Wind (3)

Document NumberREV

XPage

2/2ProjectPackageOriginatorDisciplineDoc. Type Unit n Serial nRAPIDP0022PUNJPMGXXXXXXXXXXX

DOCUMENT TITLE(REFINERY TANK FARM)

APPENDIX-X.1 : BLAST RESISTANCE DESIGN FOR WALL:

Option #1: Add Reinforce steel and fill wall cavities solid with concrete.

Total Thickness of C.M.U wall =8in.Actual thickness of C.M.U wall, tw =7.63in.(NCMA, Table 3b)Area, A =7.63in.2(NCMA, Table 3b)Moment of Inertia, Ig =36.9in.4#4, rebar area=0.2in.2rebar spacing =8in.

Calculate Bending Resistance

for dynamic flexure,(Appendix 5.A)fy =60ksi72.51885fdy = (SIF)(DIF)fy=(1.1)(1.17)(60 ksi)=77.22ksi53.2412595KNfdy =532.41Mpa

f'dm = (SIF)(DIF)f'm =(1)(1.19)(1.5 ksi)=1.785ksif'dm = 12.31Mpa

As =(rebar area)*(unit width)/(rebar spacing)=(0.2)(1)/(8)As =0.025in.2d =tw/2=3.81in. =As/bd(ACI 318, Equation 10-3) =0.0066> 200/fdy, O.K.a =As(fdy)/0.85(f'dm)(b)(as per UBC, Section 2108.2.1.2)ksimpa=0.025*(77.22/(0.85*1.785*1)16.8947590868=1.272inpsimpaMp = Mn =As(fdy)[d-a/2](as per UBC, Section 2108.2.1.2)10.0068947591=(0.025*77.22)*(3.81-(1.27/2))psiksi=6.13k-in11000

Rb =8 Mp / Lwhere,L =120inRb =(8 * 6.13) / 120=0.41kips(Table 6.1)unit resistance,Rb =(409 lb)/(120 in span)(1in width)=3.41psi=23.49kpaCalculate Shear Resistancefor dynamic flexure,(Appendix 5.A)f'dm = (SIF)(DIF)f'm =(1)(1)(1.5 ksi)91.5=1.5ksi7.625f'dm = 10.34MpaArea, A =4.187.625Vn =2 A (f'm)(UBC, Equation 8-35)/ACI 318 Eq. 11-3=323.78lbthe critical section for shear is tw from the support,Rs =Vn L / (0.5L-tw)741.84lb

unit resistance,Rs =(742 lb)/(120 in span)(1in width)=6.18psi=42.62kpaCompute SDOF Equivalent SystembecauseRb< Rs, bending controlsallowable response, a =0.75(medium range)(Table 5.B.1)masonry modulus of elasticity, (based on flexure)Em =750 f'dm=1338.75ksi=9230.36Mpamasonry modulus of elasticity, Es =29000000psi =199948.013516486Mpa

modular ratio,n =Es/Em =21.6619981326in.2cracked moment of inertia,n As =0.5415499533in.2C =-n As + [n As (nAs + 2bd)] b =-0.54 + [0.54 (0.54 + 2(1)(3.81)]1 =1.56in.Icr =b C3/3 + n As (d-C)2 =[ (1)(1.56^3 ]/3 + [(0.54)(3.81 - 1.56)^2] =4.01in.4average moment of inertia,Ia =(Ig+Icr)/2 =20.46in.4effective stiffness,K =384 EI(Table 6.1)5L3 =384 (1338.75 ksi)(20.46)5 (120^3) =1.22k/in =10.2psi/inyield deflection,ye =Ru / K =3.41 / 10.2 =0.33in.beam mass =(wall weight) / (gravity) =(0.144 kcf)(0.64 ft thick)(0.083 ft unit width)(10 ft span) / (386 in/sec^2) =0.0002k-sec2/in. =1651psi-ms2/in.Because of the espected response, use an average of values for KLM elastic KLM =0.5/0.64= 0.77(Table 6.1)plastic KLM =0.33/0.5= 0.66(Table 6.1)average KLM =(0.77+0.66)/2 =0.72Equivalent mass,Me =(KLM)(beam mass) =(0.72)(1651) =1180psi-ms2/in.period of vibration,tn =2 (Me / K)(Equation 6.8) =2 (1180.47 / 10.2) =68msChart Solutiontd/tn =90/68 =1.32Ru/Po =3.41/3.4 =1.00using the chart: d =1.8(Figure 6.9)maximum deflection,ym =(d)(ye) =1.8*0.33 =0.60support rotation,d =arctan(ym / 0.5L)(Figure 5.9) =arctan[0.6/ (0.5*120)] =0.0100190737o d =