2bex. Statics Examples • Ex.2b.1 • Ex. 2b.2 • Ex. 2b.3 • Ex. 2b.4 • Ex. 2b.5 
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Sol'n: From the FBD of the entire beam, the reaction forces at the supports are found by taking moments about each end: SM_{x=0} = 0 –(4 kN)(4 m) + R_{2}(10 m) = 0SM_{x=10} = 0 (4 kN)(6 m) – R_{1}(10 m) = 0 R_{1} = 2.4 kN R_{2} = 1.6 kN Check: SF_{y}= 2.4 kN + 1.6 kN – 4 kN = 0 OK A segment of the beam of length x (0< x < 4 m) is next isolated and treated as a FBD. The internal shear force and bending moment acting on the segment are shown in their positive directions (per this text). Applying equilibrium:
These internal loads act on any crosssection of the beam to the left of the 4 kN load (x<4m). A segment of the beam to the right of the 4kN load is next isolated in a similar manner. Equilibrium gives: S F_{y} = 0 V(x) = 1.6 kN S M = 0 M(x) = 1.6(10 – x) kN·m These internal loads act on any crosssection of the beam to the right of the point load (x>4m). 

V(x) and M(x) can be plotted on Shear Force and Bending Moment Diagrams.
The change in moment between two crosssections is the negative in area under the shear diagram. From 0 to 4 m: Area = (2.4 N)(4 m) = 9.6 kN·m]. Beams are covered in more depth in Chapter 6. 
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