1.0 1.0 TITL TITLE E
Deflection of Curved Bars 2.0 INTRODUCTION
The deflection of a beam or bars must be often be limited in order to provide integrity and stability of structure or machine. Plus, code restrictions often require these members not vibrate or deflect severely in order to safely support their intended loading. This experiment helps us to show some kind of deflection and how to calculate the deflection value by using Castigliano’s Theorem and make a comparison between result of the experiment and the theory. 3.0 OBJECTIVE
To validate the formula of deflection of curved bars based on Castigliano’s Theorem. 4.0 APPARA APPARATUS TUS
TECQUIPMENT curved bar apparatus, quarter circle and circle bars. 5.0 TH THEO EORY RY
When a force is exerted on an elastic object, work is done on it will deform. The work energy which is stored in the elastic object is called strain energy. The elastic object will return to its original shape when the force is released. Strain energy can be caused by pressure, pressure, tension, tension, bending, bending, torsion, torsion, shear force, force, etc. Castigliano’s theorem states that the deflection caused by any external load is equal to the partial differential of the strain energy with respect to that external load. At every point of action action of these these forces forces there are deflecti deflections. ons. Figure 1 shows an elastic object which is acted upon by forces W1, W2, W3, …., Wn. At every point of action of these forces there are deflections. According to Castigliano’s Theorem, the deflections at these points are:
∂U = X1 ; ∂U = X 2; ∂U = X 3; ……..; ∂U = X n ∂W1 ∂W2 ∂W3 ∂W n With
U W
= =
total strain energy force at point
X
=
deflection at respective W
The deflection of point A in the direction of the force is equal to the partial differential of the total strain energy U for the system with respect to W 1 (force acting at point A).
W3
W2 C
B
x3
x2 A
W4 x4
D
x1 x5
W1
E W5
A. Quarter Circle Bar
For conservation of energy, the total strain energy which is caused by bending moment M is, U = ∫ M2/2EI ds (1) The bending moment for a quarter circle bar, at any section dθ is, M = PvRsinθ So, the strain energy stored due to a load Pv is, U = πPv2R 3/8EI
(2) (3)
The vertical deflection is then, δv = d U/d PV = πPR 3/4EI
(4)
The horizontal deflection can be calculated by applying an imaginary force PH in the horizontal direction. From Castigliano’s Theorem, the horizontal is given by,
δH = d U/d PH = PR 3/2EI
(5)
With R = radius of quarter circle bar = 150mm E = elastic modulus of bar material = 200GN/m2 I = second moment of area of the cross section I = bh3 / 12 b = width of the cross section = 25mm d = thickness of the cross section = 3mm
Figure 2(a) Quarter Circle Bar
Figure 2(b) Circular Bar
B. Circular Bar
The deflection which is caused by force PV is, δV
=
∂U ∂ Pv
PR³
=
2 π ( - ) EI 4 π PR³ = 0.149 EI
with the values of R, E and I the as above. The cross section of the bar is 25 x 3 mm.
6.0 PROCEDURE
A. Quarter Circle Bar
(i) (ii) (iii) (iv) (v)
A hanger attached to the free end of the bar. Two dial gauges clamped at their allocated points. 0.2 kg load applied and the vertical and horizontal deflections are recorded. Loads added in increment of 0.2 kg until a maximum load of 1.6 kg. The deflections recorded for each loading.
B. Circular Bar
(i)
(ii)
The above method followed but with a load increment of 0.5 kg until a maximum load of 4 kg. The vertical displacements for each loading recorded.
6.0 RESULTS
Load 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Vertical deflection, P1 (mm) Increasing load Decreasing load Average 0.38 0.41 0.40 0.83 1.00 0.92 1.35 1.40 1.38 1.82 1.93 1.88 2.34 2.42 2.38 2.80 2.82 2.81 3.25 3.35 3.30 3.79 3.79 3.79
Horizontal deflection,P2 (mm) Increasing load Decreasing load Average 0.18 0.35 0.27 0.48 0.76 0.62 0.89 1.00 0.95 1.23 1.35 1.29 1.58 1.65 1.62 1.86 1.89 1.88 2.15 2.22 2.19 2.49 2.49 2.49
Table for quarter circular bar
Load 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Increasing load 0.20 0.41 0.62 0.83 1.03 1.22 1.43 1.64
Vertical deflection, P (mm) Decreasing load 0.21 0.42 0.63 0.84 1.05 1.26 1.46 1.64
Average 0.21 0.41 0.63 0.84 1.04 1.22 1.45 1.64
Table for circular bar Load (kg)
Vertical deflection
Horizontal deflection
0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
(mm) 0.462 0.925 1.387 1.849 2.311 2.774 3.236 3.698
(mm) 0.294 0.589 0.883 1.177 1.472 1.766 2.060 2.354
Table for deflections of quarter bar for each loading using theory analysis
Load (kg)
Vertical deflection
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
(mm) 0.219 0.438 0.657 0.876 1.095 1.314 1.534 1.752
Table for deflections of circular bar for each loading using theory analysis
6.1 Example of Calculation
a) Quarter Circle Bar i.
Second moment of area of the cross section I = bd3 / 12 = 25 x 10-3 x (3 x 10-3)3 12 = 5.625 x 10-11 m4
ii.
Vertical deflection (theory) 0.2kg, δv
= π PR 3 4 EI =
( 0.2 x 9.81 ) (0.15)3 (4) (200 x 109 ) (5.625 x 10-11) π
= 0.462 mm
iii.
Second moment of area of the cross section I = bd3 / 12 = 25 x 10-3 x (3 x 10-3)3 12 = 5.625 x 10-11 m4
iv.
Horizontal deflection (theory) 0.2kg, = PR3 2 EI
δH
=
(0.2 x 9.81) (0.15)3 (2) (200 x 109 ) (5.625 x 10-11)
= 0.294 mm
b) Circular bar i. Second moment of area of the cross section I = bh3 / 12 = 25 x 10-3 x (3 x 10-3)3 12 = 5.625 x 10-11 m4 ii.
Deflection (theory) 0.5kg, δ
= 0.149PR3 EI = 0.149 ( 0.5 x 9.81 ) (0.15)3 (200 x 109 ) (5.625 x 10-11) = 0.219 mm
7.0 DISCUSSION
Graph is straight lines with a positive slope. So, its can be said that load is linearly dependent with vertical deflection for quarter circle bar. the vertical deflection will increase when the load increases. From the graph plotted, the result from the experiment is match or same with the results that calculated using theorem castigliano's. Theoretically the straight lines obtained are through the origin point. Slope for both of the lines are almost the same value. This shows that the error occur in the experiment are very small and the results can be accepted. Graph obtained is straight line with a positive gradient. Even though the graph is straight line, but there is a difference between the experiment results and theoretical results. Besides that, theoretical result is bigger compare than experiment results. There are differences in the both results because there are some errors occurred during the experiment. The errors may be caused by the Castigliano’s Theorem formula which is not recommended to be used in the calculation for horizontal deflection for quarter circle bar. This is because the forces applied are only assumptions values and not the real values. But, it still can use for compare with theory and experiment. Graph in a straight line shape obtained for load (N) versus vertical deflection v (mm) for circular bar. The gradient for both of the lines are positive. The lines are through the origin point. There differences between the experiment results and theoretical results are very small compare to graph of load (N) versus horizontal deflection H (mm) for quarter circle bar. This shows that the experiment results and calculation results are almost the same. Even though the errors are very small, but it could not be neglected and have to be taken in consider. Fro the graphs plotted, we can conclude that deflection is linearly dependent on applied load. Deflection will be increase, when the applied load increases.
From the data that has been recorded, there are a lot of errors occurred causes the differences between the experiment results and theoretical results. The factors which cause the errors are: • • • • •
(b)
Applied load is not a standard value The bar is already in deformed condition before the experiment begin The mistakes done by the person who recording the readings Random errors caused by the equipments Dial gauge is not placed at the exact place and always dislocated The advantages of using Castigliano’s Theorem
Castigliano’s theorem can be used to measure deflection of curved beams, but other theorems such as Macaulay theorem can be used only to measure vertical or horizontal bar. This are the advantages of using Castigliano’s theorem compared to other theories. i.
This theorem can measured deflection of the bar that bend but Macaulay
methods can only measured deflection of vertical and horizontal. ii.
The value of young Modulus E is fixed according to the material been used.
iii.
Calculation that is to get the value of beam deflection is more simple if the
differences between load is solve first before we integrate or other solution. iv.
This method is using the ’dummy’ where it function is to take the point that
is not mentioned the actual load. (c)
Besides Castigliano’s Theorem, other methods that can be used to measure the beam deflection are:-
i.
Macaulay method – Need an equation for all beams. This method is using
all three basics equations to find the value of deflection and beam slope. ii.
Double integration method – This method is having 3 equations to find the
value of deflection that is overall beam moment equation, deflection equation and slope equation. If the load situation is changed along the beam, the moment equation is also changed.
Besides Castigliano’s method, other methods such as Macaulay method, Superposition method, and Moment Area method can be used to calculate beam deflection.
8.0 CONCLUSION
As a conclusion, this experiment has achieved its main objective to validate the formula of deflection of curved bars based on Castigliano’s Theorem. The formulas are:
Vertical deflection (quarter circle bar) = πPR 3/4EI Deflection (circular bar) = 0.149PR 3/EI
Even though the formula for horizontal deflection for quarter circle bar is given as H = PR3/4EI, but the results obtained from experiment is much bigger compare to the calculation done using that formula. According to reference book Mechanics of Engineering Materials, the formula for horizontal deflection for quarter circle bar is H = PR3/2EI. This formula is almost matching with results obtained from the experiment. So the more exact formula to calculate the horizontal deflection for quarter circle bar is H = PR3/2EI. The conclusion are: i.
with using the Castigliano’s Theorem Method in calculate the bar
deflection, it more easier if compared with other method. ii.
the result of deflection value is not far between experimental and theoretical.
iii.
this experiment is succesful that is the deflection bar formula according to
the Castigliano’s Theorem can be approved it punctuality for getting the beam deflection. There are some differences of results obtained from the experiment and the theorem. This is because of the errors occurred during the experiment. However the overall conclusion is, deflection is linearly dependent on the applied weight.
9.0 REFERENCES Books:
a) Benham, P.P. and Crawford, R. J., Mechanics of Engineering Materials, 1994, Longman Scientific & Technical, England. b) Beer & E.Russell Johnston, Jr.(1992). "Mechanics of Materials, Metrics Editions, McGraw-Hill Book Company c) 1. Clive L. Dyme (1997), Structural Modelling & Analysis Cambridge University Press, UK. Page: 261 d) 2. R.C Hibbeler (2011), Mechanics of Materials Eighth Edition in S.I Units Prentice Hall Inc, Page: 569 & 619. e) 3. William A. Nash (1998), Schaum’s of Theory & Problems of Strength of Materials McGraw Hill, Page : 474
Internet:
http://www.public.iastate.edu/~fanous/ce332/virtualwork/homepage.html http://en.wikipedia.org/wiki/Macaulay%27s_method http://www.codecogs.com/reference/engineering/materials/curved_beams.php#sec8
