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DROP ARM BARRIER AT ANP UP HQ MAYDAN SHAHR (W917PM PDF

20 Pages·2011·0.23 MB·English
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DROP 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 SAFE v8.0.1 File: COMBINED FOOTING KN-m Units PAGE 1 October 7,2009 3:05 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 SAFE v8.0.1 File: COMBINED FOOTING KN-m Units PAGE 2 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 SAFE v8.0.1 File: COMBINED FOOTING KN-m Units PAGE 3 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

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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.
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