RMIT CIVE1187: Forces in a Roof Truss Experiment Report

This report details an experiment conducted to analyze the forces within a roof truss. The experiment involved applying loads to the truss and measuring the resulting forces in its members. The report includes a comprehensive analysis of the experimental results, presented in tables and figures, alongside calculations using the method of joints. The analysis identifies tension and compression members, and compares experimental and analytical results, highlighting discrepancies and potential sources of error. The report also covers the dimensions of the truss and provides a detailed breakdown of the forces acting on each member under various load conditions. Finally, the report compares the measured and calculated forces and discusses the differences between the experimental and analytical results, attributing the inaccuracies in the experimental results to measurement deviations and malfunctioning equipment.

Forces in a Roof Truss Experiment 1
FORCES IN A ROOF TRUSS EXPERIMENT
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Forces in a Roof Truss Experiment 2
Forces in a Roof Truss Experiment
Analysis of Results
The results obtained for member forces from the experiment is provided in Table 1 below
Table 1: Experimental results of member forces
Loa
d
(N)
1 2 3 4 5 6 7 8 9 10 11 12 13
AH HG GF AB BC CD DE FE HB HC GC FC FD
0 3.2 -2.1 2.1 5.3 -3.2 2.7 -3.7 -3.2 2.1 0.5 1.1 0 -2.7
100 -101.3 -108.2 -101.3 95.9 85.3 77.8 89.5 -100.7 -7.5 12.8 100.7 0.5 -9.6
200 -202.5 -210 -199.8 179.
6
160.9 143.4 167.9 -201.4 -11.2 10.7 201.4 5.9 -3.7
300 -291.5 -303.8 -289.4 266.
5
240.3 218.5 247.3 -295.2 -13.3 12.8 291.5 1.1 -1.6
400 -385.8 -402.3 -394.3 344.
3
323.5 295.2 334.1 -398.6 -21.8 16.5 394.9 2.7 -5.9
1. Length of truss members
The truss with its full dimensions is shown in Figure 1 below. The dimensions are in mm.
Figure 1: Truss with full dimensions
2. Tension and compression members
140
140 140
70 70
140
121.24 121.24 121.24 121.24
140140 140

Forces in a Roof Truss Experiment 3
The tensions and compression members of the truss are shown in Figure 2 below. The tension
members are marked red while the compression members are marked green. The members in
blue do not carry any load (zero force). The tension members are the ones with positive forces
while the compression members are the ones with negative forces.
Figure 2: Tension and compression members
3. Truss analysis
The truss in this experiment has straight members, is symmetrical and connected by joints. The
truss is analyzed and its member forces calculated using trigonometry, Pythagoras theorem,
algebra, addition of vertical and horizontal forces, and simultaneous equations. The member
force represents that weight that the respective member bears (Garner, 2015).
4. Method of joints
The member forces of the truss are solved using method of joints. The force/load W is acting at
joint C (centre of the truss).
Support reactions Ax, Ay and Ey are determined using the equations of static equilibrium i.e. by
finding the sum of vertical and horizontal forces and taking moments at one of the points of the
truss as follows
Sum of forces in x-direction, ∑ Fx=0

Forces in a Roof Truss Experiment 4
→ Ax = 0
Sum of forces in y-direction, ∑ Fy=0
→ Ay – W + Ey = 0
Ay + Ey = W
Taking moments at A, ∑ MA =0
(242.48 x W) – (484.96 x Ey) = 0
484.96Ey = 242.48W
Ey = 242.48W
484.96 =0.5 W
But Ay + Ey = W
This means that Ay = W – Ey = W – 0.5W = 0.5W
Therefore support reactions are: Ax = 0, Ay = 0.5W and Ey = 0.5W
Using the method of joints, each joint is isolated and its member forces determined separately
(Moore, (n.d)), as follows
Joint A
∑ Fy=0,
Ay + AH sin 30° = 0
Ay + 0.5AH = 0 (substituting Ay = 0.5W)
0.5W + 0.5AH = 0; 0.5AH = -0.5W
AH = -W
∑ F x=0,
AB + AH cos 30° = 0; AB + 0.866AH = 0
AB = -0.866AH (substituting AH = -W)
AB = -0.866(-W)
AB = 0.866W
Joint B
∑ Fy=0,

Forces in a Roof Truss Experiment 5
HB = 0
∑ F x=0,
BC – AB = 0
BC = AB = 0.866W
Joint H
∑ Fy=0,
HG sin 30° – HB – AH sin 30° – HC sin 30° = 0
0.5HG – HB – 0.5AH – 0.5HC = 0 (substituting HB = 0 and AH = -W)
0.5HG – 0 – 0.5(-W) – 0.5HC = 0
0.5HG + 0.5W – 0.5HC = 0
0.5HG – 0.5HC = -0.5W (dividing through by 0.5)
HG – HC = -W ………………………………………… (1)
∑ F x=0,
HG cos 30° + HC cos 30° – AH cos 30° = 0
0.866HG + 0.866HC – 0.866AH = 0 (substituting AH = -W)
0.866HG + 0.866HC – 0.866(-W) = 0
0.866HG + 0.866HC + 0.866W = 0
0.866HG + 0.866HC = -0.866W (dividing through by 0.866)
HG + HC = -W …………………….………………… (2)
Adding (1) and (2) to solve them simultaneously gives
HG – HC = -W
HG + HC = -W
2HG = -2W; HG = -W
Substituting HG = -W into HG + HC = -W
-W + HC = -W
HC = -W + W; HC = 0
Joint G

Forces in a Roof Truss Experiment 6
∑ F x=0,
GF cos 30° – HG cos 30° = 0
0.866GF – 0.866HG = 0
0.866GF = 0.866HG (dividing through by 0.866)
GF =HG = -W
∑ Fy=0,
-GC – HG sin 30 – GF sin 30 = 0
-GC – 0.5HG – 0.5GF = 0 (substituting HG = -W and GF = -W)
-GC -0.5(-W) – 0.5(-W) = 0
-GC + 0.5W + 0.5W = 0
-GC + W = 0
-GC = -W
GC = W
Joint C
∑ Fy=0,
-W + GC + HC sin 30° + FC sin 30° = 0
-W + GC + 0.5HC + 0.5FC = 0 (substituting GC = W and HC = 0)
-W + W + 0.5(0) + 0.5FC = 0
0.5FC= 0
FC = 0
∑ F x=0,
CD – BC + FC cos 30° – HC cos 30° = 0
CD – BC + 0.866FC – 0.866HC = 0 (substituting BC = 0.866W, FC = 0 and HC = 0)
CD – 0.866W + 0.866(0) – 0.866(0) = 0
CD – 0.866W = 0
CD = 0.866W
Joint F

Forces in a Roof Truss Experiment 7
∑ F x=0,
FE cos 30° – GF cos 30° – FC cos 30° = 0
0.866FE – 0.866GF – 0.866FC = 0 (substituting GF = -W and FC = 0)
0.866FE – 0.866(-W) – 0.866(0) = 0
0.866FE + 0.866W – 0 = 0
0.866FE = -0.866W (dividing through by 0.866)
FE = -W
∑ Fy=0,
GF sin 30 – FC sin 30 – FE sin 30 – FD = 0
0.5GF – 0.5FC – 0.5FE – FD = 0 (substituting GF = -W, FC = 0 and FE = -W)
0.5(-W) – 0.5(0) – 0.5(-W) – FD = 0
-0.5W + 0.5W – FD = 0
0 – FD = 0
FD = 0
Joint D
∑ F x=0,
DE – CD = 0
DE = CD = 0.866W
The summary of truss member forces is provided in Table 2 below
Table 2: Truss member forces
Member Force
1 = AH -W
2 = HG -W
3 = GF -W
4 = AB 0.866W
5 = BC 0.866W
6 = CD 0.866W
7 = DE 0.866W
8 = FE -W
9 = HB 0
10 = HC 0

Forces in a Roof Truss Experiment 8
11 = GC W
12 = FC 0
13 = FD 0
5. Comparison of measured and calculated forces
The member force for each load is obtained by substituting W for each member force in Table 2
above with the value of load applied on the truss. Table 3 below shows the comparison between
measured and calculated member forces for each load. The measured and calculated member
forces are provided in blue and red colors respectively.
Table 3: Comparison of measured and calculated truss member forces
Loa
d
(N)
1 2 3 4 5 6 7 8 9 10 11 12 13
AH HG GF AB BC CD DE FE HB HC GC FC FD
0 3.2 -2.1 2.1 5.3 -3.2 2.7 -3.7 -3.2 2.1 0.5 1.1 0 -2.7
0 0 0 0 0 0 0 0 0 0 0 0 0 0
100 -101.3 -108.2 -101.3 95.9 85.3 77.8 89.5 -100.7 -7.5 12.8 100.7 0.5 -9.6
100 -100 -100 -100 86.6 86.6 86.6 86.6 -100 0 0 100 0 0
200 -202.5 -210 -199.8 179.
6
160.9 143.4 167.9 -201.4 -11.2 10.7 201.4 5.9 -3.7
200 -200 -200 -200 173.
2
173.2 173.2 173.2 -200 0 0 200 0 0
300 -291.5 -303.8 -289.4 266.
5
240.3 218.5 247.3 -295.2 -13.3 12.8 291.5 1.1 -1.6
300 -300 -300 -300 259.
8
259.8 259.8 259.8 -300 0 0 300 0 0
400 -385.8 -402.3 -394.3 344.
3
323.5 295.2 334.1 -398.6 -21.8 16.5 394.9 2.7 -5.9
400 -400 -400 -400 346.
4
346.4 346.4 346.4 -400 0 0 400 0 0
6. Differences between experimental and analytical results
From the results presented in Table 3 above, there were some differences between experimental
and analytical results obtained. The analytical results were more accurate than the experimental

Forces in a Roof Truss Experiment 9
results. For instance, it is obvious that when the load applied on the truss is zero (W = 0), all
member forces of the truss are expected to be zero. However, the experimental results show that
some values of member forces were recorded even when load applied was zero. Also, HB and
FD are expected to be zero (carry zero force) due to the principle that if a joint has three
members and two of the members are parallel and no external force is acting at the joint, then the
third member that is not parallel to the other two is said to be a zero force member (Erochko,
2016). However, the experimental results obtained show that members HB and FD carried some
forces. Therefore analytical results are more accurate than experimental results. The inaccuracy
of experimental results is caused by some errors. The major possible sources of errors in the
experimental results include: the measurement deviation of ± 3.0 N of the load applied on the
truss and use of malfunctioning equipment.
References

Forces in a Roof Truss Experiment 10
Erochko, J., 2016. Identifying zero force member. [Online]
Available at: http://www.learnaboutstructures.com/Identifying-Zero-Force-Members
[Accessed 27 May 2019].
Garner, M., 2015. Analyzing a Simple Truss by the Method of Joints. [Online]
Available at: https://www.instructables.com/id/Analyzing-a-Simple-Truss-by-the-Method-of-Joints/
[Accessed 27 May 2019].
Moore, J., (n.d). The method of joints. [Online]
Available at: http://adaptivemap.ma.psu.edu/websites/structures/method_of_joints/
methodofjoints.html
[Accessed 27 May 2019].