تصميم هامة الخوازيق

1. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Design of pile cap
Pile cap can be designed using one of the following methods:
1-Conventional design method, ‫الطريق‬‫ة‬‫التقليدية‬
2- Finite element method. ‫المحدودة‬ ‫العناصر‬ ‫نظرية‬ ‫تطبيقات‬ ‫باستخدام‬
3-Circulage method, ‫التحزيم‬ ‫طريقة‬ ‫أو‬ ‫الدائرية‬ ‫الطريقة‬
4-Rigid beam method, ‫طريق‬‫ة‬‫الجاسئة‬ ‫الكمرة‬
a) Conventional design method
‫من‬ ‫بدال‬ ‫الخوازيق‬ ‫من‬ ‫مركزة‬ ‫أفعال‬ ‫لردود‬ ‫معرضة‬ ‫مفردة‬ ‫كقاعدة‬ ‫الهامة‬ ‫تصميم‬ ‫يتم‬ ‫الطريقة‬ ‫هذه‬ ‫فى‬
‫التربة‬ ‫جهد‬
Steps of Design:
1- Where:
Pw=Working load of the column,
Qall =Allowable bearing capacity of the pile.
2- Draw pile cap and get Dimension:
Assuming that:
-Thickness of PC = 10 cm
-Take S= Smin, where:
Smin = 3xD→ for friction piles, Smin = 2.5xD → for bearing piles,
Smax = 6xD
-Take side distance (e) = (1:1.5) D, where D= diameter of the pile
A pile cap can be drawing as shown in figure (3-3)
Figure (3-3) Pile cap dimension
Where:
Q
Pw1.15
pilesNo.
all


2. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
L=2(S+e)
B=S+2e
3- Determine the pile working load
a) For concentrically loaded pile cap: ‫حم‬‫ل‬‫الهامة‬ ‫على‬ ‫متمركز‬ ‫رأسى‬
b) For eccentrically loaded pile cap: ‫حم‬‫ل‬‫الهامة‬ ‫على‬ ‫متمركز‬ ‫غير‬ ‫رأسى‬
4-Determine the ultimate pile load
Pu (pile) =1.5 Pw (pile)
5-Design of flexure
The critical section of moment is taken at the face of column as shown in
figure (3-4).
Figure (3-4) The critical section for moment
‫خازوق‬ ‫كل‬ ‫عند‬ ‫الفعل‬ ‫رد‬ ‫حاصل‬ ‫مجموع‬ = )‫العمود‬ ‫لوجه‬ ‫المماسين‬ ‫عندالقطاعين‬ ‫(يؤخذ‬ ‫االنحناء‬ ‫عزم‬
‫الحرج‬ ‫القطاع‬ ‫عن‬ ‫الخازوق‬ ‫مركز‬ ‫بعد‬ *
Mu1 = Pu1xa+Pu4 x a
Mu2 = Pu1xb+Pu2 x b + Pu3xb
Where:
pileNo.of
mn)1.1Pw(colu
Pw(pile) 

 22v
y
yMx.
x
.M
N
V-
P
Xy

3. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Mu1= the ultimate bending moment at sec 1-1
Mu2= the ultimate bending moment at sec 2-2
Pu1= the ultimate load for pile number 1
Pu2= the ultimate load for pile number 2
Pu3= the ultimate load for pile number 3
Pu4= the ultimate load for pile number 4
6-Calculate the pile cap depth
Once calculation the bending moment the pile cap depth can be determined as
follow:
By taking C1=5 , get d1,d2 then choose the biggest value.
Check d>dmin,
dmin =1.5D+10 cm , or (dmin =2D) where D= pile diameter
Take t= d+ cover , cover>700 mm
.‫سم‬ 7 ‫عن‬ ‫يقل‬ ‫ال‬ ‫غطاء‬ ‫أخذ‬ ‫مع‬ ‫األكبر‬ ‫العمق‬ ‫يؤخذ‬ ‫ثم‬ ‫حدة‬ ‫على‬ ‫اتجاه‬ ‫لكل‬ ‫العمق‬ ‫حساب‬ ‫يتم‬ ‫حيث‬
7-Check of shear (one way shear)
The critical section of shear is located at d/2 from face of column, where d is
the depth of pile cap.
:‫األتى‬ ‫مراعاة‬ ‫مع‬ ‫الهامة‬ ‫عمق‬ ‫نصف‬ ‫مسافة‬ ‫على‬ ‫يقع‬ ‫للقص‬ ‫الحرج‬ ‫القطاع‬
1-‫للقص‬ ‫الحرج‬ ‫القطاع‬ ‫خارج‬ ‫بالكامل‬ ‫الخازوق‬ ‫يقع‬ ‫عندما‬ ‫للخازوق‬ ‫الفعل‬ ‫رد‬ ‫قيمة‬ ‫كامل‬ ‫احتساب‬ ‫يتم‬
2-‫للقص‬ ‫الحرج‬ ‫القطاع‬ ‫داخل‬ ‫بالكامل‬ ‫الخازوق‬ ‫يقع‬ ‫عندما‬ ‫الخازوق‬ ‫فعل‬ ‫رد‬ ‫تجاهل‬ ‫يتم‬
3-‫من‬ ‫فقط‬ ‫الجزء‬ ‫هذا‬ ‫احتساب‬ ‫يتم‬ ‫للقص‬ ‫الحرج‬ ‫القطاع‬ ‫خارج‬ ‫الخازوق‬ ‫من‬ ‫جزء‬ ‫وجود‬ ‫حالة‬ ‫فى‬
‫من‬ ‫كنسبة‬ ‫وذلك‬ ‫القص‬ ‫قوى‬ ‫حساب‬ ‫عند‬ ‫الخازوق‬( ‫بشكل‬ ‫موضح‬ ‫هو‬ ‫كما‬ ‫الخازوق‬ ‫مساحة‬5-3)

4. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Figure (3-5) Effect of pile location on shear forces
Figure (3-6) The critical section for one way shear
The shear forces for the pile cap shown in figure (3-6) can be determined as
follow:
Qsh1-1=qu1+qu4 1-1 ‫القطاع‬ ‫خارج‬ ‫الخوزازيق‬ ‫احمال‬ ‫مجموع‬
Qsh2-2=zero ‫القطاع‬ ‫خارج‬ ‫الخوزازيق‬ ‫احمال‬ ‫مجموع‬2-2‫صفر‬ =
Where:
fcu =the concrete compressive strength.
γc =1.5
B.d
1-Qsh1
1-qsh1 

5. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
If q sh1-1 ˂ qcu safe
If qsh 1-1 > qcu un safe, increase the depth
:‫كالتالى‬ ‫تقريبيا‬ ‫القص‬ ‫قوى‬ ‫حساب‬ ‫يتم‬ ‫الحرج‬ ‫القطاع‬ ‫داخل‬ ‫يقع‬ ‫الخازوق‬ ‫من‬ ‫جزء‬ ‫حالة‬ ‫فى‬
‫الخازوق‬ ‫ومركز‬ ‫العمود‬ ‫وجه‬ ‫بين‬ ‫المسافة‬ ‫من‬ ‫أكبر‬ ‫مسافة‬ ‫على‬ ‫يقع‬ ‫الحرج‬ ‫القطاع‬ ‫يكون‬ ‫ما‬ ‫حالة‬ ‫فى‬
‫الخازوق‬ ‫ومركز‬ ‫العمود‬ ‫وجه‬ ‫بين‬ ‫المسافة‬ ‫من‬ ‫أقل‬ ‫مسافة‬ ‫على‬ ‫يقع‬ ‫الحرج‬ ‫القطاع‬ ‫يكون‬ ‫ما‬ ‫حالة‬ ‫فى‬
‫للخازوق‬ ‫الكلى‬ ‫القطر‬ / ‫القطر‬ ‫من‬ ‫القص‬ ‫فى‬ ‫المؤثر‬ ‫الجزء‬ ‫طول‬ = ‫التخفيض‬ ‫معامل‬ ‫أن‬ ‫أى‬
Where:
Fr = reduction factor
D=pile diameter
X= ‫الحرج‬ ‫والقطاع‬ ‫الخازوق‬ ‫محور‬ ‫بين‬ ‫المسافة‬
8-Check of punching (two way shear)
The critical section of punching is located at d/2 from each side of column as
shown in figure (3-7).
D
D/2)(X
Fr


rsh FxQ uP
D
X)(D/2
Fr



6. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Figure (3-7) The critical section for punching
‫مساحة‬ ‫مكونا‬ ‫ناحية‬ ‫كل‬ ‫من‬ ‫العمود‬ ‫وجه‬ ‫من‬ ‫الهامة‬ ‫عمق‬ ‫نصف‬ ‫بعد‬ ‫على‬ ‫للثقب‬ ‫الحرج‬ ‫القطاع‬ ‫يقع‬
( ‫بالشكل‬ ‫موضح‬ ‫هو‬ ‫كما‬ ‫العمود‬ ‫حول‬ ‫الثاقب‬ ‫للقص‬7-3‫الحمل‬ ‫عن‬ ‫عبارة‬ ‫الثاقب‬ ‫القص‬ ‫ويكون‬ )
:‫كالتالى‬ ‫وذلك‬ ‫الحرج‬ ‫القطاع‬ ‫داخل‬ ‫الخوازيق‬ ‫أفعال‬ ‫ردود‬ ‫منه‬ ‫مطروحا‬ ‫للعمود‬ ‫األقصى‬
Qup=Pcu - ∑ Pu(pile)
Where:
qp=punching stress
Pcu =column ultimate load ‫للعمود‬ ‫األقصى‬ ‫الحمل‬
Pu= pile ultimate load ‫الحرج‬ ‫القطاع‬ ‫خارج‬ ‫الخازوق‬ ‫فعل‬ ‫رد‬
U=perimeter of critical section
U= 2[(a+d)+(b+d)]
The allowable punching stress (qcup) is giving as the least value of the
following:
Where:
U.d
q
up
p
Q


7. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
qcup=the punching shear strength provided by concrete (allowable),
a,b=the ratio of long side to short side of column,
α=4for interior column,3 for edge column ,2 for corner column,
fcu =the concrete compressive strength.
If qp ˂ qcup safe
If qp>qcup un safe, increase the depth
: ‫كالتالى‬ ‫الثقب‬ ‫قوى‬ ‫حساب‬ ‫فيتم‬ ‫للتثقيب‬ ‫الحرج‬ ‫القطاع‬ ‫داخل‬ ‫منها‬ ‫جزء‬ ‫يقع‬ ‫التى‬ ‫للخوازيق‬ ‫بالنسبة‬
Qup=Pcu - λx No .of piles in critical section x pile load ultimate,
As shown in figure (3-7)
9-Reinforcement of the Cap Pile:
The area of steel can be determined as follow:
As 1-1= → As1 /m = As 1-1/B
Check : As 1-1> As min= (0.6/fy)x B x d
As2-2= →=As2/m = As2-2/L
Check: As2-2 > As min= (0.6/fy)x L x d
9-Details of reinforcement
‫أو‬ ‫منفصلة‬ ‫(سواء‬ ‫القواعدالمسلحة‬ ‫لتسليح‬ ‫مماثال‬ ‫التصميم‬ ‫من‬ ‫الطريقة‬ ‫هذه‬ ‫فى‬ ‫الهامة‬ ‫تسليح‬ ‫يكون‬
( ‫بشكل‬ ‫موضح‬ ‫كماهو‬ ‫وذلك‬ )‫جار‬ ‫قاعدة‬ ‫أو‬ ‫مشتركة‬8-3.)
j.d.fy
1-1Mu
j.d.fy
2-2Mu
piletheofareagross
piletheofareahatched


8. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Figure (3-8) Details of reinforcement for pile cap
‫فى‬ ‫ويراعى‬: ‫األتى‬ ‫الهامة‬ ‫تسليح‬
1-‫عن‬ ‫تقل‬ ‫ال‬ ‫لمسافة‬ ‫الهامة‬ ‫فى‬ ‫الخازوق‬ ‫تسليح‬ ‫حديد‬ ‫يمتد‬ ‫أن‬ ‫يجب‬60.‫سم‬
2-‫تمثل‬ ‫علوية‬ ‫بشبكة‬ ‫الهامة‬ ‫تسلح‬ ‫أن‬ ‫يمكن‬1‫عن‬ ‫اليقل‬ ‫بما‬ ‫الهامة‬ ‫مقطع‬ ‫مساحة‬ ‫من‬ %5‫فى‬ ‫أسياخ‬
.‫االتجاهين‬
Problem (3-2)
Design and give complete reinforcement detailing for a pile cap, knowing that
column dimension = 0.6×0. 6 m, column working load = 2000 kN
Pile diameter = 0.4 m, Pile working load =600 KN, fcu = 30 N/mm² , fy = 360
N/mm².
Solution
= 1.15x2000/600=3.83, take 4 piles
Assume S= 3D (pile diameter) = 3x0.4= 1.2 m
e=1.25xD=1.25x0.4=0.5m→ L= B= 1.2+2x0.5=2.2m
Q
Pw1.15
pilesNo.
all


9. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Pu (pile)=1.5x550=825kN
Mu1-1=Mu2-2 =Mu
Mu= No. of piles xPu xa , a=(1.2-0.6)/2= 0.3m
Mu=2x825x0.3=495 kN.m
d=5√ (495x106
/30x2200) =433 mm ≈44cm
dmin =1.5D+10 cm =1.5x0.4+0.1=0.7m
Take d=70 cm , t=d+10 = 80 cm
Check of shear:
550kN1.1x2000/4
pileN.o.of
mn)1.1Pw(colu
Pw(pile) 

10. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
‫بعد‬ ‫على‬ ‫يقع‬ ‫للقص‬ ‫الحرج‬ ‫القطاع‬ ‫أن‬ ‫حيث‬35‫القص‬ ‫حساب‬ ‫يمكن‬ ‫ذلك‬ ‫فعلى‬ ‫العمود‬ ‫وش‬ ‫من‬ ‫سم‬
: ‫كالتالى‬
Qsh=2x309.4=618.8kN
qsh = 618.8x103
/(2200x700)=0.401 N/mm²
qcu =0.16√(30/1.5)=0.715N/mm²
qsh˂qcu ok
Check of punching:
D
X)-Pu(D/2
(pile)Qsh 
B.d
Qsh
qsh 
kN4.309375.0825
0.4
0.05)-/24.825(0
 x

11. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Four hatched area is located inside the critical punching area:
Qup=Pcu - ∑ Pu(pile) - λx No .of piles in critical section x pile load ultimate
Qup=1.5x2000 - 4x 0.43x825=2081KN
U=[(0.6+0.7)+(0.6+0.7)]x2=5.2m=5200mm
qp=2081x10³/(5200x700)=0.57N/mm²
The allowable punching stress (qcup) is giving as the least value of the
following:
1-qcup= 0.316√(30/1.5)=1.41N/mm²˂1.6 N/mm²
2-qcup= 0.136(0.5+0.6/0.6)√(30/1.5)=2.12 N/mm²
3- qcup=0.8(0.2+(4x0.7/2.6)√(30/1.5)=4.57N/mm²
qcup=1.41N/mm², qp˂ qcup safe
Uxd
q
up
p
Q


12. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
Reinforcement of the pile cap
‫فقط‬ ‫واحد‬ ‫التجاه‬ ‫التسليح‬ ‫حساب‬ ‫يتم‬ ‫االتجاهين‬ ‫من‬ ‫متماثلة‬ ‫الهامة‬ ‫ألن‬ ‫نظرا‬
As= C1=5→ j=0.826
As=495x106
/(0.826x700x360)=2378.06 mm²/2.2m
As min= (0.6/fy) x B x d= (0.6/360)x2200x700=2566.6 mm²
Take As= As min=2566.6mm²/2.2m
As/m=2566.6/2.2=1166.6mm²/m=11.66cm²/m, take 6ϕ16/m
: ‫ملحوظة‬
‫محصلة‬ ‫تأثير‬ ‫نقطة‬ ‫تنطبق‬ ‫بحيث‬ ‫الخوازيق‬ ‫توزيع‬ ‫من‬ ‫البد‬ ‫مشتركة‬ ‫كقاعدة‬ ‫الخوازيق‬ ‫هامة‬ ‫تصميم‬ ‫عند‬
.‫المعتادة‬ ‫بالطريقة‬ ‫الحل‬ ‫يستكمل‬ ‫ثم‬ ‫الخوازيق‬ ‫ثقل‬ ‫مركز‬ ‫مع‬ ‫األعمدة‬ ‫أحمال‬
b) Finite element method analysis of pile cap
fdj
M
y
u

13. ‫الرءوف‬ ‫عبد‬ ‫السيد‬ ‫مؤمن‬ /‫د.م‬-‫األزهر‬ ‫هندسة‬
In some cases, the manual methods to analysis a pile cap is not available or
not accurate, the internal forces of pile cap can be determined by using a
computer program like sap. In this method the pile cap is modeled as a shell
element and pile modeled as a spring element as shown in figure (3-9). The
pile spring constant (Kpile) can be calculated as following:
Permissible settlement can be determined from pile loading test.
Figure (3-9) Modeling of the pile cap
Permissible settlement may be taken =0.01D
Where D=Pile diameter
‫العالقة‬ ‫من‬ ‫الصالبة‬ ‫معامل‬ ‫حساب‬ ‫يمكن‬ ‫كما‬:
K=EA/L
Where:
L = Pile length, (m);
E = Elastic modulus of concrete, (t/m²) = 1.40x106
t/m²=140 t/cm2
.
A = Pile cross-section area.
settlementepermissibl
loadworkingPile
Kpile 