N_-_Rigid_Body_Equilibrium_2-D

5 kn ray 12 14 18 rey 44 235 205 kn example 2

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Unformatted text preview: LE 2 Determine all the support reactions at A and B SOLUTION 2 Step 1: Draw FBD RAY RB X RB SOLUTION 2 -Continued Step 2: Write the equilibrium equations ΣFX = 0; RBX – 0.5 = 0 RAY Σ FY = 0; RAY-5-7-10-2+RBY = 0 Σ MB = 0; R B X -14(RAY)+14(5)+6(7)-6(2)+6(0.5) = 0 RB Y SOLUTION 2 -Continued Step 3: Solve the system equations RBX = 0.5 kip RAY RB X RB Y RAY = (14(5) + 6(7) – 6(2) + 6(0.5))/14 = 7.36 kips RBY = 5 + 7 + 10 + 2 – RAY = 24 – 7.36 = 16.6 kips EXAMPLE 3 Determine all the support reactions at A and C SOLUTION 3 Step 1: Draw FBD after calculating F1, F2 and F3 Note that the distributed load at the top can be broken dawn to rectangular (where F2 can be calculated) and triangular (where F3 can be calculated) distributions as shown F3 B F2 F1 RC Y RA X A RA Y SOLUTION 3 - continued F3 B F2 F1 RC Y RA Y F1 = 5(200) = 1000 N; located at Y1 = 2.5 m (from A) F2 = 6(100) = 600 N; located at X2 = 3 m (from A or B) F3 = 0.5(6)(100) = 300 N; located at X3 = 2 m (from A or B) RA X A SOLUTION 3 - continued F3 B F2 F1 RC Y RA Y Step 2: write the equilibrium equations Σ FX = 0; F1 – RAX = 1000 – RAX = 0 Σ FY = 0; RAY + RCY – F2 – F3 = RAY + RCY – 900 = 0 Σ MA = 0; F1X1 – F2X2 – F3X3 + 6(RCY) = 0 1000(2.5) – 600(3) – 300(2) + 6(RCY) = 0...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.

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