NivaarExam PrepOfficial exam papers ↗

04-For-B7 Transportation of Forest Products · May 2013

Question 5 of 6: Rail Transportation — Tractive Effort vs. Resistance

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

EGBC National Exam — Forest Engineering, 04-For-B7 Transportation of Forest Products, May 2013. Closed book; approved Casio/Sharp calculator only. 3 hours. Six question groups in two parts: Part A (Questions 1–4, choose 3 of 4, 60 points) and Part B (Questions 5–6, both compulsory, 60 points).

Reference texts: Heinimann, Forest Operations Engineering (forest transportation systems, vehicle/road interaction); Sessions, Forest Road Engineering Guidebook (road geometric design, stopping sight distance, road-surface traction); FPInnovations/FERIC reports (log-hauling vehicle configurations, tire-inflation systems, machine/vehicle productivity); Transportation Association of Canada (TAC), Geometric Design Guide for Canadian Roads (stopping sight distance formula, Canadian design practice); the exam's own “Required Propulsion Formulae” page (standard Davis-type railway tractive-effort and train-resistance equations, consistent with general railway/traction engineering practice, e.g. Hay, Railroad Engineering).

Check: the printed question text for Question 1 has three sub-parts (1.1, 1.2, 1.3), but the source's own mark-allocation table (page 4 of 4) lists only two 10-mark line items for Question 1 ("1.1 10 marks total", "1.2 10 marks total") -- no separate mark line for 1.3, even though every other Part A question group (2, 3, 4) prints exactly two sub-parts matching its two 10-mark line items. This looks like a table/print omission in the original exam rather than an instruction to skip 1.3 -- all three sub-parts are answered in full below, with the group shown as its 20 marks total per the printed table.

Question 5: Rail Transportation — Tractive Effort vs. Resistance (28 marks total, 5.1–5.2)

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.

QuantityValue
Locomotives6, each 2500 hp, 4 axles, 145 tons, cross-section 125 ft²
Freight cars22, each 4 axles, 65 tons
Transmission efficiency, $\eta$83%
Target speed, $V$45 mph
Section 5.2 grade / curve2% grade, 3° curve
Unit conversion1 ton = 2,000 lb

Propulsion: $M = 375\,P\,\eta/V$. Inherent resistance (locomotive): $R=1.3T+29N+0.03TV+CAV^2$, $C=0.0024$. Inherent resistance (freight car): $R=1.5T+72.5N+0.015TV+0.055V^2$. Grade resistance: $R_g=(Tp)/100$ ($T$ in lb, $p$ = % grade). Curvature resistance: $R_c=\varepsilon DT$, $\varepsilon=0.8\,\text{lb/ton-degree}$, $T$ in tons.

Find. Whether the train's available propulsive force at 45 mph exceeds the total resisting force at 45 mph, for (5.1) straight & level track and (5.2) the same speed on a 2% grade with a 3° curve.

Approach. Compute the propulsive force the six locomotives can deliver at $V=45$ mph from the propulsion formula, compute the total resisting force at the same speed (inherent resistance of all locomotives and all cars for 5.1; add grade and curvature resistance for 5.2), and compare the two -- the train can hold 45 mph on a section only if the available propulsive force there is at least as large as the total resistance there.

  1. Available propulsive force at 45 mph. Total locomotive power $P = 6\times2500 = 15{,}000$ hp: $$M = \frac{375\,P\,\eta}{V} = \frac{375\times15{,}000\times0.83}{45} = \boxed{103{,}750\ \text{lb}}$$ This is the tractive force available at the wheels while the train holds 45 mph, and it is the same for both 5.1 and 5.2 since $P$, $\eta$ and $V$ are unchanged.
  2. Inherent resistance of the six locomotives. Per locomotive ($T=145$, $N=4$, $A=125$, $V=45$): $$R_{loco}=1.3(145)+29(4)+0.03(145)(45)+0.0024(125)(45)^2 = 188.5+116+195.75+607.5 = 1{,}107.75\ \text{lb}$$ For all six: $6\times1{,}107.75 = 6{,}646.5\ \text{lb}$.
  3. Inherent resistance of the twenty-two freight cars. Per car ($T=65$, $N=4$, $V=45$): $$R_{car}=1.5(65)+72.5(4)+0.015(65)(45)+0.055(45)^2 = 97.5+290+43.875+111.375 = 542.75\ \text{lb}$$ For all twenty-two: $22\times542.75 = 11{,}940.5\ \text{lb}$.
  4. 5.1 — straight & level. Total resistance is just the summed inherent resistance: $$R_{level} = 6{,}646.5+11{,}940.5 = \boxed{18{,}587\ \text{lb}}$$ Since $M=103{,}750\ \text{lb} > R_{level}=18{,}587\ \text{lb}$ (a margin of $85{,}163\ \text{lb}$), yes -- the train can maintain 45 mph on straight, level track, with a very large reserve of tractive effort to spare.
  5. 5.2 — grade resistance. Total train weight $T=6(145)+22(65)=2{,}300$ tons $=4{,}600{,}000$ lb. At $p=2\%$ grade: $$R_g = \frac{Tp}{100} = \frac{4{,}600{,}000\times2}{100} = 92{,}000\ \text{lb}$$
  6. 5.2 — curvature resistance. At $D=3^{\circ}$ curvature, $T=2{,}300$ tons: $$R_c = \varepsilon DT = 0.8\times3\times2{,}300 = 5{,}520\ \text{lb}$$
  7. 5.2 — total resistance and verdict. Inherent resistance at 45 mph is unchanged by grade/curvature (same $V$), so: $$R_{grade} = R_{level}+R_g+R_c = 18{,}587+92{,}000+5{,}520 = \boxed{116{,}107\ \text{lb}}$$ Since $M=103{,}750\ \text{lb} < R_{grade}=116{,}107\ \text{lb}$ (a shortfall of $12{,}357\ \text{lb}$), no -- the train cannot maintain 45 mph on the 2% grade with the 3° curve; it would decelerate below 45 mph (where the same propulsion formula returns a larger $M$, since $M\propto1/V$) until propulsive force and resistance again balance at some lower speed.
Question 5 — final results
SectionAvailable force $M$Total resistanceCan hold 45 mph?
5.1 Straight & level$103{,}750\ \text{lb}$$18{,}587\ \text{lb}$Yes (margin $+85{,}163\ \text{lb}$)
5.2 2% grade, 3° curve$103{,}750\ \text{lb}$$116{,}107\ \text{lb}$No (shortfall $-12{,}357\ \text{lb}$)