Summary/Introduction The purpose of this project was to find at what critical weight or load being applied to one end of a column or a vertical beam will cause the beam to buckle or lose its shape. This project was composed of four different scenarios of how a column can be loaded. The four are both ends being pinned, both ends being fixed, one end being fixed and the other pinned while the beam had a variable cross section, and the beam lying on an elastic foundation. During each different scenario, we were asked to find the general solution to the differential equation (DE) that gave us the deflection of the beam and to try to solve for as many constants as we could with our given initial conditions (IC). We were also asked to find the critical load at which the beam will buckle. This project thought us that different settings will cause different deflection results on a column along with different critical loads. Results Throughout the project, we made the assumption that the beam had a negligible mass that would not affect the deflection. In real life, the weight would lower the critical load that makes the beam buckle. The solution of the DE, v(x), throughout is the path of deflection of the beam due to a load being applied. Part I: Simply Supported Ends In this scenario, the ends of the beam cannot move laterally or vertically but are allowed to bend. All calculations for this problem can be found in Appendix A. The deflection of our column could be found by solving the DE in Equation (1). EI v"+Pv=0; v(0)=v(L)=0 (1) Part I, Exercise 1: We found the general solution of the DE. In this part of the project and in the next, we assumed that the elastic modulus of the beam, E, and the area moment of inertia, ... ... middle of paper ... ...e variables r=5 cm into I and E=200*109 GPa, I, and k=1,000,000 N/m into β. We set P=1 for Equation (23) in order to graph the deflection of the beam in Figure 5. v(x)=e^(-0.710371x) {P/(8*〖0.710371〗^3 )∙cos(0.710371x)+P/(8*〖0.710371〗^3 )∙sin(0.710371x) } (23) Figure 5. Deflection of beam IV.3 which sits on a bed of springs. Part IV, Exercise 4: This exercise required us to find v(0) and v’’(0) then use MATLAB and ode45 to solve the IVP and compare it to our graph in IV.3. We get v(0)= 0.348702392610958 and v’’(0)= -0.351928938921180. Using the ode45 solver on MATLAB we acquired a graph of v(x). Figure 6. Deflection of the beam found using ode45. The two graphs do not seem to be different. They have the same shape and are situated on the same scale. They are not shown to be unstable and only small sections of the graphs actually deviate from each other.
A connecting rod subjected to an axial load F may buckle with x-axis as neutral axis in the plane of motion of the connecting rod, {or} y-axis is a neutral axis. The connecting rod is considered like both ends hinged for buckling about x axis and both ends fixed for buckling about y-axis. A connecting rod should be equally strong in buckling about either axis [8].
Both models share the turning point of (1.77, 5.044) for the larger parabola and (1.65, 3.94)
Since the potential energy of the counterweight and the kinetic energy of the projectile are the same, the equations can be rearranged to
In section II of this paper, theoretical background relevant to this problem is presented. Section III is a brief summary of the numerical data from Giorgini, Boronat, and Casulleras.
Stringent seismic criteria related to construction in the San Diego area made it difficult for Kahn's structural engineer to convince local building officials, who wanted him to use steel frame, that concrete, Vierendeel truss system would have the required flexibility. They agreed only after a 400 page report of undoubtedly integrated deflection computations that shows how post-tensioned columns would provide the main resistance to lateral seismic forces. These columns absorb both dead and live load compression plus vertical post-tensioning forces. They were also designed to maintain zero tension if subjected to lateral movements by earthquake. The trusses are 9 ft deep, spaced 20 ft on center and have a clear span of 65 ft (diagram 2). He made use of the 9 ft high resultant space as service area, allowing pipe chases to be dropped to the 65x 245 ft floor below with more latitude than before.
In the previous section, the governing equation of the dynamic and stability behavior of the nanobeam are derived. The Eq. (19) and Eq. (20) are the fourth order partial differential equations which are obtained as the governing equation of the vibration and buckling of the nanobeam, respectively. If it is not impossible to solve these equations as analytically, it is very hard to solve these equations as exact solutions. For this purpose, for computing the vibration frequencies and the buckling loads, the differential quadrature method is selected. The real reason of this selection is because that this method is one of the useful methods to solve the ordinary and partial boundary value and initial
Laws such as the lever law and Euler’s Buckling Theorem come into play when testing and competition begins. A structure of wood and glue surely has much more to offer than meets the eye.
The Volume Library, vol. I, Physics: Newton's Law of Motion. Pg. 436. The Southwestern Company, Nashville, Tennessee, 1988.
In this paper, the theory of mass moment of inertia are reviewed and discussed. The experimental results and theoretical values also included and discussed. Finally, comparison between the theoretical values and experimental values were discussed and deducted.
Investigating the Factors Which Will Affect the Stretching of a Helical Spring when Put Under a Load.
-In order to solve this differential equation you look at it till a solution occurs to you.
From the graphs, it could be determined that the results are fairly precise due to the data points being fairly close rather than scattered. This can be seen as all the points are relatively close to the average value on graphs 1 and 2, not causing the trend line to be skewed.
Angular projectile motion is used to calculate how far an object with an initial velocity that is projected at an angle to the horizontal will travel horizontally. It can also be used to calculate the maximum height reached by the object and how long it was in the air for. When solving angular projectile motion problems, one must consider the following steps. To begin with one must calculate the horizontal acceleration of the object, keeping in mind that the vertical acceleration is 9.8 m/s2 due to gravity. In most cases one is given the angle of the ramp to the horizontal, and the velocity. If not given, the velocity can be calculated using the object’s acceleration at a given moment in time. One must then calculate the vertical and horizontal components of the velocity using a vector diagram. To represent the problem, draw a rough diagram of the problem. Before one calculates the horizontal displacement of the object, one must find the time the object was in the air for.