1 / 15

Examples

Examples. Internal forces in simple structures Example: the truss Example with friction. (Serway 12.1-12.3 – see p. 372). Trusses are composed of triangles made of rigid, massless beams joined with pins. They rest on two support points attached to pins, only one of which is fixed.

aimee
Download Presentation

Examples

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Examples • Internal forces in simple structures • Example: the truss • Example with friction (Serway 12.1-12.3 – see p. 372) Physics 1D03 - Lecture 18

  2. Trusses are composed of triangles made of rigid, massless beams joined with pins. They rest on two support points attached to pins, only one of which is fixed. Pins or dowels through holes in beams do not allow any torque to be transmitted through the joint. Massless beams are a good approximation because the truss can support a load many times its own weight. With only one fixed support, all horizontal supporting forces are collected at one point. Normal force only here Roller support point at a pin Fixed support point at a pin Physics 1D03 - Lecture 18

  3. “Two-Force Members” Member in tension D B B D forces on member forces by member Member in compression D B B D forces on member forces by member In general, a member is subject to bending as well as tension or compression; the truss is a special structure in which all of the members are stressed in tension or compression only. Physics 1D03 - Lecture 18

  4. Concept: To solve for all the internal and external forces in a ideal truss, set SF = 0 and St = 0 for the external forces. Then set SF = 0 at each pin. • The torques due to internal forces are not required, because they are automatically zero when the internal forces are parallel to the beams. • Since the compression or tension forces at the end of any beam are equal and opposite, not all of these equations are required to solve for every force. Physics 1D03 - Lecture 18

  5. The Truss Examples: Steel bridges, hydro towers, crane booms, roof supports in buildings, the supports for the walkway between ABB and JHE, ... A stable truss is built from triangles. Provided the loads are applied only at the joints, each member exerts a force parallel to itself on each joint. Physics 1D03 - Lecture 18

  6. C D B A E G H I 6 kN Example 6 kN The horizontal members are 1.00 m long, and the angles are 30o, 60o, and 90o. Find the force of compression or tension in each member. (For now: in member CH). • Method: • Find the external forces on the whole truss. • Draw a free-body diagram for each joint. The members exert forces on the joints; the directions of these forces are known. • Write equilibrium equations at each joint and solve. Physics 1D03 - Lecture 18

  7. C B D A E H G I C • Questions: • If the directions shown for the forces are correct, which members would be in tension, and which in compression? • How many unknowns can be found just by using the equations of static equilibrium for joint C? FDC FBC FHC Forces on joint C Physics 1D03 - Lecture 18

  8. 6 kN C D B A E G H I 6 kN NE NA FED 30o FGE NE 12 kN 30o FGE NE Results:External forces, NA = NE = 6 kN Joint E: just two unknown forces, solve to get: FED= -12 kN, FGE = 12cos(30) = 10.4 kN The – sign just means we guessed wrong about the direction of FED. (It is actually in compression.) Which joint should be done next? Physics 1D03 - Lecture 18

  9. 6 kN C D B A E G H I 6 kN NE NA Joint G: FGD = 6 kN (tension) FGH = 10.4 kN (tension) Joint D: FDH = 6 kN (compression) FDC = 6 kN (compression) Joint C: FCH = 6 kN (tension) FBC = 6 kN (compression) Question: Can we now say we know the rest of the forces “by symmetry”? For example, can we say that FAB = FDE, and FBI = FGD? Physics 1D03 - Lecture 18

  10. Quiz A What does the force diagram for the pin at A look like? C B D A) B) C) D) Physics 1D03 - Lecture 18

  11. P a C B b A Example: rolling a spool. Inner radius a, Outer radius b = 2a ms = 0.60 at both surfaces If the spool has weight w = 100 N, what minimum force P is needed to rotate the spool? Plan: Assume the spool is not rotating (still in equilibrium) but is about to slip (so fs = ms N at each contact). Use the conditions for equilibrium, and the two friction equations, to solve for P. Physics 1D03 - Lecture 18

  12. NB w fB fA NA 1) Free-body diagram: b = 2a P y w = 100 N ms = 0.60 x a C B b A Physics 1D03 - Lecture 18

  13. P NB a C B b w fB fA A NA b = 2a 1) Free-body diagram: w = 100 N Are the friction forces in the right directions? y ms = 0.60 x Torques about centre: aP = b(fA + fB ) , so P = (b/a)(fA + fB ) = 2(fA + fB ) Forces: x-components, fA = NB y-components, NA+ fB + P = w Friction (about to slip): fA = ms NA fB = ms NB Five equations in 5 unknowns – no sweat! (but plan your solution before starting the algebra). Physics 1D03 - Lecture 18

  14. a b Quiz When the rope is pulled as shown, which way will the spool move? • roll to the left • roll to the right • it won’t rotate • depends on the angle pull Physics 1D03 - Lecture 18

  15. Summary • Internal forces may be found by considering the forces on one part of a structure. • If only two forces act on a member in equilibrium, the forces must be parallel to the member. Practice problems: 49, 57 (both versions of book) Physics 1D03 - Lecture 18

More Related