Beam Deflection

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TorqueLaboratory Three: Parallam Beam Deflection Lab Group – 1st Mondays, Late: Jesse Bertrand, Ryan Carmichael, Anne Krikorian, Noah Marks, Ann Murray Report by Ryan Carmichael and Anne Krikorian E6 Laboratory Report – Submitted 12 May 2008 Department of Engineering, Swarthmore College Abstract: In this laboratory, we determined six different values for the Elastic Flexural Modulus of a 4-by10 (100” x 3. 50” x 9. 46”) Parallam wood-composite test beam.

To accomplish this, we loaded the beam at 1/3 span with 1200 psi in five load increments in both the upright (9. 46 inch side vertical) and flat (9. 46 inch side horizontal) orientations. We then used three different leastsquare methods (utilizing Matlab and Kaleidagraph) on the data for each orientation to fit the data, resulting in the following: E: Upright Orientation Units Method One Method Two Method Three E: Flat Orientation 10 ksi 103 ksi 3 0. 981 ± 0. 100 1. 253 ± 0. 198 1. 065 ± 0. 247 1. 880 ± 0. 046 2. 080 ± 0. 083 1. 881 ± 0. 106 1 Purpose:

The purpose of this lab is to determine the flexural elastic modulus of a Parallam woodcomposite beam by examining its behavior when simply supported and under flexural stress, and to analyze deflection data using different least-squares methods to fit theoretical deflection curves. Theory: In theory, a beam’s deflection can be mapped by the governing equation of beam flexure: EI d2y/dx2 = M(x), where E is the elastic modulus, I is the second moment of inertia about the neutral axis of the beam (the value of which changes significantly according to orientation), y is deflection, and M(x) is bending moment in the beam.

This equation requires that several assumptions be made about the beam: 1) Geometric Assumption: the beam must be a straight, prismatic member with at least one axis of symmetry. 2) Material assumption: the beam must be linear, elastic, isotropic, and homogeneous, and the modulus of elasticity in tension must equal the modulus of elasticity in compression. 3) Loading Assumption: the beam must be loaded in pure moment in a plane of symmetry. 2 4) Deformation Assumption: plane sections before bending must remain in plane after bending.

Making these assumptions, we can apply the general equation for beam flexure to our experiment. Assuming we are using point loads or can model our setup with point loads, we can then use singularity functions to determine that the bending moment of the beam is: 2/3 P*x – P 1 Where P is the load applied with the UTM, L is the length of the beam, and x is the distance from the origin (defined as the end closest to the applied load). From this we get: M(x) = EI d2y/dx2 = 2/3 P*x – P 1 Taking an integral of both sides with respect to x yields: where c1 is a constant.

Taking another derivative yields: where c2 is a constant. Rearranging we get: . EI dy/dx = P/3 * x2 – P/2 * 2 +c1 y * EI = P/9 * x3 – P/6 * 3 +c1x + c2 y = Px3/9EI – P/6EI * 3 +c1x/EI + c2/EI To solve for the constants we need to make two more assumptions: that when x=0 and when x=L there will be no deflection (i. e. y=0). Using these assumptions, we can plug into our previous equation and use algebra to determine that c1 = -5PL2/81 and that c2 = 0. This gives us: 3 y = P/EI (x3/9 – 3 /6 – 5L2x/81) This is the theoretical beam deflection equation for the lab.

Then, to ease calculations, we make the previous equation non-dimensional by multiplying both sides by EI/PL3, which yields: yEI/PL3 = (x/L)3/9 – 3/6 – 5/81 (x/L) We define this dimensionless quantity as: (x/L)3/9 – 3/6 – 5/81 (x/L) = ! theoretical where: ! theoretical = f(x/L) Similarly, we define: ymeasured * EI/PL3 = ! measured. If the beam were to behave as a theoretical beam, then ! theoretical would equal ! measured. E is defined as the slope of the stress-strain curve in the elastic region. However, there is no perfect way to measure stress and strain in the loaded beam.

As a result, to determine E one must make some assumptions. For methods one and two the assumption made is that ! theoretical = ! measured. This is done because ! measured can only be calculated if the value of E is known (if E is unknown, then the equation ymeasured * EI/PL3 = ! measured has two unknowns and is thus unsolvable). For method one this assumption is used to write this equation: f(x/L) = E (I ymeasured/PL3). Manipulating this equation gives an equation in the form: P = E (I ymeasured/f(x/L)L3) 4 This equation is in the form of y=mx, the form of a line.

Thus, if it is plotted P versus (I ymeasured/f(x/L)L3) then the slope of the line will be E. In method two, the same assumption is made, resulting in the formula: E = f(x/L)PL3/ ymeasuredI From this formula E can be calculated on a point by point basis and then the values can be averaged. Method three approaches the problem in a different way. Instead of assuming that ! theoretical = ! measured, a new quantity V was defined as: V = ! theoretical – ! measured Then we make a guess for the E value and solve for the rms error, defined as: rms = sqr(1/n * sum(V2)) here V represents the difference between theoretical and measured deflection for every data point at a certain E value, and n is the total number of V values (5 loads * 4 locations = n = 20). The rms error is then plotted against the many guessed E values, and the point on the graph where the rms error is minimized is determined to best the best value of E for method three. 5 Procedure: In the lab, we tested a simply supported Parallam beam (nominal dimensions: 4 by 10) in two orientations while loaded in flexural stress from the UTM (setup shown in figure 1).

The beam’s dimensions were 100 inches span by 3. 50 inches thick by 9. 46 inches deep. Our two orientations were with the 9. 46 inch side vertical (the ‘upright’ orientation) and then with the 9. 46 inch side horizontal (the ‘flat’ orientation). For each orientation, we applied an approximate point load by placing a roller between the UTM and the beam at the point L/3 on the span. (In fact, as the roller comes into contact with a small area of the beam and not a single point, 6 describing it as a ‘point load’ is not quite accurate. We applied the load in five increments: 240, 480, 720, 960, and 1200 psi. At each of the load increments, we measured deflection at three points: L/4, L/2, and 3L/4 (the UTM recorded deflection at L/3). We also observed the deflection and the location of maximum deflection, and calculated values of I (the second moment of inertia) for each orientation. Outside of lab, we used three methods to determine E. As discussed in the theory, method one consisted of plotting the load P (lb) versus the quantity Iymeasured/f(x/L)L3 (in2). The slope of this graph was the first value for E.

Both Matlab and KaleidaGraph were used in this process. Utilizing the same theory as method one, method two used the equation E = f(x/L)PL3/ Iymeasured to solve for E for each individual point with each load. The resulting values of E were then averaged to determine the best value of E for method two. The average was found using Matlab and the error using KaleidaGraph. Method three (also as discussed in the theory section) plots rms error against many guessed E values. The best value of E (for method three) was found by determining where the rms error was minimized.

This process was done entirely in Matlab. 7 Results: E: Upright Orientation Units Method One Method Two Method Three E: Flat Orientation 10 ksi 103 ksi 3 0. 981 ± 0. 100 1. 253 ± 0. 198 1. 065 ± 0. 247 1. 880 ± 0. 046 2. 080 ± 0. 083 1. 881 ± 0. 106 Discussion: The values of E that we determined for each orientation were very close in value. The values for the upright beam all fall within error of each other, while for the flat orientation one value (while still very close) was just outside of the error of the other two, which are nearly identical.

Error in our values comes partially from universal measurement error, and from flaws and inconsistencies in the beam (i. e. , a non-isotropic and non-homogeneous beam); these types of error have a global influence on our results. Other major sources of error are method-specific. In method one, there is error from fitting a line to a set of data that is not precisely linear. As a result, we took our method one values from Kaleidagraph, which is more specifically graphing software and which provides a curve fitting error.

We also used Matlab values as a check of accuracy. In method two, error came from the variance in the E value of each data point. For this method, we used Kaleidagraph simply to determine error (having calculated the values in 8 Matlab), taking the standard deviation as representative of the variance. In method three, error comes from the lack of a perfect fit of a deflection graph to our data; our E value minimized the error between predicted and actual deflection, which was then represented as rms error.

In all of the methods, we weighted each data point equally (this will be discussed more thoroughly later in this section). Interesting to note is the difference in the value of E for each orientation. This is partially a result of the composition of the beam, as, upon inspection, the grain of the wood is pronouncedly evident (see Appendix F). We expect the material that the beam is composed of to behave more rigidly when loaded to parallel the grain (in the upright position), and to bend more easily when loaded perpendicular to the grain (in the flat position).

The grain of the wood is largely the result of the beam’s construction, as it is fabricated from strips of wood bound with glue and pressed until formed: this method of construction results in a major difference in stiffness, according to orientation. However, although the material does perform more rigidly according to grain orientation, the difference in the value of I made a more significant impact on our final values of E and, thus, the beam behaved more rigidly in the flat orientation, where the I value was significantly smaller.

Another point of interest was the location-specific variation in E: upon examination of the graph for method one (in appendix B), this variation becomes apparent. The data points 9 collected at L/4 and 3L/4 are the two points closest to the left at each load; the slope of the line formed by connecting these points is steeper than the rest, which means that the resulting E value is higher. The data points from L/3 appear at the far right at each level of load, and when connected have a lower slope and therefore smaller E value.

The data points from L/2 are in the middle, and have a slightly less steep slope than at L/4 and 3L/4, and thus a slightly smaller E value. It should be noted that the E value that differed most from the other values is the E value at L/3, which was the value determined in part by the deflection measured by the UTM. The difference in measuring methods may be the cause of this. It is possible, for instance, that the UTM measurements include deflection from the compression of the beam itself where the load was applied, or alternately that the measurements from the UTM were more accurate than those we found by manually reading gauges.

However, as we do not know whether those values are more or less accurate than the others, we have no definite reason to weight it differently one way or the other and, thus, weight all four locations equally. Having found values for E, we then decided that the best value for each orientation was that determined with method three. This is because the method three values are the median values for each orientation, and they do not rely on the assumption that we based methods one and two on. Numerically, this yields the following values: for the upright orientation, 10 (1. 65 ± 0. 247) *103 ksi, and for the flat orientation, (1. 881 ± 0. 106) * 103 ksi. Conclusion: The results of this beam testing determined six values of E for the beam, three in the upright orientation and three in the flat orientation. Of the E values, those for each orientation were very similar to the other E values for the same orientation; from these values, the method three values were deemed most reliable. Furthermore, if only one E value were to be chosen to tell a consumer, it would be (1. 881 ± 0. 106) * 103 ksi, the value from method three for the flat orientation.

This choice was made with the assumption that the consumer would know that it was only the E value for one orientation, and the flat value chosen because the flat orientation seems a more likely choice for flexural loading, and also because it is the more appealing value and so more competitive on the market. References: Figure 1 adapted from Zachary Eichenwald, Swarthmore College, May 2007. Matlab created with assistance from Noah Marks et al. Appendices H and I are the intellectual property of E. Carr Everbach, Swarthmore College, May 2008. 11 List of Appendices: A…………………………………………….

Lab Data B……………………………………………. KaleidaGraph Graphs for Method One C……………………………………………. Matlab Graphs for Method One D……………………………………………. Matlab Graphs for Method Three E…………………………………………….. Nondimensionalized Deflection Graphs F…………………………………………….. Image Depicting Grain G……………………………… …………….. Matlab Code H……………………………………………. Theory Notes I…………….. ………………………………. Lab Handout 12 Appendix A: Upright test Load- in units of lb*ft 240 480 720 960 1200 240 480 720 960 1200 Defl @ L/4 in units of . 001 inch 12 23. 5 34 44. 5 55 50 101. 5 151 201 251 Defl @ L/3 units of . 001 inch 19. 5 37. 7 54. 3 70. 2 92. 65 131. 8 197 262 332 Defl @ L/2 in units of . 001 inch 13 26 39 52 65 64 129 193. 5 257 322 Defl @ 3L/4 in units of . 001 inch 8 17 25 34 42 40. 5 82 124 165 206 Flat test 13 Appendix B: 14 15 Appendix C: 16 17 Appendix D: 18 19 Appendix E: 20 21 Appendix F: Image of Beam in Flat Orientation: Load Perpendicular to Grain 22 Appendix G: Code for Methods One, Two and Three clear defup=. 001*[12, 23. 5, 34, 44. 5, 55; 19. 5, 37. 7, 54. 3, 70. 2, 92. 5;13, 26, 39, 52, 65; 8, 17, 25, 34, 42]; defflat=. 001*[50, 101. 5, 151, 201, 251; 65, 131. 8, 197, 262, 332; 64, 129, 193. 5, 257, 322; 40. , 82, 124, 165, 206]; load = [240, 480, 720, 960, 1200]; for i = 1:100 a(i) = (i-1)/99; if a(i) < 1/3 f(i) = (1/9). *(a(i)). ^3-(5/81). *(a(i)); else f(i) = (1/9). *(a(i)). ^3-(1/6). *((a(i))-1/3). ^3-(5/81). *(a(i)); end end f1 = [f(25) f(33) f(50) f(75)]; %Method 1 Upright for i=1:4 for j=1:5 x(i,j) = -(defup(i,j). *246. 922). /(f1(i)’. *100^3); end end figure L=[240, 480, 720, 960, 1200;240, 480, 720, 960, 1200; 240, 480, 720, 960, 1200; 240, 480, 720, 960, 1200]; [e1]=polyfit(x,L,1) plot(x,load,’bo’, x, e1(1)*x+e1(2),’r-‘) title(‘Method One Upright’) %Method 1 Flat for i=1:4 for j=1:5 x(i,j) = -(defflat(i,j). 33. 7998). /(f1(i)’. *100^3); end end figure 23 L=[240, 480, 720, 960, 1200;240, 480, 720, 960, 1200; 240, 480, 720, 960, 1200; 240, 480, 720, 960, 1200]; [e2]=polyfit(x,L,1) plot(x,load,’bo’, x, e2(1)*x+e2(2),’r-‘) title(‘Method One Flat’) %Method 2 Upright for i = 1:5 for j = 1:4 Eup1(i,j) = f1(j)*(load(i)/defup(1,i))*100^3/246. 922; end end avgEup=mean(Eup1); E2up = -mean(avgEup) figure plot(a,f,’bo’) title(‘Method 2 Upright’) %Method 2 Flat for i = 1:length(load) for j = 1:length(f1) Eflat1(i,j) = f1(j)*(load(i)/defflat(1,i))*100^3/33. 998; end end avgEflat=mean(Eflat1); E2flat = -mean(avgEflat) figure plot(a,f,’bo’) title(‘Method 2 Flat’) %Method 3 Upright E = [5e5:1e3:1. 5e6]; for k = 1:length(E) accum = 0; for i = 1:5 for j = 1:4 v(i,j,k) = f1(j); vstar(i,j) = -(defup(j,i)*E(k)*246. 922)/(load(i)*100^3); result = (v(i,j,k)-vstar(i,j)). ^2; accum = accum + result; end end RMSu(k)= sqrt(accum/20); end figure plot(E,RMSu,’bo’) 24 title(‘Method 3 Upright’) [RMSmin,index] = min(RMSu); min(RMSu) E3up = E(index) %Method 3 Flat E2 = [1. 7e6:1e3:2. e6]; for k = 1:length(E2) accum2 = 0; for i = 1:5 for j = 1:4 vflat(i,j,k) = f1(j); vstarflat(i,j) = -(defflat(j,i)*E2(k)*33. 7998)/(load(i)*100^3); result2 = (vflat(i,j,k)-vstarflat(i,j)). ^2; accum2 = accum2 + result2; end end RMSf(k)= sqrt(accum2/20); end figure plot(E2,RMSf,’bo’) title(‘Method 3 Flat’) [RMSmin2,index2] = min(RMSf); min(RMSf) E3flat = E2(index2) Error Code for Method Three defup=. 001*[12, 23. 5, 34, 44. 5, 55; 19. 5, 37. 7, 54. 3, 70. 2, 92. 5;13, 26, 39, 52, 65; 8, 17, 25, 34, 42]; defflat=. 001*[50, 101. 5, 151, 201, 251; 65, 131. 8, 197, 262, 332; 64, 129, 193. 5, 257, 322; 40. , 82, 124, 165, 206]; load = [240, 480, 720, 960, 1200]; for i=1:4 for j=1:5 E(i,j)= 7. 9325e-04*load(j)*100^3/(33. 7998*defflat(i,j)) %this is E=rms*PL^3/Iy end end Eerr=mean([mean([E])]) 25 Appendix H: 26 27 28 Appendix I: E6 Mechanics Least-Squares fitting to Beam Flexure data 2008 The purpose of this lab is to examine the behavior of a simply supported timber beam under a point load at 1/3 span, and to show how the Flexural Modulus of a particular piece of timber may be found by fitting a theoretical deflection curve to experimental load-deflection data, using three different least-squares techniques.

In the Lab: Using the universal testing machine and three-point bending setup, load the nominal 4-by-10 Parallam wood-composite test beam to a maximum fiber stress of 1200 psi in five load increments in both the Upright and Flat orientations. At each load, measure and record deflections at the load, midspan and both quarter-span locations, and visually observe the deflected shape and location of maximum deflection. Beam dimensions are 100 inches span by 3. 50 inches thick by 9. 46 inches deep. In your Group Lab Report (maximum 4 people per lab report): 1.

Develop a singularity-function deflection solution of the form ! = (PL3/EI) * f(x/L) for transverse deflection of a simply-supported beam with a single point load at x/L = 1/3, and non-dimensionalize it to the form EI! /PL3 = f(x/L). 2. Code the non-dimensional polynomial singularity function f(x/L) in MATLAB and plot it versus x/L for x/L = n/24 , 0 ” n ” 24 . Note that only the term corresponding to the external load is a singularity function. 3. Find best E values for the upright and flat loading conditions by 3 different Least-Squares methods: Method #1: Use test results to determine Eslope = f(x/L) *(P/! * L3/I using the slope of a P-! plot as data for each of the four measuring locations. Determine a separate Eslope for each location and each beam orientation, and then use a weighted average (you choose the weighting factors) of the four values to estimate a “best” Eup1 and Eflat1 for the beam. Method #2: Develop a MATLAB script to determine the “best” Eup2 and Eflat2 for the beam by using each individual data point (rather than P-! slope) to calculate Eijk = f(xi/L) * Pj/! j * L3/Ik and then minimize the sum of squared differences between Eijk and Ekbest .

Here the subscript i refers to location along the beam, j to load and k to beam orientation. Method #3: Use your curve from Method #2 above, and the left-hand-side of the equation EI! /PL3 = f(x/L) from Method #1 above to plot non-dimensional deflection data points on the same axes as the curve 29 – make separate plots for the upright and flat orientations. Notice that where the points fall (above vs below the curves) depends on your choice of a value for E. Write a MATLAB script to determine the value of Ebest3 for which the sum of squared deviations between the left and right sides of the equation EI! PL3 = f(x/L) is a minimum, and run it twice, first using data for Upright and then for Flat loading, to find the best Eup3 and Eflat3 for the beam. Tabulate your results for all 3 Methods, then compare and discuss them, particularly with regard to a) variation of E with location and beam orientation (which regions of the beam are stiffest and least stiff), and b) appropriate weighting factors for quarter-span data versus midspan data in determining Ebest. Make and include whatever additional plots, figures, tables or other graphics you find most helpful to explain your results.

Finally, answer this question: If you had to furnish only one value to represent E of this beam to a buyer who might use it as a load bearing beam or column, what value would you give, and why? Use this blank table to record the experimental data from your lab session: Load – in units of lbf Upright 9. 46dx3. 50w Defl. @ L/4 in units of 0. 001 inch Defl. @ L/3 in units of 0. 001 inch Defl. @ L/2 in units of 0. 001 inch Defl. @ 3L/4 in units of 0. 001 inch Flat 3. 50dx9. 46w 30

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