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Truss, member forces and bearing loads

Task

The force F is acting under an angle on a truss. The truss has a locating bearing and a floating bearing. Calculate the bearing loads and the member forces for the truss.

F = 10 kN
α = 60°
a = 2 m
Truss with fixed bearing and floating bearing under external load
Truss with fixed bearing and floating bearing under external load

Solution

The following video is in german language.

Technische Mechanik 1, Übung 19 - Stabwerk
Solution video for exercise 19 -- truss

Sketches

The following sketches show the bearing reaction forces, the member and node numbering and the location and direction of the forces at each node.

Truss with single nodes and forces
Truss with single nodes and forces

The forces are solved symbolic first, then the real numbers are set in to calculate the values. Since the angles of the bars are always 45 °, the alternative notation of the trigonometric functions is used.

Bearing reactions

Sum of all forces in y-direction

\[\tag{1} 0=-F \sin{\left( \alpha \right) }+{F_{\mathit{By}}}+{F_{\mathit{Ay}}}\]

Sum of all forces in x-direction

\[\tag{2} 0={F_{\mathit{Bx}}}-F \cos{\left( \alpha \right) }\]

Sum of all moments around point A

\[\tag{3} 0=-2 F a \sin{\left( \alpha \right) }+F a \cos{\left( \alpha \right) }+3 {F_{\mathit{By}}} a\]

Node equations

Node I x

\[\tag{4} 0=\frac{{S_4}}{\sqrt{2}}+{S_1}\]

Node I y

\[\tag{5} 0=\frac{{S_4}}{\sqrt{2}}+{F_{\mathit{Ay}}}\]

Node II x

\[\tag{6} 0=\frac{{S_6}}{\sqrt{2}}+{S_2}-{S_1}\]

Node II y

\[\tag{7} 0=\frac{\mathit{S6}}{\sqrt{2}}+\mathit{S5}\]

Node III x

\[\tag{8} 0={S_3}-{S_2}\]

Node III y

\[\tag{9} 0={S_7}\]

Node IV x

\[\tag{10} 0=-\frac{{S_8}}{\sqrt{2}}-{S_3}+{F_{\mathit{Bx}}}\]

Node IV y

\[\tag{11} 0=\frac{{S_8}}{\sqrt{2}}+{F_{\mathit{By}}}\]

Node V x

\[\tag{12} 0={S_9}-\frac{{S_4}}{\sqrt{2}}\]

Node V y

\[\tag{13} 0=-{S_5}-\frac{{S_4}}{\sqrt{2}}\]

Node VI x

\[\tag{14} 0=-F \cos{\left( \alpha \right) }-{S_9}+\frac{{S_8}}{\sqrt{2}}-\frac{{S_6}}{\sqrt{2}}\]

Node VI y

\[\tag{15} 0=-F \sin{\left( \alpha \right) }-\frac{{S_8}}{\sqrt{2}}-{S_7}-\frac{{S_6}}{\sqrt{2}}\]

Equation (2) solved for FBx

\[\tag{16} {F_{\mathit{Bx}}}=F \cos{\left( \alpha \right) }\]

Equation (9) gives Zero-member S7

\[\tag{17} {S_7}=0\]

Equation (3) solved for FBy

\[\tag{18} {F_{\mathit{By}}}=\frac{2 F \sin{\left( \alpha \right) }-F \cos{\left( \alpha \right) }}{3}\]

FBy set in equation (1) …

\[\tag{19} 0=\frac{2 F \sin{\left( \alpha \right) }-F \cos{\left( \alpha \right) }}{3}-F \sin{\left( \alpha \right) }+{F_{\mathit{Ay}}}\]

… and solved for FAy

\[\tag{20} {F_{\mathit{Ay}}}=\frac{F \sin{\left( \alpha \right) }+F \cos{\left( \alpha \right) }}{3}\]

In the following, the forces solved so far are inserted into the remaining equations and the unknowns are solved

\[\tag{21} {S_4}=-\frac{\sqrt{2} F \sin{\left( \alpha \right) }+\sqrt{2} F \cos{\left( \alpha \right) }}{3}\]

\[\tag{22} {S_1}=\frac{F \sin{\left( \alpha \right) }+F \cos{\left( \alpha \right) }}{3}\]

\[\tag{23} {S_9}=-\frac{F \sin{\left( \alpha \right) }+F \cos{\left( \alpha \right) }}{3}\]

\[\tag{24} {S_5}=\frac{F \sin{\left( \alpha \right) }+F \cos{\left( \alpha \right) }}{3}\]

\[\tag{25} {S_8}=-\frac{{{2}^{\frac{3}{2}}} F \sin{\left( \alpha \right) }-\sqrt{2} F \cos{\left( \alpha \right) }}{3}\]

\[\tag{26} {S_3}=\frac{2 F \sin{\left( \alpha \right) }+2 F \cos{\left( \alpha \right) }}{3}\]

\[\tag{27} {S_6}=-\frac{\sqrt{2} F \sin{\left( \alpha \right) }+\sqrt{2} F \cos{\left( \alpha \right) }}{3}\]

\[\tag{28} {S_2}=\frac{2 F \sin{\left( \alpha \right) }+2 F \cos{\left( \alpha \right) }}{3}\]

Resulting member forces and bearing loads

\[\tag{29} {S_1}=4.55 \mathit{kN}\]

\[\tag{30} {S_2}=9.11 \mathit{kN}\]

\[\tag{31} {S_3}=9.11 \mathit{kN}\]

\[\tag{32} {S_4}=-6.44 \mathit{kN}\]

\[\tag{33} {S_5}=4.55 \mathit{kN}\]

\[\tag{34} {S_6}=-6.44 \mathit{kN}\]

\[\tag{35} {S_7}=0.0\]

\[\tag{36} {S_8}=-5.81 \mathit{kN}\]

\[\tag{37} {S_9}=-4.55 \mathit{kN}\]

\[\tag{38} {F_{\mathit{Ay}}}=4.55 \mathit{kN}\]

\[\tag{39} {F_{\mathit{By}}}=4.11 \mathit{kN}\]

\[\tag{40} {F_{\mathit{Bx}}}=5.0 \mathit{kN}\]