CELE Engineering Mechanics — Frames, Machines and CablesCheat Sheet
Frames, Machines and Cables cheat sheet — the reference card you wish you had on exam day. Condensed from the full study notes, this is the high-yield core of Frames, Machines and Cables for CELE Engineering Mechanics. Download, print, revise.
Exam context
On the CELE 2026, the Engineering Mechanics subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Frames, Machines and Cables lands at position 4th out of 8 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Engineering Mechanics on a typical CELE paper.
Frames, Machines and Cables - Cheat Sheet
Rapid-fire reference guide for the last 30 minutes before the exam. Master dismembering, two-force members, cable tension, and mechanical advantage. All formulas, definitions, and exam pitfalls in one place.
Sections
Formulas
Formula
∑Fx = 0, ∑Fy = 0, ∑M = 0
Meaning
Sum of forces in x and y directions; sum of moments about any point must equal zero for equilibrium of each member.
Watch Out
Do NOT apply to the whole frame first if you need internal pin forces — you must dismember to expose the pins.
When To Use
Apply to every FBD after dismembering — three equations per member.
Section Title
Frames — Fundamentals
Important Facts
- A frame ALWAYS contains at least one multi-force member; a truss contains ONLY two-force members.
- To find internal pin forces, you MUST dismember the frame — the external reactions alone are insufficient.
- Identify two-force members immediately — their force direction is known (along the member), reducing unknowns.
- After dismembering, apply ∑Fx = 0, ∑Fy = 0, ∑M = 0 to each member independently.
- Pin forces on adjacent members are equal and opposite (Newton's third law) — this is your check.
- The order of dismembering matters: start with members that have the fewest unknown forces.
- A frame reaction can be found from the whole-frame FBD; internal forces require dismembering.
- If a member has a moment applied to it (not at a pin), it is definitely multi-force.
- Assume pin forces are unknown in both x and y directions unless you can prove otherwise (e.g., if a two-force member, the force is along the member).
Key Definitions
Term
Frame
Example
A knee brace, an overhead crane support, a bracket anchoring a sign.
Definition
A stationary structure with at least one multi-force member (loaded at more than two points or carrying a moment).
Term
Multi-force member
Example
A horizontal beam with a load at midspan and pin supports at both ends.
Definition
A member loaded at three or more points or carrying an applied moment; develops shear and bending.
Term
Two-force member
Example
A strut, a tie rod, any member with pins at both ends and no other loads.
Definition
A member loaded only at two points; force must act along the line joining those two points (acts in axial tension or compression only).
Term
Dismember
Example
Cut the frame at every pin and show equal-and-opposite forces on the two adjacent members.
Definition
Separate a frame into individual members and draw a free-body diagram (FBD) for each, showing all pin forces.
Term
Pin force (Newton's Third Law at pins)
Example
If the beam pushes on the pin with 25 kN to the right, the pin pushes back on the beam with 25 kN to the left.
Definition
Force exerted by member A on member B at a shared pin is equal in magnitude and opposite in direction to the force exerted by member B on member A.
Diagrams To Know
- Free-body diagram (FBD) of the whole frame showing external loads and support reactions.
- Dismembered FBD of each individual member with all pin forces labeled at joints.
- Tension vs. compression notation (assumed positive, then solve; if negative, reverse the direction).
Formulas
Formula
M.A. = Load / Effort = Effort Arm / Load Arm
Meaning
Mechanical Advantage; Load is the output force, Effort is the input force; arms are distances from the pivot.
Watch Out
Do NOT confuse effort arm with load arm — effort arm is where you push, load arm is where the load sits. M.A. > 1 means the machine amplifies force.
When To Use
Any simple lever, pulley system, or compound machine where you know the geometry and input force.
Formula
Effort × Effort Arm = Load × Load Arm
Meaning
Moment (torque) balance about the pivot; effort arm is the distance from the pivot to the effort; load arm is the distance from the pivot to the load.
Watch Out
This equation assumes the lever is in equilibrium — it is derived from ∑M = 0 about the pivot.
When To Use
Any lever or rigid-body pivot problem; rearrange to solve for unknown force or arm length.
Formula
Power Input = Power Output (ideal machine, no friction)
Meaning
Effort × Velocity_effort = Load × Velocity_load; if M.A. is high, the load moves slower than the effort.
Watch Out
Real machines lose energy to friction — actual load is always less than theoretical; efficiency = Output/Input < 1.
When To Use
Verify energy conservation in a machine; relate linear and angular velocities in pulleys or gears.
Section Title
Machines — Mechanical Advantage & Force Transmission
Important Facts
- A machine amplifies force by sacrificing distance — high M.A. means the load moves less than the effort.
- Dismember machines the same way as frames: cut at pins and apply equilibrium to each member.
- The pivot (or fulcrum) in a lever is a zero-force point — no load acts there, only reaction forces.
- In a compound machine (gears, pulleys in series), multiply the M.A. values of each stage: M.A._total = M.A._1 × M.A._2 × ...
- Real machines are never ideal — friction losses mean actual M.A. < theoretical M.A.; efficiency accounts for this.
- A machine that increases force decreases distance and vice versa (conservation of energy).
- Common machines: lever, pulley, inclined plane, screw, wedge, gear; all follow the same moment-balance principle.
- If M.A. = 1, the machine does not amplify force (purely directional change, like a fixed pulley).
Key Definitions
Term
Machine
Example
Pliers, a hand-operated crane, a screw jack, a pulley system, a hydraulic ram.
Definition
A device with moving parts designed to transmit or modify forces; contains at least one multi-force member.
Term
Mechanical Advantage (M.A.)
Example
A lever with M.A. = 4 multiplies a 100 N hand force into a 400 N load.
Definition
The ratio of the load (output force) to the effort (input force); also equals the ratio of effort arm to load arm for a lever.
Term
Effort
Example
The 50 N push on one end of pliers.
Definition
The input force applied by the operator (e.g., hand force on a lever, pull on a rope).
Term
Load
Example
The gripping force at the business end of pliers.
Definition
The output force exerted by the machine (e.g., clamping force, lifting force).
Term
Effort arm
Example
If you push 0.4 m from a pivot, the effort arm is 0.4 m.
Definition
The perpendicular distance from the pivot to the line of action of the effort force.
Term
Load arm
Example
If the load acts 0.05 m from the pivot, the load arm is 0.05 m.
Definition
The perpendicular distance from the pivot to the line of action of the load force.
Diagrams To Know
- Lever diagram showing effort, load, pivot, effort arm, and load arm.
- Pulley system with mechanical advantage (e.g., movable pulley = M.A. 2, fixed pulley = M.A. 1).
- Compound machine (gears or cascaded levers) showing how M.A. multiplies through stages.
Reactions Or Equations
Note
This is the fundamental equation for all lever problems; rearrange as needed.
Equation
∑M_pivot = 0: Effort × Effort Arm = Load × Load Arm
Conditions
Static equilibrium of the lever; assumed no friction or weight of the lever.
Formulas
Formula
H = wL² / (8d)
Meaning
Horizontal component of tension in a parabolic cable; w = uniform load per unit horizontal length (kN/m), L = span (m), d = sag (m).
Watch Out
w is load per HORIZONTAL distance, not per cable length. Sag d must be in the same units as L. Small sag (d < L/10) ≈ parabola; larger sag approaches catenary (rare in civil exams).
When To Use
Any cable (suspension bridge, guy rope) carrying a uniform horizontal load (e.g., deck weight). Cable shape is parabolic.
Formula
T_max = √(H² + (wL/2)²)
Meaning
Maximum tension in a parabolic cable, occurring at the supports (steepest point); H is horizontal component, wL/2 is vertical component (half the total load).
Watch Out
This is NOT the tension at midspan (where it is minimum and equals H). Midspan tension is ≈ H for shallow cables.
When To Use
Find the peak cable tension for design (affects anchor strength). Occurs at the points of maximum slope (the supports).
Formula
T = H / cos(θ)
Meaning
Tension at any point where the cable makes an angle θ with the horizontal; H = constant horizontal component.
Watch Out
H is constant throughout the cable. As θ increases (cable gets steeper), T increases. At θ = 0 (flat section), T = H.
When To Use
Find tension anywhere on the cable if you know the slope angle θ at that point; applies to any cable shape.
Formula
Catenary: y = (c/a) cosh(ax/c), where a = T₀/w (T₀ = horizontal tension, w = weight per unit length along cable)
Meaning
Shape of a cable hanging under its own weight; cosh is the hyperbolic cosine function.
Watch Out
For shallow sags (d < L/10), parabola ≈ catenary — use parabola formula unless told otherwise. Catenary is hyperbolic, not polynomial.
When To Use
Only if the problem explicitly states self-weight of the cable is significant, OR sag is very large. Most PRC exams use parabola.
Formula
Cable with concentrated loads (at known points): tan(θ) = ΔVertical / ΔHorizontal (for each segment)
Meaning
Slope angle of each straight segment between point loads is found from the vertical component of the load below that segment divided by the constant horizontal tension H.
Watch Out
H is constant throughout. Vertical component at each point equals the sum of all loads below it. Sketch the polygon first.
When To Use
Cable supporting discrete loads at given spans; cable is straight between loads, bent at each load (funicular polygon).
Section Title
Cables — Tension, Sag & Shape
Important Facts
- A cable carries ONLY tension — never compression or bending.
- The horizontal component H is constant everywhere in the cable; vertical component changes.
- Tension is MAXIMUM at the steepest point (usually the supports); minimum at the flattest point (usually midspan for symmetric loading).
- Sag d and horizontal component H are inversely related: smaller sag ⇒ higher H ⇒ higher tension at supports; larger sag ⇒ lower H ⇒ lower tension.
- For a parabolic cable, the shape is parabolic only if the load is uniform per HORIZONTAL distance (not per cable length).
- If the cable has its own weight and no other load, the shape is a catenary, not a parabola — but most PRC problems ignore cable weight.
- A cable with concentrated loads at known points is straight between loads and bent (changes slope) at each load.
- The vertical component of tension at any point equals the sum of all loads (or reactions) below that point.
- Cable problems often ask for H (and thus the anchor forces) and T_max (for strength design).
Key Definitions
Term
Cable
Example
Suspension bridge cables, guy ropes, rigging on a crane, power transmission lines.
Definition
A perfectly flexible member that carries only tension; cannot resist bending or compression.
Term
Sag
Meaning
d in the formula H = wL² / (8d); larger sag ⇒ lower tension, smaller sag ⇒ higher tension.
Definition
The vertical distance from a straight line joining the two cable supports to the lowest point of the cable.
Term
Span
Example
A suspension bridge cable span might be 1386 m (Golden Gate Bridge).
Definition
The horizontal distance between the two cable supports.
Term
Parabolic cable
Example
The main cables of a suspension bridge carrying the deck weight.
Definition
Cable shape under a uniform load distributed horizontally (e.g., suspension bridge deck load); forms a parabola.
Term
Catenary
Example
A bare power transmission line or a chain hanging freely between two poles.
Definition
Cable shape under its own weight alone (distributed along the cable length, not horizontally); forms a hyperbolic cosine curve.
Term
Horizontal component (H)
Example
H = wL² / (8d) for a parabolic cable.
Definition
The constant horizontal component of cable tension throughout the cable; determined by the overall equilibrium of the cable.
Term
Funicular polygon
Example
A cable supporting three equal loads at the quarter points forms a funicular polygon with three straight segments.
Definition
The shape of a cable (or the polygon formed by its segments) when supporting concentrated loads at specific points; straight segments between loads, bent at each load.
Diagrams To Know
- Cable in profile showing sag d, span L, supports, loads, and tension direction at key points.
- Parabolic cable with uniform load: straight-line support reactions, curved cable profile.
- Funicular polygon (cable with concentrated loads): piecewise-linear cable with bends at each load.
- Catenary (cable under self-weight): smooth curve (not straight between loads).
- Tension vs. position plot showing H constant, vertical component increasing toward supports, and total tension increasing.
Reactions Or Equations
Note
At the supports, the cable reactions have both horizontal (H, constant) and vertical (load-dependent) components.
Equation
∑F_vertical = 0: (Vertical component at left support) + (Vertical component at right support) = Total load
Conditions
The cable must be in vertical equilibrium; the sum of upward and downward forces equals zero.
Note
For a symmetric parabolic cable with symmetric loading, each support carries half the total load vertically.
Equation
∑M = 0 about any point: takes into account cable slope and loads to find support reactions.
Conditions
Use moment equilibrium about a support to find the vertical component of reaction.
Section Title
Two-Force Members & Identifying Them
Important Facts
- If a member has exactly two load points (typically pins) and nothing else, assume it is a two-force member.
- The force in a two-force member acts along the member — this is the key: if the member is at an angle θ, the force makes angle θ with the horizontal.
- Two-force members are ALWAYS in either pure tension or pure compression — no bending or shear.
- Identifying a two-force member early reduces unknowns: instead of two unknowns (Fx, Fy) at a pin, you have one (magnitude F along the member).
- A two-force member at angle θ to horizontal has components F cos(θ) horizontal and F sin(θ) vertical.
- If a member is multi-force (load at a third point, or an applied moment), it is NOT a two-force member.
- In a truss, all members are two-force members (by design); in a frame, some are multi-force.
Key Definitions
Term
Two-force member (definition & identification)
Example
A diagonal strut in a frame, a tie rod, the members of a truss.
Definition
A member with loads applied at exactly two points (both usually pins) and no other loads (no applied moment, no distributed load); the force must act along the line joining the two load points.
Diagrams To Know
- Two-force member FBD showing forces at both ends, aligned along the member.
Section Title
Analysis Procedure (Frames & Machines)
Important Facts
- Step 1: Draw FBD of the entire frame/machine. Find external reactions using ∑Fx, ∑Fy, ∑M (3 equations for a 2D structure with 3 unknown reactions).
- Step 2: Identify two-force members — their direction is known, reducing unknowns immediately.
- Step 3: Dismember the frame at every pin. Draw FBD of each member separately.
- Step 4: At each pin, show equal-and-opposite forces on the two adjacent members (Newton's third law).
- Step 5: Apply ∑Fx = 0, ∑Fy = 0, ∑M = 0 to each member in sequence. Start with members having the fewest unknowns.
- Step 6: Solve the equilibrium equations to find all unknown pin forces.
- Step 7: Check: the forces on one member at a pin should be equal and opposite to forces on the adjacent member.
Diagrams To Know
- Step-by-step dismembered FBD sequence showing how to isolate each member and label pin forces.
Section Title
Cable Problems — Standard Scenarios
Important Facts
- Scenario A: Parabolic cable with uniform load — use H = wL² / (8d), then T_max = √(H² + (wL/2)²).
- Scenario B: Cable with two equal concentrated loads at symmetric positions — cable sags equally on both sides; apply ∑M about midspan.
- Scenario C: Cable with asymmetric loads — different sags on left and right; use moment balance about each support separately.
- Scenario D: Catenary (cable under self-weight only) — rare in PRC exams unless explicitly stated; use parabola as approximation if sag is small.
- Always check: as sag increases, H decreases (cable sagging more means it carries less horizontal tension).
- Maximum tension always occurs at the steepest point; for a symmetric cable, this is at the supports.
Key Definitions
Term
Suspension bridge
Example
Golden Gate Bridge (parabolic main cables), but the problem might ask for tension in the vertical suspenders or main cables.
Definition
A bridge where the deck is hung by vertical cables from main cables that span between towers; the main cable form is parabolic under deck load.
Term
Guy rope
Example
Temporary braces on a construction crane or communications tower.
Definition
An angled cable or rod used to stabilize a vertical pole or structure; typically in tension.
Diagrams To Know
- Cable profile with loads, sag, and reactions marked.
- Free-body diagram of a cable section showing internal tension and external loads.
Must Remember
- A FRAME contains at least one MULTI-FORCE member (loaded at ≥3 points or carrying a moment); a TRUSS contains ONLY two-force members.
- To find INTERNAL PIN FORCES, you MUST DISMEMBER the frame — the external reactions alone are not enough.
- At a shared pin, forces on member A are EQUAL and OPPOSITE to forces on member B (Newton's third law) — this is your check.
- Identify TWO-FORCE MEMBERS immediately — their force is ALONG the member, which removes unknowns and simplifies the solution.
- A CABLE carries only TENSION; the HORIZONTAL COMPONENT H is CONSTANT throughout; MAXIMUM TENSION is at the STEEPEST POINT (usually the supports).
- For a PARABOLIC CABLE with uniform load: H = wL² / (8d), where w is load per HORIZONTAL distance, L is span, d is sag.
- Maximum cable tension: T_max = √(H² + (wL/2)²) — occurs at the supports, not at midspan.
- MECHANICAL ADVANTAGE: M.A. = Load / Effort = Effort Arm / Load Arm — a machine with M.A. > 1 amplifies force but sacrifices distance.
- In a LEVER, ∑M about the pivot = 0, so Effort × Effort Arm = Load × Load Arm — this is the fundamental equation.
- SAG and HORIZONTAL TENSION are INVERSELY related: smaller sag → higher H → higher cable tension at supports; larger sag → lower H.
Last Minute Tips
- DISMEMBERING IS MANDATORY for frames and machines. Draw a separate FBD for each member, showing all pin forces with directions. Label everything and check Newton's third law at pins.
- IDENTIFY TWO-FORCE MEMBERS FIRST — you can immediately state their force direction (along the member), cutting unknowns in half and simplifying the equilibrium equations.
- For CABLES, always sketch the shape (parabola or funicular polygon) and mark the sag, span, and loads clearly. Then use H = wL² / (8d) for parabolic cables or moment balance for concentrated loads.
- CHECK YOUR WORK: After solving pin forces on one member, verify they are equal-and-opposite on the adjacent member. Mismatches reveal arithmetic errors or incorrect FBD assumptions.
- MAXIMUM CABLE TENSION is at the STEEPEST POINT (usually the supports), NOT at midspan. Use T_max = √(H² + (wL/2)²) at the supports for a parabolic cable; midspan tension ≈ H.
Comparison Tables
Rows
Values
- Stationary structure with at least one multi-force member.
- Device with moving parts designed to transmit/modify forces; has multi-force members.
- Rigid structure made ONLY of two-force members (pins at ends, no other loads).
Property
Definition
Values
- Mix of two-force and multi-force members.
- Primarily multi-force members.
- All members are two-force members.
Property
Member types
Values
- Axial, shear, and bending in multi-force members.
- Axial, shear, and bending depending on geometry and loads.
- Axial tension or compression only; no shear or bending (ideal).
Property
Internal forces
Values
- Dismember and apply equilibrium to each member.
- Dismember and apply equilibrium to each member (same as frames).
- Method of joints or method of sections; simpler because forces are axial only.
Property
Analysis method
Values
- Bracket, crane support, knee brace.
- Pliers, hand jack, hydraulic ram, crane linkage.
- Bridge truss, roof truss, lattice tower.
Property
Typical examples
Columns
- Characteristic
- Frame
- Machine
- Truss
Table Title
Frames vs. Machines vs. Trusses
Rows
Values
- Parabola
- H = wL² / (8d)
- Suspension bridge, any cable with deck load or uniform distributed load.
- Load is per horizontal distance (not per cable length).
Property
Uniform load (horizontal distance)
Values
- Catenary
- y = (c/a) cosh(ax/c), where a = T₀/w
- Bare power line, chain hanging freely.
- Load is per unit cable length (self-weight); usually negligible for bridges.
Property
Self-weight (cable length)
Values
- Funicular polygon (piecewise linear)
- H = constant; tan(θ) = V/H for each segment
- Cable carrying discrete point loads; straight segments between loads.
- Vertical component below each load point is the sum of loads below.
Property
Concentrated loads at specific points
Values
- Parabola ≈ Catenary (nearly identical)
- Both formulas give very close results.
- Either formula acceptable; parabola is simpler.
- Shallow sag means low curvature; shape is insensitive to load distribution.
Property
Small sag (d ≪ L)
Columns
- Loading type
- Cable shape
- Horizontal component formula
- When to use
- Key assumption
Table Title
Cable Shapes & Formulas
Rows
Values
- A member with a load at a third point is multi-force, not two-force. The force is NOT along the member; it develops shear and bending.
- Check member loading carefully. If load is at a point other than the two ends, it is multi-force. Dismember and apply ∑M and ∑F to the member itself.
Property
Treating a frame member as two-force when it carries a mid-span load
Values
- Pin forces on member A and member B must be equal and opposite. Forgetting this leads to inconsistent solutions and failed checks.
- After solving for pin forces on one member, verify they are equal-and-opposite on the adjacent member. This is a built-in check.
Property
Forgetting Newton's third law at pins when dismembering
Values
- Confusing the load w (deck weight or applied load) with cable self-weight. Most PRC exams ignore cable weight unless explicitly stated.
- Use w = deck load or applied load per unit horizontal length. Ignore cable weight unless the problem says the cable is heavy or sag is very large.
Property
Using cable weight in the formula H = wL² / (8d) when the cable weight is negligible
Values
- Maximum tension is at the steepest point (supports), not at midspan. Midspan tension is minimum (approximately H for shallow cables).
- Always find T_max at the supports: T_max = √(H² + (wL/2)²). Midspan tension ≈ H (horizontal component only).
Property
Looking for tension at midspan as the maximum
Values
- The parabolic cable formula H = wL² / (8d) assumes w is load per horizontal metre. If w is per cable length, the formula is wrong.
- Always state w clearly: 'w = 6 kN/m horizontally' or 'w = 6 kN per horizontal metre'. For a suspension bridge, w is the deck load divided by span (not by cable length).
Property
Using w as load per cable length instead of per horizontal distance in parabolic cable formula
Values
- Effort arm is where you push (input); load arm is where the load is (output). Swapping them inverts the mechanical advantage.
- Draw the lever clearly. Effort arm = distance from pivot to effort point. Load arm = distance from pivot to load point. M.A. = Effort arm / Load arm.
Property
Confusing effort arm with load arm in lever mechanics
Values
- Only members with loads at exactly two ends (pins) and no other loads are two-force. A beam with a mid-span load is multi-force.
- Check each member individually. Identify two-force members first; they simplify the problem. The rest are multi-force.
Property
Assuming all frame members are two-force members
Columns
- Mistake
- Why it is wrong
- Correct approach
Table Title
Common Mistakes & Corrections
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