1. CE 203 Structural Mechanics Week 5

2. Bending moment & Shearing force diagrams

3. What is a beam Members that are loaded in a direction perpendicular to their longitudinal axis Length is more significant than lengths in cross-section We could have: Simply supported, Cantilevered, Overhanging or Continuous beams

4. Find the internal forces at C.

5. Make section at C & Indicate internal forces

6. Sign Convention

7. Draw shear & bending moment diagrams

8. Draw shear & bending moment diagrams

10. Load-shear-moment relations dV/dx = -w(x) dM/dx = V(x)

11. Notes No load constant shear & linear moment variation Uniform load linear shear & quadratic moment variation

12. Notes Concentrated load causes jump in shear Concentrated moment causes jump in moment

13. 40 kips 1 kip/ ft 2 kip/ ft 15 ft 15 ft 5 ft 5 ft 10 ft 5 ft Draw shear force and bending moment diagrams

14. 40 kips 1 kip/ ft 2 kip/ ft 15 ft 15 ft 5 ft 5 ft 10 ft 5 ft 27.5 kips 37.5 kips Reactions

15. 40 kips 1 kip/ ft 2 kip/ ft 15 ft 15 ft 5 ft 5 ft 10 ft 5 ft 27.5 kips 37.5 kips 12.5 k 10 k -15 k -27.5 k

16. 37.5 k.ft 12.5 k 10 k -50 k.ft -15 k -150 k.ft -100 k.ft -27.5 k SFD BMD

17. Bending stresses

18. Bending of straight beams

19. What type of beams are we talking about ? Prismatic, straight. X-section has an axis of symmetry Moment on an axis perpendicular to axis of symmetry

20. Deformation due to pure bending

21. Axis x longitudinal axis y axis of symmetry in cross section z axis of bending y z

22. Assumptions No change in length of longitudinal axis but becomes a curve Cross section remain plane perpendicular to longitudinal axis Deformation of cross section is neglected

23. Deformation of a differential element along the beam Δs =  Δθ Δs’ = (-y) Δθ

24. Longitudinal strain x = (Δs’–Δs)/ Δs If  = radius of curvature (can be f(x)) Δs =  Δθ Δs’ = (-y) Δθ x = -y/  (1)

25. Strain variation along y

26. Flextural stresses Assume linear elastic material and σy , σz much less than σx then σx = E x From 1 σx = - E y/  (2)

27. Resulting stress distribution From statics Fx = ∫ σx dA And Mz = ∫ (-y) σx dA

28. Location of neutral axis But Fx = 0 ∫ σx dA = ∫ - E y/  dA = 0 = -E/  ∫ y dA = 0 ỹ dA = 0 This only true if the z-axis passes by the centroid of the cross section.

29. Flexture formula Mz = ∫ (-y) σx dA = ∫ (-y) (- E y/ ) dA (3) Mz = E/  ∫ y2 dA = E/  Iz

30. Flexture formula Using equations (2) and (3) σx = - E y/  (2) (3) Mz = E/  ∫ y2 dA = E/  Iz We come up with the bending stresses σx = - Mz y / Iz

31. Comments about stress distribution σx = - Mz y / Iz

32. Bending stresses Using the flexure formula

34. Draw BM & SF diagrams Find max tensile stress and indicate its location Find max compressive stress and indicate its location Plot stress distribution in a section through point D.

36. P a 1.5 in 50 in 6 in I=50 in4 What if we measure strain? Determine the magnitude of the force P, if the strain at point a is measured to be 7x10-5 .Take E = 30x106psi

37. P a 1.5 in 50 in P = 460 lb

38. a h b Contribution to moment resistance • What percentage of Moment is resisted by The shaded area?

39. h b h/2 If a=1/2 h only 12.5%

40. h/4 b

41. Derivation

42. Examples from text σx = - Mz y / Iz Example 6.14 page 299 Example 6.15 page 301 Example 6.16 page 303 Example 6.17 page 304