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EF 202, Module 3, Lecture 3 Frames EF 202 - Week 11.

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Presentation on theme: "EF 202, Module 3, Lecture 3 Frames EF 202 - Week 11."— Presentation transcript:

1 EF 202, Module 3, Lecture 3 Frames EF 202 - Week 11

2 EF 202, Module 3, Lecture 3 2 Multiforce Member A multiforce member has either three or more forces acting on it, or two or more forces and one or more moments acting on it.

3 EF 202, Module 3, Lecture 3 3 Frames A frame has at least one multiforce member, is stationary, and is designed to support/resist loads.

4 EF 202, Module 3, Lecture 3 4 Machines A machine has at least one multiforce member, is not stationary, and is designed to transmit loads from one thing to another.

5 EF 202, Module 3, Lecture 3 5 Why? By definition, a truss is not a frame, because.... a truss has only two-force members and a frame must have at least one multiforce member.

6 EF 202, Module 3, Lecture 3 6 Frame Examples Example 6.9, p. 288

7 EF 202, Module 3, Lecture 3 7 Frame Examples PS 2-2, Problem 4

8 EF 202, Module 3, Lecture 3 8 Frame Examples PS 2-2, Problem 5

9 EF 202, Module 3, Lecture 3 9 Frame Examples Online Homework 2-2-3

10 EF 202, Module 3, Lecture 3 10 Frame Examples Test Two, Problem 2

11 EF 202, Module 3, Lecture 3 11 Analysis of Frames Because some forces are not directed along members, the method of joints is not sufficient; the method of sections is not sufficient. But Newton’s first and third laws are still sufficient

12 EF 202, Module 3, Lecture 3 12 x y + Statically indeterminate: 3 equations 4 unknowns Let’s try to solve without recognizing that BD and ABC are a 2-force and a 3-force member, respectively.

13 EF 202, Module 3, Lecture 3 13 x y + Statically determinate: 6 equations 6 unknowns Now, let’s draw a separate FBD for each member and recount equations and unknowns.

14 EF 202, Module 3, Lecture 3 14 Observation Sometimes the only way to find all the loads on a frame is to draw a separate FBD for each member.

15 EF 202, Module 3, Lecture 3 15 ABC is now statically determinate, because it has only three unknowns. Now, let’s use the fact that BD is a 2-force member.

16 EF 202, Module 3, Lecture 3 16 Observation Using what we know about two-force and three-force members reduces the work.

17 EF 202, Module 3, Lecture 3 17 Why is this a frame, and not a truss? ADC and BGC are 3-force members. Any 2-force members? DH and GH Statically determinate? No.

18 EF 202, Module 3, Lecture 3 18

19 EF 202, Module 2, Lecture 3 19 Suppose that one pair of lines form an angle, and a second pair of lines form an angle. If each line in the second pair is perpendicular to a line in the first pair, then the two angles are equal. Nixon(?) Theorem

20 EF 202, Module 3, Lecture 3 20 Statically determinate: 6 equations 6 unknowns

21 EF 202, Module 3, Lecture 3 21 Why is this a frame, and not a truss? 3-force members Any 2-force members? No. Statically determinate? Yes. (So, what?)

22 EF 202, Module 3, Lecture 3 22

23 EF 202, Module 3, Lecture 3 23 Why is this a frame, and not a truss? 3-force members Any 2-force members? Yes. Statically determinate? Yes. (So, what?)

24 EF 202, Module 3, Lecture 3 24 Statically determinate? Yes. 10 equations 7 unknowns

25 EF 202, Module 3, Lecture 3 25

26 EF 202, Module 3, Lecture 3 26 F = 150 N (9.8) = 1470 N F = 150N (5.33) = 800 N


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