Nt1310 Unit 4 Lab

Sheridan College
Faculty of Applied Science and Technology (FAST)
APPLIED MECHANICS 2 – ENGI 13386 (FALL 2014)
Lab Report 1
Method of Joints and Sections lab
Class: 1149_40517 (1149_40518)
Authors: Paritosh Chahal and Rony George
Date of Experiment: 17th September 2014
24 September 2014 – 12pm Instructor M. Arthur

Signed Declaration: the work submitted here is my own and not copied or plagiarized from another source.

Signiture:____________________________________________________________

Contents
Executive Summary:- 1
Objective:- 2
Theory:- 2
Equipment & Materials:- 2
Procedure:- 3
DATA:- 3
Calculations: 5
RESULTS:- 11
Percentage Difference:- 11
DISCUSSION:- 12
Conclusion: 12
Real life application: 13
APPENDIX: 13 REFERENCES………………………………………………………………………………………………………………………………….......21

Executive Summary:-

Truss (fig 1) is a type
of a structure which is used in one’s everyday live. It is “composed of bars or members joined to form one or more connected triangles. Each of the members is pinned at each end and, if carrying a load, is in tension or compression” (Walker 139). Without the theory behind it, many building, bridges, roller coasters, cranes and playground swings etc would not stay stable; thus putting millions of lives in danger. In this lab, the observers will examine to test the conditions of a simple truss model whose set-up similar to a crane. After following the procedures of the different experiment, it was noted that the truss model will be in equilibrium no matter what the load on it is.

Fig 1: Truss
Objective:-
The main objectives in this experiment
Examine how increase in load affects the members in a simple truss model
Test the conditions of equilibrium of the model
Compare the calculated result of different force members with values found by the “Data Studio”
Theory:-
As mentioned in summary, truss is a type of structure which consists of members connected (by pinning to each other) to each other in one or more triangles; and the members are in compression or tension if a load is applied to it. Each member is considered to be a two force member (the two forces are applied at the ends of the member) “if we neglect the weight of the member” (Walker 139).
There are two different ways of solving truss structures that have only two force members which are:
a.
Method of joints: In this method, free body diagrams of adjacent joints of a member are seen to check the forces acting on different members. Also, “the first joint selected must have two unknown forces and one known force” (Walker 139). The unknown forces are later solved by using ∑x = 0 and ∑y = 0 (Since the whole model is at rest, it is considered that each member in the model will be at rest too). Once the two unknown forces are found, they become known forces to other adjacent joints and are used to find other unknown forces of another member. This technique is used on all the members of the model is found.
b. Method of Sections: This method is usually “used to solve for the force in a member near the middle of truss” (Walker 149). Here, the structure in cut into two pieces; one of the pieces is not used while the other has a force member which is required to find. The free body diagrams of the section which the observers needs are drawn, and after steps similar to methods of joints are
followed.
Equipment & Materials:-
1. Boom Structure (Fig 2)

Fig 2: Boom structure
2. 2 Load cell sensors: This device is used to find the forces applied on the joints of the members through the help of “Data Studio” software, the sensors are attached to the respective members.

Fig 3 Load sensors Adapter
3. “Data Studio” Software & Interface: The software which helps the observers to know the forces applied on the members over a period of time (in seconds). The load cell sensors respond to the software and show the reading in terms of a graph.
4. Tape Measure
5. A designated hanging mass: 1kg plate X 1 and 2 kg plates X 2.
Procedure:-

Method of Joints:
1. The “Data Studio- PASCO” software was downloaded to the computer.
2. The Boom structure was modeled as instructed.
3. The length of each members and angle between two members were measured.
4. The Load sensors were placed on the respective members.
5. The assigned file “Graph and Digits” was opened in Data Studio.
6. The template file “Graph and Digits” was set up and calibrated for the boom structure. With no mass hanging from the end of the boom, the display for the two sensors read zero Newton.
7. First mass plate was inserted.
8. The mass was recorded of the first assigned hanging weight.
9. The experimental internal forces (Tension and Compression) in members were recorded.
10. Steps 7-9 were repeated for 3 other masses until the overall load went from 1kg to 5kg.
11. The internal were checked again by removing respective weight plates until it went from 5kg to 1kg.
12. The findings were recorded.

Method of Sections:
1. The boom structured was modeled as instructed.
2. Steps 3 to 11 from “Methods of joint” procedures were followed.
3. The findings were recorded.

DATA:-
1. METHOD OF JOINTS
Trial
Mass
Length
Tension at BC
Compression at AB #
(KG)
(cm) T1 (N) T2 (N) 1)
1
53.34
5.05
9 2)
3
53.34
15.96
23.75 3)
4
53.34
21.33
36.71
4
5
53.34
26.57
45.83
Table 1: Method of joints Data
2. METHOD OF SECTIONS
Trial
Mass
Length
Tension at BC
Compression at AB #
(KG)
(cm) T1 (N) T2 (N) 1) 1
91.44
9.76
9.12
2)
3
91.44
29.33
27.58 3)
4
91.44
36.80
36.94
4)
5
91.44
49
46.5
Table 2: Method of Sections Data

3. Lengths and angle of members:
a. Method of joints:

Fig 4: Set-up of experiment 1
b. Method of Sections

Fig 5: Setup of fig 5

NOTE: In our Raw Data , we used different alphabets at different points, so according to the Raw Data BC = AB (From the lab format), AC = BC (From Lab Format), AB = BC (From Lab format)

Calculations:
1. Interior Angles
a. Method of joints: (NOTE: The lengths of members are converted to cm)
For ∠A, ∠B &∠C
AC(b) = 100.33 cm (Fig 6)
BC (a) = 53.34 cm (Fig 6)
AB (c) = 91.44 cm (Fig 6) Fig 6 In ∆ABC a2=b2+c2 – [2 (b X c) (cos (A))]
(53.34)2 = (100.33)2 + (91.44)2 – [2(100.33) (91.44) (cos (A))]
2845.15 = 10066.10+ 8361.27-18348.35(cos (A))
A = cos-1()
∠A=31.890

∠C= 64.80
∠B=1800 - ∠B - ∠A = 1800 – 64.80 – 31.890 = 83.310

Fig 7

a= 900 - ∠A = 900 - 31.890 = 58.110 γ= ∠C – a= 64.80 - 58.110 = 6.690 γ = 6.690

b. Method of Section: (Note the lengths are converted to cm)

Fig 8

For ∠A, ∠B &∠C
AC(b) = 100.33 cm
BC (a) = 91.44 cm
AB (c) = 91.44 cm In ∆ABC a2=b2+c2 – [2 (b X c) (cos (A))]
(91.44)2 = (100.33)2 + (91.44)2 – [2(100.33) (91.44) (cos (A))]
8361.27 = 10066.10+ 8361.27-18348.35(cos (A))
A = cos-1() = 56.730

∠B=1800 - ∠B - ∠A = 1800 – 56.770 – 31.890 = 66.54 0

Fig 9

a= 900 - ∠A = 900 – 56.730 = 33.270 γ= ∠C – a= 56.730 – 33.270 = 23.460 γ = 23.540
2. Theoretical values of Force members:
a. Method of joints: CB Free Body Diagram of point C: CB 53.34 6.69 52.97 C

49.05 100.33 85.20 CA 58.11 Fig 10
Let left and down direction be “-“ and up and right direction be “+”

Equation 1

-.0993CB - 0.579CA = 0 Equation 2
Substitute equation 2 to equation 1:

CB = 31.14 N = 31.14 N (C)
Substitute CB into equation 2

CA = 53.25 N (T)
b. Method of sections:

CB Free Body Diagram of bottom section:
` CB 91.44 36.40 83.88 C AB 49.05 A Fig 11
Let clockwise be “+”, and counter clockwise “-“
∑MA = 0

CB = 49.06 N (C)

CB Free Body Diagram of bottom section:
1.44

Fig 12

Let clockwise be “+”, and counter clockwise “-“
∑MA = 0

-
CA = 53.81 N (T)
RESULTS:-
1. Method of Joints:
Trail
#
Weight
(N)
Length
(cm)
Calculated
Value (CB)
{Compression}
Experimental Value (CB)
{Compression}
Calculated Value (CA)
{Tension}
Experimental Value (CA)
{Tension}
1)
9.81
53.34 6.24 5.05 10.68 9 2)
29.43
53.34 18.74 15.96 32.04 23.75 3)
39.24
53.34 24.91 21.33 42.91 36.71
4)
49.05
53.34
31.14
26.57
53.24
45.83
Table 3: Method of joints Results
2. Method of Sections:
Trail
#
Weight
(N)
Length
(cm)
Calculated
Value (CB)
{Compression}
Experimental Value (CB)
{Compression}
Calculated Value (CA)
{Tension}
Experimental Value (CA)
{Tension}
1)
9.81
91.44 9.87 9.76 10.84 9.12 2)
29.43
91.44 29.63 29.33 32.54 27.58 3)
39.24
91.44 39.51 36.80 43.39 36.94
4)
49.05
91.44
49.06 N
49N
53.81
46.5
Table 4: Method of Sections Results
The tables show the calculated value and experimented values of the two different methods (joints and sections) used to solve the simple truss model. Since the experimented and calculated values are very much similar, one can say that the whole system was in equilibrium.
Percentage Difference:-

1. Method of Joints: (CB)

= -17.2%
2. Method of Sections: (CB)

= -1.22 %
DISCUSSION:-
After following the procedures of the lab and looking at the results, it was observed that the truss model was in equilibrium (with ∑x = 0 and ∑y = 0), and also that “Method of joints” and “Method of Sections” are good methods which can used to solve trusses with two force members. The lab consisted of two experiments; each of which the truss models used were different and helped observers learn something important about trusses. For instance, the first experiment helped the observers know that trusses questions can be solved by using the equilibrium statement where with ∑x = 0 and ∑y = 0; one can solve it this way because the model was still throughout the experiment and did not move at all in any direction. Also, looking at the results where the experimented value and calculated value are very close to each other proves that the statement made is correct. In addition, the second experiment helped the observers realise that trusses can also be solved by using “Moments” in which you take one of the points as reference (pivot point) and see if the object moves clockwise or counter clockwise by the external forces working on it. Looking at the results of this experiment where the experiment value and calculated values are really close (difference of few decimals) can prove that the above statement is true.
Throughout the lab, the observers accounted some sources of error that did not give them accurate answers. These errors were:
1. Accuracy of the Software: The software did not give the observers exact reading of the forces on the respective members, the numbers always triggered and confused the observers to pick the correct one.
2. Measurement Error: The length of the members and angles between respective members were not measured properly. This might have affected the result too since without exact measurements, one may not find the correct angles that will be later used to find the experimental value. This error has occurred during the lab which can be seen by the difference of experimental angle measure of member and the calculated values of the members.
Conclusion:
The lab was performed to see whether the two methods (Joints and Sections) are good to use for solving trusses questions with two force members. After following the procedures, collecting data, and doing calculations; the above points were proved to be true. Two lab experiments were performed in which two different structure models were made and the load cell sensors were used to measure the force of each member at different loads of 1 kg, 3 kg, 4 kg, and 5kg. Afterwards, the force members were calculated using Method of Joints” for the first experiment and the “Method of Sections” for the second experiment. Finally, looking at the results of the lab the observers learned that all truss structures are in equilibrium since the net force acting on it is zero and that the two methods stated above are perfect for solving trusses problem with two force members only.
NOTE: For improving the lab next time, one may suggest that time given to the observers should have been little more since that way they would have properly measure the lengths of different members to reduce errors. Secondly, the instructors must have informed to class to install the “DATA STUDIO” software a day before the scheduled lab, this way less time would have been wasted among the students which would have led to lesser errors. Finally, the joints beforehand should be labelled by the instructors or the observers themselves; this way there will be less confusion while noting down the lengths of members, angle between two respective members, and the forces of respective members. Real life application:
This experiment can relate to many real life applications, but the one that observers believe will be the most relevant is how the structure designed during the lab are very similar to that of a simple crane or the demolition crane. The two machines uses trusses structures which help them maintain the load at the end of the respective member. Without properly analyzes of these structures, the machines will be a very dangerous object which can destroy a lot of property and also kill thousands of people.

APPENDIX:

1. Data Collection:

2. Peer Evaluation:

REFERENCES
1. Walker, Keith M. "Equilibrium." Applied Mechanics for Engineering Technology. Upper Saddle River, NJ: Pearson/Prentice Hall, 2004. 139-150. Print.
2. http://hitec.ca/trusses.html