U_-_Relationships_Between_Load__Shear__a

Solution draw sfd and bmd 600 lb 5 ft 350 lb 7 ft 250

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Unformatted text preview: N Draw SFD and BMD 600 lb 5 ft 350 lb 7 ft 250 lb 350 lb 0 lb SFD – 250 lb 0 + 5(350) = 1750 lb-ft BMD 1750 – 7(250) = 0 lb-ft EXAMPLE 2 EXAMPLE Draw SFD and BMD for the beam SOLUTION 2 SOLUTION Solve for support reactions RAX RAY RBY ∑ MA = 0; RBY(20) – 50(20)(10) – 200 = 0; RBY = 510 lb ∑ FY = 0; RAY – 50(20) + 510 = 0; RAY = 490 lb ∑ Fx = 0; RAX = 0 SOLUTION 2 -Contd. SOLUTION 490 lb 510 lb 490 lb SFD 0 lb 490/50 = 9.8 ft 0.5(9.8)(490) = 2401 lb-ft 490 – 50(20) = – 510 lb BMD 2401 ...
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This note was uploaded on 10/23/2012 for the course EGR 221 taught by Professor Baladi during the Spring '11 term at Michigan State University.

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