RRB JE Junior Engineer · Technical Abilities (CBT-2)
Mechanical Engineering (CBT-2)
Statics and moments, strength of materials, air-standard engines, pumps, production processes, and vapour-cycle bookkeeping at diploma depth for RRB JE CBT-2 Technical Abilities…
Six concepts. RRB JE CBT-2 Mechanical is the diploma paper a junior engineer meets in a railway diesel shed: keep a lifting beam from rotating, keep a drawbar inside its elastic line, read the loco's Diesel cycle, size a cooling-water pump, turn a journal and join a rail, then tell Rankine work from fridge heat. Each concept is one calculation family the paper repeats — forces and moments, stress and strain, air-standard engines, pump power, cutting speed and weld class, vapour-cycle bookkeeping — not a chapter dump.
- RRB JE Junior Engineer
- Medium level
- 6 concepts
- 48 practice questions
1Equilibrium of forces and moments
A rigid body is in equilibrium when it neither accelerates nor starts to spin. That is two independent statements: the vector sum of forces is zero, and the sum of moments about any point is zero. A force's turning effect is its moment, M = F \times d_{\perp}, where d_{\perp} is the perpendicular distance from the point to the line of action — not the distance along the member.
Reach for this on a simply supported beam, a jib, a concurrent-force joint, or a block about to slide. Do not use it once the body is accelerating: then \Sigma F = ma and \Sigma M = I\alpha. On a simply supported span L with a mid-span point load W, each support takes W/2 and the peak bending moment is WL/4 at mid-span. Impending sliding is F = \mu N, with N the normal reaction, not automatically the weight. The exam trap is treating equilibrium as two statements, not one: the forces close and the moments close.
Figure. Arrow lengths are to scale for the three forces: the 20 kN load is drawn twice as long as each 10 kN support. Beam thickness is schematic, not a cross-section. M_{\max} = WL/4 = 20 kN·m sits at mid-span.
How it works
- Close the forcesResolve every load and reaction into components. \Sigma F_x = 0 and \Sigma F_y = 0 fix the support magnitudes on a statically determinate beam.
- Close the momentsPick a convenient point — a support kills that reaction's moment — and write \Sigma M = 0. Moment is F times the perpendicular lever, never the slant length along the beam.
- Name the peakFor a mid-span point load on a simply supported span, M_{\max} = WL/4 at the load. WL/2 is the couple you would get if both W and L/2 were mistaken for the lever.
| Case | What closes | Watch |
|---|---|---|
| Concurrent forces | \Sigma F_x = 0, \Sigma F_y = 0 | No moment if lines meet at one point |
| Any rigid body | Those plus \Sigma M = 0 | Forces alone do not prevent spin |
| SS beam, mid-span W | R_A = R_B = W/2, M_{\max} = WL/4 | Not WL/2 |
| Impending slide | F = \mu N | N is the normal, not always mg |
Lifting beam in the shed
A 4 m simply supported lifting beam in the diesel shed carries a 20 kN point load at mid-span. Find each support reaction and the maximum bending moment.
- R_A = R_B = W/2 = 20/210 kN each
- M_max = WL/4 = 20 × 4 / 420 kN·m at mid-span
Pro tip. The two 10 kN reactions close the vertical forces. They do not by themselves close the moments: the 20 kN·m is what the beam must carry at mid-span. A cantilever with the same end load would have been M = WL = 80 kN·m — different support, different formula.
A 6 m simply supported beam carries a 24 kN point load at mid-span. What is the maximum bending moment, and why is 72 kN·m wrong?
- 36 kN·m — M_{\max} = WL/4 = 24 \times 6 / 4; 72 kN·m is WL/2
- 72 kN·m — each half-span is a 3 m lever on the full 24 kN
- 12 kN·m — that is also each support reaction, so it must be the moment
Mid-span point load on a simply supported span: M_{\max} = WL/4 = 36 kN·m, and each support is 12 kN. WL/2 = 72 kN·m uses the half-span as if the other support were absent. A reaction equal to 12 kN is a force, not a moment.
2Axial stress, strain and elasticity
Stress is how hard a load presses on the material that is actually there: axial stress \sigma = P/A is the axial force divided by the cross-section it acts on. Strain is how much the piece stretches relative to the length you started with: \varepsilon = \delta / L. Neither is 'the load' and neither carries a unit of force by itself — stress is N/mm^2 or MPa, strain is a pure number.
Young's modulus E = \sigma/\varepsilon = PL/(A\delta) is the slope of that line only while the material is elastic. Past yield the current stress-over-strain is not E. Use this on a tie rod, a hanger, a short column in pure compression. Do not use that elastic rearrangement after yield, and do not mix mm^2 in the area with metres in P without converting. Factor of safety is the named failure stress divided by the working stress — yield or ultimate, whichever the stem names. Young's modulus is the slope of the elastic line, not the current stress-over-strain after the material has yielded.
Figure. Equal-and-opposite 40 kN arrows, same drawn length, pull on a 200 mm² coupon. The 1.2 m length is to scale as a span; the 1.2 mm extension is written, not drawn — at this canvas it would be 0.1% of the rod and invisible. Not a stress–strain curve.
How it works
- Name P and AConvert the load to newtons and the area to mm^2 or m^2, not one of each. \sigma = P/A is then immediate; halving a diameter quarters the area and multiplies stress by four.
- Strain from stretch\varepsilon = \delta/L uses the original length. A 1.2 mm stretch on a 1.2 m rod is 0.001, not 1.2.
- E only while elastic\delta = PL/(AE) rearranges E = \sigma/\varepsilon. The formula is the elastic line. Past yield you need the actual curve, not this E.
| Quantity | Relation | Watch |
|---|---|---|
| Axial stress | \sigma = P/A | Halve diameter \Rightarrow 4\times stress |
| Shear stress | \tau = F/A | Same unit discipline as \sigma |
| Strain | \varepsilon = \delta/L | Dimensionless; original L |
| Young's modulus | E = \sigma/\varepsilon = PL/(A\delta) | Elastic range only |
| Bending | M/I = \sigma/y = E/R | Neutral-axis y, not the full depth |
| Factor of safety | \sigma_{\mathrm{fail}}/\sigma_{\mathrm{work}} | Name yield or ultimate |
Drawbar coupon
A 1.2 m mild-steel coupon of area 200 mm^2 carries 40 kN in tension. Take E = 200 GPa. Find the axial stress, the extension, and the strain.
- A = 200 mm² = 200 × 10⁻⁶ m²2.0 × 10⁻⁴ m²
- σ = P/A = 40000 / (2.0 × 10⁻⁴)2.0 × 10⁸ Pa = 200 MPa
- δ = σL/E = (2.0 × 10⁸ × 1.2) / (2.0 × 10¹¹)1.2 × 10⁻³ m = 1.2 mm
- ε = δ/L = 1.2 × 10⁻³ / 1.20.001
Pro tip. 200 MPa on 200 GPa is a strain of 0.001 by arithmetic, not coincidence: E = 200 GPa means 200 MPa produces \varepsilon = 0.001 on the elastic line. If the stem had said the coupon had yielded, this E would no longer give the stretch.
A mild-steel coupon is loaded past yield. A classmate computes E as the current stress divided by the current strain and uses that E in \delta = PL/(AE). What is wrong?
- Nothing — E is defined as \sigma/\varepsilon at every point on the curve
- Young's modulus is the slope of the elastic line; past yield the current stress-over-strain is not E, so \delta = PL/(AE) does not give the stretch
- Past yield the stress is zero, so the formula is undefined
E is the elastic slope. After yield the coupon's stress-over-strain is a different number and \delta = PL/(AE) no longer reconstructs the extension. Stress does not drop to zero at yield; it follows the plastic curve.
3Otto, Diesel and engine power
An internal-combustion engine is a periodic heat engine that burns fuel inside the working cylinder. The air-standard Otto cycle (spark-ignition, petrol) adds heat at constant volume; the air-standard Diesel cycle (compression-ignition) adds heat at constant pressure. Compression ratio r is the volume before compression divided by the volume after it — a geometric ratio, not a pressure ratio.
Otto's air-standard efficiency is \eta = 1 - 1/r^{\gamma-1} with \gamma = 1.4 for air. At the same compression ratio the Otto air-standard efficiency is higher than Diesel; a Diesel engine in the shed can still be the better plant because it runs a higher r than a spark engine can knock-free. Indicated power is the work at the piston; brake power is the work at the shaft; friction power is the difference; mechanical efficiency is BP/IP, never IP/BP.
Figure. Air-standard efficiencies at the same compression ratio r = 8 (\gamma = 1.4). Diesel bar uses cutoff \rho = 2. The figure compares two computed percentages; it is not a P–V diagram and it does not animate the piston.
How it works
- Name the heat-addConstant-volume heat addition is Otto (spark). Constant-pressure heat addition is Diesel (compression-ignition). The loco in the shed is Diesel; the formula still wants the cycle name, not the vehicle.
- Raise r to the γ−1For Otto, compute r^{\gamma-1} first. With r = 8 and \gamma = 1.4 that is 8^{0.4} = 2.297. Then \eta = 1 - 1/2.297.
- Split IP and BPIndicated power lives on the indicator card. Brake power is what a dynamometer reads at the shaft. IP = BP + FP and \eta_{\mathrm{mech}} = BP/IP.
| Name | Meaning | Watch |
|---|---|---|
| Otto | Heat add at constant V (SI) | Same r: Otto \eta beats Diesel |
| Diesel | Heat add at constant P (CI) | Same max P,T: Diesel can win |
| Otto \eta | 1 - 1/r^{\gamma-1} | \gamma = 1.4 for air |
| Diesel \eta at r=8, \rho=2 | 49.04% | Cutoff \rho = V_3/V_2 |
| Mechanical \eta | BP/IP | Not IP/BP; IP = BP+FP |
Otto air-standard at r = 8
An air-standard Otto cycle has compression ratio 8. Take \gamma = 1.4. Find the air-standard efficiency.
- r^{γ−1} = 8^{0.4}2.297
- 1 / 2.2970.4353
- η = 1 − 0.43530.5647 = 56.47%
Pro tip. The same r = 8 with Diesel cutoff \rho = 2 drops the air-standard efficiency to 49.04%. That is why 'Diesel is more efficient' is false as a same-r claim and only sometimes true as a same-peak-pressure claim. The shed's loco wins on the r it can actually run, not on the cycle name.
Two air-standard cycles share compression ratio 8. One is Otto, one is Diesel with cutoff 2. Which efficiency statement is true?
- Diesel is higher because compression-ignition engines are always more efficient
- Otto is higher at the same r — 56.47% against 49.04% at cutoff 2; Diesel can beat Otto only under a different constraint, such as the same peak pressure
- They are equal because \gamma and r are the same
Otto \eta = 1 - 1/r^{\gamma-1} depends only on r. Diesel pays an extra cutoff factor (\rho^\gamma-1)/(\gamma(\rho-1)), which at \rho=2 pulls 56.47% down to 49.04%. Equal \gamma and r do not equalise the cycles. 'Diesel always wins' confuses the plant with the same-r comparison.
4Pump power and turbines
A pump adds mechanical energy to a liquid so the liquid can climb a head H; a turbine does the reverse and takes energy out. The power that actually appears in the fluid is the hydraulic power \rho g Q H: density times g times volume flow rate times the head the fluid gains (pump) or loses (turbine). Q is cubic metres per second, not litres per minute, before it enters that product.
A pump's shaft power is larger than the hydraulic power because efficiency sits in the denominator: P_{\mathrm{shaft}} = \rho g Q H / \eta. A turbine's shaft power is smaller than \rho g Q H because efficiency sits in the numerator. Specific speed N_s = N\sqrt{Q}/H^{3/4} (pump) classifies the machine — radial, mixed, axial — it is not a rotational speed you set on the coupling. Cavitation begins when local pressure falls to the liquid's vapour pressure, not when the pump is merely 'running fast'.
Figure. Same pump as the worked example: hydraulic 14.715 kW versus shaft 19.62 kW at 75% efficiency. The dashed rule sits on the hydraulic value. Shaft is taller because \eta divides. Not a flow animation.
How it works
- Put Q in SILitres per second divide by 1000 to get m^3/s. Head H is metres of fluid. \rho = 1000 kg/m^3 for cold water unless the stem says otherwise; g = 9.81 m/s^2.
- Hydraulic first\rho g Q H is the power in the water. For Q = 0.06 m^3/s and H = 25 m that is 14.715 kW, before any efficiency.
- Then divide or multiplyPump: divide by \eta to get shaft input. Turbine: multiply by \eta to get shaft output. Multiplying a pump by \eta is the usual slip.
| Quantity | Relation | Watch |
|---|---|---|
| Hydraulic power | \rho g Q H | Q in m^3/s, g = 9.81 |
| Pump shaft | \rho g Q H / \eta | Larger than hydraulic |
| Turbine shaft | \rho g Q H \times \eta | Smaller than hydraulic |
| Pump specific speed | N\sqrt{Q}/H^{3/4} | Shape class, not rpm |
| Cavitation | Local p to vapour pressure | Not 'high rpm' by itself |
Cooling-water pump
A centrifugal pump delivers 0.06 m^3/s of water against a 25 m head. Overall efficiency is 75%. Take \rho = 1000 kg/m^3 and g = 9.81 m/s^2. Find the hydraulic power and the shaft power.
- ρgQH = 1000 × 9.81 × 0.06 × 2514715 W = 14.715 kW
- P_shaft = 14.715 / 0.7519.62 kW
Pro tip. 14.715 / 0.75 is 19.62, not 11.04. The 11.04 kW figure is 14.715 \times 0.75 — that is a turbine shaft, or a pump someone treated as a turbine. If the stem had given 60 L/s, the first job was still to write 0.06 m^3/s.
A pump lifts water at 0.05 m^3/s against 20 m head with overall efficiency 80%. g = 9.81 m/s^2. What shaft power does it draw, and why is 7.85 kW wrong?
- 9.81 kW — that is already \rho g Q H, so efficiency can be ignored
- 12.26 kW — \rho g Q H = 9.81 kW, then divide by 0.80; 7.85 kW is 9.81 × 0.80, the turbine slip
- 7.85 kW — efficiency reduces the power the motor must supply
\rho g Q H = 1000 \times 9.81 \times 0.05 \times 20 = 9.81 kW is hydraulic. Shaft input is 9.81/0.80 = 12.26 kW. Multiplying by 0.80 gives 7.85 kW, which would be a turbine output, not a pump input. Efficiency does not shrink the motor on a pump.
5Cutting speed and weld class
Turning removes a chip from a rotating job. Cutting speed V is the peripheral speed of the work surface, V = \pi DN/1000 metres per minute, with D the diameter in millimetres and N the spindle speed in revolutions per minute. Feed f is how far the tool advances per revolution; depth of cut d is how deep the tool bites radially. Material removal rate in turning is \mathrm{MRR} = \pi D N f d cubic millimetres per minute when D, f and d are in millimetres. Cutting speed is a peripheral speed in metres per minute, not the spindle rpm.
Joining is the other shop question. Shielded metal arc (SMAW) uses a consumable coated stick and leaves slag. MIG (GMAW) feeds a consumable wire under a shielding gas. TIG (GTAW) uses a non-consumable tungsten electrode; any filler is a separate rod. Thermit welding is aluminothermic — no electric arc — and is how running rails are joined in the field. The class is the electrode and the heat source, not the brand name on the set.
Figure. The rectangle stands in for the turned diameter — the vocabulary has no circle primitive. D and the peripheral-speed arrow are the teaching geometry; rotation itself is not drawn. Arrow length is schematic, not a scale speed.
How it works
- Peripheral speedV = \pi DN/1000 with D in mm and N in rpm gives m/min. Diploma papers take \pi = 3.14 unless they write 22/7. Leaving out the 1000 leaves an answer 1000 times too large.
- Then the chip rateOnce V is known, \mathrm{MRR} = \pi D N f d (mm³/min) uses the same D and N plus feed and depth in millimetres. It is a volume per minute, not a speed.
- Name the weld by the electrodeCoated stick and slag: SMAW. Continuous wire and gas: MIG. Non-consumable tungsten: TIG. Aluminium powder reducing iron oxide, no arc: thermit — the rail joint.
| Process | Electrode / heat | Exam peg |
|---|---|---|
| SMAW (MMAW) | Consumable coated stick | Slag; common shop weld |
| GMAW (MIG) | Consumable wire + gas | Continuous wire feed |
| GTAW (TIG) | Non-consumable tungsten | Filler is a separate rod |
| Thermit | Aluminothermic, no arc | Running-rail field joints |
Journal on the shed lathe
A journal 80 mm in diameter is turned at 250 rpm with feed 0.2 mm/rev and depth of cut 2 mm. Take \pi = 3.14. Find the cutting speed and the material removal rate.
- V = πDN/1000 = 3.14 × 80 × 250 / 100062.8 m/min
- MRR = πDNfd = 3.14 × 80 × 250 × 0.2 × 225120 mm³/min
Pro tip. 62.8 m/min is not 62.8 m/s and not 250 m/min. The 1000 in the cutting-speed formula is the millimetre-to-metre conversion riding with the per-minute clock; drop it and the figure becomes a workshop impossibility. MRR reuses \pi D N and then multiplies by feed and depth — it is not V converted into another unit.
A stem asks which process uses a non-consumable electrode. MIG is offered because 'it is also a gas-shielded arc'. Which process is correct, and why is MIG wrong?
- SMAW — the coating is not consumed, only the core wire
- TIG (GTAW) — the tungsten stays; filler, if used, is a separate rod. MIG's wire is consumable
- Thermit — there is no electrode at all, so it cannot be consumed
TIG's tungsten is non-consumable. MIG feeds a consumable wire; the gas is only the shield. SMAW's stick is consumable — coating and core. Thermit has no electrode, so it does not answer a stem that asks for a non-consumable electrode.
6Rankine plant and vapour-compression
Rankine and vapour-compression are the same four-station vapour loop pointed at opposite products. Rankine boils a liquid, expands it through a turbine, condenses it, and pumps it back: the product is net work, and the figure of merit is \eta = W_{\mathrm{net}}/Q_{\mathrm{in}}, which cannot exceed 1. Vapour-compression evaporates a refrigerant, compresses the vapour, condenses it, and drops the pressure in an expansion valve: the product is heat lifted at the evaporator, and the figure of merit is COP = Q_L/W.
COP can be larger than one because it is heat lifted per unit work, not work per unit heat. Diploma papers take 1 tonne of refrigeration as 210 kJ/min = 3.5 kW. A heat pump on the same VCR hardware has COP_{\mathrm{HP}} = Q_H/W = COP_{\mathrm{R}} + 1. Use Rankine \eta on a steam plant; use COP on a fridge or heat pump. Swapping the two is the paper's favourite thermal slip.
Figure. Topology only — four stations each, not a flowing cycle. Sage marks the product (turbine work; evaporator heat). Terracotta marks work you pay (Rankine pump; VCR compressor). Condensers dump heat in both loops and are not the product.
How it works
- Name the productSteam plant: net work W_{\mathrm{turbine}} - W_{\mathrm{pump}}. Fridge: heat lifted Q_L at the evaporator. The condenser dumps heat in both loops; dumping is not the product.
- Write the ratio that matchesRankine: \eta = W_{\mathrm{net}}/Q_{\mathrm{in}}. VCR: COP_{\mathrm{R}} = Q_L/W. Same numbers in the wrong ratio invent an 'efficiency' of 2.3 on a fridge or a COP of 0.28 on a turbine.
- Convert TR lastCapacity in TR is Q_L in kW divided by 3.5, or Q_L in kJ/min divided by 210. 210 kJ/min is exactly 3.5 kW. COP_{\mathrm{HP}} = COP_{\mathrm{R}} + 1 on the same machine.
| Station / merit | Rankine | VCR |
|---|---|---|
| Heat-in station | Boiler Q_{\mathrm{in}} | Evaporator Q_L |
| Work machine | Turbine (work out) | Compressor (work in) |
| Heat-out station | Condenser | Condenser Q_H |
| Fourth station | Pump (small work in) | Expansion valve (no work) |
| Figure of merit | \eta = W_{\mathrm{net}}/Q_{\mathrm{in}} < 1 | COP = Q_L/W (may exceed 1) |
| 1 TR | — | 210 kJ/min = 3.5 kW |
Pantry-car plant
A vapour-compression plant on a pantry car removes 210 kJ/min of heat. The compressor draws 1.5 kW. Using the diploma conversion 1 TR = 210 kJ/min = 3.5 kW, find Q_L in kW, the refrigerating COP, and the capacity in TR.
- Q_L = 210 kJ/min = 210 / 603.5 kW
- COP_R = Q_L / W = 3.5 / 1.52.333
- Capacity = 3.5 / 3.51 TR
Pro tip. 2.333 is allowed. Calling it an efficiency of 233% is the slip: \eta is work over heat-in and stays below 1; COP is heat-out-of-the-cabin over work. The same machine as a heat pump would show COP_{\mathrm{HP}} = 2.333 + 1 = 3.333.
A vapour-compression plant lifts 3.5 kW of heat with 1.0 kW of compressor work, so COP = 3.5. A classmate says a thermal efficiency cannot exceed 1, so the figure is impossible. What is the right reading?
- The classmate is right — report the plant as \eta = 1/3.5 = 0.286
- COP is Q_L/W, not W/Q_{\mathrm{in}}, so it can exceed 1; 3.5 is consistent and equals 1 TR per kilowatt of lift at the diploma conversion
- COP is Q_H/Q_L, which is 1 on any fridge, so 3.5 must be a unit error
COP is heat lifted per unit work. It is not an efficiency, so the 'cannot exceed 1' rule does not apply. 3.5 kW of lift is 1 TR; with 1 kW of work the COP is 3.5, not 1/3.5. Q_H/Q_L is not the definition of COP.
Notes
- Equilibrium is two statements: A rigid body is in equilibrium when the forces close, \Sigma F = 0, and the moments close, \Sigma M = 0. A moment is force times the perpendicular lever, never the distance along the member. A mid-span point load W on a simply supported span L gives supports of W/2 and a peak bending moment of WL/4.
- Stress, strain and the elastic slope: Axial stress is \sigma = P/A and strain is \varepsilon = \delta/L measured on the original length. Young’s modulus E = \sigma/\varepsilon is the slope of the elastic line only, so the rearrangement \delta = PL/(AE) does not survive yield. Halving a diameter quarters the area and quadruples the stress.
- Otto, Diesel and engine power: Otto adds heat at constant volume and Diesel at constant pressure, and compression ratio is a volume ratio rather than a pressure ratio. Otto’s air-standard efficiency is \eta = 1 - 1/r^{\gamma-1} with \gamma = 1.4 for air, so at the same compression ratio Otto beats Diesel. Indicated power minus brake power is friction power, and mechanical efficiency is brake over indicated.
- Pump and turbine power: Hydraulic power is \rho g Q H with Q in cubic metres per second. A pump’s shaft power divides by efficiency and is therefore larger, while a turbine’s multiplies by it and is smaller. Specific speed classifies the machine’s shape, and cavitation begins when the local pressure falls to the liquid’s vapour pressure, not when the pump merely runs fast.
- Cutting speed and weld class: Cutting speed is the peripheral speed V = \pi D N/1000 in metres per minute, not the spindle rpm, and turning removes \pi D N f d per minute. SMAW uses a consumable coated stick, MIG a consumable wire under gas and TIG a non-consumable tungsten electrode, while thermit welding is aluminothermic with no arc - the field joint for running rails.
- Rankine work against refrigeration heat: Both are four-station vapour loops pointed at opposite products. Rankine’s merit is \eta = W_{net}/Q_{in} and cannot exceed one, while vapour compression’s merit is COP = Q_{L}/W and can, because it is heat lifted per unit work. A heat pump on the same hardware has a COP one greater than the refrigerator’s, and one tonne of refrigeration is taken as 3.5 kW.
Formulas
- Equilibrium is \Sigma F = 0 and \Sigma M = 0; a moment uses the perpendicular lever. A mid-span load W on span L gives reactions W/2 and peak bending moment WL/4.
- Axial behaviour: \sigma = P/A, \varepsilon = \delta/L, E = \sigma/\varepsilon, and \delta = PL/(AE) only inside the elastic range.
- Otto air-standard efficiency \eta = 1 - 1/r^{\gamma-1} with \gamma = 1.4 for air. Indicated power - brake power = friction power; mechanical efficiency = brake / indicated.
- Hydraulic power = \rho g Q H with Q in m³/s. A pump's shaft power divides by efficiency (larger); a turbine's output multiplies by it (smaller). Cavitation starts at the liquid's vapour pressure.
- Machining: cutting speed V = \pi D N/1000 m/min, and turning removes \pi D N f d per minute. Welding electrodes: SMAW consumable coated, MIG consumable wire, TIG non-consumable tungsten, thermit no arc.
- Cycle merits: Rankine \eta = W_{net}/Q_{in} (never above 1) against refrigeration COP = Q_{L}/W (can exceed 1). Heat-pump COP = refrigerator COP + 1, and 1 tonne of refrigeration is 3.5 kW.
Exam traps & shortcuts
- A moment uses the perpendicular distance, never the length measured along the member.
- Strain is computed on the original length, and \delta = PL/(AE) stops being valid past yield.
- Compression ratio is a volume ratio. Treating it as a pressure ratio is the standard Otto-cycle error.
- Pump shaft power divides by efficiency and turbine output multiplies by it — the two corrections go opposite ways.
- Cutting speed is the peripheral speed in m/min, not the spindle rpm; check the units before substituting.
- A COP above 1 is not an efficiency violation: it is heat moved per unit of work, and a heat pump reads one higher than the refrigerator.
Reference tables
Every line is reconstructible from the concept that owns it. g = 9.81 m/s^2, \pi = 3.14, 1 TR = 210 kJ/min = 3.5 kW.
| Quantity | Relation | Watch |
|---|---|---|
| Moment | M = F d_{\perp} | Perpendicular lever |
| SS mid-span load | R = W/2, M_{\max} = WL/4 | Cantilever end load is WL |
| Axial stress | \sigma = P/A | Diameter halved \Rightarrow 4\times |
| Hooke (elastic) | \delta = PL/(AE) | Not past yield |
| Otto \eta | 1 - 1/r^{\gamma-1} | Same r: Otto > Diesel |
| Mechanical \eta | BP/IP | IP = BP + FP |
| Pump shaft | \rho g Q H / \eta | Turbine multiplies by \eta |
| Cutting speed | V = \pi DN/1000 m/min | D in mm, N in rpm |
| Turning MRR | \pi D N f d mm³/min | Not V in another unit |
| Rankine \eta | W_{\mathrm{net}}/Q_{\mathrm{in}} | Cannot exceed 1 |
| Fridge COP | Q_L/W | May exceed 1 |
| Heat-pump COP | COP_{\mathrm{R}} + 1 | Same hardware |
Recap
Read only this the night before.
- Statics
- Equilibrium is \Sigma F = 0 and \Sigma M = 0. Mid-span point load on a simply supported beam: M_{\max} = WL/4, not WL/2.
- SOM
- \sigma = P/A, \varepsilon = \delta/L, E is the elastic slope. Halve the diameter and stress quadruples. \delta = PL/(AE) dies at yield.
- Engines
- Otto adds heat at constant volume; Diesel at constant pressure. Same r: Otto \eta is higher. r=8, \gamma=1.4 \Rightarrow 56.47%. BP/IP is mechanical efficiency.
- Pumps
- Hydraulic \rho g Q H; pump shaft divides by \eta, turbine shaft multiplies. Cavitation is vapour pressure, not rpm. N_s is a shape class.
- Shop
- V = \pi DN/1000 m/min. TIG = non-consumable tungsten; MIG = consumable wire; thermit = rail joints, no arc.
- Vapour loops
- Rankine product is work (\eta < 1). VCR product is Q_L (COP may exceed 1). 1 TR = 210 kJ/min = 3.5 kW. Heat-pump COP is fridge COP plus one.
Practise Mechanical Engineering (CBT-2)
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