CE211 Study Skills
From wsuwiki
Study Skills Page for CE 211-Statics-- A page written by students for students
How to Write Unit Vectors
Being able to determine unit vectors is a great resource when trying to add vectors (Chapter 2), determine a particle at equilibrium (Chapter 3) and determine Moments (Chapter 4). Unit vectors have no units and are always equal to one. There are three different ways to write vectors: Point Coordinates, Space Angles, and Spherical Coordingates.
Point Coordinates: To determine a unit vector using point coordinates you must first make a position vector. Say I am trying to find a position vector from A to B with my two points A(3,2,0) and B(7,5,-11). You subtract [(7-3)i+(5-2)j+(-11-0)k] which gives you [4i+3j-11k]. Now to finish finding your unit vector you must divide by the magnitude of this distance. To do this take the square root of 4 squared + 3 squared + 11 squared. This value gives you 12.08, so your unit vector= [(4/12.08)i+(3/12.08)j-(11/12.08)k].
Spherical Coordinates: Spherical Coordinates are indicated by the little blue triangles in pictorial diagrams. Usually the hypotenuse of the triangle is a force being put on an object with another corresponding blue triangle situated perpendicular to the first. For example lets say we have a force pointing in the positive direction of each axis. A blue triangle's hypotenuse runs parallel to the force, one leg runs parallel to the z-axis down to the x/y plane and then runs to the origen. The angle from the hypotenuse to the leg running in the x/y plane is 45 degrees. Another blue triangle is situated so that its hypotenuse is on top of the leg of the first triangle that runs in the x/y plane. The other legs of the second blue triangle run to the positive x-axis and then down the x-axis to the origen. The angle between the hypotenuse and the x-axis is 60 degrees (This situation is similar to the picure on p. 74 as example 2-117). Using these two triangle we can determine the unit vector of this force. To determine the i,j, and k directions we now use combinations of sine and cosine functions of the angles given. The easiest part of the unit vector is the k component because it is simply the sin(45) in this situation. The i and j components are more complicated because you mus first determine the length the leg of the triangle that is in both triangles which in this case is cos(45). Now the i and j components can be completed by using the second triangle and the known length of cos(45). The resulting unit vector will look something like this [(cos60cos45)i+(sin60cos45)k+(sin45)k].
Space Angles: This coordinate system is designated by sweeping arc to each axis with the blue shaded triangles. Lets say we have a force pointing into Quadrant 1 with angles of 60 degrees to the x-axis, 30 degrees to y-axis, and 20 degrees to the z-axis. The unit vector for this situation would be given by the cosine of each angle; [(cos60)i+(cos30)j+(cos20)k].
A good check for your unit vectors is to use the components that you have found and find the magnitude of the vector--sqrt(a^2+b^2+c^2), where V=ai+bj+ck. If the magnitude doesn't come out to be 1 you know that you have done something wrong.
--bjalexander 17:17, September 26, 2007 (Pacific Standard Time)
--Ejlynch 15:24, September 22, 2007 (Pacific Daylight Time)
How to express a force vector in Cartesian form
Cartesian vector form is just another way of expressing a force vector. A force vector consists of a set of points on the xyz-plane that describe where the vector is and its directon and magnitude. Cartesian vector repersentation uses symbols to represent the unit vectors on the xyz-plane. The abreviations are i for the x-axis, j for the y axis, and k for the z-axis. So a force vector A, with its three xyz components, can be written in Cartesian vector form as follows; A=Axi + Ayj + Azk. Also the magnitude of the Cartesian vector, A, can be calculated as follows; A= sqrt(Ax^2 + Ay^2 + Az^2). It is actually a very simple concept to grasp. When I started doing Cartesian vectors all I did was add an i after the x-component, a j after the y-component, and a k after the z-component of the force vector.
--Mbofenkamp 14:44, December 5, 2006 (Pacific Standard Time)
Once you have your unit vector in Cartesian form for a force you can easily change it into a force vector. Simply multiply each component of the unit vector by the magnitude of the force and you have your force vector in Cartesian form.
Example: There is a force of 26N acting in the direction of the unit vector v=(5/13)i+(12/13)j. Express the force vector in Cartesian vector form.
The force vector, F= 26N(5/13)i + 26N(12/13)j = (10i + 24j )N
Notice that the unit vector had no units, but when we multiply by the magnitude of the force, which is in Newtons, the vector has units of Newtons, making it a force vector.
--Ejlynch 15:23, September 22, 2007 (Pacific Daylight Time)
Choosing a System
Choosing a system is the first step in creating a model and the first step in solving a problem. It's very important to choose an appropriate system from the start.
The most important factor in choosing a system is whether the number of equilibrium equations available to the system is equal to or greater than the number of unknown values in the system. This should be kept in mind at all times when considering possible system choices.
|Class of | Number of|Types of
|System | Equations|Equations Available
-------------------------------------------------------------------------------
Particle |2D| 2 |Forces in x and y
--------------------------------------------------------------------
|3D| 3 |Forces in x, y, and z
-------------------------------------------------------------------------------
Rigid Body |2D| 3 |Forces in x and y, and moments about a point
--------------------------------------------------------------------
|3D| 6 |Forces in x, y, and z, and moments about x, y, and z
-------------------------------------------------------------------------------
System Choices
|Type of |Class of |Max # of Unknowns |Special
System |Choices |Choice |to Solve |Condition
----------------------------------------------------------------------------------------------------------
Truss |Whole Truss |rigid body |3 |
--------------------------------------------------------------------------------------------
|Joints |particle |2 |Cut 2-force members
--------------------------------------------------------------------------------------------
|Group of Members |rigid body |3 |Cut 2-force members
----------------------------------------------------------------------------------------------------------
Frame/Machine |Whole Frame/Machine |rigid body |3 |
--------------------------------------------------------------------------------------------
|Each Member |rigid body |3 |Only multi-force members; cut joints
--------------------------------------------------------------------------------------------
|Groups of Members |rigid body |3 |Cut 2-force members or joints
----------------------------------------------------------------------------------------------------------
Many problems require multiple systems to be analyzed. Establish a "system path" at the beginning of the problem as it will act as a map to the solution. Think of it as: System A will yield unknowns 1 and 2, with unknowns 1 and 2 system B can be analyzed yielding unknown 3, unknown 3 will allow system C to be analyzed... and so on until the final unknown can be determined.
Remember to always indicate system choice. Literally write "system=...." next to free-body diagrams and equilibrium equations.
--Jdeleo 14:33, 29 October 2008 (UTC)
How to draw a free-body diagram
1. Draw Outlined Shape
2. Show All Forces
3. Identify Each Force
4. Identify The System
5. Identify Enviroment
6. Identify Interactions
--Robbord 23:28, October 24, 2006 (Pacific Daylight Time)
In order to draw a free-body diagram, you must first have a system. The choice of the system determines everything about the contents of the free-body diagram. Once you have chosen the system, draw the system on the free-body diagram as if it is standing alone and cut off from anything else outside of the system. After isolating the particle or group of particles, look on the original drawing for anything that directly touches the system, be it a cable, a beam, a force, etc. Draw each of these forces causing an effect on the system in the free-body diagram. This is a good time to do a quick check to see if the particle will even be in equilibrium, by making sure that there is at least one counter for every force in the system. If there is an unbalanced force that is not in equilibrium, you may have forgotten to include something. Once you have the free-body diagram draw out with the corresponding forces, you can then write equilibrium equations for the system. For example, if you are trying to find the forces acting on the hook of a crane hanging from the cable, you would have the hook as your system. The free-body diagram would include the hook, and then the cable holding up the hook; it would not include anything else on the crane. Then you would have to consider the other force acting on the hook, gravity, to balance out the upward force from the cable holding the hook in place.
--Jdmalone 18:59, December 8, 2006 (Pacific Standard Time)
How to write equilibrium equations
Initially one must place each of the forces and the object they act upon on a coordinate system. In many cases a cartesian coordinate system is the most convenient to use. Whatever coordinate system is used, one must determine all outside forces acting on your system. Assuming a standard cartesian coordinate system is used (e.g. the positive y axis is placed to the left of the orgin, the positive z axis is placed above the orgin, and the positive x axis is placed in the foreground in front of the orgin) forces that point downward are negative, forces that point upward are positive, forces that point right to left are negative, and forces that point left to right are positive.
In the event that forces are not entirely parallel or adjacent to any of the axes, one must find the individual X, Y, and Z components of the force. It may be helpful to imagine the force vector as the hypotenuse of a triangle whose other sides lie along the planes formed by any of the sets of axes in the coordinate system. With these triangles, standard trig functions can be used to find the components of forces(relative to the coordinate system), e.g. sin(theta) = Opposite/Hypotonuse, cos(theta) = Adjacent/Hypotonuse, and tan(theta) = Opposite/Adjacent.
The next step is to lie all of these equations out such that the sum of the forces along each axis collected from each force is equal to zero. Basic algerbra can then be used to solve for any unknowns.
Writing Equilibrium Equations
We can always write a set of equilibrium equations for any free body diagram. The following are the steps that helps me 1. If you have a 2D problem, the equilibrium equations results in 3 independent scalar equations. The maximun number of unknowns you can solve here is 3.
2. To make the equation simpler, I always try to select the point I take moments around such that the line of action. If you are missing this step then your algebra will be hard to solve.
3. You can sometimes take moments about two or three different points in a problem. Select each point so that you eliminate one or more unknowns from the resulting moment equation. Remember that the additional equations you generate in this way are not independent of the original equations, and, therefore, you will still have only three independent equations in 2D problems per free-body diagram and you can only solve for three unknowns per free-body diagram.
4. In 3-D problems the equilibrium equations result in six independent scalar equations (three components of force and three components of moment). Therefore, you can solve for up to six scalar unknowns per free-body diagram.
Determining 3D Equilibrium
Many problems in chapter 3 follow the idea of finding forces in cables in order to keep a hanging mass in equilibrium. To do this you first find each cable in cartesian vector notation for each cable. to do this simply calculate how many units in the x,y,&z direction one end of the cable is from the other, and find the square root of those values squared. Grouping the x,y,&z or (i,j,&k) together, solve for the variables you set for each cable.
Demand/Capacity Problems
The first thing I always forget is to do two cases! This is the most important step. Say you're given two ropes, after drawing your free body diagram, determine the i and j (2D) coordinates or unit vector for each rope. You then put those into your equilibrium equations, and since these are usually hanging capacities, the -w is included in the Y equilibrium equation. Example: Fxa(sin45)i + Fxb(sin45)i = 0 with the unit vectors in terms of the unknow value of each rope. Finally, to solve you try the two cases by applying one of the given max tensions of one of the ropes to the equilibrium equations. Solve both equations using one value and then solve them again using the other given value. Whichever value does not exceed the given values is the determined max hanging weight.
Another way to calculate capacity in a two dimension situation is by solving for the unknown
forces in terms of the known forces, then using simultaneous equations set the two unknown forces with their
coefficients equal to each other and solving for one. After you do this you will see one of your unknown forces in
terms of the other. If the first force is some fraction of the second, you know that the second force MUST
be larger than the first, so the first force cannot be equal to the capacity because the second force would then violate
that capacity condition. If the first force is, say, 1.5 times the second, you know that the first force MUST
be larger than the second force so the second force cannot be equal to the capacity because the first would violate the capacity
condition. For example, take a system in which a single mass is held up by two ropes that branch
off to form a Y-shape with the mass hanging at the bottom and you have to find the maximum weight able to be held by the ropes if
the maximum tension is 500 N. Take the angle that the tension in the left rope(T1) makes with the ceiling to be 30 degrees
and the angle the tension in the rope on the right(T2) makes with the ceiling to be 45. For the system to be in equilibrium,
you know that the X-components of the forces must add to zero. So your X-Force equation would be T1cos30=T2cos45, and since cos30
over cos45 equals about 1.5, you know that the maximum weight depends upon T1, so if you set T1 to be the maximum tension (500N)
and solve for T2, you have both the tensions and can get the maximum weight allowable.
How to calculate a moment about a point
To calculate the moment that a force creates about a certain point on a rigid body, the perpendicular distance and the magnitude of the force are needed. So, when the force is in the y direction, you will need an x distance for the moment to act on. Conversely, for forces in x, you will need a y distance. Be forewarned, this may require breaking a force into x and y components before you can find the moment. So, for example, if looking for the moment about a point lying on the y axis and a vertical (y) force of 200 N is given, you will need to find the x distance between the point (y axis) and where the force is located. If the x distance is 2 meters, then the moment would be (200)(2) = 400 N*m. When working in 3D, things become slightly more complicated. In 3D there are 3 different moments, one about x, one about y, and one about z. The moment about z is the same as the one moment in 2D, requiring a distance in x and a force in y, or vice versa. Moments about x require a distance in y and a force in z, or vice versa, and a moment about y requires a distance in x and a force in z, or vice versa.
--Jdmalone 18:59, December 8, 2006 (Pacific Standard Time)
Cartesian Vector Formulation:
In addition to calculating the moment about a point using the Principle of Moments as stated above, one can also calculate it by using vector analysis, or using the cross product with a matrix.
For a moment about a certain point O:
1. We need to find any position vector (r) from our point O to any point on the line of action of the force F.
2. We need to represent the x, y, and z components of the force vector.
Our basic equation is M = r X F: the moment about a point equals position vector (r) CROSSED with the force vector. So we need to use the cross product for this method. Our matrix should have three rows and three columns:
|i j k |
|r(x) r(y) r(z)|
|F(x) F(y) F(z)|
keeping in mind that these are the components of the position and force vectors and that the i,j, and k vectors have unit vector hats. Also remember that the position vector is from the point O TO the force.
Happy moment finding!
--jjreyes 20:43, September, 25 2007 (Pacific Standard Time)
Studying for the First Exam
If you are anything like me, you are nervous for the first exam. If you follow the outlined procedure below you will be fine and be totally ready for the first exam. Exam one covers the foundation of what the course is all about, and more importantly, it covers the basic general ideas of the idea of Statics. Here is what I did to be successful on the first exam.
1.) (recommended amount of time is 30 minutes) Three days before the exam, I began to look at what was going to be covered and began to outline major concepts on note cards from each chapter (2-4, neglecting chapter one because it was all on unit conversions). After I looked at all of the major concepts for these three chapters and putting them on note cards, I began to skim over the actually reading material in the book.
2.) (recommended amount of time is 1 hour to an hour and a half) Two days before the exam, I began the extra credit problems that our instructor assigned. I would highly recommend doing these, the instructor isn't going to waste your time by assigning things that won't be covered on the test. I would suggest that you just to half of the assigned problems, you don't want to do everything at once, its too much. After completing 1/2 of the e.c. assignment, and understanding what the problems are asking, and how to do them, I would call it good for that day after reviewing the note cards that I made the night before with the main ideas on them a few times.
3.) (recommended amount of time is 1 hour to an hour and a half) The night before the exam, I would finish the review e.c. problems and would review the problems that I did the night before as well as the note cards with the main ideas that are being covered. By now you should have these note cards pretty well memorized. After completing the e.c. assignment, I would suggest going back over the entire assignment and writing out procedures of how to solve the problems you encountered. This really helps out because it allows you to think of how to solve a problem the night before the test rather than during the test. I would review these a number of times until I had them pretty well memorized. After doing that, I suggest you just go over the e.c. assignment one last time and call it good for the night.
4.) (recommended amount of time is 30 minutes) The day of the exam (if you have time), I would review the note cards that have the procedures on them that you wrote out the night before on how to solve certain problems. I would also review one last time the e.c. assignment, and really make sure I understand what the problem was asking me to do. Then take the test!!!
If you follow the above four steps above, you will be more than ready for the first exam. Also, While you do these I also highly recommend that you VERBALIZE what you are doing. Throughout the semester, our teacher highly emphasized verbalizing what we were doing, and I didn't do it but then started doing it and let me tell you...IT WORKS! I began doing it, and I have never understood statics so well in my life! I would also recommend taking the time recommendations above. You don't want to jam it all in the day before or worse the day of the test, and doing it this way allows repetitive motion to familiarize your self with the material. Finally, I would also recommend that you take advantage of office hours if you have any questions, and other class-mates if the situation calls for it.
Good Luck and more Good LUCK!
--apeasley 13:59, April, 1 2008 (Pacific Standard Time)
NOTE: The statics exams are VERY fair. the questions are formatted EXACTLY like the homework and there are no surprises. Make yourself very comfortable with the concepts and ALL the homework problems assigned for that exam (the EC is a great summary, if you feel you struggled with one questions on the EC, look back to that chapter's earlier homework and review all of those to get the concepts down packed). If you can dominate the homework and you feel comfortable with the concepts than you're in great shape and the only thing to watch out for is the little mistakes that make a big difference (like punching things into your calculator wrong can mess up an entire question so be careful!).
--Corchilibeck 14:44, April 24, 2008 (Pacific Daylight Time)
How to evaluate cross products
Evaluating a cross product is quite simple. First, let's set up the cross product as a 3x3 matrix:
| a b c | | d e f | | g h i |
The formula to evaluate the cross product of this matrix is:
a * (e * i - f * h) - b * (d * i - f * g) + c * (d * h - e * g)
But who can remember that? Instead, think of it this way:
1) Cross out the first row and column. You will be left with this:
| e f | | h i |
2) Next, multiply the diagonals, e * i and f * h, and subtract the second from the first, or e * i - f * h.
3) Multiply #2 by a to get a * (e * i - f * h). Notice that this is the first term in the formula above.
4) Repeat steps 1-2, except crossing out the first row and second column, and then multiply the result by -b to get -b * (d * i - f * g) (the second term in the formula above).
5) Repeat steps 1-2 again, but cross out the first row and third column, and then multiplying the result by c to get c * (d * h - e * g) (the third term in the formula above).
6) Add the results in steps 3-5.
--Wlandowski 22:20, February 14, 2008 (Pacific Standard Time)
Using the crossP() function on your TI-89 for Fun and Statics Homework
If you're like me and bad with negative signs, and you have a TI-89, the crossP() function is a life saver. Here's how to use it to your advantage.
1) Start out at the TI-89's 'home' thing. This is the part that you're probably already using, so don't really worry about this step.
2) Hit 2nd->5(math)
3) Hit 4 or scroll down to 'Matrix' and open the sub-menu
4) Scroll down to 'Vector ops' (or scroll up, it will be faster) and open that sub menu
5) Hit 2 or scroll down to 'crossP(' and hit enter
6) The crossP() function takes two arguments, the two vectors you want to cross
Enter each vector (i.e. [1,2,3]) separated by a comma and close the parentheses. It should look something like this:
crossP([1,2,3],[4,5,6])
This will calculate [1,2,3] x [4,5,6] and give you the result as a vector with the i,j, and k components. To enter the brackets, hit 2nd->, ([) or 2nd->/ (])
7) Never make an answer-crippling cross-product sign error again! (well, maybe)
You'll notice that the 'Vector ops' menu has a couple other useful functions in it.
--Jdryan 20:04, March 20, 2008 (Pacific Daylight Time)
Implementation of the Right Hand Rule
In addition to the procedure of how to calculate a moment around a point, you also need to implement the right hand rule. The process described above is known as the Principle of Moments. After you have found the x and y components of your force, you must then multiply your force by a perpendicular distance. For example, in 2D the only moment to be calculated is a z-moment which is calculated by either an x-force multiplied by a y-distance, or a y-force multiplied by a x-distance. You then add those together to calculate your z-moment. However, the determination of the signs (positive and negative) is a very crucial step to the principle of moments. You must use the Right Hand Rule. This is done by imagining that you are at the origin of a 3D axis (x,y,z), where the x-axis is in front of you and extends in the positive direction in front of you and negative behind you, y-axis positive to your right, negative to your left, and positive z-axis running vertically through your body upwards, and negatively downwards towards your feet. You first point your right hand fingers in the direction of the distance, then to turn your palm towards the direction of the force. You then curl your fingers in towards your palm without moving your hand. The direction of your thumb is then the sign of that calculation in that direction of the axis. For example, if you had a 3m distance in the x-direction, and a force in the y-direction of 40N, you would multiply those, with the directional sign of +z. I hope this helps you in your Principle of Moment Calculations.
--r_vannice 14:41, September 21, 2007 (Pacific Standard Time)
Another method for the right hand rule is to point your index finger of your right hand along the first vector (position vector). Then turn your hand, keeping your index finger in the correct direction, until your middle finger is pointing along the second vector (the force vector). The direction your thumb is pointing is the resultant vector or moment direction.
--Elizabeth arment 21:04, February 18, 2008 (Pacific Standard Time)
How to calculate a moment about an axis
Moment about an axis-to determine the moment of a force about an axis we must first define what a moment of a force is. It can be simple put is that is a force that has a rotating character of it which is around a point on a object it does not lie on the same line as of the force(s). Now, we can choose a system to evaluate and determine where and account for all the forces and where to place the moment in the system. We also to need to take in account were the moment can be placed, because you want to place the moment were it will cancel out forces that are on the same line and include forces that are perpendicular to it. For example, when evaluating a moment and there is a force perpendicular in the y-direction one would use the x-direction distance and multiply it by its force. Where this equation would be used to solve for the moment about the axis by adapting the following equation, M=dF where the moment is about an arbitrary axis, as also the distance. Like stated earlier, if the force intersects the axis that is being evaluated at then the force is zero. As well as if the force is parallel to the axis then the moment of the force about the axis is then zero. Another tool that we need to evaluate the moment about an axis is the unit vector, position vector, and the force, but you also have to take in account that if you get a negative answer for your moment of the equation M=u(rxF) (to calculate the moment about an axis), then the direction of the moment is opposite of the unit vector is in.
How to determine a couple moment
"A couple is defined as tow parallel forces that have the same magnitude, have oppositie direcitons, and are seperated by a perpendicular distance d". This relationship can be expressed by the equation M=Fd where M is the resultant moment, F is the magnitude of the force, and d is the perpendicular distance between the forces. It is particularly important to remember that d is the perpendicular distance between the forces when calculating the couple moments that are not square with the coordinate system when it is easy to determine that distance. The resultant moment's sign is determined by implementing the right hand rule outlined above.
--Ejergensen 12:29, September 26, 2007 (Pacific Daylight Time)
How to determine an equivalent loading
When finding an equivalent load first find the component forces in each direction and the moment on the system. Then at the point you wish to put the equivalent load at apply the component forces you found and the couple moment. If you are trying to apply a single force to a 2D system to equate to for all the forces and moments on the system add up forces and moments as before. Next use the equation M=Fd with this solve for distance using your component force and moment. This will give the location of the equivalent force, now find the magnitude of the force and put it at this location.
--Svengasser 15:05, February 19, 2007 (Pacific Standard Time)
Calculating a point load out of a distributed load
Distributed loads are loads that act over a distance, rather than just at a concentrated point. However, when analyzing the forces in a free-body diagram, it is necessary to condense these distributed loads to a single force acting at a point. Distributed loads have units of Force/Distance. Therefore, in order to find an equivalent force, we must multiply these quantities by the distance over which they act. In this class we see three different ways in which these loads can be distributed: in a rectangular distribution, a triangular distribution, or a combination of the two. In the case of a rectangular distribution, you simply multiply the loading by the distance over which it acts. This gives you the area of the distributed force, which is also the value of the equivalent force. Because of symmetry, we then place this load at the center of the length over which it acts. It will point in the same direction as the distributed load. In the case of a triangular distribution, you simply multiply the loading by the distance over which it acts and divide it by 2. This is also equivalent to the area of the triangle, and again represents the value of the resulting concentrated force. Unlike the rectangle; however, this force is not placed in the center of the length. This makes sense, because more of the load is concentrated to one side. Therefore, the load is placed 2/3 of the entire distance from the pointed end of the triangle (or 1/3 from the tall end). Finally, we often see strange shapes that don’t seem to fit either of these categories. In this case, we try to break up the loading into rectangles and triangles and evaluate them separately using the rules above.
How to solve 2D rigid body equilibrium problems
Two-dimensional rigid body problems are not very difficult to solve, if you choose the right system that is. Just like in the 3-D problems, frame problems, and machine problems, choosing a system is critical. You need to choose the apropriate system, or systems, in order to solve the problem. Not choosing the correct system can lead to wrong answers and twice the ammount of work time on a problem. The way I choose systems is by looking at the drawings and visualizing which parts would have the least unkowns and which parts I could group together to get rid of unknowns. You will find out that for 2-D problems system choice is fairly straight forward and not to difficult. Once you choose your system you need to identify the enviroment. This means taking into account all of the external forces acting on an object. It helps me to list these in words on my assignment. Next you need to identify the interactions. These are the results of any external forces acting on the system and I find it helpful to write these next to the enviroment discriptions in the form of symbols, vectors, and numbers where nessesary. The next step is to draw the free-body diagram wich includes your system and all of the interactions, which are drawn and labled. The next step is to wirte the equilibrium equations, this includes sum of the forzes in the x and y directions, and also the moment about the z-axis. After you write the three equilibrium equations you can move on to solving them. Don’t forget to include the different cases if it is a demand/capacity problem.
--Mbofenkamp 17:26, December 5, 2006 (Pacific Standard Time)
How to solve 3D rigid body equilibrium problems
The first thing you need to do with 3D equilibrium problems, as with most other statics problems, is to choose an appropriate system to model. Once the system is chosen you should draw a free body diagram. The best free body diagrams should start with an outline of the system and include all forces and couple moments acting on the system as well as support forces. Also you should pick an origin for your axis that allows most or all of the applied forces to be parallel to them and include any dimensions given. If you draw your free body diagram well, the rest of the problem should be pretty easy. In three dimensions you have six equilibrium equations that you may apply: the sum of the forces in the x, y, and z directions as well as three moment equations. The moment equations are the sum of the moments about an axis, x,y, or z. The moment equations can be a bit tricky to the unpracticed, but here is a general guideline: For x-axis moments look at z-forces and y-distances and y-forces and z-distances. For the y-axis look at z-forces and x-distances and x-forces and z-distances. For z-axis moments we look at x-forces and y-distances and y-forces and x-distances. The easiest way to define the direction of a moment about an axis is to picture the force pulling on the object. If the object will spin counter clockwise about an axis, then it is a positive moment. If the object will spin clockwise about an axis, then we put it into the equilibrium equation as negative.
--Gfielding 18:44, May 3, 2007 (Pacific Daylight Time)
"The first and most important thing to solve for an equilibrium problem of a rigid body is to draw the free body diagram of the body. Correctly drawing the free body diagram will help you in solving the 90% of the problem because after this you only have to do correct calculations. To draw the free body diagram, show all the forces and couple moments acting on the body in x,y, and z plane and label all the loadings and specify their directions relative to x,y, and z axis. Also indicate the dimensions of the body necessary for computing the moments of forces. Once free body diagram is done we move on to writing equilibrium equations. If the x,y, and z forces and moment components are easy to determine, then apply the six scalar equations of equilibrium, otherwise use the vector equations. Choose the direction of an axis for moment summation such that it intersects the lines of action of as many unknown forces as possible. The moments of forces passing through points on this axis and forces which are parallel to the axis will then be zero. If the solution of the equilibrium equations yields a negative scalar for a force or a couple moment magnitude, it indicates that the direction you have chosen in your free body diagram is opposite to the actual direction of that perticular force or moment. So you just need to reverse the direction in your free body diagram." Karan Chaudhary (ID# 10942796)
How to solve for member forces in trusses... How to start?
I understood how to use the method of joints and sections but still got lost on where to start and would get stuck with systems that had too many variables, so this is how i made things easier
1)First things first, when you have a bridge, find the reaction forces. look at the entire bridge and look at the pins, rollers and/or supports holding the bridge up, and look at the loads applied. Then take the sum of the forces in the X and Y directions to find the reactions, if you dont have enough equations to solve, then also take the sum of the moments about a point.
2)using either the method of joints or sections (explained below) the next step is to look at one end that has the fewest members. Even when they tell you to find a member force that is in the middle of the bridge, usually you cant determine that force without finding out most of the others as well, and it doesnt take much more time to do so.
so look at one end, and find the other 2 members (maybe three) using either methods below (i liked joints) because now you have the reaction forces at the pins/rollers/supports
3) then work your way inwards to the member force in question. using joints or sections will get you there just the same, some people find joints requires more work, but i found them easier.
--Corchilibeck 14:52, April 24, 2008 (Pacific Daylight Time)
How to solve for member forces in trusses using method of joints
As far as trusses and member forces go, the method of joints is the longest method to use when it comes down to solving for the forces in truss members. The way I go about using this method is I start by making the whole truss the system and solving for the external forces, i.e. the forces at the end joints. Once I have the forces at either side of the truss then I take the first joint at either end of the truss and solve inward, either from right to left or left to right, whichever has fewer unkowns in it. When it comes to unkowns always watch out for zero-force members that you can identify quickly to get rid of some of the unknown force members. When you pick a joint you treat it as a particle equilibrium problem so first you draw a free body diagram. Remember when you draw the free body diagram draw all of the force members in tension (pointing away from the particle/joint) so that if you get a negative value you know that the member is in compression, if not it is in tension, just as you drew it. You can then use two equations, sum of the forces in the x-direction or sum of the forces in the y-direction, whichever equation has fewer unkowns in it, to solve for the unkowns in the particle. Then you just move inwards to the next joint and solve for it, and make sure to try to visually see which joint will have the least unkowns in it. This should make this method a lot easier. If you can visualize the joints and imagine which members should be in tension and compression, based on external loads, then it makes the method of joints for solving force members much easier because you can catch your own mistakes, when it comes to signs at least.
--Mbofenkamp 14:43, December 5, 2006 (Pacific Standard Time)
How to solve for member forces in trusses using method of sections
We use method of sections to find the force of selected members of a truss. To start, you must analyze the truss and solve for all external forces. These external forces will allow you to solve the forces of the members. After you have your external forces, you can "cut" the truss respectively, allowing you to find your members using this method. The cut is not a straight line down your truss, but in fact is curved and cuts the members into halves. When cutting, make sure that you can still solve for all your members without too many unknowns. Now that you have two pieces of the truss, the time comes to choose a system. The more preferred system is the easier one with more known constants. When solving for your first member, be sure to analyze every member. Some of your selected members will share a constant, so you can take a moment equation at that constant to solve for a member without that constant. Also, remember to look for zero-force members. It is possible that one of the members you must find is a zero-force. Use your equilibrium equations to solve for other members and you will have solved for the members in the truss using method of sections.
Method of Sections-Before any calculations be made we must first understand what method of sections really means. First the method of sections is a way to cut trusses to examine all of the internal forces, but also one must know why possible to cut a truss into pieces and where to do so. Now must look at why it is possible to cut the truss into pieces, simple stated by the method of sections that the truss is in equilibrium, so there for all of its members are in equilibrium which gives us the ability to cut the truss into pieces. Another hint making the cut is that you want to make it perpendicular to two of the three forces unknown forces, also when making a cut try to avoid cutting more than three forces in the truss because it will make it easier to solve. We now can set up are system and evaluate are external forces, and then we can start making the cuts using the knowledge given above. For me a good way to evaluate the internal forces due to the cut is to draw them all in tension and later go back and redraw them; if the member is tension it is pulling from the section if the member is in compression than the member would be pushing towards the section. And that should be enough the solve with the method of sections, but also take in consideration of all the assumed knowledge of summations of the forces in the x,y,(z) direction and the moment.
All about 2-force, 3-force, and zero-force members
2 Force Members These are members in which there are only two forces acting along the object. All 2 force members are either in compression or in tension.
3 Force Members These are members in which an additional force is acting on the object, creating a moment.
Zero-Force Members These can be determined by examining one joint at a time.
If the joint has exactly two members and both are two force members than both are zero-force members.
If the joint has exactly three members, two linear the third is a zero-force member unless it is loaded.
--Robbord 09:36, October 24, 2006 (Pacific Daylight Time)
Method of Sections-Before any calculations be made we must first understand what method of sections really means. First the method of sections is a way to cut trusses to examine all of the internal forces, but also one must know why possible to cut a truss into pieces and where to do so. Now must look at why it is possible to cut the truss into pieces, simple stated by the method of sections that the truss is in equilibrium, so there for all of its members are in equilibrium which gives us the ability to cut the truss into pieces. Another hint making the cut is that you want to make it perpendicular to two of the three forces unknown forces, also when making a cut try to avoid cutting more than three forces in the truss because it will make it easier to solve. We now can set up are system and evaluate are external forces, and then we can start making the cuts using the knowledge given above. For me a good way to evaluate the internal forces due to the cut is to draw them all in tension and later go back and redraw them; if the member is tension it is pulling from the section if the member is in compression than the member would be pushing towards the section. And that should be enough the solve with the method of sections, but also take in consideration of all the assumed knowledge of summations of the forces in the x,y,(z) direction and the moment.
Studying for Exam 2
If you are anything like me, you did decent on the first exam and thought the second exam would go the same way. If you follow the outlined procedure below you will be fine and ready for the second exam. Exam two covers a large portion of the course dealing with rigid bodies, and trusses. Here is what you should do to be successful on the second exam. In terms of the grand scheme of things, this exam really covers what this course is all about: static equilibrium, and choosing a system. You should really familiarize yourself to the distributed loads in chapter four, this won't be the only time you see them in the course, they come back "indefinitely" and you will see them again. In specific you should learn the reduction process of a simple distributed load, force or couple system. More importantly, you should really try to understand what the equations of equilibrium in chapter five actually are: ways for you to solve for unknown tensions/compressions and forces in frames or machines. I would also suggest that you learn how to draw free body diagrams; these are somewhat difficult to do if you have never seen them at first, but after doing them 10-20 times you should be a pro at them! Finally, i would recommend that you become VERY familiar with the method of sections and method of joints to solve for tensile and compression forces as well as zero-force members in chapter six trusses.
1.) (recommended amount of time is 30 minutes) Three days before the exam, you should begin to look at what is going to be covered and began to outline major concepts on note cards from each chapter (end of chapter 4 to the end of chapter 6). After you looked at all of the major concepts for these three chapters and putting them on note cards, you should begin to skim over the actually reading material in the book.
2.) (recommended amount of time is 1 hour to an hour and a half) Two days before the exam, I began the extra credit problems that our instructor assigned. I would highly recommend doing these, the instructor isn't going to waste your time by assigning things that won't be covered on the test. I would suggest that you just to half of the assigned problems, you don't want to do everything at once, its too much. After completing 1/2 of the e.c. assignment, and understanding what the problems are asking, and how to do them, I would suggest calling it good for that day after reviewing the note cards that I made the night before with the main ideas on them a few times.
3.) (recommended amount of time is 1 hour to an hour and a half) The night before the exam, I would finish the review e.c. problems and would review the problems that I did the night before as well as the note cards with the main ideas that are being covered. By now you should have these note cards pretty well memorized. After completing the e.c. assignment, I would suggest going back over the entire assignment and writing out procedures on how to solve the problems you encountered. This really helps out because it allows you to think of how to solve a problem the night before the test rather than during the test. I would review these a number of times until I had them pretty well memorized. After doing that, I suggest you just go over the e.c. assignment one last time and call it good for the night.
4.) (recommended amount of time is 30 minutes) The day of the exam (if you have time), I would review the note cards that have the procedures on them that you wrote out the night before on how to solve certain problems. I would also review one last time the e.c. assignment, and really make sure I understand what the problem was asking me to do. Then take the test!!!
If you follow the above four steps above, you will be more than ready for the second exam. Also, While you do these I also highly recommend that you VERBALIZE what you are doing. Throughout the semester, our teacher highly emphasized verbalizing what we were doing, and I didn't do it but then started doing it and let me tell you...IT WORKS! I began doing it, and I have never understood statics so well in my life! Another thing i would recommend is looking at this Wiki cite, it has a lot of good information on how to solve things in terms of a procedure, or step by step process. I would also recommend taking the time recommendations above. You don't want to jam it all in the day before or worse the day of the test, and doing it this way allows repetitive motion to familiarize your self with the material. Finally, I would also recommend that you take advantage of office hours if you have any questions, and other class-mates if the situation calls for it.
Good Luck and more Good LUCK!
--apeasley 12.36, April 15 2008 (Pacific Standard Time)
Using Cramer's Rule
An extremely helpful tool when solving equilibrium equations for many unknowns is Cramer's Rule. All you need is a calculator that can solve matrices (most graphing calculators can). By doing this you can save yourself a ton of time and effort of doing tedious algebra. Cramer’s rule states that there exists a unique solution to set of linear equations if the number of variables is equal to number of equations. Also the matrix of coefficients (A) must be nonsingular (has an inverse).
Ax=b (representation of a linear system)
A is the matrix containing the coefficients and x is the column vector of variables and b represents the column vector of solutions.
The column vector variables x can be represented as
x=A^(-1)b
After coming up with your equilibrium equations all you need to do is make sure the equation is simplified as much as possible. Next, place the coefficients of the unknowns in the A matrix (one row per equation). On a TI-83 press the 2nd key then X^(-1) key to create/edit a matrix.
If I was trying to solve for the following force equations (pg 252 Hibbeler)
Ax + (0.707)Tc -(3/9)Td = 0
Ay + (6/9)Td = 1000
Az - (0.707)Tc - (6/9)Td = 0
and moment equations
(-4)Td = -6000
(4.24)Tc - (2)Td = 0
My A matrix would look like
__Ax___Ay____Az___Tc____Td__(coefficients)
|_1_____0_____0___.707___-3/9___|
|_0_____1_____0____0______6/9__|
|_0_____0_____1__-.707___-6/9___|
|_0_____0_____0____0______-4___|
|_0_____0_____0__4.24_____-2___|
My b (solution) matrix would look like
|___0___|
|_1000__|
|___0___|
|_-6000_|
|___0___|
After creating both of these matrices, all you need to do is go back to the Matrix function, go to names, the [A] matrix, and then press the X^(-1) key. Press the matrix function one last time, go to names, the [B] matrix, and then just press enter. The calculator should spit out a solution of the form:
[~=0.......] solution for Az
[0.........] solution for Ay
[1500.23...] solution for Az
[707.547...] solution for Tc
[1500......] solution for Td
Louis Tibbs Spring 2007,CE211
Using MATLAB to solve Linear Equations
For those of you who don't have an awesome calculator that can compute matrixes or would prefer to use a computer, MATLAB is an extremely powerful program that can run circles around most calculators. You will most likely end up learning how to use it in higher level engineering courses anyways. So why not get a head start? There also just happens to be a free version online that is available to all students at WSU.
1. Click the above link.
2. Go to Software
3. Go to MATLAB
You should now be looking at a console and display window
The technique is the same as previous except entering the matrixes are a little different in this application.
To enter my A (coefficient) matrix I would need to simply type
A = [1 0 0 .707 -3/9; 0 1 0 0 6/9; 0 0 1 -.707 -6/9; 0 0 0 0 -4; 0 0 0 4.24 -2]
- the ';' represents starting a new column in MATLAB
My b (solution) matrix would look like
b = [ 0; 1000; 0; -6000; 0]
Rather than using A^-1*b, a better method is to use MATLAB's \ operator. The inverse method will not work with an over determined system, while \ will. Any error resulting from two equations not agreeing on a solution is automatically eliminated using the least squares method. The command to solve the system is:
A\b
Now just hit the 'Run MATLAB' button and it should spit out an anwser that looks something like
ans =
-2.3585e-01
0
1.5002e+03
7.0755e+02
1.5000e+03
- for personal preference I used 'format short e' to display my answer you can play around with the command if you don't like how MATLAB displays numerical values. Just add the format command as the first line of code. This makes the results more readable.
Louis Tibbs Spring 2007,CE211 --Bseater 12:06, April 24, 2007 (Pacific Daylight Time)
How to solve rigid body/machine problems
First off, remember when drawing the free body diagrams and forces that apply to more than one system should show up as equal but OPPOSITE forces. Also two force members are modeled as force. For example, a two force member that is bent at a 90 degree angle will create a force that can be modled as a straight line that passes through the end points of the member.
The first and most important step to solving rigid body/machine problems is to choose a system! Once you have chosen your system(s), then you can draw your free body diagram. Picking correct systems is important! worst case scenario is you can draw all of the free body diagrams and try to work through them all, you'll get the problem eventually but it will take forever. but learning how to look at the problem and pick systems that will solve easily and quickly will make your life so much easier! how to do this the simple way is below
the easiest way to pick which systems to use is to look at one end of the machine (usually an end with a load which has a known force) and work from that end piece in towards the members in question. you can choose multiple pieces of your machine as a system, but make sure you will have enough equations to solve for the variables. now, once you've identified that first piece as your system the equilibrium and moment equations can be calculated and you can then solve for reaction forces on that member (its pins, normal forces ect). then you can chose your next system, picking the piece that is connected to the one you just used works best because then you know some reaction forces already and will then be able to solve for the others. note: the reaction forces you just calculated will be of the same magnitude but opposite in direction on the other member(your new system choice). following this pattern of working from the outside in will get you to your answer in a timely fashion.
Keep in mind, that if there is a 90 degree two force member, it's dimensions are modeled in the shape of a triangle; a length in the x-direction, a length in the y-direction, and solve for the hypotenuse. Then when plugged into the equilibrium equations, the fx is multiplied by the x-distance over the hypotenuse, and fy by the y-distence over the hypotenuse.
--Corchilibeck 22:12, April 24, 2008 (Pacific Daylight Time)
How to draw shear and moment diagrams
1) Determine the forces and moments acting on the beam.
2) Cut in between the forces and moments.
3) Section the beam perpendicular to its axis at each distance and draw the FBD of on of the segments.
4) Calculate Shear force by summing the forces perpendicular to the axis.
5) Find the moment by summing the moments about the sectioned end of the segment.
Before you do anything you must determine the forces and moments acting on the beam. Find the unknowns, such as distributed loads and forces that require you to use a moment equation to find. Next you must look at the beam and cut at every point where there are forces and moments acting on them. Remember to cut in the right spot with distributed loads. For example, there is a load of 100 lb/ft over 4ft. So there is a force of 400lb 2 feet into that distributed load. Next, section or "cut" the beam to each of its *discontinuous points. Draw Free Body Diagrams of each segment. You can calculate shear force by summing the forces perpendicular to the axis. If you get an x, solve the equation for it because you will need it for the moment. Next find all the moments by summing the moments about the sectioned ends of the segments. After you have all your numbers make two graphs. It is easier to line them up if they are near/under a Free Body Diagram of your beam. Use your numbers and graph. It should look like pieces of different equations all pieced together. Make sure to shade in the graph.
Notes: *It is important to not include point force or moment force acting at the point in which your discontinuity happens in your diagram of your cut. The cut includes everything (left or right of the cut) up to where the point force is but there is not included in the equations until the next section. If there is a point for at the end of a beam it is easier to use a combination of analyzing your cut and leaving out the force in your equations till the diagram. Make sure that your diagram is in equilibrium.
--Hicks4 10:30, April 25, 2008 (Pacific Daylight Time)
Sign convention for internal forces
Before determining any internal forces, we must first establish a sign convention for each of the three internal forces (normal force, shear force, and bending moment).
1) Normal force - A normal force is considered positive when it points away from the object at the location of the cut. For instance, a horizontal beam that is cut with the left side remaining would have a positive normal force if that force points to the right.
2) Shear force - A shear force is considered positive when the force would cause the remaining member to rotate in a clockwise direction. For example, a horizontal beam that is cut with the left side remaining would have a positive shear force if that force points down.
3) Bending moment - A bending moment is considered positive when the moment would cause the remaining member to make a smiley face if the member were elastic (concave up). For example, a horizontal beam that is cut with the left side remaining would have a positive bending moment that is counter clockwise.
It is important to make sure you have established the right sign convention in your free body diagrams before you try to draw the the shear and moment diagrams!
--Wlandowski 3:50, April 14, 2008 (Pacific Standard Time)
Why we use the variables we do in Shear and Moment Diagrams
N Normal - forces colinear, on the horizontal axis of symmetry.
V Shear - forces perpendicular, on the vertical axis of symmetry.
M Moment - external forces causing a bend on the axis of symmetry.
--Robbord 00:05, October 25, 2006 (Pacific Daylight Time)
Shear and Moment Diagrams without Equations
Drawing shear and moment diagrams without using equations is relatively straight forward as long as you follow the points outlined below. First thing, it is essential to draw the beam with just its loads and moments. From there you can start from the left side of the beam and work to the right (looking at the sections between forces and moments).
Here are some important concepts:
1. If the load is concentrated, at a point, then there will be a jump in the shear curve with the same value and same direction.
2. If the moment is concentrated, then there will be a jump in the moment curve, same value, but opposite direction. Keep in mind that the moments will not affect your shear curve.
For sections of the beam in between these point loads or moments,
1. If the load is concentrated (zero, ie no distributed loads), then the shear curve will be constant, and the moment curve will be linear.
2. If the load is constant (rectangular distributed load), then the shear curve will be linear and the moment curve is quadratic.
3. If the load is linear (triangular distributed load), the shear curve will be quadratic, and the moment curve will be cubic.
For values of the curves,
1. dv/dx = -w (load) the slope of the shear curve = - value of the load
2. dM/dx = v the slop of the moment curve = value of the shear
3. area under load function = change in value of shear
4. area under shear curve = change in value of moment
NOTE: if your shear curve is linear (distributed load) it often crosses through the x-axis at an unknown x-distance and when using this method, you want to calculate the area under the shear curve to find dM/dx, but you dont know the base length of the triangle? simple! take similar triangles of the big triangle (you know the entire base lenth= length load distributed for, and the total height), and compare that ratio of b/h to the known height of your small triangle and set its uknown base length as x and solve that ratio of big triangle to small triangle for the uknown base length x. Then you can easily take the area under the small triangle to find dM/dx in that interval.
So this is another method to find the shear and moment curves of a beam. Keep in mind that this method can be used in conjunction with the calculation method and can help you make sure you're curves are correct! Happy shear and moment curve finding!
--Jjreyes 21:34, December 2, 2007 (Pacific Standard Time) --Corchilibeck 09:49, April 25, 2008 (Pacific Daylight Time)
Determine slip or tip
First thing to remember is that neither slip nor tip can be determined without solving for the other. Normaly you will be giving an pbject that has a (weight) froce, a normal force, a pulling force P, and a frictional force. The P will have an x or y value or both if it is at an angle, the normal force is a y force with no dementions, the weight force is a y force, and the friction acts as a x-force. Draw a free body diagram with all forces acting in their proper direction, then you can begin solving. TIP: The point on which the object will tip is most likely a corner on the same side of P but on the corner in contact with the surface or floor. you can take the moment about the tipping point, but because tip causes the surface along the floor to "lift" off of the floor, the normal force and friction force are no longer taken into acount. So the force P can be solved for by taking the moment about the tipping point.
SLIP: In the case of sloving for slip, take the sum of the forces in the Y direction and solve for the normal force in terms of P. Then take the sum of the forces in the X direction. Since the frictional force equals the normal force multiplied by the static friction (F=uN), the normal force multiplied by the static friction can be subsituted in for the frictional force in the x-direction, and from that should be able to solve for P
And which ever case has the lowest value of P will be your answer.
There are three cases you should be aware of when dealing with friction.
1) When the Force is static, there is no motion impending with static friction. Use f<Usn to solve.
2) When the force is dynamic(kinetic) and impending motion at all points at the same time. Use F=Usn and three equilibrium equation to solve.
3) When the force is dynamic(kinetic) and impending motion of all parts. Use F=Usn and three equilibruim equations to solve.
-Usn =static friction
--N3wdeal 13:56, April 24, 2007 (Pacific Standard Time)
Calculating slip or tip
When looking at a slip or tip problems you have to consider what the object can do. Are there multiple objects that can slide against eachother? Is one of the objects going to tip? Is the object going to move at all?
1) Draw a free body diagram of the object (or objects). Remember that the normal force is an unevenly distributed force so that means you cannot use moment equations because you don't know the location of the normal force. Friction always opposes motion so if you have a force pushing to the left you friction force will point right. You free body diagram should include a friction force, a normal force, a weight at the center of mass (usually labeled), and force(s) being applied.
2) Determine how much force it takes to cause impending motion on the object. Remember you only get two equations forces in the x and forces in the y and you also get a conditional equation for force of friction=normal x the coefficient of static friction.
3) If the amount of force applied is given check to see if it is enough to cause impending motion.
4) Make another free body diagram and assume the normal force acts at a corner. This is the tipping cases. Now you can use three equations: moment and forces x and y. Compare this force to the amount of force required to cause the object to slip. If the tipping case needs less force, then the object will tip. If not it will slip.
--Run4life 22:37, April 21, 2008 (Pacific Daylight Time)
Studying for Exam 3
If you are anything like me, you did decent on the first exam, not as good on the second exam and thought the third exam was going to be somewhere in the middle of the first two(in terms of a score). If you follow the outlined procedure below you will be fine and ready for the third exam. Exam three covers about a quarter of the course dealing with shear and moment diagrams, static friction, and internal/external problems. Here is what you should do to be successful on the second exam. In terms of grand scheme of things, some major concepts that you will really want to know are shear and moment diagrams; how to make internal forces, external; and i would also recommend that you look at the friction chapter in some detail, learning how to determine weather a system is going to slip, tip, or move together are very important to learn for this exam. However you shouldn't forget past information because this course is after all about static equilibrium, choosing a system and solving for the unknowns, so don't expect the answer to choosing a system just pops out, you should practice determining systems that will give you the right equations to solve for the unknowns.
1.) (recommended amount of time is 30 minutes) Three days before the exam, you should begin to look at what is going to be covered and began to outline major concepts on note cards from each chapter (end of chapter 6 to the end of chapter 8). After you looked at all of the major concepts for these three chapters and putting them on note cards, you should begin to skim over the actually reading material in the book.
2.) (recommended amount of time is 1 hour to an hour and a half) Two days before the exam, I began the extra credit problems that our instructor assigned. I would highly recommend doing these, the instructor isn't going to waste your time by assigning things that won't be covered on the test. I would suggest that you just to half of the assigned problems, you don't want to do everything at once, its too much. After completing 1/2 of the e.c. assignment, and understanding what the problems are asking, and how to do them, I would suggest calling it good for that day after reviewing the note cards that I made the night before with the main ideas on them a few times.
3.) (recommended amount of time is 1 hour to an hour and a half) The night before the exam, I would finish the review e.c. problems and would review the problems that I did the night before as well as the note cards with the main ideas that are being covered. By now you should have these note cards pretty well memorized. After completing the e.c. assignment, I would suggest going back over the entire assignment and writing out procedures on how to solve the problems you encountered. This really helps out because it allows you to think of how to solve a problem the night before the test rather than during the test. I would review these a number of times until I had them pretty well memorized. After doing that, I suggest you just go over the e.c. assignment one last time and call it good for the night.
4.) (recommended amount of time is 30 minutes) The day of the exam (if you have time), I would review the note cards that have the procedures on them that you wrote out the night before on how to solve certain problems. I would also review one last time the e.c. assignment, and really make sure I understand what the problem was asking me to do. Then take the test!!!
If you follow the above four steps above, you will be more than ready for the third exam. Also, While you do these I also highly recommend that you VERBALIZE what you are doing. Throughout the semester, our teacher highly emphasized verbalizing what we were doing, and I didn't do it but then started doing it and let me tell you...IT WORKS! I began doing it, and I have never understood statics so well in my life! Another thing I would recommend is looking at this Wiki site, it has a lot of good information on how to solve things in terms of a procedure, or step by step process. I would also recommend taking the time recommendations above. You don't want to jam it all in the day before or worse the day of the test, and doing it this way allows repetitive motion to familiarize your self with the material. Finally, I would also recommend that you take advantage of office hours if you have any questions, and other class-mates if the situation calls for it. Our instructor also posts a couple of review problems on the online course website, I would recommend doing these as well, they might just show up on the exam. GOOD LUCK!
Moment of inertia for composite areas
In order to find the moment of inertia for composite areas, you must first divide up the total area into smaller basic shapes. It is easiest if the area is divided up into rectangles, triangles, or circles. Once this has been done, it is necessary to find the centroid of the area. This can be done using the equations:
X = (∑xiAi)/( ∑Ai) Y = (∑yiAi)/( ∑Ai)
Once the centroid of the area is found, we can then move on to find the moment of inertia for each individual piece. The back of your textbook will give you examples to use for each type of basic shape. After finding Ix or Iy for each of the smaller areas, we then use the Parallel Axis Theorem to find the moment of inertia for the total area. The Parallel Axis Theorem states that:
Ix = Ix + A(dy)^2
And
Iy = Iy + A(dx)^2
Therefore, in order to find the total moment of inertia about the centroid, we must add up the moment of inertia for each individual area (after it has been shifted using the Parallel Axis Theorem). So,
Ixtotal = (Ix = Ix + A(dy)^2)for Area 1 + (Ix = Ix + A(dy)^2)for Area 2 + (Ix = Ix + A(dy)^2)for Area 3 + …
And
Iytotal = (Iy = Iy + A(dx)^2)for Area 1 + (Iy = Iy + A(dx)^2)for Area 2 + (Iy = Iy + A(dx)^2)for Area 3 + …
Finding the Centroid by summation
finding centroids is less difficult than you think. you can do it by summation easily and here is how to do it:
start off by sectioning your system off into pieces so you can easily find their individual areas as well as the center of each of the pieces (if its a rectangle then the center is located at half its height and half its width).
note where your origin is going to be, usually drawing the x and y axis on one edge of your system works nicely (like the base of your beam). you will need to note the distance between the location of the center of each piece and this axis (that will be your dx or dy) and chosing an edge makes it easy to calculate that distance.
Then, all you have to do to find Xc and Yc is take the sum of (area of each piece and multiply it by your dx or dy (depending if its Yc or Xc)) and divide by the sum of the areas of each piece. And you're done!
--Corchilibeck 10:11, April 25, 2008 (Pacific Daylight Time)
How to set up an integral for center of mass
Finding the integral for the center of mass:
To start lets say you have a rod with mass density Md and length L.
The total mass of the rod can be found by integrating the density over the length.
So the total mass is found to be:
Now to find the center we integrate the mass density relative to the current position.
Then substituting
We arrive at:
Proving that it is the center of the rod and there was no mistake in the derivation.
So let’s say you have a object that gets heavier as it moves from the origin. Such that the Mass density function is 2*x kg/m, where x is the distance from the origin and is of length L
The total mass is found to be:
And the CG is:
So the key to remember is that you integrate the mass density function over the length, and then divide that by the total mass. That division cancels out the units of mass, leaving units of length. That length is the location of the center of mass from the beginning of the integral.
--Bbarnett 13:18, February 22, 2007 (Pacific Standard Time)
The Parallel Axis Theorem
The Parallel Axis Theorem is a powerful tool that will enable you to find the moment of inertia of an object about an axis other then the primary x and y ones. The formula used in this theorem is Ix=[Ix]+A(dy)^2 where Ix is the moment of inertia about the axis of interest, [Ix] is the moment of inertia about the regular x (or y) axis, A is the area of the piece of the object in question, and dy is the distance of the piece's centroid from the axis of interest. Let's do an example to help clear things up: Suppose you have a rectangular plank with a length of four feet (x-axis) and a width of one foot (y-axis). The centroid of the plank is located two feet from either end of the plank along the x-axis and half a foot from either the top or bottom along the y-axis. Now suppose your statics teacher wants to know the moment of inertia about the top of the plank (the x'-axis). First you have to know the plank's moment of inertia about its centroid. To find that use the equation (1/12)*(length perpendicular to the axis of interest^3)*(length parallel to axis of interest) Since you're looking for the moment of inertia about x' you want to find the original moment of inertia about x. This means that the planks width is the perpendicular length in this case and should be cubed, and the length is the parallel side. The equation should end up looking like this:
(1/12)*(1^3)*4=(1/3)ft^4
Now that you know the moment of inertia about x (the value of [Ix]) you find the area of the piece in question (in this case the whole plank; 4ft^2) and the distance of its centroid from the x'-axis (0.5ft). Square dy and multiply these two values together. Add the result to [Ix]:
(1/3)+4(0.5)^2=(4/3)ft^4 <--This is the moment of inertia about the x'axis!
For more complicated objects you have to cut them up into smaller pieces (i.e. triangles and rectangles), find each piece's moment of inertia about the axis of interest, and sum the moments together. Also note that the equation for the moment of inertia used in this example is specific to rectangles. Triangles and other shapes have their own equations that can be found in the back of your statics text.
--Morgen anyan 20:10, December 10, 2007 (Pacific Standard Time)
Imax, Imin and THETA(p)
Unfortunately for us, objects in this world are not always perfectly oriented along the x,y and z axes. Objects tilt and/or rest on inclines and they can themselves be tilted to produce greater moments of inertia. You've probably notices if you've ever tried to balance a book on your hand. When the book is oriented so that one of it's sides rests in you palm its very easy to prevent rotation of keep it balanced. However, once you place a corner of the book on your palm the book suddenly has greater trouble staying up! But just where does the greatest and least moment of inertia occur? To find this you must first calculate the moment of inertia about the x and y axis of the object, as well as the moment of inertia about the xy axis (Ixy). Ixy can be found using a slighty different version of the parallel axis theorem: Ixy=[Ixy]+A(dx)(dy) Be very careful when finding dx and dy because the values are not squared like they are in the regular parallel axis theorem, so a missed sign will not be corrected. Always take the final position (the centroid's location) minus the original position (the piece you'r looking at's location). Also, Ixy is zero for symmetrical objects and in this course we deal with symmetrical objects more often then not. As with Ix and Iy you may have to break an object into parts, find each one's Ixy and sum them together. Once you have the values of Ix, Iy and Ixy you can find the maximum and minimum moments of inertia by using the following equation:
Imax/min=[(Ix+Iy)/2] +/- sqrt[[(Ix-Iy)/2]^2]+Ixy^2]
To find the orientation of the axes on which these values occur use the following equation and solve for THETA(p) (the orientation angle):
tan(2THETA(p))=-Ixy/[(Ix-Iy)/2]
Once you have THETA(p) you add it to the x axis to find the axis that corresponds to the maximum moment of inertia. Adding ninety degrees to that value will give you the axis the corresponds to the minimum moment of inertia.
--Morgen anyan 20:32, December 10, 2007 (Pacific Standard Time)
Using a TI-83+ to calculate Imax, Imin, and θP
How to write a program in a TI-83+ to solve for Imax, Imin, and θP:
-Hit the PRGM button on your calculator and create a new program by selecting NEW and Create New
-Name your program appropriately, (IMAX is a good name)
-Once you've named your program, enter the following code: (all the inputs can be found in the calculator's catalog menu)
- Disp "IX="
- Input I
- Disp "IY="
- Input J
- Disp "IXY="
- Input K
- ((I+J)/2)+√(((I-J)/2)2+K2)→L
- ((I+J)/2)-√(((I-J)/2)2+K2)→M
- .5*tan-1(-K/((I-J)/2)→N
- Disp "IMAX IS",L
- Disp "IMIN IS",M
- Disp "θP IS",N
-This program will ask for Ix, Iy, and Ixy as inputs and yield Imax, Imin, and θP as outputs.
--Jamesclarkii 15:02, April 22, 2008 (Pacific Daylight Time)
How to study for the final exam
Teachers often have the courtesy of recommendations and review problems when it comes to the final exam. If your professor gives review problems, do them. Only one person knows whats on that test and hey, they're giving out hints. Talk to the professor, teacher, TA...whoever you can. Ask what they recommend you look at from the course of the semester. If your teacher says "everything," suck it up and ask yourself "what have I learned over the semester and where has this already been summarized for me about three times?" Oh yeah, the midterms. They offer a look into what the teacher already thought was important, but are not a complete guide! Remember that some material hasn't been tested yet, and you can bet a chunk of that final is going concern that new stuff. However, that isn't too much to actually comprehend the old fashioned way: reading. --Jctanner05 23:59, April 18, 2007 (Pacific Daylight Time)
The final exam is a comprensive exam so, obviouly, all of the chapters covered in the class should be studied. The best way to do this is by practicing and then practicing some more. Picking two or three problems from the review section from each chapter is the best way to go about doing this. If you actually do the problems you can see your strengths and weaknesses and then study the subject matter accordingly. Looking back and reviewing lecture notes is also a very good way to review for the final exam since there are example problems from every section covered in the lecture. Also another helpful hint is that everything that needs to be known, equation wise that is, can be found inside the front cover of the textbook. Looking over the actual and sample tests from the semister will also help you study for the final exam. I found it helpful to take the problems off of the practice and actual exams and do them again just to make sure that I understood all of the material. Above all though it is practice, the more practice problems you do the better you will perform on the exam. Just remember, practice makes perfect.
To study for the final exam, I would go about first making a comprehensive note card. This would include on one side all of the formulas and tricks to certain problems you learned throughout the year. The note card would also include examples of problems. These examples wouldn't be one for every chapter, but examples that would include numerous chapters and techniques. After the note card is made, then I usually go about doing sample problems. The reason I do the note card first is because it refreshes the memory of what you have done. When going to through a sample test, I usually like to make notes about the problem, i.e. writing out the equation(s) I used or tricks that I did to solve the problem, or possible ways that they could change the that particular problem during the exam to see if you can adapt. My last bit of advice is usually to study with other people so, because their method of solving a problem might be easier than the one that you use and they can show you how to better go about solving the problem.
--Mthurber 11:37, April 13, 2007 (Pacific Daylight Time)
--Mbofenkamp 14:41, December 5, 2006 (Pacific Standard Time)
Studying for the final
The extra credit problems given to you toward the end that covers all the chapters is an excellent insight into what the teacher has decided is the most important material for the test. Do NOT make the mistake of simply copying down these problems and then going into the test unstudied, because you will get burned. Make sure that you understand how they work and how to answer similiar questions that are worded differently over the same material. A big part of the final exam will come from the previous tests throughout the semester. You should keep all of your old tests and hopefully your teacher has posted the solutions for the old tests so you can know where you went wrong when you took the test. First you should go over the problems on the old tests. You should start with the problems you got wrong and learn that information or make alterations so you can get these problems correct. Then once you think you know most of the information, go back and practice the previous three tests. If you can do the previous tests that is a good start to understanding the final. You should also see if your teacher has posted practice final exams online and try to be able to do those aswell. Also before the final make sure you get a good amount of sleep and eat a good breakfast. GOOD LUCK!
Studying for and Taking the Final
The first step in studying for the final is to leave enough time to study effectively. Give yourself about a week to review the material that will be on the exam. Crammin is NOT a good idea for any test, finals especially. Once your teacher gives you practice problems start working on them. Statics exams are not like the fill in the blank tests we had in high school. You have to actually use your knowledge andn skills to solve them. The practice problems are you best study tool because they are similar in layout and complexity to what will be on the exams. Going over old exams is also not a bad idea since they tell you which concepts your teacher thinks are important.
You are aloud to have a notecard for the final. USE IT! Keep a sheet of blank paper next to you while you study and write down any equations you might need and that you don't have commited to memory. Creating the note sheet as you go along ensures that you actually make one and it helps ingrain concepts in your mind. You may not use your note sheet durin the test, but it never hurts to have a backup in case you space out.
Once your class time turns into a review session some of your fellow students will stop coming to class. This opens the door for you to more easily get help from your teacher. Use the practice problems to identify areas where your knowledge is lacking and ask you instructor for clarification. Fewer students in class makes this easier and gives the teacher more time to work with you and make sure you understand the concepts. Don't wait until just before the exam to ask her how to calculate a unit vector.
Finally, don't freak out. Yes, Dead Week and Finals Week are hectic, but stressing out before and during the exam won't help improve your score. In fact they might do just the opposite. Try to get a good night's sleep before the test (I say try because I know from experience how finals and sleep often don't go together), have some protein before hand (that means no skipping breakfast!) to give your brain a boost, take a deep breath, and dive into the test. Complete the problems you know how to do first and then go back to the less obvious ones later. Leave some time to check back over your answers before turning it in. You might just catch something you missed the first time around.
Good luck!!!
--Morgen anyan 14:30, December 3, 2007 (Pacific Standard Time)
== Studying-find a study group! ==
Find a study group to study with. If you don't not know the answer or how to solve a problem there is a possibility that somebody does in the class as is willing to help you. Not only does this help you but it will be helping them as well. This will force them to verbalize what the basic concept are.
In the "real world" you will have to work with people and by starting now you will be more successful and confident in a group of peers.
--Hicks4 10:29, April 25, 2008 (Pacific Daylight Time)
How to succeed in statics (and possibly get an A)
I'm sure your statics teacher will tell you these things over and over again, but as a student, these are things I feel should be reiterated to my peers in order to succeed in statics.
1. Go to class. I don't know how many times I've been told this, but this will certainly help in statics. The instructor will explain the readings in another fashion and show examples. These 50 minutes will help you understand the material moreso, than just sitting there and "sponging" in the information.
2. Take advantage of extra credit opportunities. Extra credit is a godsend and taking advantage of them can help your grade.
3. Do your homework. What better way to understand the concepts in statics than to practice. Practice really does make perfect.
4. Read the book. I know that in some classes reading the books will not help. However, read the statics book. It has very good examples and explains the concepts on hand very thoroughly.
5. Lastly, if you need help, don't be scared to ask for it. These teachers are here to teach and help. If you're still too scared to ask them, go find a study buddy in class. Help is always there when you need it.
--Jjreyes 17:28, December 10, 2007 (Pacific Standard Time)
How to succeed:
What worked best for me for studying for statics was having a variety of sources to choose form. That included the statics study pack which I bought for around $20 from the bookie. It's basically a condensed version of the class book and is perfect for studying. Also inside the study pack is a student access code for www.prenhall.com/hibbeler which is an awesome resource for doing homework and studying for the exams. It contains 1000 problems and solutions for extra practice which is the key to succeeding in statics. It really helped me further understand all of the chapters.
- Josh Daniels 12/2/08
Homework Format:
Each problem has a total score composed of three equally weighted categories: Set-up, Follow through, and Communication. Most of the points for a problem can be attained through proper formatting even if the final answer is not correct. The following are format suggestions based on a semester of experience:
The Basics
-Use engineering paper (write on the backside of the grid; the lighter side)
-Write with a sharpened or mechanical pencil
-Use a straight edge when drawing anything! (exception: curves)
Header
From left to right: Name....Course....Problem Numbers....Due Date....Page Number out of Total Pages
Ex: John Doe......CE 211......#'s:105,115,117......9/22/08......1 of 2
Body
1. Begin by writing the problem number and problem statement or a summary of the statement.
Ex: 105. "Replace the loading system acting on..."
2. Draw diagram(s)! Diagrams of one kind or another (free-body, shear/moment, etc.) are required on pretty much every problem. Detail and accuracy are important!
3. List known and unknown forces/moments with position, magnitude, and direction (A table format works well here). Later in the semester this may no longer be required, however, it's a helpful practice.
4. Explain each step in solving the problem before writing out the actual step (doing the math).
Ex: I'll first solve for the resultant force through Cartesian vector addition. F=[(40-(3/5)60)i+...
5. Finish by writing a solution statement which answers the original problem statement.
Ex: The resulting force at point P is {4i-78j}lb.
Great Test Studying Tips
1. Make flash cards of key topics( mainly topics that you don't understand 100%). There's no use in making notecards for things you already know 100%.
2. First read over all practice problems from the chapters you will be tested on, then go over all the old hw that you must know for the exam. the exam questions will be very similar to the hw questions. Make sure you don't check the answer until after you have finished the problem.
3. Read over the statics study pack to see if it explains any topics better than the book.
4. Go onto www.prenhall.com/hibbeler and work on the practice problems that will be coered on the test.
5. Try looking on wikkipedia if you have any trouble understanding. Wikkipedia gives you a good general overview of exactly what statics is and how all the parts of statics such as static equilibrium and center of mass are related.
6. Try asking around to see if you can find any old tests. These tests will give you a good idea of how the statics test problems are set-up. These are great practice problems.
7. Helpful websites:
http://www.eng.iastate.edu/efmd/statics.htm http://www.glenbrook.k12.i1.us/gbssci/Phys/Class/vectors/u313c.html http://dept.Physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsection4_1_5.html http://www.aerostudents.com/files/statics/staticsFullVersion.pdf http://www.ent.ohiou.edu/~statics/
- Josh Daniels 12/2/08
How to Succeed on the Statics Project
At first you might think the project is going to take forever and be really hard. It's really not. If you follow these steps you'll be fine: 1. Find good partners. I've heard of too many people who chose people they didn't really know and they ended up doing the project last minute or almost all on their own. 2. Make a list of which days are good project meeting days. You don't want to plan a meeting last minute and then find out that only you can make it to the meeting. 3. Make a list of what needs to be done and try and make deadlines for each item on your list. This will make the project easier because you'll be working on it in small parts. 4. Make goals for your team. We made it our goal to try and make one of the cheapest bridges so that we could get one of the extra credit prizes. Don't ever think that extra credit isn't worth it. If anything, it's more practice that will help you out on the next exam. 5. Remember that you need to make at least two different bridges. Dont' be afraid to make more than that. We made a third one just to see how different it would be and it ended up cutting $1000 off of the total price. 6. Make sure you read over the instructions very well. You need to calculate the stresses for winter and summer. 7. Our group found that if you lower the angles of the beams, the tension and compression will increase, but that's ok as long as they don't exceed the max tension and max compression. This was a very important part to how we lowered the cost of our bridge. 8. Don't be afraid to talk with other groups to try and get some new ideas, a little competition is good, just don't tell them everything about your bridge. 9. Finally, keep in mind of the due dates and if you have any questions make sure to ask one of the TA's or the teacher.
-Josh Daniels 12/2/08
How to use your Graphing Calculator (TI 83) to solve quadratic equations
The following program can be entered into your calculator to solve quadratic equations.
Disp "A is"
Input A
Disp "B is"
Input B
Disp "C is"
Input C
((-B)+Sqrt((B)^2-4AC)/2A->X
((-B)-Sqrt((B)^2-4AC)/2A->Y
Disp "+IS",X
Disp "-IS",y
Stop
This can come in handy for checking your work whenever you have a quadratic function to solve for. One instance in particular is when you are finding equations for shear and moment diagrams. Sometimes it is necessary to find where a quadratic function is equal to zero. In cases like this, it is very handy to simply let the calculator do all the work.
How to use your Graphing Calculator (TI 89) in Statics
Most engineering students have a graphing calculator, and luckily you will be able to use it extensively in CE 211. Most people who have graphing calculators will not find out their full potential as a study tool until they have completed the class, and then it is too late. Here is a short list of the applications (apps) and functions I have found most useful both for homework and exams. • First off you need to have the adapter cable and TI 89 software installed on your computer, if you don’t then you are using a $140 calculator to do the same job that a $10 one can. Instructions and software are available free on the Texas Instruments website. • Programs! These can be very useful for solving problems you are very familiar with, but be cautious. You can find a ton of programs on websites such as ticalc.org that will compute everything from bending moments to products of inertia and forces in 3D. But I have found that these programs are not very useful due to complex interfaces and unfamiliar terminology, you practically need a class on how to use the programs on your calculator much less be able to rely on it for that critical answer during an exam. • Where the TI 89 and similar models does help the most is when solving multiple equations with the same set of variables, i.e. Matrices. If you aren’t familiar with linear algebra then you might not notice it but in CE 211 you will create matrix equations. It is possible to solve matrices with out a calculator but that is painful, and errors are likely to occur. Graphing calculators have matrix-solving software built into them and they work well provided you get all the signs and coefficients correct. • On the TI 89 there is a flash app called Simultaneous Equations Solver specifically designed to solve multiple equations: Effectively it does the same thing as a matrix-solver but the interface is more user-friendly and you don’t have the scroll through a million matrix functions to get the values of each variable.
--Aaron Stroup 15:34, December 12, 2007 (Pacific Standard Time)
Solving integrals on your TI-89
Earlier you learned how to set up integrals and if you are the type of person who messes up computing the integral by hand, or are too lazy to go through your graphing calculator manual to find out how to compute integrals on your calculator, then look no further than right here! This explains how to punch in your equation with your limits of integration and it will spit out your area under the curve beautifully!
1) have your equation you wish to integrate and the limits of integration ready. (see earlier section for how to set up integrals)
2) go to the Y= window (*F1 on titanium) and type in your equation and hit enter, for example:
3)hit graph (for ti-89 titanium its *F3)
4)press the correct F button that says "math" (F5 on titanium) NOT the math button you get by pressing 2nd 5!!!
5) scroll down to the integral of f(x)dx and hit enter
6) it will ask for you to input your lower limit, do it and hit enter, then input your uppper limit and hit enter and ta da! you're finished!
This is a quick way to complete the integrals if you are like most of us and make mistakes integrating. But be careful! make sure you set up your integral correctly!
--Corchilibeck 14:19, April 24, 2008 (Pacific Daylight Time)
Using Simultaneous Equation Solver on your TI-89 for Fun and Statics Homework
1) Push the Apps button on your calculator and find the Simultaneous Equation Solver
2) Go to "New"
3) Enter in the number of equations and unknowns.
For two dimensions, you ought to have at least two equations, with as many or fewer unknowns. Likewise, for three dimensions, you ought to have at least three equations, with as many or fewer unknowns.
4) Write your equations (usually the equilibrium equations) in matrix equation format.
If you remember nothing from Linear Algebra, this is most easily visually done by writing the equations with the variables and their coefficients in the same columns and the leftover numbers on the right side of the equation. For example, you might turn
-3Fc + 2Fa -20 = 0 -Fb + 300 = 0 Fa + Fb + Fc = 0
into
2Fa + 0Fb - 3Fc = 20 0Fa - 1Fb + 0Fc = -300 1Fa + 1Fb + 1Fc = 0
From this, we can transfer the equations into the following matrix
[2 0 -3] [Fa] = [ 20 ] [0 -1 0] [Fb] = [-300] [1 1 1] [Fc] = [ 0 ]
5) Plug the coefficients in the matrix into the leftmost columns and the contents of the b vector into the rightmost column within the bolded area (labeled b1)
6) Hit F5 to solve and ace your Statics homework!
Pay attention to the order that your variables are in inside your equilibrium equations. You want to know what it is you just solved for!
--Jdryan 23:34, February 13, 2008 (Pacific Standard Time)
Another way to solve a system of equations on your TI-89
There are a couple of ways to solve matrices, I had too much trouble using the method above and found an easier way if you are not math 220 friendly.
1) hit apps and scroll down to data/matrix editor and select new (these first few steps will create a general matrix you can go back and edit)
2)as type, select matrix. as folder, select new. as variable, select any letter (remember which one!) and chose row dimension as 6 and columns as 7 (usually we dont have more than 6 equations, but if you want you can chose rows 7 and columns 8 to be safe). then hit enter and you should be at a window with many rows and columns.
3)using the example equations in the above method of solving systems, you have the equations:
2Fa + 0Fb - 3Fc = 20 0Fa - 1Fb + 0Fc = -300 1Fa + 1Fb + 1Fc = 0
making sure that all first varaibles are Fa, all second variables are Fb ect and the constants are on the other side of the = sign.
4)now just punch in the coefficients of each variable into their corresponding row and column (all Fa's are in column 1, all Fb's are in column 2 ect... and make sure that the first equation occupies row one, the second equation occupies row two ect). your input should look like this:
[2 0 -3 20] [0 -1 0 -300] [1 1 1 0]
note: you keep the signs of the constants as they were when they were on the other side of the = sign.
5) once you are satisfied you input the coefficients correctly, hit quit to go to the main page, and hit 2nd 5 (or MATH) and scroll down to "matrix" and select "rref( " and hit enter.
6)enter your letter you remembered above and close the bracket and hit enter. you should get a matrix that looks like this:
[1 0 0 -176] [0 1 0 300] [0 0 1 -124]
this says that your first variable in each eqn has a value of -176 (Fa), and the second variable (Fb) equals 300 and the third variable (Fc) is -124
NOTE: if you made a mistake or wish to put in a new matrix you dont have to go in and make a new file, all you have to do is hit apps, matrix editor, CURRENT, and this will lead you back to the matrix window and you can make changes there.
--Corchilibeck 14:36, April 24, 2008 (Pacific Daylight Time)
How to determine if it is a frame/machine or a truss
Truss Description:
1)Made up of 2-force members
2)Joined by pins
3)Loaded at joints
4)Forces are axial
5)Built of triangles
Frame/Machine Description:
1)Has at least one multi-force member
2)Joined by pins
3)Can be loaded anywhere on the system
4)Frames are static
5)Machines are intended to move
--Elizabeth arment 21:25, March 20, 2008 (Pacific Daylight Time)
How to succeed on bridge project:
1: Begin by figuring out the basic concepts of how you want your bridge to be set up (including design style, bottom member spacing, etc.)
2: Find which members are in tension or compression and make sure to document them on your bridge and do account for their direction in any free body diagram
3: Calculate a preliminary load for the bridge and find out where stress is highest
4: Set where ever the highest load is to just below the fail point of the member
5: Use free body diagrams (you may need a lot of them before you find an answer without using guess and check methods) to find the force in all members
6: Try to lower the angles used to determine the height of the truss to see if any alteration causes failure at high stress points (the lower the angle, the shorter the bridge and the less expensive the bridge)
7: Ensure that the write up for the project and any other paperwork is done neatly and professionally (it's worth a lot of the possible points)
- Josh Yoder 4/30/08