Question 2 of 6: Question 2 (paper Question II) — Forces in All Members of a Truss (Part A · Statics, equal value)

Question 2 (paper Question II) — Forces in All Members of a Truss (Part A · Statics, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A planar truss with joints A, B, C, D, E, F pinned at A (fixed/pin support) and B (roller support). Reading the figure gives the joint coordinates below and nine members: AB, AC, AD, BD, CD, CE, DE, DF, EF. A 50 kN load acts at F, 60° below the horizontal, directed down and to the right.

Find. The axial force (tension or compression) in every member.

Approach. Confirm static determinacy, find the two support reactions from global equilibrium, then solve the sixteen scalar joint-equilibrium equations ($\sum F_x=0,\ \sum F_y=0$ at each of the six joints) simultaneously for the nine member forces and three reactions (method of joints, matrix form) — equivalent to, but more reliable than, working joint-by-joint by hand through C, E, D, B.

1. Determinacy check. $m=9$ members, $r=3$ reaction components (2 at the pin A, 1 at the roller B), $j=6$ joints: $m+r=12=2j=12$ — statically determinate.
2. Resolve the applied load. $F_x=50\cos60^\circ=25.0\ \text{kN}$, $F_y=-50\sin60^\circ=-43.30\ \text{kN}$ at F.
3. Support reactions (global equilibrium). Taking moments about A ($\sum M_A=0$) with only $B_y$ (roller, 1.0 m to the right of A) and the load at F contributing: $$B_y(1.0)+\big[(4.0)(-43.30)-(1.5)(25.0)\big]=0\ \Rightarrow\ B_y=210.7\ \text{kN (up)}$$ Then $\sum F_x=0\Rightarrow A_x=-25.0\ \text{kN}$ (i.e. 25.0 kN pointing left) and $\sum F_y=0\Rightarrow A_y=-(B_y+F_y)=-167.4\ \text{kN}$ (i.e. 167.4 kN downward).
4. Method of joints (representative joint F). Only members EF (horizontal) and DF (along $(2.0,-1.5)/2.5$) meet the load at F: $$\sum F_y=0:\ T_{DF}\!\left(\dfrac{-1.5}{2.5}\right)+F_y=0\ \Rightarrow\ T_{DF}=-72.17\ \text{kN (compression)}$$ $$\sum F_x=0:\ T_{EF}+T_{DF}\!\left(\dfrac{2.0}{2.5}\right)+F_x=0\ \Rightarrow\ T_{EF}=82.74\ \text{kN (tension)}$$ Proceeding the same way through joints E, C, D and B (each now a 2-equation, 2-unknown system) and solving simultaneously with the reactions above gives the complete member set in the results table; every value below satisfies both joint equilibrium at all six joints and the global checks $\sum F_x=\sum F_y=\sum M_A=0$ to within rounding.