Problem No.1

Design a load bearing wall with the following data:

DL = 45 KN/mLL = 20 KN/mqall = 100 KN/m2

Depth of top of foundation from the groundlevel Df = 1.2 m

fy = 420 Mpafc' = 24 MpaUnit weight of soil g s = 18 KN/m3Unit weight of concrete g c = 25 KEN/m3

Solution:

According to ACI-Code 7.7.1, the minimum depth of footing d = 150mmover steel, cover = 70 mm, sohmin = 150+70 = 220 mmAssume h = 250 mm,gived = 250-70 = 180 mmalso,weight of foundation = 0.25 gc = 0.25x25 = 6.25 KN/m2weight of soil above foundation = 1.2 gs = 1.2x18 = 21.6 KN/m2

Total = 27.85 KN/m2

Net allowable soil bearing capacity

qall(net) = 100-27.85 72.15 KN/m2Hence, required width of foundation

DL = 45KNLL = 20KNqallnet = 72.15KN/m2

=BDL LL

qallnet +

= 0.901 m

assume B = 1 mAccording to ACI-CodeU = 1.4 DL + 1.7 LLU = 1.4x45 + 1.7 x20 = 97 KN/m

Find ultimate pressure under footing,

=qultU

B 1.

=97

1 1

= 97.00 KN/m Investigation of shear strength of foundation:

critical shear section occur at distance d from the face of wall(ACI-Code 15.2.2 and 11.11.1.1)Vu = [0.35 -d] x qult = [0.35 -0.18 ] x 97 = 16.49 KN/m

=Vc0.17 fc' b d

1000

. . .

=0.17 24 1000 180

1000

= 149.91 KN/m

fVc = 0.85 x 149.9 = 127.4 KN/m> Vu = 16.49 KN/m .................O.K

Flexural Reinforcement:

l = 0.75mqu = 97KN/m2

=Muqu l2

2

. = 27.28 KN-m/m

=RuMu 106

fi b d2. ..

=5.94 106

0.9 1000 1802

= 0.20 Mpa

=mfy

0.85 fc'.

=420

0.85 24

= 20.588235

=Ro1m

1 12 Ru m

fy

. . - -.

=1

20.5882351 1

2 0.2 20.588235420

- -

= 0.000479

Romin = 0.0018 (ACI-Code 7.12.2)

Beta = 0.85fc' = 24fy = 420

=Romax 0.75 0.85fc'fy

Beta600

600 fy +. . . . = 0.0182

Ro > Rmax .......................O.K

As = 0.0018 x100 x18 = 3.24 cm2use f 10 - 24 cm c/c

=As"Provided"10024

3.144

1.0 2. .

=10024

3.144

1.0 2

= 3.27 cm2

Check development length:

=Ld 0.019 Abfyfc'

. .

= 0.019

= 0.00100

But not less than"

=Ld 0.058 db fy. .

= 0.058

= 0.00100

or 300 mmfor f =10 mm Ab = 78.5 mm2

=Ld 0.019 Abfyfc'

. .

= 0.019 78.5420

24

= 127.87 mm

Ld = 0.058 x db x fy = 0.058 x10 x 420 = 243.6 mm ControlHence Ld = 300 mmLd Provided = 350 -70(cover) = 280 mm < 300 mm Use standard hook

Long Direction:

Asmin = 0.0018 x100 x 18 = 3.24 Use 3 f 12mm

Sketch:

Problem No. 2

Design a square footing to support column load given the followint information:fc' = 24 Mpafy = 420 MpaDL = 1125 KNLL = 675 KNColumn size = 0.5 m x 0.5 mAllowable bearing soil-capacity qall = 300 KN/m2g c = 25 KN/m3g s = 20 KN/m3

Solution:

let us assume that the average weight of concrete and soil above the base of the foundationequal to 22.5 KN/m3, soqall(net) = qall-Df x 22.5 = 300 x 22.5 x 1.25 = 271.9 KN/m2

=AfDL LL

qallnet +

=1125 675

271.9 +

= 6.62 m2

use foundation of with dimension of B = 2.6m x 2.6mAf = 2.6 x2.6 = 6.76 m2 > 6.62 m2 ..........O.KU = 1.4 DL + 1.7 LLU = 1.4 x 1125 + 1.7 x 675 = 2722.5 KN

=qultUB2

=2 722.5

2.6 2

= 402.74 KN/M2

Check Punch Shear:

Critical section for punch shear occur at distance d/2 from the face of supportVu = qult x (critical area)Critical area = (2.6 x 2.6) - (0.5 x 0.5)2 = 5.7 m2Assume H = 60 cm

D = 60 - 7 = 53 cm B0 = (0.5 + 0.5) x 4 = 4.12 mVu = 402.74 x 5.7 = 2295.26 KN

=Vc0.34 fc' B0 D

1000

. . .

=0.34 24 4210 530

1000

= 3 716.57 KNf Vc = 0.85 x 3637.1 = 3091.5 KN

f Vc > Vu ...........O.K

Check one-way shear:

Critical section for one-way shear occur at distance d from the face of support, henceVu = qult x critical area= 402.74 x 2.6 x 0.52 = 544.5 KN

=Vc0.17 fc' b d

1000

. . .

=0.17 24 2600 530

1000

= 1 147.63 KN

Vc = 0.85 x 1147.63 = 975.5 KN

f Vc > Vu ...........O.K

Flexural Reinforcement:

=Muqult B L2

2

. .

=402.74 2 1.052

2

= 444.02 KN-m

=RuMu 106

fi b d2. ..

=577.23 106

0.9 2600 5302

= 0.88 Mpa

=mfy

0.85 fc'.

=420

0.85 24

= 20.5882

=Ro1m

1 12 Ru m

fy

. . - -.

=1

20.58821 1

2 0.88 20.5882420

- -

= 0.0021

=AsRo b d

100

. .

=0.0021 2600 530

100

= 29.5235 cm2

Use f 14 mm

=S260 14

29.51.54

- .

=260 14

29.51.54

-

= 12.84 cm

Use f 16 mm

=S260 16

29.52.0

- .

=260 16

29.52.0

-

= 16.54 cm

Use f 16 -16 cm c/c

Check development length:

=Ld 0.019 Abfyfc'

. .

= 0.019

= 0.00100

But not less than"

=Ld 0.058 db fy. .

= 0.058

= 0.00100

or 300 mm

=Ld 0.019 Abfyfc'

. .

= 0.019 200420

24

= 325.78 mm

Also,

Ld = 0.58x db x fy = 0.58 x16 x 420 = 389.76 mm ControlSo, Ld = 389.76 mm

=ActualLd260 50

250

- +

=260 50

250

- +

= 155.00 mm

155.0 cm > 38.9 cm ...........O.K

Check bearing on concrete:

=fb 0.85 phi fc' A1A2A1

. . . .

= 0.85 0.00100 0.001000.00100

= 0.00100 Mpa

with a limit of sqrt(A2/A1) 7140 KN .........O.KHence use minimum dowels with at least 4 bars same as column steel.

Reinforcement Sketch:

Problem No. 3Design concrete foundation to support two columns givrn the following informations:

fc' = 24 Mpafy = 420 MpaColumn size = 0.5 m x 0.5 m

Allowable bearing soil-capacity qall = 300 KN/m2g c = 25 KN/m3g s = 20 KN/m3Df = 1.5 m

Loading:Column 1: Pd = 1000 KN PL = 500 KNColumn 2: Pd = 2000 KN PL = 1000 KN

Solution:

Find L that the center of the resultant loads coincidewith the center of the footing

=X1500 1 3000 4

4500

. . +

=1500 1 3000 4

4500 +

= 3.00 m

L = 2 x X= 2 x 3 = 6 mLr = 6-4 = 2 m

Find BAssume the average weight of concrete and soil abovefooting is equal to gav = 22 kn/m3

qall(net) = 300-gav x Df = 300 -22 x 1.5 = 267 KN/m2

So,

=BR

Qallnet L.

=4500

267 6

= 2.81 m

Find pressure under footing:

Neglegt excentricity at ultimate load, give

=quPd 1.4 PL 1.7

B L

. . +.

=3000 1.4 1500 1.7

2.8 6 +

= 401.79 KN/m2

Draw moment and shear diagramn longitudinal direction:

=M1qu B L2

2

. .

=401.8 2.8 1.0 2

2

= 562.52 KN-m

=M3qu B L2

2

. .

=401.8 2.8 2 2

2

= 2 250.08 KN-m

M2 can be find where the shear is zero

2.8 x 401.8 (X0) - (1000 x 1.4 + 500 x 1.7) = 0X0 = 2 m

=M2qu B x02

2Pu1 Ll

. .. -

=401.8 2.8 2 2

22225 1

-

= 25.080 KN-M

Assume h = 70 cmd = 70 - 7 = 63 cm

Flexural Design:

=RuMu 106

phi b d2. ..

=2250 106

0.9 2800 6302

= 2.249577 Mpa

=mfy

0.85 fc'.

=420

0.85 24

= 20.588235

=Ro1m

1 12 Ru m

fy

. . - -.

=1

20.5882351 1

2 2.249577 20.588235420

- -

= 0.005689

For M3 = 2250.1 KN-m

Use 40 f 18 - 7cm c/c

For M1 = 562.5 KN-m

=RuMu 106

phi b d2. ..

=562.5 106

0.9 2800 6302

= 0.562394 Mpa

=mfy

0.85 fc'.

=420

0.85 24

= 20.588235

=Ro1m

1 12 Ru m

fy

. . - -.

=1

20.5882351 1

2 0.562394 20.588235420

- -

= 0.001358

=Romin1.4fy

=1.4420

= 0.003333

=Romax 0.75 beta1 0.85fc'fy

600600 fy +

. . . .

= 0.75 0.85 0.8524420

600600 420 +

= 0.018214

=As Rmin b d. .

= 0.00333 2800 630

= 5 874.120000 mm2

Use 24 f 18 - 12 cm c/c

for M3 use AsminAs = 58.8 cm2

Use 24 f 18 - 12 cm c/c

Check Shear: The critical section for shear occure at distance d from the face of support

Vu = 1260 KN

=Vc 0.17 fc' B d 10 3-. . . .

= 0.17 24 2800 630 10 3-

= 1 469.106 KN

f Vc = 0.85 x 1469.12 = 1248 KN @ 1260 KN O.K

No shear reinforcement required:

Chek flexure in transverse direction:Band width = with of the columnPlus two times effective depth

= 50 + 2x 63 = 176 cm

For left column:

=quPd 1.4 PL 1.7

B Band

. . +.

=1000 1.4 500 1.7

2.8 1.76 +

= 456.575 KN/m2

=Muqult B L2

2

. .

=456.6 1.76 1.152

2

= 531.391080 KN-m

=RuMu 106

phi b d2. ..

=531.391 106

0.9 1760 6302

= 0.845236 Mpa

=mfy

0.85 fc'.

=420

0.85 24

= 20.588235

=Ro1m

1 12 Ru m

fy

. . - -.

=1

20.5882351 1

2 0.845236 20.588235420

- -

= 0.002056

=Romin 0.0018

=Romax 0.75 beta1 0.85fc'fy

600600 fy +

. . . .

= 0.75 0.85 0.8524420

600600 420 +

= 0.018214

Use Romin = 0.002056As = Romin x b x d =0.00256 x 176 x 63 =22.7 cm2Use 12 f 16 - 15 cm c/c

For right column:

=quPd 1.4 PL 1.7

B Band

. . +.

=2000 1.4 1000 1.7

2.8 1.76 +

= 913.149 KN/m2

=Muqult B L2

2

. .

=913.149 1.76 1.152

2

= 1 062.722806 KN-m

=RuMu 106

phi b d2. ..

=1 062.72 106

0.9 1760 6302

= 1.69 Mpa

=mfy

0.85 fc'.

=420

0.85 24

= 20.588235

=Ro1m

1 12 Ru m

fy

. . - -.

=1

20.5882351 1

2 1.69 20.588235420

- -

= 0.004206

=Romin 0.0018

=Romax 0.75 beta1 0.85fc'fy

600600 fy +

. . . .

= 0.75 0.85 0.8524420

600600 420 +

= 0.018214

=As Ro b d. .

= 0.004206 1760 630

= 4 663.612800 mm2