Question 1 of 8: Statically Indeterminate Rod-Hanger Assembly
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam, May 2019 — 04-BS-06: Mechanics of Materials. Closed book, one 8.5"×11" handwritten aid sheet permitted, 3 hours, 8 questions of equal value (any 5 of 8 constitute a complete paper — all 8 are answered here as a complete study resource).
Reference texts: R.C. Hibbeler, Mechanics of Materials, 10th ed. (Pearson) — axial deformation of statically indeterminate assemblies, beam deflection by integration, shear/moment diagrams, plane-stress transformation and Mohr's circle, column buckling, torsion of circular shafts, composite (transformed-section) beams, combined axial-flexural loading, transverse shear.
Find. The maximum uniformly distributed load $w$ (lb/ft) that the rigid bar $BDF$ can carry without exceeding the allowable normal stress in any rod.
Rigid bar BDF hung from fixed bar by rods (1) A-B, (2) C-D, (3) E-F; UDL w on BDF.
Approach. The bar $BDF$ is rigid and hung by three parallel rods — two equilibrium equations are not enough to find three rod forces, so a geometric compatibility equation (rigid-bar rotation) supplies the third; each rod's stress is then checked against its own allowable stress.
Rigid-body equilibrium of bar BDF. Let $F_1,F_2,F_3$ be the tensions in rods (1),(2),(3), acting at $x=0,\,a,\,3a$ along the bar (measured from B). The UDL resultant is $w(3a)$ at $x=1.5a$.
$$F_1+F_2+F_3 = w(3a)$$
$$\sum M_B=0:\quad F_2\,a + F_3\,(3a) = w(3a)(1.5a) = 4.5\,w a^2$$
Compatibility (rigid bar ⇒ linear displacement). All three rods share the same length $L$ and hang from the same (fixed) upper bar, so each rod's elongation equals the downward deflection of its lower attachment point. Because $BDF$ stays straight, deflection is linear in $x$:
$$\delta_D=\delta_B+\tfrac{1}{3}(\delta_F-\delta_B) \quad\Rightarrow\quad \tfrac{3F_2 L}{A_2E_2}=\tfrac{2F_1L}{A_1E_1}+\tfrac{F_3L}{A_3E_3}$$
With $A_1=A_2=\tfrac{\pi}{4}(0.75)^2=0.4418\text{ in}^2$ and $A_3=\tfrac{\pi}{4}(1.25)^2=1.2272\text{ in}^2$, solving the three equations simultaneously gives
$$F_1 = 44.214\,w a,\qquad F_2 = 41.679\,w a,\qquad F_3 = 58.107\,w a$$
(with $w$ in kip/in and $a=48$ in, so $F_i$ is in kip). This is a linear system solved directly — no iteration needed.
Stress per rod, per unit $w$.
$$\frac{\sigma_1}{w}=\frac{F_1}{A_1 w}=100.08\ \text{in}^{-1},\quad \frac{\sigma_2}{w}=94.34\ \text{in}^{-1},\quad \frac{\sigma_3}{w}=47.35\ \text{in}^{-1}$$
Allowable stress and governing rod. $\sigma_{allow}=\sigma_Y/FS$: rods (1),(2) $\to 36/1.8=20.0$ ksi; rod (3) $\to 48/1.8=26.67$ ksi. Solving $\sigma_i=\sigma_{allow,i}$ for $w$ in each rod and keeping the smallest:
$$w_{1}=\frac{20.0}{100.08}=0.1998\ \tfrac{\text{kip}}{\text{in}},\quad w_2=0.2120\ \tfrac{\text{kip}}{\text{in}},\quad w_3=0.5632\ \tfrac{\text{kip}}{\text{in}}$$
Rod (1) governs.
$$\boxed{w_{max}=0.1998\ \text{kip/in} = 2.40\ \text{kip/ft} \approx 2400\ \text{lb/ft}}$$
