m smithurst advanced strength lab report

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Advanced Strength of Materials Laboratory Report

Martin Smithurst N 0201171 1 12 / 03 / 2011

Structural Engineering, Laboratory Report

Advanced Strength of Materials (DESN40113)

Submission Deadline 24 / 03 / 2011

Martin Smithurst N 0201171,

MSc in structural Engineering with Materials 2010


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Aim

The Aim of the laboratory testing is to demonstrate the practical application of the theoretical

calculations through experiment.

SHEAR CENTRE

Method

Using different sections with thin wall steel welded at each end to the section frame, resulting in the

beam being fixed. Loads will be applied to the centroid of the section at right angles to the section on

a plate (Figure 1). The load will be moved in 20mm increments from one side of the section to the

other. Dial gauges are placed at the top of the plate will read the differential vertical movement Yof

the plate to the right and left of the section (Figure 2). When the central plate is horizontal with the

load applied, torsion on the section will be equalised both left and right. The point at which this

equalisation of torsion (both +V plus -V = 0) occurs is the shear centre within limitations of the 20mm

increments, at this point the components of the force are in the Y axis (Figure 3).

Figure 1, Apparatus Martin Smithurst CAD

Figure 2, Differential movement Martin Smithurst CAD


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Figure 3, Differential movement Martin Smithurst CAD

Three sections are to be tested, Equal angle, Channel & Semicircle dimensions as below (Figure 4).

The sections are made from consistent cold formed sheet steel 1.63mm thick.

Figure 4, Sections Martin Smithurst CAD

Picture 1, Sections being tested in lab, featuring the channel Martin Smithurst


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Equal Angle,

Theory

The equal angle is symmetrical resulting in the shear centre being within the centroid of the section.

The shear centre remains unaffected by the orientation of the angle because the components of the

force change in equal magnitude between the X and Y axis transferring from one to the other (Figure

5). The theoretical shear centre should be at the centre at position 6 in Figure 6. (SEWARD, D)

Figure 5 Equal Angle section, Martin Smithurst CAD

Figure 6 Equal Angle section, Martin Smithurst CAD


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Practical Application Results

Equal Angle

Load Position (mm) from Right Left Dial reading Right Dial reading

1 0 3.68 -4.79

2 20 2.93 -2.973 40 2.18 -3.29

4 60 1.47 -1.56

5 80 0.74 -0.81

6 100 0.04 -0.06

7 120 -0.68 0.71

8 140 -1.4 1.46

9 160 -2.06 2.18

10 180 -2.72 3.92

11 200 -3.38 3.65Table 1 Equal Angle test results

Chart 1 Equal Angle

Results

From the table it can be seen that position 6 is where the load is 0.02mm from the equilibrium from

left to right about position 6. When this data is plotted into a chart the intersection of the two results is

the shear centre (Chart 1). The 0.02mm can be explained by the fixed 20mm position and the

thickness of the steel section, if this were an infinite sliding scale there would be a point of true

equilibrium. This demonstrates that the theory follows the practice.

-5

-4

-3

-2

-1

0

1

2

3

4

1 2 3 4 5 6 7 8 9 10 11

D

isplacement(mm)

Position Number

Equal Angle

Left

Right


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Channel

Theory

As this is non-symmetrical section the shear centre will fall out of the centre line of the section. The

eccentricity of this is dimension e (Figure 7). Distance e is calculated using the following equation

(Equation 1).

Equation 1 Channel

e= eccentricity

h= height or depth of the section

b= breadth of the section

This puts the theoretical shear centre distance e of18.75mm outside the channel section. This

theoretical result should result in a shear centre in the practical test somewhere between positions 6

and 7 (Figure 7).

Figure 7 Channel, Martin Smithurst CAD


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Channel


1 0 0.14 -0.28

2 20 0.1 -0.23

3 40 0.08 -0.19

4 60 0.05 -0.14

5 80 0.02 -0.095

6 100 -0.015 -0.035

7 120 -0.045 -0.005

8 140 -0.075 0.035

9 160 -0.115 0.08

10 180 -0.15 0.125

11 200 -0.185 0.175Table 2 Channel test results

Chart 2 Channel

Results

As predicted the point of equilibrium falls outside of the section between positions 6 and 7. When this

data is plotted the intersection of the two dial readings is where the load is at its greatest Y =

maximum component and X = 0 being the shear centre of the beam. This demonstrates that the

theory follows the practice. In the graph it is seen that the intersection falls below the 0 line as the

whole section is deflected by this displacement.

-0.3

-0.2

-0.1

0

0.1

0.2

1 2 3 4 5 6 7 8 9 10 11

Displacement(mm)

Position Number

Channel

Left

Right


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Semicircle,

Theory

Again this is non-symmetrical section the shear centre will fall out of the centre line of the section. The

eccentricity of this is dimension e (Figure 8). Distance e is calculated using the following equation

(Equation 2). (MEGSON, T.H.G)

( )

( )

Equation 2 Semicircle

e= eccentricity

r= radius

This puts the theoretical shear centre distance e at 13.66mm outside the Semicircle section. This

theoretical outcome should result in a shear centre in the practical test somewhere between again

positions 6 and 7 (Figure 8)

Figure 8 Semicircle, Martin Smithurst CAD


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Semicircle


1 0 0.8 -1.09

2 20 0.66 -1.07

3 40 0.51 -0.7

4 60 0.37 -0.52

5 80 0.21 -0.33

6 100 0.06 -0.15

7 120 -0.475 0.035

8 140 -0.615 0.23

9 160 -0.26 0.41

10 180 -0.5 0.585

11 200 -0.65 0.77Table 3 Channel test results

Chart 3 Channel

Results

Again as predicted the point of equilibrium falls outside of the Semicircle section between positions 6

and 7. When this data is plotted the intersection of the two dial readings is where the load is at its

greatest Ycomponent and X= 0 being the shear centre of the section. This demonstrates that the

theory follows the practice. The intersection falls below the 0 line as the whole section is deflected by

this displacement.

-1.2

-0.7

-0.2

0.3

0.8

1 2 3 4 5 6 7 8 9 10 11

Displa

cement(mm)

Position Number

Semicircle

Left

Right


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UNSYMMETRICAL BENDING

This is an investigation into the moment of inertia described as:

The moment of inertia of an object about a given axis describes how

difficult it is to change its angular motion about that axis. Therefore, it

encompasses not just how much mass the object has overall, but how far

each bit of mass is from the axis.

(http://en.wikipedia.org/wiki/Moment_of_inertia, March 2011)

Method

The intention is to demonstrate the changing moment of inertia and the states of unsymmetrical

bending from a maximum to a minimum by orientation of the section. Using a solid steel section bar,

600mm in length, with an average section of 9.65mm x 19.39mm, and a load of 20N applied (Figure9).

Figure 9 Unsymmetrical bending rig, Martin Smithurst CAD.

Picture 2, unsymmetrical being tested in lab Martin Smithurst
http://en.wikipedia.org/wiki/Moment_of_inertiahttp://en.wikipedia.org/wiki/Moment_of_inertiahttp://en.wikipedia.org/wiki/Moment_of_inertiahttp://en.wikipedia.org/wiki/Moment_of_inertia


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This will demonstrate the various states of unsymmetrical bending through the symmetrical state.

The bar is rotated though 105 of rotation from 45 to 150, this changing orientation of the section will

change the moment of inertia I in the X and Y axis. The change in moment of inertia will alter the

deflection of the bar given a constant Young's modulus (the stiffness of the material). The

components X and Y deflection can be recorded as the angle of rotation is changed (Figure 10).

Figure 10 Unsymmetrical bending rig, Martin Smithurst CAD.

Theory

The moment of inertia I for a square section is calculated by (Equation 3) (MEGSON, T.H.G)

Equation 3 moment of inertia

Ix= moment of inertia about axis X

Iy= moment of inertia aboutaxis Y

b= breadth of section

d= depth of the section

The components of the load are given by the following (equation 4) (MEGSON, T.H.G)


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Equation 4 Deflection of the beam

dx= deflection about axis X (mm)

dy= deflection about axis Y(mm)

W= Weight applied (N)

L= Effective length (mm)

=Angle of rotation (degrees)E= Module of elasticity, constant (N/mm)

Theoretical Results

Table 4 Theoretical results

Theory

bd/12 bd/12

Angle()

E = 210x 10E3

L = LengthW =

Weight (N)I x Iy Cos Sin Gauge 1 Gauge 2

45 210000 216000000 20 5860.42 5802.1422 0.707 0.707 0.827243 0.835553

60 210000 216000000 20 5860.42 5802.1422 0.5 0.866 1.013285 0.590914

75 210000 216000000 20 5860.42 5802.1422 0.259 0.966 1.130292 0.306093

90 210000 216000000 20 5860.42 5802.1422 0 1 1.170075 0

105 210000 216000000 20 5860.42 5802.1422 0.259 0.966 1.130292 0.306093

120 210000 216000000 20 5860.42 5802.1422 0.5 0.866 1.013285 0.590914

135 210000 216000000 20 5860.42 5802.1422 0.707 0.707 0.8272439 0.835553

150 210000 216000000 20 5860.42 5802.1422 0.866 0.5 0.585037 1.023464


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Chart 4 unsymmetrical bending, theory results


Actual

Angle () TOP (X) BOTTOM (Y)

unloaded Loaded Actual X Unloaded Loaded Actual Y

45 21.38 20.28 1.1 22.26 26.14 3.88

60 21.19 19.91 1.28 22.28 25.86 3.58

75 21.08 17.905 3.175 22.13 25.05 2.92

90 21.01 17.29 3.72 22.11 24.22 2.11

105 20.78 16.7 4.08 21.92 21.82 0.1

120 20.73 16.51 4.22 21.63 21.52 0.11

135 20.705 16.72 3.985 21.46 20.25 1.21

150 21.04 17.42 3.62 21.58 19.34 2.24

Chart 4 unsymmetrical bending, laboratory results

0

0.2

0.4

0.6

0.8

1

1.2

1.4

40 60 80 100 120 140 160

Deflectionatfreeen

d(mm)

Angle of rotation

Theoretical results

Gauge 1

Gauge 2

-1

0

1

2

3

4

5

40 60 80 100 120 140 160

Deflectionatfreeend

(mm)

Angle of rotation

Laboratory results

Gauge 1

Gauge 2


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The colouration of the graphs denotes a similar trend showing the moment of inertia transforming as

the beam is rotated. Taking this to 315 of rotation as shown below, this demonstrates that with

consistent dimensions there will be maximum inertia of the section twice in rotation for a rectangle

section.

Figure 11 Full rotation of the section

0

0.2

0.4

0.6

0.8

1

1.2

1.4

40 140 240 340

D

eflectionatfreeend(mm)

Angle of rotation

Laboratory results

Guage 1

Guage 2


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References:

MEGSON, T.H.G 2005, Structural and Stress Analysis, Second edition, Butterworth

Heinemann, Oxford

SEWARD, D 1998, Understanding Structures, Second Edition, Palgrave

Publishing, Basingstoke, Hampshire.

WIKIPEDIA (http://en.wikipedia.org/wiki/Moment_of_inertia, March 2011)

The work submitted is mine alone and not the product of plagiarism, collusion or other academic

irregularity, as defined by the regulations of the University.

Martin Smithurst .19/03/2011
http://en.wikipedia.org/wiki/Moment_of_inertiahttp://en.wikipedia.org/wiki/Moment_of_inertiahttp://en.wikipedia.org/wiki/Moment_of_inertiahttp://en.wikipedia.org/wiki/Moment_of_inertia

m smithurst advanced strength lab report

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