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# Mechanical Engineering Design III ME 360L

UNM

GPA 3.52

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This 15 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 Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/212222/me-360l-university-of-new-mexico in Mechanical Engineering at University of New Mexico.

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Date Created: 09/23/15

Money Engineers frequently nd themselves working as much with money and nancing as they do with technical design All engineering requires money Money is needed to pay salaries to buy materials for producing the product and to buy the manufacturing machines This money may be borrowed from banks or investors and must be repaid These loans have interest attached to them SIMPLE INTEREST Simple interest applies the interest to the original loan The interest is usually applied annually We can illustrate this with an example You borrow 5000 from you uncle to pay for college You promise to pay him back 1 year a er graduation You both agree on a simple interest rate of 8 If you repay the loan in 5 years how much will you pay F the amount to be paid back P the principle originally borrowed n the number of periods the interests is applied usually number of years 139 the interest rate per year We can create a table showing equations for the amount to be paid back each year The formulas show the year the amount due at the beginning of the year and the amount due at the end of the year Looking at the table we can create a formula for any number of years It is F P Pin 1 We can compute how much you need to repay your uncle F 5000 5000 X 008 X 5 7300000 You will owe him 7000 dollars at the end of 5 years COMPOUND INTEREST Banks and investors usually look at compound interest rather than simple interest With compound interest the interest over the rst year or period is added to the loan and the interest for the next year or period is computed using that new principle F the amount to be paid back P the principle originally borrowed n the number of periods the interests is applied usually number of years 139 the interest rate per year We can create a table for this similar to the one created for simple interest The formula for this can be written F Pl 139n 2 Example 1 The bank says it will loan you the same 5000 at 7 compounded annually Who has the best deal the bank or your uncle F 50001 0075 5000 x 1075 701276 The bank charges 1 less then your uncle but they compound the interest annually The bank will charge you 1276 more for the loan Example 2 In 10 years you would like to have 10000 in the bank to put down on a house How much do you have to put in the bank now to have 10000 in ten years The bank will give you 6 interest compounded annually Starting with equation 2 we see F Pl 139n 2 Solving for P PF1z 3 P 10000 1 0610 10000 17908 355158335 Another bank compounds the interest monthly What would you have to invest for this bank We start with equation 3 but this time we do not use the annual interest rate because the interest is compounded monthly Instead we use 139 imual 12 00612 0005 The interest is being compounded monthly so there are 12 compounding periods per year N interest periods 10 years X 12 120 Substituting this into equation 3 yields P F 1 139 10000100512 P 355349633 Example 3 The bank gives you 10 interest compounded annually How long will it take for your money to triple in value Starting with equation 2 F P1 139n 2 Divide both sides by P FP 1 139n If the money triples in value then FP1i 3 We take the log of both sides of the equation and solve for 11 Log FP Log3 nLog1z39 Or n Log3 Log1z39 Log3 Log 11 047712 004139 115 years A Uniform Series of Compound Interest De ne A the end of the period disbursement or receipt from a uniform sum of money At the end of each year you are going to deposit a xed amount of money into an account paying 139 percent interest How much will you have a er 11 year F A1z39 391 Alz393 Alz 2 Alz A 21 Multiply both sides by 1z39 F 11 A1z39 Alz39 391 Alz393 Alz 2 Alz Or F 1F A1z Alz39 391 Alz393 Alz 2 Alz Subtracting F from both sides z39F A1z39 Alz 391 Alz393 Alz 2 A1z F Now substituting equation 21 for F on the right hand side of the equation yields z39F A1z39 Alz39 391 Alz393 Alz 2 A1z A1z39 391 A1z39339 A1i239 A1z 39 A Subtracting the terms results in iFA1i 7A or F A1i 71i 22 Example 4 You deposit 500 in the bank at the end of each year for 5 years How much money will you have after the last deposit if the bank pays 5 interest compounded annually F 5000055 7 1 005 276282 F in equation 22 is the value of the money at some future time at the end of the investment period How much is that money worth now We can compute this by substituting the compound interest formula we derived in the last lecture into equation 22 The compound interest formula was F Pl 139n Substituting yields F Pl 1 A 1z n 71z39 Solving for A A Pi1i 1in 7 l 23 Example 5 You borrow 200000 from the bank How much must you pay monthly to repay the loan in 10 years The interest is 6 compounded monthly 139 00612 0005 n 10 years X 12 monthsyear 120 A 7 200000 005100512 1005120 7 1 A 222041 Example 6 Your company needs to buy a lathe for manufacturing parts The lathe costs 35000 but the company will allow you to pay 400 per month for 10 years You know you could invest the money at 6 interest Which is the best deal To solve this we will compare the present value of the loan to the 35000 Both are in present dollars so we can compare them directly You cannot compare a future value with a present value A Pi1i 1z n 71 Solving for P P A1z39rl 7 li1z39 24 139 6 12 0005 A 400 per month n 10 X 12 120 Substituting P 4001005120 710005100512 3602938 So what should you do The present value of the payment plan is 3602938 and the cost of buying the lathe outright is 35000 It appears that buying the lathe outright is the best arrangement Example 7 In Example 6 we looked at purchasing a lathe by comparing the present value of purchasing the lathe outright and making monthly payments for 5 years We could have done the comparison by looking at the future value of both We can compare present values or future values but we cannot compare a present value to a future value The future value of the 35 000 used to purchase the lathe outright is F P1 139n F 35000 1 0065 3546383790 The future value of the 400 monthly payment is F A 1z39 7111 F 4001005120 7 110005 6555174 The difference is considerably larger when we compare future values but the results are the same It will be to our advantage to purchase the lathe outright than to pay it off in monthly payments Example Problems Problem 1 If you wish to have 1000 in a saving account at the end of 5 years and the interest rate is 6 paid annually how much should you put in the saving account now We know from the lecture F P1 139n Solving this forP results in P F 1 1 F the sture value 1000 139 the interest rate 6 006 n the number of years 5 P 1000 11 0065 74726 ProblemZ Joe read that a 1 acre parcel of land could be purchased for 1000 in cash He decided to save a uniform amount at the end of each month so he would have the required 1000 at the end of one year The bank pays 6 interest compounded monthly How much should Joe save each month F A1i 71i 22 Solving this forA results in A z39F 1z n 71 139 006 12 0005 n1X1212 F1000 A 0005 x 1000 1005 12 1 8107 Problem 3 On January 1 a man deposits 5000 in a credit union that pays 8 interest compounded annually He wished to withdraw all the money in 5 equal endofyear sums beginning December 31st of the rst year How much should he withdraw each year A Pi1i 1z39n 71 P 5000 n 5 139 8 A unknown A 7 5000 008 10851 1085 71 A 125228 Problem 4 A 5000 loan was to be repaid with 8 simple annual interest A total of 5 800 was paid How long was the loan outstanding F P Pin Solvingfor n n F 7 PPz39 where 139 8 P 5000 F 5800 n 5800 7 50005000 X 008 n 2 years Problem 5 A sum of money invested at 4 interest compounded semiannually will double in amount in approximately how many years F Pl 139n Solving for 71 yields F P 1 1 Log FP nLogl 139 Or n Log FP Log1 139 where FP2 i0042002 n Log 2 Log 102 35 periods or 175 years MONEY Homework A woman borrowed 2000 and agreed to repay it at the end of three years together with 10 simple interest How much will she pay at the end of the three years D 2662 D 3000 D 2600 D 2200 D 2400 You put 6000 in the bank How long will it have to stay there to double in value if the bank pays 8 interest compounded annually D 6 Years D 4Years D 9years D 8years D 20 Years You invest 1000 in the bank each year for 3 years The bank pays 7 interest compounded annually How much money will you have in 5 years D 412536 D 382257 D 393839 D 265537 D 360004 You put 5000 in the bank and keep it there for 20 years The bank pays 65 interest compounded monthly How much money will you have after 20 years D 1828223 D 1761823 D 1150000 D 1535674 D 1272156 Each year your grandmother gives you 1000 You put the money in the bank at 6 annual interest compounded annually How much money will you have after your 10111 deposit D 10600 D 7360 D 13181 D 10000 D not listed What is the present value of the money your grandmother will give you D 10600 D 7360 D 13181 D 10000 D not listed You borrow 18000 from the bank The bank charges 8 annual interest compounded monthly What will be your monthly payment if you repay the loan in 5 years D 300 D 426 D 331 D 365 D not listed You win 10000000 future value in the lottery and choose to take the money as an annuity paid out over 50 years The interest rate on the annuity is 6 annually which is compounded monthly What will be the monthly payment to you D 2640 D 16667 D 41800 D 3264 Dnot listed Testing the Constant Strain Mesh 50 Principal Stresses Previously we developed the FEA mathematics for a constant strain triangular mesh This method is applicable to problems that can be idealized as two dimensional plates All constraints and loads must be in the plane of the plate We developed the equation c7DBq 473 orexpanding Q1 92 0x 1 v 0 yzs 0 y 0 yu 0 E l 93 7 7 2 v l 0 0 x32 0 x13 0 x21 474 y l V 1V detJ q4 TX 0 0 T x32 yzs x13 y31 x21 ylZ q 5 95 for computing the stresses for each triangular element We used a linear equation to interpolate the displacements and this yielded a constant strain or stress across each triangular element Knowing that this is only an approximation we decided to place the computed stress near the centroid of the triangle Equation 474 indicates we are computing two axial stresses and a shear stress These are illustrated in Figure 1 below y 6X F a 0X Txy Txy 6y Figure 1 Material element showing the axid and shear stresses L x The axial and shear stresses vary with the orientation of the small element and we can compute these with Mohr s circle Testing the Constant Strain Mesh Page 1 of 6 51 Mohr s Circle Mohr s circle is a graphical method for computing axial and shear stresses in the element of material shown in Figure 1 The circle is plotted on a coordinate system with the axial stresses plotted along the horizontal axis and the shear stresses plotted along the vertical axis T Figure 2 Mohr39s circle drawn from the computed stresses The stresses computed in Equation 474 are plotted by plotting the point de ned by away and 0X72 This de ned two points on the circle We know that the circle is centered point C on the s axis so we can compute the coordinates of this point by averaging the two axial stresses Ccxcy2 51 The distance R from point C to the circle can be computed with Pythagoras s formula The hypotenuse of the triangle is the distance from the center point to the circle One leg of the triangle is Ty and the other can be computed as A G 0 52 2 The distance R is computed with equation 53 R A2 I 53 The principal stresses where the circle crosses the axial stress axis s can be computed with 0391 2 C R 54 and Testing the Constant Strain Mesh Page 2 of 6 qzC R on These points are shown in Figure 3 A2 1 02 C R I I 01 CR PrInCIpaI Stresses Figure 3 Mohr39s circle with principal stresses The circle represents various orientatiom of the element shown in Figure 1 Each degree of rotation of the element is represented by two degrees on Mohr s circle When an element is oriented so that all of the stresses are axial the stresses in the element will relate to the principal stresses The maximum shear stresses occur when the element is oriented 45 degrees from the principal stress orientation On Mohr s circle this corresponds to the top and bottom of the circle The maximum shear stresses can be computed with 139 H R 56 m 52 Von Mises Stress As designers it is usually important to limit the stresses so that our designs do not deform permanently Richard von Mises and several other researchers studied this problem and determined that the material will yield when the distortion energy per unit of volume equals the tension yield stress of the material We call this measure of distortion energy the von Mises stress It can be computed with Lqq on I 00 47 58 mam i 6 Testing the Constant Strain Mesh Page 3 of 6 Equation 59 shows what is commonly called the Von Mises stress It can be computed with the axial and shear stresses computed in Equation 47 is important to keep the Von Mises stress below the yield stress of the material we are using in our design the Von Mises stresses go beyond the yield stress the object we are designing will permanently deform n example ofa stress strain curve is shown in Figure 5 The FEA method we have developed only works in the linear ran e ofthe curve I e Von Mises stress is below the Yield Point stress the material will be in the linear elastic range Figure 4 r Richand vim Mises Elastic Range Strai n Figure 5 7 Typical stress strain curve iur many metals 53 Testing the Method One way to test the method we have developed is to write a program based on the method then compare the results with that produced by other programs We will look at a rectangular plate that is xed at one end and free on the other We will place a 1000 pound load on the free end 1000 Lbs Figure6 rPlate with elemenu shuwn Testing the Constant Strain Mesh Page 4 of 6 The plate is 3 inches wide 2 inches high and 05 inches thick The modulus of elasticity is 30XlO6 and Poison s ratio is 025 We will divide the plate into 16 elements The actual data is shown in the two tables below The table labeled Nodal Coordinates gives the node numbers and the coordinates of the node The table labeled Element Connectivity de nes each triangular element and the three nodes de ning its vertices The von Mises stresses are shown below in Figure 7 below The method we developed assumed a linear shape function and this implied that the stress and strain were constant across each element We know that the stress is not constant across the entire element so we assume that it applies to the center of each triangle near where the stresses are shown in the gure 1000 Lbs 4657 3382 2592 4794 2672 2642 2218 1492 2672 2725 2454 1947 4620 3287 2124 1298 D Figure 7 Plate with von Mises stresses shown Testing the Constant Strain Mesh Page 5 of 6 The same problem was solved using ANSYS a commercial nite element program ANSYS allows the exact same shape of elements to be de ned but it uses a different algorithm ANSYS uses a quadratic shape function which should produce a more realistic solution than the method we developed in class The ANSYS results and the results from our class developed method are shown in the table below Elem Class ANSYS Elem Class ANSYS 1 2672 2672 9 2672 2672 2 4620 4620 10 4657 4620 3 2725 2725 11 2642 2725 4 3287 3287 12 3382 3287 5 2454 2454 13 2218 2454 6 2124 2124 14 2592 2124 7 1948 1948 15 1492 1948 8 1298 1298 16 4794 4467 As you can see from the table above the solutions using the method we developed compare quite well with the ANSYS solutions The solutions match very well on the left side of the cantilever but diverge from the ANSYS solutions as we move closer to the point load There will be a force concentration at this point and the quadratic elements used in ANSYS handle this stress concentration better than our constant strain elements As you move farther away from this high stress area the stresses match very closely 54 Homework Using the method we developed in class write a Matlab program to compute the displacements at each node and the stresses principal stresses and von Mises stresses for each element Test your program using the problem solved in the notes Secondly test your program using the plate problem shown above Show the results of both tests Testing the Constant Strain Mesh Page 6 of 6

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