CELE Engineering Mechanics — Frames, Machines and CablesRevision Notes

Formulas

Exam Tips

• Start every frame problem by scanning for two-force members — they are usually straight members with pins only at both ends and no loads in between. Label them immediately.
• Choose your moment equation point wisely: a pin shared by two unknown force components eliminates both with one equation.
• Check your answer: the pin forces on a dismembered member must satisfy all three equilibrium equations independently.
• In multiple-choice items, if four unknowns appear at a single pin but only three equations are available per member, look for additional two-force members or use the whole-frame FBD first.
• Boards often ask for the 'force in strut CD' — this is the magnitude of the two-force member reaction, so remember to divide by the appropriate trig function.

Key Points

• A frame is a stationary structure designed to support loads; it contains at least one multi-force member (a member loaded at more than two points, or carrying an applied moment).
• Multi-force members carry shear, bending moment, and axial force — NOT axial force alone.
• A two-force member is loaded ONLY at two points with no transverse loads between them; its resultant force must act along the line joining the two points.
• Identify two-force members first — they reduce unknowns immediately by fixing the direction of the pin force.
• Step 1: Draw the FBD of the entire frame and find external reactions using ΣFx = 0, ΣFy = 0, ΣM = 0.
• Step 2: Dismember the frame — draw a separate FBD for each individual member.
• Step 3: At every shared pin, apply Newton's third law — the force on member A from member B is equal in magnitude and opposite in direction to the force on member B from member A.
• Step 4: Apply three equilibrium equations to each member and solve for pin forces.
• The total number of unknowns must equal the total number of equilibrium equations available.

Definitions

Section Title

Frames — Stationary Multi-Force Structures

Common Mistakes

• Labeling a member as two-force when it carries a transverse load or has three pin connections — always count the number of force application points first.
• After dismembering, drawing pin forces in the same direction on both members instead of reversing (violating Newton's third law).
• Taking moments about a point that introduces two unknowns — always select a point that eliminates as many unknowns as possible (e.g., a pin location).
• Forgetting to include the weight of members if specified — self-weight acts at the centroid and makes a member multi-force.
• Treating the entire frame as a single rigid body when computing internal pin forces — you must dismember it.

Formulas

Exam Tips

• For a machine problem, first write ΣM about the pivot/pin to get the load-effort relationship directly — this is usually the fastest path to the answer.
• Identify whether the pivot is a fixed pin or a moving pin, as this affects the FBD.
• If the problem gives geometry (lengths and angles), resolve forces into components before taking moments.
• Board examiners frequently test pliers and toggle clamps — practice both by dismembering at the pivot and applying ΣM = 0 to one jaw.

Key Points

• A machine is a structure with moving parts designed to transmit or modify forces (e.g., pliers, toggle clamp, hydraulic press, crane hook, scissors, wheel-and-axle).
• Analysis method is identical to frames: dismember → FBD of each part → apply equilibrium.
• The key output of a machine analysis is the relationship between the effort (input force) and the load (output force), called the mechanical advantage (M.A.).
• Mechanical advantage > 1 means the machine amplifies force (effort arm > load arm for a lever).
• For a simple lever: M.A. = effort arm / load arm. If M.A. = 8, a 50 N effort lifts 400 N.
• For a pulley system: M.A. equals the number of rope segments supporting the load.
• In compound machines, multiply the M.A. of each stage.
• Efficiency η = (actual load) / (ideal load) — real machines have friction losses so η < 1.
• Board problems on machines most commonly involve levers, pliers, toggle clamps, and cranes.

Definitions

Section Title

Machines — Force Transmission and Mechanical Advantage

Common Mistakes

• Using the total length of the lever instead of the perpendicular moment arm — always use the perpendicular distance to the line of action of the force.
• Confusing the class of lever, which changes where the fulcrum, effort, and load are located.
• Forgetting that a machine must also be dismembered at its pins for complete internal force analysis — the overall moment equation gives the load-effort relationship, but pin forces require the full FBD of each part.
• Not accounting for efficiency when the problem states a friction coefficient or percent efficiency.
• For pliers/scissors: treating the joint as a single body instead of two separate members connected at a pivot.

Formulas

Exam Tips

• Always identify the loading type first: horizontal UDL → parabola (H = wL²/8d); self-weight → catenary (parabola approximation OK for shallow sag); point loads → funicular polygon.
• Find H first using ΣM about a support — H is the same everywhere and simplifies all subsequent calculations.
• For a parabolic cable, the vertical reaction at each support for a symmetric loading equals wL/2, making T_max computation straightforward.
• In board exam problems, the answer choices for T_max are often close — carry at least four significant figures through your calculation before rounding.
• For a cable with unequal support elevations, use a Cartesian coordinate system with the origin at one support, and write the moment equation at a point to find H — the procedure is the same but the geometry is asymmetric.

Key Points

• A cable is perfectly flexible — it has no bending stiffness. It carries ONLY tension (no compression, no shear, no bending moment).
• The tension in a cable is always directed tangentially along the cable at any point.
• The HORIZONTAL COMPONENT of cable tension H is CONSTANT throughout the cable's length (for any loading — this follows from ΣFx = 0 applied to any free-body cut of the cable).
• The maximum tension occurs at the steepest point — for symmetric loading this is at the supports, NOT at midspan.
• THREE LOADING CASES: (1) Concentrated loads → funicular polygon (straight segments between loads); (2) Uniform load per horizontal distance → parabola; (3) Uniform load along cable length (self-weight) → catenary.
• For shallow sag (sag/span < 0.1), the catenary is well approximated by the parabola.
• For a cable with concentrated loads: the sag at each load point is found by solving moment equations at the load points.
• At any load point: the change in cable slope equals the applied load divided by H.

Definitions

Section Title

Cables — Concentrated Loads and Parabolic Cables

Common Mistakes

• Assuming maximum tension occurs at midspan — it occurs at the SUPPORT (steepest point) for a symmetric parabolic or concentrated-load cable.
• Using the arc length of the cable instead of the horizontal span L in the formula H = wL²/8d — L is always the HORIZONTAL span.
• Assuming sag and H are independent — they are inversely related: deeper sag means lower H and lower tension.
• Forgetting that H is constant — some students recompute H at every point when only T (the resultant) changes along the cable.
• Using the catenary formula for a uniformly distributed deck load (which is parabolic) or vice versa.
• For cables with two loads at different sag values, incorrectly assuming the sag is at midspan — the geometry must be solved from ΣM = 0.

Formulas

Exam Tips

• Allocate 4–5 minutes per frame/cable problem in the licensure exam. If a problem takes longer, skip and return — these problems are systematic once the FBD is correct.
• In four-choice items, eliminate obviously wrong orders of magnitude first (e.g., a cable tension that is less than the applied load is physically impossible).
• Memorize H = wL²/8d and T_max = √(H² + (wL/2)²) — these two equations solve 80% of parabolic cable board problems.
• For frames, memorize the procedure: whole → dismember → Newton's third law at pins. Practice this sequence until it is automatic.

Key Points

• A typical board problem involves 4–5 steps; the most common errors occur in setting up the FBD, not in the algebra.
• Always state your sign convention clearly (positive x to the right, positive y upward, counterclockwise moments positive) and maintain it throughout.
• For frame problems: whole-frame FBD → external reactions → dismember → member FBDs → pin forces.
• For cable problems: identify cable type → find H → find T at required location using T = H/cos θ.
• For machine problems: dismember at pivot → ΣM about pivot → load-effort relationship → M.A.
• Verify: check that ΣFx = 0 and ΣFy = 0 on each FBD after solving — this is your built-in error check.
• In the board exam, most cable and frame problems can be solved in under 5 minutes if the FBD is drawn correctly on the first attempt.

Definitions

Section Title

Integrated Approach — Solving Board-Style Problems

Common Mistakes

• Rushing to write equations without first drawing a complete, correct FBD — always draw first, compute second.
• Mixing up the directions of force components when resolving at an angle — use a right-triangle diagram for each inclined force.
• Rounding intermediate values too aggressively — carry at least four significant figures until the final answer.
• Not verifying the solution against a remaining unused equilibrium equation — always use your third equation as a check.