Replace the force and couple moment system acting on the beam in Fig. 4–44a by an equivalent resultant force, and find where its line of action intersects the beam, measured from point O.

Question

Accepted Answer

Force Summation. Summing the force components,

[latex]\underrightarrow{+} (F_{R})_{x} = ∑F_{x}; (F_{R})_{x} = 8 kN (\frac{3}{5} ) = 4.80 kN →[/latex]

[latex]+ \uparrow (F_{R})_{y} = ∑F_{y}; (F_{R})_{y} = - 4 kN + 8 kN (\frac{4}{5} ) = 2.40 kN \uparrow [/latex]

From Fig. 4–44 b , the magnitude of [latex]\mathbf{F}_{R}[/latex] is

[latex]F_{R} = \sqrt{(4.80 kN)^{2} + (2.40 kN)^{2}} = 5.37 kN [/latex]

The angle [latex] heta[/latex] is

[latex] heta = an ^{- 1} \left(\frac{2.40 kN}{4.80 kN} ight) = 26.6°[/latex]

Moment Summation. We must equate the moment of [latex]\mathbf{F}_{R}[/latex] about point O in Fig. 4–44 b to the sum of the moments of the force and couple moment system about point O in Fig. 4–44 a . Since the line of action of [latex](\mathbf{F}_{R})_{x}[/latex] acts through point O , only [latex](\mathbf{F}_{R})_{y}[/latex] produces a moment about this point. Thus,

[latex]⤿+ (M_{R})_{O} = ∑M_{O}; 2.40 kN(d) = - (4 kN)(1.5 m) - 15 kN · m [/latex]

[latex]- \left[8 kN (\frac{3}{5} ) ight] (0.5 m) + \left[8 kN (\frac{4}{5} ) ight] (4.5 m)[/latex]

[latex]d = 2.25 m[/latex]

Question 4.17: Replace the force and couple moment system acting on the bea...

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