Unformatted text preview: Module 5: Par=cles in 3
D
September 6, 2010 Module Content:
1. Par=cle equilibrium analysis rests upon free body diagrams.
2. 3
D equilibrium scenarios are approaches in exactly the same way as 2
D problems.
3. Cross products are frequently used in sta=cs for calcula=ons involving moments. Module Reading, Problems, and Demo:
Reading: Chapter 3
Problems: Prob. 3.12, 3.15, 3.27 (par=cle equilibrium and FBDs)
Demo: none
Homework PlaUorm: hVp://www.masteringengineering.com
Course Blog: hVp://pages.shan=.virginia.edu/sta=cs2010 MAE 2300 Sta=cs © E. J. Berger, 2010 5
1 Theory: The Free Body Diagram
• a Free Body Diagram (FBD) is a pictorial method for represen=ng all the external loads ac=ng on a system, all the relevant reac=ons with which the system responds, plus a coordinate system and relevant distances/lengths/angles used to deﬁne the problem • pictorial method: a graphical representa=on of a system • external loads: are applied to the system in the form of forces (N), pressures (Pa), torques (N
m), moments (N
m), etc. • relevant reac=ons: the forces/pressures/torques which connect the system to ground • coordinate system: the system of reference used to deﬁne the geometry of the system • distances/lengths/angles: factual informa=on about the geometry of the problem • drawing a proper FBD is the crucial ﬁrst step to wri7ng equilibrium equa7ons MAE 2300 Sta=cs © E. J. Berger, 2010 5
2 Theory: Free Body Diagrams (FBDs)
• there are ﬁve steps for FBD construc=on:
1.
2. deﬁne a coordinate system 3. deﬁne and label other relevant quan==es 4. draw and label external forces 5.
• draw the par=cle write the equilibrium equa=ons based upon your FBD NOTE that the FBD is NOT (!!!!!!!) the problem schema=c MAE 2300 Sta=cs © E. J. Berger, 2010 5
3 Example: Schema=c and FBD
schema'c MAE 2300 Sta=cs FBD © E. J. Berger, 2010 5
4 Example: Schema=c and FBD
schema'c FBD TEC MAE 2300 Sta=cs © E. J. Berger, 2010 5
5 Example: Schema=c and FBD
schema'c FBD TAB = k (s − so ) MAE 2300 Sta=cs © E. J. Berger, 2010 5
6 Theory: Special FBD Features F = ks MAE 2300 Sta=cs © E. J. Berger, 2010 5
7 A General Rule in this Class MAE 2300 Sta=cs © E. J. Berger, 2010 5
8 A General Rule in this Class Problem solu7ons which do not include a FBD will receive substan7ally reduced credit. You will never get full credit for a problem solu7on if you don’t draw a FBD.
MAE 2300 Sta=cs © E. J. Berger, 2010 5
8 Theory: Par=cles in 3
D
• the situa=on is precisely the same as in 2
D, but there is an increased premium placed upon good procedure (follow the 5 steps) • the vector equilibrium equa=on now produces 3 scalar equa=ons: • and we simply use the components of the Cartesian vectors in the summa=ons MAE 2300 Sta=cs © E. J. Berger, 2010 5
9 Introducing Chapter 4: Force System Resultants
• Chapter 4 introduces the concept of a resultant, and deﬁnes a resultant in terms of equivalence: we can calculate the resultant of a system of forces, which represents the equivalent net eﬀect of that system of forces (but is generally a simpler to analyze) • the main focus of the Chapter is on moments produced by forces • a moment of a force about a point or axis is a measure of the tendency of the force to cause a body to rotate about that point or axis • classic example: • we will also see cross products in this Chapter MAE 2300 Sta=cs © E. J. Berger, 2010 5
10 Introducing Chapter 4: Force System Resultants
• Chapter 4 introduces the concept of a resultant, and deﬁnes a resultant in terms of equivalence: we can calculate the resultant of a system of forces, which represents the equivalent net eﬀect of that system of forces (but is generally a simpler to analyze) • the main focus of the Chapter is on moments produced by forces • a moment of a force about a point or axis is a measure of the tendency of the force to cause a body to rotate about that point or axis • classic example: • we will also see cross products in this Chapter MAE 2300 Sta=cs © E. J. Berger, 2010 5
10 Concept: Three Scenarios what do you notice? MAE 2300 Sta=cs © E. J. Berger, 2010 5
11 Concept: Three Scenarios what do you notice? MAE 2300 Sta=cs © E. J. Berger, 2010 5
11 Concept: Three Scenarios what do you notice? MAE 2300 Sta=cs © E. J. Berger, 2010 5
11 Concept: Three Scenarios what do you notice? MAE 2300 Sta=cs © E. J. Berger, 2010 5
11 Theory: Forces and Moments
• the moments resul=ng from these three loading scenarios all have both magnitude and direc=on; the direc=on is the axis around which rota=on would occur • as such, moments are vector quan==es with both magnitude and direc=on • moments have units of [force * distance], such as lb
in, N
m, etc. • what’s the distance? • it is the perpendicular distance (i.e. shortest distance) from the line of ac=on of the force to the point/axis of rota=on MAE 2300 Sta=cs © E. J. Berger, 2010 5
12 Theory: Forces and Moments
• the moments resul=ng from these three loading scenarios all have both magnitude and direc=on; the direc=on is the axis around which rota=on would occur • as such, moments are vector quan==es with both magnitude and direc=on • moments have units of [force * distance], such as lb
in, N
m, etc. • what’s the distance? • it is the perpendicular distance (i.e. shortest distance) from the line of ac=on of the force to the point/axis of rota=on MAE 2300 Sta=cs © E. J. Berger, 2010 5
12 Theory: Forces and Moments
• the moments resul=ng from these three loading scenarios all have both magnitude and direc=on; the direc=on is the axis around which rota=on would occur • as such, moments are vector quan==es with both magnitude and direc=on • moments have units of [force * distance], such as lb
in, N
m, etc. • what’s the distance? • it is the perpendicular distance (i.e. shortest distance) from the line of ac=on of the force to the point/axis of rota=on MAE 2300 Sta=cs © E. J. Berger, 2010 5
12 Theory: Cross Products
• the cross product opera=on states that for two vectors A and B: • MAE 2300 Sta=cs © E. J. Berger, 2010 5
13 Theory: Cross Products
• the cross product opera=on states that for two vectors A and B: • MAE 2300 Sta=cs © E. J. Berger, 2010 5
13 Theory: Cross Products
• the cross product opera=on states that for two vectors A and B: • MAE 2300 Sta=cs © E. J. Berger, 2010 5
13 Theory: Cross Products
• the cross product opera=on states that for two vectors A and B: • MAE 2300 Sta=cs © E. J. Berger, 2010 5
13 Theory: Right Hand Rule MAE 2300 Sta=cs © E. J. Berger, 2010 5
14 Theory: Right Hand Rule MAE 2300 Sta=cs © E. J. Berger, 2010 5
14 Theory: Right Hand Rule MAE 2300 Sta=cs © E. J. Berger, 2010 5
14 Theory: Right Hand Rule MAE 2300 Sta=cs © E. J. Berger, 2010 5
14 Theory: Moments as Cross Products
• the moment of a force around an axis can be constructed by considering two vectors:
• the force vector (Cartesian) • a (rela=ve) posi=on vector which locates the line of ac=on of the force with respect to the axis • NOTE that this posi=on vector can point to anywhere on the line of ac=on of the force vector! • HOW? Let’s derive it together... MAE 2300 Sta=cs © E. J. Berger, 2010 5
15 Theory: Moments as Cross Products
• the moment of a force around an axis can be constructed by considering two vectors:
• the force vector (Cartesian) • a (rela=ve) posi=on vector which locates the line of ac=on of the force with respect to the axis • NOTE that this posi=on vector can point to anywhere on the line of ac=on of the force vector! • HOW? Let’s derive it together... MAE 2300 Sta=cs © E. J. Berger, 2010 5
15 Theory: Moments as Cross Products
• the moment of a force around an axis can be constructed by considering two vectors:
• the force vector (Cartesian) • a (rela=ve) posi=on vector which locates the line of ac=on of the force with respect to the axis • NOTE that this posi=on vector can point to anywhere on the line of ac=on of the force vector! • HOW? Let’s derive it together... MAE 2300 Sta=cs © E. J. Berger, 2010 5
15 ...
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