CELE Engineering Mechanics — Frames, Machines and CablesStudy Notes

A frame is a stationary structure composed of members connected at pins or rigid joints, with at least one multi-force member (a member loaded at more than two points or carrying an applied moment). Unlike trusses, frames develop internal shear forces and bending moments in their members in addition to axial forces.

Key Distinction from Trusses:
Trusses are loaded only at joints and all members are two-force members; frames may have transverse loads on members themselves, creating internal moments and shears. This fundamental difference requires a different analysis approach.

Method of Analysis:
The standard approach involves two main steps:

1. External Equilibrium: Apply equilibrium equations (ΣFx = 0, ΣFy = 0, ΣM = 0) to the entire frame as a free body to find the support reactions.
2. Dismembering the Frame: Separate the frame into individual members, drawing a complete free-body diagram for each member. At each pin connection, apply Newton's third law—the force exerted by one member on an adjacent member is equal in magnitude but opposite in direction to the force exerted back.

Critical Identification: Always identify two-force members within the frame first. A two-force member (loaded only at its ends with no intermediate loads or moments) develops a force along its longitudinal axis. This immediately reduces unknowns. For example, if a frame contains a diagonal strut with pins only at its ends and no transverse loads, the force within it acts along the strut's axis.

Sign Convention and Direction: When dismembering, assume directions for pin forces (typically along positive x and y axes). Consistency is essential: if member AB exerts a rightward force on member BC at their shared pin, member BC exerts a leftward force on member AB at the same pin.

A machine is a structure or mechanism designed to transmit or modify forces and displacements. Unlike frames, which are typically stationary, machines often contain moving parts such as linkages, levers, pulleys, and ratchets. Examples include pliers, hydraulic jacks, hand-operated clamps, and mechanical advantage devices used on construction sites.

Fundamental Principle:
Machines operate on the principle of work: a small effort applied over a large distance may produce a large load over a small distance. The mechanical advantage (M.A.) is the ratio of output force (load) to input force (effort):

M.A. = Load / Effort = Effort Arm / Load Arm (for a simple lever)

Analysis Method:
Machines are analyzed exactly like frames—by dismembering and applying equilibrium to each member. However, the objective differs: instead of finding internal forces, we find the relationship between the effort (applied force) and the load (resistive force).

Simple Lever:
A lever pivots about a fulcrum. For a lever in equilibrium:
ΣM_fulcrum = 0:
Effort × Effort Arm = Load × Load Arm

M.A. = Load / Effort = Effort Arm / Load Arm

For ideal (frictionless) levers, M.A. > 1 means the lever amplifies force (but the effort moves farther than the load). M.A. < 1 means the lever amplifies displacement (useful for reaching or applying force over distance).

Compound Machines:
When multiple machines are combined (e.g., two levers in sequence), the overall M.A. is the product of individual advantages. However, real machines always lose energy to friction, so the actual M.A. is less than the theoretical value. The efficiency η is:
η = Actual M.A. / Theoretical M.A.

Practical Considerations for Civil Engineers:
In construction equipment analysis (cranes, winches, jacks), the mechanical advantage of the system determines how much load can be lifted or moved with a given effort. This is critical in design calculations per ASME standards and safe working load determinations.

A cable is a perfectly flexible structural member that can sustain only tension (no compression or bending). Its shape is determined entirely by the loading applied to it. Cables are fundamental in suspension bridges, guy-wires for towers, and sag vertical takeoff systems. Understanding cable behavior is essential for the PRC examination, particularly in problems involving suspension structures and cable-stayed systems.

Fundamental Property:
At any point along a cable, the internal force is tangent to the cable and acts in tension only. If the cable makes an angle θ with the horizontal at a point, the tension T at that point has components:

• Horizontal: T cos(θ)
• Vertical: T sin(θ)

Critical Insight: Constant Horizontal Component
For a cable in equilibrium under any loading pattern, the horizontal component of tension H is constant throughout the cable. This is a consequence of horizontal force equilibrium—if H varied, there would be a net horizontal force on the system, violating equilibrium.

This property is the key to solving cable problems: once you find H, you can find the tension T at any point using T = H / cos(θ).

Cable Under Concentrated Loads:
When a cable supports discrete point loads, it forms a polygon (funicular polygon). Between loads, the cable is straight (no distributed load along the segment). The shape is determined by the magnitudes and positions of the loads.

Analysis at a Point Load:
At a point where a concentrated load P acts downward, the cable bends, and the two segments meeting at that point make different angles with the horizontal. Vertical force equilibrium at that point gives:
T₁ sin(θ₁) + T₂ sin(θ₂) = P

where T₁ and T₂ are the tensions in the two segments and θ₁, θ₂ are their angles.

Since H is constant:
T₁ cos(θ₁) = T₂ cos(θ₂) = H

Cable Under Uniform Horizontal Load (Parabolic Cable):
When a cable supports a load uniformly distributed over the horizontal span (typical of suspension bridge decks), the cable forms a parabola. For such cables:

• Span: L
• Sag (maximum deflection at midspan): d
• Uniform load per unit horizontal distance: w

The horizontal component of tension:
H = wL² / (8d)

The maximum tension occurs at the supports (steepest point):
T_max = √[H² + (wL/2)²]

where (wL/2) is the vertical reaction at a support.

Cable Under Self-Weight (Catenary):
When the cable's own weight is significant relative to external loads (or is the primary load), the shape is a catenary (the curve of a hanging chain). The shape is given by the hyperbolic cosine function:
y = (H/w) × cosh(wx/H)

For shallow sags (d << L), the parabola is a good approximation to the catenary. However, for deep sags or where self-weight dominates, the catenary must be used. The catenary has no simple closed-form relationship for tension; numerical methods are often required.

Practical Distinction:

• Parabolic: Load applied horizontally (e.g., suspension bridge deck per meter of span)
• Catenary: Load applied along the cable length (e.g., cable's own weight)

Design Considerations per Philippine Standards:
When designing cable systems in the Philippines (e.g., guyed towers, suspension structures), engineers must account for:

1. Permanent loads (self-weight of cable, deck)
2. Live loads (traffic, wind, seismic per NSCP 2015)
3. Temperature effects (expansion and contraction)
4. Environmental factors (corrosion, UV for exposed cables)
5. Safety factors per ASME B30 and relevant codes

To consolidate understanding, this section compares the four primary structural systems encountered in engineering mechanics.

Trusses:

• Definition: Structure composed entirely of two-force members (loaded only at their ends).
• Internal forces: Axial only (tension or compression); no shear or bending.
• Analysis methods: Method of joints (equilibrium at each joint) or method of sections (cut and analyze portions).
• Common use: Roof trusses, bridge trusses, lattice towers.
• Advantage: Efficient for long spans with minimal deflection; simple force distribution.
• Limitation: Cannot easily carry transverse loads on members.

Frames:

• Definition: Stationary structures containing at least one multi-force member.
• Internal forces: Axial, shear, and bending moment all may be present.
• Analysis method: Dismembering (separate into individual members and apply equilibrium to each).
• Common use: Building frames, brackets, portal frames.
• Advantage: Can support transverse loads and moments; flexible design.
• Limitation: More complex analysis; greater deflections under load.

Machines:

• Definition: Mechanisms designed to transmit or modify forces; often contain moving parts.
• Internal forces: Same as frames (axial, shear, bending), but focus is on force multiplication (mechanical advantage).
• Analysis method: Dismembering plus calculation of M.A. = Load / Effort.
• Common use: Pliers, jacks, pulleys, levers, winches.
• Advantage: Amplify applied force to move/lift larger loads.
• Limitation: Mechanical advantage decreases with friction and deformation.

Cables:

• Definition: Perfectly flexible members carrying tension only; shape determined by loading.
• Internal forces: Tension (no compression, shear, or bending).
• Key property: Horizontal component of tension is constant throughout.
• Analysis method: Identify shape (funicular polygon for concentrated loads, parabola for uniform horizontal load, catenary for self-weight); apply equilibrium at each load point or use integration.
• Common use: Suspension bridges, guy-wires, roof support systems.
• Advantage: Extremely efficient for long spans; minimal material needed.
• Limitation: Sensitive to load distribution; cannot carry compression; large deflections under live load.

Comparison Table:

Decision Tree for Analysis:
When presented with a structural system:

1. Is the structure entirely composed of two-force members? → Truss (use method of joints or sections)

2. Are there multi-force members?

   • Stationary structure? → Frame (dismember and apply equilibrium)
   • Designed to transmit forces or move? → Machine (dismember + find mechanical advantage)
3. Is the member flexible and carries only tension? → Cable (identify shape and apply equilibrium)

This systematic approach ensures correct identification and appropriate analysis method.

Success on the PRC Civil Engineer Licensure Examination requires not only understanding concepts but also avoiding common pitfalls and applying strategic problem-solving approaches.

Common Mistakes:

1. Treating a Frame Member as Two-Force When It Carries a Transverse Load

   • Mistake: Assuming a member carrying a mid-span load is a two-force member and directing its internal force along the member axis.
   • Correction: Multi-force members develop shear and bending. Always apply full equilibrium (ΣFx, ΣFy, ΣM) to each dismembered member, not just force balance along the member axis.
   • Exam implication: A 15 kN load applied mid-span on a beam makes it multi-force, not two-force, even if it appears to be a "strut".

2. Forgetting Newton's Third Law at Pins

   • Mistake: Assigning the same direction to pin forces on both members sharing a pin.
   • Correction: At every pin, the force exerted by member A on member B is equal in magnitude but opposite in direction to the force exerted by B on A.
   • Exam implication: If your FBDs don't satisfy action-reaction pairs at shared pins, recalculate immediately.
   • Check: On the beam's FBD, if the strut pushes upward at the pin, on the strut's FBD, the beam must push downward at the same pin.

3. Mixing Up Load Arm and Effort Arm in M.A. Calculations

   • Mistake: Using M.A. = Load Arm / Effort Arm (inverted).
   • Correction: M.A. = Effort Arm / Load Arm. The longer the effort arm, the greater the mechanical advantage.
   • Exam implication: A lever with a 1.0 m effort arm and 0.1 m load arm has M.A. = 1.0 / 0.1 = 10, not 0.1.
   • Memory aid: A longer effort arm multiplies the effort force.

4. Assuming Cable Horizontal Component H Changes Along the Cable

   • Mistake: Solving for H at different points and getting different values (indicating an error in geometry or equilibrium).
   • Correction: H is constant throughout the cable. If your calculations yield different H values at different points, recheck geometry, load application, or equilibrium equations.
   • Exam implication: This is a powerful check—if H varies, your setup is wrong.

5. Confusing Parabolic (Uniform Horizontal Load) with Catenary (Self-Weight)

   • Mistake: Using the parabolic formula H = wL² / (8d) for a cable whose primary load is its own weight.
   • Correction: Self-weight (or load distributed along the cable length) creates a catenary, not a parabola. The catenary formula is more complex and often requires numerical methods.
   • Exam implication: The problem statement will indicate whether the load is "distributed horizontally" (parabola) or "distributed along the cable length" (catenary). Read carefully.
   • Rule of thumb: Suspension bridge decks (uniform horizontal load) → parabola. Power transmission lines (self-weight) → catenary.

6. Finding Tension at the Wrong Location

   • Mistake: Calculating the maximum tension at midspan when it actually occurs at the support (steepest slope).
   • Correction: Cable tension is maximum where the slope is steepest. For a hanging cable, the steepest point is always at or near the supports, never at midspan (where the cable is often horizontal).
   • Exam implication: T_max = √(H² + V²) or T_max = H / cos(θ_max), where θ_max occurs at the support.

7. Not Identifying Two-Force Members Before Dismembering

   • Mistake: Treating all members equally when dismembering, resulting in unnecessary unknowns.
   • Correction: Before applying equilibrium, identify which members are two-force members. For these, the force direction is known (along the member), reducing unknowns by one.
   • Exam implication: A frame with 4 members, 2 of which are two-force, has 2 fewer unknowns than if all were multi-force. This saves calculation time.
   • Quick check: If a member is pinned at both ends and has no intermediate loads or moments, it's two-force.

8. Forgetting Support Reactions When Dismembering

   • Mistake: Neglecting the support reactions (from the wall, ground, or fixed support) when drawing the FBD of an individual member.
   • Correction: When dismembering, include all forces acting on each member: applied loads, pin forces from connected members, and reactions from supports attached to that member.
   • Exam implication: If the frame is pinned to a wall at point A, the wall reaction must appear on the FBD of the member connected to the wall.

9. Applying Moment Equilibrium About the Wrong Point

   • Mistake: Using ΣM = 0 about a point when it's mathematically valid but not strategically helpful (results in complicated algebra).
   • Correction: Choose moment reference points strategically—usually at a point where unknown forces pass through (making them "disappear" from the moment equation).
   • Exam implication: If trying to find a pin force and you have two unknowns, take moments about the other unknown's location; the resulting moment equation contains only one unknown.

10. Ignoring Geometry in Cable Problems

    • Mistake: Not clearly defining the angles (slopes) of cable segments relative to the horizontal.
    • Correction: Always draw the cable geometry carefully, label all angles, and use sin(θ) = (vertical distance) / (segment length), cos(θ) = (horizontal distance) / (segment length).
    • Exam implication: Geometry errors cascade through cable calculations. A 2° error in θ can yield 5%+ error in T.

Strategic Problem-Solving Tips:

1. Draw Large, Clear Free-Body Diagrams (FBDs)

   • A large FBD is your best defense against sign errors and missed forces.
   • Include all forces: applied loads, pin forces, support reactions, and any distributed loads.
   • Label every force with a symbol and assumed direction (you can correct the sign later).

2. Use Consistent Sign Convention

   • Choose one consistently: e.g., rightward = +x, upward = +y, counterclockwise = +M.
   • Stick to it throughout the problem. When writing ΣF or ΣM equations, substitute forces and moments with their algebraic signs.

3. Write Equilibrium Equations Algebraically, Not Component-by-Component

   • Instead of "Horizontal forces: rightward = leftward", write "ΣFx = 0: 24 − Ax = 0".
   • This avoids the pitfall of balancing directions intuitively rather than algebraically.

4. Solve Systems of Equations Systematically

   • If you have 3 unknowns (e.g., Ax, Ay, Fs), write 3 independent equations.
   • Use substitution or elimination; don't rely on intuition or trial-and-error.
   • For the board exam, show your algebra: partial credit is awarded for correct approach even if the final number is wrong.

5. Check Your Answer by Substitution or an Independent Method

   • After finding Fs = 33.9 kN, substitute it back into one of the original equations to verify.
   • If using a different method (e.g., moment about a different point), the answer should match.

6. For Cable Problems, Always Verify That H Is Constant

   • Calculate H at different points (different load points or supports).
   • If H differs, there's an error. Recheck geometry or equilibrium.

7. Estimate Before Calculating

   • For a 12 m span with a 1.5 m sag and a 10 kN central load, estimate H ≈ (10 × 6 m) / 1.5 m ≈ 40 kN (very rough).
   • Your detailed calculation should be in the same ballpark. If it's wildly different, recheck.

8. Unit Consistency

   • Use SI units throughout: kN for forces, m for distances, kN/m for distributed loads.
   • If given mixed units (kips, feet), convert immediately at the start.

9. Margin of Safety and Realistic Values

   • In practice, cable tensions are often 2–3× the minimum theoretical value (for safety and fatigue).
   • A calculated tension that's physically unreasonable (e.g., 100 MPa for a typical structural cable) suggests an error.

10. Practice with Real Examples

    • Study actual bridge cable analyses, building frame designs, and machine sketches from civil engineering references.
    • The PRC exam often uses real-world scenarios; familiarity with practical systems aids intuition.