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# Mechanical Engineering Design III ME 360L

Marketplace > University of New Mexico > Mechanical Engineering > ME 360L > Mechanical Engineering Design III
Kiara O'Hara
UNM
GPA 3.52

Robert Greenlee

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COURSE
PROF.
Robert Greenlee
TYPE
Class Notes
PAGES
24
WORDS
KARMA
25 ?

## Popular in Mechanical Engineering

This 24 page Class Notes was uploaded by Kiara O'Hara on Wednesday September 23, 2015. The Class Notes belongs to ME 360L at University of New Mexico taught by Robert Greenlee in Fall. Since its upload, it has received 82 views. For similar materials see /class/212221/me-360l-university-of-new-mexico in Mechanical Engineering at University of New Mexico.

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Date Created: 09/23/15
Units Examining the units used in Mechanica Default ProE Units IPS inches Ibf seconds Results Elements Passes Displacement VM Stress Default 12 5 2433x10396 1060 units IPS units 12 5 939x10394 1060 What is the Difference 939gtltIO4 3864 243gtlt106 322gtlt 12 3864 g inft ftsec2 inft insec2 How Units are Selected Choose units fortime length and either force or mass Derive other units from these Use F MA to relate force and mass Use gravity for the acceleration its units are expressed in units of time and length that you selected Default System Units Selected Length in Mass Ibm Time seconds Derived Units FMA Gravity 3864in sec2 1 Ibm 1 Ibf 3864 force bf3864 Pressure psi3864 IPS System Fixed Units Length in Force Ibf Time seconds Derived Units FMA Gravity 3864in sec2 1 bf 3864 Ibm force Ibf Pressure psi Common Mechanica Units SI MNS mmNs Default Length Time Gravity 981 9810 322 3864 msec2 mmsec2 ftsec2 insec2 3864 insec2 Mass 1000kg slug Ibfszlin Force bf3864 Press psi3864 Finite Element Trusses 30 Trusses Using FEA We started this series of lectures looking at truss problems We limited the discussion to statically determinate structures and solved for the forces in elements and reactions at supports using basic concepts from statics In this section we will apply basic nite element techniques to solve general two dimensional truss problems The technique is a little more complex than that originally used to solve truss problems but it allows us to solve problems involving statically indeterminate structures 31 Local and Global Coordinates We start by looking at the beam or element shown in the diagram below This element attaches to two nodes 1 and 2 In the Figure we are showing two coordinate systems One is a one dimensional coordinate system that aligns with the length of the element We will call this the local coordinate system The other is a two dimensional coordinate system that does not align with the element We will call this the global coordinate system The x39y39 coordinates are the local coordinates for the element and x y are the global coordinates Local coordinate system Global coordinate 1 System x Figure 1 Local and global coordinate systems We can convert the displacements shown in the local coordinate system by looking at the following diagram We will let ql and q represent displacements in the local coordinate system and q1 q2 q3 and q4 represent displacements in the X y global coordinate system Note that the odd subscripted displacements are in the X direction and the even ones are in the y direction as shown in the following diagram Chapter 3 Finite Element Trusses Page 1 of 15 Un deformed elemen qzsinB Deformed element q1 Figure 2 The deformation of an element in both local and global coordinate systems We know that for small deformations in tension or compressiona beam acts like a spring The amount of deformation is linearly proportional to the force applied to the beam As the beam is stretched or compressed we are added potential energy to the beam This energy is called strain energy and it can be modeled with Hook s law The law states that the force is directly proportional to the deformation FkAx 31 We can compute the energy by integrating over the deformation k dx 1 kg 3 2 u x I 0 2 where k the element stiffness A the cross sectional area of the element E Young s modulus for the material and L the length of the element Q is the total change in length of the element Note that we are assuming the deformation is linear over the element All equal length segments of the element will deform the same amount We call this a constant strain deformation of the element We can rewrite this change in length as Q61i 61i 33 Substituting this into equation 32 gives us 1 i i u Ek Q12 34 Chapter 3 Finite Element Trusses Page 2 of 15 or expanding 1 I I I I u Ekqzz 2q2q1 q12 Rewriting this in vector form we let q qi 36 92 k 1 4 37 and L 1 l Withthis we can rewrite equation 35 as u q397k39q39 38 We can do the indicated operations in 38 to see how the vector notation works We do this by rst expanding the terms then doing the multiplication AE 1 1 q 39 u 2L 91 92 1 l E I I I I q u 2Lq1 q2 q1q2qu 310 AE uiqq qzqzqz q 311 AE H H u q12 qiqz qzz qiqz 2L AE H u 211q12 2q1q2 q22 Which is the same as equation 35 Equation 37 is the stiffness matrix for a one dimensional problem Chapter 3 Finite Element Trusses Page 3 of 15 62 Two Dimensional Stiffness Matrix We know for local coordinates that a q 92 and for global coordinates See Figure 2 36 314 We can transform the global coordinates to local coordinates with the equations q1 qlcos6q2 sine and q q3c0s9q4si119 This can be rewritten in vector notation as q39Mq where c s 0 0 le O 0 c s ccos6and ssi116 Using 1 T u t k t 2 q q we can substitute in equation 317 u qTMTk39Iq Now we will let Chapter 3 Finite Element Trusses 315 316 317 318 38 319 Page 4 of 15 k M7k39M 320 and doing the multiplication k our stiffness matrix for global two dimensional coordinates becomes c cs c cs AE cs s2 cs s k 321 L c2 cs 02 cs cs s2 cs s2 where E Young s modulus for the element material A the cross sectional area of the element L the length of the element 0 cos 6 s sin 6 33 Stress Computations The stress can be written as 0 Es 322 where s is the strain the change in length per unit of length We can rewrite this as total deformation 0 13 972 11 323 L length of element In vector form we can write the equation as g 91 a L 1 1 324 92 From our previous discussion we know that in local coordinates q qi 36 92 Chapter 3 Finite Element Trusses Page 5 of 15 and in global coordinates From equation 317 we know that q39 Mq where c s 0 O I i0 0 0 SJ Substituting this in to the equation 324 yields E fl 1 1W9 Now we multiply M by the vector Chapter 3 Finite Element Trusses 314 317 318 325 326 Page 6 of 15 34 Truss Example We can now use the techniques we have developed to compute the stresses in a truss Consider 25000 lbs qs L W 3 E295x106 T D4 qu Areal0in2 30 qz L D qs Computing Displacements There are 4 nodes and 4 elements making up the truss We are going to do atwo dimensional analysis so each node is constrained to move in only the X or Y direction We call these directions of motion degrees of freedom or M for short There are 4 nodes and 8 degrees of freedom two degrees of freedom for each node We can number the degrees of freedom with the formulas Vertical degree of freedom dof 2 node 327 Horizontal degree of freedom dof 2 node 1 328 where node is the node number We can locate each node by its coordinates The table below shows the coordinates of the nodes in the problem we are solving We can use these coordinates to determine the lengths and angles of the elements Table 1 in the truss Chapter 3 Finite Element Trusses Page 7 of 15 Each element can be described as extending from one node to another This also can be de ned in a table below Table 2 connect the truss From these two tables we can derive the lengths of each element and the cosine and sine of their orientation This is shown in the table below Table 3 stiffness matrix In the previous sections we developed the stiffness matrix for an element This is shown in equation 321 below c cs c cs AE cs s2 cs s2 k 321 L c2 cs 02 cs 2 2 This stiffness matrix is for an element The element attaches to two nodes and each of these nodes has two degrees of freedom The rows and columns of the stiffness matrix correlate to those degrees of freedom Using the equation shown in 321 we can construct that stiffness matrix for element 1 defined in the table above The stiffness matrix is Global dof 1 2 3 4 1 0 1 01 k 295x106 0 0 0 0 2 329 1 40 1 0 1 0 3 0 0 0 0 4 Chapter 3 Finite Element Trusses Page 8 of 15 Element 2 5 6 3 4 390 0 0 0 5 6 0 1 0 1 6 k2 295x10 330 30 0 0 0 0 3 0 1 0 1J4 Element3 1 2 5 6 quot64 48 64 48 1 k 295x106 48 36 48 36 2 331 3 50 64 48 64 48 5 48 36 48 36 6 Element4 7 8 5 6 391 0 1 0 7 6 0 0 0 0 8 k4 295x10 332 40 1 0 1 0 5 0 0 0 0 6 The next step is to add the stiffness matrices for the elements to create a matrix for the entire structure We can facilitate this by creating a common factor for Young s modulus and the length of the elements For element 1 we divide the outside by 15 and multiply each element of the matrix by 15 Multiplying and diViding by the same number is the same as multiplying and diViding by l 1 2 3 4 15 0 15 01 k 295x106 0 0 0 0 2 333 1 600 15 0 15 0 3 0 0 0 0 4 Chapter 3 Finite Element Trusses Page 9 of 15 We multiply and divide element 2 by 20 5 6 3 4 0 0 0 0 5 k 295x106 0 20 0 2 20 6 600 0 0 0 0 3 0 20 0 20 4 Multiply and divide element 3 by 12 1 2 5 F 768 576 768 k 295x106 576 432 576 3 600 768 576 768 576 432 576 6 57611 432 2 576 5 432 6 We do the same for element 4 by multiplying and dividing it by 15 7 8 5 6 F15 0 15 017 k 295x106 0 0 0 0 8 4 600 15 0 15 0 5 0 0 0 0J6 334 335 336 The coefficient for each stiffness matrix is the same so we can easily add the matrices We add the degree of freedom for each element stiffness matrix into the same degree of freedom in the structural matrix The resulting structural stiffness matrix is shown below 1 2 3 4 2268 576 150 0 576 432 0 0 150 0 150 0 K 295gtlt106 0 0 0 200 600 768 576 0 0 576 432 0 200 0 0 0 0 0 0 0 0 Chapter 3 Finite Element Trusses 5 768 576 2268 576 150 6 576 432 200 576 2432 7 0 0 0 0 15 0 150 0 Page 10 of 15 Remembering our basic equation K Q F 3 3 8 where K is the structural stiffness matrix Q is the displacement of each node and F is the external force matrix This results in 2268 576 l50 0 768 576 0 0 qt 0 576 432 0 0 576 432 0 0 i ii 0 l50 0 150 0 0 0 0 0 q3 20000 295x106 0 0 0 200 0 200 0 0 q4 0 339 600 768 576 0 0 2268 576 15 0 qs 0 576 432 0 200 576 2432 0 0 q6 25000 0 0 0 0 l50 0 150 0 q7 0 0 0 0 0 0 0 0 0 q8 0 We have boundary conditions at the xed supports Our assumption is that these joints will not move in the constrained direction We remove these from our matrix The constrained displacements are dof l 2 4 7 and 8 The lines in equation 340 show the rows and columns that are removed 295gtlt106 600 2268 576 576 2432 The resulting matrix is 6 15 0 0 q3 20000 0 2268 576 q 0 341 0 576 2432 qs 25000 Chapter 3 Finite Element Trusses Page 11 of 15 We can use Gaussian elimination or any number of other solution techniques to solve the system of equations shown above Doing so yields q3 2712 x10 q5 565gtlt10393 inches 342 q l 2225x10 3j Computing Stresses Previously we showed that E 326 L c s c sq We use this equation to compute the stress in each element 295 x106 0 2 l 0 l 0 343 01 40 27l2gtlt10393 3 0 4 or 71 20000psz39 344 565 x104 15 6 2225gtlt10393 6 345 62 2295X10 0 1 0 1 30 2712gtlt10393 3 0 4 02 21875psi 346 Using a similar technique we get 73 5208psi 347 and 64 4167psz39 348 Chapter 3 Finite Element Trusses Page 12 of 15 Computing the Reactions The last step is to compute the support reactions We need to determine the reaction forces along dof l 2 3 7 and 8 which correspond to the xed supports These are obtained by substituting Q into the original nite element equation R K Q F 348 We only need to use those rows of the structural stiffness matrix that correspond to the xed supports At these supports we are not supplying an external force so F0 Our equation becomes RKQ 350 or 0 0 R1 2268 576 7150 0 7768 7576 0 0 R 576 432 0 0 576 432 0 0 273912X1073 Z 6 39 39 39 39 o 351 R M 0 0 20 0 720 0 0 7 600 565x10 R7 0 0 0 0 7150 0 150 0 222510 R8 0 0 0 0 0 0 0 0 7 39OX 0 We multiply the stiffness matrix K and the deformation vector Q to get the reactions They are shown in the following equation R 158333 R2 3126 352 R4 21879 R7 4l67 R 0 Chapter 3 Finite Element Trusses Page 13 of 15 Problems 1 Element area 15 in2 E30000000 Element length 5 feet 1 Write the stiffness matrix for the structure The bar is vertical Show all work 2 8000 lbs 2 Using a different load the element shown in Problem 1 deforms by 002 inches in length What is the stress in the material Use a finite element approach to solve the problem Show all work 3 Use a finite element approach solve for the stress joint displacement and reaction force on the element shown in Problem 1 Use the 8000 lbs force as shown in the diagram Show all work 4 The structure shown in the diagram results in the stiffness matrix shown in the table Manually solve for the displacement of node 4 Show all work l0e006 3 3 2 2 4 1 1 10000 lbs Chapter 3 Finite Element Trusses Page 14 of 15 5 Element area l in2 Material steel A Find the joint displacements B Find the stress in the elements C Find the reactions 10000 lbs 6 Element area l in2 Material steel 2 4 5000 lbs D Find the joint displacements E Find the stress in the elements Find the reactions Write a Matlab program that uses the nite element technique discussed in class to solve for the displacements stresses and reactions in a nite element truss You may want to modify the static stress program you wrote earlier to create this new program The two programs should be able to use the same input le Solve the problem shown above to turn in Use both this new program and the static truss program to run the data le Compare the results Chapter 3 Finite Element Trusses Page 15 of 15

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