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

5. General solution process
External loads
Internal loads
Stresses
Deformations

6. Main difficulties Sign conventions 2-D and 3-D perception
“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. Most problems will reduce to 1, 2, or 3 DOF’s.
6 degrees-of-freedom x a y z b g
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. perspective incorrect
Drawing z z
Technically correct
but
perspective incorrect g g b b a a x x y y x a y z b g
CORRECT

9. Internal Loads N 6 internal loads: 3 forces: N, V1, V2
Local coordinate system defined by normal
Cut where interested in internal loads
Structure x y z
Cross-section
Normal N V1 V2 T M2 M1
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. Cut at C Select BC (simpler) A C B 1500 700 800
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. 3. Draw FBD TC C B 800 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. Write equations of equilibrium
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
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 Ay By +
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 = kN By = kN

19. Cut at C Select AC (simpler) 4 6 2 8 3 A Ax C B Ay By
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) 4 6 2 8 3 A Ax C B Ay By

20. 3. Draw FBD 4 NC C A MC 12.14 VC 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 A C MC
12.14 VC

21. Write equations of equilibrium
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
12.14 VC +
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. Meaning Part CB applies on part AC:
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)