c = 35 MPa.

𝑛 =

𝐸

𝑠

4.7√𝑓′𝑐

=

200 4.7√35

= 7.2 𝑠𝑎𝑦 7

Solve for A

s

and A’

s,

𝐴

_𝑠

=

𝜋

4 𝑥(36)

2

_{𝑥 3 = 3053.63𝑚𝑚}

2

𝑛𝐴

_𝑠

= 7 𝑥 3053.63 = 21375 𝑚𝑚

2

𝐴′

𝑠

=

𝜋

4 𝑥(25)

2

_{𝑥 3 = 1472.6 𝑚𝑚}

2

(2𝑛 − 1)𝐴

′ 𝑠

= [(2𝑥7) − 1]𝑥 1472.6 = 1914.4 𝑚𝑚

2

Let A = b/2, 𝐴 =

360 2

= 180𝑚𝑚

𝐵 = (2𝑛 − 1)𝐴

′ 𝑠

+ 𝑛𝐴

𝑠

= 1914.4 + 21375 = 40519

𝐶 = (2𝑛 − 1)𝐴

′_𝑠

𝑑

′

+ 𝑛𝐴

_𝑠

𝑑 = (19144 𝑥 75) + (21375 𝑥 595) = 14153925

Substitute A, B and C on the equation for c, we obtain

c =

−40519+ √405192+4(180+14153925) 2 𝑥 180

= 189.61𝑚𝑚

220 kN-m 40 kN-m 3.20 m 360 520 75 75

(10)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

Calculate I

tr

,

I

tr

=

𝑏𝑐 3 3

+ 𝑛𝐴

𝑠

(𝑑 − 𝑐)

2

_{+ ((2𝑛 − 1)𝐴}

′ 𝑠

)(𝑐 − 𝑑

′

)

2

=

360+189.61₃ 2

+ 21375(370.39)

2

_{+ 1914.4(114.61) = 4582.3𝑥10}

6

_𝑚𝑚

4

E

c

= 4700√𝑓

′

𝑐 = 4700√35 = 27806 𝑀𝑃𝑎

𝛿 =

𝑃𝐿

3

3𝐸𝐼

+

𝑊𝐿

4

8𝐸𝐼

=

220(3.2)

3

₍₁₀₎

12

3 𝑥 27.806 𝑥 4.5823𝑥10

2

+

40(3.2)

4

₍₁₀₎

12

8 𝑥 27.806 𝑥 4.5823 𝑥 10

12

𝜹 = 𝟐𝟐. 𝟗𝟕𝟒𝒎𝒎

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REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

WSD IRREGULAR

Example 1. Given

L = 10m., find max w that the beam can carry safely

kNm Mcap Therefore m kN w m kN L M w wL M Mcap MPa fs kNm Mcap MPa fc mm I mm c mm C mm therefore or mm c Assume mm nA say n tr s 667 . 86 / 9333 . 6 m -kN 122.59 667 . 86 10 667 . 86 8 8 8 1 10 ) 10 )( 56 . 173 550 ( ) 10 1 . 2364 ( 138 138 steel on Based 59 . 122 10 56 . 173 ) 10 1 . 2364 ( 9 9 20 45 . stresses allowable Concrete on Based 10 1 . 2364 ) 56 . 173 550 ( 12566 3 ) 56 . 173 ( 150 ) 56 . 173 )( 100 )( 200 ( ) 100 )( 200 ( 12 1 56 . 173 7911300 ) 550 ( 12566 2 ) 200 ( 100 32566 12566 ) 200 ( 100 B 75mm. 150/2 A section irregular : 5654700 175000 ) 100 550 ( 12566 50 100 x 350 . 100 12566 4 ) 20 ( 4 10 10 52 . 9 20 7 . 4 200 2 2 2 6 6 6 6 4 6 2 3 2 3 3 2 2 2 2                                                    



550 75 200 75 400 100 100

(12)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

USD SINGLY

USD DOUBLY

USD IRREGULAR

(13)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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.

𝐿 24

=

3100 24

= 129.17mm.

𝐿 28

=

2900 28

= 103.57 𝑚𝑚.

Req’d thickness= 129.17[0.4 +

276₇₀₀

]

T =102.60 say 110mm

b. w

u

=1.4[5+(0.11× 24)] + 1.7(3) = 15.796 kPa

M

u

=

₁₀1

(15.796)(3.1)

2

= 15.180 𝑘𝑁 (governs)

M

u

=

1 12

(15.796)(2.9)

2

_{= 11.070 d=110-20-}

10 2

= 85 𝑚𝑚

X=

_0.9×100015.18

(0.085

2

_{)276 = 0.03363}

m =

276 0.85×25

= 12.988

ρ=

1−√1−2𝑚𝑥 𝑚

= 0.0089822

𝜌

_𝑚𝑖𝑛

=

1.4 𝑓

_𝑦

𝑜𝑟

√𝑓𝑐

′

4𝑓

_𝑦

= 0.00507 𝑜𝑟 0.004289

∴ 𝜌

𝑚𝑖𝑛

= 0.00507 < 𝜌

Req’d s=

𝛱(10)2 4(0.0089822)(85)

= 102.87 (𝑅𝑂𝑈𝑁𝐷𝐷𝑂𝑊𝑁 𝑇𝑂 100𝑚𝑚)

S

max

= 3At

s

or 450= 330 (governs) > Req’d s

_{Adopt smallest S:}

S= 100 mm

(14)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 (𝑂𝐾𝐴𝑌)

7.2 7

= 1.02857 (𝑂𝐾𝐴𝑌)

8.2 7.2

= 1.0 (𝑂𝐾𝐴𝑌)

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

(15)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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

(16)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

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

(17)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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:

ℎ =

𝑙

_𝑛

(0.8 +

₁₄₀₀₎

𝑓

𝑦

36 + 9𝛽

=

4900(0.8 +

₁₄₀₀

276 )

36 + 9 (

4900

₄₂₀₀₎

= 105.08 ~110𝑚𝑚

𝒉 = 𝟏𝟎𝟓. 𝟎𝟖 ~𝟏𝟏𝟎𝒎𝒎

5.0 m 5.2 m 5.0 m 4.2 m 4.5 m 4.2 m

A

B

C

D

(18)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

SAMPLE PROBLEM 2.

1.) Determine the req’d 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

d

q'

mm

99

1.1 *

90 mm

1500

414

0.8 4400

1500

fy

0.80 ln

h

1.)

:

SOLUTIONS

























_



Re

07 .

110

10 .

1 *

5 .

3

4 .

4 *

9

36

9

36 

strip 1m m/ -kN 5020 . 6 2 . 4 * 4 . 11 * 032333 . 0 M b 032333 . 0 005 . 0 * 2 * 33333 . 0 029 . 0 Cb 83333 . 0 4.2 3.5 m 4 CASE kPa 4 . 11 W 84 . 3 4 . 2 * 6 . 1 W 6 . 1 56 . 7 3 . 6 * 2 . 1 W 2 . 1 .) 2 2 neg neg U L D            

(19)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

m/m

-kN

1.1402

3

2 *

3 5.1311

M a,

Strip)

(Column

Edge

ous

Discontinu

1311

.

5 1339

.

2 9972

.

2 M a,

1339

.

2

3.5 *

3.84 *

0.045364

LL

M a,

0.045364

0.004 *

2 *

0.4545

-0.049

LL

Ca,

9972

.

2

3.5 *

7.56 *

0.032364

DL

M a,

0.032364

0.004 *

2 *

0.4545

-0.036

DL

Ca,

8 CASE

79545

.

0

4.4

3.5 m

.)

3

neg pos 2 pos pos 2 pos pos







(20)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

SAMPLE PROBLEM 3.

Design the exterior panel of the two-way slab using direct design method.

Given Data: fy= 276 MPa f’c= 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

(21)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

Slab Thickness:

Assume conventional slab with stiff beams (𝛼𝑓 > 2.0)

= 2.8/1.95 = 1.4359 ℎ = 𝑙𝑛(0.8 + 𝑓𝑦 1400) 36 + 9𝛽 = 2.8(0.8 +₁₄₀₀₎276 36 + 9(1.4359) = 57.07~60𝑚𝑚 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 1.6𝑥109𝑚𝑚4 2.13𝑥109𝑚𝑚4 lS=bh^3/12 9x107𝑚𝑚4 1.79𝑥108𝑚𝑚4 Mo

=

𝑤𝑙_𝑛𝑙₂2 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 𝐶 = ∑ (1 − 0.63𝑥 𝑦) 𝑥3 3 𝑦 𝐶 = (1 − 0.63200 400) ( 2003_𝑥400 3 ) + (1 − 0.63 100 300) ( 1003_𝑥300 3 ) = 8.1𝑥10 8_𝑚𝑚4 𝛽𝑡 = 𝐸𝑐𝑏𝐶 2𝐸𝑐𝑠𝐼𝑠 = 1.6193

(22)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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

Column strip 0.5375m 1.075m 𝑙2⁄ 𝑙1 0.3583 0.7167 𝛼𝑙2⁄ 𝑙1 6.4 8.5333 𝛼𝑙2⁄𝑙1= 0 75 75 𝛼𝑙2⁄𝑙1> 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

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 2.1𝑥109𝑚𝑚4 2𝑥109𝑚𝑚4 lS=bh^3/12 2.3x108𝑚𝑚4 2𝑥108𝑚𝑚4 Mo

=

𝑤𝑙_𝑛𝑙₂2 8 21.48kN-m 21.48kN-m  9.3091 9.3091

(23)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES Copyright © 2013 X1 = 200mm Y1 = 400mm X2 = 100mm Y2 = 300mm 𝐶 = ∑ (1 − 0.63𝑥 𝑦) 𝑥3 3 𝑦 𝐶 = (1 − 0.63200 400) ( 2003𝑥400 3 ) + (1 − 0.63 100 300) ( 1003𝑥300 3 ) = 8.1𝑥10 8_𝑚𝑚4 𝛽𝑡 = 𝐸𝑐𝑏𝐶 2𝐸𝑐𝑠𝐼𝑠 = 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

Total Exterior Negative Moment = 0.4476 Column strip for Exterior Positive Moment

Column strip 1.075m 1.075m 𝑙2⁄ 𝑙1 0.3583 0.7167 𝛼𝑙2⁄ 𝑙1 6.4 8.5333 𝛼𝑙2⁄𝑙1= 0 60 60 𝛼𝑙2⁄𝑙1> 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

(24)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

DESIGN ALONG Z-DIRECTION

For Bottom Bars

Ext. Neg. Moment = 0.1627

d = h – cc- .5db = 100 – 20 - 1.5(12) = 62mm 𝑥 = 𝑀𝑢 𝜙𝑏𝑑2_𝑓𝑦= (0.1627𝑥 106₎ 0.9𝑥1000𝑥622_𝑥276= 0.0002 𝜌 = 1 − √1 − 2𝑚𝑥 𝑚 = 1 − √1 − 2(9.8396)(0.0002) 9.8396 = 0.0002 min= 0.0052

Since min is greater than , therefore use min 𝑠 =𝐴𝑏 𝜌𝑑= 𝜋122 4 ⁄ 0.0052(62)= 350.39~350𝑚𝑚 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.

Since 3h = 3x100, Therefore; Smax = 200mm

Since s is greater than Smax, therefore use Smax Spacing of the top bars is 200mm

(25)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

DESIGN ALONG X-DIRECTION

For Bottom Bars

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.

d = h – cc- .5db = 100 – 20 - 1.5(12) = 62mm 𝑥 = 𝑀𝑢 𝜙𝑏𝑑2_𝑓𝑦= (0.15𝑥 106) 0.9𝑥1000𝑥622_𝑥276= 0.0002 𝜌 = 1 − √1 − 2𝑚𝑥 𝑚 = 1 − √1 − 2(9.8396)(0.0002) 9.8396 = 0.0002 min= 0.0052

Since s is greater than Smax, therefore use Smax Spacing of the top bars is 200mm

(26)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 f’c = 24 MPa & Gr. 60 stirrups.

 

kN 14 . 402 V 7 500 600 7 4 250 2 7 70 V kN 86 . 337 V 7 600 500 7 3 250 2 7 70 V BA BA AB AB          

  

 





mm 248 . 44 24 2 433.54 s mm 600 mm 300 say (governs) mm 315 2 mm 630 mm 54 . 433 300 276 08 . 157 3 s kN 40 . 57 1 Vc mm 08 . 57 1 4 10 2 A 10 630 00 3 24 ) 1 ( 17 . 0 bd c f' 17 . 0 Vc min max 2 2 V 3              





(27)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE





 





m

.7807

70

207.49 -402.14

W

V

-V

x

kN

0 V

kN

86.708

630

276

157.08 s

d

f

A

V

ent

reinforcem

shear

minimum

For

kN

358.04

0.63

70 -402.14

d

W

-V

V

:

side

right

of

Design

U U BA 1 U y V S U BC max

2

49 .

207

708 .

86

40 .

157

85 .

300 







 

mm

100 @

rest

mm,

50 @

1 mm,

10 use

mm

100 say

mm

103.53

157.40 -0.85

358.04

10

630

276

157.08 V

V

d

f

A

s

:

support

from

d"

"

distance

@

3 -C U y V









(28)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

SAMPLE PROBLEM 1. Determine the 2 – legged stirrups, 10 mm ф stirrups, 2h from fixed

support.

W

DL

= 50

𝐾𝑁 𝑚2

P

LL

= 150

𝐾𝑁 𝑚2

W

LL

= 30

𝐾𝑁 𝑚2

fc’ = 27

P

DL

= 150

𝐾𝑁 𝑚2

Grade 40 rebars, cc = 40 mm.

Main bars = 20 mm. ɸ

P

W

B = 300

4m

h = 600

400 x 400

4m

5m

Solution:

I

AB

= 4

4

= 256

I

BC

= 3 × 6

3

= 648

DF

BA

=

0.75 ×256 4 0.75 × 256 4 + 648 9

= 0.4

DF

BC

= 0.60

W

U

= (1.4 × 50) + (1.7 × 30) = 121

𝐾𝑁 𝑚

P

U

= = (1.4 × 150) + (1.7 × 80) = 345

𝐾𝑁 𝑚

(29)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

Fixed End Moments

M

BC

=

121 × 92 12

+

346 × 4 × 52 92

= 1243.91 kNm

M

B

=

121 × 92 12

+

346 × 42_{× 5} 92

= 1158.48 kNm

Final Moments

M

B

= 1243.91 × 0.4 = 497.56 kNm

M

C

= 1158.48 +

0.6 2

(1243.91)

= 1531.65 kNm

Shears

V

CB

=

121 ×9 2

−

497.56 −1531.65 9

+

4 ( 346 ) 9

= 813.18 kNm

@ 2h = 2 × 60 = 1.2m from fixed support

V

C

=

1₆

√27

× 300 × 540 × 10

-3

= 140.29 kN

d = 600 – 40 – 10 –

20 2

= 540 mm

A

V

= 2 ×

𝜋 ×102 4

= 157.08 mm

2

Req’d S =

157.08 × 276 × 540667.98 0.85 − 140.3

× 10

-3

_{= 36.265 mm}

Min S =

3 × 157.08 × 276 2 × √27 × 300

= 41.717 mm > 36.265

(30)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 ØTn @ 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 Ø 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

(31)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

b. Determine if torsion can be neglected

Formula: Tu ≤Ø𝜆√𝑓𝑐′ 12 ( 𝐴𝑐𝑝2 𝑃𝑐𝑝) 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 ØTn @ 0.75h from the support

Ø𝑇𝑛 =Ø 2𝐴𝑜𝐴𝑡𝑓𝑦𝑡 𝑠 𝑐𝑜𝑡Ø Ø = 45° Ag = 0.85 Ao h = 0.85*90*290= 22185 mm² Ø𝑇𝑛 =0.75 ∗ 2 ∗ 221 ∗ 5 ∗ 3.1416 4 ∗ 102∗ 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

(32)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

a. Assuming torsion to be considered, determine if the section is adequate for shear and torsion. 𝐴𝑡 = 𝑇𝑢 Ø 2𝐴𝑜𝑓𝑦𝑡 𝑐𝑜𝑡Ø 𝐴𝑡 = 31.334 0.75 ∗ 2 ∗ 22.185 ∗ 276 = 3.41165 d = 400-50-10-20/2 =330 mm 𝑉𝑙 = 0.17𝜆√𝑓𝑐′𝑏𝑤𝑑 = 0.17 ∗ 1.0 ∗ √20 ∗ 200 ∗ 300 = 501.770 𝐴𝑣 = 𝑉𝑢 Ø − 𝑉𝑐 𝑓𝑦𝑑 𝑠 𝐴𝑣 = 83.866 0.75 − 50.171 276 ∗ 330 𝑠 = 0.67682 s 2At +Av =3.4116s +0.67682s = 7.500s Required s:

𝑠 =

4∗

3.14.16

4 ∗10²

7.50

= 41.888 <150 mm INADEQUATE

(33)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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= 22.68∗8 2 + 184 [ 4.5 8 + 3.5∗4.5 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

(34)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES Copyright © 2013 TALA GJ = TBLB GJ → TA = LB LA TB TA+ TB= TU TB = TU LB LA+ 1 = LATU LA+LB = TLA L = TA L 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 + 12.046∗8 2 =167.21 kN-m At “d” distance from A1, TU=167.21-0.5875*12.046 TU=160.74 kN-m

(35)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

SHORT COLUMNS

SAMPLE PROBLEM 1. Determine the forces Pn and Mn for c=200mm using the section shown below.

f’c=24MPa and Gr. 60 bars.

.

30 .

393

10 *

2

170

200 1734

)

70

200 (

*

)

76 .

765

71 .

725 (

0 .

95 .

1693

76 .

765

71 .

725 1734

0

71 .

25

7

10 *

25 *

4 *

24)

*

0.85 -390

(

)

'

85 .

0 (

414

390

600 *

200

70

200

600 *

765.76

10 *

25 *

4 *

390

414

390

600 *

200

330

600 *

1734

10 *

500 *

170 *

24 *

85 .

0 )

(

'

85 .

0 C

170

200 *

85 .

0

3 3 2 3 2 3 C 1

m

kN

Mn

M

m

kN

Pn

F

kN

A

c

f

C

MPa

c

d

c

f

kN

A

f

T

MPa

c

d

f

kN

ab

c

f

mm

c

a

s sc s sc s st s st















_











































   







200 tsc €st 0.85 f’c Cs Ts Cc 200 130 70

(36)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

SAMPLE PROBLEM 2. Determine the values of Pno and Pn for the column section shown below using

the following data:

f’c = 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

P

_no

= 0.85f

′_c

A

_c

+ A

_st

f

_y

P

_no

= 0.85(28) [400

2

−



4

25

2

_{(8)] +}



4

25

2

_(8)(414)

𝐏

_𝐧𝐨

= 𝟓𝟑𝟒𝟎𝟑𝟏𝟏. 𝟖𝟏𝟕𝐍𝐨𝐫𝟓𝟑𝟒𝟎. 𝟑𝟏𝟏𝟖𝟐𝐤𝐍

spiral

column w/

0.75 ties

column w/

0.65 ;

φP

1 φP

1

no ny nx n























)

!

(SAFE!

kN

P

kN

u

600

81 .

713

10 4009

.

1

312 .

5340

1 1300

1 2250

1

3













 n n n

P

x

P



kN

2027

.

3471

)

31182

.

5340

(

65 .

0 )

'

85 .

0 (

65 .

0 







no no s c no

P

fy

A

cA

f

P



400.00 400.00 8-25Ø

(37)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

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. f’c=30MPa Gr. 60 bars. ch s sp

d

A

s



4 

=

43 .

791 mm

)

550 (

01438

.

0 ]

)

10 (

4 [

4

2





791 . 33 10 791 . 43  

Req’d. s=43.791mm

014348

.

0

414

30 *

)

1

44 .

1 (

*

45 .

0

44 .

1

54 .

196349

34 .

282743

54 .

196349

)

500 (

4

34 .

282743

)

600 (

4

4 '

*

1

45 .

0

2 2 2 2 2 2 2



































s c g c c c g g ch g s

A

mm

A

d

A

mm

A

D

A

fyt

c

f

A







(38)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 . 49 5 . 112 6200 * 9 . 0   r klu kN x PC 3340.6 ) 6200 * 9 . 0 ( 10 0539 . 1 * 2 10 2  



5 .

0

2 1



M

138 . 1 6 . 3340 * 75 . 0 1620 1 4 . 0 75 . 0 1      C u m P P C



columns long r kl_u    12(0.5) 40 34

M

_u



1 .

138 

1 .

2  

40 

1 .

6  

68 



177 .

47 kN



m

4 .

0 )

5 .

0 (

9 .

0

6 .

0 



m

C

P_u 1620kN ;M_u 177.47kNm 55556 . 0 1620 ) 750 ( 2 . 1 1620 ) 450 ( 6 . 1 ) 750 ( 2 . 1      d u B kN P

GPa

E

_c



4 .

7

28 

24 .

870

2 10 4

10 0539

.

1 55556

.

0

1

870 .

24 *

345 *

12

1 *

4 .

0 mm

kN

x

EI









mm 375 mm 375

ties

lateral

mm

bars



12

32

6 

M

top

m

l

6 .

2

u



bottom

M

(39)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 = ?

𝜓

_𝐵

= 0

𝜓

_𝐴

=

3

4

5 +

3

3 3𝑥6

2

8 = 0.533

From nomograph with side sway

𝐾𝑙𝑢

𝑟

=

1.08 𝑥 4700

90 = 56.4 > 22 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑢𝑚𝑛

Column B

Lu = 9-0.3 = 7.7 m

R = 0.3 x 400 = 120 mm

K = ?

𝜓

_𝐵

= 0

𝜓

_𝐴

=

4

8 +

4

3 3𝑥6

2

7 = 1.267

From nomograph with side sway

𝐾𝑙𝑢

𝑟

=

1.2 𝑥 7700

1200

= 77 > 22 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑙𝑜𝑛𝑔 𝑐𝑜𝑙𝑢𝑚𝑛

B A 7 8 7 3 3 5

(40)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

SAMPLE PROBLEM 3.

A. Determine the design forces in the column shown f’c= 20mPa

Gr. 40

B. Determine the tie spacing requirements.

DFbc= 25𝑥503 4 25𝑥503 4

+

3504 10 = 0.83887

DFba= 0.16113 (Solving for Distribution factor to use in solving the Final Moments) FEM’s: Mbc= 42[ 180 20 + 120 30] = 𝟐𝟎𝟖 𝒌𝑵 − 𝒎 Mcb= 42[ 180 30 + 120 20] = 𝟏𝟗𝟐 𝒌𝑵 − 𝒎

(Solving the FEM’s 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

(41)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

4 = 𝟐𝟓𝟖. 𝟓𝟕 𝒌𝑵 (Solving for axial load)

lu= 10,000𝑚𝑚 −

500

2 = 𝟗𝟕𝟓𝟎𝒎𝒎 (lu must be decreased by the half of the beam thickness)

a= 0 (because it is fix) b=

354 10

25∗503_/4= 𝟎. 𝟏𝟗𝟐𝟎𝟖 (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 = 𝟓𝟎. 𝟏𝟒𝟑 (using the formula for non-sway frames)

34 − 12[𝑀2/𝑀1] = 34 − 12 (− 1

2) = 𝟒𝟎 < 𝟓𝟎. 𝟏𝟒𝟑

Therefore: LONG COLUMN Cm= 0.6 + 0.4 (−1 2) = 𝟎. 𝟒 EI= 4700√20∗ 1 12∗3504 1+0.7 ∗ 0.4 = 𝟔. 𝟏𝟖𝟒𝟔𝒙𝟏𝟎 𝟏𝟑_{𝑵 − 𝒎𝒎}𝟐 PC= 𝜋2𝐸𝐼 (𝑘𝑙𝑢)2= 𝜋2∗ 6.1846𝑥1013 (0.54∗9750)2∗ 10−3= 𝟐𝟐𝟎𝟐𝟎𝒌𝑵 𝛿 = Cm 1 −_0.75PPu c = 𝟎. 𝟒𝟎𝟔𝟑𝟔 < 1.0 Therefore: 𝛿 = 𝟏. 𝟎

Therefore the design forces are:

Pu= 258.57 kN

Mu1= 16.758 kN-m

Mu2= 33.515 kN-m

B. Required spacing of 10mmties 16db=16 ∗ 20 = 𝟑𝟐𝟎𝒎𝒎 (𝒈𝒐𝒗𝒆𝒓𝒏𝒔) 48dbt=48 ∗ 10 = 480𝑚𝑚

(42)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

DEVELOPMENT LENGTH

SAMPLE PROBLEM 1. A cantilever beam is reinforced with four 28-mm diameter bar with fy = 276 MPa.

Use f’c = 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:



_{𝑑ℎ =}𝑜.24 𝑒 𝑓𝑦 √𝑓′𝑐

∗ 𝑑𝑏

ldh = (0.24*1*276/1 √27.6)*28 ldh = 353.04mm

which ldh shall not be less than 8db which is 224mmor 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

(43)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 Ø bars extend 100 mm past the centerline of supports.

Α = 1.0 (bottom bars), β = 1.0 (uncoated reinforced), λ = 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 f’c = 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.

β = 1.0 λ = 1.0 α = 1.0 Ld = 3fy α β λ = 9 (414.7) (1) (1) (1) db 5 √f’c 10 √20.7 3-28 mm

(44)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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 = π (28)2_{(3) = 1847 mm}2

4 T=Asfy

And the, we can now compute for the value of a. T = 0.85f’cab

Asfy = 0.85f’cab

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)

(45)

REINFORCED CONCRETE DESIGN

Engr. ALBERTO C. CAÑETE

EXAMPLES

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.