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. �.,STOM DESIGN & ENGINEERING, IIVC <br /> STRUCTURAL ANALYSIS& DESIGN <br /> Check if A: �A��min <br /> A,.m;� =3(('�)'��/60000 x(bw d)=0.17 in�/ft<0.22 in�/ft <br /> A, =0.22 in2/ft <br /> As.mi� =Z00(bw)�d�/60000=0.22 in'/ft>0.22 inZ/ft <br /> A, =0.22 in�/k <br /> Retnforcement <br /> Try No 5@ 16.00 in.,Ar . =0307 sq.in. <br /> Provided,A,.o„� =A6�,(12 in/ft)/spacing=0.307(12)/16.000=0.230 in2/ft <br /> Compute a and check if f, =60000 and i(tension controls <br /> a=r,,.,,,.,e0000/(.�sp�o„,)=0.230�60000)/[0.85(2500)(12.00)]=0.541 in. <br /> a�,:=0.541/5.540=0.098 <br /> ab/d=0.582,Table A-4-Compute <br /> ab/d>a/d,Therefore f: =60000 OK <br /> Ctieck if tension cantrols <br /> a/d, =a/d=0.098(one layer) <br /> a���/d, =0.319,Table A-4-Compute <br /> a«�/d� >c/d„ Tension co�trols&�=0.90 <br /> Q M„=� [A�.o�a�60000(d-a/2)j/12=0.90(0.230)(6D000�[5.54-0.54/2J=5456 ft-Ib/ft <br /> (�M�=5456 ft-Ib/ft)>(Mu =4770 ft-Ib/fi) OK <br /> Restraint at top <br /> Bolt capacity <br /> eolt diameter=0.500 in. <br /> Main member: Concrete(Loading on plate) <br /> Thickness main member=6.00 in. <br /> Side member: Hem-Fir <br /> Thickness side member=3.50 in. <br /> Load direction main meinber=1,Theta=90.0 <br /> Fem=10186 psi <br /> Load direction side member=0,Theta=0.0 <br /> Fes=4816 psi <br /> K-Theta=1.25 <br /> k1=1.0373 <br /> k2=1.5173 <br /> k3=0.7869 <br /> 11006 60T"AVE WEST-MUKILTEO WA 98275 <br /> PHONE (425) 343-7517 - FAX (425)493-8388 <br /> � <br />