244 lab 5

Department of Building, Civil, & Environmental Engineering ENGR 244: Mechanics of Materials Experiment No.5: Deflection of beams Luis Alberto Alvarado Bravo 40193230 Group members: Erica Sciotto, Sofia Mendicino Section: CC-CM-X Summer 2022 July 25, 2022

1 OBJECTIVE The purpose of this experiment is to calculate the modulus of elasticity of different materials by analyzing the deflection of a beam created by a load onto a simply supported beam and a cantilever beam. INTRODUCTION When a transverse load is applied onto a beam, it forces the beam to deform accordingly. This behavior is called deflection. Extensive analysis of such behavior is quite important for engineers since they must follow many regulations that limit deflection of many structural components. The deflection of a beam is dependent mostly on the magnitude of the load, the location of it and the length span of the beam. When analyzing this effect, it is best to look at the curvature of the neutral axis of the beam. The curvature can be expressed using this formula 1 𝜌 = ?(?) 𝐸𝐼 where 𝜌 is the radius of curvature, ?(?) is the bending moment at distance ? , 𝐸 is the modulus of elasticity and 𝐼 is the moment of inertia. The curvature of the deflection curve in the elastic can be expressed as 1 𝜌 = 𝑑 2 ? 𝑑? 2 So, when combining the two formulas, the expression for the deflection of a beam can be found by double integrating the following formula 𝑑 2 ? 𝑑? 2 = ?(?) 𝐸𝐼

3 RESULTS Table 1 - Measured values of the beams Table 2 - Measured values of the cantilever beams As stated earlier, the expression for the deflection of a beam is the double integral of formula 𝑑 2 ? 𝑑? 2 = ?(?) 𝐸𝐼 For a simply supported beam, given that ?(?) = 𝑃 2 ? , the formula can be rewritten as follows ∫ 𝑑 2 ? 𝑑? 2 𝐸𝐼 = ∫ 𝑃 2 ? ∫ 𝑑? 𝑑? 𝐸𝐼 = ∫ 𝑃 4 ? 2 + 𝐶 1 In calculus, the first derivative is used to find maximum points, meaning points where the slope of a curve is 0. Knowing this, the maximum point of a curve for the deflection of these beams is at the center, so ?? ?? = 0 and ? = 𝐿 2 . This gives the value of 𝐶 1 = − 𝑃𝐿 2 16 . Now, performing the second integral yields the equation ? = 𝑃? 3 12𝐸𝐼 − 𝑃? 2 16𝐸𝐼 ? + 𝐶 2 𝐶 2 is equal to zero because it points at the supports where the beam hangs, meaning ? is also equal to 0. The formula can be simplified even further by replacing ? = 𝐿 2 and ? = 𝐿 4 . Dimensions BRASS STEEL ALUMINUM Span length, L 455.000 455.000 455.000 Width, w (mm) 19.370 19.157 19.227 Height, h (mm) 12.737 12.703 12.850 Simply-supported Beams Dimensions BRASS STEEL ALUMINUM Span length, L 247.500 250.000 246.000 Width, w (mm) 19.190 19.220 19.310 Height, h (mm) 3.160 3.223 3.390 Cantilever Beams

4 ? = ? 2 → ? = − 𝑃? 3 48𝐸𝐼 ? = ? 4 → ? = − 11𝑃? 3 768𝐸𝐼 For the theorical values, the published values for modulus of elasticity were used. 𝐸 ?????(𝑝???𝑖?ℎ??) = 105 𝐺𝑃𝑎 𝐸 ?????(𝑝???𝑖?ℎ??) = 200 𝐺𝑃𝑎 𝐸 ????𝑖???(𝑝???𝑖?ℎ??) = 70 𝐺𝑃𝑎 For a steel beam at ? = 𝐿 2 with a load of 400 N, the value for deflection was found to be ? = − 400 ( 455 2 ) 3 48 ( 19.57(12.703) 3 12 ) (200000) = −2.24 𝑚𝑚 Table 3 - Deflection of brass Table 4 - Deflection of steel at x = L/2 at x = L/4 at x = L/2 at x = L/4 200 -1.120 -0.770 -1.240 -0.870 400 -2.240 -1.540 -2.850 -1.960 600 -3.360 -2.310 -4.440 -3.060 800 -4.480 -3.080 -6.100 -4.190 1000 -5.599 -3.850 -7.700 -7.280 Applied Load (N) Theoretical deflection Experimental deflection Brass Deflection of Beam, δ (mm) at x = L/2 at x = L/4 at x = L/2 at x = L/4 200 -0.600 -0.413 -0.540 -0.370 400 -1.200 -0.825 -1.190 -0.840 600 -1.800 -1.238 -1.840 -1.290 800 -2.400 -1.650 -2.490 -1.750 1000 -3.000 -2.063 -3.160 -2.220 Applied Load (N) Theoretical deflection Experimental deflection Steel Deflection of Beam, δ (mm)

6 Figure 1 - Experimental deflections of beams Figure 2 - Theoretical deflections of beams -10.000 -9.000 -8.000 -7.000 -6.000 -5.000 -4.000 -3.000 -2.000 -1.000 0.000 0 200 400 600 800 1000 1200 DEFLECTION (MM) LOAD (N) EXPERIMENTAL DEFLECTIONS Brass Steel Aluminum -9.000 -8.000 -7.000 -6.000 -5.000 -4.000 -3.000 -2.000 -1.000 0.000 0 2 0 0 4 0 0 600 800 1000 1200 DEFLECTION(MM) LOAD (N) THEORETICAL DEFLECTIONS Brass Steel Aluminum

7 For a cantilever taken at distances ? = ? and ? = 𝐿 2 , the formula for the deflection is ? = 𝑃 6𝐸𝐼 (? 3 − 3?? 2 ) A brass bar at distance ? = ? and weight of 311.07g, the theoretical deflection is ? = (. 31107 × 9.81) 6(105000) ( 19.31(3.39) 3 12 ) (247.5 3 − 3(247.5) 3 ) = −2.911 𝑚𝑚 Table 6 - Deflection of cantilever brass Table 7 - Deflection of cantilever steel at x = L/2 at x = L at x = L/2 at x = L 111.3 -0.325 -1.041 -0.450 -1.330 210.25 -0.615 -1.967 -0.800 -2.410 311.07 -0.910 -2.911 -1.210 -3.620 414.38 -1.212 -3.877 -1.560 -4.680 513.45 -1.501 -4.804 -1.970 -5.900 Applied Load (gm) Theoretical deflection Experimental deflection Brass Cantilever Beam Deflection, δ (mm) at x = L/2 at x = L at x = L/2 at x = L 110.7 -0.165 -0.529 -0.210 -0.600 210.7 -0.315 -1.007 -0.400 -1.170 312 -0.466 -1.491 -0.590 -1.740 413.12 -0.617 -1.974 -0.780 -2.330 514.9 -0.769 -2.460 -0.980 -2.900 Steel Cantilever Beam Deflection, δ (mm) Applied Load (gm) Theoretical deflection Experimental deflection

9 Figure 4 - Theoretical deflections of cantilever beams DISCUSSION & CONCLUSION Comparing the theoretical values to the experimental values, we can see that the values are close to each other. However, even small differences can have significant impacts. Looking at the load vs deflection graphs, while the steel and aluminum curves resemble closely, the curve for brass is much steeper for the experimental data. It can be said that the accuracy of the data depends on the type of material a lot more rather than beam configuration and load machine. The reason is, for example, despite a beam being of the same material, say aluminum, there are many different types of aluminum, each with their own modulus of elasticity. This could explain why the theorical values for brass differ from the experimental, the bar could be a small variation of brass. As for the modulus of elasticity, for the simply supported beam, we have similar trend, where the values for aluminum and steel are much closer to the real value, as opposed to brass where the calculated modulus of elasticity has a 24.03% difference. For the cantilever beams, brass is still the lowest specimen with a difference of 21.71%, while steel and aluminum have around 18% difference each to their published values. Many of these discrepancies can be due to the amount of wear the beams have already been through. -2.000 -1.800 -1.600 -1.400 -1.200 -1.000 -0.800 -0.600 -0.400 -0.200 0.000 0 1 2 3 4 5 6 DEFLECTION (MM) LOAD (N) THEORETICAL CANTILEVER DEEFLECTIONS Brass Steel Aluminum

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