Purpose

The remit is to calculate how the setting of the hanger weight (on different lengths of the beam) would affect the deflection at mid-span. Furthermore I aimed to see how the theoretical answers compared with the experimental answers, and to discuss why this is the case.

Introduction

As an engineer it is important to know how certain properties of a material will affect the overall outcome of the structure e.g. how much a beam deflects due to additional weight. Therefore the experiments carried out were to distinguish if the theoretical values contradict or coincide with the experimental values. This experiment is also used in real life situations during the testing procedures of the Central Deflection. In order to calculate the Central Deflection the following procedure must be applied during testing. From the beams that have been produced a random one is selected on which tests are carried out and the equation is applied to in order to deduce the results. Having been instructed to carry out the above experiment my findings were noted and are used throughout the report.

Equipment

The equipment used in these experiments was as follows:

* Tecquipment Universal Beam Rig

* dead hanger weight(s)

* weights: 100g, 200g, 500g and 1000g

* Mercer Gauge

* A steel beam

Macaulay’s Method

When the loads on a beam do not conform to standard cases, the solution for slope and deflection must be found from first principles. Macaulay developed a method for making the integrations simpler.

The basic equation governing the deflection of beams is

EI d2y/dx2 = Mxx

When a beam has a number of loads it is difficult to apply this theory because some loads may be within limits of x during the derivation but not during the solution at a particular point. Macaulay’s Method makes it possible to do the integration necessary by placing all the terms containing x within a square bracket and integrating the bracket, not x. During evaluation, any bracket with a negative value is ignored. The general method of solution is conducted as follows

1. EI d2y/dx2 = Mxx = R1[x] – F1[x – a] – F2[x – b] – F3 [x – c]

2. 1st integration

EI dy/dx = R1[x] 2 / 2 – F1[x – a] 2 / 2 – F2[x – b] 2 / 2 – F3 [x – c] 2 / 2 + C1

3. 2nd integration

EI y = R1[x] 3 / 6 – F1[x – a] 3 / 6 – F2[x – b] 3 / 6 – F3 [x – c] 3 / 6 + C1x + C2

4. Use boundary conditions to solve C1 and C2

5. Solve the deflection by putting in the appropriate value of x. IGNORE any brackets containing ï¿½negative values.

Theoretical Development

Experiment 1

Theoretical Development shown for a load of 9.81N where

EI = 113.076 MNm2

Mxx = 4.905x – 9.81[x – 500]

– EI d2y/dx2 = Mxx

EI d2y/dx2 = – (4.905x – 9.81[x – 500])

1st integration = EI dy/dx = -4.905×2 / 2 + 9.81/2[x – 500]2 + C1

2nd integration = EI y = -4.905×3 / 6 + 9.81 /6[x – 500]3 + C1x + C2

Boundary Conditions at x = 0, y = 0

Substitute in to eqn 2

EI (0) = -4.905(0)3 / 6 + 9.81 /6[(0) – 500]3 + C1(0) + C2

0 = 0 + 0 + 0 + C2

Therefore C2 = 0

At x = 1,000, y = 0

EI (0) = -4.905(1,000) 3 / 6 + 9.81/6[(1,000) – 500]3 + C1(1,000) + 0

0 = – 4,905,000,000 + 1,226,250,000 / 6 +1000 C1

Therefore C1 = – 613,125,000 / 1000

C1 = 613,125

EI y = -4.905×3 / 6 + 9.81 /6[x – 500]3 + 613,125x

At x = 500

EI y = -4.905(500)3 / 6 + 9.81 /6[(500) – 500]3 + 613,125(500)

EI y = 2,043,750,000

y = 204.375 / 113.076

y = 1.8074

Alternatively

y = WL3 / 48EI

y = 9.81(1000)3 / 48EI

y = 2,043,750,000 / EI

y = 1.8074

Experiment 2

Theoretical Development shown for a load of 9.81N where

EI = 111.018 MNm2

Mxx = 4.905x – 4.905[x – 250] – 4.905[x – 750]

– EI d2y/dx2 = Mxx

EI d2y/dx2 = – (4.905x – 4.905[x – 250] – 4.905[x – 750])

1st integration

= EI dy/dx = – 4.905×2 / 2 + 4.905 /2 [x – 250]2 + 4.905 /2 [x – 750]2 + C1

2nd integration

= EI y = – 4.905×3 / 6 + 4.905 /6 [x – 250]3 + 4.905 /6 [x – 750]3 + C1x+ C2

Boundary Conditions at x = 0, y = 0

Substitute in to eqn 2

EI (0) = – 4.905(0)3 / 6 + 4.905 /6 [0 – 250]3 + 4.905 /6 [0 – 750]3 + C1(0) + C2

0 = 0 + 0 + 0 + C2

Therefore C2 = 0

At x = 1,000, y = 0

EI (0) = -4.905(1,000) 3 / 6 + 4 .905 /6 [1,000 – 250]3 + 4.905 /6 [1,000 – 750]3 + C1(1,000)

0 = – 4,905,000,000 + 2,069,296,875 + 76,640,625 / 6 +1,000 C1

C1 = 459,843

EI y = – 4.905×3 / 6 + 4.905 /6 [x – 250]3 + 4.905 /6 [x – 750]3 + 459,843x

At x = 500

EI y = – 4.905(500)3 / 6 + 4.905 /6 [500 – 250]3 + 4.905 /6 [500 – 750]3 + 459,843(500)

EI y = 2,042,731,200

y = 204.273 / 111.018

y = 1.839

Alternatively

y = WL3 / 48EI

y = 9.81(1000)3 / 48EI

y = 2,042,731,200 / EI

y = 1.839

Experiment 3

Theoretical Development shown for a load of 9.81N where

EI = 119.397 MNm2

Mxx = 3.2375x – 9.81[x – 670]

– EI d2y/dx2 = Mxx

EI d2y/dx2 = – (3.237x – 9.81[x – 670])

1st integration = EI dy/dx = – 3.237×2 / 2 + 9.81/2[x – 670]2 + C1

2nd integration = EI y = – 3.237×3 / 6 + 9.81 /6[x – 670]3 + C1x + C2

Boundary Conditions at x = 0, y = 0

Substitute in to eqn 2

EI (0) = -3.237(0)3 / 6 + 9.81 /6[(0) – 670]3 + C1(0) + C2

0 = 0 + 0 + 0 + C2

Therefore C2 = 0

At x = 1,000, y = 0

EI (0) = -3.237(1,000) 3 / 6 + 9.81/6[(1,000) – 670]3 + C1(1,000) + 0

0 = – 3,237,000,000 + 352,541,970 / 6 +1000 C1

Therefore C1 = 480,743

EI y = – 3.237×3 / 6 + 9.81 /6[x – 670]3 + 480,743x

At x = 500

EI y = -3.237(500)3 / 6 + 9.81 /6[(500) – 670]3 + 480,743(500)

EI y = 2,044,076,664

y = 204.41 / 119.397

y = 1.712

Alternatively

y = WL3 / 48EI

y = 9.81(1000)3 / 48EI

y = 2,044,076,664 / EI

y = 1.712

Procedures

Experimental Procedure

The beam was placed on the knife edge supports exactly 1m (1000mm) apart. The dead hanger weight was then (for Experiment 1) placed mid-span (50cm), the weights were then added in increments of 0.2 kg at a time up to a maximum of 1.6kg. The Central Deflection was read off the Mercer Gauge and noted down. The same method was used for Experiments 2 and 3 to calculate Central Deflection, but in experiment 2 instead of one weight being placed mid-span, two were placed 1/4 spans (25cm) from the knife support on each side. In Experiment 3 again only one weight was placed, but this time it was placed about 1/3 span (33cm) from the knife support.

This is how the Experimental calculations (values) were obtained.

Theoretical Procedure

To calculate Central Deflection you would use the equation ? = WL3/48E?I, where:

* ? is the central deflection

* W is the vertical point load

* L is the length between the two knife edge supports

* E is Young’s Modulus

* I is the second moment of area

For this equation to calculate the Deflection (?) you would have to be given the variables (and these are assumed values e.g. Young’s Modulus ï¿½ 210kN/mm2). This may result in the answers being different, from them obtained through experimental procedures. This equation was used to calculate Deflection (?) for all the different loads for the 3 different experiments. Young’s Modulus (E) was an assumed value of 210kN/mm2.

Diagrams

Below are diagrams showing how the equipment was setup for the different experiments…

Experiment 1

Experiment 2

Experiment 3

Results by Experimental Procedures

Table and Graph for Experiment 1

(Figure E.1.2) (Figure E.1.3)

Load

Central Deflection (?) in mm

Kg

N

0.2

1.962

0.36

0.4

3.924

0.73

0.6

5.886

1.05

0.8

7.848

1.46

1.0

9.810

1.80

1.2

11.772

2.20

1.4

13.734

2.55

1.6

15.696

2.95

Average E = 196.655 kN/mm2

Table and Graph for Experiment 2

(Figure E.2.2) (FigureE.2.3)

Load

Central Deflection (?) in mm

Kg

N

0.2

1.962

0.37

0.4

3.924

0.74

0.6

5.886

1.11

0.8

7.848

1.45

1.0

9.810

1.83

1.2

11.772

2.18

1.4

13.734

2.58

1.6

15.696

2.96

Average E = 193.075 kN/mm2

Table and Graph for Experiment 3

(Figure E.3.2) (Figure E.3.3)

Load

Central Deflection (?) in mm

Kg

N

0.2

1.962

0.31

0.4

3.924

0.61

0.6

5.886

0.93

0.8

7.848

1.25

1.0

9.810

1.57

1.2

11.772

1.89

1.4

13.734

2.21

1.6

15.696

2.53

Average E = 207.647 kN/mm2

Results by Theoretical Procedures

Experiment 1

Figure T.1

Load

(W)

(in N)

Central Deflection (?)

(in mm)

1.962

0.361

3.924

0.723

5.886

1.084

7.848

1.446

9.810

1.807

11.772

2.169

13.734

2.530

15.696

2.892

Experiment 2

Load

(W)

(in N)

Central Deflection (?)

(in mm)

1.962

0.368

3.924

0.736

5.886

1.105

7.848

1.473

9.810

1.841

11.772

2.209

13.734

2.577

15.696

2.945

Figure T.2

Experiment 3

Figure T.3

Load

(W)

(in N)

Central Deflection (?)

(in mm)

1.962

0.342

3.924

0.685

5.886

1.027

7.848

1.369

9.810

1.712

11.772

2.054

13.734

2.396

15.696

2.739

Discussion

After conducting both experiments I found that the theoretical and experimental results were contradicting one another. The reason for this contradiction can consist of a number of things e.g. it can be due to human error in reading off the equipment, human error in placing the equipment correctly onto the beam etc. Other reasons can include ageing and fatigue (due to constant usage). Yet still these are not the main reasons to why the results are different. This is largely due to the fact that during experimental procedures you have to put up with the real world e.g. slippage at the knife edge supports (contributing in a small increase in the deflection) etc.

The ideal values which will make a straight line on a Load/Deflection graph only occur in theoretical experiments, but as it can be seen in Figure E.1.3, Figure E.2.3 and Figure E.3.3 on page 11 closely matches the graph made using the theoretical values (Figure T.1, T.2 and T.3 on page 12). This shows that the error margin was not large, and that the experimental value average for:

* Experiment 1 was only out by 0.011 mm, (0.67%)

* Experiment 2 was only out by 0.004 mm, (0.26%)

* Experiment 3 was only out by 0.091 mm, (6.41%)

The overall error for the experiments was 1.78%.

Putting this all into account, a very limited amount of things can be done to improve the experiment e.g.

* new beams can be used instead of old ones

* the beams used can be built into the knife edge supports

* a digital gauge can be used to measure deflection

* The technician can do the experiment for us.

Even with these things improved the values through experimental procedures will not mach them obtained through theoretical calculations. This is why when designing a structure we as engineers make an F.o.S (Factor of Safety), this is equivalent to Strength of Component / Load on component.

e.g.

If a component needs to withstand a load of 100 Newtons and a F.O.S of 4 is selected then the component is designed to support 400 Newtons.

Below (on page 14) is a Stress / Strain graph showing where the F.o.S is made. This is to ensure that in no matter what situation of overloading etc… The structure will not collapse.

Stress / Strain

a = Elastic Rang

b = Plastic Range

c = Strain Hardening

d = Necking and Failure

Stress (?)

F.o.S

Strain (?)

Conclusion

To summarize the whole report, I learned that it is almost impossible to get the same values from theoretical and experimental calculations

ï¿½ Negative value means that the Load it refers to is not within the limit of x.

? Given value of I = 575 x10-12 m4