Educational · ECCENTRIC LOADING EFFECT OF LACK OF ... VARIATION OF CRIPPLING LOAD WITH SLENDERNESS...
60
@TecQuipment Ltd 1999 No part of this publicationmay be reproduced or transmittedin any form or by any means, electronicor mechanical, including photocopy, recording or any information storageand retrieval system without the express permission of TecQuipment Limited. All due care has been taken to ensure that the contentsof this manualare accurate and up to date. However, if any errorsare discovered please inform TecQuipment so the problem may be rectified. A Packing Contents Ust is supplied with the equipment. Carefully checkthe contentsof the package(s) against the list. If any items are missing or damaged, contact your local TecQuipment agent or TecQuipment ltd immediately. Educational PRODUCTS
Transcript of Educational · ECCENTRIC LOADING EFFECT OF LACK OF ... VARIATION OF CRIPPLING LOAD WITH SLENDERNESS...
@TecQuipment Ltd 1999No part of this publication may be reproduced or transmitted inany form or by any means, electronic or mechanical, includingphotocopy, recording or any information storage and retrievalsystem without the express permission of TecQuipment Limited.
All due care has been taken to ensure that the contents of thismanual are accurate and up to date. However, if any errors arediscovered please inform TecQuipment so the problem may berectified.
A Packing Contents Ust is supplied with the equipment.Carefully check the contents of the package(s) against the list. Ifany items are missing or damaged, contact your localTecQuipment agent or TecQuipment ltd immediately.
EducationalPRODUCTS
v
SMIO5 Strut Machine
Assembly Instructions
Remove all the components from their packing, taking care not to discard anyloose components with the packing material.
Layout all the components on a suitable bench area and check them offagainst the packing contents list.
Assemble front specimen cross beam assembly (item 5 on the PackingContents List) onto the two guide rods (1) with screws and nuts.
Slide the scale and support assembly (7) onto guide rods and secure withscrews through the beam and rods (1) with screws and nuts.
Slide the scale and support assembly (7) onto the guide rods and secure withscrews through the beam and rods into end support leg (6). Tighten withAllen key.
Slide on dial gauge cross bar sub-assembly (4) from rear.
Slide on end leg support cross bar (2) and secure with screws through beamand rod into end support leg (6). Tighten with Allen key.
Select strut and fit in position for test.
The small weight hanger, load cord and stirrup hook are provided to applysideways load to the strut under test. The stirrup fits over the strut and thecord passes through the end of the centre cross-member over the guide rod.This cord should already be in position.
CONTENTSPage no.
i SPECIFICATION
PART 1THEORY FOR STRUTS WITH SIMPLE GEOMETRY
1
2
4
7
8
10
13
15
INTRODUCTION
CRITICAL LOADS FOR STRUTS
BOTH ENDS PIN JOINTED (HINGED)
CLAMPED - FREE CONDITION
CLAMPt;D - CLAMPED CONDITION
ECCENTRIC LOADING
EFFECT OF LACK OF STRAIGHTNESS
LATERALLY LOADED STRUT
PJ.RT 2
EXPERIMENTS ON STRUTS WITH SIMPLE GEOMETRY
19
20
23
29
33
INTRODUCTION
DETERMINATION OF FLEXURAL RIGIDITY
DEMONSTRATION OF CRIPPLED SHAPE
DETERMINATION OF LOAD v DEFLECTION CURVES AND
CRIPPLING LOAD FOR A STRUT WITH VARIOUS END
CONDITIONS
VARIATION OF CRIPPLING LOAD WITH SLENDERNESS
RATIO (FREE-FREE END CONDITIONS)
VARIATION OF CRIPPLING LOAD WITH SLENDERNESS
RATIO (FIXED-FIXED END CONDITIONS)
LOAD v DEFLECTION CURVES FOR ECCENTRICALLY
LOADED STRUTS
SOUTHWELL PLOTS USING PREVIOUSLY OBTAINED RESULTS
EXPERIMENTS ON LATERALLY LOADED STRUTS
37
40
4751
PART "1". .I~_,"
.T.HEORY FOR STRUTS ,WITH SIMPLE GEOMETRY
INTRODUCTION
A compression member whose .length is considerably greater
than the least radius of gyration of the cross-section is called
a :«:e)i\'J~'~1 or !-~~:~!~J. The two words are now taken to have the
same meaning but originally a column was taken to be vertical.
The terms l~.t;~'.':~ and !:Y!1'.'~[~:.r6)~1 are also used to describe vert-:-
ical columns.
If it is accepted that it is. extremely difficult to ensure
concentricity of load application then it will be readily apprec-
iated that longitudinal loading of a strut will cause it to bend.
This bending materially affects the stress distribution 'throughout
the cross section of the strut. Some bending is produced in all
members subjected to longitudinal loading but in long slender
struts the bending stresses far exceed the stresses due to direct
compression
In addition to the difficulty of applying the lo,ngitudinal
load so that it acts through' the controid of the cross section it
seldom happens that a strut is absolutely straight. This lack
of straightness means that the theoretical axis of the strut is
not coincident with the axis of loading. (Figure 1.1)
4XJS
fig 
2
The above discussions arise from the fact that long, slender
members do collapse or cripple when subjected to loads above
a certain critical value. It is necessary, therefore, to take
account of this fact when designing structures which contain long,
slender components which may be subjected to compressive loading.
Even if there is no theoretical reason for a strut to bend (ie the
above three factors are absent) the slightest lateral displacement,
possibly due to vibration, can cause the strut to suddenly cripple
into a bent shape.
In the theory which follows it is assumed that the strut
is ideal which implies that it is straight, its moment of inertia
is constant along the whole length and that the material is homo-
and has the same modulus of elasticity throughout. It
assumed that the plane of bending contains one of the
axes of the cross-section which ensures uniplanar
geneous
is also
principal
bending.
2 CRITICAL LOADS FOR STRUTS"'._~ - - -- . -
The theory which follows is known as Euler's theory
is based on the following assumptions:-
i) the strut is initially perfectly straight
ii) the load is applied axially
iii) the length is large compared with the radius of gyration
iv) the assumptions made in the theory of bending apply.
3
The following nomenclature ;,is used
Symbol Representing:,I;i , Length
k', Radius of gyration of cross-section
1 Second moment of area
Cross-sectional area
Breadth of strut
Depth of strut
'Lateml. load; Axial 1oad
Euler's crippling load
Bending moment
Fixing moment
Q Co-ordinate position,- CO-ordina te deflection,
Central deflection
Stress
Young's modulus
Eccentricity of loading
Numerical constants unless defined otherwise
Unit-mm
m42m4r
b
d
F
p
P'
M
M'
mm
N
N
N
x m
y
y'f
E
mm
e
a.,C,D
I.
2.1 CRITICAL LOAD FOR A STRUT WITHBOTH ENDS PIN JOINTED(HtNGED)
This case is also referred to as that for a strut 'Position
fixed at each end'. One end, at least, must be assumed free
to move in the direction of application of the load (how else can
the strain energy be supplied?). Neither end can move laterally
but there is no rotational constraint at either end.
Assume that the strut takes up the deflected form shown
in figure 2.1. Let the deflection at a point C on the centre-line
of the strut be y.
p . p~;:~:::~- .uI Y.0
fl9~
The bending moment at Cis:
M -Py 2.1=
From the theory of bending:
4dxMET 2.2=
Substituting 2.2 in 2.1 gives:
d2
~ -a 2.3.whence:
2 24+aydx
0 2.4=2 p
ETwhere a =
This is a second
for which the solution is:
order homogeneous differential equation
C cas ax + D sin ax 2.5y =
5
Differentiating 2.5, the slope is given by
a Csin ax + a Dcos ax 2.6
from equationThe slope is zero when x is zero therefore
2.6 D must be zero and the equation reduces to:
a C sin ax 2.6a
and equation (2.5)
reduces to:
C cos axy = 2.5a
The maximum deflection y' occurs at mid-span and therefore
from equation 2.5a C = y' and hence:
y y!cr,;S&'X 2.7=
At the two ends of the strut y is zero therefore:
ycos(aL/2) 0 2.8=
TheFor y to have a real value cos(aL/2) must be zero.
only solution of practical importance is that for which
aL/2 n /2=
2n2/L2ie a 2.9=
n2EI/L2Therefore P 2.10=
If the strut bends and just remains bent under the action of
the load P then P is the EULER CRIPPLING LOAD ie:
n2EI/L2p' 2.11=
The Euler formula can be applied only to ideal struts and
one would expect the actual crippling load to be lower than Pbecause a practical strut is never absolutely straight. Also,
in practice there will always be some eccentricity of the load.
The Euler formula takes no account of direct stress, a state
of affairs which is acceptable in the case of lon_'! slender struts
6
but definitely not acceptable in the case of short stocky struts
(short and stocky is the exact opposite of long and slender) in
which case it is the compressive stress that is of prime importance.
ASSUMED DEFLECTED SHAPES
Since the ideal assumptions of the Euler deriv~tion cannot
be achieved in practice a number of so-called approximate formula
are examined. The first of these assumes that the deflected form
is the same as that of a beam subjected to a point load applied
at mid-span.
Let the mid-span load on the beam be W
let the buckling load for the strut be P
let the mid-span deflection in both cases be y.
Then if the two situations are comparable the bending
moment at mid-span must be the same in each case therefore:
WL/4 Py 2.12=
From bending theory the deflection under the theload onbeam is:
WL3/48EI 2.13y =
Substituting (2.13) in (2.12):
12EI/L2p 2.14=
If the above load is assumed
5WL3/384EI in which case:uniformly distributed then
v =
p 2.15=
It is possible to consider a number of other deflected forms
eg circular arc, parabolic arc, catenary etc., which are useful
exercises for the student.
CRIPPLING STRESS
The theoretical crippling stress is given by:
n2EI/AL2P'/Af 2.16= =
7
Slenderness ratio is defined as L/k where k is the least value, I
of the ~adius of gyration of the cross-section. For a rectangular
section k = d!3.464. Equation 2.16 can therefore be written,
for a rectangular strut, in the form:
2.17f =
beThe experimental and theoretical crippling stresses cancompared by taking the experimental direct stress to be
by:
given
PIAf =
vs L/kand plotting graphs of f
f vs L/k
This, of course, ignores the bending stresses which in
case are of paramount importance.
CRITICAL LOAD FOR STRUTS WITH ONE2.2END CLAMPED AND OTHER END FREE
Referring to figure 2.2 and assuming that P is the critical
load ie the load at which instability ~curs. the moment induced
at the" fixed end will be:. . ,
M Pe=
At a distance x from the fixed end let the deflection of the
strut be y then the bending moment at this point will be P{e-y)
and from the theory of bending:
d2
~P(e-y)
EIMEf
==
whence
2~ ~--:-2 + EIdx
PeET
=
8
solution of this second order differerttial 'e"qu-a'tion is
y e + C sin ~ ~-+ D cos ax=2.22
2where a = PIEl and C and Dare- .'" --.-
consfan""ts Qf integration0 when x
= 0 when)C
e(l - cas ax)
Now Y =Also dy/dx
y =
0 therefore D: = 0 wh~~e
= -e=
~.23
e and x L which,= =At the free end of the strut y
substituted into equation 2.23 gives:
-e cos aL 0'= 2.24
If e is non-zero then cos aL must be zero in which case:
aL n./2, 3 n/2 etc=
which gives the least value of P to be:
'P'f4. 2.25
n2EI/L2where p' =
In+1D that the crippling load given by equation 2.25 is eXactlyone quarter that given by equation 2.11. It can be deduced
therefore that equivalent length of a strut with one end hinged
and the other clamped is 21 (ie a strut with both ends hinged
and of length 21 has the same crippling load).
L/2The mid-span deflection will be given by setting xin equation 2.23:
=
y' e(l - cas aL/2)=
e(l - cos ;nn> 2.26
2.3 CRI!!CAL LOAD FOR ~~UT WITH BOTH ENDS C1~PED
Referring to figure 2.3 the theory follows the same procedureas that in section 2.1 except that in the present case a bending
moment M'is induced at each end of the strut due to the clamping.
9
At the point C the bending moment. is:
M M' - Py 2.27=
2
~dx~-~El El 2.28=
2Let PIEI
be written:
and y - M'/P Y then equation 2.28 cana= .d2~
d2y
~2
-aY 2.29= .
pp
fig~
Equation ~.29 is of exactly the same form as equation 2.3
and its solution is therefore:
y 2.30C cas ax + D sin ax=
C cos ax + D sin ax + M:IPHence y 2.31=
The constants of integration C and D are found using the
end conditions of zero slope and zero deflection and the condition
of zero slope at mid-span.
Differentiating 2.31 gives
dy/dxAt x
2.32
0 and equation 2.31
-Ca sin ax + Da cos ax
0 dy/dx = 0 .'. D
.= =
becomes
C cos ax + Mi'P 2.33y =
At x 0 y' hence:y= =
y' - M'/P 2.34c =
10
Substituting 2.34 in 2.33 gives the equation of the deflected
centre-line of the strut:
(yl - N'/P) cos ax + M'/P 2.35y =
Now since the slope at the ends must always be zero differ-
entiating 2.35 and equating to zero gives:
0 aL/2) 2.36-(y' -N'/P) sin.y' - M'/P = 0 implies that the maximum deflection is
constant for all loads. (This includes zero load and therefore
the strut remains straight!
The alternative solution of 2.36 is that sin
in which case aL/2 = n and the crippling load is(aL/2) is zero
I.U2EI/L2 2.374~.The crippling load given by equation' 2.37 is the same as
that for a pin-ended strut of length L/2.
ECCENTRICALLY LOADED STRUTS- --3
Referring to figure 3.1 the .load is applied at di starK:e e
from the centre of area on the same side of the centreline of the
strut at each end and also on a principal axis of the cross
section so. that bending is uniplanar and the assumptions of the
theory of simple bending can be applied.
~p
fi9urt 3.1
The bending moment at C is Py therefore:
2El~
dx23.1-Py
11
ci2y~
,j 2"'-
(~. ~ 0,+, 3.2=
2 fl/jE 1where a .The solution of equation 3.2 gives
C sin ax + D cos axy =
At x
At xu-.,' e e=. .
= 00 Y =L/2 dy/dx ..~ C cos (aL/2) - e sin (aL/2) 0= =
hence:
c e tan(~at.721=
Substitution in equation 3~3 gives the eq~afion for the deflected
shape of the strut:
y = e(tan(aL/2).sin ax +
The maximum deflection occurs at xcos ax)
= 1/2 and is given by:
y' e sec(aL(~) e sec ( tn{P7P') 3.6= =
The deflection of
position of its axis is:the strut measured from the original
e(sec(aL/2) - ~)y' - e .For the same strut without
denection is given by equationeccentricity ie 0 thee =
c cos (tnlP/P')C cos (aL/2)y' = =
An approximation for secant which allows equation 3.6 to
be expressed in dimensionless form for values of P /P' between 0.5
and 0.9 is:
tn/P/iJi 1.2/(1 - P/.P')sec =
-I-~ AccOtding to equation 3.5 a strut loaded eccentrically mu:s.t
bend for all values of P whereas a perfectly straight strut loaded
axially will not deflect until the Euler crippling load is reached.
12
Table 3.1 gives a comparison of the values of the two sides
of equation 3.8 for the range PIP' 0.5 to 0.9.
TABLE 3.1 APPROXIMATION OF SECANT
PIP' 6.5 0.6 '0.7 0.8 0.9
sec~ ~ 2.25 2.88 6.~3.94 12.4
1.21-P/P' 2.0 6.03.0 4.0 12.0
Substituting for the secant in equation 3.6 gives:
~'e
1.21-P /P'=
In experiment No.7 in Part 2 the deflection characteristics
are obtained for two different eccentricities and also for axially
applied loading. From these characteristics the effective eccent-
ricity of the 'axial' loading is determined. (The effect of a
lack of straightness is discussed in section 3.1.) Also the
experimental dimensionless plots are compared with those based
on equation 3.9... The effect of eccentric loading on a clamped-clamped strut
is merely to induce an additional moment at the clamp equal to
eP and has no effect on the deflection of the strut. If, however,
the clamping is not perfect a value for the effective eccentricity
must be assumed. - A typical value is:
(L + 22.5D)/SOOe =
-This equation also allows for a lack of straightness of the strut.
where D . depth of section in the plane
of bending.
113
EFFECT OF LACK OF STRAIGHTNESS
The initial shape of the strut may be assumed most conven-
iently to be sinusoidal since any shape can be represented as
a'Fourier series of sinusoidal terms. The assumption of a sinus-
oidal shape also simplifies the following analysis.
Let the shape of the centre-line of the strut be represented
by
y' Y sin (nx/L) 3.10=
where Y is the maximum displacement
the centre-line from the straight line joining the ends of theof
strut
Applying the bending equation in the form
d2
~d~'
~ ~Py 3.11EI ~
2d ~.
'd';;1~dx2
4dx
ie 3.12+ 1"ay =
n2-:z.
La;;2y 3.13Y sin (nx/tti+ =
complete solution of equation 3.13 is
n2Y/L2 sin (Ux/L) 3.14C sin ax+D cos ax -y =
a
L/_2,- dy~d~,- 0 whichWhen x = 0" y = 0 and when x~ ~ ,I ~,, JI.
renders both C and D zero and gives:
= =
y x 112/12 J .!;
D2/L2.l.£a~ x sin(nx!L 3.15Y. =
L/2:And at x .2 2 2-1
Y(1t - a L In ) ry' =
14
or approximately:-
y' =3.16
x is substituted
deflection of an
If the approximation for sec(,
3.6' we obtain for the mid-span
loaded strut:
in equation
eccentrically
y' =3.6a
Thus instead of the displacement Y an equ~valent eccentricity
can be used which is given by:
e =
Y/(l + 0.025 PL2/EI 13.17
The limits of 0.025 PL2/EI are:BD
00 when P =0.25 when P fi2EI/L2and p= .It
{when Pfollows therefore
= 0) and 0.8Ythe limits
F'>.~f ye are betweenthat
when P =
3.2 THE ~OUTHW~L PLOT
. . ':.-:U'.1t 45 assumed that the deflected form of an eccent-
rically loaded strut is sinusoidal and also that initial lack of
straightness can be expressed in the same form, a graphical
construction can be used to determine the Euler crippling load
experimentally.
Since the load applied should eventually approach the mag-
"_nitude of the Euler crippling load, the previously employed
approximation for the secant of an angle is invalid.
Using the energy method it can be shown that the mid-span
deflection of a strut is given by the equation:-
15
P'/Py' e/( - I}'=.
C + ewhere e =
Rearranging equation 3.18:
Y(y'lPy -:'e.Equation 3.19 shows that a plot of y' against y'/P should
pttxiuce a straight line whose slope is P' and whose intercept gives
the equivalent eccentricity of the loading.
I. STRUTS WITH LATERA~ LOA~~~
PIN ENDED STRUT WITH LATERAL POINT LOAD AT MID-SPAN4.1
Referring to figure 4.1 the bending moment at C due to the
axial load and the transverse load is:
- Py - W(L/2 - x)/2 4.1M =
d2EI ._~
dxW(L/2 - x)/2Py-+ =
pp
LiCJl!:!!!1
The solution of equation 4.2 is:
C sin ax + D cas ax + E +Fxy ~
/Prowhere a =
areThe values of the constants of integration C C E and F
easily shown to be:C = -W/2Pa -..
D = (W/2Pa) tan(aL/2)
E = -\'!L/4P
F = W/2P
16
equation 4.3 and setting 0x =Substituting these values in
gives the mid-span deflectton:
y' (W/2Pa tan(aL/2) - WL/4P=
The bending moment 'at mid-span is given by:
M - Py - WL/4A
-(W/2a tan (aL!Z)
Since the bending moment at mid-span when P = 0 is equal to
WL/4 then equation' 4.5 indicates that the value of the maximum
bending moment is increased, d~e to the application of P, by
a factor m given by:
m = (2/aL) tan(aL/2) ,
"'co ;c
(2i n) (/P7P) tan( n/2) VP7P)
.rP'7PThe variation of with is shown figurem in
~~
f!9~
4.2 PIN-ENDED STRUT WITH UNIFORM LATERAL LOADING
A more frequently encountered c
that of a horizontal strut, subjected
cases may involve axial loading togeth
loading (often encountered in machinery).
ase of lateral loading is
to its own weight. Other
er with transverse inertia
17
The fh't~Qt~f'is dt.:vclopt>d in the Siln-IC way a~. tll,it in article
Referring 10 figure 4.3 the bending moment at Cis:
M = - Py + wx2/2 - \OiL 2/8 4.7
. 0~~~~--1p-~~-\o/.u I I~ .::::::~~~ -p
-~~:::.,~~~~---~~~~ ~~~~ ~..I~2
LiguJ:.!!!:l
1..1.
~ l/2 ~
~~~
J
~
resulting
for the
SubstitutingEld 2y/dx2 for rv1 and solving the
second order di fferential equation gives the equation
deflected form of the strut:
y = C sin ax + D cos ax + E + Fx + Gx2 4.8
setting x = 0integration andofEvaluating the constants
gives the mid-span deflection
2(w/Pa)(sec(aL/2) - 1) - wL lap 4.9
by:bending moment at mid-span is
2/8M
~
=
4.10-
Mie
4.11(Binm
The variation of
~
101
B
6
~
,
~~
~
. I 'f f II
0-2 0-4 0-6 0-8 1-0
/PIp.
~
-
~[KJ~
18
PART 2
EXPERIMENTS ON STRUTS WITH SIMPLE G£OMETRY
~,I; r ~~
~ 23
29
33
37
40
4751
INTRODUCTION
DETEJ(MINATION OF FLEXURAL RIGIDITY,. , ,
DEMONSTRATION 'OF CRIPPLED SHAPE"-
DETERMINATION OF lOAD v DEFLECTION CURVESAND CRIPPLING
LOAD FOR A STRUT wITH VARIOUS END CONDITIONS. '-"
VARIATION OF CRIPPLING LOAD WITRSLENDERNESS RATIO
(FREE-FREE END CONDITIONS)
VARIATION OF CRIPPLING LOAD WITH SLENC'ERNESS RATIO
(FIXED-FIXED END CONDITIONS)LOAD v DEFLECTION Cl:RVES FOR ECCEN~"}.')(ALLY LOAD2D
STRUTS
SOUTHWELL PLOTS USING PREVIOUSLY OBTAINED RESULTS
EXPERIMENTS ON LATERALLY LOADED STRUTS
.."
/'/' ~
FIGURE E2.1-
5 LH cross member 9 Vertical pillar for dial gauge 14 Dowel pins6 Load cell and cross support 10, 11 Beam knife edges 15 50g. Hanger? ~e~s 12 Dial gauge 16 Packing piece
GlJide rodsRH cross supportRH cross ~r
19
INTRODUCTION
The SM 105 apparatus is supplied with a set of struts of
rectangular cross section. The following 9 experiments have been
carried out using these struts of simple geometry but the range
of experiments can be extended using the additional struts avail-
able (SM lO,5a).
The basic difference between t~e theoretical and experimental
approach to the design and testing of struts lies in the fact that
in the theoretical approach the natural concern is with the
stresses induced in the strut when subjected to loading, whilst
in the experimental approach it is natural to observe and measure
deflection
The above difference in approach is further complicated
because, theoretically, a strut will remain straight until the Euler
crippling load is reached. In pract.ice the strut invariably.,deflects during loading and one has to use the fact that as tile
deflection becomes large the load reaches a constant value. BUT
if the strut is to be re-usable the yield stress ~ust not be
exceeded. In this case the maximum experimental load must
always be less than the Euler crippling load.
A rea~<?nable i~~}~~ti9n of the crippling load can be obtained
by reversing the natural direction of crippling when the strut
starts to deflect in the rig. This is easily done provided that
the deflection is still fairly small when the strut is pushed over.
False values of load will be obtained if the struts are
subjected to too large a deflection ~n the pinned end condition
because the side of the knife edge' on the strut can press again~t
the side of the specimen holder thus altering the end condition.
The values recorded in the following pages are truly typical
values and only where obvious errors occurred were tests re-run
and fresh sets of results obtained. It should be possible there-
fore for students to obtain results showing the same order of
agreement between theory and experiment.
20
2 DETERMINATION ~f_~EXURAL RIGIDITY
2.1 INTRODUCTION
Even a cursory alance thrpugh the equations governing the. " ,
beha~iour of a strut under: v~r~ous "ponditions of loading will
convince the reader that one of the principal variables with which
we are concerned is the flexural rigidity, EI, of the strut. One
of the first requirements in experimental work is therefore to
determine the flexural rigidity and one of the most convenient
methods is to measure the mid-span deflection of the strut when
it is mounted as a simply supported beam and subjected to a
central load.
2.2 PROCEDURE
1 Having selected the required strut, set the knife edge, 10,
to the correct position by moving the rear specimen beam,
3, to a position so that the strut can rest on the knife
edges near to its ends.
2 Attach the dial gauge to the vertical pillar, 9. Remove
the ball end from the dial gauge stem and exchange it with
the flat end which will be found screwed into the top of
the stem.
3' Release the central
position so that the
the knife edges.
member, 4, and slide
stem is mid-way
cross
dial gauge
it to a
between
4 Rest the strut on the knife edges with equal overhang at
each end and the dial gauge positioned on its centre-line.
:5 Open the latch of the stirrup, 8, and thread it onto the
strut. Close the latch and position the stirrup so that the
foot of the dial gauge rests on the flat c:.ut-out'in the top
of the latch.
6 Attach the weight hanger to the stirrup.
21
Measure span carefully and check that the dial gauge
is at mid-span.
the7
8 Adjust the dial gauge to read zero then attach loads to the
hanger reading the dial gauge after each increment. Tap
the top of the dial gauge stem gently before taking each
reading.
Plot a graph of load against dial gauge reading. (Re-9
10 From the slope of graph of load vs deflection, which should
be linear passing through the origin, determine the value
of EI and compare this with the value obtained assuming11 2E = 2 x 10 N/m.
TYPICAL RESULTS2.3
STRUT No.6 widt~2Omm, thickness 3.3mm, length 550mm
span 4SOmm.
RESULTS
W {f)BOO}-
7001
6001
5001
I.OO~
300~ 1 div = 0.1
2001
1001
member that each division on the dial gauge scale represents
O.lmm deflection.)
22
CALCULATIONS-
c For a simply supported beam with a point load at mid-span.1 ,
~he deflection under the load is given by:
WL 3/48E1v .therefore the flexural rigidity is given by:
E1 W/y x=
From the graph:
wy
860 x'9.81,1.71. " 4.59 kN/m= =
x 0.453/48 8.71 Nm2EI = =
bd3/12 4m=
1.94 x 1011 N/m2E871 ~ 1 _11 / L" 5. "'x W ...= .
Using a curve fitting program on a Commodore PET Computer the
value of W/y was 4.65 kN/m and the corresponding value of E
was 1.96 x 1011 N/m2
COMMENTS
The value of E obtained. here is slightly lower than might
be expected but is never-the-less quite acceptable for a structural
steel.
23
3 DEMONSTRATION OF CRIPPLED SHAPE
j!lTRODUCTION
Two assumptions are fundamental in the Euler Theory. One,
that the strut is ideal the other, closely associated with the
ideal strut, is that the mid-span deflection suddenly increases
from zero to infinity. It is impossible for the latter to happen
in practice and the former can be approached only in the case
of very carefully prepared specimen struts.
is
The above mentioned factors combine to render it impossible
to determine directly the Euler crippling load one has to look
elsewhere, therefore, to gain confidence in the theory. Equation
2.7 indicates that a crippled strut will have the shape of half
a cosine wave. It would be reasonable to expect a strut which
cripples gradually to do so with a shape not vastly different
from the theoretical shape if the theory is valid.
In'this experiment
fixings are examined.
the shapes of struts with different end
PROCEDURE
1 Ensure that the clamps of the specimen holders are tightened
2 Select the required strut and adjust the rear specimen beam
to the correct position and insert the dowel pins.
~ Turn the dial gauge so that the stem does not impede the
insertion of the strut.
4 Examine the strut and ~m.1'J straighten it if necessary.
Insert the strut with its ends in the vee groov,es. of -the specimen
holders. The edge of the strut will rest agai~st the stops
at the bottom of the holders. It may be necessary to
unscrew the loading knob, 17, to reduce the load to zero
after the strut is inserted.
5 in order
the dial
Lay the
find
rule across the
point on
holders
Turn
metre
the mid-spantospecimen
the strut.
24
gauge so that its stem is perpendicular to the strut and
thp~foot' (ball end) is on the axis of'the strut. Release thelock on the central cross member and set it so that
the dial gauge will measure the mid-span deflection of the
strut.
6 Adjust the metre rule so that the mid-span reading is a
convenient whole number (eg 30cm). It might be convenient
to 'fix' the rule with adhesive tape so that the original
position is not accidentally 'lost'.
7 Adjust the bezel of the dial gauge to indicate zero.
~c Slacken the lock on the c~~tral cross member, 4, and slide
it carefully towards one end by 2cm increments. At each
increment read the dial gauge. BD It is not possible
to take measurements close to the ends of the strut.
9 Repeat 8 sliding the cross member towards the other end.
This establishes the original shape of the strut and allows
for any mis-alignment of the specimel;1 holders.
10 Apply a load to the strut biasing the deflection away from
the dial gauge. A suitable load is one which gives a
central deflection of about 60 divisions (6mm).
11 and 9 checking frequently the load'thatRepeat steps 8remains constant
12 Repeat the experiment for the strut with the right hand end
'pinned' and the left hand end clamped. It is necessary
to adjust the rear cross member, 3, in order to use the
clamp.
13 Repeat the experiment for the strut with both ends clamped.
It is again necessary to adjust the rear cross member.
14 Plot graphs and analyse the results as indicated below.
25
TYPICAL RESULTS
STRUT NO.6 NOMINAL SIZE 20mm x 3mm x 5SOmm
1 Pinned ends. Initial anddeflection loaded deflection
One end ptnned other clamped. Loaded deflection
measured
TEST 2
recorded.
3 Both ends clamped. Loaded deflection recorded.
RESULTS See Table 3.1
GRAPHS
Figures E3.1, E?>-2, E3.3 show the plotted results.
In figure E3.1 a cosine curve is also plotted.
In figure E3.2 the curve for the strut with the clamp not fully
tightened is also shown.
COMMENTS
Figure E3.1 shows good agreement between a cosine curve and
the corrected deflection curve for the strut. The shape of the
strut is a half sine wave.
Figure E3.2 shows clearly that there is an initial lack of straight-
ness in the strut. This is indicated by the offset to the left
of the origin. The curve for the strut with imperfect clamping
shows that the slope of the strut at the clamp is not zero. The
shape of the strut is a three quarter sine wave. A good fit to
a sine wave is shown.
Figure E3.3 shows the shape of the strut to correspond 'to" a
fulrsine wav,e" and the quality of fit is seen in the figure.
'8Hi4H~
2i4~NOi4.
."NN
0('8
~
10~
..Po4
N
9
...
\0
~
~
0
~
~
\D
..
0~
,..Po4
..
w~
m
0N
NN
..~
~~a
...~
In
..~
...~
't\
.'"~
~
N
Po4
~
()
~
r)
0
~
()
a
~
~
~
..,
'"
r)
~
~
""
an...
IWt~
G\~
In~
c'"
..an
\II.~
N~
c~
Q
~g
"
".,....
on
lit...
roton
an..
~...
10M
0'"'
In~
e~
N~
-
.~~
~...
Pot..
on..,
0~
~N
~,...
~~
r4...
~
..
~
~
')
ft
'4
c
N...
\0
.-:~
\DN
01"\
10'"'
rot.~.
~
.~
f"\~
'"''"'
/WIN
.~""
Po4
lOt.
10...
G~
.~roo.
...
t)
\0
...
...
rotc-.
~~
,.....
\0~
II".tit
.~tot~
G\"'"
10'"
...0
26
~~
~
,
~~n
r-\4
Pot~
M~
In..
0~
~
In~
~
.~"()
-"'
.94
\0
tOt~
~~
GM
'D~
,.,~
0~
In.'"
"too
U)~:sSo.
~
{/)
'CQ)
...
Q.Q.
-Pol
So.
CJ,
§
C0
-Pol~-Pol
U)0
Co
.c~-Pol
):
C0
-Pol
'It}Q)
ft.o4Q)
0
f+o4
0
g-Pol
~~
-PolSo.~
>
.-t.t9)
~~~
27
~
,
d(div)
28
figure [3.3. Sh\1pe of loaded strut. Test No.3.d=dialgaUge readingx=distance from centre
29
4 DETERMINATION OF LOAD~DEFLECTION CURVES~
~~~~_I_PP~ING LOAD .FOR VARIOUS END CONDITIONS
INTRODUCTION
According to the theory of section 1 an ideal strut will
remain straight until the Euler crippling load is reached.
Equation 3.5 indicates that any eccentricity of loading will cause
the strut to deflect for all values of the axial load. Likewise
equation 3.15 shows that any lack of straightness produces the
same effect. It was shown in the theory that it is possible to
encounter a situation in which one effect offsets the other but
in practice this is a rare occurance and it is usual for a pract-
ical strut to deflect laterally under the action of an axial load.
In the experiments it is necessary to avoid plastic deformation
of the struts and it is therefore wise to limit the lateral deflection
of the struts. It~ is a~vised.-th.~t~ thf9~ilm~t 'be set at ISO dial
gauge divisions for the steel struts supplied.
PROCEDURE
1..4 Follow the procedure of experiment no. 3'.
5 Adjust the position of the central cross member so that the
dial gauge foot (ball end) rests on the centre-line of the
strut at mid-span.
6 Apply a load to the strut and ensure that the deflection
is away from the dial gauge. If deflection is towards the
dial gauge remove the load, turn the strut over and reload
to give a central deflection of about lOmm (100 divisions).
7 adjust the
the dial
bezel of the dial\
gauge stem gently
Reduce the load to zero and
gauge to indicate zero. Tap
and check the zeros.
8 Apply load by increments tapping the dial gauge stem very
gently whilst applying the load and record the load and
dial gauge reading for each increment of load.
9 Repeat the above procedure for the various end conditions
30
required and for the selection of struts being tested
10 Plot graphs of load v deflection and extrapolate the curves
to obtain the experimental crippling load.
experimental
Euler equatic
experimental,
crippling loads
ons. Determine
crippling loads
with those pre-
the relationship
for the various
1_\ the
~he
Comparedicted by
between the
end conditions
~.3 TYPICAL RESULTS
20mm x 3mm x 550mmSTRUT No.6 NOMINAL SIZE
MATERIAL MILD STEEL
Pinned EndsTest No.1. Test. No.2Pinlied...clamped
~ 1 division ;;0.1 mm.
31
Test No: 3 Clamped EndsTest No. 2a Imperfect Clamping
4.4 COMMENTS
The graphs plotted in .figure E4 show the typical load v
deflection ~urves for struts. The extrapolated experimental crip- .
pIing loads are in the ratio 4 : 2 : 1 for tests 3, 2 and 1 res-
pectively. This is exactly as predicted by the theory but the
values are as expected much lower than the Euler crippling loads.
32
q{fr-J'
l
33
$ VARIATION OF CRIPPLING LOAD WITH.§h~~DERNESS RATIO (PINNED ENDS)
INTRODUCTION
The'~Eul~r Theory shows that for pin ended struts the theor-
etical cripplifig load is:
n2EI/L2p' =
Since the
axial loads up
stress will be:
strut, again theoretically, remains straight for
to the crippling load then the axial compressive
n2Ek2/L2f P'/A= =
The slenderness ratio of a strut is defined by L/k therefore
the theoretical compressive stress at the Euler crippling load is
inversely proportinal to {slenderness ratio)2.
There are two ways of examining the experimental results.
One is to plot the experimental and theoretical curves of crippling
load against slenderness ratio which produces a curve. This is
useful in itself but it is difficult to pin-point errors. The other
way is to plot the apparent compressive stress PIA against
(k/L)2. The theoretical pl9t is, of course, linear.
The plot of experimental results shows which of the struts
tested show the greater deviation from theoretical behaviour.
The results recorded below have been processed both ways
PROCEDURE- -
Follow the procedure of experiment No.4.
9 Load the struts so that crippling occurs away from the dial
gauge. Load until the load cell reading remains constant.
Tap the strut gently in the direction of crippling. Recordthe crippling load. C .
~
10 Repeat the procedure for the desired selection of struts
34
11 If the struts are all of the same nominal cross section there
is no need to work out the value of PIA for each strut.
Plot P against Llk and against (k/L) 2. If the struts are
not of the same cross sectional area then the individual
values of PIA are required and the appropriate graphs are
those of PIA against Llk and against (k/L)2.
.12 In the cases of the aluminium and brass struts the differentvalues of E must be taken into account. Since P is propor-
tional to E then in order to compare the behaviour of the
brass and aluminium struts with steel struts the values of
L/k must be multiplied by (modulus ratio) t.
13 Compare the theoretical and experimental curves and comment
on the discrepancies between the two sets of curves.
5.3, m_ICA.L' RE5\JL T5
Tests carried out on all of the struts supplied with the-5-M
:1~ gave the results tabulated below. All struts were carefully
straightened prior to testing. -- --
pi
NrSTRVTINO
Lmm
dmm
L/k (k/L)'" 6OP/AN
PN
APPROXAREA
2mm
i:~~3:;f()4-6
h~~ 10~
~.7~~lO~,1.92.102.06.10~12.48.1-0-6
-8'3.34.10:
-63~~,,"~O '",,~ -6
2.37.10-6
3.31.10-
866808750722693635547
,559( 1 r.,"'-
I .'-"
650
545
1
2;;.( 134
$,67*81
9
,~O,
'I"J0 . f:J0.70f1 ? l ;', ;,
0.65J . ~ ~"0.62
;0.60~
0.550.750.750.75.0,75
3.'0"
~~O-"',' '...' 0
3.0s or t) :0
3.03 n"
1 .\,(,
3.0
4.75
4.65. ":4.76
4.77
150
~10200
215
230
270280
180~ ',,; :270
300
155177206223242288303186277317
60~6060606090906048
150170
~215230270187120
270375
-J i; c, J l. '" 10 2 .1Brass strut ~t:);,)Ib9'7i;JO10 N/m2 (E(steel)/E) 2 = 1.42Aluminium strut E = 6.7.10 N/m (E(steel)/E)t = 1.71
.I,
1
.15
0 500 600
£lqure £5.1.
700 800 900
P v Ltk for strutswith pinned ends
1000 .~
36
GRAPHS
The plot of P and p' v L/k is shown in figure E5.1. The
points for the aluminium and brass struts are indicated in two
positions. The positions away from the general curve are the'raw' results and th . . on the curve are obtained by
multiplying the L/kc; the: root of the modulus
e posltlons
values by squareratio.
DISCUSSION
The curve of figure ES.l is the one usually depicted in text-
books and it is seen that the experimental results lie quite close
to the 'Euler' curve.
The plots in figure E5.2 are straight lines and are therefore
much easier to deal with. It is seen that the experimental results
lie very close to the line. This is due in no small way to care-
ful straightening of the struts prior to testing.
It is seen in both of the figures ES.! and ES.2 that the
transposition of the points for the brass and .aluminium struts
bring them into ~ood agreement with the other results.
37
~ VARIATION OF CRIPPLING LOAD WITHSLENDERNESS RATIO FOR STRUTS WITH CLAMPED ENDS
6.1 INTRODUCTION
The theory of section 1 indicates that the Euler crippling
load of a strut depends on the fixing condition at the ends of
the strut. The theory also shows that the crippling load can
be related to that of a strut with both ends pinned. The results
of experiment 3 showed that the shape of a deflected strut is
approximately sinusoidal and that the number of half sine waves
representing the shape depends on the type of end fixings.
The aim of this experiment is to show that the shape of
the crippling load v slenderness ratio curve is the same as that
obtained in experiment 5 and also that the results can be pres-
ented in the same way for the load v (slenderness ratio)2.
6,2 ~RQCEDUR~
The difference in procedure between this experiment and
experiment 5 lies mainly in the fact that the ends are clamped.
It is necessary to ensure that the clamps are tight and that the
struts are initially straight in order to obtain good results.
Test the strut before recording the crippling
ensure that the deflection is away from the dial gauge.
deflection is in the wrong direction turn the strut over.
~ is required in the loading of the struts if they are not to
be permanently deformed. The experimental crippling load is
indicated by a constant reading on the load cell for increasing
deflection. (Tap the strut gently at mid-span whilst loading.)
load and,,"
If the
Process the results in the same way as those for experiment 5.
6.-3 TYPICAL ~_~TS
~11 of the strutsTable 6.1 shows the results of tests on
supplied with SM 105.
38
~
(~jli}2-6
10
STRUTNO
Lmm
drom
L/k PN
.'4PN
6OP/AN
1
2
3
I.
5'6
7;8"
9oj ,
10
700650
600
575
550
500
700
700
700
700
3.0:~.O3.03.03.03.04.754.65
4.764.77
808
751693664635577510521606508
620
780
960
9801000
1240
1250
7801100
1230
725
840
987
1075
1175
1470
1394
854
1272
1457
1.53
14n2.08
2.27
2.48
3.00
3.84
3.68
2.72
3.88
833
520
1540
GRAPHS
The Plot of P v is shown in figure E6.1 and that ofP v (k/L)2 is shown in figure E6.2. The scale for P is one
quarter of that in figures ES.! and, ES.2.
L/k
OlSCU5Sl0N
The curve of figure EG.l is identical to that of figure E5.1
except that the ordinate scale is four times as large. This isin agreement with the theory that the crippling load of a strut
with clamped ends is four times as great as that of a strut with
pinned ends.
The.. line of figure E6.2 is again identical to that of figureE5.2.
It is seen that the experimental curve deviates slightly
from the theoretical curve. This is due in part to a desire not
to damage the strut and therefore to under estimate the crippling
load, and in part to the imperfections of the struts and their
clamping.
40
7 LOAD v DEFLECTION CURVES FOR~ -
ECCENTRICALLY LOADED STRUTS-
INTRODUCTION
The SM 105 apparatus is designed to allow eccentricities
of loading of 5mm and 7.5mm. Additionally if the right hand
end of the strut is clamped eccentricities of 1.25 and 3mm can
be obtained by inserting the packing pieces (item 16) behind the
left hand specimen holder. The 5mm and 7.5mm eccentricities are
obtained by attaching the offset knife edges (item 14) to the ends
of the strut. Attaching the offset knife edges one way round
gives a eccentricity of loading of 5mm whilst reversing the offset
knife edges gives an eccentricity of 7.5mm. Only struts numbered1 to 6 * are supplied with drilled ends these being the struts
having a nominal section 2Omm x 3mm.
Almost any combinations of eccentricity and end fixings can
be examined. The theory for loading with two different eccent-
ricities is developed in exactly the same way as the theory
theory of part 1 section 3. The theory for struts loaded eccent-rically at one end with the other end clamped is easily developed
and, as indicated in the theory, there is no real purpose served
in experimenting with struts eccentrically loaded but clamped at
both ends.
7.2 PROCEDURE
1 Select the strut to be tested, check for straightness, and
then attach the offset knife ed.Qes to the ends usin.Q the Philips
.I, IItm Strut No.1 cannot be used with eccentric loading atboth ends unless the ends are clamped. Since the offset knifeedges are not part of the strut it is necessary either to correctthe dial gauge readings in order to obtain the deflection of thestrut itself, (see section 7.4 for details of the correction) or toconsider the strut extended by SOmm.
41
headed screws. the offset knife edges to the same
side of the strut unless intended otherwise, check
the eccentricity is the same at each end.
Attach
~ and.
2 Adjust the right hand cross member so as to accomodate the
strut and insert the strut in the rig with the strut offset
towards the front of the rig.
3 Adjust the centre cross member so that the dial gauge stem
is at mid-span and on the axis of the strut.
4 Load up the strut so that the central
150 divisions then unload to zero.
deflection is about
5' ~t the dial gauge bezel to read zero and then apply the
load gradually; tap the top of the dial gauge stem gently
all the while. When the required load has been reached
record the load and the dial gauge reading. The maximum
load should be abOut 90% of the Euler crippling load.
,6. Repeat
tested.
-
the above procedure 1 ~~5~lor all the struts being
7 If more. .
than one experiment on eccentric loading is being
carrlea out repeat the above procedure 1 - 6 for the
remaining end conditions.
8 Plot graphs of load v corrected mid-span d,eflection for the
various end conditions examined,
9 Plot graphs of PIP' v (y + e)/y and compare with the curve
for equation -E7.1.
7.3 TYPICAL RESULTS
The results recorded during the experiment
ented in tabular lomas shown.are best pres-
1.2
Table 7.1 Experimental Results for eccentricity 7.5mm
LOAD (N) 10 20 30 40 50 60 70 80 90 100 110 120 130
STRUT NO DIAL GAUGE READING
2 3 7 20 28 4S 66 92 _11,3 1.62
3 6 14 23 38 43 61 74 95 118
3 13 22 JJ4 46 62 75 99 J1.~ 147
.. 12 185 26 35 40 55 73 87 110 131
3 9 "16 22 286 36 43 ~5 ~7 80 91 107 124
!a~!~ 1.2 Experimental Results fortc'ien,ti:i~!ty 5m~
LOAD (N) lq 20 30 40 so 60 70 80 90 lcx> ~lO 120 130 140
STRUT NO DIAL GAUGE READING
2 5 142 20 31 46 66 90 111 154
5 1.0 15 22 303 40 50 63 81 98 125
6 11 1.7 24 31 834 40 52 65 103 126 156
5 7 13 185 23 30 37 46 58 74 86 112
i 163 11 20 25 32 39 47 566 63 14 88 107
CORRECTION OF RECORDED RESULTS7.4
The deflection at mid-span under eccentric loading conditions
can be obtained by correcting the dial gauge reading to take
account of the addition to the length of the strut caused by the
offset knife edges. If it is assumed that the whole strut assembly
deflects to form a circular arc then the deflection of the centre-
line of the strut from its original {unloaded} position can be
calculated with reference to figure E7.1.
43
"QI
c.:N-1~.u'
~ t~
flgure £7.1
Using the sagital equation:-
2 (2R - c)ca =
b2 (2R - d)d=
b2
~dcApproximately =
= 350mm and a = 375mm, and
of 10mm (100 divisions on the dial
For the longest strut tested b
for a measured deflection, c,
gauge) then:-
8.71mmd =
For the shortest strut b
c = lOmm d = 8.40mm.
300mm then if275mm and a= =
An alternative method is to use a second micrometer to
measure the lateral displacement of the end of the strut itself
and subtract this deflection from the observed result. All of the
above observed results have been reduced by a factor of 0.85
before plotting the graphs in figures E7.2 and E7. 3
7.5 DIMENSIONLESS PLOT
Equation 3.9 (p 12) showed that, using an approximation
for the secant of an angle, the deflection of the strut (measured
from the load axis) is related to the load by the equation
y'e.
1.21 - PIP'=
E7.1
44
,[\9ure E7-2~ Load eccentricity 7-5mm.
li9ure E7-3_. Load e«.entricity Smm.
45
The experimental results for two struts with different eccen-
tricities of loading are reproduced below and processed for the
dimensionless plot of figure E7.4.
RESULTS
STRUT NO 6 Total length = 598mm pI = 248N
Y = deflection measured from axis of strut
P (N)PIP'
1200.48
1300.52
1400.56
1600.65
1500.61
1700.69
e = Smm
y(div)
yYe
562.1
82
2.6
70
2.4
102
3.0120
3.4
145
3.9
e=7.5mm
y(div)
y'le
105
2.4
150 186
3.5
233 270123
2.6 3.0 4.64.1
STRUT NO 2 Total length 750mm 158Np= =
P(N 60 8070 90 100 110 120
e = 5mmy(div)y'/e
582.2
78
2.6
42
1.8104
3.1
135
3.7168 225
5.54.4
e = 7.5rnrn
y(div)
y'/e
69
1.9
932.2
160
3.1205
3.7255120
2.6 4.4
Note 1 division = 0.1 mm
4:;
£i9ure E7.1.. Dimensionless plot forfcuntrically loaded struts
DISCUSSION OF DIMENSIONLESS PLOTS
The experimental results plotted in figure £7.4 show accept-
agreement with the curve of equation E7.1 (equation 3.9).
The main conclusion of this part of the experiment is that
it is possible to obtain satisfactory approximations to complicated
equations by the adjustment of constants, eg a better fit to the
experimental results would be obtained for the higher values of
y'/e by using a factor 1.4 in place of factor 1.2 but this would
be a much higher factor than. is normally accepted.
47
8 SOUTHWELL PLOTS
8.1 INTRODUCTION
The previous experiments will have demonstrated that it is
difficult to obtain reproducible results when testing struts. The
theory of part 1 section 3.2 showed that a Southwell Plot
(yIP v y) should produce a straight line of slope equal to the
crippling load and intercept equal in the eccentricity of loading.
It follows from the above that a Southwell Plot can be used
to rationalise the experimental results and at the same time deter-
mine the equivalent eccentricity of loading.
8.2 ILLUSTRATIONS
The Southwell Plot will be illustrated by using results from
the previous experiments as follows:
1
2
3
4
56
Strut 6 with pinned ends
Strut 6 with clamped ends
Strut 6 with one end pinned other clamped
Strut I. with pinned ends
Strut I. loaded eccentrically e = Smm
Strut I. loaded eccentrically e = 7.Smm
8'.3 RtSULTS AND GRAPHS
results will be found according to the following table:
The Southwell Plots are shown in figures E8.1 ~ E8. 3
NOTE 1 div = 0.1 mm
0
1.9
Table of -Results for Southwell Plots for Strut No I.
Smme =
7.5mme =
0e .-
180
12
0.066
p 20
0.5
0.025
1SO
7.5
0.05
100
3.5
0.035
120
I.
0.033
y
yIP
401
0.025
60
2
0.033
80
30.0375
8.4 DISCUSSIONS
Plots for strut 61
The values of the parameters plotted differ so greatly
between the 'pinned ends' condition and the other two conditions
that it is not possible to make ~ll the Southwell Plots on the
same graph.
It is seen from figures 8.1 and 8.2 that it is possible to
draw straight lines through the plotted points and comparing the
slopes of these lines gives the experimental crippling loads in
the ratios 1 : 1.54 : 3.93 for the conditions Pinned-Pinned:
Pinned-Clamped : Clamped-Clamped compared with the theoretical
values 1 : 2 : 4.
so
The plots also show that the equivalent eccentricity is not
the same for each test which is easily accounted for by the
changes in end conditions during the experiments.
foundagree with thoselo:~d~The values of the crippling
in the previous experiments.
Plots for strut 42
Figure 8.3 shows the plots for zero eccentricity and eccen-
tricities of 5mm and 7.5mm and from the graphs the following
observations can be made:
three sets of experi-to allfittedStraight lines
mental results.
bea) can
w If straight lines are drawn 'by eye' then the experimental
crippling load given by the Southwell Plot is higher for
the larger eccentricity than for the lower which is not a
satisfactory conclusion. If, however, straight lines are
fitted so that the intercept agrees with the eccentricity then
the experimental crippling loads are approximately the same
for the two eccentricities and the equivalent eccentricity
due to the crookedness of the strut is the same for all
three tests. This is a much more satisfactory conclusion.
51
9 EXPERIMENTS ON LATERALLY LOADED STRUTS
I~TRODUCTI~
The theory of part 1 section 4.1 illustrates the complex
behaviour of a strut subjected to lateral loading at mid-span.
In practice what is required is information relating the deflec-
tion characteristics of a strut with and without lateral loading.
If the lateral loading is considered as producing an initial
deflected shape (assumed sinusoidal), then lateral loading can
be considered as a special case of initial crookedness (section
3.1). If the strut is considered to be a simply supported beam
loaded at mid-span then the equivalent eccentricity is:
0.75 WL3~
e =
and the theoretical
under axial loading will be given bymid-span deflection
e sec(aL/2) e sec (n/2 x {""P7P')y = =
The secant can be approximated to:
sec (11/2 x ",P7P"J 1.2/( 1 - PIP,)=
for P /pI O~5 to-'cf.'9=
Hence:
p'(l - (O.75L3/40Elyp x W) E9.1
Obviously a real strut will not behave in perfect accord-
ance with equation E9.1 and one of the main purPoses of the exp-
eriment is to investigate the quality of the agreement between
actual behaviour and that predicted by equations such as
equation E9.1.
In the following experiment an investigation is made into
the relationship between the axial loads required to produce th~
same mid-span deflection under the action of various lateral
loads applied at mid-span for various struts. Also investigated
52
is the relationship between axial
lateral load for a single strut.
load, mid-span deflection and
There which be carriedare many other investigations canout.
9.2 PROCEDURE
The student should be familiar by now with the process of
setting up the experiment and of loading the struts, measuring
deflectionet. It should only be necessary therefore to outline the
techniques peculiar to this particular experiment.
item 8)The
as follows:
lateral load appliedis by means of the stirrup
1 Ensure that the nylon
central cross member.
line the thepasses ~ guide on
z Open the latch on the stirrup and thread the stirrup
through the ring on the end of the nylon line then thread
over the strut. Close the latch .and arrange for the dial
gauge foot to rest on the flat machined in the latch of the
stirrup.
3 Attach the load carrier
nylon line which should
the central cross member.
(50g) to the ring at the end of the
now be hanging vertically below
4 Load the strut axially to obtain a mid-span deflection of
about 30 divisions and then remove the load.
So Adjust the bezel on the dial gauge to read zero tapping
the strut at mid-span whilst carrying out the zeroing.
.~ Load up
uous.ly,
the strut, tapping the strut at mid-span contin-
until the required deflection is obtained.
"~ Read the load cell and record the results.
8 Repeat this procedure
been obtained.
until a1\ the results required have
53
9 Process the results as indicated in the following section.
TYPICAL RESULTS
9~1 LATERAL LOADING 'OF STRUTS WITH PINNED ENDS
STRUTS USED NOS. 2, 3, 4, 5 and 6
10mm (100 divisions)LATERAL DEFLECTION AT MID-SPAN =
~ The loads supplied are labelled in grams and are recorded
Tabl~2.1 Axjal Lo!Q (N) !!~qy;!r~g tE ~!:2duceMid-s'Dan Deflection of lOmm
These results are presented in graphical form in figure E9.1
TEST 9.2 LATERAL LOADING OF STtUT NO 5 WITH PINNED ENDS
P' = 247 NLATERAL DEFLECTIONS AT MID-SPAN: 2.5, 5, 7.5, 10, 12.5 mm.
Lateral Deflection under the Action of Lateral Loading of Strut ;No 5
LATERAL LOAD (gf)200 400 600
y(div BOO50 OOfE 1 div = O. 1 Imn
85
120
143
85115130
72
110
125
255075
102
157182
100
138
152
as such in the table. It is necessary to convert the values to
Newtons in the subsequent calculations.
54
results are presented in graphic~l form in figure E9.2
DISCUSSION -
The graphs in figure E9.1 are linear indicating that the
relationship between P and W can be expressed in the form:
p P (1 - aW)0
where P the value of p whenis9
w ~..=
According to eq~ation E9.l, a should be proportional to
L3. If a graph of a v L3 is plotted using the results given it
is not easy to obtain a suitable straight line. There are num-
erous reasons for this difficulty excluding human error. (eg What
role does friction play at the supports? Was lOmm a suitable
choice for the lateral deflection?)
The graphs
the above problem.
in figure E9.2 go some way towards solving
Again the results allow straight line graphs to be drawn
In this case since L is not varied the lines should be parallel
and from the slopes of the individual lines an average slope is
obtained of 5.36 (when the units for lateral force are corrected).
12.Smm the followingTaking the results for y =rel.a~ionship between P and W is obtained:
p 192(1 - 0.0287 w=
rigidi ty is found to beUsing equation E9.! the flexural
Nm2 which is some 25% high.
EXTENDING THE INVESTIGATION
Equation E9.1 can be recast in the form:
L3m-r;r
1.Z '1 - PIP'~ x=
and for strut no. 5 this becomes (with EI = 9.0)
.f,
~I
.119u[t E9.3. Dimensionless plot for
loteralloadinqofstrut No.5
~w E9.2=
Figure E9. 3 shows plots of Y IW v PIP' for eq ua tion E9. 2 and for
strut no. 5 with deflection of 7.5mm, lOmm and 12.5mm.
9.6 FURTHER DlSCUS$"lON.
It now becomes evident that the experimental curves
plotted in figure E9.3 approach the curve of equation E9.2 as
the mid-span deflection increases. This is exactly what 'Would
be expected and obviously if one is prepared to subject the strut
to permanent deformation it is possible to obtain much closer
agreement between theoretical and experimental results.
NOTE
Whilst every attempt is made to ensure that photographs and text
match the equipment, the policy of continuous product improvement maymean some dissimilarity. No responsibility is accepted for such
dissimilarity.
