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Beam Loading Example

Beam Loading Example. To determine the reaction forces acting at each support, we must determine the moments (represented as torques).

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Beam Loading Example

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  1. Beam Loading Example • To determine the reaction forces acting at each support, we must determine the moments (represented as torques). • Static Equilibrium: We can use this fact to find the conditions for "static equilibrium": the condition an object is in when there are forces acting on it, but it is not moving. • The conditions for static equilibrium are easy to state: the sum of the (vector) forces must equal zero, and the sum of the torques must equal zero: • Σ F = 0 and Σ τ = 0. • In this example, because the beam is in static equilibrium, the sum of the sum of the forces in the y-direction is zero and the sum of the moment torques about one end must also equal zero.

  2. Forces Acting Upon the Beam • The first step is to identify the forces acting upon the beam • In this example, you have the downward force of the block (C) the downward force of the beam (D), and the two reaction forces of the supports (A and B) • In this example, the beam weighs 500 pounds and the block weighs 750 pounds.

  3. Forces Acting Upon the Beam • Because the system is in static equilibrium, the upward and downward forces must sum to be 0 • Therefore, -A + -B + C + D = 0 • We’ll use indicate downward forces as “+” and upward as “-” • Therefore, -A + -B + 500 lbs + 750 lbs = 0 • We need to determine the reaction forces at A and B, but we have two unknowns and the block is not at the center of the beam

  4. Using Moments (Torques) • Because the system is in static equilibrium, the sum of the moment torques must also sum to be 0 • Sum of the clockwise torques + sum of counterclockwise torques = 0 • Working from the A support, we have the following torques: • A clockwise torque at C • A clockwise torque at D • A counterclockwise torque at B

  5. Using Moments (Torques) • To calculate the moments, we need to know the distances • (A moment is force X distance) • Working from the A support, we have the following moments: • SMA = 0 • 0 = MC + MD + MB • A , there is no moment at A since it is our analysispoint • C = (2 ft X 750 lbs) = +1,500 ft-lbs • D = (5 ft X 500 lbs) = +2,500 ft-lbs • B = - (10 ft X ?? Lbs)

  6. Using Moments (Torques) • Replacing the moments in the equation: • SMA = 0 • 0 = MC + MD + MB 0 = (1,500 ft-lbs)+(2,500 ft-lbs)-(10 ft)(??lbs) • Remember, the moment at B is negative because it is counterclockwise. • Using algebra, we get: (1,500 ft-lbs)+(2,500 ft-lbs)=(10 ft)(??lbs) • We can solve for the force at B ( 4,000 ft-lbs / 10 ft) = 400 lbs • So, the reaction force at B is 400 lbs

  7. Beam Forces • Going back to the static equilibrium formula for forces, we can now take care of one of the two unknowns we had in the formula: -A + -B + 500 lbs + 750 lbs = 0 -A + -400 lbs + 750 lbs + 500 lbs = 0 Using algebra, we can determine the reaction force at A -A = 400 lbs + -750 lbs + -500 lbs A = 850 lbs

  8. Final Diagram • If the structural supports at A and B had a breaking strength of 900 pounds, what conclusions would you reach? • If the design criteria called for a safety factor of 4 for the supports, what conclusions would you reach? • Assuming the forces generated at A are worst case possible, to meet the safety factor of 4, what would the minimum breaking strength have to be for the supports?

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