Chapter 2: Load Analysis

Source note: Source is Juvinall and Marshek, Fundamentals of Machine Component Design, 5th ed., Chapter 2 (pp. 45 to 88). The project copy of the book is a plain-text extraction, so figures are not available and some equation layouts are scrambled. Numbers below come from the book’s text and figure labels, cross-checked by equilibrium. Worked examples: Sample Problems 2.1 to 2.10 are included in short form. End-of-chapter problems are skipped. The vectorial treatment of equilibrium (Appendix G) and Castigliano’s method for redundant reactions (§5.9) are referenced by the chapter but not covered here. Where a value is rebuilt from figure labels rather than stated in the text, the guide says so. Added background is in blockquotes marked as such.

0. Overview

Chapter 2 is about getting the loads right before any stress analysis. A sophisticated stress or deflection analysis is of little value if it starts from wrong loads (§2.1). The chapter covers equilibrium and free-body diagrams at every level (whole machine, subassembly, single part, internal section), beam load, shear, and moment diagrams, the force flow concept for finding critical sections, load division between redundant supports, and failure of redundant ductile structures.

Core definitions and vocabulary

• Equilibrium equations: $\sum F = 0$, $\sum M = 0$ for a nonaccelerating body (2.1). $\sum F = ma$, $\sum M = I\alpha$ for an accelerating body (2.2) (§2.2).
• Free-body diagram: a member, or a portion of it cut at a section, drawn in equilibrium under all loads external to it. At a cut, these include the internal loads applied by the removed portion (§2.2, SP 2.5, Fig. 2.12 comment).
• Internal loads: forces and moments acting at a section inside a member, found from a free body of one side of the section (SP 2.5).
• Three-force member: member loaded by three forces. For equilibrium their lines of action meet at one point (SP 2.6).
• Torque ratio $R$: $T_{out}/T_{in}$. Equal to the speed ratio $\omega_{in}/\omega_{out}$ only without friction losses (SP 2.3, SP 2.4).
• Beam loading: lateral loading of members that are long compared with their cross-section. Torsion, axial load, or both may also be present (§2.3).
• Positive shear $V$: the portion left of the section tends to move up relative to the portion right of it (§2.3, Fig. 2.10a).
• Positive bending moment $M$: top of a horizontal beam in compression, bottom in tension. The beam “smiles” (§2.3).
• Critical section: section subjected to the most severe loading (§2.4, SP 2.8).
• Force flow concept: follow the “lines of force” (approximate force paths found by inspection) through the parts and mark each section suspected of being critical (§2.4).
• Double shear: a pin loaded across two surfaces in parallel, each carrying half the load (SP 2.9).
• Hoop tension: circumferential tensile loading around a hole caused by distributed bearing loads (SP 2.9, Fig. 2.15).
• Redundant support: a support that could be removed while the member stays in equilibrium (§2.5).
• Spring rate (stiffness): force per unit deflection. For an axial bar, $AE/L$ (§2.5).
• Tangent modulus: slope of the stress-strain diagram at a particular stress level (§2.6, footnote 2).

Key relations (all in book)

• Equilibrium: (2.1), (2.2), on each of three mutually perpendicular axes (§2.2).
• Beam rules: shear = slope of moment diagram, load intensity = slope of shear diagram, change in shear = area under load diagram, change in moment = area under shear diagram (§2.3).
• Redundant supports share load in proportion to their stiffness (§2.5).
• Ductile redundant structures yield generally only when all parallel paths reach their yield strengths (§2.6).

1. Introduction: Why Loads Come First (§2.1)

Physical picture: machine and structural components carry loads, so load analysis is the first step. A component cannot be satisfactory unless its design is based on realistic operating loads (§2.1).

Situation	How loads are found	Book example
Known operating torque and speed	Readily determined	Engines, compressors, electric generators
Loads depend on environment and operator	Difficult to determine	Automotive chassis (road surface, driving practice), airplane structure (turbulence, pilot decisions)
Uncertain service loads	Experimental methods give a statistical definition of applied loads	Not specified
Service failures available	Failure records plus strength analysis infer reasonable load estimates	Not specified

2. Equilibrium Equations and Free-Body Diagrams (§2.2)

Physical picture: once some applied loads are known or estimated, equilibrium gives the loads at every other point. The same method works at every level: a whole automobile, a transmission inside it, a countershaft inside the transmission, and an internal section of a member (§2.2, SP 2.4 comment 1).

Governing equations

Nonaccelerating body:

$$\sum F = 0 \quad \text{and} \quad \sum M = 0 \tag{2.1}$$

Accelerating body:

$$\sum F = ma \quad \text{and} \quad \sum M = I\alpha \tag{2.2}$$

$F$ force, $M$ moment, $m$ mass, $a$ linear acceleration, $I$ mass moment of inertia, $\alpha$ angular acceleration.

Assumptions and validity range

• The equations apply to each of any three mutually perpendicular axes (X, Y, Z). Many problems involve only one or two axes (§2.2).
• In SP 2.5 the book assumes deflections do not change the geometry significantly. Deflection of the member in Fig. 2.5a would reduce the moment $aF$, usually insignificantly.

Solution workflow (FBD method)

1. Isolate the body (or the part on one side of a cut) and draw every external load on it.
2. Use the action-reaction principle at every interface (force of A on B equals and opposes force of B on A).
3. Take moments about a point that removes unknowns (book choices: rear tire contacts, bearing I, the pivot pin, the intersection point of two lines of action).
4. Apply force sums for the remaining unknowns.
5. Check that every free body, including housings and frames, is balanced in forces and moments.

Sample Problem 2.1, Car at Constant Speed (pp. 46 to 47)

Given: 3000-lb car, 60 mph, aerodynamic drag 16 hp. Wheelbase 100 in, CG 50 in ahead of the rear axle and 20 in high, center of pressure (CP) 25 in high (Fig. 2.1). Find ground reactions on front and rear wheels.

Assumptions: constant speed, rear-wheel drive, no vertical aerodynamic forces, no tire rolling resistance.

1. Drag force: $\text{hp} = F_d V/33{,}000$ with $V$ in ft/min. $16 = F_d(5280)/33{,}000$, so $F_d = 100$ lb.
2. $\sum F = 0$ along the motion: road thrust on the tires $F_t = 100$ lb forward, split between the rear wheels.
3. Moments about the rear tire contacts: $(3000)(50) - (100)(25) - W_f(100) = 0$, so $W_f = 1475$ lb.
4. Vertical forces: $W_r = 3000 - 1475 = 1525$ lb.

Comments: parked, $W_f = W_r = 1500$ lb. At 60 mph the drag and thrust form a front-lifting couple of $100 \times 25$ lb·in, balanced by 25 lb shifted to the rear wheels. Vertical aerodynamic forces matter at high speed (spoilers, wings). Thrust is not $\mu W$ on the driving wheels, but it cannot exceed it. Traction holds while $\mu \ge 100/1525 = 0.066$.

Sample Problem 2.2, Car Under Acceleration (pp. 47 to 48)

Given: same car at 60 mph, full throttle, 96 hp. Find wheel reactions and acceleration.

Assumptions: rotational inertia of wheels, flywheel, and other rotating parts is equivalent to 7 percent more car weight (high gear). Rear wheels develop the needed traction.

1. 16 hp maintains speed, so 80 hp is available for acceleration.
2. Only 100/107 of it accelerates the car linearly: $80(100/107) = 74.8$ hp.
3. By proportion with 16 hp giving 100 lb: extra thrust $= 467.5$ lb. Total $F_t = 567.5$ lb.
4. From Eq. (2.2): $a = F/m = Fg/W = (467.5)(32.2)/3000 = 5.0$ ft/s².
5. The 467.5-lb inertia force acts rearward at the CG and shifts 93.5 lb from front to rear (Fig. 2.2): $W_f = 1381.5$ lb, $W_r = 1618.5$ lb.

Check of step 5 (the book leaves it to the reader): $467.5 \times 20/100 = 93.5$ lb.

Comment: traction holds while $\mu \ge 567.5/1617 = 0.351$ (the book’s denominator is 1617 lb, Fig. 2.2 shows 1618.5 lb, same result).

Sample Problem 2.3, Power Train Components (pp. 48 to 49)

Given: engine delivers torque $T$, transmission torque ratio $R$ (Fig. 2.3, one line: front-engine, rear-drive engine, transmission, and propeller shaft with moments about the X axis). Find loads, excluding gravity.

Assumptions: engine supported at two points, component weights neglected, transmission friction neglected.

Member	Torques acting on it
Transmission	$T$ in from the engine, reaction $RT$ from the propeller shaft, so the engine structure must apply $RT - T$ to the housing
Engine	$T$ and $RT - T$ from the transmission (action-reaction), so the frame applies $RT$ through the engine mounts
Propeller shaft	Equal and opposite torques $RT$ at its two ends

Sample Problem 2.4, Transmission Components (pp. 50 to 52)

Given: $T = 3000$ lb·in input, low gear $R = T_{out}/T_{in} = 2.778$ (Fig. 2.4, one line: (a) whole transmission, (b) mainshaft, (c) countershaft, (d) housing). Gears A and C are 2¼ in diameter, B and D are 3¾ in. Input gear A drives countershaft gear B, countershaft gear C drives output gear D. Mainshaft on bearings I and II, countershaft on III and IV.

Assumptions: tooth forces tangential only (radial and axial components neglected, covered in Ch. 15 and 16). Steady input and output torques.

1. Whole transmission (Fig. 2.4a): output $RT = 8333$ lb·in, so the housing must react $RT - T = 5333$ lb·in. This holds for any transmission of $R = 2.778$, including hydraulic or electric ones. A design with no path for this reaction torque to the housing cannot work.
2. Gear A force: $3000/1.125 = 2667$ lb. Countershaft torque: $2667 \times 1.875 = 5000$ lb·in. Gear C force: $5000/1.125 = 4444$ lb.
3. Output torque: $4444 \times 1.875 = 8333$ lb·in. Check by diameter ratios: $3000 \times (3\tfrac{3}{4}/2\tfrac{1}{4}) \times (3\tfrac{3}{4}/2\tfrac{1}{4}) = 8333$ lb·in.
4. Bearing II, moments about bearing I: $(2667)(2) - (4444)(7) + F_{II}(9) = 0$, so $F_{II} = 2864$ lb.
5. Bearing I from $\sum F = 0$ (or $\sum M_{II} = 0$): 1087 lb. Countershaft bearing reactions follow the same way.
6. Housing (Fig. 2.4d): loaded only by the four bearings and the bolts to the engine. It is a free body in equilibrium.

Note: with friction, the speed ratio $\omega_{in}/\omega_{out}$ is slightly greater than the torque ratio (SP 2.4 statement).

Sample Problem 2.5, Internal Loads (pp. 52 to 53)

Cut each member at section AA and take one side as a free body.

Member	Loads at section AA
Fig. 2.5, force $F$ applied at offset $a$ from the section	Force $F$ and bending moment $Fa$
Fig. 2.6, bracket from a fixed support, arms $a$ and $b$	Shear force $F$, bending moment $Fa$, torque $Fb$

Assumption: deflections do not change the geometry significantly.

Sample Problem 2.6, Three-Force Member, Bell Crank (pp. 54 to 55)

Given: bell crank (link 2) pivots on fixed frame (link 1). Horizontal rod (link 3) applies $F_{32} = 40$ lb at the top. Rod at 30° (link 4) applies $F_{42}$ at the bottom. Subscript convention: $F_{32}$ is the force applied by link 3 to link 2. Find $F_{42}$ and $F_{12}$.

Assumptions: frictionless pins, no acceleration.

Analytical:

1. Moments about the pivot: $40 \times 1\tfrac{1}{2} = F_{42} \times 1$, so $F_{42} = 60$ lb.
2. $V_{12} = 60 \sin 30° = 30$ lb. $H_{12} = 40 + 60\cos 30° = 92$ lb.
3. $F_{12} = \sqrt{30^2 + 92^2} = 97$ lb, upward and to the right at $\tan^{-1}(30/92) = 18°$.

Graphical:

1. Moments about point 0 (intersection of the two known lines of action) must vanish, so $F_{12}$ must also pass through 0.
2. Draw $F_{32}$ to scale, draw the $F_{12}$ direction through one end and the $F_{42}$ direction through the other, and scale off the magnitudes (force polygon, Fig. 2.8).

Sample Problem 2.7, Human Finger as Three-Force Member (pp. 55 to 56)

Given: 10-lb pinch at the fingertip, moment arm 3 in about pivot 0. Tendon force $F_t$ acts at 0.5 in (Fig. 2.9).

Assumptions: load carried only by tendon and bone, no acceleration, finger weight neglected.

1. Tendon arm is one-sixth of the pinch arm, so $F_t = 60$ lb.
2. Force polygon: bone compression (proximal phalanx on metacarpal) about 55 lb, enough to crush deteriorated arthritic bone.

Decision rules

Situation	Method	Why
Unknowns at several points	Moments about a point that eliminates the most unknowns	One equation, one unknown
Three-force member, one force known fully, one in direction only	Analytical (3 equations, 3 unknowns) or graphical force polygon	Concurrency fixes the third line of action
Load inside a member	Cut at the section, free body of one side	Internal loads become external to the free body
Device with internal parts	FBD of the whole unit first	Equilibrium of the whole is independent of what is inside (SP 2.4)

3. Beam Loading (§2.3)

Physical picture: a beam carries lateral loads. Each case uses three diagrams: external loads, internal transverse shear $V$, and internal bending moment $M$. Reactions come first from $\sum M = 0$ and $\sum F = 0$, with a distributed load $w$ over length $b$ replaced by a concentrated load $wb$ at the middle of $b$ (§2.3).

Sign convention (recommended, arbitrary) (§2.3)

• Draw the shear diagram from left to right, following the direction of the applied loads. An upward load steps $V$ up, a downward load steps it down.
• The shear diagram must return to zero at the right-end reaction.
• Positive $V$: left portion tends to move up relative to the right portion.
• Positive $M$: top fiber in compression, bottom in tension (“smile”).

The four load-shear-moment rules (§2.3)

1. Shear $V$ at any point = slope of the moment diagram: $V = dM/dx$.
2. Local load intensity at any point = slope of the shear diagram. Point supports give a theoretically infinite force intensity, so the shear diagram is vertical there.
3. Change in shear between two points = area under the load diagram between them.
4. Change in moment between two points = area under the shear diagram between them.

Per-case formulas (Fig. 2.10)

Case	Reactions	Maximum moment
(a) Concentrated $F$ at distance $a$ from left support, $b$ from right, span $L$	$R_1 = Fb/L$, $R_2 = Fa/L$	$Fab/L$ under the load
(b) Uniform $w$ over length $b$ ending at one support, starting $a$ from the other ($L = a + b$)	Far support: $wb^2/2L$. Support at loaded end: $wb(a + b/2)/L$	$\dfrac{wb^2}{2L}\left(a + \dfrac{b^2}{4L}\right)$ at zero shear, $b^2/2L$ into the loaded length

Case (b) moment and zero-shear location are rebuilt from the Fig. 2.10b labels. The book asks the reader to verify all Fig. 2.10 expressions as a review exercise.

Sample Problem 2.8, Transmission Countershaft (pp. 58 to 60)

Given: countershaft of SP 2.4 (Fig. 2.11). Gear loads 2667 lb at B and 4444 lb at C, bearing reactions 1087 lb and 2864 lb. Spacing 2 in, 5 in, 2 in along the shaft. Find the critical section and its loads.

Assumptions: uniform rotation. Transverse shear stresses negligible compared with bending and torsional shear stresses.

1. Load diagram is in equilibrium in the plane of the paper.
2. Shear diagram values: 1087 lb, 1580 lb, 2864 lb. Moment diagram peaks: 2174 lb·in at B and 5728 lb·in at C, of opposite sign.
3. Torque: $(4444)(2.25/2) = (2667)(3.75/2) = 5000$ lb·in, acting only between the gears. Zero outboard of the gears (bearing friction neglected).
4. Critical section: just beside gear C on the side toward gear B, where maximum torque meets essentially maximum bending.
5. Loads there (Fig. 2.12): $T = 5000$ lb·in, $M = 5728$ lb·in (actually slightly less, depending on gear C width), $V = 1580$ lb.

Comment: the left portion of the shaft is a free body in equilibrium under the external loads plus the internal loads from the right portion. Transverse shear is unimportant compared with bending except for extremely short shafts.

Limiting cases and sanity checks

• Shear diagram closes to zero at the last support. Moment diagram is zero at simple supports.
• Moment peaks where shear crosses zero (rule 1).
• Torque is zero outboard of the torque-transmitting elements.

4. Locating Critical Sections: Force Flow Concept (§2.4)

Physical picture: in simple members the critical section is obvious. In complex parts several sections may be critical. Follow the approximate force paths through each part and flag every bottleneck, turn, and contact surface on the way (§2.4).

Sample Problem 2.9, Yoke Connection (pp. 60 to 62)

Given: fork (two prongs, thickness $b$ each), blade (thickness $2b$), pin, loaded in tension by $F$ (Fig. 2.13). Find critical sections (Fig. 2.14, one line: dashed force flow lines with circled critical areas 1 to 7 along fork, pin, and blade).

Assumptions: weight ignored. Load splits equally between prongs and equally between the two sides of each hole. Distributed loads shown as concentrated. Deflections do not change load distribution. Snug pin fit.

Area	Location	Loading	Note
1	Fork body	Tension	First section on the path
2	Fork across the holes	Tension on reduced area	$F$ splits into four identical paths, each of area $\tfrac{1}{2}(m - a)b$
3	Fork end beyond the hole	Shear	Turning flow tends to “push out” the end segments
4, 4’	Fork-hole and pin surfaces	Bearing	Equal bearing loads also at pin and blade-hole surfaces
5	Pin, two surfaces	Direct shear, double shear	Each surface carries $F/2$
6	Pin center	Maximum bending moment	Bearing loads load the pin as a beam
4’, 3, 2, 1	Blade	Same as fork	Flow leaves the pin through like-numbered blade sections
7	Around the hole	Hoop tension	Bearing load is distributed over the hole surface (Fig. 2.15)

Caveats on the simple assumptions (SP 2.9 comments)

• Section 2 is not in uniform tension. Bending adds tension at the hole surface and subtracts at the outer surface.
• Bearing pressure on 4 and 4’ is not uniform. It depends on pin fit and member rigidity. Pin bending puts the highest bearing load near the fork-blade interfaces, and a close fit restrains pin bending.

Three needs this example illustrates (§2.4)

1. Make reasonable simplifying assumptions and get usable answers quickly.
2. Remember those assumptions and interpret results accordingly.
3. Judge whether the simple solution is adequate or a more advanced analysis or experiment is justified.

5. Load Division Between Redundant Supports (§2.5)

Physical picture: with redundant supports there are more unknowns than equilibrium equations, so statics alone cannot give the support loads. Each redundant support adds stiffness and carries load in proportion to its stiffness (§2.5).

Governing relations

For supports deflecting together by $\delta$:

$$P_i = k_i \delta, \qquad \delta = \frac{P}{\sum k_i}, \qquad P_i = \frac{k_i}{\sum k_i} P$$

Axial bar stiffness: $k = AE/L$.

The book states the proportionality in words and works it with numbers. The symbolic form above is a restatement of that example.

Book example, weight on two springs (Fig. 2.17)

1. 100-lb weight, top spring in tension $k_1 = 10$ lb/in, bottom spring in compression $k_2 = 40$ lb/in, both stress-free without the load.
2. Both deflect by $\delta$: $\sum F = 0$ gives $10\delta + 40\delta = 100$, $\delta = 2$ in.
3. Spring forces: 20 lb (top) and 80 lb (bottom), in proportion to stiffness.

Adding a stiff wire (same example)

1. Steel wire inside the top spring: 10 in long, 0.020 in diameter, $A = 0.000314$ in².
2. Wire rate $AE/L \approx 942$ lb/in. Total stiffness 992 lb/in.
3. Wire load: $(942/992)(100) = 95$ lb. Stress $P/A = 95/0.000314 \approx 303{,}000$ psi, above its probable tensile strength, so it breaks or yields.

Rule: make the strengths of redundant load-carrying members roughly proportional to their stiffnesses (§2.5).

Angle iron with welded web (Fig. 2.18)

• A welded triangular web reduces angular deflection but may add stiffness far out of proportion to its strength.
• Cracks at the weld remove the added stiffness and can propagate into the main angle, so the “reinforcement” weakens the part.
• Converse fix: removing stiff but weak portions (thin webs in a casting) can cure failures, if the remaining material can carry the extra load and the extra deflection is acceptable.

Methods for redundant reactions

Case	Method	Where
Completely elastic system	Castigliano’s method	§5.9
Ductile failure of redundant supports	Force flow with load redistribution	§2.6

6. Force Flow Concept Applied to Redundant Ductile Structures (§2.6)

Physical picture: parallel redundant paths share load by stiffness. What happens near failure depends on ductility (§2.6).

Path material	Behavior as load rises	Why	Failure of whole structure
Brittle	One path fractures first and dumps its load on the others, which then fracture in turn	No yielding to relieve the stiffest path	Cascade to failure of all paths
Ductile	One path yields first, its stiffness drops to the tangent modulus, and load moves to other paths	Yielding lowers stiffness of the overloaded path	General yielding only when all parallel paths reach their yield strengths

Sample Problem 2.10, Triple-Riveted Butt Joint (pp. 64 to 67)

Given: two plates butted together, top and bottom straps each two-thirds of plate thickness, same material. Three rows of rivets per plate (outer, middle, inner), pattern repeats every pitch (Fig. 2.19, one line: (a) whole joint, (b) to (f) force flow through plate, rivets, straps, and the complete flow diagram).

Assumptions: weight ignored, load even across the width (no misalignment), snug rivet fit.

1. The structure is redundant: any one row could satisfy statics.
2. Plate force flow: the outer row section (one hole per pitch) carries the full load. The middle row section (two holes) carries all load not taken by the outer row. The inner row section carries only the inner row’s share.
3. Each rivet path crosses five critical sections in series: reduced plate tensile area, rivet bearing on plate, rivet shear, rivet bearing on strap, reduced strap tensile area.

Critical area	Location	Why critical
1	Plate, reduced section at outer row	All three paths cross it. Failure severs every path
2	Plate, reduced section at middle row	Two paths cross it, on less area than at 1
(none)	Plate at inner row	One path on an area equal to that at 2, so no failure possible
3	Strap, reduced section at outer row	Only place strap tension can fail, given plate and strap thicknesses
4 to 9	Rivets	Each rivet fails in shear or at its weaker bearing area. Outer row: single shear, strap bearing (strap thinner than plate). Middle and inner rows: double shear, plate bearing (two straps together are thicker than the plate)

4. Ductile material lets slight local yielding redistribute load, so the joint fails only when the load exceeds the combined capacity of all three paths. Three ways:
   • (a) Tensile failure at 1.
   • (b) Simultaneous failure of the weakest link in each of the three paths.
   • (c) Simultaneous failure at 2 and at the weakest link of the outer path (3, 4, or 5).

7. Common Exam Traps

• “Wheel thrust equals the coefficient of friction times the weight on the driving wheels.” False. Thrust comes from equilibrium. $\mu W$ is only the upper limit for traction (SP 2.1 comment 2).
• “Front and rear axle loads at speed equal the parked loads.” False. Drag at CP height shifts 25 lb rearward at 60 mph, and acceleration shifts another 93.5 lb (SP 2.1, SP 2.2).
• “All extra engine power goes into linear acceleration.” False. Rotating parts act like 7 percent more weight in high gear, so only 100/107 of the excess power accelerates the car (SP 2.2).
• “Vertical aerodynamic forces can always be ignored.” Only in the simplified analysis. They matter at high speed (spoilers, wings) (SP 2.1 comment 1).
• “A transmission works if its gearing is right.” It must also pass the reaction torque $RT - T$ to the housing. The whole-unit FBD is independent of the internals (SP 2.4).
• “The engine mounts react the engine torque $T$.” False. The frame applies $RT$ to the engine through the mounts (SP 2.3).
• “Torque ratio equals speed ratio.” Only without friction. With friction losses the speed ratio is slightly greater (SP 2.4).
• “Gear tooth forces are purely tangential.” A simplification. Radial and axial components exist (Ch. 15 and 16) (SP 2.4).
• “The three forces on a three-force member can act anywhere.” False. Their lines of action must meet at one point (SP 2.6).
• “The critical section of a shaft is where shear is largest.” False. It is where maximum torque meets near-maximum bending. Transverse shear matters only for extremely short shafts (SP 2.8).
• “Torque acts along the whole shaft.” False. It is zero outboard of the gears when bearing friction is neglected (SP 2.8).
• “Section 2 of a yoke is in uniform tension and bearing pressure on the pin is uniform.” Both are simplifications. Bending adds tension at the hole surface, and bearing load peaks near the fork-blade interfaces (SP 2.9 comments).
• “The yoke pin is in single shear with load $F$.” False. It is in double shear, $F/2$ on each surface (SP 2.9).
• “Redundant support loads follow from statics.” False. More unknowns than equations. Load divides in proportion to stiffness (§2.5).
• “Adding a stiff member always strengthens a structure.” False. The stiff wire took 95 lb and failed at about 303,000 psi. A welded web can crack and weaken an angle iron. Strength should be proportional to stiffness (§2.5).
• “A ductile redundant structure fails when its first path yields.” False. General yielding needs all parallel paths at yield strength. Brittle paths cascade instead (§2.6).
• “A riveted joint fails at its single weakest section.” Ductile joints redistribute load. Failure needs all three paths to go: tension at 1, all three weakest links, or 2 plus the outer-path weak link (SP 2.10).
• “Member deflection never changes the moment arm.” It is an assumption. Deflection reduces $aF$ in Fig. 2.5a, usually insignificantly (SP 2.5).

8. Quick-Reference Summary

Start every load problem with a free-body diagram of the largest unit, then work inward: subassembly, part, internal section. Use (2.1) for steady operation and (2.2) when the body accelerates, with $a = Fg/W$ in gravitational units. Convert power to force with hp $= FV/33{,}000$ (V in ft/min) or to torque with Ch. 1 Eqs. (1.2), (1.3). For three-force members use concurrency. For shafts and beams draw load, shear, moment, and torque diagrams and look for the section where torque and bending peak together. For complex parts follow the force flow and list every tension, shear, bearing, bending, and hoop-tension area. When supports are redundant, statics is not enough: split load by stiffness for elastic supports (Castigliano in §5.9 for general elastic cases), and for ductile structures check combined path capacity.

Workflow A, vehicle axle loads (SP 2.1, SP 2.2)

1. Drag force from drag power: $F_d = 33{,}000\,\text{hp}/V$ (V in ft/min).
2. Thrust from $\sum F$ along the motion (add inertia force $ma$ when accelerating).
3. Front axle load from moments about the rear tire contacts (drag at CP height, inertia at CG height).
4. Rear axle load from vertical $\sum F$.
5. Traction check: $\mu \ge F_t / W_{drive}$.

Workflow B, acceleration from excess power (SP 2.2)

1. Excess power = available power minus road-load power.
2. Scale by 100/107 for rotational inertia (high gear).
3. Extra thrust by proportion with the known thrust-power pair at that speed.
4. $a = F g / W$.

Workflow C, gearbox loads (SP 2.3, SP 2.4)

1. Whole unit: output torque $RT$, housing reaction $RT - T$.
2. Tooth force = torque / pitch radius at each mesh.
3. Shaft torque between gears from the driven gear’s force times its radius.
4. Bearing reactions: moments about one bearing, then $\sum F$.
5. Check that the housing is a balanced free body.

Workflow D, shaft critical section (SP 2.8)

1. Load diagram in equilibrium.
2. Shear diagram left to right, closing at zero.
3. Moment diagram by shear areas.
4. Torque diagram between torque-transmitting elements.
5. Pick the section with max torque and near-max moment. Record $T$, $M$, $V$.

Workflow E, redundant elastic supports (§2.5)

1. Find each support stiffness ($k$, or $AE/L$ for a bar).
2. Common deflection $\delta = P/\sum k$.
3. Support loads $k_i\delta$.
4. Stress in each support, compare with its strength.
5. If one support is overloaded, match strengths to stiffnesses or remove the stiff weak member.

Key Takeaways

• Correct loads come before any stress analysis (§2.1).
• Free-body diagrams work at every level, from a whole car to an internal section (§2.2).
• A transmission must pass reaction torque $RT - T$ to its housing (SP 2.4).
• Three-force members have concurrent lines of action (SP 2.6).
• Shear is the slope of moment, load intensity the slope of shear, and diagram changes equal areas (§2.3).
• Shaft critical section: max torque together with near-max bending (SP 2.8).
• Redundant supports share load by stiffness, so strengths should follow stiffnesses (§2.5).
• Ductile redundant structures fail only when all paths reach capacity (§2.6).

Quick-Reference Table

Quantity	Formula	Units	When to use
Static equilibrium	$\sum F = 0$, $\sum M = 0$ (2.1)	lb, N, lb·in, N·m	Nonaccelerating body, each axis
Dynamic equilibrium	$\sum F = ma$, $\sum M = I\alpha$ (2.2)	lb, N, lb·in, N·m	Accelerating body
Acceleration from force	$a = F/m = Fg/W$	ft/s², m/s²	Weight given instead of mass (SP 2.2)
Force from power	$\text{hp} = FV/33{,}000$	hp, lb, ft/min	Drag or thrust from power (SP 2.1)
Traction limit	$\mu \ge F_t / W_{drive}$	dimensionless	Wheel slip check (SP 2.1, 2.2)
Transmission output torque	$RT$, housing reaction $RT - T$	lb·in, N·m	Any transmission, any internals (SP 2.3, 2.4)
Tooth force	$F = T / r$	lb, N	Tangential gear load (SP 2.4)
Two-stage gear output torque	$T_{in}(D_B/D_A)(D_D/D_C)$	lb·in, N·m	Countershaft gearbox (SP 2.4)
Beam, concentrated load reactions	$R_1 = Fb/L$, $R_2 = Fa/L$	lb, N	Fig. 2.10a
Beam, concentrated load max moment	$M_{max} = Fab/L$	lb·in, N·m	Fig. 2.10a
Beam, partial uniform load reactions	$wb^2/2L$ (far support), $wb(a + b/2)/L$ (loaded end)	lb, N	Fig. 2.10b
Shear-moment relation	$V = dM/dx$	lb, N	Rule 1, §2.3
Load-shear relation	load intensity = slope of $V$ diagram	lb/in, N/m	Rule 2, §2.3
Diagram areas	$\Delta V$ = load area, $\Delta M$ = shear area	lb, lb·in	Rules 3 and 4, §2.3
Bar stiffness	$k = AE/L$	lb/in, N/m	Redundant axial member (§2.5)
Load share, redundant supports	$P_i = (k_i/\sum k)P$	lb, N	Parallel elastic supports (§2.5)
Axial stress	$\sigma = P/A$	psi, Pa	Check each support (§2.5)
Pin in double shear	$F/2$ per shear surface	lb, N	Yoke and clevis pins (SP 2.9)