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Lecture 5-6 Beam Mechanics of Materials Laboratory Sec. 3-4 Nathan Sniadecki

Mechanics of Materials Lab. Lecture 5-6 Beam Mechanics of Materials Laboratory Sec. 3-4 Nathan Sniadecki University of Washington. $100. Answer: What is the moment of inertia (Second Moment of Area) with respect to the x axis. $100.

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Lecture 5-6 Beam Mechanics of Materials Laboratory Sec. 3-4 Nathan Sniadecki

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  1. Mechanics of Materials Lab Lecture 5-6 Beam Mechanics of Materials Laboratory Sec. 3-4 Nathan Sniadecki University of Washington

  2. $100 Answer: What is the moment of inertia (Second Moment of Area) with respect to the x axis

  3. $100 Answer: What is the moment of inertia (second moment of area) with respect to the y axis

  4. $200 Answer: What is the moment of inertia for a rectangle

  5. $200 Answer: What is the moment of inertia for a circle

  6. $300 • The straight line that defines a surface where ex and sx are zero Answer: What is the neutral axis

  7. $300 Answer: What is the bending stress in the x-direction at a distance y from the origin of the coordinate system due to the loading of a couple vectorMx acting in x-direction

  8. $400 • The point on the stress-strain curve where the material no longer deforms elastically, but also plastically. Answer: What is the proportional limit

  9. $400 • The theorem that expresses that the moment of inertia Ix of an area with respect to an arbitrary x axis is equal to the moment of inertia Ixc with respect to the centroidal x axis, plus the product Ad2 of the area A and the square of the distance d between the two axis? Answer: What is the Parallel Axis Theorem Ix = Ixc + Ad2

  10. $500 • The principle that states that the effect of a given combined loading on a structure can be obtained by determining the effects of each load separately and then combining the results obtained together as long as 1) each effect is linearly related to the load that produces it and 2) the deformation resulting from any given load is small and does not affect the conditions of application of the other loads Answer: What is the Principle of Superposition

  11. $500 • The principle that states that except in the immediate vicinity of the point of loading, the stress distribution may be assumed to be independent of the actual mode of loading, i.e. for axial loading, at a distance equal to or greater than the width of a member, the distribution of stress across a given section is the same. Answer: What is Saint-Venant’s Principle

  12. FINAL JEOPARDY

  13. FINAL JEOPARDY Answer: What is the angle of the neutral axis for an asymmetrically loaded beam

  14. Inclined Load Notice the sign convention: positive Mz compress upper part, negative stress; positive My extend front part, positive stress!

  15. Inclined Load Stress b Neutral axis

  16. Example

  17. Stress Distribution

  18. Asymmetrical Beam The centroid of the area A is defined as the point C with coordinates (yc,zc) which satisfies If the origin of y and z axes is placed at centroid C (orientation is arbitrary.)

  19. Pure Bending • Consider of beam segment AB of length L • After deformation, length of neutral surface DE remains L, but JK becomes

  20. Asymmetric Beam When z axis is the neutral axis; If z is a principal axis (symmetry), the product of inertia Iyz is zero  My = 0, bending in x-y plane, analogous to a symmetric beam

  21. Asymmetric Beam When y axis is the neutral axis; If y is a principal axis, the product of inertia Iyz is zero  Mz = 0, bending in x-z plane, analogous to a symmetric beam

  22. Asymmetric Beam • When an asymmetric beam is in pure bending, the plane in which the bending moments acts is perpendicular to the neutral surface if and only if (iff) the y and z axes are principle centroidal axes and the bending moments act in one of the two principle planes. In such a case, the principle plane in which bending moment acts becomes the plane of bending and the usual bending theory is valid

  23. Analysis of Asymmetric Beam • Locating the centroid, and constructing a set of principal axes • Resolving bending moment into My and Mz • Superposition

  24. Principle Axes

  25. Analysis of Asymmetric Beam A channel section C10  15.3 M = 15 kips-in Iy=2.28 in4, Iz=67.4 in4 Location of Point C c=0.634 in Location of Point A yA=5.00 in zA=-2.6+0.634=-1.966 in Calculate bending stress Locate neutral axis

  26. Analysis of Asymmetric Beam

  27. Normal Stress in Beam

  28. Curved Beams • What if the beam is already ‘bent’? • Where will the beam likely fail?

  29. Bending Stress for Curved Beam • #1: Neutral surface remains constant: • #2: Deformation at JK:

  30. Bending Stress for Curved Beam • #3: Strain: • #4: Stress:

  31. Bending Stress for Curved Beam • #5: Neutral Axis: • #6 Centroid: Aside: R = rn • #7: N.A. Location: Since Dq > 0 for M > 0

  32. Location of N.A. in Curved Beam • Cross-sectional dimensions define neutral axis location for a curve beam about C

  33. Curved Beams Neutral axis is no longer the centroidal axis Positive M decreases curvature

  34. Curved Beam

  35. Curved Beams Recover to straight beam Curvature is small, e is small, rn is close to rc

  36. Curved Beam Pay attention to the sign of s

  37. Curved Beam Pay attention to the sign of s

  38. Assignment • Read Mechanics of Materials Lab Sec. 4 • 4.26(e), 4.72 posted online

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