cantilever beam
TITLE: CANTILEVER BEAM
OBJECTIVE:
To determine the deflection of cantilever beams
APPARATUS:
| Num | Part Name | Num | Part Name |
| 1 | Fixed Support | 4 | Dial Gauge |
| 2 | Cantilever beam | 5 | Weight (W) |
| 3 | Clamping lever | 6 | Base Plate |
L
4
1 2
5
6
Figure E2
THEORY:
The deflection at free end of a cantilever beam is given by this relation
y max = 
where F=
PROCEDURE:
- The apparatus is set up as shown in Figure E2.
- The distance of L is measured.
- The scale on the dial gauge is set to zero
- The beam is loaded with 5 N.
- The dial gauge is read and recorded.
- The load is increased and the reading is recorded.
- Step 6 is repeated to get at least four reading.
- Step 1 to 7 is repeated by using the beam with the different cross-sectional area.
RESULT:
Beam 1: I- Cross sectional
L = 550 mm I = 12736 mm4
E = 68000 N/mm2 Beam dimension: 25 mm x 25 mm x 3 mm
| W (N) | Yb (mm) Exp | Yb (mm) Theory |
| 5 | 0.300 | 0.352 |
| 10 | 0.670 | 0.704 |
| 15 | 1.032 | 1.057 |
| 20 | 1.450 | 1.409 |
| 25 | 1.690 | 1.761 |
Beam 2: L - Cross sectional
L = 550 mm I = 13030 mm4
E = 68000 N/mm2 Beam dimension: 25 mm x 25 mm x 3 mm
| W (N) | Yb (mm) Exp | Yb (mm) Theory |
| 5 | 0.290 | 0.344 |
| 10 | 0.620 | 0.689 |
| 15 | 0.970 | 1.033 |
| 20 | 1.320 | 1.377 |
| 25 | 1.650 | 1.721 |
Beam 3: U- Cross sectional
L = 550 mm I= 19970 mm4
E = 68000 N/mm2 Beam dimension: 25 mm x 25 mm x 3 mm
| W (N) | Yb (mm) Exp | Yb (mm) Theory |
| 5 | 0.260 | 0.225 |
| 10 | 0.470 | 0.449 |
| 15 | 0.700 | 0.674 |
| 20 | 0.830 | 0.898 |
| 25 | 1.140 | 1.123 |
The graph of deflection against weight.
A) Beam 1: I- Cross sectional
Graph Deflection (y) versus Weight (F) – Experiment
 |
Graph Deflection (y) versus Weight (F) – Theory
 |
B) Beam 2: L - Cross sectional
Graph Deflection (y) versus Weight (F) – Experiment
 |
Graph Deflection (y) versus Weight (F) – Theory
 |
C) Beam 3: U- Cross sectional
Graph Deflection (y) versus Weight (F) – Experiment
 |
Graph Deflection (y) versus Weight (F) – Theory
 |
DISCUSSION:
- Sample of calculation (theory)
- for I-cross sectional
ymax = FL3
3EI
Where, F = 1.1(W)
ymax = 1.1(W) (550)3
3(68000) (12736)
Where, W = 5N
ymax = 1.1(5) (550)3
3(68000) (12736)
ymax = 0.352 mm
Percentage of error, %δ
when W = 5N
%δ = yv (theory) – yv (exp) x 100%
yv (theory)
= 0.352 – 0.320 x 100%
0.352
= 9.1%
- Compare the theoretical and experimental results.
From the experiments results, we can see they are more different than the theoretical values we get from the calculation. Some factors of errors were determined in this experiment:
- The measuring instrument that is not perfect – errors when reading the scale on the measuring instrument such as meter ruler. In addition, the zero errors where the tools are damage and ‘zero’ point cannot be seen. So, take the errors and correct it. A few instruments which is too old should be replace with the new one or recondition to remove the corrosive surfaces and make the scale more clear as they will cause the imperfect experiment.
- Parallax error due to reading taken from the experiment.
- Inadequate pre-load made between the dial gauge and the load
- The workbench might not in a flat position which contributes to unbalanced position of specimen and as a result the readings obtained were not accurate and precise as expected.
- The percentage of error.
% of error =
when W = 5N
= 0.300 – 0.352 x 100%
0.352
= -14.77%
CONCLUSION:
It can be concluded that the objective of the experiment have been achieved.
However, the values from the experiment not accurate because the percentage error is more than 5%, but we can minimize these errors by avoid relax parallax and make several adjustment on the apparatus and also replaced old apparatus.
REFFERENCE:
1. Advanced Mechanics of Materials, Robert D. Cook, Warren C. Young, Prentice Hall, Second Edition, 1999.
2. Mechanics of materials, Ferdinand P. beer, E. Russell Johnton, Jr, John T. deWolf, Mc Graw Hill, Third edition in SI units.
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