Superfluous Reinforcement Undertakings Fundamental Engineer - Research Paper Example

Redundant Truss Experiments Name Institution Table of Contents Introduction 4 Objective 4 Learning Outcome 4 Procedure 4 Results and Calculations 5 EXPERIMENT 1 5 Experimental forces 9 Member one 9 Member two 9 Member three 10 Member four 10 Member five 10 Member seven 10 Member eight 11 Theoretical force 11 Joint A 11 Joint D 12 Joint C 12 Joint B 13 EXPERIMENT 2 13 Experimental forces 17 Member one 17 Member two 17 Member three 17 Member Four 18 Member five 18 Member six 18 Member seven 18 Member eight 18 Discussion 19 Question one 19 Question two 20 Question three 20 Question four 20 Question five 20 Conclusion 21 References 22 Redundant Truss Experiments Introduction In the experiment of a redundant truss a framework used as a test frame has two members; one is the redundant member that can be removed from the framework without compromising the safety and stability of the structure. The other member has central ring that allows the redundant member to place itself into the framework. The frame work has two support one pinned and the other free or roller support. Objective The objectives of the experiment were to observe the effects of redundant members of a structure and to understand the method used to analyze such type of structure Learning Outcome Learn how to apply structural engineering knowledge in practical situations To use laboratory applications to enhance structural engineering technical competency Procedure Before the start of the experiment the thumbwheel up the boss on the redundant member was hand-tighten after being wind. No tools were used on the thumbwheel during the tightening process A 100N per-load was applied downwards, the load cell was re-zeroed and the digital indicator was then carefully zeroed. A 250 load was applied carefully and checked to ensure the frame was secure and stable. The load leaving the preload of 100Nn to zero was returned. The digital indicator was rechecked and then re-zeroed. Loads of higher values than the specified (250N) were never applied. Load increments were applied as shown in table 1, and the indicator and strain readings were recorded. The initial strain readings(zero) were then subtracted and recorded in table 2 The equipment member force at 250N were then calculated and entered in the table 3 The procedure was repeated for task A A graph of load vs. deflection with reference from table 1 was then plotted The flexibility method was then used to calculate the theoretical forces while Young’s Modulus was to calculate the equivalent member forces at 250N and table three was then filled. Results and Calculations EXPERIMENT 1 Member Strains (µɛ) Load Displacement Experimental Strain 1 Experimental Strain 2 Experimental Strain 3 Experimental Strain 4 Experimental Strain 5 Experimental Strain 6 Experimental Strain 7 Experimental Strain 8 Rod Diameter Material N mm me me me me me me me me mm GPa 0 0.001 -0.2 0.0 0.1 0.1 -0.1 --- -0.1 -0.2 6.00 210.0 50 0.024 8.8 -8.8 -8.7 -16.9 -0.6 --- 12.1 12.9 6.00 210.0 99 0.040 17.4 -17.6 -17.6 -33.7 -0.2 --- 24.3 24.8 6.00 210.0 151 0.056 26.9 -26.9 -26.7 -51.9 -0.9 --- 37.5 37.9 6.00 210.0 201 0.073 36.6 -36.4 -36.2 -70.1 -0.8 --- 51.0 51.0 6.00 210.0 251 0.084 44.4 -45.0 -44.5 -86.1 -1.3 --- 62.9 62.4 6.00 210.0 Table 1 A: Strain readings and the frame deflection The strain readings and the digital indicator readings were recorded after the load increments were applied Member Strains (µɛ) Load Displacement Experimental Force 1 Experimental Force 2 Experimental Force 3 Experimental Force 4 Experimental Force 5 Experimental Force 6 Experimental Force 7 Experimental Force 8 Rod Diameter Material N mm N N N N N N N N mm GPa 0 0.001 -0.9 -0.1 0.4 0.8 -0.5 --- -0.5 -1.4 6.00 210.0 50 0.024 52.3 -52.5 -51.5 -100.6 -3.5 --- 72.0 76.5 6.00 210.0 99 0.040 103.3 -104.7 -104.6 -200.1 -1.2 --- 144.4 147.0 6.00 210.0 151 0.056 159.7 -160.4 -159.0 -308.0 -5.6 --- 224.3 225.3 6.00 210.0 201 0.073 216.7 -215.5 -214.4 -415.2 -5.2 --- 302.4 303.5 6.00 210.0 251 0.084 263.6 -266.2 -263.3 -511.0 -7.8 --- 373.6 369.8 6.00 210.0 Table 2 A: True strain readings The true strain readings were calculated by subtracting the recorded readings in table one from the initial readings of each member. It is a graph-showing load vs. strain and true strain readings for member number one It is a graph-showing load vs. strain and true strain readings for member number four E = E = Young’s Modulus = Stress in the member = Displayed strain = F = Force in member (N) A = cross sectional of member E = F = EA A = A = = 28.274 E = 2.10 Experimental forces Member one = 263.6 F = 2.10 28.274 263.6 F = 1565.13 N Member two = -266.2 F = 2.10 28.274 -266.2 F = -1579.38 N Member three = -263.3 F = 2.10 28.274 -263.3 F = -1563.35 N Member four = -511.0 F = 2.10 28.274 -511.0 F = -3034.08 N Member five = -7.8 F = 2.10 28.274 -7.8 F = -46.31 N Member seven = 373.6 F = 2.10 28.274 373.6 F = 2218.26N Member eight = 369.8 F = 2.10 28.274 369.8 F = 2195.70N Theoretical force Virtual work method =0 =Fx =Fy Consider moment at B B =0 -HA (1) + 251(2) = 0 HA = 502 N x =0 HA + HB = 0 HB = -502 N y =0 -VB -250 = 0 VB = 251 N Joint A y = y FAB = 0 x = x FAC = -502 KN (C) Joint D y = y FyED = 250 x = x FCD = -FxED -250 KN (C) FXED = 250 FED = 355 KN Joint C y = y FCE = FyBC = 0 = -251 KN x = x FCD = -FAC = FYbc - 251 N = - 502 + FxBC FxBC = 251 N FBC = 355 KN Joint B y = y 250 = FBC + FAB FXbc = 250 x = x FBE = 500 – FxBC FBE =500 – FxBC = 250 KN Member Experimental force Theoretical force 1 1565.13 N 251 2 -1579.38 N -251 3 -1563.35 N -251 4 -3034.08 N -502 5 -46.31 N 0 7 2218.26 N 355 8 2195.70 N 355 EXPERIMENT 2 Load Displacement Experimental Strain 1 Experimental Strain 2 Experimental Strain 3 Experimental Strain 4 Experimental Strain 5 Experimental Strain 6 Experimental Strain 7 Experimental Strain 8 Rod Diameter Material N mm me me me me me me me me mm GPa 1 0.000 0.0 0.0 -0.2 -0.1 0.1 -0.1 0.1 -0.1 6.00 210.0 49 0.019 12.0 -5.0 -8.6 -13.1 3.4 -5.0 11.9 7.3 6.00 210.0 99 0.034 24.4 -10.0 -17.2 -26.1 6.7 -10.5 24.1 14.2 6.00 210.0 150 0.047 37.6 -15.2 -26.4 -39.8 10.4 -16.6 37.2 21.6 6.00 210.0 201 0.062 51.8 -20.4 -35.9 -53.9 14.1 -22.5 50.8 28.6 6.00 210.0 250 0.072 63.4 -25.0 -44.1 -66.3 17.4 -28.1 62.5 35.1 6.00 210.0 Load Displacement Experimental Force 1 Experimental Force 2 Experimental Force 3 Experimental Force 4 Experimental Force 5 Experimental Force 6 Experimental Force 7 Experimental Force 8 Rod Diameter Material N mm N N N N N N N N mm GPa 1 0.000 -0.3 -0.4 0.2 -0.1 0.8 -0.5 0.4 -1.0 6.00 210.0 49 0.019 71.8 -30.4 -50.7 -78.1 20.2 -29.8 70.4 43.2 6.00 210.0 99 0.034 145.1 -59.3 -102.0 -155.1 39.8 -62.1 143.2 84.3 6.00 210.0 150 0.047 224.3 -90.2 -156.8 -237.1 61.1 -97.8 220.7 128.7 6.00 210.0 201 0.062 308.1 -121.3 -212.9 -321.0 83.2 -133.4 301.4 169.2 6.00 210.0 250 0.072 377.0 -149.4 -261.3 -394.2 103.4 -166.5 371.1 208.2 6.00 210.0 It is a graph-showing load vs. strain and true strain readings for member number two It is a graph-showing load vs. strain and true strain readings for member number five It is a graph-showing load vs. deflection that occurring in the members of the framework. E = E = Young’s Modulus = Stress in the member = Displayed strain = F = Force in member (N) A = cross sectional of member E = F = EA A = A = = 28.274 E = 2.10 Experimental forces Member one = 377 F = 2.10 28.274 377 F = 2238.45 N Member two = -149.4 F = 2.10 28.274 -149.4 F = -887.06 N Member three = -261.3 F = 2.10 28.274 -261.3 F = -1563.35 N Member Four = -394.2 F = 2.10 28.274 -394.2 F = -2340.57 N Member five = 103.4 F = 2.10 28.274 103.4 F = 613.94 N Member six = -166.5 F = 2.10 28.274 -166.5 F = 988.60N Member seven = 371.1 F = 2.10 28.274 371.1 F = 2203.42N Member eight = 208.2 F = 2.10 28.274 208.2 F = 1236.19N Member Experimental force Theoretical force 0 312.597 375 0.075 -141.553 -125 0.08 -259.514 -250 0.254 -330.29 -375 0.33 141.553 125 0.404 -182.839 -177 0 383.373 354 0.075 241.820 177 Discussion Question one The experimental values greatly differ from the theoretical values. There could be a number of reasons as to why the two sets of values differ but the main reason could be due to parallax. Equipment’s were not functioning well on that particular day of the experiment or the equipment have not been undergoing proper maintenance, another reason could be that that particular day of the experiment the lab conditions were not conducive to conduct an accurate experiment. Question two Both the theoretical and experiment values agree as to which members of the trust are in compression and which are in tension. From our values we can see that members one, seven and eight are in tension while members two three four and five are in compression. While using the theoretical method we can establish by looking at the at the free body diagram during calculations. If the reaction is facing vertically upwards and it is in tension all the reactions or forces facing downwards are opposing forces and are in compression. Question three While using the theoretical method the reading at member number five is equal to zero. From the free body diagram we can see that both sides of member five are fixed meaning that there is the forces at the joint have to be in equilibrium thus it is equating to zero. Question four Yes the strain gauges can be adequately used to effectively trace the measurement forces in the frame work. There are several mathematical methods adapted to accurately calculate for forces in the framework using strains. These methods give accurate answers like the other methods that do not use strains; hence strains can be adequately utilized. Question five Yes the frame work complies with the pin joint theory despite the fact that the joints are not truly pin jointed. The theory only hold true as long as the connections in the joints are ductile enough to successfully transmit the moment through an alternative way or path. This is the case with the framework the connection behave as though they are pin with springs but the theory only holds when ductility is present in the members but once the maximum load has been reached the theory is no log valid. Conclusion The most important aspect of the experiment was to investigate a structures ability to safely carry loads the limit of the equipment used was 350 N, which was not exceeded. Safety evaluation was done mathematically and then compared to the theoretically obtained values. The differences in the values can be attributed to paradox errors and mistakes in reading the values. The experiment has shown that structural engineers do not have to build life like prototype when evaluating the safety and durability of the structures they are designing. It would be uneconomical unsafe and impractical. With every experiment small differences in the value, because the equipment used and the people conducting the experiment are not one hundred percent accurate. In conclusion, the equipment should be properly maintained to reduce the errors of future experiments. References Beaufait, F. W. (1977). Basic concepts of structural analysis. Englewood Cliffs, N.J: Prentice-Hall. Bhavikatti, S. S. (2010). Structural analysis. New Delhi: Vikas Publishing House Chajes, A., & Chajes, A. (1990). Structural analysis, second edition, solutions manual. Englewood Cliffs, N.J: Prentice Hall. Leet, K., Uang, C.-M., & Gilbert, A. M. (2011). Fundamentals of structural analysis. New York, NY: McGraw-Hill Rajan, S. D. (2000). Introduction to structural analysis & design. New York: John Wiley. Roy, S. K., & Chakrabarty, S. (2003). Fundamentals of structural analysis: With computer analysis & applications. New Delhi: S. Chand. . Read More