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

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?

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

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

General solution process External loads Internal loads Stresses Deformations

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

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

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

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

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

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

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

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

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 -800 + Tc = 0 Tc = 800 lb.ft TC C + 800 B

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

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

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

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 = 12.14 kN By = 10.86 kN

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

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

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

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)