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Determine the resultant internal normal force acting on the cross section through point | 1-1 Determine the resultant internal normal force acting on the cross section through point A in each column. In a segment BC weighs 180 Ib ft and segment CD weighs 250 Ib ft. In ft the column has a mass of 200 kg m. 5 kip 8 kN a b ÎÏF 0 1.0 3 3 1.8 5-0 Fa 13.8 kip Ans 5 Fa b 8M î IFy 0 -4.5-4.5-5.89-6-6-8 0 FA 34.9 kN Ans a From Mechanics of Materials Sixth Edition by R. C. Hibbeler ISBN 0-13-191345-X. 2005 R. C. Hibbeler. Published by Pearson Prentice Hall Pearson Education Inc. Upper Saddle River NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher. 1-2 Determine the resultant internal torque acting on the cross sections through points C and D. The support bearings at A and B allow free turning of the shaft. 250 Nm 0 Tc - 250 0 Tc 250 N m Ans 0 To 0 Ans From Mechanics of Materials Sixth Edition by R. C. Hibbeler ISBN 0-13-191345-X. 2005 R. C. Hibbeler. Published by Pearson Prentice Hall Pearson Education Inc. Upper Saddle River NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher. 1-3 Determine the resultant interna torque acting on the cross sections through points B and C. MX 0 Tb 350 - 500 0 7g 150 lb ft Ans 2M 0 Tc-500 0 Tc 500 lb ft Ans From Mechanics of Materials Sixth Edition by R. C. Hibbeler ISBN 0-13-191345-X. 2005 R. C. Hibbeler. Published by Pearson Prentice Hall Pearson Education Inc. Upper Saddle River NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced in any form or by any means without permission in writing from the .
