9FORCE IN A STATICALLY DETERMINATE CANTILEVER TRUSS 1.0 OBJECTIVE 1.1 To observe the effect of redundant member in a structure in a structure and understand the method of analysis type of this structure. 2.0 2.1 2.2 LEARNING OUTCOME The application of the engineering knowledge in practical application To enhance technical competency in structural engineering through laboratory application 3.0 THEORY 3.1 In a statically indeterminate truss, static equilibriumalone cannot be used to calculated member force. If we were to try, we would find that there would be too many “unknowns” and we would not be able to complete the calculations. 3.2 Instead we will use a method know as the the flexibility method, which uses an idea know as strain energy. 3.3 The mathe,atical approach to the flexibility method will be found in the most appropriate text books. Basically the flexibility method uses the idea that energy stored in the frame would be the same for a given loa for a given weather load wheather or not the redundant member wether or not. In the other word, the external energy = internal energy In practice, the loads in the frame are calculated in its “released” from (that is, without the redundant member) and then calculated with a unit load in place of the redundant member and remaining members. There redundant member load in given by: P= ∑ The remaining member force are then given by: Member force = Pn+f Where, P= Redundant member load (N) L=Length of member (as ratio of the shortest) N=Load in each member due to unit load in place of redundant member (N) . Force in a statically determinate truss Method of Joints .F=Force in each member when the frame is release (N) Figure shows the force in the frame due to the load of 250N . .This method entails the use of a free body diagram of joints with the equilibrium equation ΣFx = 0 and ΣFy = 0. 4.Suitable to use in calculating all of the member forces for a truss. You should be able to calculate these values from experiment.0 Apparatus and Equipments .Calculation only can be started for joint where the numbers of unknown are two or less. The load is applied carefully of 250 N and the frame is stable and secure. 2. 8. and a load of 250N is carefully applied and the frame is checked for it’s stability and security. the thumwheel on the “redundant” member is unscrewed. Loads greater than those specified on the equipment should never be applied. Load in the increment shown in Table 1 is applied recording the strain readings and the digital indicator readings. Table 4 is refered and the value is entered and carefully calculated the other terms as required. Just as illustrated in the idealized diagram. A graph of Loa vs Deflection from table 1 is plotte on the same axis as Load vs deflection when the redundant removed. . and the digital indicator is rechecked and re-zero it. The load is returned to zero (leaving the 100N preload). 10. First. 7.indicator The strain reading is subtracted from the initial (zero). 4. Be careful with the sign. 6.0 PROCEDURES 1. the pre-load of 100N is applied downward.5. Calculation for redundant truss is made simple and easier if the tabular method is used to sum up all of the Fnl and n²l terms. 3. 9. the load cell is re-scaled to zero. Then. The equipment member force at 250N is calculated and is entered into Table 3. 5. the initial (zero) strain readings is subtraced to complete Table 2. 11. it shouldn’t be effectively part of the structure any longer. The result is entered in to Table 3. 057 . Load (N) 0 50 100 150 1 131 144 157 170 2 215 210 206 201 3 -40 -49 -56 -66 Strain Reading 4 -75 -89 -103 -115 5 100 105 109 113 6 -35 -42 -49 -55 7 12 25 38 51 8 11 18 25 32 DIR* (mm) 0 -0.046 -0.ΣFAY A ΣFAX 1 B 8 5 ΣFEX E 4 D 3 2 7 C 6.026 -0.0 RESULT 6.1 Tabulate data Table 1: Strain Readings and Frame deflection for Experiment 1. 098 * Digital Indicator Reading Table 2: True Strain reading for Experiment 1 Load (N) 1 0 13 26 39 53 66 2 0 -5 -9 -14 -18 -22 3 0 -9 -16 -26 -35 -43 4 0 -14 -28 -40 -53 -67 5 0 5 9 13 17 21 6 0 -7 -14 -20 -27 -33 7 0 13 26 39 -4 64 8 0 7 14 21 29 35 0 50 100 150 200 250 Using the Young’s Modulus relationship. and the result obtained from the calculation were recorded in the Table 3 (below).88 Theorectical Force (N) 250 . Member 1 Experimental Force (N) 297. E = Young’s Modulus (Nm-2) σ = Stress in the member (Nm-2) ε = Displayed strain andσ = F / A Where.081 -0. calculate the equivalent member force. Complete the experimental force in Table 3. (ignore member 6 at this stage) E=σ/ε Where . F = Force in member (N) A = cross section area of the member (m2) Teorectical Force (N) was calculated by using the Method of Joint.200 250 184 197 197 193 -75 -83 -128 -142 117 121 -62 -68 64 77 40 46 -0. 250) = Strain reading (load50.200.55 Table 3 : Measured and Theorectical Force in the Cantilever Truss Data Analysis Calculation For True Strain Reading.200.60 -591.356) = .150. 100.18 -250 -250 -500 0 0 353.351 – (.150.55 353.215 – (. True Strain Reading (load50.250) – Strain reading (load0) Load. N = 250 N Example for True Strain Reading (load = 250 N) True Strain Reading (member 1) = 263 – 215 = 44 = .72 0 407.311) = .93 = 178 – 185 =2 True Strain Reading (member 7) = 159 – 94 = 64 True Strain Reading (member 2) True Strain Reading (member 3) True Strain Reading (member 4) True Strain Reading (member 5) .56 -12.10 426.60 -292.397 – (.46 = . 100.2 3 4 5 6 7 8 -292.46 = .135) = . True Strain Reading (member 8) = 43 – (. E = Young’s Modulus (Nm-2) σ = Stress in the member (Nm-2) ε = Displayed strain F = Force in member (N) A = cross section area of the member (m2) Given.29 x (2.10 x 105) x (-46) ] x 10-6 = -292. from equation:E = σ / ε σ=Eε Thus . From σ=F/A F=Aσ But. Esteel = 2.29 x (2. Diameter.21 mm From equation . A = D 2 4 = 30. D = 6.60N Calculation for Member 3: F=AEε = [30.29 mm 2 Calculation For Experimental Force (N).10 x 105) x (-46)] x 10-6 = -292.29x (2.10 x 105 N/mm2 Calculation for Member 1: F=AEε = [30.88 N Calculation for Member 2: F=AEε = [30.F = A E ε Where.10 x 105) x (44)] x 10-6 = 279.60 N .17) = 67 Calculation for Cross Section Area of the member (m2) . 10 N Calculation for Member 8 : F=AEε = [30.29x (2.10 x 105) x (67)] x 10-6 = 426.56 N Calculation for Member 5 : F=AEε = [30.29 x (2.72 N Calculation for Member 7: F=AEε = [30.10 x 105) x (-2)] x 10-6 = -12.Calculation for Member 4: F=AEε = [30.10 x 105) x (64)] x 10-6 = 407.18N Calculation For Theoretical Force (N): VA A HA 1 E 8 5 2 7 HB B 4 C 3 D 250 N ΣMA =0 250 (2) – HB = 0 HB = 500 N( ) .29x (2.29x (2.10 x 105) x (-93)] x 10-6 = -591. ΣFY =0 =0 VA = – 250 N = 250 N ( ) – VA– 250 ΣFX =0 =0 ) HA+ HB HA = 500 N ( Internal forces of members: θ = = At point B: ΣFY = ΣFY FAB = 0 ΣFX = ΣFX FBC + 500 = 0 FBC = 500 N (C) At Point A : ΣFY = ΣFY 250 = FAC sin FAC= 353.55 N (T) ΣFX = ΣFX FAE + FAC cos = 00 FAE = 250 N (T) . At Point E: ΣFY = ΣFY 0 = FDE cos FCD FCD = 250N(C) At Point D: ΣFY = ΣFY FDE sin = 0 FDE = 353.200N A HA = .250 N 353.55 N(T) ΣFX = ΣFX FDE sin FCD= 250 N(C) = 0 Theoretical Result VA = .500N 250 N E 353.55 N 0N HB = 500N .55 N . B .3 6. 6.4 On another graph.500 N C . Comment on your graph.2. Plot a separate graph of deflection (mm) against Load (N).2.2.2 6.2 Graphs 6. do the same for a different member (non member 6). Strain (με) against Load (N) for member 3 80 70 60 Strain (με) 50 40 30 20 10 0 0 50 100 150 Graph 1 200 250 Recorded Strain True Starin Load (N) . and on the same axis plot a graph of Recorded Strain με against Load (N) and True Strain με against Load (N).1 Choose a member (except member 6).250N N 250 D 6.2. Strain (με) against Load (N) for member 7 160 140 120 Strain (με) Recorded Strain True Starin 100 80 60 40 20 0 0 50 100 150 Graph 2 200 250 Load (N) Deflection (mm) against Load (N) 18 Deflection x 10-2 (mm) 16 14 12 10 8 6 4 2 0 0 50 100 Graph 2 150 200 250 Load (N) Comment on the graph: i) For Graph of Strain versus load for member . From the graphs. The positive values for the force obtain in member 1. These differences exist because the forces created in the truss are different at the point of the joint when the load is applied on it.55 N E 2 . For member 5. However. From your result and the theoritical member force. This might be caused by some extra load that we cannot avoid it. Values for Recorded strain are higher than the True Strain.From the Graph 1 and Graph 2 plotted for member 3 and 7. in order to find the forces in the truss for each member. This was caused by the compression and tensioned in the member of truss when the loading is applied on it. Compare the experimental and theoretical result. the deflection also increased. these members are being compressed. From the table 3. we obtain a linear graph. we noticed that. identify which members are in compression and which member are in tension. 7 and 8shows that these members are in tensioned. so it still acceptable. there are only small differences. Besides. The experimental values of every member have a greater value than that of the theoretical values. we can notice that there is a difference between the values of strain for the strain we recorded in the experiment and the true strain. We have calculated the experimental forces using the formula and theoretical forces by the method of joint. The same go to the graph of Strain versus load for member 8. we obtain a linear graph for both members. when the there are no load acted on it. 3 and 4. When the values of loading apply increase.200N A HA = . which means both of the graph are parallel to each other . This was caused by the compression and tensioned in the member of truss when the loading is applied on it. 7. when the values of strain increase. From the graphs. the deflection also can happen due to the high temperature or an error occurs when recording the reading during the experiment. ii) For Graph of Deflection versus Load . VA = .250 N 353.0 DISCUSSION 1. From the Graph 3 plotted.This is because the true strain is the comparison value with the initial strain readings. it is a zero member force. The values of the graph I the readings we obtain from the digital indicator reading of the machine used. The arrangements of the member in the truss also cause the difference of the force in the truss. Explain your choice.500N 5 0N 8 250 N 1 353. In this experiment.55 N 7 . the deflection created is proportional to the loading apply. While for the negative values for the force obtain in member 2. Therefore. we can notice that there are slight differences between the values for Experimental Force and Theoretical Force. we have to use equilibrium equation for xaxis and y-axis equal to zero and the calculation is done part by part for each member. there are no deflection occur on the frame. From Graph of Strain versus load for member 3. The compression and tensioned created will cause the deflection inthe member of truss. 2. The pattern of the graph are similar and the gradient of the recorded strain and true strain should be the same. the loading apply also increased. we noticed that the reading force is almost zero. the compressed member has a negative force values while tensioned member has a positive force values. 3. Explain why the reading is amost zero.250N N 250 N D From the results for Experimental Force and the Theoretical member force. Therefore. two forces acting towards it on the horizontal axis and vertical axis. From the framework. the deflection also increase due to the compression and tensioned. Method of Joints is suitable to be used in calculating all the member forces for .HB = 500N B 4 . we noticed that the value of forces obtains from the Experimental and Theoretical (calculation) are almost equivalent. we also noticed that the framework comply with the pin joint theory even though the joint are not truly pin joint. our group managed to examine a statically determinate frame and to analyze the frame using simple pin joint theory. the framework comply with the pin joint theory even though the joint are not truly pin joint.it gives us an accurate reading of works. we also conclude that. member 5 is attached by a pin joint and a roller joint at both ends. On the roller joint. Therefore. The same goes to member 4 which was compressed by member 5 and 8. There are only slight differences between the values for Experimental Force and Theoretical Force. We conclude that. 5. the reading force of member 5 is almost zero due to these three forces. In this experiment. This is because we know that. we also noticed that member 2 is being burdened and compressed by both member 3 and 7.001mm which is rarely can be calculated as well as spotted the displacement with naked eyes.500 N C 3 . the strain gauges are an effective tranducers for the measurement of forces in the framework. Observe the reading of member 5.it gives reading up to 0. 8. We know that.thus. we noticed that the members of 2. it should be zero for its internal force. this indicates that. By the theoretical aspect. From the results obtain by observing the member 5. Does the framework comply with pin joint theory even though the joint are not truly pin joint? From the results. On the pin joint. when the loading apply on the member in the truss is increase. Therefore. when the loading apply is increase. 4. there is only one force acting towards it on the horizontal axis. Therefore. 7 and 8 are in tensioned. Are the strain gauges are an effective tranducers for the measurement of forces in the framework? Yes.0 CONCLUSION From this experiment. 3. the stain for both recorded and true strain value will increase. and 4 are in compression while the members of 1.it is sensitive with the displacement of deflection of the truss. This method entails the use of a free body diagram of joints with the equilibrium equations ΣFx = 0 and ΣFy = 0. we get very large .0 DISCUSSION AND CONCLUSION 1.edu/people/williar4/html/haped/nsf/stat/truss.pdf http://www. 9. Many group used this experiment equipment before our group used it. it show that the result are not accurate to the theoretical.0 REFRENCES http://www.html 6. so an error may happens when our group used it again.com/essays/Truss-Report-694655.a truss in this experiment. All members give the different value caompared to the theoretical. Zero value doesn’t appear at the digital display. It’smean that our result is not closed to the theoretical. From the experiment that we have done.ohio. This happens because the problem from the facts that we used. Compare all of the member forces and the deflection to those from statically determinate frame. The comparison is very different. Comment on the accuracy of your result From the result shows that there are a lot of different values between experimantal and theoretical.studymode. Comment on them in terms of economy and safety of the structure. The different between both is very appreciable. compare your answer to the experimental value. Comparison all of the memberforces give a different value compare to the theoretical. From table 3. So that the deflection gives a very unfavourable value because from results. 2.
Report "Lab 5-Force in a Statically Determinate Cantilever Truss UTHM"