Frames and Machines: FBDs

Frames and Machines: FBDs [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Reaction forces: one or two unknowns needed? Itcanbeshown.com Sirajuddin, David • How to draw FBDs of the machine? Dissemble into convenient parts to analyze. Draw an FBD of each • Each part is pin-connected  • reaction forces R in general: R = Rxi + Ryj • The reaction R points in some direction, we can always write R = Rxi + Ryj • But, these are two unknowns (Rx , Ry)! It is in our best interest to avoid doing this unless we have to. We would prefer only one unknown. • When do we need two unknowns? • only when we have no idea which way the “resultant” R actually points • When may we use only one unknown? • 2-force members! • Normal forces (e.g. rollers)

Reaction forces: one or two unknowns needed? Itcanbeshown.com Sirajuddin, David • When may we use only one unknown? • 2-force members! • Normal forces (e.g. rollers)

Reaction forces: one or two unknowns needed? • When may we use only one unknown? • 2-force members! • Normal forces (e.g. rollers) • For example, remember this problem [P5.85, Plesha, Gray, Costanzo]? Itcanbeshown.com Sirajuddin, David

Reaction forces: one or two unknowns needed? • When may we use only one unknown? • 2-force members! • Normal forces (e.g. rollers) • For example, remember this problem [P5.85, Plesha, Gray, Costanzo]? One Unknown A = -Acos30oi + Asin30oj OR Two Unknowns A = -Axi + Ayj (No need to model the force like this if we can only use one unknown!) Itcanbeshown.com Sirajuddin, David

Reaction forces: one or two unknowns needed? • When may we use only one unknown? • 2-force members! • Normal forces (e.g. rollers) Neverdo this. This is a 2-force member, but there are 4 unknowns. On this HW assignment, you can model them like this and it works out, but it will not always. You must make an insight. This gives a lot of unknowns. If you do not have a lot of leeway, this can put you in trouble of not being able to solve problems 2-Force member!, one unknown, We know the direction Itcanbeshown.com Sirajuddin, David

Frames and Machines: FBDs • By the way, remember zero-force members in trusses? • That is also key to solving some problems, and you must use physical insight to determine these. • Like zero-force members, if you cannot identify a two-force member you may not be able to solve the problem • Friendly off-topic analog: • (1) When given a “frames and machines” problem, your first step is find 2-force members • (2) When given a truss, your first step is to find zero-force members Itcanbeshown.com Sirajuddin, David

Frames and Machines: FBDs Itcanbeshown.com Sirajuddin, David What do we mean by “reactions”? For rollers and normal forces it is obvious For pins, it can sometimes be confusing when many parts are all connected at one point…

Frames and Machines: FBDs [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David • What do we mean by “reactions”? • For rollers and normal forces it is obvious • For pins, it can sometimes be confusing when many parts are all connected at one point… • “Which reactions go on which part?” Is the major question (see point C)

Reaction forces occur between each part and the pin to which it is connected Itcanbeshown.com Sirajuddin, David To answer this question, we need to understand exactly where the forces come from: Reaction forces are present because a particular part is pushing on a pinAND because the pin is pushing back on the part This is facilitated by Newton’s third law For now, all we need to know is that if these forces are present on an FBD of a machine part, then those same force will be present pushing back on the pin (meaning, wherever we put the pin physically in the picture [our choice] is the only place where we draw the equal and opposite forces “connecting,” nowhere else) The example problem will clarify

Frames and Machines: FBD Procedure [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David • General procedure • Identify 2-force members first • Definition of a 2-force member • A part that has only two connection points (where pins were) • The FBD of the part itself cannot have a pin on it if one end is connected to more than one member • For this problem: • Cannot have a pin at point C of member CD (we will see why later, C is a trisection of parts) • Canhave a pin at point D on member CD (only two parts for the one pin at D, no problem) • If the above is true, then the lines of actions of each force acting at one connection point, intersects the other connection point. • For a straight member, the forces follow the geometry of the member • For any curved or odd shaped member, draw a line connecting the two connection points, this is the line of action of the forces at each connection point. (This tends to seem strange to students. Just try and trust yourself using the methods).

Frames and Machines: FBD Procedure [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David • General procedure • Identify 2-force members first • Connect these forces to where you choose to put the pins in your FBD (pins can be left on some of the parts themselves, or a point FBD can be drawn of any pin) • I draw circles to help me keep track of where the pins are • Fill in the remaining forces in any order you wish

(1) Dissemble the machine into parts [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(2) Identify two-force members 2-force members! [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(3) Draw the reaction forces on each two-force member These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(4) Choose where the pins physically are These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(5) “Connect” reactions from 2-force members to the pins These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(6) Fill in any other forces whose direction is known These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this) Fill in the roller force next if you like, it is convenient [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(7) Choose where the remaining pins are physically located The rest of the forces, we no nothing about which way they actually point  x,y components Remember where your pins are! We have more to figure out [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(7) Choose where the remaining pin A is physically located The rest of the forces, we no nothing about which way they actually point  x,y components Remember where your pins are! We have more to figure out [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

(8) Draw in the remaining forces The rest of the forces, we no nothing about which way they actually point  x,y components for all remaining forces Remember where your pins are! We have more to figure out By definition (see the figure), there is no pin at E, or B (no connecting part present) Can choose pin locations for A on the dump or part ABC (no difference in forces) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Choice of pin position of A does not change anything The rest of the forces, we no nothing about which way they actually point  x,y components for all remaining forces Remember where your pins are! We have more to figure out By definition (see the figure), there is no pin at E, or B (no connecting part present) Can choose pin locations for A on the dump or part ABC (no difference in forces) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Choice of pin position of D does not change anything The rest of the forces, we no nothing about which way they actually point  x,y components for all remaining forces Remember where your pins are! We have more to figure out By definition (see the figure), there is no pin at E, or B (no connecting part present) Can choose pin locations for A on the dump or part ABC (no difference in forces) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Changes due to choosing a different location for pin C We could have chosen pins to be at different locations. How does this change things? Try moving the pin at C on FBD ABC to FBD CD New unknowns arise, and CD ≠ 2-force member anymore (4-force member) [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Changes due to choosing a second different location for pin C Move pin at point C to FBD CF New unknowns arise, and CD != 2-force member anymore (4-force member)  Leaving the pin at FBD ABC is probably the best strategy [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Using a point FBD of pin C Can also isolate pins, draw point FBDs If we take off pin C, then… Pin C But now we have one more FBD to work with to find the additional C reactions Sometimes this actually is useful Note: point FBD of pin C  no distances  only force eqns. can be used [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Final FBD used for solution: Find FCF Note: we may use 3 equations for each FBD [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Final FBD used for solution: Find FCF Start here, find Ax, Ay, FG [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Final FBD used for solution: Find FCF FBD of EDG  FCD [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Final FBD used for solution: Find FCF Then, FBD of ABC has three unknowns, Find FCF [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

Final FBD used for solution: Find FCF Then, FBD of ABC has three unknowns, Find FCF,  FCF= 27,172 lb [Plesha, Gray, Costanzo, Engineering Mechanics: Statics] Itcanbeshown.com Sirajuddin, David

References • M. E. Plesha, Gray, Costanzo. Engineering Mechanics: Statics. McGraw-Hill Higher Education. 2010.

Frames and Machines: FBDs