Behaviour of Structures: Deflection of Pin-jointed Structures - Assignment Example

Name: Course: Title: Deflection of Pin-jointed Structures Task: Date: Deflection of Pin-jointed Structures Task 1 Introduction Energy methods assume that the total work in a system is equal to the sum of internal work and the external work. Suppose that the rigid body is given small, imaginary (virtual) displacement Δv, then the total virtual work done , is equal to the sum of virtual work done by internal forces and virtual work done by external forces, . Therefore, (Equation 1) The total work is assumed to be zero, thus it is taken that the sum of internal work and external work is zero. Defining work as the force applied at the same direction of displacement. (Equation 2) Where Fe is the external force applied to the system, δ is the total deflection of the system due to and in the direction with Fe, ni is the internal force applied to member and Δi is the deflection of member due to and in the same direction not with ni, but rather with the original load, If an external load W is gradually applied on the pin frame, it produce a displacement of distance δ, the load moves a distance y. the internal force P produce an extension δl in the frame. The external work done must be equal to the internal energy stored in the structure. Then ∑Wy/2 = ∑Pδl/2 (Equation 3) To the unloaded structure a unit virtual\load is applied in the direction of δ, resulting in the force u in any member of the frame. If a real load W is applied to the structure gradually, and equate external work to the internal energy, ∑Wy/2 + 1 x δ= ∑Pδl/2 + ∑ul/δl (Equation 4) Subtracting the above equations, 1 x δ = = (Equation 5) Where P is the internal force applied by loads and A, l and E are the area, length of the frame and the modulus of elasticity respectively and u is the internal force due to virtual load. For a pin jointed frame, the virtual\load is applied to the frame expecting a deflection which is positive. If the force is applied in the opposite direction to the deflection, the deflection will be negative (Williams, 2009). The disadvantage of work-energy method, of equating the internal strain energy to external work is that normally only the deflection due to a single force can be obtained. Virtual work method gives procedure of determining deflections and rotations (slopes) at any point in the truss or other structures subjected a number of loadings (Williams, 2009). Objectives: 1. To determine the deflection of a loaded pin-jointed structure experimentally and compare it with the theoretical deflection calculated using virtual work analysis. Materials needed Dial gauge Load hanger 5 – 20 weights Pin – join structure like as shown in figure 1 Methodology 1. The experiment begins with calibration of the gauge by setting it at zero. A very light tap on the front of the screen may be help in calibration. 2. After assembling the load hanger on joint G, the gauge reading was recorded. This was treated as no load. 3. The load equal to 20N was added to the load hanger and the reading recorded. 4. The load of 20N was increased to 100N and the deflection reading for 20, 40, 60, 80 and 100N was record. 5. The deflections of joint G were then plotted against the load. Using linear equation, the deflection of a load of 100N was calculated. 6. The theoretical value of deflection was calculated using virtual work method. The cross section is 20mm2 and E is 205kN/mm2. Results and Discussion Average deflections per weight are given in table 1 Table 1: average deflection per weight Load (N) Loading deflection (mm) Unloading deflection (mm) Average deflection (mm) 0 0 0 0 20 0.031 0.035 0.033 40 0.072 0.067 0.0695 60 0.099 0.098 0.0985 80 0.128 0.129 0.1285 100 0.157 0.157 0.157 As shown in table 1 above, the load increases as the deflection increase. The graph of deflections against the load applied gives a linear curve as shown below. Figure 2. A graph of displacement of trust against the load applied As shown on the graph obtained, the curve obeys a linear function, thus shows that the displacement of trust is still elastic. The line passes through 0.002 mm of y axis. This error is small, and may originate from the measuring instrument. The deflection corresponding to a load of 100N is 0.001x100+0.002 = 0.102 mm. The theoretical value of deflection is obtained a self- developed spread sheet using trust analysis method (Johnston and Beer, 2003) as shown in the table below. A table of the theoretical values and the practical values together with the percentage errors are recorded in the table 3 below. Table 3: Experimental and computed deflection values Load (N) Displacement (mm) Theoretical Experimental 0 0.00 0.000 20 0.073 0.074 40 0.146 0.139 60 0.220 0.189 80 0.293 0.229 100 0.366 0.279 From the table we can see that the error increase with the increase in load. This shows that the energy method is more accurate with small loads. Comparing the theoretical value of deflection at a load of 100N, the theoretical deflection is 0.366mm and the practical value is 0.102 mm which gives an error of27.9 %. The graph obtained from theoretical and practical values, as shown in figure 3 below, shows that the error is quite large at higher loads. This can be assign of the presence of systematic errors. The sources of errors can also be due to the presence of modulus of elasticity and assumptions. Figure 3: A displacement graph for actual and theoretical trust Conclusion The deflection of loaded pin-jointed structure can be determined practically or through a computation using virtual work analysis. The values from the experiment differ from the theoretical virtual work values. For example, at a load of 100 N the error experimental value is less by 27.9%. The errors increase with increase in load, which shows that systematic errors are present. This practical also shows that the deflections are more reliable with small loads. Task 2 Assume the initial shape of the strut can be described by the mathematical expression: , whereis the maximum central amplitude. Figure 4: A strut with a maximum central amplitude The elastic bending behaviour of the strut can be expressed mathematically with a differential equation, . Where P is the force applied. With the boundary condition, the solution of the equation would be; (Equation 6) This equation can be rearranged and substituted with y0 to give: (Equation 7) And (Equation 8) If y* is the additional deflection due to compressional load equals to ymax-a0, then: Or (Equation 9) The slope of a graph of y* against y*/P is equal to Euler buckling load and the y intercept gives the amplitude of the initial curve a0. Using the boundary conditions and the differential equation, the additional central deflection of the strut can be: (Equation 10) Where e is the eccentricity of the load. The straight line gives Euler buckling load, and intercept is equal to (Chilver et al, 1999) Objectives: 1. To experimentally investigate the behaviour of eccentrically loaded struts with initial curvature. 2. To experimentally determine Euler buckling load using Southwell plot. 3. To compare the experimental and theoretical values of the Euler buckling load Materials Strut with the measurement of 600 x 25.4 x 1.6 mm HST Buckling equipment as shown in figure 4 below: Methodology 1. Set-up the strut in the frame as in figure 5. Parts HST 1509, the horizontal HST1511 and HST 1506 are not included in the experiment. 2. Make sure the two knife edge brackets (HST.1505) in the groove corresponding to zero eccentricity. 3. Set the counterweight on the loading arm (HST1501) such that the weight of the arm is not acting on the strut. 4. Attach the 100g (0.98) load hanger. The weight 0.98N should be added to weights recorded for the experiment. 5. Set the scale (HST 1503) to zero. 6. Add weights of order 20N+5+5+1+1+1+1+0.5N to the hanger and record the corresponding deflections. 7. Compute the actual load acting on the strut. Note that the lever arm magnified the weights and the force acting on the strut = the weight on the hanger x (1000/750). 8. For maximum eccentricity repeat steps 3 -7 while ensuring that the knifes edges (HST 1505) are in the groove correspond to maximum eccentricity. Results The table 4 below shows the theoretical Euler buckling load values for the strut using equation 7. The moment of inertia is and for the weak axis. Therefore, the theoretical Euler buckling load is taken as 52.47N. Table 4: Euler buckling load for Euler buckling load computed using Axis PE Strong axis 12,279.79 Weak axis 52.47 The data obtained in the zero – eccentricity experiment are as shown in table 5 below. Table 5: data for zero eccentricity experiment Load, N Displacement, mm Y*/P 20.1 1.0 0.05 25.1 1.5 0.06 30.1 2.5 0.08 31.1 4.5 0.14 32.1 6.0 0.19 33.1 9.0 0.27 34.1 17.0 0.5 34.6 43.0 1.24 From the table, a graph of y* against y*P is obtained as in figure 6 below. It can be seen from the linear equation that the actual Euler Bulking is equal to 35 N, which has an error of 33.3% with the theoretical value of 52.47 N. Figure 6: A graph of y* verses y*/P for the Zero-eccentricity experiemnt A graph of y* vs P as shown in figure 7 below. The line produced in the graph is more like an exponential than linear. Figure 7: A graph of y* vs P The data obtained for maximum eccentricity experirmnt are recorded in table 6. The figure 7 below shows a graph of y* against y*/P for the experiment. Table 6: Data for maximum eccentricity experiemnt Load, N Displacement, mm ymax - a y*/P 0 7 0 0 10.1 10 3 0.3 15.1 13 6 0.40 20.1 19 12 0.60 25.1 30 23 0.91 26.1 33 26 1.00 26.6 36 29 0.81 Figure 7: A graph y* vs p From the graph, the Euler Buckling load for the maximum eccentricity experiment is 33.1 N, which has an error of 36.9% compared to the theoretical buckling load. The Euler buckling for loads with eccentricity should however be lower, because they develop moments increase the bulking of the column (Chilver et al, 1999). As shown from figure 8, it can be seen that the eccentrically loaded column has higher displacement at a lower load. This means that maximum deflection may be reached at lower loads. Figure 8: A graph of deflection against the applied load Conclusion The Euler Bulking derived from obtained from the experiment differs at maximum of 33.3% with the theoretical value. The presence of errors can be from measurements. The experiment confirms that the eccentric load beams deflect more than the axial load beams. Bibliography Chilver H. C., Case J., Baron and Ross C T F, (1999). Strength of materials and structures, Oxford; New York: Butterworth-Heinemann Fenner R. T., Reddy J.N., (2012). Mechanics of Solids and Structures, Second Edition Computational Mechanics and Applied Analysis, CRC Press Pandit, (2001). Structural Analysis: a Matrix Approach, Tata McGraw-Hill Education Ramaswamy G. S., Eekhout M., Suresh G. R., et al, (2002). Analysis, design and construction of steel space frames, London Thomas Telford Williams A., (2009). Structural analysis: in theory and practice, Amsterdam; Boston: Elsevier/Butterworth-Heinemann; [Washington, D.C.]: International Code Council Appendix Culculation for virtual work Read More