simple euler strut
DESCRIPTION
For DuduTRANSCRIPT
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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.
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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
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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.
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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 .
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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, sinL 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 crc
(2.06)
This can be rearranged to give
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aP
vPv ccrc
(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 Youngs 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
tan=deflection. (where =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.
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Let,
OX and OY Perpendicular axis through the centroid. OU and OV Principal axis. A 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 + y sin and v = y cos x sin Thus giving (14)
The condition for a principle axis state that IUV = 0, Thus
(15)
Tan2 =
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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
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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 Youngs 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
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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
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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
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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
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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
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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:
Tan2=1.933 2=62.65 = 31.33
By theoretical result 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
Tan2=dy/dx=3.571 =37.17
0
2
4
6
8
10
12
17.5 18 18.5 19 19.5
Deflections
Deflections
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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 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 was found out to be 1.124 and the angle of deflection was found out to be 48.331o.
The theoretical value of 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 was found to be 31.33oand an error of 5.84o was found out.
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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