MECHANICS AND MATERIALS LABORATORY
SEM 2 2012/13
EXPERIMENT 6: THIN CYLINDER
DATE PERFORMED: 13TH DECEMBER 2012
DUE DATE: 20TH DECEMBER 2012
SECTION: 2
GROUP NUMBER: 6
GROUP MEMBERS: a) MUHAMAD HADI BIN MOHAMED RADZI (ME087932) b) THINES A/L MURUGAN (ME086895) c) MUHAMMAD HASRUL BIN ROSLI (ME087000) d) HAIZUM AMALINA BINTI A. WAHID (ME087898)
LAB INSTRUCTOR: MADAM NOLIA HARUDIN
TABLE OF CONTENT
No. | Content | Page | 1. | Summary / Abstract | 3 | 2. | Statement of Purpose / Introduction / Objectives | 4 | 3. | Theory | 4-10 | 4. | Equipment / Description of Experimental Apparatus | 11-13 | 5. | Procedure | 14-15 | 6. | Data and Observations | 16-17 | 7. | Analysis and Results * …show more content…
Considering an element of material:
H will cause strains of:-
H1 = H / E and L1 = - H / E
And these are the two principal strains. As can be seen from the equation, in this condition L will be negative quantity, i.e. the cylinder in the longitudinal direction will be in compression.
ii) Closed End Condition
By constraining the ends, a longitudinal as well as circumferential stress will be imposed upon the cylinder. Considering an element of material:
H will cause strains of:-
H = H / E and L1 = - H / E
L will cause strains of:-
L = L / E and H = - L / E
The principal strains are a combination of these values i.e.
H=(H - L)/E and L=(L - H)/E
The principal strains may be evaluated and a Mohr Strain Circle constructed for each test condition. From this circle the strain at any position relative to the principal axes may be determined. iii) To determine a value for Poisson’s Ratio
Dividing the equations give;
L1 / H1 = - This equation only applicable to the open ends