7.2 Plane Trusses, Friction Mechanics & Screws/Wedges

6.2 Plane Trusses, Friction Mechanics & Screws/Wedges

In mining engineering and heavy industrial plant infrastructure, structural trusses support mine headframes, conveyor gantries, and coal handling plant (CHP) roofs, while friction mechanics governs haul road vehicle stability, wedge-based coal splitters, and power screw hoists. Understanding the governing equations for pin-jointed frameworks and friction-driven mechanical drives is critical for CIL MT aspirants.

---

1. Plane Trusses: Structural Fundamentals

A truss is a triangulated structure composed of slender structural members interconnected at their ends by joints. A plane truss lies entirely in a two-dimensional plane and carries in-plane loads.

Core Engineering Assumptions for Ideal Trusses

1. Pin-Jointed Connections: All joints are treated as frictionless, smooth cylindrical pins.
2. Two-Force Members: All members are straight, slender, and subjected to loads applied exclusively at the joints (nodes). Consequently, every member carries only axial tension ($T$) or axial compression ($C$), with zero bending moment and zero transverse shear force.
3. Negligible Member Weight: The self-weight of members is negligible compared to applied external loads (or split equally as lumped loads at bounding joints).

               Tension Member: Pulled at ends, pulls on joints
               [Joint A] ◄───(Member in Tension)───► [Joint B]

               Compression Member: Pushed at ends, pushes on joints
               [Joint A] ───►(Member in Compression)◄─── [Joint B]

---

2. Determinacy and Stability of Plane Trusses

Let $m$ be the number of members, $j$ be the number of joints, and $r$ be the number of external support reactions (for a standard statically determinate support system, $r = 3$):

Plane Truss Determinacy Criteria ($m = 2j - 3$)

Note for Space Trusses (3D): Determinacy condition is $m = 3j - 6$.

---

3. Analysis Methods for Plane Trusses

                      ┌────────────────────────────────────────┐
                      │       Truss Analysis Methodology       │
                      └───────────────────┬────────────────────┘
                                          │
               ┌──────────────────────────┴──────────────────────────┐
               │                                                     │
    ┌──────────▼──────────┐                               ┌──────────▼──────────┐
    │   Method of Joints  │                               │  Method of Sections │
    └──────────┬──────────┘                               └──────────┬──────────┘
               │                                                     │
    • Isolates single nodes                                • Cuts across ≤ 3 members
    • Applies ΣFx = 0, ΣFy = 0                             • Applies ΣFx=0, ΣFy=0, ΣMO=0
    • Solves ≤ 2 unknown forces/node                       • Direct solution for specific member
    • Best for ALL member forces                           • Best for targeted internal forces

Method of Joints

1. Solve for external support reactions using global equilibrium ($\Sigma F_x = 0, \Sigma F_y = 0, \Sigma M = 0$).
2. Select a joint with at most two unknown member forces and at least one known applied load or reaction.
3. Draw the joint FBD, assuming all unknown member forces are in tension (pointing away from joint). A negative result indicates compression.
4. Apply $\Sigma F_x = 0$ and $\Sigma F_y = 0$. Progress sequentially across the truss.

Method of Sections (Ritter's Method)

1. Pass an imaginary cutting line through the truss, slicing not more than three non-concurrent, non-parallel members including the target member.
2. Isolate one side of the severed truss (left or right portion).
3. Apply the three rigid body equilibrium equations: $\Sigma F_x = 0, \Sigma F_y = 0, \Sigma M_O = 0$.
4. Taking moments about the intersection point of two unknown severed members yields the force in the third member directly in a single step.

---

4. Zero-Force Member Inspection Rules

Identifying zero-force members by inspection saves immense calculation time in competitive exams:

Rule 1: Two non-collinear members, no load       Rule 2: Three members, two collinear, no load

            Member 1                                        Member 1 ────┬──── Member 2 (Collinear)
           /                                                             │
          /  (θ ≠ 180°)                                                  │ Member 3 (F3 = 0)
         * Joint (No external load)                                      ▼
          \
           \ Member 2                                    Result: F3 = 0, F1 = F2

   Result: F1 = 0 and F2 = 0

1. Rule 1 (Two-Member Joint): If only two non-collinear members meet at an unloaded joint, the force in both members is zero ($F_1 = 0, F_2 = 0$).
2. Rule 2 (Three-Member Joint with Collinear Pair): If three members meet at an unloaded joint and two members are collinear, the third non-collinear member is a zero-force member ($F_3 = 0$), and the two collinear members carry equal forces ($F_1 = F_2$).
3. Rule 3 (Two Members with External Load): If two non-collinear members meet at a joint with an external load aligned collinearly with one member, the other non-aligned member is a zero-force member.

---

5. Coulomb's Laws of Dry Friction

Dry friction (Coulomb friction) arises from microscopic surface asperities resisting tangential sliding motion between contacting solid bodies.

       Friction Force (F)
               ▲
               │                     Limiting Friction (F_lim = μs * N)
               │                            /│
               │                           / ┼───────────────────────── Kinetic Friction (Fk = μk * N)
               │                          /  │
               │   Static Region         /   │      Dynamic / Kinetic Region
               │   (F ≤ μs * N)         /    │      (F = μk * N, nearly constant)
               │   F = Applied Force P /     │
               │                      /      │
               └─────────────────────┴───────┴────────────────────────► Applied Tangential Force (P)
                                  Impending Motion

Key Parameters

• Static Friction Force: $0 \le F_s \le F_{\text{lim}}$, where $F_s$ is self-adjusting to match applied shear.
• Limiting Static Friction: $F_{\text{lim}} = \mu_s N$.
• Kinetic (Dynamic) Friction: $F_k = \mu_k N$, where $\mu_k < \mu_s$.
• Angle of Friction ($\phi$): The angle between the resultant contact reaction $\vec{R}$ and the surface normal $\vec{N}$ at impending motion: $\\tan\\phi = \\frac{F_{\\text{lim}}}{N} = \\frac{\\mu_s N}{N} = \\mu_s \\implies \\phi = \\tan^{-1}(\\mu_s)$
• Cone of Friction: A cone of apex angle $2\phi$ generated around the normal. If the resultant contact force lies inside the cone of friction, motion cannot occur.
• Angle of Repose ($\theta_{\text{repose}}$): The maximum inclination angle of a rough plane with the horizontal at which a block rests in limiting equilibrium under gravity alone without sliding down: $mg \\sin\\theta_{\\text{repose}} = F_{\\text{lim}} = \\mu_s (mg \\cos\\theta_{\\text{repose}}) \\implies \\tan\\theta_{\\text{repose}} = \\mu_s \\implies \\theta_{\\text{repose}} = \\phi$

---

6. Slipping vs. Tipping (Overturning) Mechanics

For a rectangular block of width $b$, height $h$, and mass $m$ on a horizontal rough surface subjected to a horizontal force $P$ applied at height $y$:

                         ┌─────────── b ───────────┐
                         │                         │
                         │         ┌───┐           │
             P ──────────┼─────────┤ G ├───────────┤ (Height y)
                         │         └───┘           │
                         │           │             │ Height h
                         │           │ mg          │
                         └───────────┼─────────────┘
                                     │       ▲ N
                               ◄─────┴───────┤ (At corner for impending tipping)
                                 F_friction  │

1. Condition for Impending Slipping: $P_{\text{slip}} = \mu_s mg$.
2. Condition for Impending Tipping: Taking moments about the bottom tipping edge: $P_{\\text{tip}} \\cdot y = mg \\left(\\frac{b}{2}\\right) \\implies P_{\\text{tip}} = mg \\left(\\frac{b}{2y}\\right)$
3. Governing Failure Mode:
   • If $P_{\text{slip}} < P_{\text{tip}} \implies \mu_s < \frac{b}{2y}$, the block slips before tipping.
   • If $P_{\text{tip}} < P_{\text{slip}} \implies \frac{b}{2y} < \mu_s$, the block tips before slipping.
   • For force applied at top edge ($y = h$), slip occurs first when $\frac{h}{b} < \frac{1}{2\mu_s}$.

---

7. Wedge Mechanics

A wedge is a simple machine used to lift heavy loads or tighten machine components via small driving forces.

                      ┌──────────────────────┐
                      │      Block (W)       │ ◄─── Constrained vertically
                      │                      │
                      └──────────┬───────────┘
                                /│ R2 (Reaction at inclined face, inclined by φ2)
                               / │
                     ┌────────/  │
     Drive Force P ──┼► Wedge │ θ│ (Wedge taper angle θ)
                     └────────┴──┘
                     ───────────────────────── R1 (Floor reaction, inclined by φ1)

Analytical Equilibrium of Wedge and Block

At impending motion (wedge driven forward to raise block of weight $W$):

1. Isolate the Block: $\\sum F_x = 0, \\quad \\sum F_y = 0$ Total reaction $R_2$ on inclined face makes angle $(\theta + \phi_2)$ with vertical, where $\phi_2 = \tan^{-1}\mu_2$.
2. Isolate the Wedge: Horizontal drive force required: $P = R_1 \\sin\\phi_1 + R_2 \\sin(\\theta + \\phi_2)$ For equal friction coefficient $\phi_1 = \phi_2 = \phi_3 = \phi$ at all surfaces: $P = W \\tan(\\theta + 2\\phi)$

---

8. Power Screws & Thread Mechanics

Power screws convert rotary motion into high-force linear translation (e.g., screw jacks, machine tool lead screws, sluice gates).

     Developed Screw Thread on Inclined Plane
           ▲
           │              /│
           │             / │
      Lead │            /  │ Lead Angle α = tan⁻¹(L / π*dm)
       (L) │           /   │
           │          /    │
           └─────────┴─────┴────────────────►
                Mean Circumference (π * dm)

Kinematics and Geometry

• Lead ($L$): Axial distance moved in one complete revolution. For single-start thread, $L = p$ (pitch); for $n$-start thread, $L = n \cdot p$.
• Mean Diameter ($d_m$): $d_m = d - \frac{p}{2} = \frac{d + d_c}{2}$ (where $d$ is nominal diameter, $d_c$ is core diameter).
• Lead (Helix) Angle ($\alpha$): $\\tan\\alpha = \\frac{L}{\\pi d_m}$

Torque Equations for Square Threads

1. Torque Required to Raise Load ($W$): $T_{\\text{raise}} = W \\frac{d_m}{2} \\tan(\\alpha + \\phi) = W \\frac{d_m}{2} \\left[\\frac{\\tan\\alpha + \\mu}{1 - \\mu \\tan\\alpha}\\right]$
2. Torque Required to Lower Load ($W$): $T_{\\text{lower}} = W \\frac{d_m}{2} \\tan(\\phi - \\alpha) = W \\frac{d_m}{2} \\left[\\frac{\\mu - \\tan\\alpha}{1 + \\mu \\tan\\alpha}\\right]$

Self-Locking vs. Overhauling Condition

• Self-Locking Screw: When the load remains stationary and does not descend spontaneously when the driving torque is removed. Requires: $\\phi \\ge \\alpha \\iff \\mu \\ge \\tan\\alpha$
• Overhauling (Backdriving) Screw: When $\alpha > \phi$, $T_{\text{lower}}$ becomes negative, meaning the load descends spontaneously unless a restraining torque is applied.

Power Screw Efficiency ($\eta$)

• Efficiency ($\eta$): Ratio of ideal torque without friction ($\phi = 0$) to actual torque with friction: $\\eta = \\frac{T_0}{T_{\\text{raise}}} = \\frac{W \\frac{d_m}{2} \\tan\\alpha}{W \\frac{d_m}{2} \\tan(\\alpha + \\phi)} = \\frac{\\tan\\alpha}{\\tan(\\alpha + \\phi)}$
• Maximum Efficiency Condition: Differentiating $\eta$ with respect to $\alpha$ gives the optimum helix angle: $\\alpha_{\\text{opt}} = 45^{\\circ} - \\frac{\\phi}{2}$ $\\eta_{\\text{max}} = \\frac{1 - \\sin\\phi}{1 + \\sin\\phi}$
• Efficiency of Self-Locking Screws: Since $\alpha \le \phi$, $\\eta_{\\text{self-lock}} = \\frac{\\tan\\alpha}{\\tan(\\alpha + \\phi)} < \\frac{\\tan\\phi}{\\tan(2\\phi)} = \\frac{\\tan\\phi}{\\frac{2\\tan\\phi}{1 - \\tan^2\\phi}} = \\frac{1 - \\tan^2\\phi}{2} < 0.50$ Rule: A self-locking power screw always operates at an efficiency of less than 50%.

---

9. Step-by-Step Worked Engineering Calculations

Worked Example 6.2.1: Method of Sections on a Warren Truss

Problem: A symmetric plane Warren truss spans $L = 12\text{ m}$ between a pinned support at $A$ and a roller support at $B$. The bottom chord consists of three $4\text{ m}$ panels ($AC, CD, DB$). The top chord consists of joints $E$ and $F$ at height $h = 3\text{ m}$. A vertical downward load of $60\text{ kN}$ acts at bottom joint $C$, and $40\text{ kN}$ acts at joint $D$. Calculate the internal force in the top chord member $EF$ and bottom chord member $CD$.

Step-by-step Solution:

1. Calculate External Support Reactions:

   • Span = $12\text{ m}$. Coordinates from $A$: $C(4\text{ m}), D(8\text{ m}), B(12\text{ m})$.
   • $\Sigma M_A = 0 \implies R_B \times 12 - (60 \times 4) - (40 \times 8) = 0$
   • $12 R_B = 240 + 320 = 560 \implies R_B = 46.67\text{ kN}$
   • $R_A = (60 + 40) - 46.67 = 53.33\text{ kN}$

2. Select Cut Section: Pass a vertical section cut $1-1$ through top chord $EF$, diagonal $ED$, and bottom chord $CD$. Isolate the left portion (Joints $A, C, E$).

3. Calculate Force in Top Chord $EF$: Take moments about joint $D(x = 8\text{ m}, y = 0)$ where unknown diagonal $ED$ and bottom chord $CD$ intersect: $\\Sigma M_D = 0 \\implies (R_A \\times 8) - (60 \\times 4) + (F_{EF} \\times h) = 0$ $(53.33 \\times 8) - 240 + (F_{EF} \\times 3) = 0$ $426.64 - 240 + 3 F_{EF} = 0 \\implies 186.64 + 3 F_{EF} = 0$ $F_{EF} = -62.21\\text{ kN} \\implies 62.21\\text{ kN} \\text{ (Compression)}$

4. Calculate Force in Bottom Chord $CD$: Top chord joint $E$ is located at $x = 2\text{ m}, y = 3\text{ m}$. Take moments about joint $E$: $\\Sigma M_E = 0 \\implies (R_A \\times 2) - (F_{CD} \\times 3) = 0$ $(53.33 \\times 2) - 3 F_{CD} = 0 \\implies 106.66 = 3 F_{CD}$ $F_{CD} = +35.55\\text{ kN} \\implies 35.55\\text{ kN} \\text{ (Tension)}$

---

Worked Example 6.2.2: Power Screw Torque and Efficiency

Problem: A single-start square-threaded screw jack has a nominal diameter $d = 50\text{ mm}$, pitch $p = 10\text{ mm}$, and coefficient of thread friction $\mu = 0.15$. Determine (a) the lead angle $\alpha$, (b) whether the screw is self-locking, (c) the torque required to raise a load of $W = 25\text{ kN}$, and (d) the screw efficiency.

Step-by-step Solution:

1. Mean Diameter and Lead Angle:

   • Mean diameter $d_m = d - \frac{p}{2} = 50 - 5 = 45\text{ mm} = 0.045\text{ m}$.
   • Lead $L = 10\text{ mm}$.
   • $\tan\alpha = \frac{L}{\pi d_m} = \frac{10}{\pi \times 45} = \frac{10}{141.37} \approx 0.07073 \implies \alpha = \tan^{-1}(0.07073) \approx 4.05^{\circ}$.

2. Friction Angle and Self-Locking Verification:

   • Friction angle $\phi = \tan^{-1}(\mu) = \tan^{-1}(0.15) \approx 8.53^{\circ}$.
   • Since $\phi = 8.53^{\circ} > \alpha = 4.05^{\circ}$ (i.e., $\mu > \tan\alpha$), the screw is strictly self-locking.

3. Torque Required to Raise $W = 25\text{ kN}$: $\\tan(\\alpha + \\phi) = \\frac{\\tan\\alpha + \\tan\\phi}{1 - \\tan\\alpha \\tan\\phi} = \\frac{0.07073 + 0.15}{1 - (0.07073 \\times 0.15)} = \\frac{0.22073}{0.98939} \\approx 0.2231$ $T_{\\text{raise}} = W \\frac{d_m}{2} \\tan(\\alpha + \\phi) = (25000) \\times \\left(\\frac{0.045}{2}\\right) \\times 0.2231 = 25000 \\times 0.0225 \\times 0.2231 = 125.50\\text{ N}\\cdot\\text{m}$

4. Screw Efficiency: $\\eta = \\frac{\\tan\\alpha}{\\tan(\\alpha + \\phi)} = \\frac{0.07073}{0.2231} \\approx 0.3170$ (Consistent with the fundamental rule that $\eta < 50\%$ for self-locking screws, giving an efficiency of 31.70%).