Lesson 2.2: Bending Stresses in Simple Beams

• Jaw failure during heavy object manipulation
• Plastic deformation reducing gripping accuracy
• Fatigue cracking from repeated loading cycles
• Insufficient grip force due to excessive flexibility

Benefits of Proper Bending Analysis:

• Reliable gripping under maximum load conditions
• Optimized jaw geometry for strength and weight
• Predictable deflections for control system compensation
• Extended service life through proper stress management

1. Identify critical section:

   The maximum moment occurs at the fixed support for a cantilever beam.

2. Calculate maximum moment:

3. Verify using beam theory:

   From shear force and moment diagrams, the moment increases linearly from 0 at the tip to maximum at the support. ✅

Cross-sectional properties for rectangular section (15 mm × 8 mm):

Second moment of area:

Distance to extreme fiber:

Section modulus:

1. Apply flexural formula:

2. Alternative using section modulus:

3. Check safety factor:

1. Required maximum stress with SF = 3.0:

2. Required section modulus:

3. Calculate required height (width = 15 mm):

   Design Decision: Use 15 mm × 14 mm cross-section for standard sizing.

1. ✓ Calculate section properties

   For solid circular cross-section:

   • Area: A = πd²/4 = π(25)²/4 = 491 mm²
   • Second moment of area: I = πd⁴/64 = π(25)⁴/64 = 19,175 mm⁴
   • Section modulus: S = I/c = I/(d/2) = 19,175/12.5 = 1,534 mm³

2. ✓ Apply flexural formula

   Maximum bending stress occurs at extreme fiber: σ_max = M/S = (180 × 10³)/1,534 = 117.3 MPa

3. ✓ Check safety factor

   SF = σ_yield/σ_max = 300/117.3 = 2.56 Required SF = 2.5, Actual SF = 2.56 ✓

4. ✓ Stress distribution analysis

   • Maximum tensile stress: +117.3 MPa (bottom fiber)
   • Maximum compressive stress: -117.3 MPa (top fiber)
   • Neutral axis stress: 0 MPa

5. ✓ Design verification

   The 25 mm diameter shaft is adequate with SF = 2.56 > 2.5 required

1. Identify loading and support configuration:

   Cantilever beam with concentrated load P = 800 N at free end

   Fixed support at train roof connection provides both vertical reaction and moment reaction

2. Apply equilibrium equations:

   Vertical force equilibrium:

   (upward reaction at fixed support)

3. Moment equilibrium about fixed support A:

   (reaction moment at fixed support)

4. Verify equilibrium:

   • Vertical forces: ✅
   • Moments about any point: ✅

• Calculate shear forces at critical points:

1. Shear force distribution:

   For cantilever with concentrated end load:

   • At any point along the beam: V = -P = -800 N
   • Constant shear force throughout the cantilever length
   • Negative sign indicates internal shear acts downward

2. Verify shear force values:

   • At fixed support (x = 0): V = -800 N
   • At free end (x = 1200 mm): V = -800 N
   • No variation along beam length for single concentrated load ✅

Shear Force Diagram

Key observations: Constant shear force V = -800 N throughout the cantilever, Negative sign indicates downward internal force, No variation along beam length for concentrated end load

• Calculate bending moments at critical points:

• Method: For cantilever beam, moment at distance x from fixed support = -P × (L - x)

1. At fixed support (x = 0):

   (maximum moment)

2. At quarter span (x = 300 mm):

3. At midspan (x = 600 mm):

4. At free end (x = 1200 mm):

   (free end boundary condition)

Note: For this cantilever configuration, negative moments indicate tension on the top fiber and compression on the bottom fiber (beam curves downward).

Bending Moment Diagram

Key observations:

• Linear moment variation: M(x) = -P(L-x) = -800(1.2-x) N⋅m
• Maximum moment at fixed support (x=0): M_max = -960 N⋅m
• Zero moment at free end (x=1.2m)
• Negative moment causes tension on top fiber (cantilever curves downward)

1. Identify loading configuration:

   Cantilever beam with concentrated load P = 800 N at free end

   Length L = 1.2 m = 1200 mm ✅

2. Apply cantilever moment equations:

   For cantilever with end load, maximum moment occurs at fixed support:

   ✅

3. Verify moment distribution:

   • At fixed support (x = 0): M = -960,000 N·mm (maximum, negative indicates compression on top fiber)
   • At free end (x = L): M = 0 N·mm (free end boundary condition)
   • Moment varies linearly along the beam length ✅

1. Apply flexural formula:

2. Alternative using section modulus:

   ✅

3. Stress distribution:

   • Maximum tensile stress: +9.80 MPa (top fiber at fixed end)
   • Maximum compressive stress: -9.80 MPa (bottom fiber at fixed end)
   • Neutral axis stress: 0 MPa ✅

1. Calculate actual safety factor:

2. Compare with required safety factor:

   Required SF = 3.0 Actual SF = 25.5 >> 3.0 ✅

3. Design adequacy assessment:

   The pantograph arm is significantly over-designed with respect to static bending stress.

   This high safety margin is intentional for dynamic loading conditions. ✅

1. Dynamic amplification considerations:

   Current analysis uses static load of 800 N which already includes vibration effects.

   Actual contact force may vary: 200-1200 N depending on wire tension and vehicle dynamics. ✅

2. Sudden contact loss scenarios:

   Risk 1: Rapid load removal

   • Stored elastic energy in cantilever suddenly released
   • Can cause violent upward motion and potential wire damage
   • Damping mechanisms essential for controlled motion ✅

   Risk 2: Recontact impact

   • Higher impact forces during wire recontact
   • May exceed static analysis by 2-4× due to dynamic effects
   • Still within safety margin: 4 × 9.80 = 39.2 MPa 250 MPa ✅

3. Fatigue considerations:

   Continuous contact/loss cycles create fatigue loading. High safety factor provides adequate fatigue life margin. ✅

1. Account for distributed loading effects:

   Distributed load total force: W = w × L_total = 800 × 4.0 = 3200 N

   This acts at the centroid of the beam: x̄ = 2.0 m from A ✅

2. Apply dynamic amplification factor:

   Applied loads with dynamic effects:

   • P₁_dynamic = 1.4 × 5000 = 7000 N
   • P₂_dynamic = 1.4 × 3000 = 4200 N
   • W remains at 3200 N (self-weight not amplified) ✅

3. Calculate reaction at support B using moment equilibrium about A:

   ✅

4. Calculate reaction at support A using force equilibrium:

   ✅

5. Verify equilibrium check:

   ✅

• Calculate shear forces at critical points:

1. At x = 0 (point A): V = +3167 N

2. Just before P₁ (x = 1.5⁻): V = 3167 - 800(1.5) = 1967 N

   Just after P₁ (x = 1.5⁺): V = 1967 - 7000 = -5033 N ✅

3. Just before B (x = 3.0⁻): V = -5033 - 800(1.5) = -6233 N

   Just after B (x = 3.0⁺): V = -6233 + 11233 = 5000 N ✅

4. At overhang end (x = 4.0): V = 5000 - 800(1.0) - 4200 = 0 ✅

• Calculate bending moments at critical points:

• Method: Cut beam at calculation point, sum moments of all forces to the LEFT of the cut

1. At support A (x = 0 m):

   Forces to the left: None (this is the starting point)

   ✅

2. At midspan (x = 1.5 m):

   Forces to the left: R_A and partial distributed load

   • R_A = 3167 N at distance 1.5 m from cut
   • Distributed load from 0 to 1.5 m = 800(1.5) = 1200 N at centroid distance 0.75 m from cut

   ✅

3. At support B (x = 3.0 m):

   Forces to the left: R_A, distributed load from 0 to 3.0 m, and P₁

   • R_A = 3167 N at distance 3.0 m from cut
   • Distributed load from 0 to 3.0 m = 800(3.0) = 2400 N at centroid distance 1.5 m from cut
   • P₁ = 7000 N at distance (3.0 - 1.5) = 1.5 m from cut

   ✅

4. At overhang end (x = 4.0 m):

   Forces to the left: R_A, full distributed load, P₁, and R_B

   • R_A = 3167 N at distance 4.0 m from cut
   • Full distributed load = 800(4.0) = 3200 N at centroid distance 2.0 m from cut
   • P₁ = 7000 N at distance (4.0 - 1.5) = 2.5 m from cut
   • R_B = 11233 N at distance (4.0 - 3.0) = 1.0 m from cut
   • P₂ = 4200 N at distance (4.0 - 4.0) = 0 m from cut

   ✅

Note: The moment at the free end is approximately zero, confirming equilibrium.

3. Identify maximum moment locations:

   • Positive moment maximum: M = +3850.5 N·m at x = 1.5 m
   • Negative moment maximum: M = -4599 N·m at support B (x = 3.0 m) ✅

1. Calculate section modulus using flexural formula:

   Flexural Formula:

   For maximum stress:

   Section Modulus:

   Therefore:

   Given I = 8.2 × 10⁶ mm⁴ and c = 75 mm: ✅

2. Calculate stress at positive moment maximum (x = 1.5 m):

   ✅

   Stress distribution: Compression on top fiber, tension on bottom fiber

3. Calculate stress at negative moment maximum (x = 3.0 m):

   ✅

   Stress distribution: Tension on top fiber, compression on bottom fiber

4. Determine controlling stress:

   Comparing the two maximum stresses:

   • Positive moment stress: 35.2 MPa
   • Negative moment stress: 42.1 MPa

   Maximum bending stress = 42.1 MPa occurs at support B ✅

1. Calculate safety factor against yielding:

   ✅

2. Check against required safety factor:

   Design Criterion: If the required safety factor is 2.5, then your design safety factor must be ≥ 2.5

   Required SF = 2.5, Actual SF = 5.94 > 2.5 ✅

   Design Status: ADEQUATE with substantial margin

3. Assess critical design considerations:

   • Support B region experiences highest stress (42.1 MPa)
   • Negative bending creates tension on top flange at support B
   • Dynamic loading significantly increases applied loads (40% amplification applied) ✅
   • Design is conservative with SF = 5.94, well above required SF = 2.5

1. ✓ Calculate maximum bending moment

   For simply supported beam with uniform load + point load at center:

   • From uniform load: M₁ = wL²/8 = 500 × 0.4²/8 = 10 N·m
   • From point load: M₂ = PL/4 = 2000 × 0.4/4 = 200 N·m
   • Total: M_max = M₁ + M₂ = 210 N·m

2. ✓ Calculate I-beam properties

   • Total height: h = 180 mm
   • Area: A = 2(80×12) + (156×8) = 3,168 mm²
   • I ≈ 15.6 × 10⁶ mm⁴ (using parallel axis theorem)
   • Distance to extreme fiber: c = 90 mm

3. ✓ Calculate maximum bending stress

   σ_max = Mc/I = (210 × 10³ × 90)/(15.6 × 10⁶) = 1.21 MPa

4. ✓ Check both failure modes

   • Tensile stress (bottom): σ_t = +1.21 MPa < 150 MPa ✓
   • Compressive stress (top): σ_c = -1.21 MPa < 600 MPa ✓
   • Tensile failure governs (lower strength)

5. ✓ Safety factor assessment

   SF_tensile = 150/1.21 = 124 (very conservative design) SF_compressive = 600/1.21 = 496