1 / 11

Shear & Moment Diagrams- Cont.

Shear & Moment Diagrams- Cont. Problem 7-57 The beam will fail when the maximum shear force is V max = 800 lb or the maximum moment is M max=1200 lb-ft. Determine the largest intensity w of the distributed loading it will support. 18 ft. Shear & Moment Diagrams- Cont. Problem 7-78

jirair
Download Presentation

Shear & Moment Diagrams- Cont.

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. Shear & Moment Diagrams- Cont. Problem 7-57 The beam will fail when the maximum shear force is Vmax = 800 lb or the maximum moment is Mmax=1200 lb-ft. Determine the largest intensity w of the distributed loading it will support. 18 ft

  2. Shear & Moment Diagrams- Cont. Problem 7-78 The beam will fail when the maximum moment is Mmax=30 kip-ft or the maximum shear is Vmax= 8 kip. Determine the largest distributed load w the beam will support. 6 ft 6 ft

  3. FRICTION (Sections 8.1 - 8.2) Today’s Objective: Students will be able to: a) Understand the characteristics of dry friction. b) Draw a FBD including friction. c) Solve problems involving friction. • In-Class Activities: • Check homework, if any • Reading quiz • Applications • Characteristics of dry friction • Problems involving dry friction • Concept quiz • Group problem solving • Attention quiz

  4. READING QUIZ • 1. A friction force always acts _____ to the contact surface. • A) normal B) at 45° • C) parallel D) at the angle of static friction • 2. If a block is stationary, then the friction force acting on it is ________ . • A) s N B) = sN • C)  sN D) = k N

  5. APPLICATIONS In designing a brake system for a bicycle, car, or any other vehicle, it is important to understand the frictional forces involved. For an applied force on the brake pads, how can we determine the magnitude and direction of the resulting friction force?

  6. APPLICATIONS (continued) Consider pushing a box as shown here. How can you determine if it will slide, tilt, or stay in static equilibrium? What physical factors affect the answer to this question?

  7. CHARACTERISTICS OF DRY FRICTION (Section 8.1) Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body. Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion. For the body shown in the figure to be in equilibrium, the following must be true: F = P, N = W, and Wx = Ph.

  8. CHARACTERISTICS OF FRICTION (continued) To study the characteristics of the friction force F, let us assume that tipping does not occur (i.e., “h” is small or “a” is large). Then we gradually increase the magnitude of the force P. Typically, experiments show that the friction force F varies with P, as shown in the left figure above.

  9. FRICTION CHARACERISTICS (continued) The maximum friction force is attained just before the block begins to move (a situation that is called “impending motion”). The value of the force is found using Fs = sN, where sis called the coefficient of static friction. The value of sdepends on the materials in contact. Once the block begins to move, the frictional force typically drops and is given by Fk = k N. The value ofk (coefficient of kinetic friction) is less than s.

  10. The inclination, s, is noted. Analysis of the block just before it begins to move gives (using Fs = s N): +  Fy = N – W cos s = 0 +  FX = S N – W sin s = 0 DETERMING s EXPERIMENTALLY A block with weight w is placed on an inclined plane. The plane is slowly tilted until the block just begins to slip. Using these two equations, we get s = (W sin s) / (W cos s) = tan sThis simple experiment allows us to find the S between two materials in contact.

  11. PROCEDURE FOR ANALYSIS (Section 8.2) Steps for solving equilibrium problems involving dry friction: 1. Draw the necessary free body diagrams. Make sure that you show the friction force in the correct direction (it always opposes the motion or impending motion). 2. Determine the number of unknowns. Do not assume F = S N unless the impending motion condition is given. 3. Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns.

More Related