1. STRUCTURAL ENGINEERING: FORCES AND STRUCTURES APPLICATIONS OF TECHNOLOGY

2. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K A B SITUATION A beam is supported by two columns (A & B) and a concentrated load has been placed on the beam. HOW DO YOU DETERMINE HOW MUCH WEIGHT EACH COLUMN IS SUPPORTING?

3. DETERMINING RESULTANT FORCES What is it called when you have to solve for 2 variables within an equation in mathematics? System of Equations What steps do you take solve for both variables? x+y=20 2x+y=30 -2(x+y=20) 2x+y=30 ▪ Isolate one of the variables ▪ Solve for the value of the isolated variable ▪ Substitute the value of the isolated value back into one of the original equations ▪ Solve for the other value -2x-2y=-40 2x+y=30 -y=-10 y=10 x+y=20 x+(10)= 20 -10 = -10 x = 10

4. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K A B To solve for the resultant forces at columns A and B, we use the formula M=fd and isolate one of the columns. ▪ M stands for moment (a rotational force) ▪ f= force (magnitude and direction) ▪ d=perpendicular distance (distance from the object being isolated

5. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K A Y B Y When using this formula we must take Newton’s 3 rd Law of Motion into account: Every action has an equal and opposite reaction. ALL THE FORCES IN THE Y- DIRECTION MUST BE IN EQUILIBRIUM

6. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K Here is how we us the formula: We will take the moments about Column A and analyze all the forces that would cause the beam to rotate. Chose one of the columns to be isolated (A or B) ΣM A = 0 AY AY B Y

7. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K Based on the diagram, how many forces would cause the beam to rotate about column A? ΣM A = 0 ΣM A = 0= (f)(d)+(f)(d) TWO AY AY B Y

8. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K What is the magnitude and direction of each force? ΣM A = 0 ΣM A = 0= (f)(d)+(f)(d) ΣM A = 0= (-10K)(d)+(B Y )(d) negative AY AY BY BY positive negative positive

9. DETERMINING RESULTANT FORCES 12 FT 18 FT -10K 12 FT + 18 FT = 30 FT How far is each force from the point of rotation? ΣM A = 0 ΣM A = 0= (f)(d)+(f)(d) ΣM A = 0= (-10K)(12ft)+(B Y )(30ft) AY AY BY BY negative positive

10. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K 12 FT + 18 FT = 30 FT How is much weight is B Y supporting? ΣM A = 0 ΣM A = 0= (f)(d)+(f)(d) ΣM A = 0= (-10K)(12ft)+(B Y )(30ft) ΣM A = 0= -120Kft+(B Y )(30ft) +120Kft= +120Kft 120Kft= (B Y )(30ft) 30ft 30ft 4K= B Y AY AY BY BY negative positive

11. DETERMINING RESULTANT FORCES 12 FT 18 FT 10K 12 FT + 18 FT = 30 FT How is much weight is A Y supporting? ΣF Y = 0=-10K+A Y +B Y ΣF Y = 0= -10K+A Y +4K ΣF Y = 0= A Y -6K +6K= +6K 6K= A Y AY AY B Y= 4K negative positive REMEMBER ALL THE FORCES MUST BE IN EQUILIBRIUM.

12. DETERMINING RESULTANT FORCES The weight of a uniform load is distributed evenly along a distance. Uniform loads should be treated as a concentrated load applied at half the length of the load. 20 FT 10 FT