Mos unit iii

Mechanics of Solids (NME-302) Unit-III
Unit-III Page 1
UNIT-III
Helical and Leaf Springs:
i) Deflection of springs by energy method.
ii) Helical springs under axial load and under axial twist (respectively for circular and square
cross sections)
iii) Axial load and twisting moment acting simultaneously both for open and closed coiled
springs.
iv) Laminated springs.
Columns and Struts:
i) Buckling and stability.
ii) Slenderness ratio.
iii) Combined bending and direct stress,
iv) Middle third and middle quarter rules.
v) Struts with different end conditions.
vi) Euler’s theory for pin ended, columns.
vii) Effect of end conditions on column buckling.
viii) Ranking Gordon formulae.
ix) Examples of columns in mechanical equipments and machines.

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Helical and Leaf Springs
Helical Springs:
When the axis of the wire of the spring forms a helix on the surface of a right circular cylinder or a right circular
cane, a helical spring is obtained.
Figure 13.1 Coil of a helical spring Figure 13.2
Figure 13.1 shows a coil of a helical spring subjected to an axial load W or an axial moment M. Then ABCDE is the
helix made by the spring wire. The axis of the spring wire forms a right circular cylinder of radius R and axes of
the cylinder is OO′, ∠BPL = α, helix angle.
R = mean radius of the coil of spring (radius of right circular cylinder)
α = helix angle
n = number of coils in the spring
Length of spring wire in one coil = 2πR/cos α
Length of spring wire in n coils = 2πnR/cos α
Therefore, the length of spring wire = 2πnR
Close-Coiled Helical Spring:
When the coils of the spring are very close to each other then the coils are regarded as lying in planes at right
angle to the axis of the helix, α very small (Fig. 13.2).
Say, wire diameter = d
Polar moment of inertia of section, J= πd4/32
Direct shear force at any section = W
Direct shear stress, τ0=4W/αd2
Torsional shear stress due to twisting moment WR,τ = 16WR /αd3
As shown in Fig. 13.3, direct shear stress τ0 is added to τ at inner radius and is subtracted from τ at outer radius.
So, the resultant shear stress at the inner radius of the coil is more than the resultant shear stress at the
outer radius of the coil.
Wahl’s Factor:
The coil of a spring is not contained in a plane as shown in Fig. 13.2, but the coil is curved throughout its length.
Shearing strain at the inner surface of the coil is more than the shearing strain at the outer surface of the coil.
Wahl studied the effect of continuous curvature on spring and presented a research paper in a conference and
observed that the maximum shear stress in a spring is given by the equation,

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where 16WR/αd3 = τ, torsional shear stress
Close-Coiled Helical Spring Subjected to an Axial Load:
Consider a small element mnn′m′ of the spring wire subtending an angle δø at the centre of the spring axis. Say,
under the action of twisting moment T = WR (acting on every wire section along this coil), angular twist is δθ as
shown in Fig. 13.4.
Figure 13.3 Resultant shear stress in spring wire section Figure 13.4
Vertical deflection, dδ = Rδθ
Angular twist, δθ = TRd/GJ
Because Rdφ = dL = length of wire considered cc′
T, twisting moment at any section is same, that is, WR
n = number of turns of the spring
Total angle subtended by coils at the axis of the spring, Ø = 2πn
Polar moment of inertia of wire, J = πd4/32
So, axial deflection,
Stiffness of the spring, k =W/δ = Gd4/64nR3, i.e. an axial force required to extend or compress the spring by a
unit axial deflection.
Closed-Coiled Helical Spring Subjected to an Axial Moment:
Figure 13.5 shows a close-coiled helical spring subjected to an axial couple M. The effect of the couple is to rotate
end B of the spring with respect to end A.
Helix angle α is very small.
Ø is the total angle through which one end of the spring is turned relative to the other, when couple M is applied.
Work done on spring = (1/2)MØ
Figure 13.6 shows a bar AB of length or length of spring wire subjected to bending moment M.
ρØ = L

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Ø = L/ρ, where ρ is radius of curvature
From Flexure formula,
So, Ø = ML/EI
L = 2πnR, where n = number of coils, R = radius of spring
I = πd4/64 for spring wire
Maximum stress in spring wire due to axial couple M is
Open-Coiled Helical Spring:
In the case of open-coiled helical spring, helix angle α is sufficient and effect of α on twisting moment or bending
moment cannot be neglected. We have to clearly identify the following four parameters in open-coiled helical
spring.
1. Spring axis and coil axis, both inclined at helix angle, α.
2. Plane of spring and plane of coil, both are inclined to each other at helix angle, α as shown in Fig. 13.7.
Say, an open-coiled helical spring of mean coil radius R is subjected to an axial load W (along axis YY of spring)
WR = total moment on spring wire.
WRcos α = T ′ = Twisting moment on a small element of length, δL at the centres of coil (about y′y′ axis).
WRsin α = M ′, bending moment on small element of length, δL in the plane of coil.
Due to twisting moment T ′, there will be an angular twist δø′ and due to bending moment M′, there will be an
angular rotation δø′. δθ′ is shown by ln and it has two components δθ′cos α and δθ′sin α.
Figure 13.7
Figure 13.5
Figure 13.6

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δØ′= shown by pq has two components,
δØ′ cos α and δφ′ sin α.
δ θ = angular twist about XX axis = δθ′ cos α + δØ′ sin α
δ Ø = angular rotation about YY axis = mm − qr = δθ′ sin α − δØ′ cos α
Moreover, δθ′ = T′δL/GJ and δØ′ = M′δL/EI
Total angular twist θ, about XX axis,
When L = length of spring wire = 2πnR sec α
Axial deflection, δ = Rθ
Total angular rotation about YY axis,
Substituting the values of T ′ and M ′
Length of spring wire,
Strain Energy Method:
Total axial deflection of a spring can also be obtained by strain energy principle.
As
Open-Coiled Helical Spring Subjected to Axial Moment:
An open-coiled helical spring of mean coil radius R, helix angle α and number of turns n is subjected to an axial
moment M as shown in Fig. 13.8. Components of M are M′ = M cos α = bending moment and T′ = M sin α =
twisting moment. Take a small length, δL along coil of the spring.

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Figure 13.8
δθ′ = angular twist due to T ′
δθ′ = T′δL/GJ, about x′x′
δø = angular rotation due to M′ about y′y′ axis = M′δL/EI
Taking components of δθ′ and δØ′ along xy direction
δθ = angular twist about XX axis = δØ′ sin α − δθ′ cos α, substituting the value of δθ′, δØ′, T′, M
δØ′ = angular rotation about YY axis = δØ′ cos α + δθ′ sin α
where L = spring wire length = 2πnR sec α
as L = 2πnR sec α
Axial deflection,
Substituting the value of L = 2πnRsec α
Open-Coiled Helical Spring — Stresses Developed in Spring Wire:
(a) Axial load W on spring:
Total torque, T = WR

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Twisting moment, T ′ = WR cos α
Bending moment, M ′ = WR sin α
Wire diameter = d
Maximum torsional shear stress, τs = 16T ′/πd3
(b) Axial couple M acting on spring:
Bending moment, M′ = M cos α
Twisting moment, T ′ = M sinα (Fig. 13.10)
Torsional shear stress on spring wire, τs = 16 M sin α/πd3
Figure 13.9 Figure 13.10
Leaf Spring:
These springs are used in vehicles such as cars, trucks, railway wagons, trolley, etc. and are connected between
chassis and wheels. They are also termed as Carriage springs.
Generally they are of two types: (1) semi elliptic and (2) quarter elliptic shapes. A number of plates of
rectangular section of same width but different lengths are clamped together with the help of bolts and nuts. At

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the ends of largest leaf, there are eyes which connect the chassis to the spring with the help of shackles in
between.
Figure 13.14 Equivalent carriage spring semi elliptical spring Figure 13.15
A simple theory for deflection, stress and radius of curvature of a leaf spring is developed taking the following
assumptions:
1. Centre line of all the leaves is initially a circular arc of same radius R, so that contact between the plates is only
at the ends.
2. Each plates is of uniform thickness and each plate overlaps the plate below it by an amount p = L/2n,
where L is the length of longest leaf, n is the number of plates.
3. The overlaps are tapered in width to a triangular shape. Figure 13.14 shows the Equivalent Carriage spring of
width B = nb.
Since each plate is initially of same radius of curvature, each plate will touch the one above it only at its ends
when unloaded. After the application of a central load, the change in radius of curvature is uniform and radius of
curvature becomes the same for all leaves (plates). They will continue to contact at the ends only.
Consider only two plates, top one of length L and bottom one of length (L – 2p), loaded at ends by W/2 load.
Bending moment diagram varies linearly from A to B and C to D, while bending moment is uniform from B to C
and equal to (W/2)p. This shows that each triangular overhanging end (refer to Fig. 13.15) is loaded as a
cantilever while portion of uniform width carries uniform bending moment Wp/2. So on this tapered
portion, M and I (moment of inertia) vary linearly and proportional to this distance from end, so M/I remains
constant. Moreover in the central portion M and I are constant. This proves that M/I remain constant throughout,
the length of the plate (with triangular ends as shown).
Now, M/EI = 1/R′–1/R, change in curvature. If M/EI is constant, then 1/R is also constant. This 1/R′ is also
constant with changed radius of curvature R′, contact between the leaves continues to be at ends only. Taking the
friction between the plates as negligible, each plate is of same radius of curvature. So, these plates can be
considered to be arranged side by side forming a beam of same thickness t with width B = nb at centre as shown
in Fig. 13.14.
In a carriage spring as shown in Fig. 13.14
Maximum bending moment, Mmax = WL/4
Maximum width, B = nb
Moment of inertia,

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Bending stress,
Initial central deflection, y0 = L2/8R (using properties of a circle)
Final radius of curvature,
R′ is constant, if all the plates are bent into the arcs of a circle of radius R′.
Say R′ = ∝ (infinity) plates will become straight under Proof load, W0
Cantilever Leaf Spring (Quarter Elliptic Spring):
Bending moment = WL
Moment of inertia, I = nbt3/12
Bending stress, ς = 6WL/nbt2
Figure 13.16 Equivalent cantilever spring
Say, R′ = final radius of curative
R = initial radius of curative
y0 = L2/2 (using property of a circle)
y = final deflection under load
or,
Say, 1/R′=α (infinity)

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Springs in Series:
If two springs of different stiffness are joined end on and carry a common load W, they are said to be connected
in series and the combined stiffness and deflection are given by the following equation.
or
Springs in Parallel:
If the two spring are joined in such a way that they have a common deflection ‘x' ; then they are said to be
connected in parallel. In this care the load carried is shared between the two springs and total load W = W1 + W2
Thus
Further
W = W1 + W2
Thus, k = k1 + k2
Columns and Struts
Eccentric Axial Thrust on a Column:
A column of rectangular section B × D is shown in Fig. 10.1. G is the centroid of the section abcd. A vertical
load P is applied at G along xx axis such that GG′ = e, eccentricity. If this load acts at the CG of the section, then a
direct compressive stress is developed in section. Effect of the eccentric load is to produce bending
moment, M = Pe, on the section producing tensile and compressive stresses in the section. Along the edge ad,
there will be maximum tensile stress due to bending and along edge bc, there will be maximum compressive
stress due to bending.
Figure 10.1 Column under eccentric load
Say, Iyy = moment of inertia of section about yy axis
Direct compressive load = P
In series In parallel

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Maximum bending stresses developed in section,
Resultant stress at edge bc
Resultant stress at edge ad
If ςb > ςd, then resultant stress at edge ad will be tensile. Resultant stress distribution along xx axis is shown in
the Fig. 10.2.
Along edge bc, ςb + ςd = compressive stress
Along edge ad
Note that if
ςb − ςd = 0, no tensile stress is developed along edge ad
Therefore,
eccentricity must be less than B/6 for no tensile stress to develop in section, when load is applied
along xx axis
Load Eccentric to Both Axes (Rectangular Section):
Consider a column of rectangular section B × D as shown in Fig. 10.4. Centroid of the section abcd is at G but a
vertical load P is applied at G′ such that GG′ = e, eccentricity. There are two components of this eccentricity
along xx and yy axes, that is, ex and ey as shown is the Fig. 10.4.
Direct vertical load on column = P
Bending moment in xx plane, Mx = Pex
Bending moment in yy plane, My = Pey
Bending stresses along xx axis, edges ad and bc
Bending stresses along yy axis,
Direct compressive stress,
Figure 10.2

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Figure 10.4 Load eccentric to both axes
Resultant stresses at corners
Note that we have taken compressive stress as positive stress in a column as the tensile stress is undesirable.
Core of Rectangular Section:
The core of a rectangular section is the part of the column section in which the load can be applied without
causing tensile stress anywhere in the section. Since the columns are made of cast iron, brick work, concrete etc.,
which are not supposed to withstand tensile stresses because tensile stress may develop cracks in column
leading to its failure.
Consider a rectangular section B × D of a column. Say a load P is applied at point G′, such that GG′ = ex (Fig. 10.6).
Bending moment, Mx = Pex
Stress due to bending,
If the resultant stresses σd ± σbx has to be only compressive and no tensile stress is permitted, then

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or ,
Similarly, load can be considered on the other side of G, and
In the figure, dimension
If the load is applied within the line ln along x axis, then there will not be any tensile stress anywhere in the
section.
Eccentricity along yy axis = ey
Bending moment, My = Pey
If no tensile stress is permitted in the section, then
line mo shows that
If load is applied anywhere within line mo, no tensile stress is produced in the section. Joining the
ends l, m, n and o makes a rhombus of diagonals B/3 and D/3. This rhombus is termed as core of a rectangular
section.
Figure 10.6 Core of rectangular section Figure 10.7 Core of a circular section
Core of Circular Section:
Let us consider that the section of a column is circular of diameter d as shown in Fig. 10.7, xx and yy are
centroidal axes, with G as centroid of the section. Say load P is applied along xx axis at G′, such that GG′ = e,
eccentricity
Bending moment, Mx = Pe

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If the resultant stress is not to be tensile anywhere in the section, then:
Core of the circular section is a circle of radius d/8 or diameter d/4 as shown in figure. Area covered by a circle
of diameter d/4 is called CORE or KERNEL of circular section. If a load on column is applied within the core, then
no tensile stress will be developed anywhere in the section.
Core of Any Section:
For any section, core can be marked if we know the area of section, Zx and Zy of the section. Figure 10.8 shows an
I-section, D × B, with moment of inertia Ixx and Iyy about xx and yy axes. Section modulus,
Maximum eccentricity along x axis,
where A = area of only section
Similarly, eccentricity along yy axis,
Elastic Stability of Columns:
Structural members which carry compressive loads may be divided into two broad categories depending on their
relative lengths and cross-sectional dimensions.
Columns:
Short, thick members are generally termed columns and these usually fail by crushing when the yield stress of
the material in compression is exceeded.
Struts:
Long, slender columns are generally termed as struts; they fail by buckling some time before the yield stress in
compression is reached. The buckling occurs owing to one the following reasons.
(a) the strut may not be perfectly straight initially.
(b) the load may not be applied exactly along the axis of the Strut.
(c) one part of the material may yield in compression more readily than others owing to some lack of uniformity
in the material properties through out the strut.
In all the problems considered so far we have assumed that the deformation to be both progressive with
increasing load and simple in form i.e. we assumed that a member in simple tension or compression becomes
progressively longer or shorter but remains straight. Under some circumstances however, our assumptions of
progressive and simple deformation may no longer hold good and the member become unstable. The term strut
and column are widely used, often interchangeably in the context of buckling of slender members.]
At values of load below the buckling load a strut will be in stable equilibrium where the displacement caused by
any lateral disturbance will be totally recovered when the disturbance is removed. At the buckling load the strut
is said to be in a state of neutral equilibrium, and theoretically it should than be possible to gently deflect the
strut into a simple sine wave provided that the amplitude of wave is kept small.
Theoretically, it is possible for struts to achieve a condition of unstable equilibrium with loads exceeding the
buckling load, any slight lateral disturbance then causing failure by buckling, this condition is never achieved in
practice under static load conditions. Buckling occurs immediately at the point where the buckling load is
reached, owing to the reasons stated earlier.
The resistance of any member to bending is determined by its flexural rigidity EI and is the quantity I may be
written as I = Ak2,

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written as I = Ak2,
Where I = area of moment of inertia
A = area of the cross-section
k = radius of gyration.
The load per unit area which the member can withstand is therefore related to k. There will be two principal
moments of inertia; if the least of these is taken then the ratio is called the slenderness ratio. Its numerical value
indicates whether the member falls into the class of columns or struts.
Euler's Theory:
The struts which fail by buckling can be analyzed by Euler's theory. In the following sections, different cases of
the struts have been analyzed.
Case A: Strut with pinned ends:
Consider an axially loaded strut, shown below, and is subjected to an axial load ‘P' this load ‘P' produces a
deflection ‘y' at a distance ‘x' from one end.
Assume that the ends are either pin jointed or rounded so that there is no moment at either end.
Assumption:
The strut is assumed to be initially straight, the end load being applied axially through centroid.
In this equation ‘M' is not a function ‘x'. Therefore this equation can not be integrated directly as has been done
in the case of deflection of beams by integration method. Thus
Though this equation is in ‘y' but we can't say at this stage where the deflection would be maximum or minimum.
So the above differential equation can be arranged in the following form
Let us define a operator
D = d/dx
(D2 + n2) y = 0 where n2 = P/EI

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This is a second order differential equation which has a solution of the form consisting of complimentary
function and particular integral but for the time being we are interested in the complementary solution only[in
this P.I = 0; since the R.H.S of Diff. equation = 0]
Thus y = A cos (nx) + B sin (nx)
Where A and B are some constants. Therefore
In order to evaluate the constants A and B let us apply the boundary conditions,
(i) at x = 0; y = 0
(ii) at x = L ; y = 0
Applying the first boundary condition yields A = 0. Applying the second boundary condition gives
if B = 0, then y = 0 for all values of x hence the strut has not buckled yet. Therefore the solution is
From the above relationship the least value of P which will cause the strut to buckle, and it is called the “ Euler
Crippling Load ” Pe from which we obtain
It may be noted that the value of I used in this expression is the least moment of inertia. It should be noted that
the other solutions exists for the equation.
The interpretation of the above analysis is that for all the values of the load P, other than those which make sin
nL = 0; the strut will remain perfectly straight since
y = B sin nL = 0
For the particular value of
Then we say that the strut is in a state of neutral equilibrium, and theoretically any deflection which it suffers
will be maintained. This is subjected to the limitation that ‘L' remains sensibly constant and in practice slight
increase in load at the critical value will cause the deflection to increase appreciably until the material fails by
yielding.
Further it should be noted that the deflection is not proportional to load, and this applies to all strut problems;
like wise it will be found that the maximum stress is not proportional to load. The solution chosen of nL = π is

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just one particular solution; the solutions nL= 2π, 3π, 5π etc are equally valid mathematically and they do, in
fact, produce values of ‘Pe' which are equally valid for modes of buckling of strut different from that of a simple
bow. Theoretically therefore, there are an infinite number of values of Pe , each corresponding with a different
mode of buckling.
The value selected above is so called the fundamental mode value and is the lowest critical load producing
the single bow buckling condition.
The solution nL = 2π produces buckling in two half – waves, 3π in three half-waves etc.
If load is applied sufficiently quickly to the strut, then it is possible to pass through the fundamental mode and to
achieve at least one of the other modes which are theoretically possible. In practical loading situations, however,
this is rarely achieved since the high stress associated with the first critical condition generally ensures
immediate collapse.
Case b: One end fixed and the other free:
Writing down the value of bending moment at the point C
On rearranging
Hence in operator form, the differential equation reduces to
The solution of the above equation would consist of complementary solution and particular solution, therefore
ygen = A cos(nx) + sin(nx) + P. I
where
P.I = the P.I is a particular value of y which satisfies the differential equation

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Hence yP.I = a
Therefore the complete solution becomes
Y = A cos(nx) + B sin(nx) + a
Now imposing the boundary conditions to evaluate the constants A and B
(i) at x = 0; y = 0
This yields A = -a
(ii) at x = 0; dy/dx = 0
This yields B = 0
Hence
y = -a cos(nx) + a
Futher, at x = L; y = a
Therefore a = - a cos(nx) + a or 0 = cos(nL)
Now the fundamental mode of buckling in this case would be nL = π/2
Case 3: Strut with fixed ends:
Due to the fixed end supports bending moment would also appears at the supports, since this is the property of
the support.
Bending Moment at point C = M – P.y
Hence the general solution would be
Boundary conditions relevant to this case are at x = 0, y = 0
Also at x = 0; dy/dx = 0, hence A = 0
Further, it may be noted that at x = L ; y = 0
Then
Obviously (1 – cos nL) = 0 i.e. cos nL = 1
Hence least solution would be nL = 2π

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Case 4: One end fixed, the other pinned
In order to maintain the pin-joint on the horizontal axis of the unloaded strut, it is necessary in this case to
introduce a vertical load F at the pin. The moment of F about the built in end then balances the fixing moment.
With the origin at the built in end, the B.M. at C is given as
In the operator form the equation reduces to
Therefore the full solution is therefore
The boundary conditions are at x = 0; y = 0
Also at x = 0; dy/dx = 0
Also when x = L ; y = 0
Therefore
nL Cos nL = Sin nL or tan nL = nL
The lowest value of nL (neglecting zero) which satisfies this condition and which therefore produces the
fundamental buckling condition is nL = 4.49 radian
Equivalent Strut Length:
Having derived the results for the buckling load of a strut with pinned ends the Euler loads for other end
conditions may all be written in the same form.
Where L is the equivalent length of the strut and can be related to the actual length of the strut depending on the
end conditions.

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The equivalent length is found to be the length of a simple bow (half sine wave) in each of the strut deflection
curves shown. The buckling load for each end condition shown is then readily obtained. The use of equivalent
length is not restricted to the Euler's theory and it will be used in other derivations later.
The critical load for columns with other end conditions can be expressed in terms of the critical load for a hinged
column, which is taken as a fundamental case.
For case(c) see the figure, the column or strut has inflection points at quarter points of its unsupported length.
Since the bending moment is zero at a point of inflection, the free-body diagram would indicates that the middle
half of the fixed ended is equivalent to a hinged column having an effective length Le = L / 2.
The four different cases which we have considered so far are:
(a) Both ends pinned (b) Both ends fixed (c) One end fixed, other free (d) One end fixed and other pinned
(a) (b) (c) (d)
Limitations of Euler's Theory:
In practice the ideal conditions are never [i.e. the strut is initially straight and the end load being applied axially
through centroid] reached. There is always some eccentricity and initial curvature present. These factors need to
be accommodated in the required formulas.
It is realized that, due to the above mentioned imperfections the strut will suffer a deflection which increases
with load and consequently a bending moment is introduced which causes failure before the Euler's load is
reached. In fact failure is by stress rather than by buckling and the deviation from the Euler value is more
marked as the slenderness-ratio l/k is reduced. For values of l/k < 120 approx, the error in applying the Euler
theory is too great to allow of its use. The stress to cause buckling from the Euler formula for the pin ended strut
is
But I = Ak2
A plot of σe versus l/k ratio is shown by the curve ABC. Allowing for the imperfections of loading and strut,
actual values at failure must lie within and below line CBD. Other formulae have therefore been derived to
attempt to obtain closer agreement between the actual failing load and the predicted value in this particular
range of slenderness ratio i.e. l/k = 40 to l/k = 100.

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(a) Straight – Line Formulae: The permissible load is given by the formulae
Where the value of index ‘n' depends on the material used and the end conditions.
(b) Johnson parabolic Formulae: The Johnson parabolic formulae is defined as
where the value of index ‘b' depends on the end conditions.
(c) Rankine Gordon Formulae :
Where Pe = Euler crippling load; Pc = Crushing load or Yield point load in Compression; PR = Actual load to
cause failure or Rankine load
For a very short strut Pe is very large hence 1/Pe would be large so that 1/Pecan be neglected.
Thus PR = Pc for very large struts, Pe is very small so 1/Pe would be large and 1/Pc can be neglected , hence PR =
Pe
The Rankine formulae is therefore valid for extreme values of 1/k. It is also found to be fairly accurate for the
intermediate values in the range under consideration. Thus rewriting the formula in terms of stresses, we have
For struts with both ends pinned
Where and the value of ‘a' is found by conducting experiments on various materials. Theoretically, but
having a value normally found by experiment for various materials. This will take into account other types of end
conditions.
Note: a = 4 x (a for fixed ends)
Since the values of ‘a' are not exactly equal to the theoretical values , the Rankine loads for long struts will not be
identical to those estimated by the Euler theory as estimated.
Strut with initial Curvature:
As we know that the true conditions are never realized , but there are always some imperfections. Let us say that
the strut is having some initial curvature. i.e., it is not perfectly straight before loading. The situation will
influence the stability.
Let us analyze this effect. By differential calculus
But for this case

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Since strut is having some initial curvature, now putting
Where ‘ y0' is the value of deflection before the load is applied to the strut when the load is applied to the strut
the deflection increases to a value ‘y'. Hence
If the pinned strut is under the action of a load P then obviously the BM would be (-Py)
The initial shape of the strut y0 may be assumed circular, parabolic or sinusoidal without making much
difference to the final results, but the most convenient form is
where C is some constant or here it is amplitude which satisfies the end conditions and corresponds to a
maximum deviation ‘C'. Any other shape could be analyzed into a Fourier series of sine terms. Then
The computer solution would be therefore
Boundary conditions which are relevant to the problem are
at x = 0 ; y = 0 thus B = 0
Again when x = l ; y = 0 or x = l / 2 ; dy/dx = 0 the above condition gives B = 0
Therefore the complete solution would be
Again the above solution can be slightly rearranged since,
Hence the term after multiplying the denominator and numerator by EI is equal to
since n2 = P/EI where Pe =Euler’s load; P = applied load

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Thus,
The crippling load is again
Since the BM for a pin ended strut at any point is given as
M = -Py and Max BM = P ymax
Now in order to define the absolute value in terms of maximum amplitude let us use the symbol as ‘^'.
sin (πx/I)= 1 when (πx/I) = π/2
Strut with Eccentric Load:
Let ‘e' be the eccentricity of the applied end load, and measuring y from the line of action of the load.
or (D2 + n2) y = 0 where n2 = P / EI
Therefore
ygeneral = ycomplementary = Asin nx + Bcos nx
applying the boundary conditions then we can determine the constants i.e. at x = 0 ; y = e thus B = e
at x = l / 2 ; dy / dx = 0
Therefore
Hence the complete solution becomes
y = A sin(nx) + B cos(nx)
substituting the values of A and B we get
Note that with an eccentric load, the strut deflects for all values of P, and not only for the critical value as was the
case with an axially applied load. The deflection becomes infinite for tan (nl)/2 = ∞ i.e. nl = π giving the
Same crippling load However, due to additional bending moment set up by deflection, the strut will
always fail by compressive stress before Euler load is reached.
Since
Hence maximum BM would be

Unit-III Page 24
Now the maximum stress is obtained by combined and direct strain
The second term is obviously due the bending action. Consider a short strut subjected to an eccentrically applied
compressive force P at its upper end. If such a strut is comparatively short and stiff, the deflection due to bending
action of the eccentric load will be neglible compared with eccentricity ‘e' and the principal of super-imposition
applies.
If the strut is assumed to have a plane of symmetry (the xy - plane) and the load P lies in this plane at the
distance ‘e' from the centroidal axis ox. Then such a loading may be replaced by its statically equivalent of a
centrally applied compressive force ‘P' and a couple of moment P.e
1. The centrally applied load P produces a uniform compressive ς1 = P/A stress over each cross-section as shown
by the stress diagram.
2. The end moment ‘M' produces a linearly varying bending stress ς2 = My/I as shown in the figure. Then by
super-impostion, the total compressive stress in any fibre due to combined bending and compression becomes,

Mos unit iii

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