Question 2 (paper Question II) — Compound Frame, Method of Joints (Part A · Statics, equal value): 04-BS-3, Undated paper, Q2

Question 2 (paper Question II) — Compound Frame, Method of Joints (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. Plane frame, pin support at $A(0,0)$, roller support at $D(3,7)$ (horizontal reaction only — see check note). Joints $B(3,4)$, $C(6,4)$. A 250 kN load acts down at C; a 300 kN load acts down at E$(9,7)$. Members: AB, AC, BC, BD, CD, CE, DE (7 members, 5 joints, pin+roller = 3 reactions; $m+r=2j=10$, determinate).

Find. The force in every member (tension T / compression C) and the support reactions.

Figure 2. Five-joint frame: A(0,0) pin, D(3,7) roller, B(3,4), C(6,4) with 250 kN, E(9,7) with 300 kN.

Approach. Find the reactions from global equilibrium (roller at D gives a horizontal-only unknown), then work joint-by-joint, starting where only two member forces are unknown.

Support reactions (moment about A): $-7D_x-250(6)-300(9)=0\Rightarrow D_x=-600$ kN (i.e. 600 kN pointing $-x$). Then $\Sigma F_x=0\Rightarrow A_x=600$ kN; $\Sigma F_y=0\Rightarrow A_y=550$ kN.
Joint E (members DE horizontal, CE diagonal to C, 300 kN down): $\Sigma F_y=0\Rightarrow F_{CE}=-300\sqrt2=-424.3$ kN (424.3 kN C); $\Sigma F_x=0\Rightarrow F_{DE}=300$ kN (T).
Joint D (DE known, DB vertical, DC diagonal, reaction $D_x=-600$): $\Sigma F_x=0$: $300+\tfrac{1}{\sqrt2}F_{DC}-600=0\Rightarrow F_{DC}=424.3$ kN (T); $\Sigma F_y=0\Rightarrow F_{DB}=-\tfrac{1}{\sqrt2}(424.3)=-300$ kN (300 kN C).
Joint B (BD known, BC horizontal, BA diagonal to A(0,0)): $\Sigma F_y=0$: $-300-0.8F_{BA}=0\Rightarrow F_{BA}=-375$ kN (375 kN C); $\Sigma F_x=0\Rightarrow F_{BC}=0.6(-375)=-225$ kN (225 kN C).
Joint C — check: with $F_{CD}=424.3$ (T), $F_{CE}=-424.3$ (C), $F_{CB}=-225$: $\Sigma F_x=-424.3(0.7071)+424.3(0.7071)-(-225)-0.8321F_{CA}=0\Rightarrow F_{CA}=-450.7$ kN (450.7 kN C) ✓; $\Sigma F_y$ also closes exactly to zero ✓.

Final Results — Question 2 (member forces, T = tension, C = compression)
Member	Force	Sense
AB	375.0 kN	C
AC	450.7 kN ($125\sqrt{13}$)	C
BC	225.0 kN	C
BD	300.0 kN	C
CD	424.3 kN ($300\sqrt2$)	T
CE	424.3 kN ($300\sqrt2$)	C
DE	300.0 kN	T
$A_x,A_y$	600 kN, 550 kN	—
$D_x$	600 kN ($-x$)	—

Topic: Plane-frame analysis by the method of joints, with a determinacy check.

Key relations: at each pin joint $\Sigma F_x=0$, $\Sigma F_y=0$; determinacy $m+r=2j$; member components use run/length, rise/length.

Why this works. Every member here carries load only at its pinned ends, so each is a two-force member; solving joints in an order that never leaves more than two unknowns propagates known forces through the whole frame, exactly as for a simple truss.

Common pitfall: reading D as a second full pin (as its hatched-wall symbol might suggest) gives $m+r=11>2j=10$ — one redundant reaction, unsolvable by statics alone; recognizing D as a roller (only $D_x$) restores determinacy and returns clean, exactly-closing reactions, which is itself strong evidence for the correct reading.

Source: Hibbeler, Engineering Mechanics: Statics, Ch. 6 (Structural Analysis — method of joints).

Question 2 of 6: Question 2 (paper Question II) — Compound Frame, Method of Joints (Part A · Statics, equal value)

Question 2 (paper Question II) — Compound Frame, Method of Joints (Part A · Statics, equal value)