Transcript Download

Chapter 1 :
Introduction
Introduction:
 Al-Nour building is 8 stories reinforced
concrete building ,located in Nablus city and
used as residential building.
 The first story is used as garages with plan
area of 700 m^2 and the above 7 stories used
as residential apartment (two apartments per
floor) with plan area of 490 m^2 due to the
setback.
 The soil bearing capacity = 400 KN/m2
Introduction :
 The Following slides shows :
 1. columns centers plan.
 2. 3D model of the building.
Structural System :
 The structural system used is on way ribbed slab
with load path in x-direction.
Materials:
 - Concrete :
- f’c= 320 kg/cm²( 32 MPa) For columns.
- f’c= 240 kg/cm²( 24 MPa) for others.
- The concrete unit weight = 25 (KN/m3).
 - Reinforcing Steel: The yield strength of steel is equal to 4200
Kg/cm2 (420 MPa).
 -Others :
Unit weight
Material
Reinforced concrete
Plain concrete
Sand
Aggregate
Y-tong
Blocks
Polystyrene
Masonry stone
Light weight block
Tile
(KN/m3)
25
23
18
17
5
12
0.3
27
6
26
Design loads :
 - Dead loads in addition to slab own weight :
1. Superimposed dead load = 4.5 KN/m2
2. Partition load = 1 KN/m2 .
3. Masonry wall weight = 21.22 KN/m.



- Live load = 2 KN/m2 .
-Water tanks load = 1.14 KN/m2
- Seismic loads : shown later.
Design codes and load combinations:
- The following are the design codes used :
1. ACI – code 2008 .
2. IBC 2009 .
3. ASCE for design loads.

The following are the load combinations used :
1. Wu = 1.4DL.
2. Wu = 1.2DL + 1.6LL .
3. Wu = 1.2DL + LL ± E.
4. Wu = 0.9DL ± E
•
Chapter 2 :
Preliminary Design
Preliminary design
 We performed a preliminary design for all structural
elements conceptually.
 The story height is 3.12 m.
 The following are the preliminary dimensions :
 Slab :
- depth = 25 cm (based on deflection criteria) .
- web width = 12 cm.
- slab own weight = 4.55KN/m².
- Ultimate load = 14.06KN/m².
Preliminary Design
 Beams :
Since the structural system is one way ribbed slab (load path in
x-direction) we have :
1. Main beams in y-direction : 30x60 cm.
2. Secondary beams in x-direction : 40x25 cm.
 Columns :
Take a sample columns ( B3) :
Area carried by column = 28 m2
Ultimate slab load = 14.06KN/m²
Pu = 3769.6 KN.
Ag = 2326.9 cm2.
→ Use columns of 40x60cm2.
Preliminary design and checks
 Footing :
we performed an preliminary design for footing of the
previous column as single footing.
with dimensions of 2.9x2.7x0.7 m.
Chapter 3 :
Static Design
Static design :
 Final dimensions :
 1. frame sections :
Member
Depth(cm)
Width(cm)
Col.
80
40
Main interior beams
70
40
Main exterior beams
75
30
Secondary beams
25
40
Tie beams
50
30
Static Design:
 The new web width (bw) = 15 cm.
 Area sections dimensions :
Area section name
Thickness (cm)
Actual Slab
25
Equivalent Slab thickness
19.45 (in SAP model)
Shear wall
30 (initially)
 The story height = 3.5 m
Static design:
 Verification Of SAP model:
- We perform the verification for SAP models( one and
eight stories and it was OK) the following is verification
for eight stories :
- 1. Compatibility satisfied :
Static Design
2.Equilibrium Satisfied :
Load type
Hand results(KN)
SAP results (KN)
Error %
Dead load
76262.44
76273.132
0.01
Live load
8407.24
8407.24
0
3.Stress -Strain relationship satisfied:
Taking beam C in second story (taking 8 m span) :
Load`
M-ve
(left)(KN.m)
M+ve
(KN.m)
M-ve(right)
(KN.m)
Total moment
(KN.m)
SAP Result
325.13
208.63
371.41
556.9
1D Result
0
341.40
433.61
558.21
Error
2.3%
Static Design :
 Slab design :
1. Check slab deflection :
So, ∆dead = 2.92 mm.
∆Live = 0.78mm.
Δ long term = 7.16mm.
The allowable deflection = 4000 /240 = 16.67 mm.
 So the slab deflection = 7.16mm < allowable long term def. OK.
2. Design for shear :
The rib shear strength = 23.2KN.
The max shear = 36.75 KN/m.
shear per rib = 0.55*36.75 = 20.2 KN.
 So 23.2 ≥ 20.2 OK
 So the slab is Ok for shear.
Static Design :
 3. Design for bending moment :
 The moments are read from SAP using section cut :
Point location/term
Moment(KN.m)
As(mm2)
Bars
A1
6.3
113
2 Φ 12
A1- B1
8.6
113
2 Φ 10
B1
10.48
125
2 Φ 12
B1- C1
7.25
113
2 Φ 10
C1
10.29
122
2 Φ 12
C1 – D1
8.1
113
2 Φ 10
D1
9
113
2 Φ 12
D1 – E1
8.1
113
2 Φ 10
E1
13.55
163.3
2 Φ 12
Static Design :
 Design of beams in y-direction :
 Taking a sample beam (beam B in the first floor) :
- The beam section dimensions are :
- Total depth (h) = 700 mm.
- The effective depth (d) = 650 mm.
-Beam width (bw) = 400 mm.
- min reinforcement ratio = 0.0033.
- As min = ρbd = 0.0033*400*650 = 858 mm2
- φVc = 159.1 KN.
- (Av/s)min = 0.333.
Static Design :
 - Design information :
point
As(mm)
As min
(Av/s)
(Av/s) min
PI(m)
Length(m)
B1
700
858
0.333
0.333
0.6
2
B1-2
854
858
0.333
0.333
--
5.9
B2(L)
1676
858
0.333
0.333
1.3
B2(R)
1676
858
0.55
0.333
B2-3
1044
858
0.333
B3(L)
1581
858
B3(R)
1581
B3-4
B4
bars
5Φ16
stirrups
1Φ8 @30 cm
5Φ16
1Φ8 @30 cm
4.8
6φ20
1Φ8 @30 cm
1.5
4.8
6φ20
1Φ8 @15 cm
0.333
_
7.9
5Φ16
1Φ8 @30 cm
0.48
0.333
1.5
4.8
6φ20
1Φ8 @20 cm
858
0.333
0.333
1.3
4.8
6φ20
1Φ8 @30 cm
772
858
0.333
0.333
_
5.9
854
858
0.333
0.333
1
2.3
5Φ16
5Φ16
1Φ8 @30 cm
1Φ8 @30 cm
Static Design :
 Design of secondary beams:
 Total depth(H) = 25cm.(hidden beam)
 d= 21cm (cover =4cm)
 Width = 40cm.
 The following are the values of min reinforcement:
 (As) min = 0.0033*b*d=0.0033*400*210= 277.2 mm (3Φ12).
 Vc = 0.75*0.167**400*210/1000= 68.58KN.
Static Design :
 Design information :
Point
As(mm)
As min
(mm)
(Av/s)
(Av/s)min
PI(m)
bars
stirrups
A4
469
277.2
0.333
0.333
0.88
5Φ12
1@10cm
A4-B4
277
277.2
0.333
0.333
0.88
3Φ12
1@10cm
B4
429
277.2
0.333
0.333
0.88
4Φ12
1@10cm
B4-C4
259
277.2
0.333
0.333
0.88
3Φ12
1@10cm
C4
424
277.2
0.333
0.333
0.88
4Φ12
1@10cm
C4-D4
260
277.2
0.333
0.333
0.88
3Φ12
1@10cm
D4
425
277.2
0.333
0.333
0.88
4Φ12
1@10cm
D4-E4
265
277.2
0.333
0.333
0.88
3Φ12
1@10cm
E4
432
277.2
0.333
0.333
0.88
4Φ12
1@10cm
E4-F4
147
277.2
0.333
0.333
0.88
3Φ12
1@10cm
F4
432
277.2
0.333
0.333
0.88
4Φ12
1@10cm
Design of columns:
 Column grouping, Area of steel& stirrups:
column Distributio
group
n of steel
Floor no.
Columns
As(mm2)
All floors
except
No.8
All
Columns
3200
C1
D2 & G2
3766
C2
A2, B2, C2,
H2, I2 & J2
4774
C3
Floor
No.8
Stirrups spacing
(mm)
16φ16
3φ10 @250 mm
20φ16
3φ10 @300 mm
16φ20
3φ10 @300 mm
Manual design
Check slenderness ratio for corner column
C.A-1
From the graph
 K = 3.5
𝐾𝐿𝑢
3.5x2.8
=
= 40.83 >22 ⟹Column is long.
rx
0.3x0.8
𝐾𝐿𝑢
ry
=
3.5x2.8
0.3x0.4
= 81.66 >22 ⟹Column is long.
Pu=3034.75KN
MY = 11.02 KN.m(maximum value)
MX = 153.1 KN.m( maximum value)
Mc = ns*M2 = 1.67(153.1) = 255 KN.m
 From the interaction diagram:
≤1% use minimum steel ratio use =1%.
 As =0.01xAg =0.01x40x80 =3200mm2
 Same as SAP value.
Tie beam design
Minimum area of steel = 0.0033*b*d =436 mm2.
Use 4Ф12mm bottom steel.
Use 4Ф12mm top steel.
Shear design :
Vu at distance (d = 44cm) = 16.35KN,
ФVc = 80.83KN.
Use 1Ф8 mm@200mm.
Footing design
Single footing:
 Is one of the most economical types of footing and is
used when columns are spaced at relatively long distances
.
 Bearing capacity of the soil=400 KN/m2.
Footing grouping
Group
Name
Columns
Service
load(KN)
F1
A1, B1,C1,D1,E1,F1,G1,H1,I1,J1
305.6
Ultimate
load
(KN)
375.7
F2
A2, J2,A4 B4,C4,D4,G4,H4,I4,J4,
2329.95
2878.51
F3
B2,C2,
D2,G2,H2,I2,A3,B3,C3,D3,G3,H3,
I3,J3
E4 & F4
3667.66
4599.75
2218.20
2741.80
Combined
Footing grouping according to column’s ultimate load.
footing details
Group
Steel
Area of
distribution/
Dimensions Depth
footing
m’
(m)
(m)
(m2)
(Both
directions)
Area of
shrinkage
steel
(mm2)
in each direction
Distribution
F1
0.96
1.2x0.8
0.3
4 Ø 14mm
no
No
F2
6.21
2.7x2.3
0.55
5 Ø 18mm
𝟓𝟖𝟓
1 Ø 14@300mm
F3
9.57
3.3x2.9
0.75
6 Ø 18mm
𝟔𝟕𝟓
1 Ø 14@200mm
Combined
13.11
5.7x2.3
0.55
1Ø25
/200mm
778.2
1 Ø 14@200mm
Design of Stairs
 hmin =
𝑙
20
= 0.187m ⇒ Use h = 0.20m.
 Own Weight=0.2x25=5KN/m2.
 Live load = 5KN/m2.
 Superimposed loads =2.14 KN/m2
 Superimposed loads of extra of stairs=1.76 KN/m2
 Wu=1.2(5+3.9) +1.6(5) =18.68KN/m2.
 Ø𝑉𝑐 = 110.23KN > 𝑉𝑢 = 20.82𝐾𝑁
Verification of SAP model:
Compatibility:
“Compatibility Satisfied”
Cont.
Equilibrium:
Load type
Dead load
Live load
Hand
results(KN)
SAP
results(KN)
293.04
181.24
Error %
292.82
181.12
0.08
0.07
“Equilibrium Satisfied”
Stress-Strain Relationships:
Load
3D SAP
Result
1D Result
Error
Mve(left)
(KN.m)
M+ve
(KN.m)
M-ve(right)
(KN.m)
22.1
11.9
19.46
21.72
10.86
21.72
Total
moment(
𝐖𝐥𝟐
𝟖
(KN.m)
32.68
32.58
0.3%
“Stress–strain relationship satisfied”
M = 22.1 KN. m/m
𝐴𝑠 𝑚𝑖𝑛 = 𝜌𝑠ℎ𝑟𝑖𝑛𝑘𝑎𝑔𝑒 𝑏ℎ = 360𝑚𝑚2 > 𝐴𝑠
𝜌 = 0.00184
𝐴𝑠 = 331𝑚𝑚2 /m.
⇒ 𝑢𝑠𝑒 1∅12/250𝑚𝑚 𝑖𝑛 𝑎𝑙𝑙 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑠.
)
Chapter 4 :
Dynamic Design
Dynamic Design:
 Methods for dynamic analysis:
1. Equivalent static method.
2. Time history method.
3. Response spectrum analysis.

Input parameters in dynamic analysis :
- Importance factor (I) = 1 .
- Peak ground acceleration (PGA) = 0.2g .
- Area mass = 0.458 ton/m2
- Soil class = Class B.
- Spectral accelerations : Ss = 0.5 .
S1 = 0.2 .
- response modification factor R = 3 in x-direction.
R = 4.5 in y-direction.
Dynamic Design :
 Modal information :
- For eight stories before enlarging beams in x-direction :
Mode No.
Direction
Period (sec.)
MMPR %
1
X
2.55
77
2
RZ(Torsion)
1.707
67
4
y
0.972
65
 - Enlarge the beams 2 &4 to 30x70 (width*depth)
Dynamic Design:
- For eight stories after enlarging beams in x-direction :
Mode No.
Direction
Period (sec.)
MMPR %
1
X
1.58
80
2
RZ(Torsion)
1.5
64
3
Y
0.795
65.4
4
X
0.497
11
- Comparison with manual results :
Mode direction
SAP result(sec)
Manual result(sec) Error %
x- direction
1.58
1.45
8%
y- direction
0.795
0.73
8%
Dynamic Design :
 Response spectrum analysis :
We will perform the dynamic design using response spectrum
method:
Define two response spectrum load cases one in x-direction and the
another in y-direction :
- For response-x: * Scale factor = 3.27.
*Scale factor = 0.654.
- For response-y: * Scale factor = 2.18.
*Scale factor = 0.981.
 Perform design using envelope combination and check
whether static or dynamic combination controls .
Dynamic Design :
 Slab design :
The comparison is performed.
Static design controls
Dynamic Design :
 Design of beams in y-direction :
- Reinforcement from envelope combination:
- Reinforcement from static combination:
Static design Controls
Dynamic Design :
 Design of beams in X-direction :
- Reinforcement from envelope combination is considered
since the dimensions are increased:
Dynamic design Controls
Dynamic Design :
 Design of columns :
Three representative columns are selected :
1. Interior column B3.
2. Edge column B2.
3. Corner column A4 .
The comparison is performed and static design controls
for all columns.
The following table shows the comparison for column
B3 :
Dynamic Design :
The following table shows the comparison for column B3
(M3, V2 ):
floor/ter
m
Envelope combination
Static combination
moment
shear
axial
As(mm2)
moment
shear
axial
As(mm2)
1
68.04
31.34
4553.25
3200
5.54
1.678
4553.3
3200
2
45.78
22.15
3982.31
3200
6.97
3.7
3982.2
3200
3
49.81
22.82
3409.11
3200
4.78
2.73
3409.1
3200
4
47.33
20.07
2841.94
3200
4.62
2.55
2841.9
3200
5
44.51
18.49
2277.7
3200
3.78
2.12
1716.5
3200
6
42.58
16.78
1716.51
3200
3.71
2
1699.8
3200
7
39.19
13.98
1156.2
3200
2.88
1.36
1156.2
3200
8
26.75
8
600.378
3200
6.82
3.18
603.8
3200
Static design OK for columns.
Chapter 5 :
Structural Modeling Of One Way
Ribbed Slab
Structural Modeling Of One Way Ribbed Slabs
 The ribbed slabs can be represented by one of the
following ways :
1. Equivalent stiffness method : find the equivalent
thickness of a solid slab that can achieve the same rib
stiffness.
2. Represent it as separate ribs (T-section).
1.
Represent the ribs by rectangular ribs and flange.

The main objective is to prove that three models give the
nearly the same results.
Structural Modeling Of One Way Ribbed Slabs
 Model 1 : Equivalent stiffness method :
- Equivalent slab thickness (t) = 19.45 cm.
- I T-sec = I rec →( 0.55*h3eq /12) = 3.371*10-4
h3eq = 19.45 cm.
γeq = 23.87 KN/m^3 . . . . .. .to achieve the same weight.
- Stiffness modifiers :
M11 = 0.35 .
M22 = 0.0244
M1-2 = 0.0244.
Structural Modeling Of One Way Ribbed Slabs
 Model 2 : Representation as separate ribs :
- Stiffness modifiers :
I 3-3 = 0.35 .
I 2-2 = 0.35 .
Torsional constant(J)= 0.35 .
- The loads are inserted as line
Loads on ribs.
- Substitute the weight of blocks.
Structural Modeling Of One Way Ribbed Slabs
 Model 3 : the slab is represented as rectangular ribs and flange.
1. The rectangle section should satisfy
The actual T-section.
2. Stiffness modifiers :
I 3-3 = 0.6 .
I2-2 = 0.6 .
J = 0.52 .
3. Weight modifier = 0.68 .
Flange Modifiers :
- M11 = 0.0001 .(almost zero).
- M 22 = 0.25 .
- M 1-2 = 0.0001(almost zero).
→ we have to substitute the weight of blocks.
55 cm
8 cm
25 cm
15 cm
Structural Modeling Of One Way Ribbed Slabs
 All the previous models are verified according to manual
solution.
 Static analysis is performed for the three models and we
read the moment and shear at point C3 in span C3-4 for
beam C in the second floor :
Moment
(KN.m)
Shear
(KN)
(mm2)
( (mm2/mm)
SAP Model 1
Results
371.41
268.63
1615
0.545
SAP Model 2
Results
377.87
286.43
1645.1
0.634
SAP Model 3
Results
373.56
274.13
1625
0.573
load
Structural Modeling Of One Way Ribbed Slabs
 also dynamic analysis and design performed for the
three models and the following are the modal
information :
Model 1 :
Mode
Direction
Period
MMPR
Mode 1
x-direction
1.581
80
Mode 2
Torsion(Rz)
1.503
64
Mode 3
y- direction
0.796
65.4
Model 2 :
Direction
Period
MMPR
Mode 1
x-direction
1.622
77.0
Mode 2
Torsion(Rz)
1.555
79.5
Mode 3
y- direction
0.924
70.9
Direction
Period
MMPR
Mode 1
x-direction
1.542
80.3
Mode 2
Mode 3
Torsion(Rz)
y- direction
1.456
0.769
65.5
65.7
Mode
Model 3 :
Mode
Structural Modeling Of One Way Ribbed Slabs
 Also the same span taken in the beam C and the following are the values of
moment and shear from envelope combination :
Model 1 :
Point
Envelope combination
moment
shear
As(mm2)
C2
C2-C3
C3
316.0
208.6
371.1
1359.4
1053
1613
243.7
16.3
268.5
Av/s
(mm2/mm)
0.422
0
0.545
Model 2 :
Point
Envelope combination
moment
shear
As(mm2)
C2
C2-C3
C3
329.7
211.0
377.9
1422
899.1
1645
Point
Envelope combination
moment
shear
As(mm2)
C2
C2-C3
C3
317.6
209.1
373.6
1367
891.3
1625
Model3:
259.6
17.7
286.4
243.0
16.2
274.1
Av/s
(mm2/mm)
0.5
0
0.634
Av/s
(mm2/mm)
0.418
0
0.573
Structural Modeling Of One Way Ribbed Slabs
 Final conclusion :
from the previous data shown in tables we note that all the
three models give near results.
So, we can represent the slab model by any of the previous
models but perform the changes in loads assignment and
stiffness modifiers.