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PR�.�CT NAME: FIAKE BLDG TI <br /> JOB: N9•1624 <br /> m V.= 147677 Ibs <br /> Controlling yh V�= 147677 Ibs <br /> (Q V�= 147677Ib� > (V�= -574Ib) OK <br /> Check bending <br /> � C21nCA�.S��o�J <br /> foR v�a.�9.tT <br /> � TRi6Jt'f�EY AREk <br /> B � // fo�µo.�E�1T <br /> � I / <br /> � <br /> i <br /> P��b <br /> �.b <br /> L <br /> Distance to criUcal section tor hending, L, <br /> Lo= L/2 - ci/2 = 2b0 - 2400/12 � 12� = 025ft <br /> p�= p„max�L-x - L,) /(L-x ) =2688(250 • 125 - 025� /(250 - 1251 = 2150psf <br /> P�a.rr.=2668 -2150= 536 pSf <br /> Mu� = 1/2(Lo)�p�(B �(Le)I = �/2(025)(2150(2 50)(025)� = 168 fl-Ib <br /> Muz = 2l3(Lo)It1zP�-0�n,(B �(Lo 1]= 2�3�0 25)(1l2(538)(2 50�(0 25) = 28 ft-Ib <br /> Mu= (168 + 28�/2 50 = 78 ft-Ib/ft of width <br /> Reqwred steel area, A:= Mu � [¢�f��d�) <br /> Assume j d= d-a/2 = 0 875d= 0 875(f 50) = 5 69 in <br /> A,= 78(12)/(0 90(60000)(5 691J = 0 003 in�/fl <br /> ChCCk d As'Asm:� <br /> A:.m�= 3(f�1'''/f� • �b„d) = 0 21 in�Ift '0 00 in�/fl <br /> Use A,=021 in�/ft <br /> A,-m,�= 200(b„ 1�d)/f,= 0 26 in�/ft > 0 21 in�Ifl <br /> A,= 026 in�/ft <br /> Reinforcement <br /> TryNo4@900in , Ao.-= 0196sqm <br /> Provided, As oA�= Ae..(12 inlfl�/spacing = 0.196�12)�9 000 = 0262 m7/R <br /> � <br /> CUSTOM DESIGN & ENGINEERING, INC. / � l <br /> November 11, 2013 ( Page 12/��; .� <br /> \ � <br />