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Light Structures Part-I: Simple Euler strut Student preparation Required reading prior to the laboratory session: Background notes below Experimental procedure — background When a load-deflection...

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Light Structures
Part-I: Simple Euler strut
Student preparation
Required reading prior to the laboratory session:
Background notes below
Experimental procedure — background
When a load-deflection curve is obtained experimentally for a loaded strut, the magnitude of the critical load Pcris usually obtained by drawing the horizontal asymptote to the curve. This method has several drawbacks. In practice the curve does not flatten, so that the value taken for Pcris usually based on the last reading taken and the rest of the experimental values are not used. A useful method for determining the critical load from test data within the elastic region has been suggested by R.V. Southwell.
Consider a pin-ended strut such that in its unloaded configuration the strut has a small initial curvature, such that the displacement at any value of x is vo.

Assume that ,
in which a is the initial displacement at the centre of the strut. This satisfies the boundary conditions of vo=0 at x=0, and x=L, and also dvo/dx=0 at x=L/2; the assumed deflected shape is therefore reasonable, particularly since the buckled shape of a pin-ended strut is also half sine wave.
Since the strut is initially curved, an axial load P immediately produces bending and therefore further displacements v measured from the initial displaced position. The bending moment M at any section is M=P(v + vo). If the strut is initially unstressed the bending moment at any section is proportional to the change in curvature at that section from its initial configuration and not its absolute value.
Thus M=EId2v/dx2, and hence . (Note that P is not the buckling load for the strut.)
Substituting for vo, we get.

The solution of this equation is

in which m2=P/EI.
If the ends of the strut are pinned, v=0 at x=0 and x=L. The first of these boundary conditions gives A=0 while from the second we have 0 = B sinmL.
Although this equation is identical to that derived from the boundary conditions of an initially straight, buckled, pin-ended strut the circumstances are now different. The strut in this case is in stable equilibrium since only bending is involved; the displacements are therefore determinate as must then be the arbitrary constant B. It follows that sinmL cannot be zero. Hence B=0 and

Hence . But m2=P/EI and a sinpx/L = vo, hence
.
Now p2EI/L2=Pcr, the buckling load for a perfectly straight pin-ended strut. Hence
.
It can be seen that the effect of the compressive load P is to increase the initial deflection voby a factor
.
Cleary as P approaches Pcr, v tends to infinity. In practice this is impossible since material breakdown would occur before Pcris reached. Consider the displacements at the mid-point of the strut
.
Hence . This represents a linear relationship between vcand vc/P. Thus in an actual test on an initially curves strut a graph of vcagainst vc/P will be straight line as the critical condition is approached. The slope of the liner is Pcrand the intercept on the vcaxis is equal to a, the initial displacement at the mid-point of the strut. This graph is known as a Southwell plot.

Experimental procedure — apparatus and method
The strut is of aluminium extruded section with Young’s Modulus = 70 GN/m2(10 x 106lb/sq.in)
(1) Set the dial gauge to measure the central deflection of the strut, and plot a graph of load P against central deflection, and hence determine the buckling load.
(2) From the above results of P and vc, plot a graph of vc/P (horizontal) against vc(vertical). Hence determine the buckling load and the initial deflection.
(3) For a given end load P, the strut should now be loaded centrally W, and a graph plotted of W against deflection.
When P=0, the strut becomes a simply supported beam with a central load,
Hence, deflection = . Plot a graph of load against deflection. This should be a straight line, and . Measure q from the graph, calculate 48EI/L3and compare with tanq .
Repeat for P=20 kg (200 N)
Health and safety information The activity takes place in a working laboratory and care is needed. Please stick to clear access ways when walking between activity locations. Please be aware of trip hazards and sharp edged equipment. Please do not lift heavy objects in the laboratory.
Report requirements
Long reports must be prepared in accordance with the guidance given in the Laboratory Reports section above. The report should cover the results from parts (1), (2) and (3).
Length of Strut = 992mm
Width of Strut = 25mm
Thickness of Strut = 6.54mm
Second Moment of Area of Strut = ??
Euler Strut: to obtain the buckling force in two ways:
End Load (P)Central Deflection (mm)
W (N)LoadUnloadMean
040.2440.1840.21
10039.3539.2039.28
20037.4937.1637.33
25036.7935.1835.99
30034.6531.5133.08
31031.6130.5931.10
32030.6029.2029.90
33029.7428.5629.15
34028.2226.4727.35
34527.4825.7126.60
35026.5625.6526.11
35525.6825.6825.68
Combined Axial compression and transverse bending
P = 0P = 20 kg
LoadDeflection (mm)Deflection (mm)
W (N)LoadUnloadMeanLoadUnloadMean
040.1840.2440.2137.3837.8337.61
139.7339.6339.6836.4735.9936.23
239.1239.1739.1535.4235.1935.31
338.7238.6538.6934.6434.3334.49
438.2738.2238.2533.7433.4333.59
537.6837.6937.6932.6432.3932.52
637.2337.2437.2431.8031.5931.70
736.7436.7036.7230.8430.6930.77
836.2436.2436.2430.0629.7929.93
935.7335.7435.7429.1828.8829.03
1035.2635.2635.2628.1927.9428.07

Part-II: Unsymmetrical bending
Background
In simple bending theory several assumptions are made, one of which is that the cross-section of the beam is symmetrical about an axis through the centroid parallel to the plane of bending. Such an axis is defined as a principal axis. Principal axes of any area are those about which the rectangular second moment of area is zero. Axes of symmetry through the centroid are automatically principal axes, since the rectangular second moments of area for opposite quadrants cancel each other out.
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 be perpendicular axes through the centroid, and OU and OV be the principal axes. Let dA be the area of an element with coordinates u and v relative to the OU and OV axes, and coordinates x and y relative to the OX and OY axes. The angle UOX=q. Then:

The rectangular second moment of area IUVis given by

Now, the condition for principal axes is IUV=0. Hence .
Reminder on second moment of area for rectangular sections:

Experimental procedure
Measure the cross-sectional dimensions of the angle at three places between the supports and obtain mean dimensions. Calculate:
1. and
2. IXand IY
3. IXY
Hence calculate angle q.
Place dial gauges in position at centre of span to read horizontal and vertical deflections of beam. With no load on beam adjust both dial gauges to zero. Add an equal load (1 lb or 0.5 kg) to each end of the beam and read off the horizontal and vertical deflections. Add further equal loads to each end of the beam and record the horizontal and vertical deflections in each case. Obtain about eight values.
Now remove equal loads from each end of the beam and again record the deflections in each case. Calculate the mean deflection for loading and unloading for both the horizontal and vertical deflections. Plot the horizontal deflections (X-axis) against the vertical deflections (Y-axis) and determine the slope of the resulting straight line.
Health and safety information The activity takes place in a working laboratory and care is needed. Please stick to clear access ways when walking between activity locations. Please be aware of trip hazards and sharp edged equipment. Please do not lift heavy objects in the laboratory.
Results
Cross section Dimensions
Width =38 mm
Height=25 mm
Thickness=3.25 mm
x =y=Ix=Iy=Ixy=
? =
LoadDeflections (mm)
X - DirectionY - Direction
W (N)LoadUnloadMeanLoadUnloadMean
017.7717.6917.7311.7812.0011.89
517.3417.2217.2812.9913.3313.16
1016.8316.7016.7714.3614.6814.52
1516.3216.2116.2715.6715.9215.80
2015.7515.7415.7516.9916.9916.99

Part-III: Shear centre
Background
Consider the cantilevered beam PQ shown. AB is a rod, attached to the end P of the cantilever, to which a vertical load W can be applied. If W were applied at A, then in addition to bending as a beam, the cantilever would twist so that the end P would rotate anti-clockwise. On the other hand if W were applied at B, the bending would be accompanied by a clockwise rotation of P. Thus there is some point S, between A and B, at which the application of W will produce bending but not twist. S is called the shear centre of the beam.

If the cross-section of the beam has a vertical axis of symmetry (as in the case of a T or I section), the shear centre will lie on that axis, but if there is no vertical axis of symmetry (as in the case of the channel section used in the experiment), the position of the shear centre has to be determined.
Consider the channel section under the action of a shearing force F at a distance d from the centre of the web. The shearing stress at any point on the cross-section of the channel is then given by the usual equation
.
The distribution in the rectangular web will be parabolic, but will not reduce to zero at each end because of the presence of the flange areas. When the stress in the flange area is being determined the breadth b is replaced by the thickness t, but I and still refer to the N.A.
Thus and since area beyond A=0.
Between A and B the distribution is linear, since the area A is directly proportional to the distance along AB (F, t, h and I all being constant). An exactly similar distribution will be obtained for CD.
The stresses in the flanges give rise to forces represented by
average stress x area =
These produce a torque about 0 which must equal the applied torque. (Stresses in the web producing forces which have no moment about 0.)
Equating torques about 0 for equilibrium:

Thus if a force acts on the axis of symmetry, distance d from the centre of the web, there will be no tendency for the section to twist since the moments will be balanced. The point E is then termed the shear centre of the section.
Experimental procedure
The apparatus consists of an aluminium alloy channel built in at one end. At the free end is attached a horizontal bar, at right angles to the channel. The channel can be loaded by adding weights to this bar.
Two dial gauges should be placed in contact with the top surface of the channel. Set the dial gauges so that one gauge rests near each edge of the upper flange at the free end of the cantilever. Set the gauges to zero. Place a weight (half-kilogram) at one end of the horizontal bar and note the change in dial gauge readings. Repeat this along the entire length of the horizontal bar moving the weight 3cm each time.
The angle through which the beam is twisted in indicated by the difference in the dial readings divided by the distance between the gauges. Plot the angle against the distance of the weight from the vertical flange of the channel. The weight is acting through the shear centre when the channel is not twisted.
Measure the dimensions of the cross-section of the channel. Calculate the distance of the shear centre from the vertical web at 0, using the dimensions of the channel section. Compare the theoretical result with that obtained by experiment.
Note: As the weight is moved along the bar from A to B the angle of twist decreases to zero and then becomes negative.

Health and safety information The activity takes place in a working laboratory and care is needed. Please stick to clear access ways when walking between activity locations. Please be aware of trip hazards and sharp edged equipment. Please do not lift heavy objects in the laboratory.
Results
Results
t=3.28 mm
k=36.26 mm? = (d2-d1)/L
h=34.72 mm
L=82 mm
Distance Of W from center of WebDial Gauge (1) d1 (mm)Dial Gauge (2) d2 (mm)Angle of Twist ? (Radians)
-1505.462.63 XXXXXXXXXX
-1205.692.62 XXXXXXXXXX
-905.872.61 XXXXXXXXXX
-606.112.59 XXXXXXXXXX
-306.292.55 XXXXXXXXXX
06.522.55 XXXXXXXXXX
306.732.54 XXXXXXXXXX
606.932.48 XXXXXXXXXX
907.022.45 XXXXXXXXXX
1207.092.36 XXXXXXXXXX
1507.362.32 XXXXXXXXXX
Answered Same DayDec 29, 2021

Solution

Robert answered on Dec 29 2021
82 Votes
1
Index
Chapter Page No.
1. Abstracts 2


2. Simple Euler strut 3-8
3. Unsymmetrical bending 9-11

4. Shear centre 12-15


5. Reference 16
6. Appendix 17
2
Abstracts
Euler’s buckling; unsymmetrical Bending and Shear centre experiments are ca
ied out.
All experiments are ca
ied out with the help of proper guidelines and standards. Results are
analysed and compared with theoretical calculations.

3
Title :Simple Euler strut

1.0 Introduction:
The objective is to determine the Buckling load Pcr by plotting central deflection vc VS
vc/P(where P is load applied) and observe the combine effect of axial compression and
transverse bending by applying loads gradually.

2.0 Theory:

When a load-deflection curve is obtained experimentally for a loaded strut, the magnitude of
the critical load Pcr is usually obtained by drawing the horizontal asymptote to the curve.
This method has several drawbacks. In practice the curve does not flatten, so that the value
taken for Pcr is usually based on the last reading taken and the rest of the experimental
values are not used.

Consider a pin-ended strut such that in its unloaded configuration the strut has a small initial
curvature, such that the displacement at any value of x is vo.


Assume that ,
in which a is the initial displacement at the centre of the strut. This satisfies the boundary
conditions of vo=0 at x=0, and x=L, and also dvo/dx=0 at x=L/2
Since the strut is initially curved, an axial load P immediately produces bending and
therefore further displacements v measured from the initial displaced position. The bending
moment M at any section is M=P(v + vo).
Thus M=EI d
2
v/dx
2
, and hence .
Substituting for vo, we get.

P P
L
x
a
vc
vo
v
vo  asin 
x
L
d2v
dx
2  
P
EI
v  vo 
4

The solution of this equation is
in which 
2
=P/EI.





If the ends of the strut are pinned, v=0 at x=0 and x=L. The first of these boundary
conditions gives A=0 while from the second we have 0 = B sinL.

The strut in this case is in stable equili
ium since only bending is involved; the
displacements are therefore determinate as must then be the a
itrary constant B. It follows
that sinL cannot be zero. Hence B=0 and
Hence . But 
2
=P/EI and a sinx/L = vo, hence
.
Now 
2
EI/L
2
=Pcr, the buckling load for a perfectly straight pin-ended strut. Hence
.
It can be seen that the effect of the compressive load P is to increase the initial deflection vo
y a factor
.
Cleary as P approaches Pcr, v tends to infinity. In practice this is impossible since material
eakdown would occur before Pcr is reached. Consider the displacements at the mid-point
of the strut
.
Hence . This represents a linear relationship between vc and vc/P. Thus in
an actual test on an initially curves strut a graph of vc against vc/P will be straight line as the
critical condition is approached. The slope of the liner is Pcr and the intercept on the vc axis
is equal to a, the initial displacement at the mid-point of the strut. This graph is known as a
Southwell plot.
d2v
dx
2 
P
EI
v  
P
EI
asin 
x
L
v  Acosx  Bsin x 
2a
2
L2
 2
sin
x
L
v 
2a
2
L2
 2
sin
x
L
v 
a
 2
 2L2
1
sin 
x
L
v 
vo
2EI
PL2
1
v 
vo
Pc
P
1
1
Pc
P
1
vc 
a
Pc
P
 1
vc  Pc
vc
P
 a
5



3.0 Experiment procedure and results with tables:
The strut is of aluminium extruded section with Young’s Modulus = 70 GN/m2 (10 x
10
6
l
sq.in)
 Set the dial gauge to measure the central deflection of the strut and apply the load
gradually.
 From the above results of P and vc, plot a graph of vc/P (horizontal) against vc
(vertical). Hence determine the buckling load and the initial deflection.

 For a given end load P, the strut should now be loaded centrally W, and a graph
plotted of W against deflection.

 When P=0, the strut becomes a simply supported beam with a central load,
Hence, deflection = . Plot a graph of load against deflection. This should be a
straight line, and . Measure  from the graph, calculate 48EI/L3 and
compare with tan .

 Repeat for P=20 kg (200 N)
Length of Strut =

992 mm
Width of Strut =

25 mm
Thickness of Strut =

6.54 mm
Second Moment of Area of Strut = 582.76 mm^4
Euler Strut: to obtain the buckling force in two ways:
Table 1
End Load (P) Central Deflection...
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