Introduction:
It is proposed to check the adequacy of the sections provided for the VDGS Gantries.
A modulus of elasticity of only 2.05e+008 kN/sq.m. is used in the analysis of stresses,
in structural steel and C40 grade is assumed for pedestal and footing.
Moments and shears arrived at using STAAD.Pro analysis package.
Density of in fill concrete considered for analysis is 25 kN/cu.m.
A safe bearing capacity of 150 kPA is used in the analysis of foundation.
Sections are assigned as provided in the drawings.
Methodology:
The 8.75 m span VDGS gantry is modeled in the STAAD.Pro Package along with the pedestal
upto the bottom of the isolated footing. The support conditions are assigned as fixed at footing.
The self weight is assigned vide the staad input file itself.
The wind load is separately calculated as per BS 6399-Part 2 for curved surfaces and enhanced
with appropriate coefficients.
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 1
with appropriate coefficients.
The action of assumed parameter and loadings are modeled in analysis package and the
results are tabulated as shown in the following sections. The summary of results are
used in the excel spread sheets to check the adequacy of proposed Pipe section sizes.
Conclusion:
The results hence available prove the sections provided and proposed are found to be safe.
INPUT FILE
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 02-Feb-14
JOB NAME Air side Works
JOB CLIENT ADIA
JOB REV 0
JOB PART VDGS GANTRIES
JOB REF CSM-ADAC-L&T-002-0
END JOB INFORMATION
UNIT METER KN
JOINT COORDINATES
1 0 0 0;
2 0 7 0;
3 0.875 8 0;
4 8.125 8 0;
5 9.00 7 0;
6 9.00 0 0;
DATEDOCUMENT NOABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTIONPROJECT:
2DIVMER
SHEETCHECKEDPREPAREDVDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1TITLE:
05-03-2014Rev.-1
6 9.00 0 0;
8 0 -1.9 0;
9 9.00 -1.9 0;
MEMBER INCIDENCES
1 1 2; 2 2 3; 3 3 4; 4 4 5; 5 6 5; 6 8 1; 7 9 6;
MEMBER CURVE
2 RADIUS 1 GAMMA 0 PRESSURE -9999
4 RADIUS 1 GAMMA 0 PRESSURE -9999
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+008
POISSON 0.3
DENSITY 78.5
ALPHA 1.2e-005
DAMP 0.03
ISOTROPIC CONCRETE
E 2.8e+007
POISSON 0.2
DENSITY 25
ALPHA 1e-005
DAMP 0.05
END DEFINE MATERIAL
MEMBER PROPERTY BRITISH
1 3 5 TABLE ST PIPE OD 0.273 ID 0.2476
2 4 TABLE ST PIPE OD 0.273 ID 0.2476
MEMBER PROPERTY BRITISH
6 7 PRIS YD 1
CONSTANTS
MATERIAL STEEL MEMB 1 TO 5MATERIAL STEEL MEMB 1 TO 5
MATERIAL CONCRETE MEMB 6 7
SUPPORTS
*1 6 FIXED BUT MZ
8 9 FIXED
********************************************
DEFINE UBC LOAD
*
ZONE 0.15 I 1 RWX 5.5 RWZ 5.5 STYP 4 CT 0.073
*
SELFWEIGHT 1
*
********************************
LOAD 1 UBC SEISMIC EQ-X+VE
*
UBC LOAD X 1
*
**********************************
LOAD 2 UBC SEISMIC EQ-Z+VE
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 3
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
LOAD 2 UBC SEISMIC EQ-Z+VE
*
UBC LOAD Z 1
*
**********************************
LOAD 3 DEAD LOAD
*
SELFWEIGHT Y -1 LIST 1 1 TO 7
*
****************
LOAD 4 WIND LOAD Z+VE
*
* 1.221 x 1.2 x 0.273 ~ 0.4 kN/m
*
MEMBER LOAD
1 TO 5 UNI GZ 0.4
3 CON GZ 1.5 2.35
3 CON GZ 3 3.5
*
****************
LOAD 5 WIND LOAD X+VE
*
MEMBER LOAD
1 2 4 5 UNI GX 0.4
*
****************ULS
*
LOAD COMB 6 1.4 DL + 1.4 WL
3 1.4 4 1.4 3 1.4 4 1.4
*
****************ULS
*
LOAD COMB 7 1.0 DL + 1.4 WL
3 1.0 4 1.4
*
****************ULS
*
LOAD COMB 8 1.4 DL + 1.4 WL
3 1.4 5 1.4
*
****************ULS
*
LOAD COMB 9 1.0 DL + 1.4 WL
3 1.0 5 1.4
*
****************ULS
*
LOAD COMB 10 1.4 DL + 1.4 SEIS-X
DATE
Rev.-1 05-03-2014
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 4
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO
LOAD COMB 10 1.4 DL + 1.4 SEIS-X
3 1.4 1 1.4
*
****************ULS
*
LOAD COMB 11 1.0 DL + 1.4 SEIS-X
3 1.0 1 1.4
*
****************ULS
*
LOAD COMB 12 1.4 DL + 1.4 SEIS-Z
3 1.4 2 1.4
*
****************ULS
*
LOAD COMB 13 1.0 DL + 1.4 SEIS-Z
3 1.0 2 1.4
*
****************SLS
LOAD COMB 14 1.0 DL + 1.0 WL
3 1.0 4 1.0
LOAD COMB 15 1.0 DL + 1.0 WL
3 1.0 5 1.0
LOAD COMB 16 1.0 DL + 1.0 SEIS-X
3 1.0 1 1.0
LOAD COMB 17 1.0 DL + 1.0 SEIS-Z
3 1.0 2 1.0
************************************
PERFORM ANALYSIS
UNIT MMS NEWTON
LOAD LIST 6 TO 17
PARAMETER 1
CODE BS5950
PY 275 MEMB 1 TO 5
TRACK 0 MEMB 1 TO 5
CHECK CODE MEMB 1 TO 5
UNIT METER KN
FINISH
Passing Stress Ratios:
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 5
Note that all elements are have passing ratios less than 1 hence safe.
Summary of Beam End Froces for the bolted base connection:
Summary of Deflection at the top of the portal:
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 6
Check for Deflection at the top of Gantry:
As per 26 56 25 Poles and Masts 20130401 v10.0, specifications maximum deflection
allowable = 0.6 degrees.
Height of the gantry including pedestal = 9.50 m
Limting horizontal deflection 9.50 x tan 0.6 = mm
Actual deflection from the above summary table = 51 mm
hence safe in deflection
99.49
Check for 860 mm diameter base plate stresses:
Referring to the summary of beam end forces P = 17 kN
Moment M = 63 kNm
Diameter of the base plate D = m
Thickness of the base plate tp = 0.04 m
Section Modulus of the base plate in plan Zplan = m^3
Section Modulus of the base plate in section Zsection= m^3
Area of the plate in plan A = m^2
Bearing stress of the plate on concrete pedestal P/A+M/Zplan = MPa
Hence Safe
1.038
0.86
0.062
2E-04
0.581
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 7
Hence Safe
Allowable Ultimate Bearing stress as per BS 8110 Part-1 Clause 5.2.3.4 = 16 MPa
(0.4 x fck)
Moment in the base plate 1.031*((0.86-0.273)/2)^2/2 = kNm
Bending Stress in the baseplate f = M/Zsec = MPa
Hence Safe
Allowable bending stress in the baseplate fy = 275 MPa
Bolts PCD (assume only grade 4.6 bolts for design) pcd = 0.71 m
Tension in anchor bolt p = M/pcd = kN
Diameter of anchor bolt d = 50 mm
Area of the anchor bolt a = sq.mm.
Allowable tensile strength as per Table 34 of BS 5950 Part 1 pt = 240 MPa
Hence Safe
Actual tensile stress in the bolt provided p/a = MPa
There is no tension in summary of reactions (beam end forces).
Actual shear stress in the bolts pbbact = kN
Allowable tensile strength as per Table 31 of BS 5950 Part 1 d tp pbb = kN
Interaction check for a single loaded bolt =
< 1.4 hence safe.
0.237
1963
56
88.73
1.016
736
44.71
195
< 1.4 hence safe.
Summary of forces for field splice:
Diameter of bolts provided M = 16 mm
No. of bolts provided n = 20 no.s
Maximum shear at the section Fy = 7 kN
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 8
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
Fz = 5.5 kN
There is no axial tension in the field section Bolts pcd = 336 mm
Maximum Moment at the section Mz = 12.1 kNm
My = 2.1 kNm
Push and Pull due to the moments Mz/pcd = kN
My/pcd = kN
Assuming area of cross section of bolts as 80 % of gross area = sq.mm.
Resultant Shear stress due to actual shear = 55 MPa
Sum of axial tensile stress due to push and pull = MPa
Assuming grade 4.6 bolts for design purposes
Allowable shear stress in bolt as per Table 30 of BS 5950-1 = MPa
Allowable shear stress in bolt as per Table 34 of BS 5950-1 = Mpa
Interaction check for a single loaded bolt =
< 1.4 hence safe.
240
160
1.386
Abolt
36.0
6.3
160.8
249.6
Summary of forces for foooting design
Provide 3500 x 2500 x 500 mm thick square footing in plan, with T12 @ 200 mm c/c bothways Top & Bottom
Summary of forces for pedstal design
TITLE: VDGS Gantry Design - 8.75 m span, CSM-ADAC-L&T-002-1PREPARED CHECKED SHEET
MER DIV 9
PROJECT:ABU DHABI INTERNATIONAL AIRPORT-MIDFIELD
TERMINAL COMPLEX- AIRSIDE CONSTRUCTION
DOCUMENT NO DATE
Rev.-1 05-03-2014
Provide 10 # T20 diameter bars as Main vertical bars for 1000 mm diameter pedestal
Provide T 10 circular Hoops at 150 mm c/c as stirrups.
SheetJob Number
Job Title
Client
Calcs by Checked by Date
Software Consultants (Pty) Ltd
Internet: http://www.prokon.com
E-Mail : [email protected]
1E18.1
Fire Pump Room - Al Jurf-3
FEWA
MKN Milind 16-Feb-14
C12VDGS 8.75 m Span Gantry Pedestal - ADAC
Circular column design by PROKON. (CirCol Ver W1.7.01 - 5 Oct 2000)
Design code : BS8110 - 1997
Input tables
LoadCase Description
Ultimate Limit State Design Loads
P (kN) Mx top (kNm) My top (kNm) Mx bot (kNm) My bot (kNm)
1 1.4DL+1.4WL 68.879 -62.85 -20.64 82.2 35.25
2 1.0DL+1.4SEIS- 6.982 0 -20.64 0 35.25
General design parameters and loads:
0 250
500
750
1000
1000
750
500
250
0
X X
Y
Y
General design parameters:Given: dia = 1000 mm d' = 95 mm Lo = 2.000 m fcu = 35 MPa fy = 420 MPa
Therefore: Ac = pi⋅ d˛/4 = 785398.16 mm˛ diax' = dia - d' = 905 mm diay' = dia - d' = 905 mm
Assumptions: (1) The general conditions of clause 3.8.1 are applicable. (2) The section is symmetrically reinforced. (3) The specified design axial loads include the self-weight of the column. (4) The design axial loads are taken constant over the height of the column.
Design approach:The column is designed using the following procedure: (1) The column design charts are constructed. (2) The design axis and design ultimate moment is determined . (3) The steel required for the design axial force and moment is read from the relevant design chart. (4) The area steel perpendicular to the design axis is read from the relevant design chart. (5) The procedure is repeated for each load case. (6) The critical load case is identified as the case yielding the largest steel area about the design axis.
Through inspection: Load case 1 (1.4DL+1.4WL) is critical.
Check column slenderness:End fixity and bracing for bending about the X-X axis: At the top end: Condition 1 (fully fixed). At the bottom end: Condition 1 (fully fixed). The column is unbraced. Code suggests ßx = 1.20 Table 3.22
Designer specified ßx = 2.20
SheetJob Number
Job Title
Client
Calcs by Checked by Date
Software Consultants (Pty) Ltd
Internet: http://www.prokon.com
E-Mail : [email protected]
2E18.1
Fire Pump Room - Al Jurf-3
FEWA
MKN Milind 16-Feb-14
End fixity and bracing for bending about the Y-Y axis: At the top end: Condition 1 (fully fixed). At the bottom end: Condition 1 (fully fixed). The column is unbraced. Code suggests ßy = 1.20 Table 3.22
Designer specified ßy = 2.20
Effective column height: lex = ßx⋅ Lo = 4.400 m ley = ßy⋅ Lo = 4.400 m
Check if the column is slender: 3.8.1.3
lex/dia = 4.4 < 10 ley/dia = 4.4 < 10∴ The column is short.
Initial moments:The column is bent in single curvature about the X-X axis: M1 = Smaller initial end moment = 62.9 kNm M2 = Larger initial end moment = 82.2 kNm
The initial moment near mid-height of the column : 3.8.3.7
∴ Mi = -0.4M1 + 0.6M2 ≤ 0.4M2 = 82.2 kNm
The column is bent in single curvature about the Y-Y axis: M1 = Smaller initial end moment = 20.6 kNm M2 = Larger initial end moment = 35.3 kNm
The initial moment near mid-height of the column : 3.8.3.7
∴ Mi = -0.4M1 + 0.6M2 ≤ 0.4M2 = 82.2 kNm
Design ultimate load and moment:Design axial load: Pu = 68.9 kN
Moment distribution along the height of the column for bending about the X-X: At the top, Mx = 62.9 kNm Near mid-height, Mx = 74.5 kNm At the bottom, Mx = 82.2 kNm
Mxtop=-62.9 kNm
Mxbot=82.2 kNm
Moments about X-X axis( kNm)
Initial Additional Design
Mx=82.2 kNm
Mxmin=0.1 kNm
+ =
SheetJob Number
Job Title
Client
Calcs by Checked by Date
Software Consultants (Pty) Ltd
Internet: http://www.prokon.com
E-Mail : [email protected]
3E18.1
Fire Pump Room - Al Jurf-3
FEWA
MKN Milind 16-Feb-14
Moment distribution along the height of the column for bending about the Y-Y: At the top, Mx = 20.6 kNm Near mid-height, Mx = 29.4 kNm At the bottom, Mx = 35.3 kNm
Mytop=-20.6 kNm
Mybot=35.3 kNm
Moments about Y-Y axis( kNm)
Initial Additional Design
My=35.3 kNm
Mymin=0.1 kNm
+ =
Check for miminum eccentricity: 3.8.2.4
For bi-axial bending, it is only necessary to ensure that the eccentricity excceeds the minimum about one axis at a time.
For the worst effect, apply the minimum eccentricity about the minor axis: Use emin = 20mm∴ Mmin = 1.4 kNm about the Y-Y axis.
Design of column section for ULS:Through inspection: The critical section lies at the bottom end of the column.
The column is bi-axially bent and may be designed to withstand an increasedmoment about a single axis: 3.8.4.5
Mx/dia = 69.4 > My/dia = 22.8
The effective uniaxial design moment about the X-X axis: ß = 1 - 7/6⋅ N/(Ac⋅ fcu) = 0.997 Table 3.24
∴ M'x = Mx + ß⋅ dia/dia⋅ My = 117.3 kNm
SheetJob Number
Job Title
Client
Calcs by Checked by Date
Software Consultants (Pty) Ltd
Internet: http://www.prokon.com
E-Mail : [email protected]
4E18.1
Fire Pump Room - Al Jurf-3
FEWA
MKN Milind 16-Feb-14
For bending about the design axis:
Column design chart
Mom
ent m
ax =
5690kN
m @
6340kN
-18E3-16E3-14E3-12E3-10E3-8000-6000-4000-2000
200040006000800010E312E314E316E318E320E322E324E326E328E330E332E3
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
Axia
l Load (
kN
)
Bending Moment (kNm)
6%5%4%3%2%1%0%
Reinforcement required about the X-X axis: From the design chart, Asc = 3142 mm˛ = 0.40%
For bending perpendicular to the design axis:
Column design chart
Mom
ent m
ax =
5690kN
m @
6340kN
-18E3-16E3-14E3-12E3-10E3-8000-6000-4000-2000
200040006000800010E312E314E316E318E320E322E324E326E328E330E332E3
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000
Axia
l Load (
kN
)
Bending Moment (kNm)
6%5%4%3%2%1%0%
Reinforcement required about the Y-Y axis: From the design chart, Asc = 3142 mm˛ = 0.40%
SheetJob Number
Job Title
Client
Calcs by Checked by Date
Software Consultants (Pty) Ltd
Internet: http://www.prokon.com
E-Mail : [email protected]
5E18.1
Fire Pump Room - Al Jurf-3
FEWA
MKN Milind 16-Feb-14
Summary of design calculations:
Design results for all load cases:
Load case Axis N (kN) M1 (kNm) M2 (kNm) Mi (kNm) Madd (kNm) Design M (kNm) M' (kNm) Asc (mm˛)
1
2
X-XY-Y 68.9
-62.9 -20.6
82.2 35.3
82.2 35.3
0.0 0.0
X-XBottom
62.9 20.6
0.0 35.3
3142 (0.4%) 3142 (0.4%)
X-XY-Y 7.0
0.0 -20.6
0.0 35.3
0.0 35.3
0.0 0.0
Y-YBottom
0.0 20.6
0.0 0.0
3142 (0.4%) 3142 (0.4%)
Load case 1 (1.4DL+1.4WL) is critical.
Project Date Ref. No.
Building Designed By Checked By
Title Rev. No. Approved By
Location Drg. No.
CHECK FOR BEAM:-
Beam Data :- Equivalent Square of side
Characterstic Strength of Concrete ƒcu = 40 N/mm2
Breadth of Beam b = 885 mm
Yeild Strength of Steel ƒy = 460 N/mm2
Depth of Beam D = 885 mm
Effective Depth to top steel d1 = D-Ct-Øs-Øt/2 Dia of stirrups Øs = 10 mm
= 885-75-10-20/2 Concrete Top cover Ct = 75 mm
= 790.0 mm Concrete Bot cover Cb = 75 mm
Effective Depth to bott. steel d2 = D-Cb-Øs-Øb/2 Spacer dia (Top) St = 25 mm
= 885-75-10-20/2 Spacer dia (Bot) S St = 25 mm
= 790.0 mm
For Maximum -ve moments :-
Maximum Bending Moment M = 83.00 kN-m
M / ƒcu b d² = k = 83×10^6 /40×885×790²
= 0.0038
z = 0.5 + (0.25-k/0.9) Compression Steel
= 0.950 d1 d' = 75+10+20/2
Area of Steel Ast, reqd = M / 0.95 ƒy z = 95.0
= 83×10^6 / (0.95×460×0.95×790) As', reqd =
= 1018.19 mm2
((0.0038-.156) x40x885x790 ²)
No of Layers = 1 (0.95x460x ( 790-95))
1st layer 4 Nos T 20 Bars = -11071.48
2nd layer 0 Nos T 20 Bars Provide 4 Nos T 20 Bars
3rd layer 0 Nos T 12 Bars Area provided = 1257 mm², is O.K.
Area of reinforcement provided = 1256.64 mm2, is
L & T
ABU DHABI INTERNATIONAL AIRPORT 05-Mar ***
MIDFIELD TERMINAL COMPLEX- AIRSIDE MER DIV
Pedestal Design as Beam (PDSTL) -
Beam along Grid - CSM-ADAC-L&T-002-1
(k-.156) ƒcu bd² / .95 ƒy (d-d')
Error, Type zero in the layers which are no needed
OK
For Maximum +ve moments :-
Maximum Bending Moment M = 36.00 kN-m
M / ƒcu b d² = k = 36×10^6 /40×885×790²
= 0.0016
z = 0.5 + (0.25-k/0.9) Compression Steel= 0.950 d2 d' = 75+10+20/2
Area of Steel Ast, reqd = M / 0.95 ƒy z = 95.0
= 36×10^6 / (0.95×460×0.95×790) As', reqd =
= 1018.19 mm2 ((0.0016-.156) x40x885x790 ²)
No of Layers = 1 (0.95x460x ( 790-95))
1st layer 4 Nos T 20 Bars = -11231.52
2nd layer 0 Nos T 16 Bars Provide 4 Nos T 20 Bars
3rd layer 0 Nos T 25 Bars Area provided = 1257 mm², is Ok
Area of reinforcement provided = 1256.64 mm2, is
For Shear Force :-Min Torsional Resistance = 92.42 kNm
Maximum Shear Force F = 15.00 kN (SRSS) Torsion, T = 4.00 kNm (MAX)
Shear Stress v = 15×10³/ (885×790) Torsional Shear Stress,
= 0.02 N/mm² vt = 2×T/(hmin²×(hmax-hmin/3))
Area of main reinfn. As = 1256.64 mm² = 2×4×10^6/(885²×(885-885/3))
100A s / bd 1 = 100×1256.64 / 885×790 = 0.017 N/mm²
= 0.18 Min Torsional Shear Stress = 0.40 N/mm²
Design Shear Strength of conc. vc = 0.79×[100 As / b d1]¹/³×1.17/1.25 Torsional Reinforcement not Required
= 0.79×[0.18]¹/³ × 1.17/1.25 Area of Torsion Shear Reinforcement
= 0.418 N/mm² Asv = T Sv / (.8 x1 y1 (.95×fy))
Shear resistance of conc. Vc = vc b d = 4×10^6×150
= 0.418×885×790/1000 (0.8×725×725×0.95×460= 292.24 kN = 3.27 mm²
Assign -2 L-T 10 Bars @ 150 mm c/c. (Asv Provided =157.08 mm² ) Area of Longitudinal Torsion Reinforcement
As = Asv (x1+y1) / SvArea of Shear Reinforcement Reqd. Asv = Sv×b(v-Vc) / 0.95 fy = 3.27× (725+725)/150
as ==> (v-vc) < 0.4, (v-vc) = 0.4 = 150×885(0.4) / 0.95 ×460 = 31.61 mm²
= 121.51 mm²
Total Area of Shear Reinforcement Reqd. Asv = 121.51 mm², < Asv Provided, OK
(k-.156) fcu bd² / .95 fy (d-d')
Error, Type zero in the layers which are no needed
OK
Beam in Seismic zone
Ast Provided on Each Face = 452.16 mm² is OK
Design of Isolated Footing Node 8
LC 14 - SLS 1
Bearing Pressure = 36.5 KN/m2 < 150 Safe
overburden Pressure = 0 KN/m2
Net Design Pressure = 37 KN/m2
Load factor = 1.4 (DL+LL)
Factored Design Pressure = 51.15 KN/m2
Footing L = 3.50 m
Footing B = 2.50 m
Col a = 1 m
Col b = 1 m Span
fcu = 40 N/mm2 of
Max Moment along longer dirn = 40 KNM Gantry
Max Moment along Shorter dirn = 14 KNM
b = 1000 mm
Depth of Footing = 500 mm
Clear Cover = 75 mm
Reinft dia along longer dirn = 12 mm B = 2.5 m
Reinft dia along Shorter dirn = 12 mm
d shorter = 407 mm Pedestal is treated as equivalent
d longer = 419 mm square of size 1.00 m
Reinft along Longer direction (B1)
K = Mu / fcu b d^2 = 0.0057
Zd = 0.994
Z = 0.95
Ast = 255.32 mm2
L =
3.5
m
a
b
1 of 2
Ast = 255.32 mm2
Spacing of bar = 442 mm
Spacing provided = 200 mm
Ast Provided = 565.49 mm2
Provide T12 at 200mm C/C along Longer direction
Reinft along Shorter direction (B2)
K = Mu / fcu b d^2 = 0.0022
Zd = 0.998
Z = 0.95
Ast = 93.28 mm2
Spacing of bar = 1227 mm
Spacing provided = 200 mm
Ast Provided = 565.49 mm2
Provide T12 at 200mm C/C along Shorter direction
Check for Oneway Shear:
At face of the column:
Max Shear force at the face of the column Vx = 63.95 KN
Max Shear force at the face of the column Vy = 38.36 KN
Shear Stress vx = 0.153 N/mm2
Shear Stress vy = 0.094 N/mm2
At distance "d" from face of the column:
42.51 KN
17.55 KN
Max Shear force at a distance "d" from the face
of the olumn Vx =Max Shear force at a distance "d" from the face
of the olumn Vy =
1 of 2
Shear Stress vx = 0.102 N/mm2
Shear Stress vy = 0.044 N/mm2
Pt Provided = 0.135 %
Allowable Shear Stress, τc =0.379 N/mm
2
Enhanced allowable Shear Stress = 2 x τc =0.758 N/mm
2
At distance "2d" from face of the column:
21.07 KN
-3.27 KN
Shear Stress vx = 0.051 N/mm2
Shear Stress vy = -0.009 N/mm2
Check for Punching Shear:
Average eff. Depth to resist punching Shear = 413 mm LF Axial punching Force
Factored Max Axial Load from super structure = 73.864 KN 1.4 52.76
< Allowable Enhanced Shear Stress
and Hence Safe
< Allowable Enhanced Shear Stress
and Hence Safe
Max Shear force at a distance "2d" from the face
of the olumn Vx =Max Shear force at a distance "2d" from the face
of the olumn Vy =< Allowable Shear Stress and Hence
Safe
< Allowable Shear Stress and Hence
Safe
2 of 2
Factored Max Axial Load from super structure = 73.864 KN 1.4 52.76
4000 mm
Punching Shear Stress at face of the Column = 0.045 < 0.8 x Sqrt(fcu), Hence Safe
Punching Shear Force at "1.5d" distance = -182.57 KN
8956 mm
-0.05 N/mm2
V.Load B.Press. Area Reqd. Length Breadth
158.575 187.5 0.85 1.25 0.68
Moments 36.5 3.500 2.500
58.71 -0.3
25.2
Punching Shear Stress at "1.5d" distance from
Column face =
< Allowable Shear Stress and Hence
Safe
Perimeter to resist Punching Shear at face of the
Column =
Perimeter to resist Punching Shear at "1.5d"
distance from Column face =
b
a
1.5 d
2 of 2
Project :
Title :
Doc. No. : REV-1
Description: Sheet No.
Location:
Concrete
Grade of concrete ( 28 Days Cube strength) 40 N/mm2
Modulus of elesticity of concrete (Long term) 13500 N/mm2
Steel
Yield strength of reinforcement 460 N/mm2
Design yield strength of reinforcement 460 N/mm2
Modulus of elesticity of steel 2.00E+05 N/mm2
Force and reinforcement data
Maximum working moment in the section 78.34 KN-m (Actual factored moment in footing is 40 kNm)
Thickness of the section (h) 500 mm
Dia of main bar 12 mm
Spacing of main bar (s) 200 mm
Clear cover to reinforcment (cmin) 100 mm
Factor for ε2 calculation as per B.4 BS 8007:1987 1.00
Limiting crackwidth 0.30 mm
Calculation:
Neutral axis:
Modulur ratio m = E /E = 14.81
Location:- Crack Width Calculation for Isolated Footing:
ABU DHABI INTERNATIONAL AIRPORT
L&TMIDFIELD TERMINAL COMPLEX- AIRSIDE
Crack width check for Isolated Footing
h
acr
cmin
s
Modulur ratio m = Es/Ec = 14.81
fc x = kd where k is given by
As = 565.49 mm2/m 12@200ST
d = h -cl cover -φ/2 = 394 mm
p = As/bd = 0.001435 ( b=1000mm)
fs/m Hence k = 0.186 x = 73.30 mm
Surface crack width
εm = ε1 - ε2 ε1 = average strain at the level where crack is considered
ε2 =
374.9 N/mm2
ε1 = 2.49E-03
a' = h bt = 1000 mm
ε2 = 0.001673288
εm = ε1 - ε2 = 8.21E-04
= 139.73 mm
Hence calculated crack width = 0.290000 mm <0.3mm, Hence Safe
x
d
bt * (h-x) * (h-x)
3 * Es * As* (d-x)
pmmppmk −+= )2(22
−
−+
=
xh
ca
aw
cr
mcr
min21
3 ε
s
s
E
f
xd
xh×
−
−=
=
−
=
dk
A
Mf
st
s
*)3
1(
ϕϕ 5.0)5.0(25.02
min
2−++= csacr
Project :
Title :
Doc. No. : REV-1
Description: Sheet No.
Location:
Concrete
Grade of concrete ( 28 Days Cube strength) 40 N/mm2
Modulus of elesticity of concrete (Long term) 13500 N/mm2
Steel
Yield strength of reinforcement 460 N/mm2
Design yield strength of reinforcement 460 N/mm2
Modulus of elesticity of steel 2.00E+05 N/mm2
Force and reinforcement data
Maximum working moment in the section : My 59.29 KN-m
Thickness of the section (h) 883 mm
Dia of main bar 20 mm
Spacing of main bar (s) 200 mm
Clear cover to reinforcment (cmin) 75 mm
Factor for ε2 calculation as per B.4 BS 8007:1987 1.00
Limiting crackwidth 0.30 mm
Calculation:
Neutral axis:
Modulur ratio m = E /E = 14.81
Crack width check for Pedestal
Location:- Crack Width Calculation for Pedestal:
ABU DHABI INTERNATIONAL AIRPORT
L&TMIDFIELD TERMINAL COMPLEX- AIRSIDE
h
acr
cmin
s
Modulur ratio m = Es/Ec = 14.81
fc x = kd where k is given by
As = 1570.80 mm2/m 20@200ST
d = h -cl cover -φ/2 = 798 mm
p = As/bd = 0.001968 ( b=1000mm)
fs/m Hence k = 0.214 x = 170.85 mm
Surface crack width
εm = ε1 - ε2 ε1 = average strain at the level where crack is considered
ε2 =
50.9 N/mm2
ε1 = 2.89E-04
a' = h bt = 1000 mm
ε2 = 0.000858028
εm = ε1 - ε2 = -5.69E-04
= 121.24 mm
Hence calculated crack width = -0.183129 mm <0.3mm, Hence Safe
x
d
bt * (h-x) * (h-x)
3 * Es * As* (d-x)
pmmppmk −+= )2(22
−
−+
=
xh
ca
aw
cr
mcr
min21
3 ε
s
s
E
f
xd
xh×
−
−=
=
−
=
dk
A
Mf
st
s
*)3
1(
ϕϕ 5.0)5.0(25.02
min
2−++= csacr
VDGS gantries Page: 1
CSM-ADAC-L&T-002-0 Made by: GLN
8.75 M SPAN Date: 02.02.14
ADAI Ref No:
Office: 1328
WIND LOADS: in accordance with BS6399 : Part 2: 1995
Location: Wind loads to BS6399: Part2 : 1995
Dynamic Augmentation Factor.
Chosen site is the country or terrain not defined as sea or town
Building Height H=10.00 m
Effective Height He=10 m
Building type factor Kb=4
Chosen building type is a bolted steel or R.C. unclad frame
Dimensionless constant ho=0.1 m
Logarithm of height factor lh=LOG(H/ho)/LOG(10)
=LOG(10/0.1)/LOG(10)
=2
Dynamic Augmentation Factor Cr=(Kb*(H/ho)^0.75)/800/lh
=(4*(10/0.1)^0.75)/800/2
=0.07906
Since Dynamic Augmentation Factor Cr is less than or equal to 0.25
then this structure is not dynamic.
Standard wind loads.
Basic wind speed Vb=24.5 m/sec
Structure is not located at the crest of a hill or escarpment and the
topography is not significant.
Site altitude above mean sea level deltaS=25 m
Altitude factor Sa=1+0.001*deltaS
=1+0.001*25
=1.025
Direction factor Sd=1
The building is permanent or exposed to the wind for a continuous
period of more than 6 months.
Seasonal factor Ss=1.0
The basic wind speed has an annual risk of being exceeded of Q=0.02
Probability factor Sp=1
Since site is in the country with the closest distance to the sea
being 2.5 Km, then
From Table 4 Terrain & building factor Sb=1.777
Site wind speed @ height He Vs=Vb*Sa*Sd*Ss*Sp
=24.5*1.025*1*1*1
=25.11 m/sec
Effective wind speed Ve=Vs*Sb=25.11*1.777=44.62 m/sec
VDGS gantries Page: 2
CSM-ADAC-L&T-002-0 Made by: GLN
8.75 M SPAN Date: 02.02.14
ADAI Ref No:
Office: 1328
Dynamic pressure at height He qs=0.613*Ve^2/1000
=0.613*44.62^2/1000
=1.221 kN/m2
Effective height 10 m
DESIGN Altitude factor Sa 1.025
SUMMARY Direction factor Sd 1
Seasonal factor Ss 1
Probability factor Sp 1
Dynamic wind pressure qs 1.221 kN/m2
No702