solution: fbd for link ab

2
M(x) 1 -10 V(x) 5 10 2.5 -7.5 Example 9.14 Draw the shear force and bending moment diagrams in the plot axes below for the loaded beam shown. Solution: FBD for Link AB: ! = 0: ! 2 10 1 = 0 => ! = 5 ! = 0: ! + ! 10 = 0 => ! = 5 FBD for link BD: ! = 0: 5 1 + ! = 0 => ! = 5 ! = 0: ! + ! ! 10 = 0 => ! = 20 Shear Force Diagram and Bending Moment Diagram

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M(x)

1

-10

V(x)

5

10

2.5

-7.5

Example 9.14 Drawtheshearforceandbendingmomentdiagramsintheplotaxesbelowfortheloadedbeamshown.

Solution:FBDforLinkAB:

𝑀! = 0:𝐵! 2 − 10 1 = 0 => 𝐵! = 5 𝑘𝑁

𝐹! = 0:𝐴! + 𝐵! − 10 = 0 => 𝐴! = 5 𝑘𝑁FBDforlinkBD:

𝑀! = 0: 5 1 + 𝐷! = 0 => 𝐷! = −5 𝑘𝑁

𝐹! = 0: 𝐶! + 𝐷! − 𝐵! − 10 = 0 => 𝐶! = 20 𝑘𝑁ShearForceDiagramandBendingMomentDiagram