This Demonstration calculates the forces on the members of a planar truss.
Use the sliders to set the two point loads at the center and the right (forces in negative y direction), FD and FF. Check “show member forces” to show the truss member forces. Check “make members under tension red and under compression green” to show the red and green arrows.
Arrows that point outward (green) represent the member response to compression forces, and arrows that point inward (red) represent the member response to tension forces. Compression acts to shorten the member and tension acts to lengthen the member.
The black members are zero members; that is, these members are neither in tension nor in compression, so the force is 0 kN. The purpose of zero members is to provide stability and extra support to the structure in the event of another member failing.
When the boxes “make members under tension red and under compression green” and “show member forces” are checked, move the mouse over a pivot point to show arrows that represent forces (and the values of the forces) in the x and y directions at that point. This breaks the forces in the diagonal members into their x and y components.
The method of joints is used to calculate the forces on each member of the truss. This is done by solving force balances around individual joints. First, calculate the reactions at the supports. Taking the sum of the moments at the left support:
ΣMA = 0 = 3wFD + 5wFF − 6wRG,where MA is the moment around support A (left support), w is the width of one member, Fi < 0 is the force applied to the joints at i = (D, F), and RG is the reaction at support G (right). The first two terms (3wFD and 5wFF) in the moment balance are positive because they would cause a clockwise rotation, and the third term (−6wRG) is negative because it would cause a counterclockwise rotation.
Next, take the sum of the forces in the y direction to get RA:
ΣFy = 0 = RA + RG − FD − FF.Begin solving for the forces of the members by doing force balances at the joints. The order of the balances around the joints listed here is the order in which they should be solved. Force balances are done assuming we can use logic to figure out which members are under tension and which are under compression. Starting at joint A, there is a reaction force pushing up, so F0 must be pushing down (under compression). A labeled truss is shown in Figure 1.
Joint A
ΣFy = 0 = −F0 sinθ + RA, ΣFx = 0 = −F0 cosθ + F2.Joint L
ΣFy = 0 = F1, ΣFx = 0 = −F2 + F3.Joint B
ΣFy = 0 = F0 sinθ − F1 − F7 sinθ, ΣFx = 0 = F0 cosθ − F5 + F7 cosθ.Joint C
ΣFy = 0 = F6, ΣFx = 0 = F5 − F8.Joint K
ΣFy = 0 = F6 + F7 sinθ − F12 sinθ, ΣFx = 0 = −F3 − F7 cosθ − F12 cosθ + F4.Joint J
ΣFy = 0 = F9, ΣFx = 0 = −F4 + F11.Joint D
ΣFy = 0 = F9 + F10 sinθ + F12 sinθ − FD, ΣFx = 0 = F8 − F10 cosθ + F12 cosθ − F13.Joint E
ΣFy = 0 = F14, ΣFx = 0 = F13 − F15.Joint I
ΣFy = 0 = −F10 sinθ + F14 + F16 sinθ, ΣFx = 0 = F10 cosθ − F11 + F16 cosθ + F17.Joint H
ΣFy = 0 = F18, ΣFx = 0 = −F17 + F20.Joint G
ΣFy = 0 = −F19 sinθ + RG.Note that all the vertical members are zero members, which means they are neither under tension nor under compression (force is 0 kN).
Figure 1.
All forces are in kN.
Reference
[1] SkyCiv Cloud Engineering Software. “Tutorial to Solve Truss by Method of Sections.” (Aug 29, 2017) skyciv.com/tutorials/tutorial-to-solve-truss-by-method-of-sections.
This simulation was created in the Department of Chemical and Biological Engineering at University of Colorado Boulder for LearnChemE.com by John L. Falconer using Claude AI. It is a JavaScript/HTML5 implementation of a Mathematica simulation by Rachael L. Baumann. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988) in collaboration with Washington State University. Address any questions or comments to LearnChemE@gmail.com