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Find the support reactions for the beam loaded and supported as shown in the figure.

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enter image description here

F.B.D :-

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Applying COE,

ΣFy(+ve)=080+VB10060+RFcos30=0VB+RFcos30=240(I)ΣMA0(+ve)80×2+VB×3100sin45×55060×7+RFcos30×9=03VB+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°

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