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
LEARNING OUTCOME
2.1 2.2
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)
F=Force in each member when the frame is release (N) Figure shows the force in the frame due to the load of 250N . You should be able to calculate these values from experiment. Force in a statically determinate truss
Method of Joints - Suitable to use in calculating all of the member forces for a truss. - This method entails the use of a free body diagram of joints with the equilibrium equation ΣFx = 0 and ΣFy = 0. - Calculation only can be started for joint where the numbers of unknown are two or less.
4.0
Apparatus and Equipments
5.0
PROCEDURES 1. 2.
3. 4.
5.
6. 7. 8. 9. 10. 11.
First, the thumwheel on the “redundant” member is unscrewed. Just as illustrated in the idealized diagram, it shouldn’t be effectively part of the structure any longer. Then, the pre-load of 100N is applied downward, the load cell is re-scaled to zero, and a load of 250N is carefully applied and the frame is checked for it’s stability and security. The load is applied carefully of 250 N and the frame is stable and secure. The load is returned to zero (leaving the 100N preload), and the digital indicator is rechecked and re-zero it. 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. the initial (zero) strain readings is subtraced to complete Table 2.indicator The strain reading is subtracted from the initial (zero). Be careful with the sign. The equipment member force at 250N is calculated and is entered into Table 3. A graph of Loa vs Deflection from table 1 is plotte on the same axis as Load vs deflection when the redundant removed. 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. Table 4 is refered and the value is entered and carefully calculated the other terms as required. The result is entered in to Table 3.
ΣFAY
A
1
B
ΣFAX 8
7
5
2
ΣFEX E
6.0
4
D
C
3
RESULT 6.1
Tabulate data Table 1: Strain Readings and Frame deflection for Experiment 1. Strain Reading
Load
DIR*
(N)
1
2
3
4
5
6
7
8
(mm)
0
131
215
-40
-75
100
-35
12
11
0
50
144
210
-49
-89
105
-42
25
18
-0.026
100
157
206
-56
-103
109
-49
38
25
-0.046
150
170
201
-66
-115
113
-55
51
32
-0.057
200
184
197
-75
-128
117
-62
64
40
-0.081
250
197
193
-83
-142
121
-68
77
46
-0.098
* Digital Indicator Reading
Table 2: True Strain reading for Experiment 1 Load (N)
1
2
3
4
5
6
7
8
0
0
0
0
0
0
0
0
0
50
13
-5
-9
-14
5
-7
13
7
100
26
-9
-16
-28
9
-14
26
14
150
39
-14
-26
-40
13
-20
39
21
200
53
-18
-35
-53
17
-27
-4
29
250
66
-22
-43
-67
21
-33
64
35
Using the Young’s Modulus relationship, calculate the equivalent member force. Complete the experimental force in Table 3. (ignore member 6 at this stage) E=σ/ε E = Young’s Modulus (Nm-2)
Where ;
σ = Stress in the member (Nm-2) ε = Displayed strain andσ = F / A 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, and the result obtained from the calculation were recorded in the Table 3 (below).
Member
Experimental Force (N)
Theorectical Force (N)
1
297.88
250
2
-292.60
-250
3
-292.60
-250
4
-591.56
-500
5
-12.72
0
6
0
0
7
407.10
353.55
8
426.18
353.55
Table 3 : Measured and Theorectical Force in the Cantilever Truss
Data Analysis Calculation For True Strain Reading; True Strain Reading (load50, 100,150,200,250) = Strain reading (load50, 100,150,200,250) – Strain reading (load0) Load, N = 250 N Example for True Strain Reading (load = 250 N) True Strain Reading (member 1)
= 263 – 215 = 44
True Strain Reading (member 2)
= - 351 – (- 311) = - 46
True Strain Reading (member 3)
= - 397 – (- 356) = - 46
True Strain Reading (member 4)
= - 215 – (- 135) = - 93
True Strain Reading (member 5)
= 178 – 185 =2
True Strain Reading (member 7)
= 159 – 94 = 64
True Strain Reading (member 8)
= 43 – (- 17) = 67
Calculation for Cross Section Area of the member (m2) ; Diameter, D
= 6.21 mm
From equation ; A
D 2 4
= = 30.29 mm
2
Calculation For Experimental Force (N); σ=F/A
From
F=Aσ But, from equation:E = σ / ε σ=Eε Thus ,F = A E ε Where;
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, Esteel
= 2.10 x 105 N/mm2
Calculation for Member 1: F=AEε = [30.29 x (2.10 x 105) x (44)] x 10-6 = 279.88 N Calculation for Member 2: F=AEε = [30.29 x (2.10 x 105) x (-46)] x 10-6 = -292.60N Calculation for Member 3: F=AEε = [30.29x (2.10 x 105) x (-46) ] x 10-6 = -292.60 N
Calculation for Member 4: F=AEε = [30.29x (2.10 x 105) x (-93)] x 10-6 = -591.56 N Calculation for Member 5 : F=AEε = [30.29x (2.10 x 105) x (-2)] x 10-6 = -12.72 N
Calculation for Member 7: F=AEε = [30.29x (2.10 x 105) x (64)] x 10-6 = 407.10 N Calculation for Member 8 : F=AEε = [30.29 x (2.10 x 105) x (67)] x 10-6 = 426.18N
Calculation For Theoretical Force (N): VA
A
1
E
HA 8
7
5
2
HB B
4
C
3
D
250 N ΣMA
=0
250 (2) – HB = 0 HB
= 500 N(
)
ΣFY
=0
– VA– 250
=0
VA = – 250 N = 250 N ( ) ΣFX
=0
HA+ HB
=0
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 = 250N(C)
At Point D:
ΣFY = ΣFY FDE sin
=
ΣFX = ΣFX
0
FDE sin
FDE = 353.55 N(T)
=
0
FCD= 250 N(C)
Theoretical Result VA = - 200N
A
250 N
E
HA = - 500N 353.55 N 0N HB = 500N
353.55 N - 250 N
FCD
B
- 500 N
C
D
- 250N N 250
6.2 Graphs 6.2.1
Choose a member (except member 6), and on the same axis plot a graph of
Recorded Strain με against Load (N) and True Strain με against Load (N). 6.2.2
On another graph, do the same for a different member (non member 6).
6.2.3
Plot a separate graph of deflection (mm) against Load (N).
6.2.4
Comment on your graph.
Strain (με) against Load (N) for member 3 80 70
Strain (με)
60 50 40 Recorded Strain
30
True Starin
20 10 0 0
50
100
150 Graph 1
200
250
Load (N)
Strain (με) against Load (N) for member 7 160 140
Strain (με)
120
Recorded Strain
100
True Starin
80 60 40 20
0 0
50
100
150
200
250
Load (N)
Graph 2
Deflection (mm) against Load (N) Deflection x 10-2 (mm)
18 16 14 12 10
8 6 4 2 0
0
50
100
150 Graph 2
Comment on the graph: i)
For Graph of Strain versus load for member
200
250
Load (N)
From the Graph 1 and Graph 2 plotted for member 3 and 7, we obtain a linear graph for both members. The pattern of the graph are similar and the gradient of the recorded strain and true strain should be the same, which means both of the graph are parallel to each other .This is because the true strain is the comparison value with the initial strain readings. From the graphs, when the values of strain increase, the loading apply also increased. This was caused by the compression and tensioned in the member of truss when the loading is applied on it. From Graph of Strain versus load for member 3, 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. Values for Recorded strain are higher than the True Strain. The same go to the graph of Strain versus load for member 8. 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. The arrangements of the member in the truss also cause the difference of the force in the truss. Therefore, in order to find the forces in the truss for each member, 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.
ii)
For Graph of Deflection versus Load ;
From the Graph 3 plotted, we obtain a linear graph. The values of the graph I the readings we obtain from the digital indicator reading of the machine used. From the graphs, we noticed that, when the there are no load acted on it, there are no deflection occur on the frame. When the values of loading apply increase, the deflection also increased. This was caused by the compression and tensioned in the member of truss when the loading is applied on it. The compression and tensioned created will cause the deflection inthe member of truss. In this experiment, the deflection created is proportional to the loading apply. Besides, the deflection also can happen due to the high temperature or an error occurs when recording the reading during the experiment. 7.0 DISCUSSION 1.
Compare the experimental and theoretical result. We have calculated the experimental forces using the formula and theoretical forces by the method of joint. From the table 3, we can notice that there are slight differences between the values for Experimental Force and Theoretical Force. The experimental values of every member have a greater value than that of the theoretical values. This might be caused by some extra load that we cannot avoid it. However, there are only small differences, so it still acceptable. The positive values for the force obtain in member 1, 7 and 8shows that these members are in tensioned. While for the negative values for the force obtain in member 2, 3 and 4, these members are being compressed. For member 5, it is a zero member force.
2.
From your result and the theoritical member force, identify which members are in compression and which member are in tension. Explain your choice.
VA = - 200N A
250 N 1
HA = - 500N
353.55 N 5 0N
8
E
2 - 250 N
353.55 N 7
4
HB = 500N B
- 500 N
3 C
D
- 250N N 250 N
From the results for Experimental Force and the Theoretical member force, we noticed that the members of 2, 3, and 4 are in compression while the members of 1, 7 and 8 are in tensioned. This is because we know that, the compressed member has a negative force values while tensioned member has a positive force values. From the framework, we also noticed that member 2 is being burdened and compressed by both member 3 and 7. The same goes to member 4 which was compressed by member 5 and 8. 3.
Observe the reading of member 5. Explain why the reading is amost zero. From the results obtain by observing the member 5, we noticed that the reading force is almost zero. By the theoretical aspect, it should be zero for its internal force. We know that, member 5 is attached by a pin joint and a roller joint at both ends. On the pin joint, two forces acting towards it on the horizontal axis and vertical axis. On the roller joint, there is only one force acting towards it on the horizontal axis. Therefore, the reading force of member 5 is almost zero due to these three forces.
4.
Are the strain gauges are an effective tranducers for the measurement of forces in the framework? Yes, the strain gauges are an effective tranducers for the measurement of forces in the framework.it is sensitive with the displacement of deflection of the truss.it gives reading up to 0.001mm which is rarely can be calculated as well as spotted the displacement with naked eyes.thus,it gives us an accurate reading of works.
5.
Does the framework comply with pin joint theory even though the joint are not truly pin joint? From the results, we noticed that the value of forces obtains from the Experimental and Theoretical (calculation) are almost equivalent. There are only slight differences between the values for Experimental Force and Theoretical Force. Therefore, this indicates that, the framework comply with the pin joint theory even though the joint are not truly pin joint.
8.0 CONCLUSION From this experiment, our group managed to examine a statically determinate frame and to analyze the frame using simple pin joint theory. We conclude that, when the loading apply on the member in the truss is increase, the deflection also increase due to the compression and tensioned. Therefore, we also conclude that, when the loading apply is increase, the stain for both recorded and true strain value will increase. In this experiment, we also noticed that the framework comply with the pin joint theory even though the joint are not truly pin joint. Therefore, Method of Joints is suitable to be used in calculating all the member forces for
a truss in this experiment. This method entails the use of a free body diagram of joints with the equilibrium equations ΣFx = 0 and ΣFy = 0.
9.0 REFRENCES http://www.ohio.edu/people/williar4/html/haped/nsf/stat/truss.pdf http://www.studymode.com/essays/Truss-Report-694655.html 6.0
DISCUSSION AND CONCLUSION 1.
From table 3, compare your answer to the experimental value. Comment on the accuracy of your result From the result shows that there are a lot of different values between experimantal and theoretical. The different between both is very appreciable. All members give the different value caompared to the theoretical. This happens because the problem from the facts that we used. Zero value doesn’t appear at the digital display. Many group used this experiment equipment before our group used it, so an error may happens when our group used it again. From the experiment that we have done, it show that the result are not accurate to the theoretical.
2.
Compare all of the member forces and the deflection to those from statically determinate frame. Comment on them in terms of economy and safety of the structure. Comparison all of the memberforces give a different value compare to the theoretical. The comparison is very different. It’smean that our result is not closed to the theoretical. So that the deflection gives a very unfavourable value because from results, we get very large
