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T or C = 0.7(E)(h)Po ± .6D <br />C 1 = 0.7(2104)(9.00)/(20.00) +.6(:000) =1263 Ib <br />C2 = 0.7(1683�(9.00�/(20.00) r .6(1000) =11301b <br />C =12n3 Ib CONTROLS <br />T1 = •0.7(2104)�9.00)/(20.00) r .6(10G0) _ -63 Ib <br />T2 = •0.7(1683)(9.00)/(20.00) +.6(1000) = 701b <br />T = 701b CONTROLS <br />Wind <br />T or C = W(h) ± .6D <br />CW = 651b/11�9.00) +.6(1000) =1182 Ib <br />TW = -65Ib/fl(9.00) +.6(1000) =18Ib <br />Design forces: <br />Shear = 66.31b/it <br />Uplift = 701b � 0.00' <br />Compression =12631b � 20.00' <br />E� 16•50 <br />E016•51 <br />E� 16-50 <br />E� 16-51 <br />E� 16•11 <br />E� 16-11 <br />Note: Compression capacity for 2 x is 3341 Ibs. <br />Stress perpendicular to grain controls. <br />Drag strutforees <br />Delta shear flow between sw and diaphragm, dq = v• q <br />d= 6628 • 66.28 = 0.00 Ib/fl <br />Se mei�, Shearflow Width Strul <br />Iblft fl Force Ib <br />1 0.0 20.0 0.0 <br />Check drift <br />Panel No 1, <br />Ultimate shear = 66.3/071b/ft = 947 Iblft <br />� e= Bv„h �l E A b= 8(94.7)(9.00')�/ 1600G:10(16.50 sq.in.)(20.00') = 0.001 in. <br />�,= v„ h I Gt =(94.7)(9.OD') I 25000 = 0.034 in. <br />Compute 0 � <br />Bd Nails � 6.00 o.c., total nails/12' =1216 = 2 <br />Force per nail = v� l2 Nails = 47.31bs <br />For Bd nails, e� =1.20(V� I616)� 01e =1.20(47/616)� D1B = 0.0005 <br />0 �= 0.75h e� = 0.75(9.0)(0.0005) = 0.004 in. <br />p,= Anch defl + crushing = 0.000 + 0.020 = 0.020 in. <br />For anchor design, slip = 0.000 in. (Per manuf.) <br />Per NDS, under max F'c•perp =.020 in. <br />.� a= h/b (0.125) = 9.0120.0(0.125) = 0.009 in. <br />StorydriitOs=�o+�.+0�+�1a=0.001+0.034+0.004+0.009=0.048in. <br />� =Ce�,/l =4(0.048)I1.0=0.2in. <br />Ai i=00251hx1�1=002519.Ox121=2.7in.>0.2in.OK <br />End Check drifl <br />�USiOMDESIGN Date: <br />& ENGINEERING, INC 6/17/2005 <br />pNCNPAt �NI'YSIf � W LGN <br />Analysis Daument: FOUNDATION REPAIR <br />Job Number. E6�427 <br />Page: 9 <br />