1. SAFE 605: Application of Safety Engineering Principles
Week 4: Beam Loading Example
SAFE 605: Application of Safety Engineering Principles
Week 4: Torques and Beam Loading

2. Objectives Quizzes and assignments review
Demonstrate how to calculate the moments and forces

3. Quizzes and assignments
No assignment for week 3
Week 4 assignment is due next week
Quiz 2 is open today for a week.

4. Beam Loading Example Y F X
To determine the reaction forces acting at each support, we must determine the moments (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 M = 0.
In this example, because the beam is in static equilibrium, 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. Y F X

5. 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.

6. 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 indicate downward forces as “+” and upward as “-”
Therefore,
-A + -B lbs 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

7. 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

8. 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
MA = 0 , there is no moment at A since it is our analysis point (Distance = 0)
MC = (2 ft X 750 lbs) = +1,500 ft-lbs
D = (5 ft X 500 lbs) = +2,500 ft-lbs
B = (10 ft X ?? Lbs)

9. Using Moments (Torques)
Week 4: Beam Loading Example
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

10. Beam Forces -A + -B + 500 lbs + 750 lbs = 0
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 lbs lbs = 0
-A lbs lbs lbs = 0
Using algebra, we can determine the reaction force at A
-A = 400 lbs lbs lbs
A = 850 lbs

11. 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?

12. Activity 4 from the module:
What force in each rope if both workers (250 lbs each) stood in the center of the 8 ft scaffold (weighing 750 lbs)?
What is the required breaking strength if a Safety Factor of 6 is specified?
Two workers = 500 lbs
Scaffold = 750 lbs
| l = 8 ft |
A B C D