24/7 writing help on your phone

Essay,
Pages 5 (1184 words)

Views

129

“Beams are long straight members that are subjected to loads perpendicular to their longitudinal axis and are classified according to the way they are supported”[1]. When a beam is subjected to an external load there are unseen internal forces within the beam that one must be aware of when implementing it into any design or structure. These internal forces create stress and strain that could result in failure or deformation. This lab looked at how an aluminum cantilevered beam performed under symmetric and unsymmetrical bending as well as the stresses and strains developed as a result.

Objective

“To study the stress and strain induced in an I-beam under symmetric and unsymmetrical bending” [2]. Theory: ? – Normal stress (Mpa) ? – Strain (mm/mm) M – Moment (kN•m) I – Moment of inertia (mm^6) E – Modulus of elasticity (Mpa) G – Modulus of elasticity (Mpa) v – Poisson’s ratio. L – Length (m) *Subscripts x, y, z indicate plane of reference. The strain rosettes are orientated so that ? b = 0, ? c = -45, and ? a = 45.

The strain gauge equations then simplify to ?x = ? b, ? y= ? c+ ? a- ? b, and ? xy = ? c- ? a Using Hooke’s Law: ?x= ? xE, ? y= -v ? x, ? xy=? xyG This Experiment consisted of symmetric and unsymmetrical bending.

For symmetric bending the relevant theory is as follows: Because the moment about the z-axis here is zero the equation equates to: Where: My = PLA. When rotated 45 degrees: My = PLA Cos(45) and Mz = PLA Sin(45) there is compressive stress along the y-x axis The moment of inertia about the y-axis is found by determining the inertia of the shape and subtracting the imaginary parts as shown: The max normal stress with be at the furthest distance from the neutral axis which is h/2 therefore: (? x)max = The strains can be found by implementing Hooke’s Law: Since ? y and ? z are zero in symmetric loading, the two equations simplify to:

Because the there is no shear stress in the x-y plane when the normal stress is at maximum the shear strain will also be zero.

The vertical displacement of the end of the beam is determined by multiplying the area under the moment diagram and the distance between the end and the centroid of the diagram. This equates to: For unsymmetrical bending the theory is the same however there is a moment about the y-axis and z-axis. This will affect the calculation of the normal stress and the strain in the x and y plane. Also the moment of inertia in the z-direction will need to be determined.

Procedure (a) *Mount the I-beam on to the support frame. Make sure the mounting screws are tight. (b) Measure the dimensions of the I-beam including its components. (c) Mount the magnet bases of the dial gauges at appropriate positions to permit the measurements of the deflections at the free end of the beam in the vertical and the horizontal directions. (d) *Connect properly the wires from the strain gauges to the readout unit. (e) Place weights to the hanger in increments: 4, 6, 10, 26, and 42 kg. (f) Unload the hanger in increments in the reversed order as for loading.

(g) For each increment, measured the strain readings at the given locations and the vertical and horizontal deflections at the free end of the beam. (h) Repeat steps (a) to (g) by rotating the beam with the following angles: 45°. [3] Results *Refer to appendix for sample calculation and calculated results. Part 1: I-beam at 0o Loading Loading (Kg) 4 6 10 26 42 Strain Gauge 1 (? ) 1 2 4 12 20 Strain Gauge 2 (? ) 6 10 16 43 69 Strain Gauge 3 (? ) 3 4 7 18 29 Displacement 1 (mm) 0. 09 0. 15 0. 23 0. 44 0. 5 Displacement 2 (mm) -0. 19 -0. 34 -0. 55 -1. 4 -2. 25 Load (N) 39. 2 58. 5 97. 9 255. 5 413. 1

Unloading Loading (kg) 42 26 10 6 4 Strain Gauge 1 (? ) 20 10 -3 -5 -7 Strain Gauge 2 (? ) 69 42 19 11 9 Strain Gauge 3 (? ) 29 18 6 3 2 Displacement 1 (mm) 0. 5 0. 49 0. 25 0. 16 0. 07 Displacement 2 (mm) -2. 25 -1. 46 -0. 59 -0. 37 -0. 23 Load (N) 413. 1 255. 6 96. 4 58. 7 39. 2 Part 2: I-Beam at 45o Loading Loading (kg) 4 6 10 26 42 Strain Gauge 1(? ) 1 2 2 7 13 Strain Gauge 2 (? ) 5 9 14 36 54 Strain Gauge 3 (? ) 1 1 2 8 13 Displacement 1 (mm) -0. 33 -0. 50 -0. 79 -1. 88 -2. 75 Displacement 2 (mm) -0. 66 -1. 02 -1. 69 -4. 23 -6. 40 Load (N) 39. 4 58. 7 98. 2 256. 5 413. 6 Unloading Loading (kg)

42 26 10 6 4 Strain Gauge 1 (? ) 13 4 -22 -25 -26 Strain Gauge 2 (? ) 54 38 22 20 17 Strain Gauge 3 (? ) 13 6 2 0 0 Displacement 1 (mm) -2. 75 -1. 95 -0. 92 -0. 62 0. 46 Displacement 2 (mm) -6. 40 -4. 46 -2. 17 -1. 51 -1. 15 Load (N) 413. 6 256. 3 98. 1 58. 7 39. 4 Discussion For both the symmetric and unsymmetrical bending the theoretical stresses and strains were greater than experimentally determined ones. However the experimental displacement was much higher than the theoretical displacement. These two factors can lead one to believe the I-beam has undergone this procedure many times before.

Another interesting point to note is that the stresses and strains are higher at equivalent loads when unloading demonstrating that there is residual stress in the I-beam even after it has been fully unloaded. For the most part however the measured and theoretical values are very close. It is to be expected that the theoretical stresses would be higher than the experimental values. The theoretical calculations rely on a ‘perfect’ material. The modulus of elasticity and cross-sectional are said to remain the same through the length of the beam which is rarely the case.

Minor imperfections in the beam would result in a weaker beam and less stress is required to deflect the beam. This is exactly what has been observed in this experiment. For the symmetric and bending theoretically there would be no horizontal displacement however some horizontal displacement was shown on the readouts. This is most likely due to the slight swaying of the weights. Since the scale of this experiment was relatively small a lot of the sources of error are pretty large. Just by not having the readout computer not calibrated properly or zeroed all the way would cause pretty large discrepancies.

Even the measuring or millimeters by eye caused some error. Rounding errors would be relatively small for this experiment. Conclusion In conclusion theoretical and experimental values for stress and strain are very similar to the values observed in experimental conditions. The theoretical and experimental displacements were pretty far off and at larger scales the theoretical values would not be of much use. Closer results could have been obtained by collecting more accurate measurements or by collecting multiple sets of data using a series of strain rosettes. APPENDIX I Sample Calculations Iy= = (Mz)a =(4kg)(9. 81m/s2)(0.

77m) =30. 215 Nm (Mz)b =(4kg)(9. 81m/s2)(0. 33m) =12. 95 Nm (? x)a = = = 1. 259 Mpa (? x)b = 0. 5397 Mpa (? b)v = = – = -0. 0902 mm ?xy = = = 0. 0398mm (? x)a = = =17. 22*10^-6 (? y)a = -0. 35*(? x)a = 6. 027*10^-6 Experimental – Symmetric Mass (Kg) 4 6 10 26 42 26 10 6 4 ?x (E-6) 6 10 16 43 69 42 19 11 9 ?y (E-6) -2 -4 -5 -13 -20 -14 -16 -13 -14 ?xy (Mpa) 2 2 3 6 9 8 9 8 9 (? x) (Mpa) 0. 438 0. 731 1. 17 3. 14 5. 04 3. 07 1. 39 0. 804 0. 657 (? y) (Mpa) -0. 146 -0. 292 -0. 365 -0. 950 -1. 46 -1. 02 -1. 17 -0. 950 -1. 02 ?xy (Mpa) 0. 054 0. 054 0. 081 0. 162 0. 243 0. 216 0. 243 0. 216 0. 243 Theoretical – Symmetric

Mass(Kg) 4 6 10 26 42 (Mz)a (N•m) 30. 2 45. 3 75. 5 196 317 (Mb)b (N•m) 12. 9 19. 4 32. 3 84. 1 135 (? x)a (Mpa) 1. 25 1. 88 3. 12 8. 13 13. 1 (? x)b (Mpa) 0. 536 0. 804 1. 34 3. 48 5. 62 ?xy (Mpa) 0. 0398 0. 0598 0. 0996 0. 258 0. 418 (? x)a (E-6) 17. 1 25. 7 42. 8 111 179 (? x)b (E-6) 7. 33 11. 0 18. 3 47. 6 77. 0 (? y)a (E-6) -5. 99 -8. 98 -14. 9 -38. 9 -62. 8 (? y)b (E-6) -2. 57 -3. 85 -6. 41 -16. 6 -26. 9 ?a (mm) 0. 0902 0. 135 0. 225 0. 586 0. 947 ?b (mm) 0. 00710 0. 0106 0. 0177 0. 0461 0. 0745 Experimental –Unsymmetrical Bending Mass (Kg) 4 6 10 26 42 26 10 6 4 (? x) (E-6) 5 9 14 36 54 38 22 20 17 (? y) (E-6)

-3 -6 -10 -21 -28 -28 -42 -45 -43 ?xy (E-6) 0 -1 0 1 0 2 24 25 26 (? x) (Mpa) 0. 366 0. 658 1. 02 2. 63 3. 95 2. 78 1. 61 1. 46 1. 24 (? y) (Mpa) -0. 219 -0. 439 -0. 731 -1. 54 -2. 05 -2. 05 -3. 07 -3. 29 -3. 14 Theoretical – Unsymmetrical Bending Mass (Kg) 4 6 10 26 42 (Mz,y)a (N•m) 21. 3 32. 0 53. 4 138 224 (Mz,y)b (N•m) 9. 15 13. 7 22. 9 59. 5 96. 1 (? x) (Mpa) 0. 381 0. 572 0. 954 2. 48 4. 00 (? y) (Mpa) -1. 40 -2. 10 -3. 51 -9. 12 -14. 7 (? x) (E-6) 5. 22 7. 83 13. 1 33. 9 54. 8 (? y) (E-6) 1. 83 2. 74 4. 57 11. 9 19. 2 ?x-y (mm) 0. 0902 0. 135 0. 225 0. 586 0. 946 ?x-z (mm) 0. 391 0. 587 0. 978 2. 54 4. 11

👋 Hi! I’m your smart assistant Amy!

Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.

get help with your assignment