محاضرات هندسة الاساسات د. طارق نجيب P3

April 13, 2023 Shallow Foundations, P3, Tarek Nageeb 1

SHALLOW FOUNDATIONSSHALLOW FOUNDATIONSPART 3PART 3

الضحلة الضحلة األساسات األساسات

الثالث الثالث الجزء الجزء


DESIGN OF STRAP BEAMDESIGN OF STRAP BEAM الشداد الشداد تصميم تصميم

April 13, 2023 3Shallow Foundations, P3, 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


Strap Beam

لربطعمود الشداد

الجار

CC22CC11 Strap BeamStrap Beam

جار جار قاعدة قاعدة

محور العمود

محور المسلح

ة

محور العادية

PropertyLine

الملكية حد

Pc2

Strap BeamStrap Beam

Pc1


DESIGN OF STRAP BEAMDESIGN OF STRAP BEAMالشداد الشداد تصميم تصميم

Strap beam connects neighbor footing (at the Strap beam connects neighbor footing (at the property line) with the adjacent inner footing.property line) with the adjacent inner footing.على ( الجار قاعدة ربط فى الشداد على ( يستخدم الجار قاعدة ربط فى الشداد يستخدم

( داخلى عمود أقرب قاعدة مع الملكية ) حد داخلى عمود أقرب قاعدة مع الملكية حدلها.لها.

Resultant, R = PResultant, R = Pc1c1 + P + Pc2c2

PPtotaltotal = 1.10*(P = 1.10*(Pc1c1 + P + Pc2c2))

April 13, 2023 Shallow Foundations, P3, Tarek Nageeb 6c2c1

1c2

11c

P P

L2a

*P 2a

* P y

To determine the resultant (R) point of application from the property line, take moments at point (O)وأحمال المحصلة عزوم أخذ يتم المحصلة مكان لتحديد

نقطة حول الملكية (O)القواعد حد عند

Pc2


PropertyLine

الملكية حد

Pc1R

L

y

2*y

O

a1


Dimensions of Plain Concrete Footings:

Assume the thickness of the plain concrete footings, tp.c.

العادية الخرسانة فرضتخانة يتم

Assume the lengths of footings F1 to be (Lp.c.1) and F2 to be (Lp.c.2), such that the resultant would approximately lie at the middle of the outer plain concrete edges.

القاعدتين إفتراضأطوال تكون (F1, F2)يتم بحيثللقاعدتين الخارجية الحدود منتصف فى المحصلة

تقريبا<Determine the reactions under the footings, R1 and R2

الفعل رد قيم تحديد فى (R1 & R2)يتم ومكانهقاعدة منتصفكل



To determine R1, take moments about Point (2):قيمة نقطة )(R1)لتحديد حول العزوم أخذ (:2يتم

L*10.1*P 2

L

2

aL*R c1

1.c.p11

Pc2


PropertyLine

الملكية حد

Pc1

O LR1 R2

2

Lp.c.1 Lp.c.2

a1



2a

LL

L*10.1*P R

11.c.p

c11

soil all

11.c.p q

R A

1.c.p

1.c.p1.c.p L

A B

تزيد أن الحالة هذه فى تزيد يسمح أن الحالة هذه فى ( ( LLp.c.1p.c.1))عن عن ((BBp.c.1p.c.1))يسمح

RR22 = P = Ptotaltotal – R – R11

soil all

22.c.p q

R A

2.c.p

2.c.p2.c.p L

A B

- Assume t- Assume tp.c.p.c. , and x = t , and x = tp.c.p.c.

- Reinforced Concrete Dimensions:- Reinforced Concrete Dimensions:LLR.C.1R.C.1 = L = Lp.c.1p.c.1 – x, – x, BBR.C.1R.C.1 = B = BR.C.1R.C.1 – 2 x – 2 x

LLR.C.2R.C.2 = L = Lp.c.2p.c.2 – 2 x, – 2 x, BBR.C.2R.C.2 = B = BR.C.2R.C.2 – 2 x – 2 x


Calculation of Ultimate Footing LoadsCalculation of Ultimate Footing Loads

PPuc1uc1 = 1.4 * (P = 1.4 * (Pc1c1+W+Wst.b./2st.b./2)) Dead + 1.6 * P Dead + 1.6 * Pc1c1 Live Live

PPuc2uc2 = 1.4 * (P = 1.4 * (Pc1c1+W+Wst.b./2st.b./2)) Dead + 1.6 * P Dead + 1.6 * Pc2c2 Live Live

Strap beam weight:Strap beam weight:WWst.b.st.b. = b = bst.b.st.b.* t* tst.b.st.b.*L**L*cc

إلى نصفه وإضافة الشداد وزن حساب إلى يتم نصفه وإضافة الشداد وزن حساب يتمعمود لكل الميت عمود الحمل لكل الميت الحمل

2L

2a

L

L*P R

1.C.R1

uc11u

RRu2u2 = P = Puc1 uc1 ++ P Puc2uc2 – R – Ru1u1

1.C.R

u11un L

R /m'f

2.C.R

u22un L

R /m'f

To get RTo get Ru1u1 take take

moments at point 2*moments at point 2*


Contact Pressure between P.C. and R.C. Footings

Pc1 Pc2

fun1/m' fun2/m'

Pc2


PropertyLine

الملكية حد

Pc1

L

Ru1 Ru2

LR.c.1LR.c.2

a1

2*


Shear Force Diagram of the Strap BeamShear Force Diagram of the Strap Beam

Pc1 Pc2

fun1/m' fun2/m'


Bending Moment Diagram:Bending Moment Diagram:Pc1 Pc2

fun1 fun2

Mmax1

Mmax2


Design of Strap beam for Flexure

Pc2


PropertyLine

الملكية حد

Pc1

L

LR.c.1LR.c.2

a1

Mmax2

M1M2

Mmax1

Two sections are designed for Mmax1 and the larger of M1 and M2


Design of Strap Beam for Shear Pc2


PropertyLine

الملكية حد

Pc1

L

LR.c.1LR.c.2

a1

Maximum shear took place at the strap beam free span.

Qmax1


Design of Strap Beam Footings Pc2


PropertyLine

الملكية حد

Pc1

LLR.c.1

LR.c.2

a1

Critical section for moment at the strap beam faceعلىسطح القواعد فى للعزوم الحرج القطاع

الشدادF1

CC22CC11 Strap BeamStrap BeamF2

d/2Critical Section for Shear

Critical Section for Moment

Critical section for shear at d/2 from the strap beam faceبعد على القواعد للقصفى الحرج سطح d/2القطاع

الشداد


Longitudinal Reinforcements for the Strap Beamللشداد الطولى التسليح

Strap Beam Reinforcement


Reinforcement of Strap Beam Footings

F1

CC22CC11 Strap BeamStrap BeamF2

طولى بحديد القواعد تسليح يتمفقط وعرضىسفلى


Example (5): Design of Strap BeamExample (5): Design of Strap BeamFrom Dr. Mashour Ghonaim BookFrom Dr. Mashour Ghonaim Book

Design a strap beam to connect the two footings shown in figure knowing that:Allowable soil pressure = 150 kN/m2

Concrete, fcu = 25 N/mm2; Steel, fy = 360 N/mm2


DESIGN OF STRAP BEAMDESIGN OF STRAP BEAMResultant, R = PResultant, R = Pc1c1 + P + Pc2c2

PPc1c1 = 380 + 305 = 685.0 kN = 68.50 ton = 380 + 305 = 685.0 kN = 68.50 ton

PPc2c2 = 820 + 450 = 1270.0 kN = 127.00 ton = 820 + 450 = 1270.0 kN = 127.00 ton

R = 68.50 + 127.00 = 195.50 tonR = 68.50 + 127.00 = 195.50 ton

c2c1

1c2

11c

P P

L2a

*P 2a

* P y

m 3.43

127.0 68.50

90.425.0*127.0 0.25 *68.50 y


Choose the footings such that:Lp.c.2 = 2*[6.85 – (4.90 + 0.25)]

≈ 3.40 m, to be taken 3.35 mLp.c.1 assumed to be taken 2.20 m

Pc2=127 t


PropertyLine

الملكية حد

Pc1 =68.50 t

O

R

4.90

3.43

2*y ≈ 6.85 m


Design of Plain Concrete Footings:

To determine R1, take moments about Point (2):قيمة نقطة )(R1)لتحديد حول العزوم أخذ (:2يتم

127.0 t


PropertyLine

الملكية حد

68.50 t

O 4.90 mR1 R2

2

2.20 m 3.35 m

a1

2

L

2a

L

L*10.1*P R

1.c.p1

c11 ton 91.16

220.2

250.0

90.4

90.4*10.1*50.68 R1


Design of Plain Concrete Footings:

هذه فى هذه يسمح فى يسمحتزيد أن تزيد الحالة أن الحالة

((BBp.c.1p.c.1)) عن عن((LLp.c.1p.c.1 ) )

RR22 = P = Ptotaltotal – R – R1 1 = 215.05 - 91.16 = 123.89 ton= 215.05 - 91.16 = 123.89 ton

PPtotaltotal = 1.10 *(P= 1.10 *(Pc1c1 + P + Pc2c2) )

= 1.10*(68.50+127.0) = 215.05 ton= 1.10*(68.50+127.0) = 215.05 ton

2

soil all

22.c.p m 8.26

0.15

89.123

q

R A

m 2.60 2.47 35.3

26.8

L

A B

2.c.p

2.c.p2.c.p

2

soil all

11.c.p m 6.08

0.15

16.91

q

R A

m 2.80 2.76 20.2

08.6

L

A B

1.c.p

1.c.p1.c.p


Design of Reinforced Concrete Footings:

- Assume t- Assume tp.c.p.c. = 40 cm, and x = t = 40 cm, and x = tp.c.p.c. = 40 cm = 40 cm

- Reinforced Concrete Dimensions:- Reinforced Concrete Dimensions:LLR.C.1R.C.1 = L = Lp.c.1p.c.1 – x – x = 2.20 – 0.40 = 1.80 m= 2.20 – 0.40 = 1.80 m

BBR.C.1R.C.1 = B = BR.C.1R.C.1 – 2 x – 2 x = 2.80 – 0.80 = 2.00 m= 2.80 – 0.80 = 2.00 m

LLR.C.2R.C.2 = L = Lp.c.2p.c.2 – 2 x – 2 x = 3.35 – 0.80 = 2.55 m= 3.35 – 0.80 = 2.55 m

BBR.C.2R.C.2 = B = BR.C.2R.C.2 – 2 x – 2 x = 2.60 – 0.80 = 1.80 m= 2.60 – 0.80 = 1.80 m


Calculation of Ultimate Footing LoadsCalculation of Ultimate Footing Loads

PPuc1uc1 = 1.4 * 41.15 + 1.6 * 30.50 = 106.41 ton = 1.4 * 41.15 + 1.6 * 30.50 = 106.41 ton

Strap beam weight:Strap beam weight:WWst.b.st.b. = b = bst.b.st.b.* t* tst.b.st.b.*L**L*cc = 2.5*0.4*1.3*4.9 = 2.5*0.4*1.3*4.9 6.3 t 6.3 t

PPc1DLc1DL = 38.0 + 6.3/2 = 41.15 ton = 38.0 + 6.3/2 = 41.15 ton

PPc2DLc2DL = 82.0 + 6.3/2 = 85.15 ton = 82.0 + 6.3/2 = 85.15 ton

PPuc2uc2 = 1.4 * 85.15 + 1.6 * 45.0 = 191.21 ton = 1.4 * 85.15 + 1.6 * 45.0 = 191.21 ton

ton 122.67

280.1

25.090.4

90.4*40.106

2L

2a

L

L*P R

1.C.R1

uc11u


Calculation of Ultimate Contact PressureCalculation of Ultimate Contact Pressure

t/m'68.15 80.1

67.122

L

R /m'f

1.C.R

u11un

t/m' 68.61 55.2

95.174

L

R /m'f

2.C.R

u22un

RRu2u2 = P= Puc1 uc1 ++ P Puc2uc2 – R – Ru1u1

= 106.41 + 191.21 – 122.67 = 174.95 ton = 106.41 + 191.21 – 122.67 = 174.95 ton

Design of Strap Section for Maximum Moment:


d = t – 7 cm = 130.0 – 7.0 = 123.0 cm

Mmax = 56.46 t.m2

scu

u

d*b*f

M R

0.037 123*40*250

10*46.56 R 2

5

Assume the strap beam has the following dimensions:Assume the strap beam has the following dimensions:ttst.b.st.b. = 130 cm = 130 cm bbst.b.st.b. = 40 cm = 40 cm

ddst.b.st.b. = 130.0 – 7.0 = 123.0 cm = 130.0 – 7.0 = 123.0 cm

From the R- chart, R = 0.037, then = 0.045


d*b*f

f * A s

y

cus

2s mm1537.0 1230 *400*

360

25 *0.045 A

For such large concrete section and to make sure that the failure will not be brittle, the Egyptian Code states that the minimum area of steel will be:

كبيرة أبعاد ذو خرسانى قطاع وجود حالة فىمفاجئ إنهيار حدوث عدم من التأكد فيجب

نسبة زيادة يتم وبذلك الخرسانة، بالضغطفىحدوث من للتأكد المصرى للكود طبقا< الحديد

ممطولى إنهيار

s

y

cu

min s

A 3.1

d bf

f0.225

of smaller A


steel)grade (high d b 100

0.15

steel) (mild d b 100

0.25

than largerbe houlds A min s

2s

2

y

cu

min s

mm 1998 1537.0 *1.30 A 3.1

mm 15361230*400*360

25225.0d b

f

f0.225

of smaller A

2min s mm 738 1230*400*

100

0.15 d b

100

0.15 A

As = 15.36 cm2, use 816 = 16.08 cm2


Design of Strap Sec. 2: for M = 233.3 kN.m

From the R- chart, R = 0.0154, then = 0.019

0.0154 123*40*250

10*3.332 R 2

5

2s mm649.0 1230 *400*

360

25 * 0.019 A

2s

2

y

cu

min s

mm844.0 649.0 *1.30 A 3.1

mm 15361230*400*360

25225.0d b

f

f0.225

of smaller A

2min s mm 738 1230*400*

100

0.15 d b

100

0.15 A

As = 8.44 cm2, use 4-18 = 10.18 cm2

Design of Strap Beam for Shear:


Qu = 16.27 ton

23

uu kg/cm 3.31

123*40

10*27.16

d * b

Q q

22cucu kg/cm9.80 N/mm 0.98

1.5

250.24

1.5

f0.24 q

2

ymin st mm 33.133200*400*

240

0.4 s*b*

f

0.4 A

Use four branches stirrups 58/m'

Ast = 4 * 50 = 200 mm2 > Ast min

Assume that the footing thickness t = 50 cm, and the depth d = 43 cm

For a 1.00 m strip at Sec. (1-1):


2

R.C.1

u11u t/m 34.08

1.80*2.0

67.122

A

R f

t.m 10.91

8

4.00.2*08.34

8

b -B*f M

2

2strap1.C.Ru1

1

Design of Footings: Design for MomentFooting F1, Exterior (neighbor) Footing

Ru1 = 122.67 ton

fu1 = 34.08 t/m2

2.00

0.800.80 0.40

1

1

R = 0.024, w = 0.0284

As = 8.48 cm2

As = 516/m'


Design of Footings: Design for MomentFooting F2, Interior Footing

2

R.C.2

u22u t/m 38.12

1.80*2.55

95.174

A

R f

t.m9.34

8

4.080.1*12.38

8

b -B*f M

2

2strap2.C.Ru2

2

fu2 = 38.12 t/m2

1.80

0.700.70 0.40

1

1

For a 1.00 m strip at Sec. (1-1):

Ru2 = 174.95 ton

R = 0.02, w = 0.0235

As = 7.16 cm2

As = 516/m'


Design of Footings: Design for Shear

Critical section for shear at d/2 from the strap beam faceبعد على القواعد للقصفى الحرج سطح d/2القطاع

الشداد

CC22CC11 Strap BeamStrap Beam

F2

d/2 = 0.215 m Critical Section for Shear

2.551.80

1.80

F1

2.00

0.58

5

0.48

5

0.700.80

m0.585 2

0.43 -

2

0.40 2.0 -

2

d -

2

b -B x strapR.C.1

2 For Footing F1

For Footing F2m0.485

2

0.43 -

2

0.40 -1.8

2

d -

2

b -B x strapR.C.2

2


The shear is safe, then check the shear for the second footing


Qu1 = fu1*x2*BR.C.1 = 34.08*0.585*1.80 = 35.89 ton

For Footing F1

23

R.C.1

u11u kg/cm4.64

43*180

10*89.35

d * B

Q q

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

1.5

25 0.16

1.5

f 0.16 q

qu1 (4.64 kg/cm2) < qcu (6.50 kg/cm2) O.K.


The shear is safe


Qu2 = fu2*x2*BR.C.2 = 38.12*0.485*2.55 = 47.14 ton

For Footing F2

23

R.C.2

u2uu kg/cm 4.87

43*225

10*14.47

d * B

Q q

qu1 (4.87 kg/cm2) < qcu (6.50 kg/cm2) O.K.

Reinforcement Details of the Strap Beam


Reinforcement Details of the Footings


محاضرات هندسة الاساسات د. طارق نجيب P3

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