Reader note:
While it is difficult to explain this procedure in paragraph format, every effort was made to be clear, exact, and concise.
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We will calculate the forces in members required by the problem statement. We will also summarize the forces in ALL members. All numerical force values shown in calculations below are “kN.”
1. Draw FBD of entire truss with applied joint point loadings labeling unknown reactions. Determine truss reactions at Pts. D (Dh and Dv) and E (Eh only).
2. Sum system moments about Pt. E; Dh = -7.94 (to the left).
3. Sum system vertical forces;…………...Dv = +7.5 (up).
4. Sum system horizontal forces;..….….Eh = +7.94 (to the right).
5. The force in member CE = -7.94 (compression) by inspection.
Using the Method of Sections, we strategically place complete vertical sections cutting through members in several locations (using the FBD truss portion to left of section cut in each case) and sum moments about strategically selected points to determine unknown member forces (or component forces in diagonal members). We will add additional construction points to our diagram in an effort to “zero out” values acting in those “lines of force” as we attempt to limit calculations to solving for only one unknown force value at a time.
6.
The first vertical section is cut immediately to the LEFT of Pt. C, cutting through members BD, CD, and CE. Add Pt. P located 5m right of Pt. E and 5.77m above Pt. C, on member BD. Place component forces for member BD at added Pt. C. Sum moments about Pt. C to determine component forces in member BD; (BD)h = +5.42 (to the right) and (BD)v = -3.13 (down). Using the Pythagorean Theorem, the resultant force in member BD = +6.25 (tension). We will use these component values again in another section below.
7.
In the same section cut as Item#6, sum moments about added Pt. P to calculate components in member CD: (CD)h = +2.52 (to the right). By similar triangles (or summing vertical forces), (CD)v = -4.37 (down). Using the Pythagorean Theorem, the resultant force in member CD = +5.05 (tension).
8.
The second strategic vertical section is cut immediately to the RIGHT of Pt. B, cutting through members AB and AC. Sum moments about an added Pt. R located 2.5m right of Pt. C, on AC. Solve for the force in member AB: (AB)h = +4.33 (to the right). Summing vertical forces, (AB)v = -2.5 (down). Using the Pythagorean Theorem, the resultant force in member AB = +5.0 (tension). Summing horizontal forces, (AC)h = -4.33 (to the left). The force in member AC = -4.33 (compression).
9.
A third strategic vertical section is cut immediately to the LEFT of Pt. B, cutting through members BD, BC, and AC. Place component values for BD found in Item#6 above. Sum moments about an added Pt. R located 2.5m right of Pt. C on member AC; (CB)h = -1.08 (to the left), (CB)v = -1.87 (down). Using the Pythagorean Theorem, the resultant force in member BC = -2.16 (compression).
In summary:
Force in members:
AC = -4.33 (compression)
BC = -2.16 (compression)
CD = +5.05 (tension)
Additionally, force in members…..
AB = +5.0 (tension)
BD = +6.25 (tension)
CE = -7.94 (compression)
Reaction at Pt. E…..Eh = +7.94 (to the right),
Reaction at Pt. D…..Dh = -7.94 (to the left)
Dv = +7.5 (up)
Please contact me for a more detailed explanation, setup and walkthrough of the strategic section placement, force component placement diagrams at section cuts, and FBD calculation of unknown forces at each section cut.
