Truss, member forces and bearing loads • pickedshares

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

Solution

The following video is in german language.

Technische Mechanik 1, Übung 19 - Stabwerk

Watch this video on YouTube

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.

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 F_Bx

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

Equation (9) gives Zero-member S₇

$\tag{17} {S_7}=0$

Equation (3) solved for F_By

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

F_By 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 F_Ay

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

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

$\tag{29} {S_1}=4.55 \mathit{kN}$

$\tag{30} {S_2}=9.11 \mathit{kN}$

$\tag{31} {S_3}=9.11 \mathit{kN}$

$\tag{32} {S_4}=-6.44 \mathit{kN}$

$\tag{33} {S_5}=4.55 \mathit{kN}$

$\tag{34} {S_6}=-6.44 \mathit{kN}$

$\tag{35} {S_7}=0.0$

$\tag{36} {S_8}=-5.81 \mathit{kN}$

$\tag{37} {S_9}=-4.55 \mathit{kN}$

$\tag{38} {F_{\mathit{Ay}}}=4.55 \mathit{kN}$

$\tag{39} {F_{\mathit{By}}}=4.11 \mathit{kN}$

$\tag{40} {F_{\mathit{Bx}}}=5.0 \mathit{kN}$