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Laboratory 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-by-

10 (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 least-

square methods (utilizing Matlab and Kaleidagraph) on the data for each orientation to fit the

data, resulting in the following:

E: Upright Orientation E: Flat Orientation

Units 103 ksi 10

3 ksi

Method One 0.981 ± 0.100

1.880 ± 0.046

Method Two 1.253 ± 0.198

2.080 ± 0.083

Method Three 1.065 ± 0.247 1.881 ± 0.106

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Purpose:

The purpose of this lab is to determine the flexural elastic modulus of a Parallam wood-

composite 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/dx

2 = 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.

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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 <x-L/3>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/dx

2 = 2/3 P*x – P <x-L/3>

1

.

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:

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 * <x-L/3>

2 +c1

y * EI = P/9 * x3 – P/6 * <x-L/3>

3 +c1x + c2

y = Px3/9EI – P/6EI * <x-L/3>

3 +c1x/EI + c2/EI

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y = P/EI (x3/9 - <x-L/3>

3 /6 – 5L

2x/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 - <x/L – 1/3>

3/6 – 5/81 (x/L)

We define this dimensionless quantity as: (x/L)3/9 - <x/L – 1/3>

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)

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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))

where 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.

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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,

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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 (in

2). 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.

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Results:

E: Upright Orientation E: Flat Orientation

Units 103 ksi 10

3 ksi

Method One 0.981 ± 0.100

1.880 ± 0.046

Method Two 1.253 ± 0.198

2.080 ± 0.083

Method Three 1.065 ± 0.247 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

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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

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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,

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(1.065 ± 0.247) *103 ksi, and for the flat orientation, (1.881 ± 0.106) * 10

3 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.

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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

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Appendix A:

Load- in

units of

lb*ft

Defl @ L/4 in

units of .001

inch

Defl @ L/3

units of .001

inch

Defl @ L/2 in

units of .001

inch

Defl @ 3L/4

in units of

.001 inch

Upright

test

240 12 19.5 13 8

480 23.5 37.7 26 17

720 34 54.3 39 25

960 44.5 70.2 52 34

1200 55 92.5 65 42

Flat test 240 50 65 64 40.5

480 101.5 131.8 129 82

720 151 197 193.5 124

960 201 262 257 165

1200 251 332 322 206

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Appendix B:

15

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Appendix C:

17

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Appendix D:

19

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Appendix E:

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Appendix F:

Image of Beam in Flat Orientation: Load Perpendicular to Grain

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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; 40.5, 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

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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))*100^3/33.7998;

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')

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title('Method 3 Upright')

[RMSmin,index] = min(RMSu);

min(RMSu)

E3up = E(index)

%Method 3 Flat

E2 = [1.7e6:1e3:2.1e6];

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; 40.5, 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])])

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Appendix H:

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28

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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

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- 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

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

Upright 9.46dx3.50w

Flat 3.50dx9.46w