Lesson 2.2: Bending Stresses in Simple Beams

• Structural failure due to stress exceeding material strength
• Excessive deflection affecting functionality and aesthetics
• Fatigue failure from repeated loading cycles
• Over-design leading to unnecessary material costs and weight

Benefits of Proper Bending Analysis:

• Safe, reliable structures designed for actual loading conditions
• Optimized cross-sections balancing strength, weight, and cost
• Predictable performance under service loads
• Extended service life through proper stress management
• Informed material selection based on stress requirements

1. Identify loading and support configuration:

   Pin joint at A (x = 0): Provides vertical reaction R_A only (no moment resistance!)

   Spring mechanism at x = 300 mm: Provides upward force F_spring

   Wire contact at P₁ (x = 1200 mm): Downward reaction force = 800 N

2. Calculate required spring force using moment equilibrium about pin A:

   (pin joint cannot resist moment)

   Taking counterclockwise moments as positive:

   ✅

   Physical meaning: Spring must provide 4× the contact force due to 4:1 lever arm ratio (1200/300)

3. Calculate pin reaction using vertical force equilibrium:

   Negative sign means R_A acts downward (pin pulls down on the arm) ✅

4. Verify equilibrium:

   • Vertical forces: ✅
   • Moments about A: ✅

5. Key insight - comparison with fixed cantilever:

   Fixed cantilever (incorrect model): Maximum moment (M = 960 N·m) at support A

   Pin + Spring (correct model): No moment at A (pin cannot resist moment). Maximum moment will occur at the spring location where internal forces change!

1. Region 1: Pin joint to spring (0 ≤ x < 0.3 m):

   Applying force equilibrium on a section between pin and spring:

   • Negative indicates downward internal shear force
   • No applied loads in this region, so V remains constant ✅

2. Region 2: Spring to wire contact (0.3 m < x ≤ 1.2 m):

   After passing the spring, the upward spring force adds to the reaction:

   • Positive indicates upward internal shear force
   • Remains constant until the wire contact point
   • At wire contact (x = 1.2 m), the 800 N downward force brings V back to zero ✅

3. Shear force jump at spring location (x = 0.3 m):

   • Just before spring:
   • Just after spring:
   • Jump magnitude: (equals spring force) ✅

4. Verify shear at wire contact:

   At x = 1.2 m (after wire contact): ✅

Shear Force Diagram

Key observations: Two-region shear distribution: V = -2400 N from pin to spring, then V = +800 N from spring to wire Discontinuity at spring: 3200 N upward jump where spring force is applied Unlike fixed cantilever: Shear is NOT constant—changes sign at spring location

• Method: Cut the beam at position x, sum moments of forces to the LEFT of the cut

1. At pin joint (x = 0):

   Forces to the left: None (starting point)

   Key insight: Pin joint cannot resist moment! ✅

2. Region 1 (0 < x < 0.3 m): Pin to spring

   Forces to the left: R_A only

   (in N·m when x in meters)

   • Linear increase in negative moment
   • At x = 0.3 m (spring location): ✅

3. Region 2 (0.3 m < x < 1.2 m): Spring to wire

   Forces to the left: R_A and F_spring

   (N·m)

   • At x = 0.3 m: (continuous at spring)
   • At x = 0.6 m (midspan):
   • At x = 0.9 m:
   • At x = 1.2 m (wire contact): ✅

4. Verify moment equilibrium:

   At wire contact (x = 1.2 m), moment should equal zero (free end condition):

   ✅

Critical observation: Negative moments throughout indicate tension on top fiber and compression on bottom fiber. The beam curves downward (pantograph arm droops under wire contact force).

Bending Moment Diagram

Key observations:

• Region 1 (0 to 0.3m): M(x) = -2400x (steep negative slope)
• Region 2 (0.3m to 1.2m): M(x) = 800x - 960 (gentler positive slope)
• Maximum moment magnitude: |M_max| = 720 N·m at spring location (x = 0.3 m)
• Critical design location: Spring attachment point experiences highest bending stress

1. Apply flexural formula at critical section (spring location):

2. Alternative using section modulus:

   ✅

3. Stress distribution at spring attachment (x = 0.3 m):

   • Maximum tensile stress: +7.35 MPa (top fiber)
   • Maximum compressive stress: -7.35 MPa (bottom fiber)
   • Neutral axis stress: 0 MPa ✅

1. Calculate actual safety factor:

2. Compare with required safety factor:

   Required SF = 3.0 Actual SF = 34.0 >> 3.0 ✅

3. Design adequacy assessment:

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

   This high safety margin (>10× required) is intentional for several reasons:

   • Dynamic loading: Sudden wire contact/loss creates impact forces
   • Vibration effects: Train motion causes cyclic loading
   • Fatigue resistance: Millions of contact cycles over service life
   • Electrical safety: Must not fail near high-voltage overhead wire
   • Wear tolerance: Spring force increases as contacts wear ✅

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 (dewirement)

   • Stored elastic energy in beam suddenly released
   • Spring force causes violent upward motion if not damped
   • Can damage overhead wire or contact strip
   • 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 × 7.35 = 29.4 MPa 250 MPa (SF = 8.5 even with 4× dynamic amplification) ✅

3. Fatigue considerations:

   Continuous contact/loss cycles create fatigue loading:

   • Typical train: 10,000+ contact cycles per day
   • Over 20-year service life: ~70 million cycles
   • Endurance limit for steel: ~125 MPa (50% of yield for high-cycle fatigue)
   • Maximum stress (7.35 MPa) is far below endurance limit
   • High safety factor (SF = 34) 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:

   ✅

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 ✅

• 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. Define system variables:

   Let a = position of print head from left support A (0 ≤ a ≤ L = 1200 mm)

   Print head load: P1 = 250 N (downward concentrated load at position a)

   Supports: Simply supported at A (x = 0) and B (x = L = 1200 mm) ✅

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

   (counterclockwise positive)

   Taking moments about point A:

   (in N when a in mm)

   Or: ✅

   Physical meaning: As print head moves away from A toward B, reaction at B increases linearly from 0 to P

3. Calculate reaction at support A using vertical force equilibrium:

   (in N when a in mm) ✅

   Physical meaning: As print head moves away from A, reaction at A decreases linearly from P to 0

4. Verify equilibrium:

   • Vertical forces: ✅
   • Moments about A: ✅

5. Special/Critical positions:

   • Print head at A (a = 0): N, N ✅
   • Print head at midspan (a = 600 mm): N, N ✅
   • Print head at B (a = 1200 mm): N, N ✅

The bending moment varies with both:

• x = position along beam where we calculate moment (0 ≤ x ≤ L)
• a = position of print head load (0 ≤ a ≤ L)

Method: Cut beam at position x, sum moments of forces to the LEFT of the cut

1. Case 1: Section is to the LEFT of print head (0 ≤ x < a)

   Forces to the left of cut at x: Only

   for ✅

   Substituting P1 = 250 N, L = 1200 mm:

   (N·mm)

2. Case 2: Section is to the RIGHT of print head (a < x ≤ L)

   Forces to the left of cut at x: and P

   for ✅

   Substituting P1 = 250 N, L = 1200 mm:

   (N·mm)

3. Complete bending moment function:

   At the load location (x = a):

   From left:

   From right:

   Both expressions give the same value (moment is continuous) ✅

4. Verify boundary conditions:

   • At support A (x = 0): ✅
   • At support B (x = L): ✅
   • At free ends of simply supported beam, moment must be zero ✅

Shear Force and Bending Moment Diagrams (at critical position a = 600 mm)

Key observations:

• Shear distribution: V = +125 N left of load, V = -125 N right of load
• Discontinuity at load: 250 N downward jump where print head is located
• Moment distribution: Triangular shape, peaking at load position with M = 75 N·m
• Worst case: Maximum moment occurs when print head is at midspan (a = 600 mm)

1. Critical insight: Maximum moment occurs AT the load location

   For a simply supported beam with a single concentrated load, maximum moment occurs directly under that load.

   At x = a:

   ✅

2. Find load position ‘a’ that maximizes the moment:

   To find maximum, take derivative with respect to a and set equal to zero:

   ✅

   Critical position: Maximum moment occurs when print head is at midspan (a = L/2 = 600 mm)

3. Calculate absolute maximum moment:

   Substituting a = L/2 into moment equation:

   N·mm = 75 N·m ✅

4. Physical interpretation:

   • When print head is at midspan, both supports share load equally (125 N each)
   • Moment diagram is triangular with peak at center
   • This represents the worst-case bending condition for the gantry rail
   • As print head moves away from center toward either support, maximum moment decreases

5. Moment variation with print head position:

   At various positions:

   • a = 0 mm (at A): N·m ✅
   • a = 300 mm: N·mm = 56.25 N·m ✅
   • a = 600 mm (midspan): N·mm = 75 N·m ✅
   • a = 900 mm: N·mm = 56.25 N·m ✅
   • a = 1200 mm (at B): N·m ✅

1. Apply flexural formula at critical section (midspan, a = 600 mm):

2. Alternative using section modulus:

   ✅

3. Stress distribution at midspan (worst case):

   • Maximum tensile stress: +0.852 MPa (bottom fiber when load above)
   • Maximum compressive stress: -0.852 MPa (top fiber)
   • Neutral axis stress: 0 MPa ✅

   Note: For downward load, bottom of beam is in tension, top is in compression

4. Stress at other critical locations:

   Using :

   • a = 300 mm: MPa ✅
   • a = 600 mm: MPa (maximum) ✅
   • a = 900 mm: MPa ✅

1. Calculate actual safety factor:

2. Compare with required safety factor:

   Required SF = 3.0 Actual SF = 322.8 >> 3.0 ✅

3. Design adequacy assessment:

   The gantry rail is extremely over-designed with respect to static bending stress.

   This enormous safety margin (>100× required) is intentional for several reasons:

   • Dynamic effects: Print head acceleration/deceleration creates higher dynamic loads
   • Vibration control: High stiffness (not just strength) is critical for print quality
   • Deflection limits: Deflection (not stress) likely controls design for positioning accuracy
   • Long-term reliability: Aluminum fatigue under continuous cyclic loading
   • Multiple simultaneous loads: Additional accessories, cables, and mounting hardware ✅

4. Design is controlled by stiffness, not strength:

   Maximum deflection at midspan for simply supported beam with center load:

   Where E = 69 GPa = 69,000 MPa for aluminum

   mm ✅

   For precision 3D printing:

   • Typical acceptable deflection: < 0.1 mm
   • Calculated deflection: 0.059 mm ✅
   • Deflection limit is satisfied, confirming stiffness-controlled design

1. Dynamic loading at high print speeds:

   At 200 mm/s print speed with rapid acceleration/deceleration:

   • Typical acceleration: 1-3 m/s² (1000-3000 mm/s²)
   • Dynamic force: F_dynamic = F_static + ma
   • Print head mass: m = P/g = 250/9.81 ≈ 25.5 kg
   • Additional dynamic force: ma = 25.5 × 3 = 76.5 N
   • Total dynamic load: P_total ≈ 250 + 76.5 = 326.5 N (30% increase) ✅

2. Dynamic stress assessment:

   With 30% dynamic amplification:

   MPa

   Safety factor with dynamic effects: (still extremely safe) ✅

3. Print quality implications:

   Positioning accuracy factors:

   • Beam deflection: 0.059 mm is acceptable for typical FDM printing (layer height ≥ 0.1 mm)
   • Vibration damping: Aluminum’s damping properties help reduce oscillations
   • Resonant frequencies: Beam natural frequency should be >> print motion frequency

   Critical for print quality:

   • Excessive deflection causes dimensional inaccuracy
   • Vibration creates surface defects (ringing, ghosting)
   • Dynamic deflection variation during acceleration creates layer inconsistencies ✅

4. High-speed reliability considerations:

   Challenges at 200 mm/s:

   • Vibration amplitudes increase with speed²
   • Resonance risk if motion frequency matches natural frequency
   • Bearing wear accelerates with speed and cyclic loading
   • Belt/linear rail precision becomes critical

   Design adequacy:

   • High safety factor (247×) provides margin for unforeseen loads
   • Stiffness adequate for deflection control
   • Should verify natural frequency >> operating frequency to avoid resonance ✅

5. Recommendations for improved performance:

   • Add stiffening ribs to increase natural frequency
   • Use active damping for high-speed applications
   • Monitor bearing wear for long-term position accuracy
   • Consider vibration isolation for base mounting
   • Implement motion profiling with S-curve acceleration to reduce jerks ✅