CELE Engineering Mechanics — Frames, Machines and CablesDetailed Explanation

Examples

Applications

• Structural analysis of roof trusses with knee braces (the knee brace is typically a two-force member).
• Analysis of crane booms, derricks, and loading frames — multi-force members under combined axial and transverse loads.
• Bracket connections on building facades where the bracket plate is a multi-force member.
• Analysis of formwork falsework frames per NSCP 2015 provisions on construction loads.
• Portal frame structures — a staple of Philippine low-rise industrial buildings analyzed as frames with rigid joints.

Misconceptions

• WRONG: 'The diagonal brace in a frame is always a two-force member.' — CORRECT: Only if it has exactly two pin connections and zero intermediate loads. A diagonal with a load at its midpoint is a multi-force member.
• WRONG: 'Pin forces on connected members point in the same direction.' — CORRECT: Newton's 3rd Law — equal in magnitude, OPPOSITE in direction.
• WRONG: 'I can skip Phase 2 (whole-body reactions) and just dismember.' — CORRECT: Whole-body reactions provide essential boundary conditions. Without them, you have too many unknowns.
• WRONG: 'A frame is only statically determinate if it looks symmetric.' — CORRECT: Determinacy depends on the equation count vs. unknown count, not geometry or symmetry.
• WRONG: 'The pin force at a joint equals the sum of the forces on both members.' — CORRECT: The pin force on each member equals the force that pin exerts on that member; equal and opposite forces on each side.

Related Concepts

• Truss analysis (method of joints, method of sections) — trusses are a special case with all two-force members
• Static equilibrium — ΣFx=0, ΣFy=0, ΣM=0
• Newton's Third Law applied at internal pins
• Free body diagrams (FBD) construction
• Moment of a force about a point

Common Exam Questions

Key Points To Remember

• A frame has at least one multi-force member — if all members are two-force, it is a truss.
• Always identify two-force members first — their force direction is known (along the member axis), eliminating unknowns.
• Phase 1: External reactions on the whole frame. Phase 2: Dismember, apply Newton's 3rd Law at pins.
• At each shared pin, forces on the two connected members are equal in magnitude and opposite in direction.
• Choose the moment center at a point where the most unknowns have zero moment arm — usually a pin with two unknowns.
• A frame is stationary (not moving); a machine has moving parts. Both use the same analysis method.
• Internal pin forces are NOT reactions — they are internal to the structure but external to each individual member FBD.

Examples

Applications

• Construction equipment: hydraulic jacks, manual hoists, chain blocks — all machines analyzed by equilibrium and M.A.
• Scaffolding clamps and formwork tightening devices — toggle-action clamps provide high clamping force from modest hand effort.
• Rebar benders and cutters on Philippine construction sites — lever machines with defined M.A.
• Crane hook blocks with multiple sheave pulleys — pulley system M.A. calculation is used in rigging design.
• Traffic barrier deployment arms — linkage mechanisms analyzed as machines.

Misconceptions

• WRONG: 'A machine with M.A. > 1 gives free energy.' — CORRECT: Work input = Work output (ideal). Higher force means smaller displacement: W = F × d is conserved.
• WRONG: 'The pivot force in a lever is zero.' — CORRECT: The pivot reaction equals the sum of all other forces on the lever; take ΣF = 0 to find it.
• WRONG: 'Pulley M.A. always equals the number of pulleys.' — CORRECT: M.A. = number of rope segments supporting the load, not number of pulleys.
• WRONG: 'A machine analysis requires dynamics because parts move.' — CORRECT: In CE Board problems, machines are analyzed in STATIC equilibrium at any given position — forces are balanced, no acceleration.

Related Concepts

• Simple lever classes (Class 1, 2, 3)
• Virtual work principle — alternative method for machines
• Friction in machines — efficiency η = M.A._actual / M.A._ideal
• Hydraulic press — Pascal's Law as a fluid 'machine'
• Pulley systems and mechanical advantage

Common Exam Questions

Key Points To Remember

• Machines have moving parts; frames are stationary. Same analysis method applies to both.
• M.A. = load/effort = effort arm/load arm (for a lever, from ΣM = 0 about fulcrum).
• An ideal machine has 100% efficiency: effort × effort arm = load × load arm.
• Real machines have M.A. < theoretical due to friction — but CE Board problems usually assume frictionless (ideal).
• For toggle mechanisms, M.A. → ∞ as toggle angle → 0° — the toggle is at dead center.
• Identify the pivot/fulcrum first — this is the moment center that decouples effort from load.
• Compound machine M.A. = product of individual M.A. values.

Examples

Applications

• Suspension bridges (e.g., conceptual analysis of the Pasig River cable-stayed/suspension schemes)
• Cable-supported roofs — convention centers, sports arenas
• Power transmission lines — sag calculation for clearance to ground (PEC requirements)
• Aerial tramways and ski lift cables — segment-by-segment tension analysis
• Overhead cranes with hanging cables — maximum sag check under load

Misconceptions

• WRONG: 'The cable tension is the same everywhere.' — CORRECT: Only the HORIZONTAL COMPONENT H is constant. The total tension T = H/cos θ varies with the slope.
• WRONG: 'Maximum cable tension is at midspan.' — CORRECT: Maximum is at the steepest point — at the supports (for cables with supports at same or different elevations, it's at the higher/steeper end).
• WRONG: 'A slack cable can carry compression under load.' — CORRECT: A cable can NEVER carry compression — it simply goes slack (loses tension and changes shape).
• WRONG: 'Cable geometry can be chosen freely.' — CORRECT: Under a given set of loads, there is only one equilibrium shape for a given H. The shape is determined by the loads and H.
• WRONG: 'For two equal symmetric loads at third-points, all segments have the same tension.' — CORRECT: The middle horizontal segment has T = H; the outer inclined segments have T = H/cos θ > H.

Related Concepts

• Funicular polygon (string polygon) — graphical method for cable shape
• Trusses — comparison of pure tension (cables) vs axial-only members
• Resultant of concurrent forces — cable equilibrium at each load point
• Vector resolution of forces
• Three-force body equilibrium

Common Exam Questions

Key Points To Remember

• A cable carries tension ONLY — no compression, shear, or bending.
• The horizontal component of tension H is CONSTANT throughout the cable (for vertical loads).
• Maximum tension occurs at the steepest segment — at the highest support.
• Cable geometry is determined by loads, not by the cable itself (no bending stiffness).
• For a single concentrated load at midspan: T = P / (2 sin θ), where θ = arctan(2d/L).
• For multiple loads: use funicular polygon — cable is straight between loads.
• Always find H first (using moment equation about a load point), then compute tensions.

Examples

Applications

• Suspension bridge cable design — DPWH bridge design guidelines require computation of cable tension under dead load + live load.
• Electrical transmission lines — PEC (Philippine Electrical Code) requires minimum ground clearance; sag calculation uses catenary formulas.
• Cable-stayed roof structures — parabolic cable shape under ponding load.
• Ropeway and aerial tramway cables — Tmax must be checked against cable rated breaking strength.
• Pipeline river crossings — the pipeline acts as a catenary/parabolic cable under its own weight in water.

Misconceptions

• WRONG: 'A cable loaded by its own weight forms a parabola.' — CORRECT: Self-weight (per unit cable length) → catenary. Deck load (per unit horizontal distance) → parabola. For small sag, they are nearly equal.
• WRONG: 'H increases with sag for fixed span and load.' — CORRECT: H = wL²/(8d). LARGER sag d → SMALLER H. The cable is 'less stressed' horizontally when it hangs lower.
• WRONG: 'T_max is at midspan.' — CORRECT: T_max is at the supports (steepest point). T_min = H at the crown.
• WRONG: 'For a level cable (d → 0), tension approaches zero.' — CORRECT: For d → 0, H → ∞. A perfectly horizontal loaded cable would require infinite tension — physically impossible.
• WRONG: 'The parabolic cable formula applies only to suspension bridges.' — CORRECT: It applies to any cable with load uniform per horizontal distance — e.g., horizontal conveyor cables, horizontal loading.

Related Concepts

• Catenary curve — mathematical definition using hyperbolic cosine
• Parabola — conic section, shape of uniformly horizontally loaded cable
• Virtual work for cables — alternative derivation of cable equations
• Suspension bridge design — main cable, hangers, stiffening truss/girder
• Cable sag and clearance requirements — PEC, DPWH bridge standards

Common Exam Questions

Key Points To Remember

• Uniform load per horizontal distance → Parabolic cable. H = wL²/(8d).
• Uniform load per cable length → Catenary. For shallow cables (d/L < 0.1), use parabolic formula as approximation.
• H = wL²/(8d): memorize this. It appears directly in CE Board problems.
• T_max at supports = √[H² + (wL/2)²] — the Pythagorean combination of H and V.
• Minimum tension is at the crown (lowest point) = H (horizontal component only).
• Cable length: S ≈ L[1 + (8/3)(d/L)²] for approximate calculation.
• For the CE Board: READ the problem carefully — 'per meter of span' (parabola) vs 'per meter of cable' (catenary).
• Sag-to-span ratio d/L = wL²/(8H·L) = wL/(8H): larger load or longer span → more sag for given H.

Problem

FRAME PROBLEM: A triangular frame consists of members AB (horizontal, 4 m long), BC (vertical, 3 m tall), and AC (diagonal). The frame is pinned at A (bottom left) and B (bottom right). A vertical load of 20 kN is applied at joint C (top left). Members AB is the base, BC is the left vertical, AC is the diagonal from top-left to bottom-right. Identify which members are two-force members and find the force in each member.

Solution

Step 1 — Identify two-force members: • Member AB: Connected at A and B. The pin at A is also a support — two reaction components. The load at C is NOT on AB. However, A is a support with two unknown reaction components, so AB has forces at A, B, and potentially a reaction at A — this needs more careful inspection. • Actually, redraw: A(0,0) is pin support. B(4,0) is pin support. C(0,3) has 20 kN downward load. Members: AB (A to B, horizontal), BC (B to C, diagonal from (4,0) to (0,3)), AC (A to C, vertical from (0,0) to (0,3)). Step 2 — Whole-body equilibrium: ΣMA = 0: By(4) − 20(0) = 0 → By = 0 (The 20 kN load is directly above A, so no moment about A from it.) ΣFy = 0: Ay + By − 20 = 0 → Ay = 20 kN ΣFx = 0: Ax + Bx = 0 Step 3 — Member AC (vertical, A(0,0) to C(0,3)): Loaded at C with 20 kN. Also connected at A. This is a TWO-force member? No — C has an external load, so AC is NOT a two-force member if we consider the 20 kN as applied at C. However, in the context of the frame, C is a JOINT. At joint C, the 20 kN is applied, and forces from members AC and BC meet. So AC and BC are two-force members (each connected at two joints with no intermediate loads); the 20 kN is at the JOINT, not ON a member. In this case, ALL three members are two-force members (this is actually a truss!): • AC: vertical, F_AC = internal axial force along AC direction (vertical) • BC: diagonal from B(4,0) to C(0,3), length = √(16+9) = 5 m, direction: (-4/5, +3/5) • AB: horizontal Joint C — ΣFy = 0: F_AC + F_BC·(3/5) − 20 = 0 Joint C — ΣFx = 0: F_BC·(-4/5) = 0 → F_BC = 0 From ΣFy: F_AC = 20 kN (compression, pushing C downward) Joint B — F_BC = 0, so: Bx = 0, By = 0 ✓ Joint A — F_AC = 20 kN (acts along AC): Ax = 0, Ay = 20 kN ✓ Answer: F_AC = 20 kN compression; F_BC = 0; F_AB = 0 Board-exam lesson: When the load is at a joint and all members are two-force, use method of joints from truss analysis.

Problem

FRAME PROBLEM (MULTI-FORCE): A horizontal beam AD (length 6 m) is pinned at wall at A(0,0). It carries a 15 kN load at D(6,0). A strut connects C(4,0) on the beam to B(0,−3) on the wall. Strut BC is a two-force member. Find: (a) force in strut BC, (b) pin reaction at A.

Solution

Step 1 — Geometry of strut BC: From B(0,−3) to C(4,0). Length = √(4²+3²) = 5 m. Direction cosines: l = 4/5 = 0.8 (x-direction), m = 3/5 = 0.6 (y-direction) Strut BC is in compression (pushes C toward beam at 45° from geometry). Step 2 — FBD of beam AD: At A: Ax (→) and Ay (↑) At C: strut force F_BC pushes beam at angle: horizontal component = F_BC×(4/5) pointing left (−x), vertical component = F_BC×(3/5) pointing upward (+y). [Strut pushes UP and to the LEFT on the beam.] At D: 15 kN downward (−y) Step 3 — ΣMA = 0 (eliminate Ax and Ay): (F_BC × 3/5)(4) − 15(6) = 0 F_BC × (12/5) = 90 F_BC = 90 × 5/12 = 37.5 kN Step 4 — ΣFy = 0: Ay + F_BC×(3/5) − 15 = 0 Ay = 15 − 37.5×0.6 = 15 − 22.5 = −7.5 kN → 7.5 kN downward Step 5 — ΣFx = 0: Ax − F_BC×(4/5) = 0 Ax = 37.5×0.8 = 30 kN (→, toward the beam, i.e., away from wall) Step 6 — Resultant at A: RA = √(30² + 7.5²) = √(900 + 56.25) = √956.25 = 30.9 kN Answer: F_BC = 37.5 kN (compression); RA = 30.9 kN at A (directed at arctan(7.5/30) = 14.04° below horizontal)

Problem

CABLE PROBLEM: A cable spans 30 m between two supports at the same elevation. A single concentrated load of 25 kN hangs at a point 10 m from the left support. The sag at the load point is 3 m. Find: (a) H, (b) tension in the left segment, (c) tension in the right segment, (d) support reactions.

Solution

Step 1 — Geometry: Left support A(0,0), Load point C(10,−3), Right support B(30,0). Step 2 — FBD of entire cable: At A: H → (rightward) and VA ↑ At B: H → (leftward) and VB ↑ Load: 25 kN ↓ at x = 10 m Step 3 — ΣMB = 0: VA(30) − 25(30−10) = 0 VA(30) = 25(20) = 500 VA = 500/30 = 16.67 kN Step 4 — ΣFy = 0: VA + VB = 25 → VB = 25 − 16.67 = 8.33 kN Step 5 — Find H using geometry at load point C: Consider the left portion of cable (from A to C): ΣM at C = 0 for left portion: VA(10) − H(3) = 0 [H acts horizontally, moment arm = sag = 3 m] H = VA×10/3 = 16.67×10/3 = 55.57 kN Step 6 — Tension in left segment AC: Slope of AC: from A(0,0) to C(10,−3) tan θ_L = 3/10 = 0.3; θ_L = 16.70°; cos θ_L = 0.9578; sin θ_L = 0.2873 T_AC = H/cos θ_L = 55.57/0.9578 = 58.02 kN Check: T_AC × sin θ_L = 58.02 × 0.2873 = 16.67 kN = VA ✓ Step 7 — Tension in right segment CB: Slope of CB: from C(10,−3) to B(30,0) Horizontal: 20 m; Vertical: +3 m rise tan θ_R = 3/20 = 0.15; θ_R = 8.53°; cos θ_R = 0.9889; sin θ_R = 0.1490 T_CB = H/cos θ_R = 55.57/0.9889 = 56.19 kN Check: T_CB × sin θ_R = 56.19 × 0.1490 = 8.37 ≈ 8.33 kN = VB ✓ (rounding) Answer: H = 55.6 kN; T_AC = 58.0 kN (left, steeper); T_CB = 56.2 kN (right, shallower); VA = 16.67 kN; VB = 8.33 kN

Problem

PARABOLIC CABLE: A suspension bridge cable spans 200 m between towers of equal height. The roadway load is 12 kN/m of horizontal span. The maximum allowable cable tension is 2000 kN. Determine the minimum sag required.

Solution

Given: L = 200 m, w = 12 kN/m, T_max ≤ 2000 kN Step 1 — Express T_max in terms of d: H = wL²/(8d) = 12(200²)/(8d) = 12×40000/(8d) = 480000/(8d) = 60000/d V = wL/2 = 12×200/2 = 1200 kN T_max = √(H² + V²) = √[(60000/d)² + 1200²] Step 2 — Set T_max = 2000 kN (limiting condition): √[(60000/d)² + 1200²] = 2000 (60000/d)² + 1440000 = 4000000 (60000/d)² = 2560000 60000/d = √2560000 = 1600 d = 60000/1600 = 37.5 m Step 3 — Verify: H = 60000/37.5 = 1600 kN T_max = √(1600² + 1200²) = √(2560000 + 1440000) = √4000000 = 2000 kN ✓ Answer: Minimum sag d = 37.5 m Note: Larger sag → smaller H and smaller T_max. The 37.5 m sag is the MINIMUM sag (corresponding to maximum allowable tension). Any sag less than 37.5 m would result in T_max exceeding 2000 kN.

Problem

MACHINE PROBLEM: A toggle press mechanism has two equal links, each 200 mm long. The links are connected to a central pin that slides vertically, and their outer ends are pinned horizontally 300 mm apart (so each link makes an angle θ with the horizontal). A horizontal effort F = 500 N is applied at the central pin pushing it downward. When the links make θ = 20° with the horizontal, find the compressive force P on the workpiece below the central pin.

Solution

Step 1 — Geometry: Each link has length 200 mm. Horizontal spread = 300 mm total (150 mm each side from center). cos θ = 150/200 = 0.75, so θ = 41.41° Wait — re-read: links make θ = 20° with horizontal (given directly). So: θ = 20° sin 20° = 0.342, cos 20° = 0.940 Step 2 — FBD of the central pin (joint): The 500 N force pushes DOWN (this is the effort). Two link forces (compression) act along each link — each at 20° above horizontal (pushing outward and upward on the central pin from the fixed outer pins). Wait — if the outer pins are fixed and the effort pushes the central pin downward, the links push UPWARD on the fixed pins and the reaction on the central pin from each link acts at 20° UPWARD from horizontal. Actually, for a toggle press: effort is typically HORIZONTAL to push links toward horizontal. Let me use the standard toggle: For a TOGGLE CLAMP with two links at angle θ to horizontal: Relating vertical output force P to horizontal input force F: FBD of central joint: Two link forces T at angle θ from horizontal. Vertical: 2T sin θ = P Horizontal: The input is applied as the links are pushed toward flat (but for this problem, let F = 500 N be the downward force). For downward input F pushing central pin down, equilibrium of central pin: 2T sin θ = F (vertical equilibrium, links slope upward from center to outer pins) T = F/(2 sin θ) = 500/(2×0.342) = 500/0.684 = 730.9 N The output compressive force on the workpiece = P = 2T cos θ × tan(90°−θ)... let me redo with correct model. Standard toggle press: Input = horizontal force H_in at the elbow. Output = vertical force P at central pin (downward). This gives P = H_in/(2 tan θ). But if F is already defined as the downward force: Actually simplest approach — for a symmetric toggle with 2 links of equal length, each at angle θ to horizontal, with outer pins fixed horizontally: If effort P (downward) at elbow, it drives vertical displacement. At elbow joint: 2T sin θ = P (vertical), so T = P/(2 sin θ) Horizontal spread reaction at outer pins = 2T cos θ = P cos θ/sin θ = P/tan θ If instead F = 500 N is a DOWNWARD applied force at the central pin and we want the HORIZONTAL REACTION at the outer pins: Horizontal force per outer pin = T cos θ = 500cos20°/(2sin20°) = 500(0.940)/(2×0.342) = 470/0.684 = 687 N For a toggle clamp where the OUTPUT is horizontal clamping force: P_clamp = F/(2 tan θ) = 500/(2×tan20°) = 500/(2×0.364) = 500/0.728 = 686.8 N ≈ 687 N Answer: Clamping/reaction force = 687 N. M.A. = P/F = 687/500 = 1.37 at θ = 20° Note: As θ → 0° (links become horizontal), P → ∞ — this is the toggle action principle. At θ = 5°: M.A. = 1/(2tan5°) = 1/(2×0.0875) = 5.7; at θ = 1°: M.A. = 28.6.