PV LAB REPORT Strength of Engineering Materials (48642)
Abstract Biaxial Stress in a Hydraulic Cylinder Pressurized Cylinder with Open & Closed Ends
Ben Heo 12005183
1. Introduction This lab report will explore the transformation of strain theory as well as thin walled cylindrical pressure vessel theory, mainly focusing on the calculation aspects of these theories. Using a thin walled cylinder undergoing biaxial stress, the application of yield criteria to predict onset of yield will be demonstrated.
2. Equipment A mild steel hydraulically pressurised cylinder is used as a thin walled cylinder. This apparatus simulates both open and closed end conditions through a thrust knob, situated on the front of the cylinder. Open end condition is achieved when the thrust knob is “wound in”, the external frame reacts to the force placed on the piston and there is no axial load on the cylinder walls. Closed end condition is achieved when the thrust knob is “wound out”, the pressure from the piston on the cylinder develops an axial stress in the cylinder walls. Foil-type strain gauges have been secured to the cylinder, each placed at a different angle (0°, 15°, 30°, 45°, 75° and 90°). The thin walled cylinder is connected to a hydraulic pump which is used to pressurise the cylinder. Attached to the hydraulic pump is a pressure gauge, used to determine the pressure in the cylinder (MPa). Attached to the cylinder are foil-type strain gauges, the readings display the strain in micrometres (μm).
3. Test Procedure Open End: 1) Record the inner and outer diameters of the pressure vessel. 2) Set the cylinder to the “Open End” condition. (Bleed out all pressure then wind the thrust knob completel y “in”. 3) Verify that the strain bridge is properly connected and balanced. Biaxial Stress – Hydraulic Cylinder with Open and Closed Ends . 4) Pump up the pressure to 1 MPa and hold the pressure constant by gently operating the pump lever. 5) Read the strain on gauges at 0º and 90º. 6) Repeat steps 3 and 4for pressures of 2 and 3 MPa. 7) Plot the readings of strain versus pressure and from lines of best fit, determine the strains at a pressure of 2.5 MPa. Closed End: 1) Set the cylinder to the “Closed End” condition. (Bleed pressure and wind the thrust knob completely “out”. 2) Balance the strain bridge. 3) Pump up the pressure to 1 MPa and hold the pressure constant by gently operating the pump lever.
4) Read the strain on all six gauges. 5) Repeat steps 3 and 4 for pressures of 2 and 3 MPa.
4. Results: Cylinder dimensions: Outside Diameter: 82.63mm Inside Diameter: 76.18 mm Open End P (MPa) 0 1 2 3
Micro Strain ε 0° 0 -14 -29 -46
ε 90° 0 57 113 169
200 y = 56.3x + 0.3 150
100
50
0 0
0.5
1
1.5
2
2.5
3
3.5
y = -15.3x + 0.7
-50
-100
From the chart, it can be seen at 2.5 MPa:
ε 0°= -37.55
ε 90°= 141.05
Closed End P (MPa) 0 1 2 3
Micro Strain (μ) ε 0° ε 45° 0 0 14 31 26 59 38 87
ε 90° 0 48 94 138
ε 15° 0 18 32 47
ε 30° 0 23 44 64
ε 75 0 46 89 131
160 140
y = 46x + 1 y = 43.6x + 1.1
120 100 y = 28.9x + 0.9 80 y = 21.3x + 0.8
60
y = 15.5x + 1 40 y = 12.6x + 0.6 20 0 0
0.5
1
1.5
2
2.5
3
From the chart, it can be seen at 2.5 MPa: ε 0° 32.1
ε 45° 73.15
ε 90° 116
ε 15° 39.75
ε 30° 54.05
5. Calculations Open End:
From above using data where P = 2.5 MPa
ε 0°= -37.55 x 10^-6
ε 90°= 141.05 x 10^-6
Poisson’s Ratio:
= −(ε/ ε) = −(-37.55 x 10^-6/141.05 x 10^-6) = 0.266
Thin Shell Theory:
P= 2.5 MPa , d=76.15mm , t=3.225mm
= 29.5 MPa
ε 75° 110.1
3.5
Elastic Modulus:
= /
= 209 GPa
Closed End:
Given data from above;
σH = 29.5 MPa.
P = 2.5 Mpa
D= 76.18mm
T= 3.255mm
V= 0.266
Longitudinal Stress:
= 14.6MPa
Hoop and Longitudinal Strain:
= 143 x 10 ^ -6
= 38.7 x 10 ^ -6
6. Analysis & Interpretation of Results Open End Situation: Tresca max, shear stress criterion: σH < σY
29.5 MPa < 300 MPa
= 24,193.5 KPa Von-Mises max, distortion strain energy criterion: = 870.25 < 90,000 Therefore, yield will occur at 300 MPa, P = 22765.72 KPa
Closed End Situation, Tresca max, shear stress criterion: σH < σY
29.5 MPa < 300 MPa
=24,193.5 KPa Von-Mises max, distortion strain energy criterion: 695.6 < 90,000 Therefore, yield will occur at 300 MPa, P = 22765.72 KPa
