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Find the support reactions for the beam loaded and supported as shown in the figure.
1 Answer
written 8.7 years ago by |
F.B.D :-
Applying COE,
∴ΣFy(↑+ve)=0∴−80+VB−100−60+RFcos30=0∴VB+RFcos30=240…………(I)∴ΣMA≅0(+ve)∴−80×2+VB×3−100sin45×5−50−60×7+RFcos30×9=0∴3VB+RFcos30×9=983.553…………….(II)
Solving (I) and (II) simultaneously,
∴ V_B=196.07 kN ,R_F=50.72kN \\ ∴Σ F_x (→ +ve)=0 \\ ∴H_B-100 \cos45- R_F \sin30=0 \\ ∴H_B=100 \cos45+50.72 \sin30 \\ ∴H_B=96.07 kN \\ R_B= \sqrt{H_B^2+V_B^2} =218.342 kN \\ θ= \tan^{-1}(\dfrac {V_B}{H_B })= \tan^{-1}(\dfrac {196.07}{96.07})= 63.90°