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Bearing loads and hinged rod force

Task

A beam is hold by a fixed bearing on the left and a hinged rod (pendulum support). The beam itself is stiff. The force F acts on the beam.

Determine the bearing loads in A, B and C and the force in the hinged rod.

F = 15 kN
α = 75°
Beam with pendulum support
Beam with pendulum support

Solution

The following video is in German language, but English subtitles are available.

Lagerreaktionen und Stabkraft - Technische Mechanik 1, Übung 21

Step by step

The whole beam is divided into three sections to determine the reaction forces.

Free cut of the beam
Free body diagram of the beam
Reaction forces of joint B
Reaction forces of joint B
Reaction forces of bearing C
Reaction forces of bearing C
tm1-21

Section I: Balance of forces in x-direction

\[\tag{1} 0={F_{\mathit{Bx}}}+{F_{\mathit{Ax}}}\]

Section I: Balance of forces in y-direction

\[\tag{2} 0={F_{\mathit{By}}}+{F_{\mathit{Ay}}}-F\]

Section I: Balance of moments around A

\[\tag{3} 0={F_{\mathit{By}}} l-2 F l\]

Section II: Balance of forces in x-direction

\[\tag{4} 0=-{F_S} \cos{\left( \alpha \right) }-{F_{\mathit{Bx}}}\]

Section II: Balance of forces in y-direction

\[\tag{5} 0={F_S} \sin{\left( \alpha \right) }-{F_{\mathit{By}}}\]

Section III: Balance of forces in x-direction

\[\tag{6} 0={F_S} \cos{\left( \alpha \right) }+{F_{\mathit{Cx}}}\]

Section III: Balance of forces in y-direction

\[\tag{7} 0={F_{\mathit{Cy}}}-{F_S} \sin{\left( \alpha \right) }\]

Step by step solution of the found equations

\[\tag{8} {F_{\mathit{By}}}=2 F\]

\[\tag{9} {F_{\mathit{By}}}=30 \mathit{kN}\]

\[\tag{10} {F_{\mathit{Ay}}}=F-{F_{\mathit{By}}}\]

\[\tag{11} {F_{\mathit{Ay}}}=-15 \mathit{kN}\]

\[\tag{12} {F_S}=\frac{30 \mathit{kN}}{\sin{\left( \alpha \right) }}\]

\[\tag{13} {F_S}=31.06 \mathit{kN}\]

\[\tag{14} {F_{\mathit{Bx}}}=-{F_S} \cos{\left( \alpha \right) }\]

\[\tag{15} {F_{\mathit{Bx}}}=-8.04 \mathit{kN}\]

\[\tag{16} {F_{\mathit{Ax}}}=-{F_{\mathit{Bx}}}\]

\[\tag{17} {F_{\mathit{Ax}}}=8.04 \mathit{kN}\]

\[\tag{18} {F_{\mathit{Cx}}}=-{F_S} \cos{\left( \alpha \right) }\]

\[\tag{19} {F_{\mathit{Cx}}}=-8.04 \mathit{kN}\]

\[\tag{20} {F_{\mathit{Cy}}}={F_S} \sin{\left( \alpha \right) }\]

\[\tag{21} {F_{\mathit{Cy}}}=30.0 \mathit{kN}\]