محاضرات هندسة الاساسات د. طارق نجيب p2
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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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 3
ECCENTRICALLY LOADED ECCENTRICALLY LOADED FOOTINGSFOOTINGS
بحمل المنفصلة القواعد بحمل تصميم المنفصلة القواعد تصميممركزى مركزى غير غير
1
2
Soil Pressure
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 4
12
P
MP
M e tyEccentrici ,
L
e*6 1
L*B
P
21
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 5
Soil Stress DistributionSoil Stress Distribution
+
=
OR
OR
P M
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 6
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 7
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 8
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 9
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.
April 13, 2023 11Shallow Foundations, P1, Tarek Nageeb
0.40
April 13, 2023 12Shallow Foundations, P1, Tarek Nageeb
PM
F.L.
0.80
HG.L.
2.0 m
L
B
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 13
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 ..
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 14
-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
April 13, 2023 15Shallow Foundations, P1, Tarek Nageeb
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:
April 13, 2023 16Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 17Shallow Foundations, P1, Tarek Nageeb
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'
April 13, 2023 18Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 19Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 20Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 21Shallow Foundations, P1, Tarek Nageeb
Check of ShearCheck of Shear
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.
April 13, 2023 22Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 23Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 24Shallow Foundations, P1, Tarek Nageeb
Check of PunchingCheck of Punching
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
April 13, 2023 25Shallow Foundations, P1, Tarek Nageeb
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.(.
April 13, 2023 26Shallow Foundations, P1, Tarek Nageeb
0.40
April 13, 2023 27Shallow Foundations, P1, Tarek Nageeb
PM
F.L.
0.80
HG.L.
2.0 m
L
B
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 28
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 29
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 30
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 31
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
April 13, 2023 32Shallow Foundations, P1, Tarek Nageeb
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 33
- 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
April 13, 2023 34Shallow Foundations, P1, Tarek Nageeb
0.80
4.20 m2.
20 m
5.00 m
3.00
m
R.C.
P.C.
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 35
-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
April 13, 2023 36Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 37Shallow Foundations, P1, Tarek Nageeb
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'
April 13, 2023 38Shallow Foundations, P1, Tarek Nageeb
Critical Section for Moment in Short DirectionFlexural Design
y 1
ab
LR.C.
1.0
BR
.C.
April 13, 2023 39Shallow Foundations, P1, Tarek Nageeb
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
Critical Section for Moment in Short DirectionFlexural Design
April 13, 2023 40Shallow Foundations, P1, Tarek Nageeb
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.
Check of ShearCheck of Shear
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
April 13, 2023 41Shallow Foundations, P1, Tarek Nageeb
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*
Check of ShearCheck of Shear
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.
April 13, 2023 42Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 43Shallow Foundations, P1, Tarek Nageeb
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.
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 [)1.40( + )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
April 13, 2023 44Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 45Shallow Foundations, P1, Tarek Nageeb
Check of PunchingCheck of Punching
qqupup )6.12 kg/cm )6.12 kg/cm22( < q( < qcupcup 14.10 kg/cm 14.10 kg/cm22 ( O.K. ( O.K.
2
punch
upup kg/cm 6.12
60 *480
1000 * 176.32
d * b
Q q
April 13, 2023 46Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 47Shallow Foundations, P1, Tarek Nageeb
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 48
DESIGN OF COMBINED DESIGN OF COMBINED FOOTINGSFOOTINGS
المشتركة القواعد المشتركة تصميم القواعد تصميم
April 13, 2023 49Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 50
Combined FootingCombined Footing
المشتركة المشتركة القواعد القواعد
Plain Concrete
Reinforced ConcreteReinforced Concrete
Pc2
Pc1
April 13, 2023 51Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 52Shallow Foundations, P1, Tarek Nageeb
المل~كيةالمل~كيةحد حد جارجارعمودعمود
Property Property LineLine
Neighbor Neighbor ColumnColumn
April 13, 2023 53Shallow Foundations, P1, Tarek Nageeb
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:
April 13, 2023 54Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 55Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 56Shallow Foundations, P1, Tarek Nageeb
April 13, 2023 57Shallow Foundations, P1, Tarek Nageeb
RPuc1 Puc2
xr
S
LR.C.
fun1 )t/m'(
Mmax. 1
Mmax. 2Mmax. 3
April 13, 2023 58Shallow Foundations, P1, Tarek Nageeb
RPuc1 Puc2
xr
S
LR.C.
fun1 )t/m'(
d/2d/2
Qsh1Qsh2
April 13, 2023 59Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 60Shallow Foundations, P1, Tarek Nageeb
Longitudinal Reinforcementsالطولى التسليح
Pc1Pc2
April 13, 2023 61Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 62Shallow Foundations, P1, Tarek Nageeb
Hidden Beams
d/2d/2 d/2d/2
Puc2Puc1
April 13, 2023 63Shallow Foundations, P1, Tarek Nageeb
Reinforcements in the Short Directionالقصير االتجاه فى التسليح
Puc1Puc2
d/2 d/2d/2 d/2
April 13, 2023 64Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 Shallow Foundations, P1, Tarek Nageeb 65
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
April 13, 2023 66Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 67Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 68Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 69Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 70Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 71Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 72Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 73Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 74Shallow Foundations, P1, Tarek Nageeb
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'
April 13, 2023 75Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 76Shallow Foundations, P1, Tarek Nageeb
t/m'110.94 2.65
294.0
B
P f
R.C.
uc2'2u
Flexural Design in the Short Direction
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
April 13, 2023 77Shallow Foundations, P1, Tarek Nageeb
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
Flexural Design in the Short Direction
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
April 13, 2023 78Shallow Foundations, P1, Tarek Nageeb
Flexural Design in the Short Direction
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
April 13, 2023 79Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 80Shallow Foundations, P1, Tarek Nageeb
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
April 13, 2023 81Shallow Foundations, P1, Tarek Nageeb
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
Check of PunchingCheck of Punching
April 13, 2023 82Shallow Foundations, P1, Tarek Nageeb
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
Check of PunchingCheck of Punching
qqupup )6.72 kg/cm )6.72 kg/cm22( < q( < qcupcup 12.00 kg/cm 12.00 kg/cm22 ( O.K. ( O.K.
April 13, 2023 83Shallow Foundations, P1, Tarek Nageeb
2
punch
upup kg/cm 6.72
73 * 492
1000 * 241.42
d * b
Q q
April 13, 2023 84Shallow Foundations, P1, Tarek Nageeb
Longitudinal Reinforcements, التسليحالطولى
April 13, 2023 85Shallow Foundations, P1, Tarek Nageeb
Reinforcements in the Short Directionالقصير االتجاه فى التسليح
Under Column C2
Under Column C1
top related