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Structural Damage Detection Using Modal Strain Energy - Assignment Example

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"Structural Damage Detection Using Modal Strain Energy" paper argues that the deformation ability of a structure affected by the material length, cross-sectional area, and the type of the material. Some of the applications of strain energy include the application in uniaxial loads and torsion loads…
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Introduction Strain energy can be defined as the energy a body poses due to its deformation that is either plastic or elastic. The best way to express this energy is to express it per unit volume. Releasing of the force applied to a body makes it to regain its original shape. It results to lose of strain energy from the body. Strain energy is denoted by U (Cha & Buyukozturk 2015, 350). The general formulae for determining strain energy is given by U = Fδ / 2 Where δ = compression, F = force applied. Considering the young’s modulus of elasticity the formulae for strain energy is given as follows. U = σ2 / 2E × V. Where, E = young’s modulus σ = stress, V = volume of body. Task Discussion Question 1 g i. Strain energy applied in uniaxial loads In this type of application, one can consider a load that is prismatic which is subjected to a tensile force P with a longitudinal length L. the load is uniformly applied to the load ensuring there is no motion of the body during force application process. This type of load is known as a static load. After full loading, the value of P the bar elongates to a distance given by l+d (Cha & Buyukozturk 2015, 350). During the process, tensile force P moves along the bar doing some work over a distance given by d. Therefore the amount of work done in the distance d is given by the product of the applied force P by the distance through which it travels d as summarized in the equation below. Where W= work done by the tensile force P F= force applied d= distance of application of the force. However, we realize that the force magnitude varies from a value of zero to p. To be able to find the amount of work done in this range, a displacement diagram is used to determining the variations in the force applied (Cha & Buyukozturk 2015, 351). The work projected to the load is given by the area under the curve. Application of force increases strain within the bar causing a lot of energy to be forced out of the bar. Strain energy can be defined as the type of energy absorbed by the bars as a result of the load being deformed. However, we know that the principle of conservation of energy that energy is neither created o nor destroyed but converted from one form to the other. Our energy, in this case, was equal to the work done on the load assuming there are no energy losses within the system. (Cha & Buyukozturk 2015, 353). This type of energy can be defined as internal work. The units of internal energy and that of work are the same. Gradual reduction of the applied force will result in the shortening of the bar as well as reduction of strain energy being absorbed into the bar. However, the bar may only return to its original limitations if it had not reached its elastic limit during deformation beyond which it never fully recovers its original dimensions. Basing on Hooke’s law in the displacement curve the strain energy within the material is given by The complimentary energy is applicable in Castiglione’s theorems applicable to most of the linear elastic systems. It can also be used in analyzing energy systems. To determine the total strain energy value in the bar on has to sum the strain values at the different sections of the bar. However one should remember that summation of strain energy at a different section of a bar under loads is not equal to the individual weights, strain energy values. When using the curve to determine the value of strain energy, the value obtained may not reflect the properties of the material since the curve is dictated by the size of the specimen. To do away with the size factor strain energy is analyzed per the volume and is given by the equation Toughness can be defined as the area under the stress-strain curve representing the limits through which a material can absorb strain energy before fracturing. The energy may only be recovered on condition that the material was stressed to a given point A resulting to the triangle ABC. AB and OD are parallel lines because most of the materials behave similarly once the stress is released. OABO represents the inelastic (dissipated) strain energy. ii. Strain energy being applied in torsion Considering a prismatic type of bar under the influence of torsion forces. The bar is being acted upon by a torque of value T. application of the load makes the bar to twist with one end that is free rotating at a certain angle. If the material is found to be elastic and obeys hooks law, then the relationship between the torque T and the force F will be a linear one (Muvdi & McNabb, 2012). But Then U = SL T(x)2/2GIp dx = T2L/2GIp = GIpf2/2L Under nonuniform torsion action the bar the total value of strain is given by the summation of the strain energies at the torque acting sections of the bar. However, the summation of individual strain energy loads is not equal to the total value of strain that is exerted at different sections of the bar. iii. Strain energy applied in pure shear Considering the above element with the dimensions as x,y,z under the influence of a shear load whose value is given by t. During the deformation process, the top side reaches a maximum value that is given by txz (Muvdi & McNabb, 2012). The total displacements for such a small force is given by gy therefore W = U = 1/2 t xz * gy = 1/2 tg V = 1/2 t2V/G The strain energy density in this situation is given by the following equation u = 1/2 tg = 1/2 t2/G iv. Strain energy applied in bending The material obeys Hooke`s law with small rotations. The variation of the normal stress concerning the neutral axis is a linear one. This can be expressed in the equation below. U = M2L/2EI But q = ML/EI U = EIq2/2L The total strain energy will be given by the following equation U = SL M(x)2/2EI dx = SL EI/2 (d2v/dx2)2 dx The introduction of shear forces allows the beam to absorb more strain energy required to contain the shear forces action. However in a situation where the value of L is greater than t the amount of strain energy absorbed is neglected since it is small. Question 3 b (i) and (ii) The experiment was set up as shown in the figure below. In this experiment, the original location of each point was determined from a reference point. The reference points used in this experiment included: i. The edge bench in which the frame was clamped, see figure 1 below ii. Original location of the points A and B Figure 1: Experimental set up that was used to measure deflection Apparatus Cantilever beam that is L shaped, a clamp, weights of 20 N, micrometer screw gauge, stainless steel meter rule iii. Measurement deflection at points A and B Initial location of points A and B (Location before loading) The original (initial) location of point A from the edge of the workshop bench was measured. Its location from point B was also measured. Original (Initial) location of A from the edge of the workshop bench = 0.4 m. Original location of A from point B = 0.2 m The original location of point B from the edge of the workshop bench was measured. Its location from point A was also measured. Original location of A from edge of workshop bench = 0.4 m. Original location of A from point B = 0.2 m Location of Points A and B when loading the frame Original location of Points A and B after loading the frame as shown in figure 1 above. The frame was then loaded and the new locations of points A and B measured. The results for measurement of new locations of point A during loading were as follows. Final location of A from edge of workshop bench (horizontal measurement) = 400.00357mm The new location of point B from point A was also measure, the results are shown below. Location of B from point A during loading = 200.00392 mm Calculating deflection from the measurement Deflection at each of the locations was calculated as follows: Deflection = Final Location from reference point – Initial (original) location from the reference point Horizontal deflection at point A, y1 = 400.00357 – 400 = 0. 00357 mm. Vertical deflection at point B, y2 = 200.00392 – 200 = 0.00392 mm Question 3 c Procedure a. The horizontal and vertical dimensions of the beam were measured with the help of the stainless steel rule. This was done to confirm whether the frame had the indicated dimension specifications b. The individual weights were confirmed by an electronic weighing scale to cross check for any inaccuracies that may be carried into the experiment. All the weights were confirmed to be either 20 N. c. The frame was then clamped to the working bench with the help of a clamp provided in the laboratory. d. In this experiment, the maximum value for the force 2 F2 was at 200N while that of force one F1 was taken to be 180N. e. Loadings were done t both ends of the frame with a range of 20 N being added at both points A and B simultaneously. f. When the weights were being added at each given time the system was allowed to settle with the most consistent reading of the measuring device is noted. The settled point provided the actual deflection caused by both the weights on the two dimension of the frame. g. The deflection readings both on the vertical axis (y) and the horizontal axis ( l ) were read by at least three different people to help in minimizing the errors due to parallax. h. The readings were tabulated in the table for all the loadings until the maximum loading for both ends had been obtained. A/A Diameter of the frame (mm) Force 1 (F1) Loading in Newton Horizontal reading L (mm) Horizontal deflection l (mm) Force 2 (F2) Loading in Newton Vertical scale reading y (mm) Vertical deflection y (mm) 1 15 20 20 2 15 40 40 3 15 60 60 4 15 80 80 5 15 100 100 6 15 120 120 7 15 140 140 8 15 160 160 9 15 180 180 Results A/A Diameter of the frame (mm) Force 1 (F1) Loading in Newton Horizontal reading L (mm) Horizontal deflection l (mm) Force 2 (F2) Loading in Newton Vertical scale reading y (mm) Vertical deflection y (mm) 1 15 20 4.99 4.99 20 5.21 5.21 2 15 40 10.12 5.13 20 10.98 5.77 3 15 60 15.9 5.78 20 16.2 5.22 4 15 80 21.1 5.2 20 21.6 5.4 5 15 100 26.7 5.6 20 26.95 5.35 6 15 120 31.8 5.1 20 31.4 4.45 7 15 140 36.4 4.6 20 36.2 4.8 8 15 160 41.3 4.9 20 41.6 5.4 9 15 180 47.2 5.9 20 46.7 5.1 Calculations and discussions Average deflection in the horizontal axis of the frame = 5.24444444 Average deflection in the vertical axis of the frame = 5.188889 The total deflection of the frame in the horizontal axis l was 47.2 mm = 0.472m whereas the total deflection on the vertical axis was at 0.467 m. The frame made an average deflection of 0.4695 meters for the force of 180N acting horizontally and vertically. The graphs summarizing the two deflections are as shown below. Graph showing the deflection of the frame along the horizontal axis. From the trend of the graph of deflection of the frame along the horizontal axis, we realized that it obeyed Hooke`s law. This is because the deflection was directly proportional to the force acting in the same direction. An increase in force applied resulted in a corresponding increase in the value of deflection reading on the measurement scale. Graph for deflection of the frame along the vertical axis The frame also obeyed Hooke`s law as the deflection was directly proportional to force applied to the frame. Deflection of a Beam will be dependent on its length, the material as well as the shape of the beam. In the experiment, the horizontal deflection value was higher than that of the vertical axis deflection. This is because it was longer than the latter. The longer the beam, the greater the degree of deflection of the work being done on the beam increases as the distance of action of the force increases. The vertical axis of the beam was shorter compared to the horizontal axis. Therefore, the work being done on the beam was less having in mind that work is a product of force by the distance involved. The material of the frame was responsible for the averagely high value of deflection that was obtained from the experiment. The experimental value of 0.4695 slightly differed from the theoretical value obtained from calculation of. This may be attributed to the experimental sources of errors (Taranath 2016). Conclusion At the end of the experiment, the primary objective of relating force applied to the beam and deflection was obtained. The two variables were directly proportional to each other. The value of deflection increased with a corresponding increase in loading. The three factors affecting deflection of the frame were the length dimensions, its material as well as the crossectional area of the frame. This explains why the longitudinal dimension had a higher degree of deflection compared to the vertical dimension. The variations between the experimental and theoretical values obtained by mathematical computation were as a result of experimental sources of errors. Question 3d The slight variation in the values of deflection between theoretical calculations and the experiment was attributed to the various sources of experimental errors as discussed below. Some of the weights were either slightly above or below the required 20N for the experiments. The measurement device had a slight negative error which had to be carried forward in the experiment. Errors during recording and taking the different measurements during the experiment could have occurred. An example is parallax error when reading from the measurement scale. Ignoring the experimental procedure where the system was to settle before the reading was taken completely. The other possible source of error could have occurred during data values calculation and analysis. Avoiding these sources of errors is instrumental in obtaining future accurate results (Langhaar 2016). Conclusion Strain energy is the energy stored by a body under deformation. Deformation ability of a structure is affected by length of the material, cross sectional area as well as the type of the material. Some of the applications of strain energy include the following application in uniaxial loads as well as torsion loads. Reference Cha, Y.J. and Buyukozturk, O., 2015. Structural damage detection using modal strain energy and hybrid multiobjective optimization. Computer‐Aided Civil and Infrastructure Engineering, 30(5), pp.347-358. Muvdi, B.B. and McNabb, J.W., 2012. Engineering mechanics of materials. Springer Science & Business Media. Taranath, B.S., 2016. Structural analysis and design of tall buildings: Steel and composite construction. CRC Press. Langhaar, H.L., 2016. Energy methods in applied mechanics. Courier Dover Publications. Read More

This type of energy can be defined as internal work. The units of internal energy and that of work are the same. Gradual reduction of the applied force will result in the shortening of the bar as well as reduction of strain energy being absorbed into the bar. However, the bar may only return to its original limitations if it had not reached its elastic limit during deformation beyond which it never fully recovers its original dimensions. Basing on Hooke’s law in the displacement curve the strain energy within the material is given by The complimentary energy is applicable in Castiglione’s theorems applicable to most of the linear elastic systems.

It can also be used in analyzing energy systems. To determine the total strain energy value in the bar on has to sum the strain values at the different sections of the bar. However one should remember that summation of strain energy at a different section of a bar under loads is not equal to the individual weights, strain energy values. When using the curve to determine the value of strain energy, the value obtained may not reflect the properties of the material since the curve is dictated by the size of the specimen.

To do away with the size factor strain energy is analyzed per the volume and is given by the equation Toughness can be defined as the area under the stress-strain curve representing the limits through which a material can absorb strain energy before fracturing. The energy may only be recovered on condition that the material was stressed to a given point A resulting to the triangle ABC. AB and OD are parallel lines because most of the materials behave similarly once the stress is released.

OABO represents the inelastic (dissipated) strain energy. ii. Strain energy being applied in torsion Considering a prismatic type of bar under the influence of torsion forces. The bar is being acted upon by a torque of value T. application of the load makes the bar to twist with one end that is free rotating at a certain angle. If the material is found to be elastic and obeys hooks law, then the relationship between the torque T and the force F will be a linear one (Muvdi & McNabb, 2012).

But Then U = SL T(x)2/2GIp dx = T2L/2GIp = GIpf2/2L Under nonuniform torsion action the bar the total value of strain is given by the summation of the strain energies at the torque acting sections of the bar. However, the summation of individual strain energy loads is not equal to the total value of strain that is exerted at different sections of the bar. iii. Strain energy applied in pure shear Considering the above element with the dimensions as x,y,z under the influence of a shear load whose value is given by t.

During the deformation process, the top side reaches a maximum value that is given by txz (Muvdi & McNabb, 2012). The total displacements for such a small force is given by gy therefore W = U = 1/2 t xz * gy = 1/2 tg V = 1/2 t2V/G The strain energy density in this situation is given by the following equation u = 1/2 tg = 1/2 t2/G iv. Strain energy applied in bending The material obeys Hooke`s law with small rotations. The variation of the normal stress concerning the neutral axis is a linear one.

This can be expressed in the equation below. U = M2L/2EI But q = ML/EI U = EIq2/2L The total strain energy will be given by the following equation U = SL M(x)2/2EI dx = SL EI/2 (d2v/dx2)2 dx The introduction of shear forces allows the beam to absorb more strain energy required to contain the shear forces action. However in a situation where the value of L is greater than t the amount of strain energy absorbed is neglected since it is small. Question 3 b (i) and (ii) The experiment was set up as shown in the figure below.

In this experiment, the original location of each point was determined from a reference point. The reference points used in this experiment included: i. The edge bench in which the frame was clamped, see figure 1 below ii. Original location of the points A and B Figure 1: Experimental set up that was used to measure deflection Apparatus Cantilever beam that is L shaped, a clamp, weights of 20 N, micrometer screw gauge, stainless steel meter rule iii.

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