Table Of ContentDROP ARM BARRIER AT ANP UP HQ MAYDAN SHAHR (W917PM-09-C-0072)
DESIGN COMPUTATION
DESIGN CODE REFERENCES:
International Building Code (IBC 2003)
Minimum Design Loads for Building and Other Structures (ASCE 7-02)
Uniform Building Code 1997
ACI
ASTM
DESIGN CRITERIA & MATERIALS SPECIFICATION:
Concrete Compressive Strength (f’c) at 28 days = 28 MPa
ReBars Minimum Yield Strength (fy) = 414 MPa
Cold-drawn Steel Unit Mass (γ ) = 7,833.413 kg/m
s
Allowable Soil Bearing Capacity (q) = 72 kPa
Vehicle Mass (m) = 6,800 kg
Vehicle Velocity (v) = 48 kph = 13.333 m/s
Acceleration due to Gravity (g) = 9.81 m/s2
Modulus of Elasticity of Concrete (E ) = 24,870.06 MPa
c
Concrete column width = 500mm
Concrete column depth = 250mm
Concrete cover for column = 40mm
Concrete cover for footing = 75mm
Soil density = 15.7 KN/m3
DESIGN ANALYSIS:
The design analysis was guided by the fundamental laws of physics and concrete design.
DESIGN LOADS:
Using the materials conforming the drop arm barrier, the mass/weight is approximately
equal to:
W = 641.828 kg or 6.296 kN.
G
Using work and energy principle, the kinetic energy generated by the 6,800 kg vehicle
moving at a constant velocity of 48 kph is equal to:
1
KE = mv2
2
1
KE = (6800)(13.333)2
2
KE =604.444 kNm
Assuming the vehicle slams at the drop arm barrier and able to deflect the arm by d =
g
1m. And consequently, the vehicle will also be damaged (shortened) by d = 1m.
c
d = d +d
g c
d =1+1
d = 2 m
Using work principle:
W = F(d)
604.444−6.296= F(2)
F = 295.926 kN (force generated by the vehicle upon impact)
Therefore, a total of 295.926 KN is generated by the vehicle unto the arm upon impact at
a height of 1.2 m. This force will be transferred to the base of the drop arm barrier
column where shear will occur due to car impact.
For the computation of additional shear bars due to impact of car, from ACI 318M-99 Chapter
11 Section 11.1, use equations
ØVn ≥ Vu,
Vn = Vc + Vs, Vn = nominal shear strength
For members subject to shear and flexure only, ACI 318M-99 Section 11.3.1.1
concrete shear capacity
For shear-friction calculation, ACI 318M-99 Section 11.7.4, use Equation 11-25
reinforcement shear capacity
Where , Section 11.7.4.3 ACI 318M-99
where 2 is the minimum number of bars perpendicular to shear plane
110.24+166.479)
, therefore provide shear bars at the base of column
Computation for number of shear bars:
remaining shear to be compensated with additional shear bars
= 154.42 KN
Using diameter 16mm bars, solve for number of shear bars, x
x = 2.18, therefore use 4- 16 shear bars for symmetrical orientation of bars (placing of
bars shall be two (2) bars each face (see detail of drop arm footing).
Design calculation for drop arm barrier footing
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A R E A O B J E C T D A T A
AREA JNT-1 JNT-2 JNT-3 JNT-4 SECTION SUPPORT X-STRIP Y-STRIP
AREA
5 16 17 18 19 cf SOIL1 NO NO 1.500
CSX1 1 2 3 4 YES NO 0.094
CSX2 5 6 7 8 YES NO 0.250
CSX3 9 10 11 12 YES NO 0.281
CSX4 13 20 21 22 YES NO 0.126
CSY1 23 24 25 26 NO YES 0.188
CSY2 27 28 29 30 NO YES 0.375
CSY3 31 32 33 34 NO YES 0.188
MSX1 4 3 6 5 YES NO 0.189
MSX2 8 7 10 9 YES NO 0.311
MSX3 12 11 20 13 YES NO 0.251
MSY1 24 27 30 25 NO YES 0.375
MSY2 28 31 34 29 NO YES 0.375
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October 7,2009 3:05
P O I N T O B J E C T D A T A
POINT GLOBAL-X GLOBAL-Y SUPPORTSPRING RESTRAINT RES DIM X
RES DIM Y
1 0.000 0.000
2 1.000 0.000
3 1.000 0.094
4 0.000 0.094
5 0.000 0.283
6 1.000 0.283
7 1.000 0.532
8 0.000 0.532
9 0.000 0.843
10 1.000 0.843
11 1.000 1.124
12 0.000 1.124
13 0.000 1.375
14 0.500 0.998
15 0.500 0.377
16 0.000 0.000
17 1.000 0.000
18 1.000 1.500
19 0.000 1.500
20 1.000 1.375
21 1.000 1.500
22 0.000 1.500
23 0.000 0.000
24 0.125 0.000
25 0.125 1.500
26 0.000 1.500
27 0.375 0.000
28 0.625 0.000
29 0.625 1.500
30 0.375 1.500
31 0.875 0.000
32 1.000 0.000
33 1.000 1.500
34 0.875 1.500
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October 7,2009 3:05
SAFE v8.0.1 File: COMBINED FOOTING KN-m Units PAGE 1
October 7,2009 3:06
S O I L P R E S S U R E
AREA GRID I GRID J LOAD PRESSURE
5 1 1 DL 19.130
5 1 1 LL 0.000
5 1 1 SERV1 19.130
5 1 1 ULT 26.783
5 2 1 DL 19.131
5 2 1 LL 0.000
5 2 1 SERV1 19.131
5 2 1 ULT 26.784
5 3 1 DL 19.133
5 3 1 LL 0.000
5 3 1 SERV1 19.133
5 3 1 ULT 26.786
5 4 1 DL 19.133
5 4 1 LL 0.000
5 4 1 SERV1 19.133
5 4 1 ULT 26.786
5 5 1 DL 19.133
5 5 1 LL 0.000
5 5 1 SERV1 19.133
5 5 1 ULT 26.786
5 6 1 DL 19.131
5 6 1 LL 0.000
5 6 1 SERV1 19.131
5 6 1 ULT 26.784
5 7 1 DL 19.130
5 7 1 LL 0.000
5 7 1 SERV1 19.130
5 7 1 ULT 26.783
5 1 2 DL 18.870
5 1 2 LL 0.000
5 1 2 SERV1 18.870
5 1 2 ULT 26.418
5 2 2 DL 18.871
5 2 2 LL 0.000
5 2 2 SERV1 18.871
5 2 2 ULT 26.419
5 3 2 DL 18.872
5 3 2 LL 0.000
5 3 2 SERV1 18.872
5 3 2 ULT 26.421
5 4 2 DL 18.873
5 4 2 LL 0.000
5 4 2 SERV1 18.873
5 4 2 ULT 26.422
5 5 2 DL 18.872
5 5 2 LL 0.000
5 5 2 SERV1 18.872
5 5 2 ULT 26.421
5 6 2 DL 18.871
5 6 2 LL 0.000
5 6 2 SERV1 18.871
5 6 2 ULT 26.419
5 7 2 DL 18.870
5 7 2 LL 0.000
5 7 2 SERV1 18.870
5 7 2 ULT 26.418
5 1 3 DL 18.348
5 1 3 LL 0.000
5 1 3 SERV1 18.348
5 1 3 ULT 25.688
5 2 3 DL 18.349
5 2 3 LL 0.000
5 2 3 SERV1 18.349
5 2 3 ULT 25.689
5 3 3 DL 18.351
5 3 3 LL 0.000
5 3 3 SERV1 18.351
5 3 3 ULT 25.692
5 4 3 DL 18.352
5 4 3 LL 0.000
5 4 3 SERV1 18.352
5 4 3 ULT 25.692
5 5 3 DL 18.351
5 5 3 LL 0.000
5 5 3 SERV1 18.351
5 5 3 ULT 25.692
5 6 3 DL 18.349
5 6 3 LL 0.000
5 6 3 SERV1 18.349
5 6 3 ULT 25.689
5 7 3 DL 18.348
5 7 3 LL 0.000
5 7 3 SERV1 18.348
5 7 3 ULT 25.688
5 1 4 DL 18.088
5 1 4 LL 0.000
5 1 4 SERV1 18.088
5 1 4 ULT 25.323
5 2 4 DL 18.089
5 2 4 LL 0.000
5 2 4 SERV1 18.089
5 2 4 ULT 25.324
5 3 4 DL 18.091
5 3 4 LL 0.000
5 3 4 SERV1 18.091
5 3 4 ULT 25.327
5 4 4 DL 18.091
5 4 4 LL 0.000
5 4 4 SERV1 18.091
5 4 4 ULT 25.328
5 5 4 DL 18.091
5 5 4 LL 0.000
5 5 4 SERV1 18.091
5 5 4 ULT 25.327
5 6 4 DL 18.089
5 6 4 LL 0.000
5 6 4 SERV1 18.089
Description:Oct 7, 2009 For members subject to shear and flexure only, ACI 318M-99 For shear-friction
calculation, ACI 318M-99 Section 11.7.4, use Equation 11-25.