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

We’ll analyze the beam with the print head at three different positions and calculate the support reactions for each case.

General formulas for simply supported beam with point load at distance ‘x’ from left support:

1. Position 1: Print head at left quarter-point (x = 300 mm)

   

   Calculate R_B using moment equilibrium about A:

   (taking moments about A)

   N ✅

   Calculate R_A using vertical force equilibrium:

   N ✅

   Verify: N ✅

2. Position 2: Print head at midspan (x = 600 mm)

   

   Calculate R_B:

   N ✅

   Calculate R_A:

   N ✅

   Note: At midspan, both supports share the load equally ✅

3. Position 3: Print head at right quarter-point (x = 900 mm)

   

   Calculate R_B:

   N ✅

   Calculate R_A:

   N ✅

   Observation: Notice the symmetry - reactions at Position 1 and Position 3 are mirror images ✅

4. Summary of reactions:

   Position	Distance x	R_A (N)	R_B (N)
   Left quarter	300 mm	187.5	62.5
   Midspan	600 mm	125.0	125.0
   Right quarter	900 mm	62.5	187.5

For a simply supported beam with a point load, the maximum moment occurs directly under the load at the load position.

General formula:

1. Position 1: Load at left quarter-point (x = 300 mm)

   Maximum moment occurs at x = 300 mm (where the load is):

   N·mm = 56.25 N·m ✅

2. Position 2: Load at midspan (x = 600 mm)

   Maximum moment occurs at x = 600 mm (where the load is):

   N·mm = 75.0 N·m ✅

3. Position 3: Load at right quarter-point (x = 900 mm)

   Maximum moment occurs at x = 900 mm (where the load is):

   N·mm = 56.25 N·m ✅

4. Summary and comparison:

   Position	Distance x	Maximum Moment M_max
   Left quarter	300 mm	56.25 N·m
   Midspan	600 mm	75.0 N·m ⭐
   Right quarter	900 mm	56.25 N·m

   Key Observations:

   • ⭐ Maximum moment occurs at midspan (x = 600 mm)
   • Symmetry: Quarter-point positions give identical moments
   • Midspan moment is 33% higher than quarter-point moments
   • This confirms midspan is the worst-case position for bending stress

Bending Moment Diagram for midspan position (worst case):

Now we’ll use the flexural formula to calculate the bending stress for each load position.

Flexural formula: or where

Given: I = 2.2 × 10⁶ mm⁴, c = 25 mm, S = 88,000 mm³

1. Position 1: Left quarter-point (M = 56,250 N·mm)

   Stress distribution:

   • Bottom fiber (tension): +0.639 MPa
   • Top fiber (compression): -0.639 MPa ✅

2. Position 2: Midspan (M = 75,000 N·mm)

   Stress distribution:

   • Bottom fiber (tension): +0.852 MPa
   • Top fiber (compression): -0.852 MPa ✅

3. Position 3: Right quarter-point (M = 56,250 N·mm)

   Stress distribution:

   • Bottom fiber (tension): +0.639 MPa
   • Top fiber (compression): -0.639 MPa ✅

4. Summary and worst-case identification:

   Position	Maximum Moment	Maximum Stress
   Left quarter	56.25 N·m	0.639 MPa
   Midspan	75.0 N·m	0.852 MPa ⭐
   Right quarter	56.25 N·m	0.639 MPa

   Conclusion: Maximum bending stress occurs when the print head is at midspan: σ_max = 0.852 MPa

1. Calculate actual safety factor at worst case (midspan):

2. Compare with required safety factor:

   • Required SF: 3.0
   • Actual SF: 323
   • Result: 323 >> 3.0 ✅ Design is SAFE

3. Design adequacy assessment:

   The gantry rail has an extremely high safety factor (>100× required).

   Why such a large safety margin?

   • ✅ Stiffness, not strength, controls the design - the beam must be rigid enough to prevent deflection that affects print quality
   • ✅ Dynamic loads - print head acceleration/deceleration creates additional forces
   • ✅ Vibration control - high stiffness prevents resonance and vibration during printing
   • ✅ Long-term reliability - aluminum experiences fatigue under repeated cyclic loading

4. Quick deflection check (stiffness verification):

   Maximum deflection at midspan:

   For precision 3D printing:

   • Typical acceptable deflection: < 0.1 mm
   • Calculated deflection: 0.059 mm ✅
   • Stiffness requirement is satisfied

   Conclusion: The large cross-section is needed for stiffness (deflection control), not strength. Bending stress is very low compared to material strength.