Lesson 2.1: Shear Force and Bending Moment in Beams

• Arm deflection compromises load positioning accuracy and operator safety
• Structural failure causes expensive downtime and potential injury hazards
• Excessive arm weight reduces payload capacity and increases energy consumption
• Insufficient stiffness creates handling difficulties during manual guidance

Benefits of Proper Beam Analysis in Material Handling Systems:

• Optimized structure for handling 120kg payloads safely with minimal deflection
• Predictable behavior under loading conditions for operator confidence
• Reliable operation throughout full payload range with appropriate safety margins
• Cost-effective design balancing structural requirements with manufacturing economics

1. Sum of vertical forces (↑ positive):

   Reasoning:

   • = upward reaction (positive)
   • = total distributed load (downward, negative) ✅
   • = point load (downward, negative) ✅

   ✅

2. Sum of moments about base (counterclockwise positive):

   Reasoning:

   • = base reaction moment (counterclockwise, positive)
   • = distributed load moment (clockwise, negative) ✅
   • = point load moment (clockwise, negative) ✅

   ✅

3. Verify equilibrium:

   Forces: ↑1387.5 N = ↓(187.5 + 1200) N ✅ Moments: 3234.375 N·m = 3234.375 N·m ✅

Method: Cut the beam at distance x from base and analyze equilibrium.

For 0 ≤ x ≤ 2.5 m:

Applying the cantilever shear force equation (see theory section):

Key points:

• At x = 0: V(0) = -1387.5 N (negative, as expected for downward loads) ✅
• At x = 2.5 m: V(2.5) = -1387.5 + 75(2.5) = -1200 N ✅

Jump at tip from point load:
Just before tip: V = -1200 N
Just after tip: V = -1200 + 1200 = 0 N ✅

Method: Integrate shear force or use moment equilibrium.

For 0 ≤ x ≤ 2.5 m:

Applying the cantilever bending moment equation (see theory section):

Key points:

• At x = 0: M(0) = -3234.375 N·m (negative = sagging moment at fixed end) ✅
• At x = 2.5 m: M(2.5) = -3234.375 + 1387.5(2.5) - 37.5(2.5)² = -3234.375 + 3468.75 - 234.375 = 0 N·m ✅

Free end boundary condition: At the free end (x = 2.5 m), the internal moment must be zero since there are no applied moments. Our equation gives M(2.5) = 0, confirming equilibrium at the tip ✅

1. Maximum shear location:

   Maximum |V| = 1387.5 N at x = 0 (fixed support) ✅

2. Maximum moment location:

   For cantilever beams with downward loading, the maximum moment typically occurs at the fixed support. Let’s verify by checking the boundary values:

   • At x = 0 (support): |M(0)| = 3234.375 N·m = 3,234,375 N·mm ✅
   • At x = 2.5 m (tip): |M(2.5)| = 0 N·m ✅

   Maximum |M| = 3,234,375 N·mm at x = 0 (fixed support) ✅

3. Maximum bending stress calculation:

   Using the flexure formula:

   Where:

   • N·mm
   • mm (distance from neutral axis to extreme fiber)
   • mm⁴

   ✅

4. Safety factor against yield:

   ✅

   Engineering Assessment: Safety factor of 11.3 is very conservative and indicates over-design. For robotic applications, SF = 2-4 is typically adequate.

5. Design recommendations:

   Option 1: Reduce cross-section to 60 mm × 8 mm for weight optimization while maintaining SF ≥ 3

   Option 2: Use hollow rectangular section (80×10 mm outer, 60×6 mm inner) for 40% weight reduction

   Option 3: Implement variable cross-section with maximum at base, tapering toward tip

1. Sum of vertical forces (↑ positive):

   ✅

2. Sum of moments about A (counterclockwise positive):

   ✅

3. Calculate R_A:

   ✅

4. Verify by symmetry:

   Since loads are symmetric about centerline (x = 1000 mm), reactions are equal: N ✅

Method: Track cumulative loads from left support.

Key regions and jumps:

For 0 ≤ x < 200 mm: ✅

At x = 200 mm (first load): Jump down by 400 N: V = 1000 - 400 = 600 N ✅

For 200 < x < 600 mm: ✅

At x = 600 mm (second load): Jump down by 400 N: V = 600 - 400 = 200 N ✅

For 600 < x < 1000 mm: ✅

At x = 1000 mm (center load): Jump down by 400 N: V = 200 - 400 = -200 N ✅

For 1000 < x < 1400 mm: ✅

At x = 1400 mm (fourth load): Jump down by 400 N: V = -200 - 400 = -600 N ✅

For 1400 < x < 1800 mm: ✅

At x = 1800 mm (fifth load): Jump down by 400 N: V = -600 - 400 = -1000 N ✅

For 1800 < x ≤ 2000 mm: ✅

Maximum shear force: |V_max| = 1000 N at both supports ✅

Method: Integrate shear force or use equilibrium of cut sections.

For 0 ≤ x ≤ 200 mm:

For 200 ≤ x ≤ 600 mm:

For 600 ≤ x ≤ 1000 mm:

For 1000 ≤ x ≤ 1400 mm:

For 1400 ≤ x ≤ 1800 mm:

For 1800 ≤ x ≤ 2000 mm:

Key moment values:

• At x = 0: M(0) = 0 N·mm ✅
• At x = 200 mm: M(200) = 200,000 N·mm ✅
• At x = 600 mm: M(600) = 400,000 N·mm ✅
• At x = 1000 mm: M(1000) = 520,000 N·mm ✅ ← Maximum
• At x = 1400 mm: M(1400) = 400,000 N·mm ✅
• At x = 1800 mm: M(1800) = 200,000 N·mm ✅
• At x = 2000 mm: M(2000) = 0 N·mm ✅

Maximum bending moment: M_max = 520,000 N·mm = 520 N·m at x = 1000 mm (center) ✅

1. Maximum moment location:

   Center (x = 1000 mm): M = 520 N·m ✅ Under loads (x = 600, 1400 mm): M = 400 N·m ✅

   Critical location: Center of beam (midspan) despite no load at this point ✅

2. Maximum bending stress:

   ✅

3. Safety factor:

   ✅

4. Engineering insight - Midspan vs Under-Load Comparison:

   Why midspan is critical:

   • In simply supported beams with multiple point loads, moment builds up between loads
   • Midspan moment (520 N·m) > Under-load moment (400 N·m)
   • This occurs because the center experiences cumulative effect of loads on both sides

   Design recommendation: Size beam based on midspan moment, not just load locations ✅

5. Practical implications:

   • Deflection will also be maximum at center
   • Lateral-torsional buckling check needed at center span
   • Consider intermediate supports if moment is excessive

1. Sum of vertical forces (↑ positive):

   ✅

2. Sum of moments about A (counterclockwise positive):

   ✅

3. Calculate R_A:

   ✅

4. Verification by moments about B:

   Let me recalculate: ✅

Method: Track loads from left support, accounting for distributed load accumulation.

For 0 ≤ x ≤ 2.5 m (between supports):

For 2.5 < x ≤ 3.0 m (overhang region):

Key points:

• At x = 0: V(0) = 240 N ✅
• At x = 0.8 m: V(0.8) = 240 - 300(0.8) = 0 N ✅ (zero shear point)
• Just before support B (x = 2.5⁻): V(2.5⁻) = 240 - 300(2.5) = -510 N ✅
• Just after support B (x = 2.5⁺): V(2.5⁺) = -510 + 1260 = 750 N ✅
• At tip before point load (x = 3.0⁻): V(3.0⁻) = 1500 - 300(3.0) = 600 N ✅
• At tip after point load (x = 3.0⁺): V(3.0⁺) = 600 - 600 = 0 N ✅

Maximum shear forces:

• Positive: V_max = 750 N just after support B ✅
• Negative: V_min = -510 N just before support B ✅

Method: Integrate shear force or use moment equilibrium of cut sections.

For 0 ≤ x ≤ 2.5 m:

With boundary condition M(0) = 0: C₁ = 0

For 2.5 < x ≤ 3.0 m:

With continuity at x = 2.5 m: M(2.5) = 240(2.5) - 150(2.5)² = -337.5 N·m

Key moment values:

• At x = 0: M(0) = 0 N·m ✅
• At x = 0.8 m (zero shear): M(0.8) = 240(0.8) - 150(0.8)² = 192 - 96 = 96 N·m ✅ ← Max positive
• At x = 2.5 m (support B): M(2.5) = 240(2.5) - 150(2.5)² = 600 - 937.5 = -337.5 N·m ✅ ← Max negative
• At x = 3.0 m (tip): M(3.0) = 1500(3.0) - 150(3.0)² - 3150 = 4500 - 1350 - 3150 = 0 N·m ✅

Maximum moments:

• Positive: M_max = +96 N·m at x = 0.8 m ✅
• Negative: M_min = -337.5 N·m at support B (x = 2.5 m) ✅

1. Maximum moment comparison:

   Positive moment: M_max = +96 N·m at x = 0.8 m ✅ Negative moment: |M_min| = 337.5 N·m at support B ✅

   Critical location: Support B due to larger moment magnitude ✅

2. Maximum bending stress at critical section:

   ✅

3. Safety factor:

   ✅

4. Engineering assessment - Positive vs Negative Moment Regions:

   Why support B is most critical:

   • Negative moment (337.5 N·m) >> Positive moment (96 N·m)
   • Support experiences both high shear force and maximum negative moment
   • Design implication: Beam capacity controlled by support B, not midspan

   Stress distribution at critical section:

   • Top fiber: Tension (negative moment causes beam to curve upward)
   • Bottom fiber: Compression
   • Design consideration: Check lateral-torsional buckling of compression flange

5. Practical design recommendations:

   • Reinforce support B region with stiffeners or larger section
   • Consider moment redistribution with semi-rigid connections
   • Check fatigue due to cyclic wind loading and tracking motion
   • Verify deflection limits especially in overhang region