Experiment: S2 – The Roof Truss Experiment
Aims: To gain practical experience of strain measurement using rain gauges, to compare experimentally determined forces in the roof truss with those predicted theoretically
Learning Outcomes
The evaluation of stresses and forces from experimental measurements of strain.
The determination of axial forces in the members of the roof truss with those predicted using method of joints.
Apparatus: Bench mounted model roof truss. Micrometer.
Procedure: The applied load was set to zero at joint A and measurements were taken of all the strain gauges on the roof truss model. The applied load on joint A was increased to 100N and measurements were taken of all the strain gauges on the roof truss. The applied load on joint A was then increased in an increment of 100N until it reached 500N. At each load, measurements of the strains of all the strain gauges on the roof truss were taken. The load on joint A was then decreased by decrements of 100N until it reached zero. At each load reduction, measurements of the strains on all strain gauges on the roof truss model were taken. The diameter of the roof truss members were then measured using the micrometer.
Results: For each strain gauge, the average strain gauge reading was calculated and strain for each load increment at joint A as well. Graphs of load against average strain for each gauge were plotted to check linearity of the gauges. See comments on the same. For a load of 350N at joint A, the stresses in each member of the roof truss were calculated using the equation ϭ= Es, and hence the forces in each member of the roof truss were determined using the equation F= ϭA. Using the method of joints all the member forces for a load of 350N at joint A were determined.
Figure 1: Roof Truss
Calculation of member forces using method of joints
W=350N and W/2 = 175N
Joint B
Fy = FBF sin 300
FBF = Fy/sin300
= 175/sin300 = 177.12N (Compression member)
FBD = FBD cos300
= -177.12 cos300 = 27.32N (Tension member)
Joint D
∑Fx = 0
-27.32 + FDA = 0, FDA = 27.32N
∑Fy = 0
FDA = 0N
Joint F
∑Fy = 0
-FBFsin 300 –FDF – FFAsin 300 = 0
-(-177.12)sin30 – 0 - FFAsin 300 = 0
FFA = -175/sin30 = 177.12N (Tension member)
Joint H
∑Fx = 0
FHA –(-FHF sin 300 ) – (-FFAsin 300) = 0
FHA - (-177.12 sin30) – (-177.12 sin30) = 0
FHA = 350N (Tension member)
Laboratory Results
Applied Load at A
Strain Gauge Readings (x10-6)
Applied Load at A
Strain Gauge Readings (x10-6)
Experimental Graphs of Applied Load against Strain on a truss
NB: The loading is in the opposite direction (-ve) n gauge 2
NB: The loading is in the opposite direction (-ve) also on gauge 3
NB: The loading is in the opposite direction (-ve) also on gauge 5
NB: The loading is in the opposite direction (-ve) also on gauge 6
NB: Since it’s a symmetrical truss, it will be assumed that gauge 8, 9, 10, 11, and 13 have similar loading arrangements to the above.
Relevant Theory: E= ϭ/s
Where s = Strain
Ϭ = Stress
E = Young’s Modulus for the material
Member stress calculations
Take E = 210 GPa
Member sizes = 500mm
Area of members = 500 x 500 = 250,000 mm2
For member CG and GH
Ϭ = Es
=210x2 x10-6 = 4.2 x10-4
F = ϬA = 4.2 x10-4 x250,000 = 105N
For member CE and EA
Ϭ = Es
=210x60 x10-6 = 1.26 x10-2
F = ϬA = 1.26 x10-2x250, 000 = 315N
For member GA
Ϭ = Es
=210x58 x10-6 = 1.218 x10-2
F = ϬA = 1.218 x10-2x250, 000 = 304.5N
This member stresses and forces are a mirror image to the left.
Work cited
Bolton, William. Mechanical Science. Malden, MA: Blackwell Publishers, 2005. Print.
