McGill University Department of Civil Engineering and Applied Mechanics
Solid Mechanics CIVE-207
Beams Flexure Experiment Laboratory Write-Up Report 33 Pages
Group 34 Chemen, Uyirendiren
260732722
Earle, Stuart
260735993
Nottegar, Alex
260707220
Zhang, Qitong
260732944
Sabourin, Christophe
260744128
March 16th, 2018
Table of Contents List of Graphs, Drawings and Figures, Photographs and Videos, Tables …................. p.2 Graphs …........................................................................................................................ p.4 Photographs …............................................................................................................... p.13 p.13 Drawings and Figures ………………………..………..................................................p.21 ………………………..………..................................................p.21 Tables …........................................................................................................................ ... p.26 Discussion ………………………… ……………………………………………………… ……………………………………………...….. ………………...….. p.26 Appendix …................................................................................................................... p.28 p.28 References ……………………………… ………………………………………………………… ………………………………………...…. ……………...…. p. 33
1
List of Graphs, Drawings and Figures, Photographs Photographs and Videos, Tables Graphs
Influence of Young’s Modulus of Elasticity 1. 2. 3. 4.
Applied load (N) vs. Mid-span Mid-span deflection, δmid-span, (mm) Applied load (N) vs. Angle of end rotation, θend, (rad) Applied load (N) vs. 48δ0.5LI/L3 (mm2) Flexural stiffness, k, (N∙mm) vs. Moment of inertia, I, (mm4)
Influence of Support Conditions 5. 6. 7. 8. 9.
Applied load (N) vs. Maximum deflection, δmax, (mm) Applied load (N) vs. Free-end Free-end deflection, δfree-end, (mm) Applied load (N) vs. Angle of rotation, θ, (rad) 1/δmax (mm-1) vs. EI/PL3 (mm-1) Flexural stiffness, k, (N∙mm) vs. Maximum deflection, δmax, (mm)
Photographs
Influence of Young’s Modulus of Elasticity Mild Steel 1. Mild steel test before adding load 2. Mild steel test at maximum load Aluminum 3. Aluminum test before adding load 4. Aluminum test at maximum load Brass 5. Brass test before adding load 6. Brass test at maximum load Wood 7. Wood test before adding load 8. Wood test at maximum load Acrylic 9. Acrylic test before adding load 10. Acrylic test at maximum load 2
Influence of Support Conditions Propped Cantilever (fixed-roller) 11. Propped cantilever test before adding load 12. Propped cantilever test at maximum load Cantilever (fixed-free) 13. Cantilever test before adding load 14. Cantilever test at maximum load Fixed Ended (fixed-fixed) 15. Fixed ended test before adding load Drawings and Figures
Influence of Support Conditions Propped Cantilever (fixed-roller) 1. 2. 3. 4.
Line diagram of beam configuration Elastic curve profile Shear force diagram Bending moment diagram
Cantilever (fixed-free) 5. 6. 7. 8.
Line diagram of beam configuration Elastic curve profile Shear force diagram Bending moment diagram
Fixed Ended (fixed-fixed) 9. Line diagram of beam configuration 10. Elastic curve profile 11. Shear force diagram 12. Bending moment diagram Tables
1. Influence of Young’s modulus of elasticity theoretical vs. experimental values 2. Influence of support conditions theoretical vs. experimental values
3
Graphs
1 h p a r G
4
2 h p a r G
5
3 h p a r G
6
4 h p a r G
7
5 h p a r G
8
6 h p a r G
9
7 h p a r G
10
8 h p a r G
11
9 h p a r G
12
Photographs Influence of Young’s Modulus of Elasticity Mild Steel
Photograph 1: Mild steel test before adding load
Photograph 2: Mild steel test at maximum load
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Aluminum
Photograph 3: Aluminum test before adding load
Photograph 4: Aluminum test at maximum load
14
Brass
Photograph 5: Brass test before adding load
Photograph 6: Brass test at maximum load
15
Wood
Photograph 7: Wood test before adding load
Photograph 8: Wood test at maximum load
16
Acrylic
Photograph 9: Acrylic test before adding load
Photograph 10: Acrylic test at maximum load
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Influence of Support Conditions Propped Cantilever (fixed-roller)
Photograph 11: Propped cantilever test before adding load
Photograph 12: Propped cantilever test at maximum load
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Cantilever (fixed-fixed)
Photograph 13: Cantilever test before adding load
Photograph 14: Cantilever test at maximum load
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Fixed Ended (fixed-fixed)
Photograph 15: Fixed ended test before maximum load
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Drawings and Figures Propped Cantilever (fixed-roller)
1Figure
1: Propped cantilever line diagram of beam configuration
2Figure
2: Propped cantilever elastic curve profile
1
Goodno, Gere; 2013. Gere, Timoshenko; 1997.
2
21
2Figure
2Figure
3: Propped cantilever shear force diagram
4: Propped cantilever bending moment diagram
Cantilever (fixed-free)
Figure 5: Cantilever line diagram of beam configuration
22
3
Figure 6: Cantilever elastic curve profile
4Figure
7: Cantilever shear force diagram
3
Beer, Dewolf, Johnston; 2002. http://output.to/sideway/default.asp?qno=120800023
4
23
4Figure
8: Cantilever bending moment diagram
Fixed Ended (fixed-fixed)
5Figure
5
9: Fixed ended line diagram of beam configuration
http://www.structx.com/Beam_Formulas_016.html
24
Figure 10: Propped cantilever elastic curve profile
5Figure
5Figure
11: Fixed ended shear force diagram
12: Fixed ended bending moment diagram
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Tables Table 1: Influence of Young’s Modulus of Elasticity Theoretical vs. Experimental Values Mild Steel T E E (GPa) Pmax (N)
Wood
T
E
T
E
Acrylic T E
178
70
63.5
100
103
12
5.18
2.8
2.90
20
20
20
20
20
20
20
20
20
20
2.30
2.58
6.41
7.09
5.20
5.02
2.86
6.60
3.53
3.00
0.0172
0.0184
0.0480
0.0710
0.0390
0.0324
0.0570
0.0908
0.106
0.0604
2000
2000
2000
2000
2000
2000
750
750
500
500
-
57.15
-
56.20
-
61.94
-
27.09
-
16.32
(mm) (rad) Mmax (N∙mm) σmax (MPa)
Brass
200
δmax θmax
Aluminum T E
Table 2: Influence of Support Conditions Theoretical vs. Experimental Values
Pmax (N) δmax (mm) θmax (rad) α
Mmax (N∙mm)
Propped Cantilever (fixed-roller) T E 20 20 1.07 1.24 0.0138 0.0102 95.45 1285.2
Cantilever (fixed-free) T
E
20 4.60 0.0345 4000
20 4.76 0.0487 29.11
Fixed Ended (fixed-fixed) T 20 0.58 1000
E 20 0.69 182.81
* Theoretical values obtained from Material Properties Table.
Discussion Comparison of Experimental vs. Theoretical Results One can notice that the experimental results for the wood are double what they should be which can be explained by the fact that the Young's Modulus is halved. This is most likely due to the wide range of Young's Moduli for different types of wood. Also, there is a l ot of variance in wood from sample to sample due to its organic nature. For all the different support conditions for mild steel, the theoretical value is lower than the experimental value. This may be explained by the fact that the beam had been used by previous teams and had become slightly less resistant to flexure than initially. 26
The experimental angle for acrylic is lower than expected, this can be explained by the fact that the dial gauges were pushing downward against the beams, lowering the angle. Having the lowest Young's Modulus, the acrylic's angle should be that which is most affected. The angle would also have been increased by the dial gauge at the center, for all beams but the acrylic, the middle gauge should overcome the effect of the edge gauge due to its bigger distance from the pivot, explaining higher angles for these materials.
Sources of Error This experiment was carried out by following a precise procedure in order to minimize potential error. Many sources of error were accounted for by takin g the initial deflections of the beams, and the results were zeroed to these values. Despite this, some sources of error will be acknowledged and explained here. Those discussed in this section are the defects in beam materials and frequent use of certain materials.
Defects in the materials can range from manufacturer mishaps to damages sustained in the laboratory, and can be very difficult to diagnose. For that reason, material defects has been included as a primary source of error. In reference to results received in this experiment, material defects is the outstanding reason for the large difference between the experimental and theoretical value for the Young’s Modulus of the wooden specimen. As the procedure was followed in a nearly identical manner for all beams, the main reason for this large difference can be deduced as a defect in the wood. This is the rational because wood as a material is cut and sanded to size from trees, which contain knots and ruts, while the other materials are manufactured to size.
Frequent use of the particular beams used in this experiment is also a source of error. This experiment was performed after a series of groups had already completed the same experiment, thus each beam had already experienced the loading several times. After several uses, these materials will not perform as adequately as when they were produced. This means the beams will most likely deflect more than they would have in the first trial, which can be linked to the values of the experimental Young’s Moduli being slightly less than the theoretical values for the materials.
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Appendix Derivation of Fixed-Fixed elastic curve equation To calculate elastic curve v(x) and θ(x), we start by dividing the displacement into one force induced displacement and two moment induced displacements. Suppose ve ctors going up and right is positive. The displacement v induced by a downward force P along x axis is:
4 3 = 48 ( 0 < < 2) For x>L/2, substitute x = (L-x) into the equation above we get:
84 = (2 < < ) 48 Then M is the same on both side because P acts at the middle point, so we can have v induced by the left side moment is:
3 2 = 6 The displacement, v, induced b y the right side moment is obtained by substitute x with (L-x):
= 6 Then we add all equations up and get:
4 2 22 3 ) = 48 ( 0 < < 6 2 84 2 22 = (2 <<) 48 6 Then we take the derivatives to get θ(x):
2 = 16 4 3 3 ( 0 < < 2) 2 = 48 3 4 3 3 (2 < < )
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Derivation of shear stresses The maximum load value was constant for each material as no beam failed at a loading of 20 N. All values below that were experimentally gathered can be seen in original data recording sheets. The mid-span deflections were experimentally gathered from dial readings. The maximum deflection was calculated by subtracting the initial deflection from the deflection when the 20 N load was applied:
, = , , The end rotation values were also experimentally gathered and the maximum value was calculated by making the small angle approximation:
=
Where is the angle of rotation, calculated from the division of the end deflection by the rigid arm horizontal distance, which in each case was 50 mm. The full equation for the max angle of rotation is as follows:
, = ,50 The formula used to calculate the Young’s modulus for each material was derived by re-arranging the deflection equation for a simply – supported beam:
=
= ,
Where P is the loading (20 N), L is the span length, and I is the second moment of inertia, which for all beams was equal to
.
The maximum bending moment was derived using beam-analysis of a simply – supported beam, where the maximum moment would occur halfway along the span and thus be equal to:
= 4 Finally, the maximum flexural stress derived from the equation:
= Where M is the bending moment, and y is the distance from the neutral axis. In order for the stress to be maximized, both of these values must be maximized as I is a constant. The equation for the maximum bending moment for each material is derived above, so we’ll return M as M max. The neutral axis for these rectangular beams is located directly in the middle of the cross-section and since they are being rotated with the long side parallel to the ground, the maximum distance from the neutral axis is half of the short side, or the height, h, of the beam. Thus our maximum flexural stress equation is: 29
ℎ = 2 Derivation of max deflection and max slope To calculate the deflection and slope of the beams theoretically, one must start from the FBD of the beam as a whole. In the case of the simply supported beam with a load applied at the middle of the beam, one can deduct that the reaction forces at both ends are .. To derive the equations for the maximal deflection and slope on the beam, one must look at the FBD of the first and second half of the beam separately. Respectively the sum of moments in each FBD is:
= = /2
=/2 and =/2 /2 (i) Because the beam is in equilibrium:
= 0 < < /2 and = /2 /2 /2 << (ii) Since the moment is product of the second derivative of the deflection and the flexural rigidity, we can integrate once to get:
′ = /4 and = /4 /2/2 (iii) Integrate once more to get:
= /12
and
= /12/2/6 (iv)
The equations of continuity for the beams are:
/2− = /2+
and
/2− =/2+ (v)
By applying these equations, we get that:
= and =
(vi)
The boundary conditions for the simply supported beams are:
0 = = 0
(vii)
Using the boundary conditions in (iv), we get that:
= 0 and = /16
(viii)
We can deduce and observe that the maximal deflection occurs at the middle of the beam, where the load is applied. Subbing in (iv) and using the constants (viii), we get that:
= /2 = /48
(ix)
Similarly, for the maximum slope, we can observe that it occurs at the ends of the beams. Subbing in (iii) and using the constants (viii), we get that:
=0
30
= = /16
(x)
For the cantilever beam, we are only interested in the deflection and slope at the middle of the beam, where the load is applied because no forces act on the rest of the beam. Using a FBD of the beam from the free end at any arbitrary point, we get that the sum of the moments equals:
=
(xi)
Since the beam is in equilibrium:
=
(xii)
Since the moment is product of the second derivative of the deflection and the flexural rigidity, we can integrate once to get:
=/2
(xiii)
Integrate once more to get:
= /6
(xiv)
The boundary conditions of a cantilever beam are:
= 0 and = 0
(xv)
Subbing these boundary conditions into (xiv), we get that:
= /2
and
= /3
(xvi)
The maximum moment and slope of a cantilever beam occurs where the load is applied (x = 0) in this case. Using (xiii) and (xiv) as well as the con stants (xvi), we get that:
= /3
and
= = /2
(xvii)
To calculate the maximum slope and deflection of a propped cantilever beam, we must combine a simple cantilever and two cantilever beam where one has our load applied to it and the other one has the reaction force at the end acting on it. To calculate the deflection of a cantilever beam where the load is not applied at the end, we must add the deflection at the load and the slope times the remaining length. Assuming the load is applied at point a and beam has length L, we get:
= /3 /2 = 3/6 To calculate the slope and max deflection, we must first calculate the reaction force at the roller end. This can be done by adding a the two deflections and setting the sum to 0 as the roller is level with the other end. The only remaining variable is R:
=0= 3 /6 /3 Isolating R gives 7.41N. Now that we have the forces acting on the beam, we can determine the slope and max deflection by integrating the equation for the moment twice then using boundary 31
= 0, 0 = 0 as
condition Resulting equations are:
well as continuity equations to calculate the constants.
= /2 /2/2 = /2 /2/2 /2 Where b = x = 180mm For the fixed-fixed beam, there is no slope at the edges. To calculate the maximal theoretical deflection, we must use the elastic curve equation calculated above and sub x = L/2 which is where the maximal deflection occurs.
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