Structures - · PDF fileAn experiment to find the characteristics of flow under a sluice gate with a ... In theory before loading the beams ... 2.2 Unsymmetrical Bending

Structures

ST

Shainal Sutaria

Student Number: 1059965

Wednesday, 14

th Jan, 2011

Abstract

An experiment to find the characteristics of flow under a sluice gate with a

hydraulic jump, also known as a standing wave is to be concluded. To acquire

the properties of the hydraulic jump, a graph of the energy line in a rectangular

channel through regions of gradually varied flow and rapidly varied flow is to

be plotted. The value of the Froude number is also to be determined upstream

and downstream of the hydraulic jump from the results. Thus the energy head

loss can be calculated due to the jump. Then the theoretically estimated and

measured values of downstream depth and energy head can be compared.

Contents

Page

Number

1. Introduction 3

2. Theory 4

3. Experimental Procedure and Results

3.1.1 Experimental Procedure – Objectives 5

3.1.2 Experimental Procedure – Apparatus and Method 5

3.2 Results

3.2.1 Test Data Tables 6

3.2.2 Example Calculations 6

3.2.3 Graph of Energy Head against Depth 7

3.2.4 Calculation of Loss of Specific Energy Head 8

3.2.5 Hydraulic Jump Classification 8

4. Discussion 9

5. Conclusions 10

6. References 10

1. Introduction

An experiment to find the buckling loads of a simple Euler strut is to be carried

out. To acquire the properties of the strut, values of the deflection of the beam

will be measured for various loads at the end of the strut. Also the deflection for

central loading of the strut for three different end loads also calculated.

Therefore we are able to calculate the objective of the experiment, the buckling

loads of the strut when loaded centrally and end loaded.

When calculating the deflection and buckling load in theory, the strut is

assumed to be initially straight. In reality there is and initial deflection and

curve to the beam. This makes the values obtained to differ from theoretical

solutions and more accurate.

2. Theory

For a loaded strut the magnitude of the critical load Pcr can be acquired

experimentally by considering a load-deflection curve, it is typically obtained

by drawing the horizontal asymptote to the curve. However this procedure has

numerous drawbacks. In reality the curve of the beam does not flatten, so the

calculated value of Pcr is generally measured on the final reading obtained and

the rest of the experimental values are not taken into account, thus an error is

incurred. A more accurate procedure for determining the critical load from the

experimental data within the elastic region was introduced by R.V. Southwell.

(Richard Vynne Southwell (born: Norwich 1888, died 1970) was an English

mathematician who specialized in applied mechanics. He studied in Cambridge

and in 1925, he returned to Cambridge University as a mathematics lecturer.

Then In 1929, he became a professor of engineering at Oxford University. In

1942, he became rector of Imperial College. He retired in 1948.) [2]

Consider an unloaded pin-ended strut. In theory before loading the beams

remains straight; however in reality this is not true. The strut initially has a

small curvature, such that at any value of the displacement, x the curvature is v0.

Fig. 1.

Assume that L

xav sin0

where a is the initial central displacement of the strut. This equation complies

with the boundary conditions that v0=0 when x=0, and x=L, and also d v0/dx=0

at x=L/2. Thus the assumed deflection is therefore practical.

Due to the initial curvature in the strut, a axial load P instantly create bending of

the beam and consequently additional displacements, v occur which is measured

from the original displaced position. The bending moment, M at any point in the

section is )( 0vvPM . (2.01)

When the strut is initially unstressed the bending moment at any section is

proportional to the change in curvature at that section from its initial

arrangement and is not its absolute value.

Thus2

2

dx

vdEIM , and hence )( 02

2

vvEI

P

dx

vd . (Note that P is not the buckling

load for the strut.)

Substituting for v0, we get.

L

xa

EI

Pv

EI

P

dx

vdsin

2

2

(2.02)

The solution of this equation is,

L

x

L

axBxAv

sinsincos

2

2

2

2

,

(2.03)

in which, EI

P2 .

From the initial boundary conditions in equation (2.01), where v=0 at x=0, we

get A=0 and v=0 x=L we get LB sin0

The strut in this case is in stable equilibrium as only bending is involved, sinμL

cannot be zero.

Therefore B=0, which gives the equation,

L

x

L

av

sin

122

2

,

(2.04)

Since A=0 and B=0

Substitute from above equations thatEI

P2 and

L

xav sin0 into equation

(2.04) we get the equation

12

2

PL

EI

vv o

Now crPL

EI

2

2, is the buckling load for a perfectly straight pin-ended strut.

Hence

1

0

P

P

vv

cr

(2.05)

The effect of the load P will increase the initial deflection by a factor of

1

1

P

Pcr

As P approaches Pcr, v tends to infinity. This is not possible in practice as the

strut would breakdown before reaching Pcr. Due to this we consider the

displacement at the mid-point of the strut

1P

Pav cr

c

(2.06)

This can be rearranged to give

aP

vPv c

crc

(2.07)

This is in the form of a straight line equation cmxy , which represents a linear

relationship between vc and vc/P. A graph of vc against vc/P will produce a

straight line as the critical condition is approached. The gradient of the straight

line represents Pcr and the intercept on the vc axis is equal to a (initial central

displacement). This graph is acknowledged as a Southwell plot.

When a strut is loaded centrally and without an end load, the strut becomes a

simply supported beam. Thus the mean deflection can be calculated as,

EI

WLdeflection

48

3

, where W is the weight of the central load, L is the length of the

beam, E the Young’s Modulus and I the second moment of area. Due to the

graph of load against deflection being a straight line, EI

WL

48tan

3

. Therefore

tanθ=deflection. (where θ=angle between deflection angle and the line of the

graph).

Fig. 2.

Part 2:Unsymmetrical Bending

2.2 Unsymmetrical Bending

If the plane containing the applied bending moment is not parallel to the

principal axis of the section, the simple bending formula cannot be applied to

find the bending stress. For unsymmetrical sections the direction of the principal

axes has first to be determined. In this experiment a beam made up of angle

section is used and it is required to find the principal axes.

Let,

OX and OY – Perpendicular axis through the centroid.

OU and OV – Principal axis.

δA is a elemental area at distance u and v from the OU and OV axis

respectively.

The rectangular second moment of area IUV is given by (13)

Where,

u = x cos θ + y sin θ

and v = y cos θ – x sin θ

Thus giving (14)

The condition for a principle axis state that IUV = 0, Thus

(15)

Tan2θ =

The second moment of area for the X and Y axis has to be determined. This can

be done by the use of the following equation: (16)

Ix=

Iy=

Ixy=bdxy

Where x and y – in this case is the distance from the surface of the beam to the

neutral axis.

b – The breadth of the strut

d – The Depth of the strut

Ix and Iy – Second moment of area.

2.3 Shear centre

When a beam or section does not have a vertical axis of symmetry, then the

horizontal shear stressed on the beam are not in complete equilibrium, they will

create a couple causing the section to twist. The twist leads to the development

of torsional stresses in the member. The shear centre is the point where the

torsional stresses are zero, and only bending occurs. In a channel section, there

is no vertical axis of symmetry, and so, the shear centre needs to be determined

For a channel section under a shearing force, F, at a distance d from the centre

of the web, the shearing stress at any point is given by:

=

(3)

This can help derive the equation for d, the distance from the centre of the web,

which will give the position of the shear centre [at d, there will be no torque,

just pure bending]:

d

3. Experimental Procedure and Results

3.1.1 Experimental Procedure – Apparatus and Method

(i) For the first experiment a strut of aluminium extruded section was set up,

with one end pinned and the other end free for loading weights.

(ii) The length of the strut was measured three times and the average

recorded. The width and thickness were taken in three different positions

using a micrometer, and then the mean calculated for each.

(iii) A dial gauge was placed at the centre of the strut to measure the central

deflection of the beam (the dial gauge had an accuracy of 0.01mm).

(iv) The strut used had a Young’s Modulus = 70 GN/m2 (10 x 106 lb/sq.in).

(v) The dial gauge was set to measure the central deflection of the strut,

taking readings of the deflection for loads of 0-380N, in increments of

100N from 0-200N, 5N from 200-300N, 20N from 300-360N and 10N

from 360-380N. Then the deflection was also recorded on unloading in

the same increments. From this the mean deflection for each load could

be found.

(vi) Now a graph of load P against central deflection can be plotted, and

hence determine the buckling load.

(vii) From the calculated results of P and vc, plot a graph of vc/P against vc.

Hence determine the buckling load and the initial deflection.

(viii) For the second experiment the strut was now loaded centrally and an end

load P applied, the strut.

(ix) For the first central loading no end load was applied, therefore P=0kg.

Loads were then applied centrally for loads of 0-12N in increments of 1N

each time, and the deflection noted on loading and then on unloading

also. From this the mean deflection for each load could be found.

(x) Now the procedure will be repeated a further two times for end loads of

10kg and 20kg.

(xi) A graph of W against deflection can now be plotted for all three loads

(0,10,20kg) on the same graph

3.2 Results

3.2.1 Test Data Tables

Length of strut, L (mm)

1 2 3 Mean

990 990 990 990

Width of strut, b (mm)

1 2 3 Mean

25 25 25 25

Thickness of strut, d (mm)

1 2 3 Mean

6.40 6.37 6.38 6.38

Table 3.2.1.1

Second moment of Area of Strut, I (mm4) = 541.03

E = 70x109 N/m2

381.37

N

Table 3.2.1.2

End Load (P) Central Deflection (mm) Deflection, Vc

(mm) vc/P (mm/N)

(N) Load Unload Mean

0 45.00 45.96 45.48 0.00

100 43.64 44.78 44.21 1.27 0.01270

200 42.65 41.37 42.01 3.47 0.01735

250 39.70 39.12 39.41 6.07 0.02428

300 36.36 35.28 35.82 9.66 0.03220

320 35.19 34.25 34.72 10.76 0.03363

340 33.56 33.54 33.55 11.93 0.03509

360 33.01 32.08 32.55 12.94 0.03593

370 31.24 30.47 30.86 14.63 0.03953

380 29.28 29.28 29.28 16.20 0.04263

12

38.625

12

33

bd

2

1292

2

2

99.0

)1003.541()1070(

L

EIPcr

End Load, P = 0 kg

Load Central Deflection (mm) Deflection, Vc (mm)

W (N)

Loading Unloading Mean

1 44.41 44.47 44.44 0.00

2 44.04 44.00 44.02 0.42

3 43.49 43.50 43.50 0.94

4 43.10 42.96 43.03 1.41

5 42.62 42.47 42.55 1.90

6 42.05 42.04 42.05 2.40

7 41.61 41.51 41.56 2.88

8 41.15 41.12 41.14 3.31

9 40.64 40.54 40.59 3.85

10 40.10 40.12 40.11 4.33

11 39.70 39.62 39.66 4.78

12 39.15 39.15 39.15 5.29

End Load, P = 20 kg

Load Central Deflection (mm) Deflection, Vc (mm)

W (N)

Loading Unloading Mean

1 41.22 40.80 41.01 0.00

2 40.32 39.99 40.16 0.85

3 39.50 39.00 39.25 1.76

4 38.35 38.11 38.23 2.78

5 37.47 37.20 37.34 3.68

6 36.67 36.28 36.48 4.54

7 35.75 35.59 35.67 5.34

8 34.72 34.60 34.66 6.35

9 33.75 33.84 33.80 7.22

10 32.88 32.80 32.84 8.17

11 32.11 32.64 32.38 8.64

12 31.30 31.30 31.30 9.71

Table 3.2.1.3

0

50

100

150

200

250

300

350

400

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00

Lo

ad

, P

(N

)

Deflection, vc (mm)

Graph of P against vc

-5.00

0.00

5.00

10.00

15.00

20.00

0.000 0.010 0.020 0.030 0.040 0.050

Defl

ecti

on

, V

c (

mm

)

Vc/P (mm/N)

Graph of Vc against Vc/P

0

2

4

6

8

10

12

14

-2.00 0.00 2.00 4.00 6.00 8.00 10.00 12.00

Lo

ad

, W

(N

)

Deflection, vc (mm)

Graph of Weight against Deflection P = 0 kg

P = 20 kg

Fig. 5.

3.2.5 Example Calculations

From the theory in the previous section it can be shown that:

Length of strut, L= (990 + 990 + 990) / 3 = 990 mm

Width of strut, b = (25 + 25 + 25) / 3 = 25 mm

Thickness of strut, d = (6.40 + 6.37 + 6.38) / 3 = 6.38 mm

For End Load, P = 300 N:

Central Deflection = (36.36 + 35.28) / 2 = 35.82 mm Deflection, Vc = 44.44 – 35.82 = 9.66 mm vc/P = 4.68 / 300 = 0.0322 mm/N

For End Load, P = 20 kg, Load, W = 5 N:

Deflection, Vc = 41.22 – 37.47 = 3.68 mm

4.2

Unsymmetrical Bending

b = 36.3 mm

d = 30.5 mm

t1 = 3.25 mm

t2= 3.2 mm

The centroid of the beam is determined:

y = ((36.3 x 3.2 x 1.6) + ((30.5-3.2) x 3.25 x 15.25) / ((36.3 x 3.2) + (27.3 x

3.25))

= 7.311 mm

x = ((30.5 x 3.25 x 1.625) + (33.05 x 3.2 x 18.15) / ((33.05 x 3.2) + (30.5 x

3.25))

= 10.149 mm

433

mm 541.0312

38.625

12 I Strut, of Area ofmoment Second

bd

N 381.3799.0

)1003.541()1070(2

1292

2

2

L

EIPcr

The second moment of area (Ix, Iy and Ixy) is calculated where the values of x

and y is the one calculated above:

Ix =1.483* mm4

Iy =2.356* mm4

Ixy =8.440* mm4

The angle is calculated by:

Tan2θ=1.933

2θ=62.65

θ = 31.33

By theoretical result θ was obtained to be 31.33

Deflections

X-Direction Y-Direction

Load(N) Load Unload Mean Load Unload Mean

0 18 18.005 18.0025 6 6.04 6.02

5 18.34 18.38 18.36 7.33 7.42 7.375

10 18.69 18.69 18.69 8.8 8.8 8.8

15 18.94 18.98 18.96 9.07 9.19 9.13

20 19.24 19.24 19.24 10.55 10.55 10.55

Tan2θ=dy/dx=3.571

Θ=37.17

0

2

4

6

8

10

12

17.5 18 18.5 19 19.5

Deflections

Deflections

4.3 Shear centre

Two dial gauges had been kept in contact with the surface, each gauge is rested

near the edge of the upper flange.

The initial readings of the dial gauge were then taken.

Then A weight was placed along the horizontal bar that was attached to the

channel.Changes in the dial gauge deflections were then recorded as the weights

were moved through set distances on the horizontal bar.

t=3.25 mm

h= 34.75 mm

k= 37.40 mm

L= 86 mm

The angel of twist is given by the equation:

Angel of twist = (d2-d1)/l

The above equation is applicable as at the point of zero torsion, the dial gauge

reading for both the dial gauge are the same, i.e. d1=d2.

-150 2.82 1.66 0.007733

-120 2.565 1.9 0.005542

-90 2.48 2.25 0.002556

-60 2.23 2.47 -0.004

-30 2.21 3.04 -0.02767

0 2.06 3.5 0

30 1.92 3.875 0.065167

60 1.88 4.38 0.041667

90 1.6 4.635 0.033722

120 1.55 5.06 0.02925

150 1.42 5.145 0.024833

5. Discussion

For the axial compression and transverse bending experiment the value of tan θ

was found to be 2.107 and the angle of deflection was determined to be 64.611o,

and for P =20 kg with the application if central loading, the value of tan θ was

found out to be 1.124 and the angle of deflection was found out to be 48.331o.

The theoretical value of θ was also determined to be 63.54o.

There was an error of 1.071o and 15.209o when value from graph was found our

of p = 0 and p = 20 .

The error may be due to defect in the dial gauge or the reading may have been

miss read. For the experiment on unsymmetrical bending the The slope was

found out to be 37.17o when the deflection in y axis was plotted against

deflection in x axis.the theoretical value of θ was found to be 31.33oand an error

of 5.84o was found out.

There maybe many reasons for the errors such as failure to apply load in both

the ends at the same time,error in dial gauge, etc.

For the experiment on shear centre he theoretical value for d was found out to

be 16.721 mm. From the graph plotted the experimental value of d was

determined and was 15 mm.

There was an error of 1.721 mm .there may be several reasons like the incorrect

measurement of the point where the load has to be applied,defect in dial gauge ,

etc.

6. Conclusion

The experiment was carried out successfully and the expected results were

achieved.

The behavior of action of load on a strut was studied . The values of bulking

load and deflection angle were calculated and were found to be correct.

The experimental values explained the effect of load on the deflection angel so

if the load increases so does the deflection angel.

The value of buckling load was calculated using two different ways and was

compared to its theoretical values which were found to have a slight error.

These errors could be caused due to various reasons such as defect in dial gauge

, defect in weights , incorrect method of applying weights.

7. References

Lab report notes

www.wikipedia.org