Moments in Engineering

Moments

Definition

  • The moment of a force is a measure of the tendency of the force to rotate the body upon which it acts.

Terminology

  • Lever Arm: The perpendicular distance from the pivot point to the line of action of the force.
  • Pivot: The point around which rotation occurs.
  • dd: Distance, which must be perpendicular to the force.

Moments Formula

  • Moment MM is calculated as: M=d×FM = d \times F, where
    • dd is the distance (lever arm)
    • FF is the force.

Units for Moments

English Customary Units
  • Force: Pound-force (lbf)
  • Distance: Foot (ft)
  • Moment: lbf-ft
SI Units
  • Force: Newton (N)
  • Distance: Meter (m)
  • Moment: N-m

Rotation Direction

  • Counterclockwise (CCW) is considered positive.
  • Clockwise (CW) is considered negative.

Right-Hand Rule

  • To determine the direction of the moment, curl your fingers in the direction of rotation.
    • Thumb pointing up or toward you = Positive moment.
    • Thumb pointing down or away from you = Negative moment.

Moment Calculations Examples

Example 1
  • Force F=20 lbF = 20 \text{ lb}
  • Distance d=9.0 in.=0.75 ftd = 9.0 \text{ in.} = 0.75 \text{ ft}
  • M=d×F=−(20 lb×0.75 ft)=−15 lb-ftM = d \times F = -(20 \text{ lb} \times 0.75 \text{ ft}) = -15 \text{ lb-ft} (15 lb-ft clockwise)
Example 2
  • Force F=20 lbF = 20 \text{ lb}
  • Distance d=1.0 ftd = 1.0 \text{ ft}
  • M=d×F=−(20 lb×1.0 ft)=−20 lb-ftM = d \times F = -(20 \text{ lb} \times 1.0 \text{ ft}) = -20 \text{ lb-ft}
Example 3
  • Force F=20 lbF = 20 \text{ lb}
  • Distance d=3 in.=0.25 ftd = 3 \text{ in.} = 0.25 \text{ ft}
  • M=d×F=−(20 lb×0.25 ft)=−5 lb-ftM = d \times F = -(20 \text{ lb} \times 0.25 \text{ ft}) = -5 \text{ lb-ft}
Example 4
  • Force F=20 lbF = 20 \text{ lb}
  • Distance d=8 in.+10 in.=1.5 ftd = 8 \text{ in.} + 10 \text{ in.} = 1.5 \text{ ft}
  • M=d×F=−(20 lb×1.5 ft)=−30 lb-ftM = d \times F = -(20 \text{ lb} \times 1.5 \text{ ft}) = -30 \text{ lb-ft}
Example 5 (Wheel and Axle)
  • Force F=100 NF = 100 \text{ N}
  • Radius r=d=50 cm=0.50 mr = d = 50 \text{ cm} = 0.50 \text{ m}
  • M=d×F=100 N×0.50 m=50 N-mM = d \times F = 100 \text{ N} \times 0.50 \text{ m} = 50 \text{ N-m}
Example 6 (Wheel and Axle with Angle)
  • Force F=100 NF = 100 \text{ N}
  • Radius r=d=50 cm=0.50 mr = d = 50 \text{ cm} = 0.50 \text{ m}
  • Fy=Fsin⁡(50∘)=(100 N)(0.766)=76.6 NF_y = F \sin(50^\circ) = (100 \text{ N})(0.766) = 76.6 \text{ N}
  • M=d×Fy=76.6 N×0.50 m=38.3 N-mM = d \times F_y = 76.6 \text{ N} \times 0.50 \text{ m} = 38.3 \text{ N-m}

Equilibrium

  • Equilibrium is the state of a body or physical system with unchanging rotational motion.
  • Two conditions:
    1. Object is not rotating.
    2. Object is spinning at a constant speed.
  • The sum of all moments about any point or axis is zero: ΣM=0\Sigma M = 0 , which means M<em>1+M</em>2+M3+…=0M<em>1 + M</em>2 + M_3 + … = 0

Moment Calculations - See-Saw

  • ΣM=0\Sigma M = 0 implies M<em>1+M</em>2=0M<em>1 + M</em>2 = 0
  • Using the right-hand rule to determine positive and negative moments.
  • M<em>1=−M</em>2M<em>1 = -M</em>2 or d<em>1×F</em>1=d<em>2×F</em>2d<em>1 \times F</em>1 = d<em>2 \times F</em>2
  • Given: F<em>1=25 lbF<em>1 = 25 \text{ lb}, F</em>2=40 lbF</em>2 = 40 \text{ lb}, d1=4.0 ftd_1 = 4.0 \text{ ft}
  • 25 lb×4.0 ft−40 lb×d2=025 \text{ lb} \times 4.0 \text{ ft} - 40 \text{ lb} \times d_2 = 0
  • 100 lb-ft=40 lb×d2100 \text{ lb-ft} = 40 \text{ lb} \times d_2
  • d2=100 lb-ft40 lb=2.5 ftd_2 = \frac{100 \text{ lb-ft}}{40 \text{ lb}} = 2.5 \text{ ft}

Moment Calculations - Loaded Beam

  • ΣM=0\Sigma M = 0 implies M<em>B+M</em>C=0M<em>B + M</em>C = 0
  • M<em>B=−M</em>CM<em>B = -M</em>C which means d<em>AB×R</em>By=d<em>AC×F</em>Cd<em>{AB} \times R</em>{By} = d<em>{AC} \times F</em>C
  • Given: d<em>AB=10.00 ftd<em>{AB} = 10.00 \text{ ft}, d</em>AC=3.00 ftd</em>{AC} = 3.00 \text{ ft}, FC=35.0 lbF_C = 35.0 \text{ lb}
  • 10.00 ft×RBy=3.00 ft×35.0 lb10.00 \text{ ft} \times R_{By} = 3.00 \text{ ft} \times 35.0 \text{ lb}
  • 10.00 ft×RBy=105 lb-ft10.00 \text{ ft} \times R_{By} = 105 \text{ lb-ft}
  • RBy=105 lb-ft10.00 ft=10.5 lbR_{By} = \frac{105 \text{ lb-ft}}{10.00 \text{ ft}} = 10.5 \text{ lb}
  • R<em>Ay+R</em>By=35.0 lbR<em>{Ay} + R</em>{By} = 35.0 \text{ lb}
  • RAy=35.0 lb−10.5 lb=24.5 lbR_{Ay} = 35.0 \text{ lb} - 10.5 \text{ lb} = 24.5 \text{ lb}
  • Select A as the pivot location to solve for RByR_{By}.

Moment Calculations - Truss

  • Given: F<em>C=600 lbF<em>C = 600 \text{ lb}, F</em>B=500 lbF</em>B = 500 \text{ lb}, d<em>AC=24 ftd<em>{AC} = 24 \text{ ft}, d</em>CD=8 ftd</em>{CD} = 8 \text{ ft}, d<em>CB=12 ftd<em>{CB} = 12 \text{ ft}, d</em>AD=32 ftd</em>{AD} = 32 \text{ ft}
  • ΣM=0\Sigma M = 0 implies M<em>D−M</em>B−MC=0M<em>D - M</em>B - M_C = 0
  • M<em>D=M</em>B+M<em>CM<em>D = M</em>B + M<em>C which means d</em>AD×R<em>Dy=(d</em>CB×F<em>B)+(d</em>AC×FC)d</em>{AD} \times R<em>{Dy} = (d</em>{CB} \times F<em>B) + (d</em>{AC} \times F_C)
  • 32 ft×RDy=(12 ft×500 lb)+(24 ft×600 lb)32 \text{ ft} \times R_{Dy} = (12 \text{ ft} \times 500 \text{ lb}) + (24 \text{ ft} \times 600 \text{ lb})
  • RDy×32 ft=6000 lb-ft+14400 lb-ftR_{Dy} \times 32 \text{ ft} = 6000 \text{ lb-ft} + 14400 \text{ lb-ft}
  • RDy×32 ft=20400 lb-ftR_{Dy} \times 32 \text{ ft} = 20400 \text{ lb-ft}
  • RDy=20400 lb-ft32 ft=637.5 lbR_{Dy} = \frac{20400 \text{ lb-ft}}{32 \text{ ft}} = 637.5 \text{ lb}