Experiment #6

Deflection of Curved Beams

Stephen Mirdo

Performed on November 8, 2010

Report due November 15, 2010

Table of Contents Object ………………………………………..………………………….………….…. p. 1 Theory …………………………………………………………………………..…pp. 1 - 4 Procedure ………………………….…………………………………...……..……..... p. 5 Results …................................................................................................................ pp. 6 - 8 Discussion and Conclusion …………………….......…………………….....…... pp. 9 - 10 Appendix ……………………………………..…………………..….………... pp. 11 - 13

1

Object The object of this experiment was to determine the deflections in the horizontal and vertical directions under loading of a semicircular beam and a davit by means of experiment and compare the experimental values of deflection to calculated, theoretical values.

Theory

Castigliano’s theorem can be employed to determine the deflection incurred by a force of loading in a curved beam. Castigliano’s theorem states that the component in a given direction of the deflection caused by an external force on an elastic body is equivalent to the partial derivative of the work of deformation with respect to the component of the force in the given direction. The work of deformation in this case is a moment induced by a loading force on the beam. The general expression of Castigliano’s theorem is as follows:

δ = o∫s M/EI * dM/dW * ds 1/EI * o∫s M * dM/dW * ds (Equation 1) where M is the moment induced by the force of loading, E is the elastic modulus of the beam material, I is the moment of inertia of the beam, dM/dW is the change in moment with respect to the force of loading and ds is the finite quantity of the beam over which integration is to take place. Because the modulus E and the moment of inertia I are constants, they are factored out of the integral. The work of deformation, or the moment, can be expressed as the product of the loading force, P, the radius from the center of curvature of the beam R and the sine of the angle of curvature. The moment can be expressed by the following equation:

M = PRsinθ (Equation 2)

The integrating factor ds of the general Castigliano equation can be expressed as follows:

ds = R dθ (Equation 3)

The partial derivative of the work of deformation with respect to the component of the force is expressed as a function of the radius of the beam and angle of the deflected beam. For the vertical deflection, the partial derivative is written as:

(dM/dW)V = Rsinθ (Equation 4) and for the horizontal deflection of a curved beam, the partial derivative is written as:

(dM/dW)H = R(1 – cosθ) (Equation 5)

2

Figure 1: Diagram of the semicircular beam of the testing apparatus featuring forces and finite quantities. The light gray member of the diagram is the semicircular beam before loading and the dark gray member is the beam after the load is applied.

To calculate the vertical deflection of a semicircular beam, substitute Equations 2, 3 and 4 into the general expression of Castigliano’s theorem (Equation 1). The integration is bounded by zero and π because the beam is a semicircle. This process will yield the following equation:

ΔV = 1/EI * o∫π PRsinθ * Rsinθ * R dθ PR3/EI o∫π sin2θ dθ (Equation 6)

The loading force P, the radius R, the elastic modulus E and the moment of inertia I, are all constants and can be factored out of the integral. Integrating with respect to theta yields the following equation for the vertical deflection of semicircular beam:

ΔV = [πPR3] / [2EI] (Equation 7)

To calculate the horizontal deflection of a semicircular beam, a dummy variable H must me employed as seen in Figure 1. H represents a fictitious loading force in the horizontal direction. Inserting the dummy variable allows for the integration in the horizontal direction. Substituting Equations 2, 3 and 5 into the general expression of Castigliano’s theorem yields the following expression:

ΔH = 1/EI * o∫π [PRsinθ + HR(1-cosθ)][R(1- cosθ)] R dθ (Equation 8)

Factoring out the constants P and R, and letting H equal zero, Equation 7 becomes:

ΔH = 1/EI * o∫π PR3[sinθ - sinθcosθ)] dθ PR3/EI o∫π [sinθ - sinθcosθ)] dθ (Equation 9)

Integrating Equation 9 with respect to theta yields the equation for the horizontal deflection of a curved beam.

ΔH = 2PR3/EI (Equation 10)

3

Figure 2: Diagram of the davit of the testing apparatus featuring forces and finite quantities. The light gray member of the diagram is the davit before loading and the dark gray member is the davit after the load is applied.

The calculations for the vertical and horizontal deflection of the davit differ slightly from those of the semicircular beam. The davit consists of a quarter circle curved beam and a straight leg that connects to the base as seen in Figure 2. This means that the calculations of deflection must be broken into two parts: one integral for the curved section of the beam and another for the straight leg of the beam. The integration of the curved section of the davit is bound by zero and π/2 because it is a quarter-circle and the integration of the leg is bound by zero at the base of the beam and the length L of the straight segment of the beam.

To calculate the vertical deflection caused by a force of loading for a davit, the

general equation of Castigliano’s theorem is modified to account for the straight segment of the beam. Substitute Equations 2, 3 and 4 into the general Castigliano equation and append an integral that expresses the moment endured by the straight segment.

ΔV = 1/EI * [o∫π/2 PRsinθ * Rsinθ * R dθ + o∫L PR * Rdy] (Equation 11) Factoring out the constants P and R yields the following expression:

ΔV = 1/EI * [PR3 o∫π/2 sin2θ dθ + PR2 o∫L dy] (Equation 12) Integrating with respect to theta and the y direction yields the following expression:

ΔV = 1/EI * [PR3(π/4) + PR2L] (Equation 13)

4

and can be tidied up a little and the equation for the vertical deflection of the davit can be written as follows:

ΔV = [π PR3] / [4EI] + [PR2L] / [EI] (Equation 14)

The straight segment of the davit must be accounted for in much the same way as it was for the vertical deflection in the formulation of the horizontal deflection calculation. Substituting Equations 2, 3 and 5 into the general expression of Castigliano’s theorem and appending an integral to describe the deflection in the straight segment of the davit yields the following:

ΔH = 1/EI*[ o∫π/2[PRsinθ + HR(1-cosθ)][R(1- cosθ)] R dθ + o∫L[PR + H(R + y)][R + y] dy (Equation 15)

Factoring out the constants P and R and letting the dummy variable H equal zero, Equation 15 becomes the following: ΔH = 1/EI*[PR3 o∫π/2 sinθ – sinθcosθ dθ + P o∫L (R2 + Ry) dy (Equation 16) Integrating for the curvature and straight segment yields the following expression: ΔH = 1/EI * [PR3/2 + PR2L + PRL2/2] (Equation 17) Distributing the modulus of elasticity E and moment of inertia I into Equation 17 yields: ΔH = [PR3/2EI] + [PR2L/EI] + [PRL2/2EI] (Equation 18) Tidied up a little further, the equation for the horizontal deflection of a davit can be written as follows: ΔH = [PR3/2EI] + [PRL/2EI][2R + L] (Equation 19)

5

Procedure Equipment: Hi-Tech Scientific Deflection of Curved Beams Apparatus HSM.10 (SN: 92725) Weights in Half Pound Increments up to 3.5 lbs Experiment:

1) Position the empty load hangers on the semi-circular beam of the testing apparatus (Figure 1).

2) Zero the dial indicators. 3) Place a 0.5 lb weight on the hanger. 4) Record both the horizontal and vertical deflection of the beam. The indicated

strain will be x 0.01mm. 5) Add 0.5 lb additional weight to the weight hanger. Again, record the

horizontal and vertical deflection of the beam. 6) Repeat Step 5 until a total weight applied to the weight hanger is equal to 3.5

lbs. 7) Repeat Steps 1 through 6 for a total of two trials. 8) Move the weight hanger and deflection indicator dials to the davit of the

testing apparatus and repeat the process for the davit.

Figure 3: Diagram of the deflection of curved beam apparatus. (Adapted from Materials Laboratory Manual, Fall 2010, University of Memphis, Department of M.E.)

6

Results

Table 1: Dimensions and properties of the curved beam and davit with the calculated moment of inertia used for both beams of the testing apparatus.

Radius R of Semicircular Member (in) 5.9 Radius R of Davit (in) 5.9 Length L of Vertical Leg of Davit (in) 2.95 Modulus E of Beams' Material (psi) 3.00E+07 Width of Beams (in) 1.0 Thickness T of Beams (in) 0.12 Moment of Inertia I of Beams (in4) 1.44E-04

Table 2: Experimental data of the curved beam, including the applied load, horizontal deflection and vertical deflection, paired with the calculated horizontal and vertical deflection.

P (lbf) Experimental ΔV x 0.01 in

Theoretical ΔV x 0.01 in

Experimental ΔH x 0.01 in

Theoretical ΔH x 0.01 in

0.5 2.1 3.7 1.3 4.8 1.0 5.9 7.5 6.1 9.5 1.5 9.6 11.2 11.6 14.3 2.0 13.1 14.9 15.7 19.0 2.5 19.5 18.7 17.6 23.8 3.0 22.1 22.4 20.6 28.5 3.5 24.9 26.1 26.0 33.3

Table 3: Experimental data of the davit, including the applied load, horizontal deflection and vertical deflection, paired with the calculated horizontal and vertical deflection.

P (lbf) Experimental ΔV x 0.01 in

Theoretical ΔV x 0.01 in

Experimental ΔH x 0.01 in

Theoretical ΔH x 0.01 in

0.5 2.4 3.1 1.9 2.7 1.0 5.6 6.1 4.5 5.3 1.5 8.2 9.2 7.0 8.0 2.0 11.1 12.2 9.5 10.7 2.5 14.1 15.3 12.2 13.4 3.0 17.2 18.3 15.0 16.0 3.5 20.4 21.4 17.7 18.7

7

Vertical Deflection of Semicircular Beam due to Loading

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Applied Load (lbf)

Ver

tical

Def

lect

ion

(x 0

.01

in)

Experimental ΔV Theoretical ΔV

Figure 4: Graph of the experimental vertical deflection due to applied load of the semicircular beam. Superimposed are the theoretical values of the vertical deflection of the semicircular beam.

Horizontal Deflection of Semicircular Beam due to Loading

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Loading Force (lbf)

Horiz

onta

l Def

lect

ion

(x 0

.01

in)

Experimental ΔH Theoretical ΔH

Figure 5: Graph of the experimental horizontal deflection due to applied load of the semicircular beam. Superimposed are the theoretical values of the horizontal deflection of the semicircular beam.

8

Vertical Deflection of Davit due to Loading

0.0

5.0

10.0

15.0

20.0

25.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Loading Force (lbf)

Ver

tical

Def

lect

ion

(x 0

.01

in)

Experimental ΔV Theoretical ΔV

Figure 6: Graph of the experimental vertical deflection due to applied load of the davit. Superimposed are the theoretical values of the vertical deflection of the davit.

Horizontal Deflection of Davit due to Loading

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Loading Force (lbf)

Hor

izon

tal D

efle

ctio

n ( x

0.0

1 in

)

Experimental ΔH Theoretical ΔH

Figure 7: Graph of the experimental horizontal deflection due to applied load of the davit. Superimposed are the theoretical values of the horizontal deflection of the davit.

9

Discussion & Conclusion

It was noted that the calculation and experimental values of the horizontal and vertical deflections for both the davit and semicircular beam were not in complete agreement. Thus, a percent error analysis was performed to determine the degree of this discrepancy. Table 4: Percent error analysis of the experimental and calculated vertical and horizontal deflections of the curved beam of the testing apparatus.

Experimental ΔV x 0.01 in

Theoretical ΔV x 0.01 in

% Error in Vertical

Deflection ΔV

Experimental ΔH x 0.01 in

Theoretical ΔH x 0.01 in

% Error in Horizontal

Deflection ΔH

2.1 3.7 44.6% 1.3 4.8 73.1% 5.9 7.5 21.2% 6.1 9.5 35.4% 9.6 11.2 14.1% 11.6 14.3 19.0%

13.1 14.9 12.1% 15.7 19.0 17.4% 19.5 18.7 4.5% 17.6 23.8 25.8% 22.1 22.4 1.3% 20.6 28.5 28.0% 24.9 26.1 4.8% 26.0 33.3 21.8%

Table 5: Percent error analysis of the experimental and calculated vertical and horizontal deflections of the davit member of the testing apparatus.

Experimental ΔV x 0.01 in

Theoretical ΔV x 0.01 in

% Error in Vertical

Deflection ΔV

Experimental ΔH x 0.01 in

Theoretical ΔH x 0.01 in

% Error in Horizontal

Deflection ΔH

2.4 3.1 21.4% 1.9 2.7 28.6% 5.6 6.1 9.2% 4.5 5.3 15.7% 8.2 9.2 10.7% 7.0 8.0 12.2%

11.1 12.2 9.2% 9.5 10.7 10.9% 14.1 15.3 7.5% 12.2 13.4 9.0% 17.2 18.3 5.9% 15.0 16.0 6.5% 20.4 21.4 4.6% 17.7 18.7 5.4%

Upon reviewing the results of the error analysis, it was determined that the initial

loading of the beam is where the highest percentage of error occurred. This was due to the fact that a small difference of values of one order of magnitude yields a high percentage of error. It was noted that as the loading increased, the percentage of error trended downwards and the theoretical and experimental results were more in agreement.

It was observed that the experimental values of horizontal and vertical deflection

for both beam specimens are lower in value than the theoretical values for each. This was due to error in the experiment. One source of error in this experiment is due to faulty

10

deflection indicators. It can be seen in Figures 4 and 5 that the experimental deflections as a function of loading force form a polynomial curve fit, where as the theoretical values of deflection form a linear function. The error in the experimental values was caused by the deflection indicators of the apparatus sticking and exerting a force on the beam, thereby preventing the beam from deflecting to the full extend the loading force would cause. It can be seen in Figures 6 and 7 that the experimental data took the shape of a linear function resembling the theoretical data. After it was noted that the indicators were inhibiting the deflection of the beam, a new experimental technique was adopted. This technique consisted of lightly hitting the table the apparatus was placed upon so as to vibrate the beam and counteract the friction present inside the deflection indicators. This technique seemed to allow for more accurate values of deflection to be recorded as the percent error analysis and graphs show.

Improvements to this experiment are in order. One such improvement would be

to use digital deflection indicators. The deflection indicators currently on the test apparatus are clearly faulty and require mental math on the fly to determine the actual values. The use of digital equipment would remove a lot of the guesswork in dealing with this particular testing apparatus. Another improvement would be use weights in kilograms instead of pounds so that unit conversion is not necessary to arrive at the theoretical and experimental values of deflection.

11

Appendix Data Usage Sample calculation of theoretical vertical deflection of the semicircular beam at a loading force of 0.5 lbf:

[π * 0.5 lbf * (5.9 in)3] / [2 * 30 x 106 psi * 0.000144 in4] = 0.037 in = 3.7 x 0.01 in Sample calculation of theoretical horizontal deflection of the semicircular beam at a loading force of 0.5 lbf:

[2 * 0.5 lbf * (5.9 in)3] / [30 x 106 psi * 0.000144 in4] = 0.048 in = 4.8 in x 0.01 in Sample calculation of percent error of the experimental and theoretical values of the horizontal loading force of 0.5 lbf for the semicircular beam:

[|0.048 in – 0.013in| / 0.048 in] * 100 = 73.1%

12

Bibliography

Energy Methods in Structural Analysis Version 2 CE IIT, Kharagpur, (2002)

Materials Laboratory Manual, Fall 2010

University of Memphis, Department of Mechanical Engineering