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LEC 5
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   EXAMPLE: Calculating internal forces and momentsStep 1: Find reactions at supports. F  x  = 0 R A x = 0 F  y  = 0 R A y − P   = 0 R A y =  P M  A  = 0 M  A − Pa  = 0 M  A  =  Pa 5  Step 2: If you have a multi-component structure, you will need to find jointforces too.Step 3: Find internal forces and moments.1. Set up local coordinates.2. Identify interesting points.3. Cut beam and do FBDs.6   For 0  < x < aF  x  = 0 N   = 0 F  y  = 0 P   − V   y  = 0 P   =  V   y M  A  = 0 Pa + M  z − Px  = 0 M  z  =  Px − Pa  = − P  ( a − x )OR (From the other end)FBD F  x  = 0 − N   = 0 N   = 0 F  y  = 0 − P   + V   y  = 0 V   y  =  P M  ∗  = 0 − M  z − P  ( a − x ) = 07   M  z  = − P  ( a − x )Note: This is the same as before.For  a < x < LF  x  = 0 − N   = 0 N   = 0 F  y  = 0 V   y  = 0 M  ∗  = 0 − M  z  = 0 M  z  = 0Step 4: Plot result and sanity check.8
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