L1 Cantilevered Beams-2022

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Queens University *

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Mechanical Engineering

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Dec 6, 2023

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L1. Cantilevered Beams Purpose The goals of this lab are to examine the deflection of cantilevered beams in response to point loads and to use a graphical analysis to determine the material properties and model parameters of a test beam. Two different graphical analysis approaches will be examined. Prelab Checklist Prelab Quiz Intro Video(s) Review Caliper Intro Theory A "cantilevered" beam is one that is rigidly supported at one end, and free to deflect at the other, as shown in Figure 1 below. In Figure 1, the beam is fixed (rigidly supported) at the right-hand side so that at this end it cannot move horizontally or vertically, nor can it rotate. A point load is a force that acts at a precise point on an object rather than being spread out across the object's surface. Figure 1 shows a point load P acting downward at the free end of this cantilevered beam. Figure 1 Considering the simple structure of a cantilevered beam, there are a remarkable number of factors that interact to determine its bending behaviour. The deflection at the free end ( d ) depends on the beam material, the cross-sectional shape of the beam (e.g. square, circular), the magnitude of the applied load ( P ), and the beam length (l ) . For a beam with a rectangular cross- section these quantities are related by Equation 1: Equation 1

Where d is the deflection, P is the magnitude of the point load, l is the unsupported length of the beam, E is Young's modulus (essentially the stiffness) of the material, w is the width of the beam and t is the thickness (thinner dimension) of the beam. The equation above is also known as the mathematical "model" that theoretically describes the behaviour of this simple beam in response to the point load P . In this lab you will use graphical analysis to experimentally determine the value of the power n , and Young’s modulus, E , for a wood beam. The power n can be determined by first creating a graph of d versus l . If your d versus l graph yields a straight line then the power n equals one. If the d versus l plot is not linear, there are a couple of ways of determining the power n graphically. 1. Plot d versus various powers of l until you obtain a straight-line result (e.g. try d versus l 2 , l 3 , etc). The power that linearizes the graph is the correct value for n . 2. Plot ln( d) versus ln( l) . The power n is the slope of the graph. To understand why this works consider a hypothetical case where y = Ax 4 . If we take the ln of both sides of this equation we get ln y = 4ln x + ln A . So if we plot ln y versus ln x we will get a straight line with slope=4, intercept = ln A . Once a linear plot is obtained using one of the methods above, the slope (m) of the line can be used to determine the value of Young's modulus ( E ) for the beam material. Equipment Supplied A wooden beam (metre stick), a beam support that allows the unsupported length (l) of the beam to be varied, a meter stick for measuring deflection, Vernier calipers, masking tape, a set of metric masses and a mass hanger. For this experiment, you may assume that the masses have zero error. Procedure: measuring deflection ( d ) as a function of unsupported length (l) • Measure the width and thickness of the beam using Vernier calipers. In the provided Excel spreadsheet, record these values in Table 1(b) in Sheet 1 (note that Table 1(b) is set up as a Python input table). As you work through the procedure below, remember to determine and include all measurement errors into the appropriate tables (Tables 1b and 1c) in Excel. a. Determining the power n in equation 1 using two different methods. i. Begin by securing the beam into the clamp at l = 20 cm and specifying your beam material below. The point load ( P ) will remain constant by using a constant mass M= 700g to give a force of 6.867N. Deflection ( d ) is found by measuring the heights h 0 and h up from the floor. Varying the unsupported length l from 20cm to 90cm at intervals of 10cm, measure the initial height h 0 (ensure the clamp is tight and beam is secure before measuring), then add the mass M= 700g and measure the height h . Carefully measure and record the value of l

every time you increase it. Record all data into Table 1(a), and the measurement errors into Table 1(c) in Excel Sheet 1. Once you finish collecting data, create appropriate formulas to fill the remaining columns of Table 1(a). Paste the completed Table 1(a) and Table 1(c) below under the headings provided. Table 1(a) Displays key measurements of beam (m) as well equations involving these variables. The natural logs of I and d, along with I 2 and I 3 . The error for each equation is also included. Unsupport ed length of beam l [m] Initial heigh t of beam h 0 [m] Height of deflecte d beam h [m] Deflectio n d ( h 0 - h ) [m] ln(l) ln(d) l 2 l 3 δln(l)* * δln(d)* * δl 2 ** * δl 3 ** * 0.200 0.871 0.868 0.003 - 1.60 9 - 5.80 9 0.04 0 0.00 8 0.003 0.333 0.000 0.000 0.300 0.871 0.862 0.009 - 1.20 4 - 4.71 1 0.09 0 0.02 7 0.002 0.111 0.000 0.000 0.400 0.871 0.850 0.021 - 0.91 6 - 3.86 3 0.16 0 0.06 4 0.001 0.048 0.000 0.000 0.500 0.871 0.831 0.040 - 0.69 3 - 3.21 9 0.25 0 0.12 5 0.001 0.025 0.001 0.000 0.600 0.871 0.804 0.067 - 0.51 1 - 2.70 3 0.36 0 0.21 6 0.001 0.015 0.001 0.001 0.700 0.871 0.769 0.102 - 0.35 7 - 2.28 3 0.49 0 0.34 3 0.001 0.010 0.001 0.001 0.800 0.871 0.723 0.148 - 0.22 3 - 1.91 1 0.64 0 0.51 2 0.001 0.007 0.001 0.001 0.900 0.871 0.667 0.204 - 0.10 5 - 1.59 0 0.81 0 0.72 9 0.001 0.005 0.001 0.001

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Table 1(c) Extension of Table 1(a). Displays height and initial height errors, as well as length and deflection errors ( ± m). Initial height error δ h 0 [±m] Height error δ h [±m] Unsupported length error δl [±m] Deflection error δ d [±m]* 0.0005 0.0005 0.0005 0.001 Method 1 to determine exponent n: Plot d vs l n using different values for n until you get a straight line To determine the value of the exponent n from equation 1, plot your data with l on the x-axis and d on the y-axis. If the plot of d vs. l is a straight line then n = 1. If this is not the case, then determine the value of n by plotting d vs. l 2 , and also d vs. l 3 . The data required is in Table 1(a). The plot that produces a straight line will give you the value of the exponent n. For the plot that gives you the best linear fit, include error bars and a line of best fit. Add the equation of the line. Paste your graph (Figure 1a) below. Perform a regression analysis in order to determine the slope (m) and its error δ m. Note that the regression slope should be the same as the one from the line of best fit. Also produce a residual plot and include it (Figure 1b) below. Figure 1a: Scatter plot comparing Unsupported Length of Beam Cubed "I 3 " (m 3 ) and Deflection "d" (m). Line of best fit is included with its equation and R 2 value. Both vertical and horizontal error bars are included but are barely visible. d = 0.2794(I 3 ) + 0.0036 R² = 0.9988 0.000 0.050 0.100 0.150 0.200 0.250 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 Deflection "d" (m) Unsupported Length of Beam Cubed "I 3 " (m 3 )

Figure 1b: Residual plot showing the even distribution of markers. Shows a strong positively correlated linear relationship for the data set. n = 3 (that produces a straight line for a d vs. l n plot) m = 0.279 ±0.004 (slope ± error of the straight line produced from the d vs l n plot) Method 2 to determine the exponent n: Plot a ln-ln graph of d versus l and determine the power n from the slope of the graph Plot your ln-ln data with ln(l) on the x-axis and ln(d) on the y-axis (the data is in Table 1a). Include error bars and a line of best fit. Add the equation of the line. Paste your graph (Figure 2) below. -0.004 -0.003 -0.002 -0.001 0 0.001 0.002 0.003 0.004 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 Residuals X Variable 1

Figure 2: : Scatter plot comparing the natural logarithm of the unsupported length of beam"ln(l)" (m) and the natural logarithm of deflection "ln(d)" (m). Line of best fit is included with its equation and R 2 value. Both vertical and horizontal error bars are included but are barely visible. The value of n will be the slope ( m ) of this plot. If you'd like to better understand how this works, refer to the Theory section of this lab. Perform a regression analysis in order to determine the slope (m, which also = n) and its error δ m. Note that the regression slope should be the same as the one from the line of best fit. m = n = 2.82 ± 0.01(method 2) Report n to 2 decimal places. For the remainder of the lab, use the nearest integer value when performing calculations. (nearest integer value is 3) Part (a) Questions: • The estimates of n found using Method 1 and Method 2 should agree. Discuss the agreement of your results. In method one n was found to be 3. In method 2 n was found to be 2.82. There is a 0.18 difference between these values which is not a major error, and both are the same when looking at the nearest integer value. This nearest integer value is 3 and will be used for the remainder of the lab. Therefore, the results for the n value found in method 1 and method 2 agree with each other. ln(d) = 2.8229(ln(l)) - 1.2783 R² = 0.9999 -7.000 -6.000 -5.000 -4.000 -3.000 -2.000 -1.000 0.000 -1.800 -1.600 -1.400 -1.200 -1.000 -0.800 -0.600 -0.400 -0.200 0.000 ln(d) [m] ln(l) [m]

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• What are the advantages and disadvantages of Methods 1 and 2 for determining the power n ? Which is preferred in your opinion? The advantages of using Method 2 compared to method 1 is that only one scatter plot is needed to be made and one regression analysis needs to be preformed to find the n value. This makes Method 2 much more efficient in finding the n value. Method 2 is also reported to two decimal places whereas Method 1 is an integer value, meaning Method 2 is more accurate. The disadvantage is the need for a regression analysis in Method 2 whereas in method one the n value can be found by only comparing the lines of best fit, although this does take more time. I prefer method two as it is more accurate and more efficient. Before proceeding onto the next section, ensure your n value is correct. b. Calculate the value of Young's modulus (E) from Figure 1a i. The slope of the straight line plot of d vs. l n in Figure 1a is a function of E. (review Tutorial Lab 3 if you have forgotten why we do this) m = ?𝒑 𝑬𝒘𝒕 ? (algebraic expression) Rearrange this expression to solve for E in terms of your slope m . E = ?𝒑 𝒎𝒘𝒕 ? (algebraic expression) Now you will use the Python error propagation code to calculate the value of E with error: • Fill in all of the values of the variables into Excel Table 1b. Paste this table into the box below and add an appropriate caption • Use the Python error propagation code to determine the value of E with error. • Record the value of E and error δ E below • Paste the Sensitivity Analysis figure into Figure 3 below and add an appropriate caption. Table 1(b) Table displaying needed variables and their associated errors for the Python error propagation code. “ w ” , “ t ” , “ m ” and “ p ” are included along with their names and units in the Description column. Description Value Error w Width of beam [m] 0.02658 0.00001

t Thickness of beam [m] 0.00762 0.00001 m Slope of d vs. l n 0.279 0.004 P Point load (force) ( M*g) [N] 6.867 N 0 Figure 3: Sensitivity analysis that shows distribution of error between the m, w and t values. Shows that most error is from the m value. Record your value of E and error δ E below. E = 84 ± 1 * 10 8 Pa Part (b) Questions: • Which parameter contributed the most to the uncertainty in E ? Why? The “ m ” value contributed the most to uncertainty in E. This is because it has the greatest error value in comparison to the other variables “ w ” and “ t ” . This large error value resulted in “ m ” contributing the most to the uncertainty in E. • Based on this how could you improve your determination of E ? To improve the determination of “ E ” the large error associated with “ m ” needs to be reduced. To reduce the error in “ m ” the multiple variable values that are inputted into the formula to calculate “ m ” should also have reduced errors. This reduction of error at all stages will help in the improvement of the determination of “ E ” .

• Conduct research on E values of for the wood beam you tested. Note that the published values should be reported as a range since there are many types of wood. Provide a reference for the published values. The range of “ E ” values is 8100-9500 MPa according to the reference used. Reference: https://amesweb.info/Materials/Youngs-Modulus-of-Wood.aspx • Write down your research values and compare them with your calculated value. Do your values for E seem reasonable compared with published values? The range of values for “ E ” in wood is 8100-9500 MPa. The calculated value for “ E ” in this lab is 8400 MPa which falls within this range. Therefore, since the calculated value falls within the researched range, the value does seem reasonable. Assume E is the ONLY variable with error for the remaining questions. • Use your calculated E value to determine the deflection of a beam with a cross section of 23 mm by 15 mm and a point load of 12 N at an unsupported length of 1.0 m. Show all your calculations. ? = 4𝑝𝑙 ? 𝐸𝑤𝑡 3 ? = 4(12)(1) 3 (84 ∗ 10 8 )(0.023)(0.015) 3 ? = 0.0736 Deflection = 0.0736 ± 0.0005 m • Considering the Pythagorean theorem ( ? 2 + ? 2 = ? 2 ) do you believe that your answer to the above question is reasonable? Explain why or why not. The deflection is less than 1 and greater than 0. This will ensure the hypotenuse is longer than the length of the beam being studied. If it is not longer the system would not be possible. Therefore the calculated value for deflection is reasonable at 0.0736 ± 0.0005 m. • You are working for a shelving manufacturer and have been tasked with determining the maximum load which can be applied to a wood cantilever beam (use your calculated E value). The beam in question has a width of

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3.5 inches, a thickness of 1.5 inches (1 inch ≅ 25.4 mm), an unsupported length of 1.75 m, and has a maximum allowable deflection of 0.15 m. What is the maximum allowable point load for this beam in Newtons? Show all calculations. 𝑝 = ?𝐸𝑤𝑡 3 4𝑙 3 𝑝 = (0.15)(84 ∗ 10 8 )(0.0889)(0.031) 3 4(1.75) 3 𝑝 = 288.98353 𝛿𝑝 = ?𝑤𝑡 3 4𝑙 3 𝛿𝑝 = 0.000003 𝑃 ?𝑎𝑥 = 288.983530 ± 0.000003 N. NOW UPLOAD YOUR FILE TO THE ONQ DROPBOX FOR LAB 1. IT MUST BE IN PDF FORMAT. TO DO THIS - BEFORE YOU LEAVE THE LAB - • GO TO “FILE” AND CLICK EXPORT . o CLICK “CREATE PDF” o SAVE IT (PUBLISH IT) TO YOUR LAB 1 SUBFOLDER. • THEN UPLOAD YOUR PDF FILE TO THE ONQ DROPBOX. DO NOT UPLOAD THE WORD FILE TO THE DROPBOX - MAKE SURE IT IS A PDF. WE WILL NOT MARK A WORD FILE.