EM 304 Lecture 1

1. EM 304 Lecture 1 Syllabus What is “Mechanics of materials”? Main difficulties Degrees of freedom Internal loads How to determine internal loads Examples: Problem 1.1 Problem 1.3 Problem 1.5

2. Syllabus • Instructor: Vincent Blouin (www.clemson.edu/~vblouin) • Office hours: T & Th 11:00-12:00AM or by appointment • Grading: • HW 10% • Tests 60% (3 tests in class) • Final 30% • Homework: • Everyday • Due in class before lecture • Two days for each homework • Questions?

3. What is Mechanics of Materials? • Statics (EM 201) • Rigid bodies (no deformation) • External forces and moments • Reactions R M

4. R R M M • Mechanics of Materials (EM 304) • Flexible bodies (deformation) • Internals forces and moments • Stresses

5. General solution process External loads Internal loads Stresses Deformations

6. Main difficulties • Sign conventions “Is this stress positive or negative?” • 2-D and 3-D perception “This is too complicated!” Remedy: • Understand the 6 degrees of freedom • Systematic procedures with steps • Clear notations • Clear drawings

7. z g b a x y 6 degrees-of-freedom Right-hand rule always applies to determine positive directions Positive translation 3 translations: x, y, z 3 rotations: a, b, g Most problems will reduce to 1, 2, or 3 DOF’s. Positive rotation

8. Drawing z g b a y x z z Technically correct but perspective incorrect g g b b a a x x y y CORRECT

9. z Normal Cross-section x y N T M1 M2 V1 V2 Internal Loads Local coordinate system defined by normal Cut where interested in internal loads Structure • 6 internal loads: • 3 forces: N, V1, V2 • 3 moments: M1, M2, T • 4 kinds of loads: • Normal force: N • Shear forces: V1, V2 • Bending moments: M1, M2 • Torque: T

10. How to determine internal loads Example: Problem 1-1 • Determine external loads (reactions) • Cut at C and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results

11. Determine external loads (reactions) • Cut at C and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results • Determine external loads (reactions) Free rotation at A and B: TA = TB = 0 A 1500 700 C 800 B

12. Determine external loads (reactions) • Cut at C and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results A 1500 • Cut at C Select BC (simpler) 700 C 800 B

13. Determine external loads (reactions) • Cut at C and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results 3. Draw FBD TC C 800 B

14. Determine external loads (reactions) • Cut at C and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results • Write equations of equilibrium 1 degree of freedom 1 equation -800 + Tc = 0 Tc = 800 lb.ft TC C + 800 B

15. Determine external loads (reactions) • Cut at C and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results A 1500 • Meaning Part CA applies a torque of 800 lb/ft on part BC in the direction shown on FBD 700 TC TC C 800 B

16. General procedure to determine internal loads • Determine external loads (reactions) • Note that all reactions may not be needed • This step may be done later since steps 2 and 3 usually dictate which reactions to calculate • Cut and select one side • The direction of the cut is generally dictated by the geometry of the part • One side is generally easier that the other • Draw FBD - This is the most important step • Write equations of equilibrium and solve • Understand the meaning of the results

17. Example: Problem 1-3 • Determine external loads (reactions) • Cut and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results

18. Determine external loads (reactions) • Cut and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results • Determine external loads (reactions) 4 6 2 8 3 A B Ax By Ay Ax = 0 + Ay + By – 4 – 6 – 2 – 8 – 3 = 0 + -By(7)+ 6(1.5) + 2(3) + 8(5) + 3(7) = 0 + Ax = 0 Ay = 12.14 kN By = 10.86 kN

19. Determine external loads (reactions) • Cut and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results • Cut at C Select AC (simpler) 6 2 8 3 4 A C Ax B By Ay

20. Determine external loads (reactions) • Cut and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results 3. Draw FBD 4 NC C A MC 0 VC 12.14

21. Determine external loads (reactions) • Cut and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results • Write equations of equilibrium 3 DOF’s 3 equations 4 NC C A MC 0 VC 12.14 0 + NC = 0 + 12.14– VC – 4 = 0 + VC(0.75) – MC = 0 + NC = 0 VC = 8.14 kN MC = 6.11 kN.m

22. Determine external loads (reactions) • Cut and select one side • Draw FBD • Write equations of equilibrium and solve • Understand the meaning of the results • Meaning 4 6 2 8 3 A C B Ay By Part CB applies on part AC: • no load horizontally • a shear force of 8.14 kN vertically downward • a bending moment of 6.11 kN.m counterclockwise Part AC applied on part CB the opposite (action and reaction)