Question 1 of 6: Question 1 (paper Question I) — 3D Pipe Assembly: Pin Reaction and Cable Tension (Part A · Statics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-3 / May 2019 — Statics and Dynamics. Closed book; one 8½″×11″ self-prepared note sheet permitted; approved Casio/Sharp calculator. Candidates were instructed to complete 5 of 6 questions (3 of 3 Part A, 2 of 3 Part B) — all 6 are solved below. Reference texts: Hibbeler, Engineering Mechanics: Statics, 14th ed.; Hibbeler, Engineering Mechanics: Dynamics, 14th ed.
Two genuinely under-dimensioned figures required a stated assumption per the exam's own Note 1 ("if doubt exists… submit a clear statement of any assumption made"): Question 1's pin at A is taken as free to rotate about the pipe's own axis A–E (so $M_{Ay}=0$), and cable anchor C is read at the same plan position as A but 1 m higher; Question 2's support at D is read as a roller (horizontal reaction only) rather than a second pin, since two full pins on 7 members (m+r=11 vs 2j=10) is statically indeterminate and unsolvable by first-year statics — the roller reading returns exact, self-consistent reactions. Question 6's rod angle was measured directly from the printed figure (≈31.4° below horizontal) since it is not given numerically.
Question 1 (paper Question I) — 3D Pipe Assembly: Pin Reaction and Cable Tension (Part A · Statics, equal value)
Given. Rigid pipe assembly, weightless. Pin at $A=(0,0,2)$ m. Pipe runs from A, 6 m along $+y$, to junction $E=(0,6,2)$; a perpendicular cross-pipe runs from E, 4 m along $+x$, to $B=(4,6,2)$, where cable BC attaches. The 300 kN load hangs from $D=(2,6,2)$ m (2 m from E toward B). Cable anchor $C=(0,0,3)$ m.
Given data
Quantity
Value
Load at D
300 kN, $-z$ direction
A, B, C, D, E (m)
(0,0,2), (4,6,2), (0,0,3), (2,6,2), (0,6,2)
Find. $A_x,A_y,A_z,M_{Ax},M_{Ay},M_{Az}$ at the pin, and the cable tension $T_{BC}$.
Figure 1. Pipe assembly (isometric schematic): pin A, cable BC, 300 kN load at D.
Approach. Take position vectors from A, form the cable unit vector, then write vector equilibrium $\Sigma\vec F=0$ and $\Sigma\vec M_A=0$; the pin permits free rotation about its own barrel axis A–E ($y$), so $M_{Ay}=0$, which isolates $T_{BC}$ from the $y$-moment equation alone.
Position vectors and cable direction. $\vec r_{B}=(4,6,0)$ m, $\vec r_D=(2,6,0)$ m (both relative to A). $\vec u_{BC}=\dfrac{C-B}{|C-B|}=\dfrac{(-4,-6,1)}{\sqrt{53}}$.
Moment of the applied load about A. $\vec M_D=\vec r_D\times(0,0,-300)=(6(-300)-0,\ 0-2(-300),\ 0)=(-1800,\ 600,\ 0)$ kN·m.
Moment of the cable tension about A (per unit $T$). $\vec r_B\times\vec u_{BC}=\dfrac{1}{\sqrt{53}}(6,\,-4,\,0)$ kN·m per kN of $T$.
Isolate $T$ from the $y$-moment equation (using $M_{Ay}=0$): $-\dfrac{4}{\sqrt{53}}T+600=0\ \Rightarrow\ \boxed{T_{BC}=150\sqrt{53}=1092.0\text{ kN}}$.
Back-substitute for $M_{Ax}$ and $M_{Az}$. $M_{Ax}=1800-\dfrac{6}{\sqrt{53}}(1092.0)=1800-900=\boxed{900\text{ kN}\cdot\text{m}}$; $\ M_{Az}=0-0=\boxed{0}$.
Force equilibrium at A. $\vec A=-(T\vec u_{BC}+\vec F)=-\big[1092.0\cdot\tfrac{(-4,-6,1)}{\sqrt{53}}+(0,0,-300)\big]=\boxed{(600,\ 900,\ 150)\text{ kN}}$.
