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Biaxial Stress in a Hydraulic Cylinder Cory Bostock - 97117269 Aaron Huynh - 99137869 Joel Kubo - 11964956 Les Marshall-Peel - 98123793
Biaxial Stress in a Hydraulic Cylinder Part 1 - Hydraulic Cylinder with Pressure and Torsion 1. Objectives 1.1 To verify pressure in thin walled cylinder and torsion in hollow section theories. 1.2 To determine E, G and for an extruded aluminium alloy. 1.3 To verify calculation of stress from strain gauge measurements. 1.4 To demonstrate the validity of superposition of stress. 1.5 To apply appropriate yield criteria to predict onset of yield for a general state of stress. 2. Apparatus An extruded aluminium alloy cylinder, inside diameter 95mm, wall thickness 3.3mm, is subjected to internal pressure and a torque. As shown in Figure 1, the torque is developed by the application of two weights (W) to a 2m long loading bar. Two rectangular strain gauge rosettes are bonded to the cylinder as shown in Figure 2.
3. Theory 4. Procedure Note the true gauge factors and the strain amplifier gauge factor setting and then measure the strain on each gauge for the following load cases: i) T = 0, p = 0 2 MPa (in increments of 0.5 MPa) ii) p = 0, T = 0 300 N.m (in increments of 100 N.m) iii) T = 200 N.m, p = 1.5 MPa.
5. Results 5.1. Data Processing The results are processed using a spreadsheet which corrects for gauge factor and carries out a linear regression to give the six strains for: i) T = 0, p = 1.5 MPa ii) T = 200 N.m, p = 0 MPa iii) T = 200 N.m, p = 1.5 MPa The spreadsheet also calculates the magnitude and orientation of the principal strains. This is done by calculating the average of the Mohr’s circle parameters, C, R and θ for each rosette – see Figure 3. The ε x reference gauges are taken to be gauge (i) for rosette A and gauge (iv) for rosette B – see Figure 2. Angle θ is therefore calculated with reference to axis x A for rosette A and axis x B for rosette B. The transformation equations for C, R and θ are given in Appendix B. Note that for pressure without torsion, the hoop and longitudinal strains are the principal strains with ε 1 = ε H and ε 2 = ε L. Torsion without pressure represents pure shear with ε 1,2 = ± ε xy .
6. Calculations Full calculations can be found in the appendix section: 6.1. Given: P = 1.5MPa d = 95mm t = 3.3mm T = 0Nm ε H = 243.3x10 -6 This yields: v = 0.345273 E = 73421.82 MPa = 73.42GPa 6.2. Given: T = 200Nm d = 95mm t = 3.3mm P = 0MPa T xy = 1.112MPa G = 35.73GPa 6.3. The results for tests [T = 0, p = 1.5 MPa] and [T = 200N.m, p = 0] are added together 6.4. Please see appendix section for details.

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