محاضرات هندسة الاساسات د. طارق نجيب p2

April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 1

SHALLOW FOUNDATIONSSHALLOW FOUNDATIONSPART 2PART 2

الضحلة الضحلة األساسات األساسات

الثانى الثانى الجزء الجزء

WallWall

b/4

b

Design of Wall FootingDesign of Wall Footing تصميم تصميمالحوائط الحوائط قواعد قواعد

Due to the lower strength of brick than concrete, the critical section for continuous wall is at b/4 within the wall widthأقل الطوب الحائط مقاومة ألن نظرا%

إفتراض فيتم الخرسانة مقاومة منبعد على يقع الحرج القطاع 4/1أن

الحائط داخل فى الحائط A strip of 1.0 m in the shortعرضdirection is to be designed for moment and shear. Minimum steel is to be taken in the longitudinal direction, punching will be

safe .عرضها شريحة تصميم يتم

القصير 1 االتجاه فى مترأما والقص، العزوم على

فيتم الطولى االتجاه فىتسليح حديد نسبة أقل أخذ

بأن العلم مع ممكنة،. آمن يكون سوف االختراق

April 13, 2023 2Shallow Foundations, P1, Tarek Nageeb


ECCENTRICALLY LOADED ECCENTRICALLY LOADED FOOTINGSFOOTINGS

بحمل المنفصلة القواعد بحمل تصميم المنفصلة القواعد تصميممركزى مركزى غير غير

1

2

Soil Pressure


12

P

MP

M e tyEccentrici ,

L

e*6 1

L*B

P

21


Soil Stress DistributionSoil Stress Distribution

+

=

OR

OR

P M


Soil Stress – Soil Stress – التربة التربة إجهادات إجهادات11 ≤ q ≤ qallall soil soil ( نقطة عند اإلجهاد نقطة )أن عند اإلجهاد عن( 11أن يزيد عن( ال يزيد ال

التربة تحمل التربة قدرة تحمل قدرة3 Cases may be encountered3 Cases may be encountered عند إحتماالت ثالثة عند توجد إحتماالت ثالثة توجد

((22نقطة )نقطة ) Case )1(: Case )1(: 22 compression compression ( نقطة عند اإلجهاد نقطة )أن عند اإلجهاد ( ( 22أن

ضغطضغطCase )2(: Case )2(: 22 = zero = zero ( نقطة عند اإلجهاد نقطة )أن عند اإلجهاد صفر( = صفر( = 22أنCase )3(: Case )3(: 22 tension tension ( نقطة عند اإلجهاد نقطة )أن عند اإلجهاد شد( شد( 22أن

Cases )1( and )2( are acceptable, Cases )1( and )2( are acceptable, ( ( 22و( )و( )11الحالتين )الحالتين )مقبولتينمقبولتينCase )3( needs special treatmentCase )3( needs special treatment

الحالة ) الحالة )تحتاج خاصة( 33تحتاج معالجة خاصة( إلى معالجة إلىOption )1(: Offsetting the column distance )e(Option )1(: Offsetting the column distance )e(Option )2(: Calculate the actual soil pressure, and Option )2(: Calculate the actual soil pressure, and design the footing for such case.design the footing for such case.

مسافة(: 11االختيار )االختيار ) العمود مسافة(: ترحيل العمود ((ee))ترحيلالحالى(: 22االختيار )االختيار ) الوضع على التربة ضغط الحالى(: حساب الوضع على التربة ضغط حساب

عليه عليه والتصميم والتصميم


Soil stresses under eccentrically loaded footings

التربة إجهادات توزيعغير بحمل محملة قاعدة أسفل

محورى

P M

e

e < L/6

ORe = L/6

ORe > L/6

P/A

=

+M/Z


If the resultant lies within core of section, no tension, if it lies outside core there will be tension

داخل فى المحصلة وقعت إذاشد يحدث فلن القطاع قلب

وعند الأركان، عندالقلب خارج خروجهاالجانب فى شد المقابليحدث

P

e

L/6

B/6

L

B


PM

e

Moving the footing under the column, used with the moment is fixed in value and direction

العمود أسفل القاعدة تحريكمكان) نقل إمكانية لعدم

) وتستخدم ، Z معماريا العمودثبات حالة فى الطريقة هذه

العزم وإتجاه قيمة

B

LApril 13, 2023 Shallow Foundations, P1, Tarek Nageeb 10

رقم ) (: 2الاختيارفى لعزوم المعرضة للقاعدة

إزالة يتم الاتجاهينمن للشد المعرضة المساحة

للقاعدة الفعالة المساحةL'

B'

eL

eB

For footing subjected to Mx and My:Option )2(: Removing the tension zone from the footing area to be L' * B'L' = L – 2 eL

B' = B – 2 eB

Example )2(: Eccentrically Loaded Isolated Footing

Design a reinforced concrete isolated footing to support the followings column loads at the ground surface:D.L. = 800 kN, L.L. = 600 kNMD.L. = 250 kN.m, ML.L. = 150 kN.mHD.L = 50 kN, HL.L=25 kN. Column dimensions, 80 x 40 cmqall soil = 1.50 kg/cm2, fcu = 30 N/mm2, fy = 360 N/mm2.F.L. = - 2.00 m from the ground level )G.L.(.Design a footing such that the soil pressure will be approximately uniform.


0.40


PM

F.L.

0.80

HG.L.

2.0 m

L

B


SolutionSolution

AAR.C.R.C. = P / q = P / qallall = 151.20 / 15.0 = 10.08 m = 151.20 / 15.0 = 10.08 m22

-Column Service Loads PColumn Service Loads Pcc = P = PD.L.D.L. + P + PL.L.L.L.

- PPcc = 800 + 600 = 1400 kN = 140.0 ton = 800 + 600 = 1400 kN = 140.0 ton

-P at F.L. = 1.10 x PP at F.L. = 1.10 x Pcc = 140.0 x 1.08 = 151.2 ton = 140.0 x 1.08 = 151.2 tonDimensions of P.C. footing Dimensions of P.C. footing القاعدة القاعدة أبعاد أبعادالعاديةالعادية

m 3.40 3.37 2

0.40 -0.8 10.08

2

b-a A L p.c. cp ..

m 3.00 2.97 2

0.40-0.8 -10.08

2

b-a -A B p.c. cp ..


-M at F.L. = MM at F.L. = MD.L.D.L. + M + ML.L.L.L. + M + MHH

-MMHH = )H = )HD.L.D.L. + H + HL.L.L.L. (* D (* DF.L.F.L.

-MMHH = )5.0 + 2.5( * 2.0 = 15.0 t.m = )5.0 + 2.5( * 2.0 = 15.0 t.m-M at F.L. = 25.0 + 15.0 + 15.0 = 55.0 t.mM at F.L. = 25.0 + 15.0 + 15.0 = 55.0 t.m

m 0.36 0.364 151.20

55.0

P

M e

L

e*6 1

L*B

P

21

(64.021

)1*14.82

3.40

0.36*6 1

3.4*3.0

151.20

11 = at F.L. = M = at F.L. = MD.L.D.L. + M + ML.L.L.L. + M + MHH

0.40

0.80

3.40 m

3.00

m0.36

PM

0.36

Uniform soil pressureالتربة لضغط منتظم توزيع

2

R.C.n

t/m 14.82 3.0*3.4

151.20

A

P f


2

R.C.

uun

t/m 22.02 3.0*3.4

224.64

A

P f

0.40

0.80

3.40 m

3.00

m0.36

1.66 m0.94 m

-Ultimate Column Load Ultimate Column Load PPcucu = 1.4*P = 1.4*PD.L.D.L. + 1.6*P + 1.6*PL.L.L.L.

- PPcucu = 1.4*80.0 + 1.6*60.0 = 208.0 ton = 1.4*80.0 + 1.6*60.0 = 208.0 ton

-PPuu at F.L. = 1.08 x P at F.L. = 1.08 x Pcc = 208.0 x 1.08 = 224.64 ton = 208.0 x 1.08 = 224.64 ton

-Ultimate Contact Pressure:Ultimate Contact Pressure:


x1

Critical Section for Moment Long Direction

e

a -L x CR

2..

1 ab

I

I

LR.C.

Flexural Design

1.0

2

0.1*'/

21x

*f mM unu

m 1.66 0.36

0.80 3.40 - x

21

m.t/m' 30.34 * 22.02 mMu 2

0.1*66.1'/

2


2s

ymin s mm 917 550 *1000 * d b

f A

360

6.06.0

Flexural Design

2scu

u

d*b*f

M R

d*b*f

f * A s

y

cus

From Chart, for R = 0.033, = 0.04

0.033 55*100*300

10* R 2

5

34.30

2s mm 1833.33 550 *1000*

360

30 *0.040 A

As = 18.33 cm2

As = 1016 /m'


Critical Section for Moment in Short Direction

m1.30

0.40 3.00 - y

21

Flexural Design

2

0.1*'/

21y

f mMunu

y 1

ab

LR.C.

1.0

BR

.C.

m.t/m' 18.61

* 22.02 mMu

2

0.1*30.1'/

2


0.021

55*100*300

10* R 2

5

61.18

From Chart, for R = 0.021, = 0.025

Critical Section for Moment in Short DirectionFlexural Design


2s mm962.50 550 *1000*

360

30 * 0.021 A

As = 9.63 cm2

As = 516 /m'

fun

x2

BR.C.

LR.C.

Check of ShearCheck of Shear

e

d a - -L x CR

2..

2

ab

QQshsh = 22.02* 1.39 * 3.00 = 91.82 ton = 22.02* 1.39 * 3.00 = 91.82 ton

d/2Check of Shear:Check of Shear:

cuR.C.

shu q

d * B

Q q

m 1.39

0.55 0.80 - 3.40 - x 385.136.0

22

2u kg/cm 5.56

55*300 q

1000*82.91



cc = 1.50 = 1.50

2

2cu

kg/cm7.20

N/mm 0.72 30

0.16 q

50.1

c

cucu

f 0.16 q

Shear Allowable

qquu )5.56 kg/cm )5.56 kg/cm22( < q( < qcucu )7.20 kg/cm )7.20 kg/cm22( O.K.( O.K.


BR

.C.

LR.C.

Check of PunchingCheck of Punching

QQpp=f=funun [L [LR.C.R.C. B BR.C.R.C. –a –app*b*bpp ] ]

d/2

d/2

Punching Section Perimeter: Punching Section Perimeter: bbpunchpunch = 2 [)a+d( + )b+d(] = 2 [)a+d( + )b+d(]

bbpunchpunch = 2 [)1.35( + )0.95(] = 4.60 m = 2 [)1.35( + )0.95(] = 4.60 m

aapp = )a+d( = 0.80 + 0.55 = 1.35 m = )a+d( = 0.80 + 0.55 = 1.35 m

bbp p = )b+d( = 0.40 + 0.55 = 0.95 m= )b+d( = 0.40 + 0.55 = 0.95 m

aapp

bbpp

QQpp= 22.02*[3.40*3.00 – = 22.02*[3.40*3.00 –

1.35*0.95] = 196.36 t1.35*0.95] = 196.36 t


Check of PunchingCheck of Punching2

c

cucup N/mm 1.6 1.41

30 0.316

f 0.316 q

5.1

c

cu

p

pcup

f

b

a 0.50 0.316 q

c

cu

pcup

f

b

d 0.20 0.80 q

2cup N/mm 1.41

30

0.80 0.50 0.316 q

5.1

40.0

2cup N/mm 2.43

30.0

0.20 0.80 q

50.160.4

55.0*4



qqupup )7.76 kg/cm )7.76 kg/cm22( < q( < qcupcup 14.10 kg/cm 14.10 kg/cm22 ( O.K. ( O.K.

2

punch

upup kg/cm 7.76

55 *460

1000 * 196.36

d * b

Q q


For footing subjected to large lateral loads, check of For footing subjected to large lateral loads, check of sliding should be performed. Sliding resistance will sliding should be performed. Sliding resistance will be: Footing base resistance with underlying soils.be: Footing base resistance with underlying soils.Passive earth pressure due to wall embedmentPassive earth pressure due to wall embedmentحيث كبيرة، أفقية لقوى المعرضة للقواعد الانزلاق اختبار عمل حيث يجب كبيرة، أفقية لقوى المعرضة للقواعد الانزلاق اختبار عمل يجب

: المقاوم التربة وضغط أسفلها التربة مع القاعدة إحتكاك الانزلاق : يقاوم المقاوم التربة وضغط أسفلها التربة مع القاعدة إحتكاك الانزلاق يقاوم

Example )3(: Isolated Footing Under Moment and Horizontal Forces

Design a reinforced concrete isolated footing to support the followings column loads at the ground surface:D.L. = 800 kN, L.L. = 600 kNMD.L. = 250 kN.m, ML.L. = 150 kN.mHD.L = 50kN, HL.L=25 kN. Column dimensions, 80 x 40 cmqall soil = 1.50 kg/cm2, fcu = 30 N/mm2, fy = 360 N/mm2.F.L. = - 2.00 m from the ground level )G.L.(.


0.40


PM

F.L.

0.80

HG.L.

2.0 m

L

B


SolutionSolutionDimensions of P.C. footing to avoid tension in Dimensions of P.C. footing to avoid tension in the soil:the soil:L ≥ 6 e L ≥ 6 e to avoid tension on the soilto avoid tension on the soilL = 6*e + aL = 6*e + aOr to be calculated by the usual method, Or to be calculated by the usual method, whichever is largerwhichever is larger

عن يزيد لكى القاعدة طول اختيار عن يجب يزيد لكى القاعدة طول اختيار 66يجبحسابها يتم أو المركز عن البعد حسابها مرات يتم أو المركز عن البعد مرات

أكبرهما ويؤخذ العادية أكبرهما بالطريقة ويؤخذ العادية m 0.36بالطريقة151.20

55.0

P

M e

LL1p.c.1p.c. = 6*e + a = 6*e + a = 6 * 0.36 + 0.80 = 2.96 ≈ 3.00 m = 6 * 0.36 + 0.80 = 2.96 ≈ 3.00 m


AAR.C.R.C. = P / q = P / qallall = 151.20 / 15.0 = 10.08 m = 151.20 / 15.0 = 10.08 m22

-Column Service Loads PColumn Service Loads Pcc = P = PD.L.D.L. + P + PL.L.L.L.

- PPcc = 800 + 600 = 1400 kN = 140.0 ton = 800 + 600 = 1400 kN = 140.0 ton

-P at F.L. = 1.10 x PP at F.L. = 1.10 x Pcc = 140.0 x 1.08 = 151.2 ton = 140.0 x 1.08 = 151.2 tonDimensions of P.C. footing Dimensions of P.C. footing القاعدة القاعدة أبعاد أبعادالعاديةالعادية

Therefore, LTherefore, Lp.c.p.c. is chosen to be 3.40 m is chosen to be 3.40 m

m 3.00 2.97 2

0.40 -0.8 -10.08

2

b-a -A B p.c. cp ..

m 3.40 3.37 2

0.40 -0.8 10.08

2

b-a A L p.c. cp ..2


11 = 14.82*)1 + 0.635( = 14.82*)1 + 0.635(

= 24.23 t/m= 24.23 t/m22 > 15.0 t/m > 15.0 t/m22, unsafe, unsafe

3.4

0.36*6 1

3.4*3.0

151.20

L

e*6 1

L*B

P

21

22 = 14.82*)1 - 0.635( = 14.82*)1 - 0.635(

= 5.41 t/m= 5.41 t/m22 > 0.0 )no tension(, safe > 0.0 )no tension(, safe

The footing length is increased to 5.00 m to The footing length is increased to 5.00 m to reduce the soil stress reduce the soil stress 11. For such large footing, . For such large footing,

increase the footing weight to be 0.15 of the increase the footing weight to be 0.15 of the column load.column load.

إلى القاعدة طول زيادة إلى يتم القاعدة طول زيادة لتخفيض 5.005.00يتم لتخفيض متر مترالقيم إلى التربة على الواقعة القيم اإلجهادات إلى التربة على الواقعة اإلجهادات

. القاعدة وزن زيادة كذلك ويتم بها . المسموح القاعدة وزن زيادة كذلك ويتم بها المسموح.0.150.15إلى إلى العمود حمل .من العمود حمل من


11 = 10.73*)1 + 0.408( = 10.73*)1 + 0.408(

= 15.11 t/m= 15.11 t/m22 ≈ 15.0 t/m ≈ 15.0 t/m22, considered safe, considered safe22 = 10.73*)1 – 0.408( = 10.73*)1 – 0.408(

= 6.35 t/m= 6.35 t/m22 > 0.0 )no tension(, safe > 0.0 )no tension(, safe

-P at F.L. = 1.15 x PP at F.L. = 1.15 x Pcc = 140.0 x 1.15 = 161.0 ton = 140.0 x 1.15 = 161.0 ton

m0.34 161.0

55.0

P

M e

5.0

0.34*6 1

5.0*3.0

161.0

L

e*6 1

L*B

P

21

161

55.0

Soil Pressure under the P.C. footingالعادية القاعدة أسفل التربة على اإلجهادات توزيع

15.11 t/m2

6.35 t/m2

R.C.

P.C.

Column



- Assume t- Assume tp.c.p.c. = 40 cm = 40 cm

- Reinforced Concrete Dimensions:- Reinforced Concrete Dimensions:LLR.C.R.C. = L = Lp.c.p.c. – 2 x = 5.00 – 2 * 0.40 = 4.20 m – 2 x = 5.00 – 2 * 0.40 = 4.20 m

BBR.C.R.C. = B = Bp.c.p.c. – 2 x = 3.00 – 2 * 0.40 = 2.20 m – 2 x = 3.00 – 2 * 0.40 = 2.20 m

AAR.C.R.C. = L = LR.C.R.C. * B * BR.C.R.C. = 4.20 * 2.20 = 9.24 m = 4.20 * 2.20 = 9.24 m22

R.C. Footing Design R.C. Footing Design القاعدة القاعدة تصميم تصميم Forces and Moments at the interface between Forces and Moments at the interface between المسلحةالمسلحةreinforced and plain concrete footings:reinforced and plain concrete footings:-Ultimate Column Load Ultimate Column Load PPcucu = 1.4*P = 1.4*PD.L.D.L. + 1.6*P + 1.6*PL.L.L.L.

- PPcucu = 1.4*80.0 + 1.6*60.0 = 208.0 ton = 1.4*80.0 + 1.6*60.0 = 208.0 ton

0.40


0.80

4.20 m2.

20 m

5.00 m

3.00

m

R.C.

P.C.


-MMuu at R.C. top level = 1.4 M at R.C. top level = 1.4 MD.L.D.L. + 1.6 M + 1.6 ML.L.L.L. + M + MHH

-MMHuHu = )H = )HD.L.D.L. + H + HL.L.L.L. (* (* DDR.C.R.C.

-MMHH = )1.4*5.0 + 1.6*2.5( * = )1.4*5.0 + 1.6*2.5( * 1.601.60 = 17.60 t.m = 17.60 t.m-M at R.C.-P.C. InterfaceM at R.C.-P.C. Interface

= 1.4*25.0 + 1.6*15.0 + 17.6 = 76.60 t.m= 1.4*25.0 + 1.6*15.0 + 17.6 = 76.60 t.m

(53.021

)1* 22.51

4.20

0.37*6 1

2.20*4.2

208.0 fu

DDR.C.R.C. = D = DF.L.F.L. – t – tp.c.p.c. = 2.00 – 0.40 = 1.60 m = 2.00 – 0.40 = 1.60 m

m 0.37 0.368 208.0

76.60

P

M e

u

u 1

ffu1u1 = 22.51*)1 + 0.53( = 34.44 t/m = 22.51*)1 + 0.53( = 34.44 t/m22

ffu2u2 = 22.51*)1 – 0.53( = 10.58 t/m = 22.51*)1 – 0.53( = 10.58 t/m22

208 t

76.6

Contact Pressure between R.C. & P.C.والعادية المسلحة القاعدة بين التالمس ضغط توزيع

34.4 t/m2

10.58 t/m2

fu3 =24.76 t/m2

x1


2u t/m 24.76 10.58 34.40 - 34.40 - f

20.4

70.1*3

Critical Section for Moment Long Direction

2

..1

a -L x CR

2uu1avgun t/m 29.58

22

ff f

76.2440.342

Flexural Design

m.t/m'42.75 x

xfmM avgunu 2

7.1*7.1*58.29

2**'/ 1

1


m1.70

0.80 4.20 - x

21

2scu

u

d*b*f

M R 0.047

55*100*300

10* R 2

5

75.42

2s

ymin s mm 917 550 *1000 * d b

f A

360

6.06.0

Flexural Design

d*b*f

f * A s

y

cus

From Chart, for R = 0.047, = 0.058

2s mm 2658.33550 *1000*

360

30 * 0.058 A

As = 26.58 cm2 As = 722 /m'



y 1

ab

LR.C.

1.0

BR

.C.


2uu1avguns t/m 31.57

22

ff f

73.2840.342

m0.90

0.40 2.20 - y

21

2

0.1*'/

21y

f mMunu

m.t/m' 12.79

* 31.57 mMu

2

0.1*90.0'/

2

2u t/m 28.73 10.58 34.40 - 34.40 - f

20.4

00.1*4



0.014 55*100*300

10* R 2

5

79.12

From Chart, for R = 0.014, = 0.020

d*b*f

f * A s

y

cus

As = 9.17 cm2, As = 516 /m'

2s mm 916.67 550 *1000*

360

30 *0.020 A

x2

BR.C.

LR.C.


2..

2

d a - -L x CR

ab

QQshsh = 30.35* 1.43 * 2.20 = 95.48 ton = 30.35* 1.43 * 2.20 = 95.48 ton

d/2Check of Shear:Check of Shear:

cuR.C.

shu q

d * B

Q q

m

0.55 0.80 - 4.20 - x 43.1

22


2shearuu1avgunshear t/m 30.35

22

ff f

29.2640.34

2shearu t/m 26.29 10.58 34.40 - 34.40 - f 20.4

43.1*


cc = 1.50 = 1.50

22cu kg/cm7.20 N/mm 0.72

30 0.16 q

50.1

c

cucu

f 0.16 q :Shear Allowable

qquu )7.89 kg/cm )7.89 kg/cm22( > q( > qcucu )7.20 kg/cm )7.20 kg/cm22( Unsafe.( Unsafe.


2u kg/cm 7.89

55*220 q

1000*48.95

Increase d to 60 cm, and t to 65 cm, then xIncrease d to 60 cm, and t to 65 cm, then x22 = 1.40 = 1.40

m, fm, fu-shear-avgu-shear-avg = 30.43 t/m = 30.43 t/m22, Q, Qshsh =93.72 ton, =93.72 ton,

qquu = 7.10 kg/cm = 7.10 kg/cm2 2 < 7.20 kg/cm< 7.20 kg/cm22, O.K., O.K.

d/2208

76.6

Contact Pressure for Punchingاالختراق الختبار التالمس ضغط توزيع

34.4 t/m210.58 t/m2

fu5 =26.46 t/m2fu6 =18.52 t/m2

1.40 m1.40 m


2u t/m 26.46 10.58 34.40 - 34.40 - f

20.4

40.1*5

2u t/m 18.52 10.58 34.40 - 34.40 - f

20.4

80.2*6

BR

.C.

LR.C.


QQpp=f=funun [L [LR.C.R.C. B BR.C.R.C. –a –app*b*bpp ] ]

d/2

d/2

Punching Section Perimeter: Punching Section Perimeter: bbpunchpunch = 2 [)a+d( + )b+d(] = 2 [)a+d( + )b+d(]

bbpunchpunch = 2 [)1.40( + )1.00(] = 4.80 m = 2 [)1.40( + )1.00(] = 4.80 m

aapp = )a+d( = 0.80 + 0.60 = 1.40 m = )a+d( = 0.80 + 0.60 = 1.40 m

bbp p = )b+d( = 0.40 + 0.60 = 1.00 m= )b+d( = 0.40 + 0.60 = 1.00 m

aapp

bbpp

QQpp= 22.49*[4.20*2.20 – = 22.49*[4.20*2.20 –

1.40*1.00] = 176.32 t1.40*1.00] = 176.32 t


ffu-punch-avgu-punch-avg = )26.46 + 18.52(/2 = 22.49 t/m = )26.46 + 18.52(/2 = 22.49 t/m22

Check of PunchingCheck of Punching2

c

cucup N/mm 1.6 1.41

30 0.316

f 0.316 q

5.1

c

cu

p

pcup

f

b

a 0.50 0.316 q

c

cu

pcup

f

b

d 0.20 0.80 q

2cup N/mm 1.41

30

0.80 0.50 0.316 q

5.1

40.0

2cup N/mm 2.50

30.0

0.20 0.80 q

50.180.4

60.0*4




2

punch

upup kg/cm 6.12

60 *480

1000 * 176.32

d * b

Q q


Check of footing sliding, F.S. = FCheck of footing sliding, F.S. = FResistingResisting/F/FSlidingSliding::FFSlidingSliding = Horizontal forces= Horizontal forces

FFResistingResisting = P tan = P tan for sand, for sand, = ½ up to ¾ = ½ up to ¾ FFResistingResisting = C= Caa * L for clay, C * L for clay, Caa ≈ 0.80 C, ≈ 0.80 C,

L = footing lengthL = footing lengthIgnore passive resistance for shallow depths, and Ignore passive resistance for shallow depths, and take it for relatively large depths.take it for relatively large depths.F.S. > 1.50F.S. > 1.50

Check of Development LengthCheck of Development LengthLLdd = 60 = 60 for deformed high grade bars for deformed high grade bars

)long direction()long direction(LLd1d1 = 60*2.20 = 132 cm = 60*2.20 = 132 cm

xx11 = 170 cm > L = 170 cm > Ldd O.K. O.K.

)short direction()short direction(LLd2d2 = 60*1.60 = 96 cm )short direction( = 60*1.60 = 96 cm )short direction(

yy11 = 90 cm < L = 90 cm < Ld2d2, bend bars upward , bend bars upward

516 /m'722 /m'

المسلحةالعادية



DESIGN OF COMBINED DESIGN OF COMBINED FOOTINGSFOOTINGS

المشتركة القواعد المشتركة تصميم القواعد تصميم


FOUNDATION ENGINEERINGFOUNDATION ENGINEERING1- Site Investigation1- Site Investigation2- Design of Shallow Foundations:2- Design of Shallow Foundations:

a- Design of Isolated Footings.a- Design of Isolated Footings.b- Design of Isolated Footings under b- Design of Isolated Footings under

Eccentric Loads.Eccentric Loads.c- Design of Combined Footings.c- Design of Combined Footings.d- Design of Strap Beam.d- Design of Strap Beam.e- Design of Raft Foundationse- Design of Raft Foundations


Combined FootingCombined Footing

المشتركة المشتركة القواعد القواعد

Plain Concrete

Reinforced ConcreteReinforced Concrete

Pc2

Pc1


Combined Footings are used in:Combined Footings are used in:1- Two adjacent columns with interfering footings.1- Two adjacent columns with interfering footings.2- To connect neighbor column with an interior 2- To connect neighbor column with an interior columncolumn

: عندما المشتركة القواعد :تستخدم عندما المشتركة القواعد تستخدملعمودين -1 -1 المنفصلة القواعد لعمودين تداخل المنفصلة القواعد تداخل

..متجاورينمتجاورينداخلية -2 -2 بقاعدة الجار قاعدة داخلية لربط بقاعدة الجار قاعدة ..لربط


المل~كيةالمل~كيةحد حد جارجارعمودعمود

Property Property LineLine

Neighbor Neighbor ColumnColumn


c2c1

cr P P

S * P x

2Resultant: R = Pc1 + Pc2

LP.C. = 2 * xr + C1/2 + C2/2 + )1 ~ up to 2 m(

RPc1 Pc2

xr

S

L

For the plain concrete footing:


soil-all

c2cCP q

P P A

1

..

*10.1

P.C.

CPCP L

A B ..

..

Dimensions of Plain Concrete = LP.C. * BP.C.

Choose Plain Concrete thickness, tP.C.

Projection distance, x = tP.C.

العادية الخرسانة تخانة اختيار يتمالمسلحة القاعدة عن العادية القاعدة رفرفة تساوى وهى

Dimensions of Reinforced Concrete:

LR.C. = LP.C. – 2 * x BR.C. = BP.C. – 2 * x

Assume the thickness of R.C. المسلحة للقاعدة تخانة إفتراض يتم


For Columns C1 and C2 respectivelyللعمودين ) (2و( )1وذلك

Ultimate Column Loads

Puc1 = 1.4 * Pc1 Dead + 1.6 * Pc1 Live

Puc2 = 1.4 * Pc2 Dead + 1.6 * Pc2 Live

القاعدتين بين التالمس ضغط حساب يتمالمتر على موزع كحمل والمسلحة العادية

يلى كما المسلحة القاعدة طول من الطولى t/m'

L

P P f

R.C.

uc2ucun

1

1

Combined Footing Pressure and Deflections



RPuc1 Puc2

xr

S

LR.C.

fun1 )t/m'(

Mmax. 1

Mmax. 2Mmax. 3


RPuc1 Puc2

xr

S

LR.C.

fun1 )t/m'(

d/2d/2

Qsh1Qsh2


Design for Flexure in the Longitudinal Directionالطولى االتجاه فى العزوم نتيجة التصميم

Reinforced concrete depth and main upper Reinforcements are determined using Mmax1

الخرسانية القاعدة تخانة تحديد يتمالتسليح وكذلك المسلحة

العزم طريق عن الرئيسى Mmax1Lower longitudinal reinforcements underالعلوىcolumn )2( are determined using Mmax2

أسفل السفلى الطولى التسليح تحديد يتمالعزم( 2العمود ) طريق Mmax2عن


Longitudinal Reinforcementsالطولى التسليح

Pc1Pc2


Design for Flexure in the Short Directionالقصير االتجاه فى العزوم نتيجة التصميم

Assume that all the column load is distributed under the column as a hidden beam on width only with extension of d/2 from each side

على يوزع العمود حمل أن إفتراض يتممع مدفونة ككمرة المسلحة القاعدة عرض

تصميم d/2مسافة يتم حيث جانب، كل منالعزوم هذه على القصير االتجاه


Hidden Beams

d/2d/2 d/2d/2

Puc2Puc1


Reinforcements in the Short Directionالقصير االتجاه فى التسليح

Puc1Puc2

d/2 d/2d/2 d/2


d/2d/2

QQp1 p1 =P=Puc1uc1 - f - funun *[a *[ap1p1*b*bp1p1 ] ]

QQp2 p2 =P=Puc2uc2 - f - funun *[a *[ap2p2*b*bp2p2 ] ]

ap1

bp1bp2

d/2d/2

ap2

Check of Punchingاالختراق اختبار


Example )4(: Design of Combined Footing Example )4(: Design of Combined Footing From Dr. Mashour Ghonaim BookFrom Dr. Mashour Ghonaim Book

Design a combined footing to support the two columns shown in Figure. Column C1, 0.30 * 0.40 m, working load = 1320 kNColumn C2, 0.30 * 0.70 m, working load = 1960 kNAllowable soil pressure = 175 kN/m2

Concrete, fcu = 25 N/mm2; Steel, fy = 400 N/mm2

S = 3.60 m0.30 0.30

0.700.40

C1 C2


Resultant: R = Pc1 + Pc2 = 132.0 + 196.0 = 328.0 ton

LP.C. = 2 * xr + C1/2 + C2/2 + )1 ~ up to 2 m(

RPc1=132t Pc2=196t

xr =2.15

S = 3.60 m

L

For the plain concrete footing:

m 2.15 196132

3.60* 196

P P

S * P x

c2c1

cr

2

Solution


Plain concrete footing dimensions:Lp.c. = 6.00 m, Bp.c. = 3.45 m, tp.c. Taken 40 cm

LP.C. = 2 * 2.15 + 0.20 + 0.35 + 1.15 = 6.00 m

2

soil-all

c21c.C.P m 20.62

17.50

196132*10.1

q

P P*10.1 A

m 3.45 3.44 6.00

62.20

L

A B

P.C.

.C.P.C.P

S = 3.60 m0.300.30

0.700.40R

3.0 m 3.0 m

2.15 m

3.45


Dimensions of Reinforced Concrete:

LR.C. = LP.C. – 2 * x = 6.00 – 2 * 0.40 = 5.20 m

BR.C. = BP.C. – 2 * x = 3.45 – 2 * 0.40 = 2.65 m

Assume the thickness of R.C. = 80 cm

Projection distance x = tp.c. = 40 cm

Puc1 = 1.5 * Pc1 = 1.50 * 132.0 = 198.0 ton

Puc2 = 1.5 * Pc2 = 1.50 * 196.0 = 294.0 ton

t/m' 94.62

5.20

294.0198.0

L

P P /m'f

R.C.

uc21uc1un

Ultimate contact pressure per meter run


Point of zero shear:Puc1 = fun1 * x , 198 = 94.62 * x, x = 2.09 mMoment at point of zero shear )Mmax1(:

m.t 118.072

2.09*94.62 -0.45-2.09*198.0 M

2

1max

198 294

x=2.09

3.60

5.20 m94.62 t/m'

0.45 1.15

1234 P1P2


Moments at different points

Point M )m.t(

(1) 30.27

(2) 3.55

(3) -19.61

(4) 2.96

m.t 57.622

1.15*94.62 M

2

2max

m.t 58.92

0.45*94.62 M

2

3max


Bending Moment Diagram

R198 2942.15

3.60

5.20 m94.62 t/m'

0.45 1.15

1234

118.07

62.57

9.58 30.273.55

19.61

2.96

19.61


R198 2942.15

3.60

5.20 m94.62 t/m'

0.45 1.15

1234 P1P2

d/2d/2

Shear Force Diagram

108.81

185.19

155.42

42.58117.54

41.16101.96

Flexural Design in the Long Direction


m.t/m'44.55 2.65

118.07

B

M 'm/M

R.C.

1maxmax

2scu

u

d*b*f

M R 0.033

73*100*250

10*55.44 R 2

5

For R = 0.033, = 0.04 d*b*f

f * A s

y

cus

'm/mm1825.0 730 *1000*400

25 *0.040 A 2

s

As = 620/m'

For a concrete cover of 7 cm, d = 80 – 7 = 73 cm

Flexural Design in the Long Direction


Under column C2, M1 = 30.27 m.t

m.t/m' 11.43 2.65

30.27

B

M 'm/M

R.C.

11

R = 0.009, take = 0.02, As = 9.13 cm2

Therefore, for moments at points )2(, )3(, )4( take the minimum area steel of As = 12.00 cm2 = 616/m'

2s

ymin s mm1095 730 *1000 *

400

6.0 d b

f

6.0 A

As min = 10.95 cm2, As min = 6 16 mm/m'


In the short direction, d1 = d – 2 = 73 – 2 = 71 cmbc1 = ac1 + 0.25 + d1/2

= 0.40 + 0.25 + 0.71/2 = 1.005 m ≈ 1.01 mbc2 = ac2 + 2*)d1/2( = 0.70 + 2*)0.71/2( = 1.41 m

198 294

bc1=1.01

0.25 d1/2

bc2=1.41

d1/2d1/2

رقم = )bc1حيث العمود حمل عليها سيوزع التى الشريحة ،( 1عرضbc2( = رقم العمود حمل عليها سيوزع التى الشريحة (2عرض

Flexural Design in the Short Direction


t/m'110.94 2.65

294.0

B

P f

R.C.

uc2'2u


t/m' 74.72 2.65

198.0

B

P f

R.C.

uc1'1u

294 t

f'u2 = 110.94 t/m'

2.65

198 t

f'u1 = 74.72 t/m'

2.65

1.18


m.t 52.02 2

1.18*72.74

2

yf M

221'

uc1'

1u

m.t77.24 2

1.18*94.110

2

yf M

221'

uc2'

2u


For R = 0.043, = 0.0522

2scu

u

d*b*f

M R 0.043

71*141*250

10*24.77 R 2

5

d*b*f

f * A s

y

cus

2s mm 0.3267710*1410*

400

25 * .0520 A

Under column C2, M'u2 = 77.24 m.t



For R = 0.041, = 0.04920.041 71*101*250

10*02.55 R 2

5

2s mm 0.2194710*1005*

400

25 * .04920 A

Under column C1, M'u1 = 52.02 m.t

As = 32.67 cm2 = 922 رقم ) العمود أسفل سفلى كغطاء الحديد هذا وضع ( 2يتم

مقداره عرض على متر 1.41موزع

As = 21.94 cm2 = 622 رقم ) العمود أسفل سفلى كغطاء الحديد هذا وضع ( 1يتم

مقداره عرض على متر 1.01موزع


R198 2942.15

3.60

5.20 m94.62 t/m'

0.45 1.15

1234 P1P2

d/2d/2

Shear Force Diagram

108.81

185.19

155.42

42.58117.54

41.16101.96

Design for Shear


QQshsh = 117.54 ton = 117.54 ton

Check of Shear:Check of Shear:

cuR.C.

shu q

d * B

Q q

cc = 1.50 = 1.50

qquu )6.08 kg/cm )6.08 kg/cm22( < q( < qcucu )6.50 kg/cm )6.50 kg/cm22( Safe.( Safe.

2u kg/cm 08.6

73*265

1000*54.117 q

c

cucu

f 0.16 q :Shear Allowable

22cu kg/cm6.50 N/mm0.65

50.1

25 0.16 q


ap2

bp

2

ap1

bp

1

Check of Punching

aap2p2 = )a = )a22+d( = 0.70 + 0.73 = 1.43 m+d( = 0.70 + 0.73 = 1.43 m

bbp2 p2 = )b= )b22+d( = 0.30 + 0.73 = 1.03 m+d( = 0.30 + 0.73 = 1.03 m

QQp2p2=P=Puc2uc2 - f - funun [a [ap2p2*b*bp2p2 ] = 294.0 – 35.70*[1.43*1.03] ] = 294.0 – 35.70*[1.43*1.03]

= 241.42 ton= 241.42 ton

Punching Section Perimeter: Punching Section Perimeter: bbpunch2punch2 = 2 [)1.43( + )1.03(] = 4.92 m = 2 [)1.43( + )1.03(] = 4.92 m

2

R.C.

2cc1un

t/m35.70 65.2*20.5

294198

A

PP f



2

c

cucup N/mm 1.6 1.29

5.1

25 0.316

f 0.316 q

c

cu

p

pcup

f

b

a 0.50 0.316 q

c

cu

pcup

f

b

d 0.20 0.80 q

2cup N/mm1.20

5.1

25

0.70

30.0 0.50 0.316 q

2cup N/mm 2.59

50.1

25.0

92.4

73.0*4 0.20 0.80 q




2

punch

upup kg/cm 6.72

73 * 492

1000 * 241.42

d * b

Q q


Longitudinal Reinforcements, التسليحالطولى


Reinforcements in the Short Directionالقصير االتجاه فى التسليح

Under Column C2

Under Column C1

محاضرات هندسة الاساسات د. طارق نجيب p2

Documents

shallow foundations

tarek nageeb p

tarek nageeb critical

tarek nageeb check of

direction b

e e l6

p qall

p e l6 b6