Department of Aeronautical Engineering
School of Mechanical Engineering
Vel Tech Dr RR & SR Technical University
Course Material
U4AEA10 AIRCRAFT STRUCTURES I
1
U4AEA10 AIRCRAFT STRUCTURES – I
LTPC
3104
OBJECTIVE
To study different types of beams and columns subjected to various types of loading and
support conditions with particular emphasis on aircraft structural components.
UNIT I Statically Determinate and Indeterminate Structures
9
Analysis of 2 D, 3 D trusses- Frames-Composite beams, Propped cantilever- fixed-fixed
beams- Clapeyron's Three Moment Equation - Moment Distribution Method, Super position
method (brief).
UNIT II Energy Methods
9
Strain Energy due to axial, bending and torsional loads – Castigliano’s theorems- Maxwell's
and Betis Reciprocal theorem, UNIT I load method - application to beams, trusses, frames,
rings, etc.
UNIT III Columns
9
Columns with various end conditions – Euler’s Column – Rankine’s formula - Column with
initial curvature - Eccentric loading – Southwell plot – Beam column, Short column, Long
column, Stability of columns.
UNIT IV Failure Theory
9
Maximum Stress theory – Maximum strain theory – Maximum shear stress theory –
Distortion Theory – Maximum strain energy theory and simple problems of shaft under
combined loading.
UNIT V Introduction to Theory of Elasticity
9
Equilibrium and Compatibility conditions for elastic solids. 2D elasticity equations for plane
stress, plane strain and generalized plane strain cases Airy’s stress function. Simple problems
in plane stress / plane strain using Cartesian and polar coordinates. Super position techniques.
Examples include (a) panels subjected to a Generalized plane strain Biaxial loading (b)
Uniform/Linearly varying edge loads on elastic half plane (c) Thick cylindrical shells.
TOTAL: 45+15(Tutorial) = 60 periods
TEXT BOOK
1.Donaldson, B.K., “Analysis of Aircraft Structures – An Introduction”, McGraw-Hill, 1993.
2.Bruhn.E.F.”Analysis and design of flight vehicle structures” Tri set of offset company,
USA, 1973.
REFERENCE BOOKS
1.Timoshenko, S., “Strength of Materials”, Vol. I and II, Princeton D. Von Nostrand Co,
1990.
2.Peery, D.J., and Azar, J.J., “Aircraft Structures”, 2nd edition, McGraw–Hill, N.Y., 1993.
3.Megson, T.M.G., “Aircraft Structures
for Engineering Students”, Edward Arnold,
2
1995.
UNIT – I
Analysis of 2 D, 3 D trusses
Frames
Composite beams,
Propped cantilever
Fixed-fixed beamsClapeyron's Three Moment Equation
Moment Distribution Method, Super position method (brief).
STATICALLY DETERMINATE AND INDETERMINATE STRUCTURES
Statically determinate structure.
If the structure can be analyzed and the reactions at the support can be determined by
using the equations of static equilibrium such Fx = 0 and Fy = 0 and M = 0, then it is
called as a statically determine structure. Example: Simply supported beam, pin jointed truss
or frame.
Truss and Frame
Truss
Frame
Truss is defined as number of members Frame is defined as number of members
riveted together to carry the horizontal, together to carry the horizontal.
vertical and inclined loads in equilibrium.
Vertical loads in equilibrium.
Types of Frames
Frames are classified into two types.
1. Perfect
2. Imperfect
(i) Deficient frame
(ii) Redundant frame
perfect and imperfect frames
Sl.No Perfect frame
Imperfect frame
3
1.
2
Perfect frames have sufficient or Imperfect frames have less or more
enough members to carry the load.
members to carry the load than the required
numbers.
It satisfies the formula n = 2j-3
It does not satisfy the formula n = 2j-3
Sl. No
Deficient frame
Redundant frame
If the number of members are If the number of members are
1.
less than the required number of more than the required number of
members n < 2j-3
members n > 2j-3
Eg. Triangular frame
Eg: Square frame
n = 3, j = 3
n = 4, j = 4
3
n = 2j-3
n = 2j-3
3 = 2×3-3,
4 = 2×4-3,
3=3
4≠5
Where, n = number of members, j = number of joints.
conditions of equilibrium used in the method of joints
The conditions of equilibrium used in the method of joints are,  Fx = 0,  Fy = 0.
One of the assumption is all the joints are pin jointed, there is no moment. The equilibrium
condition Mx = 0 is not used.
Pin-joined plane frame.
Pin-joined plane frames (also known as trusses) are commonly used in structures to
span large distances where constructing beams is uneconomical. They are common as roof
structures in industrial buildings, and large assembly building and bridges.
Assumptions made in the analysis of a pin-jointed plane frame.
The structural action of a frame is derived from the following assumptions to get an
ideal frame.
The frame has perfect hinge joints. For practical purposes, this assumption gives
reasonable results and makes the actual frame more stable.
The frame is loaded only at the
4
joints and not in between the joints. The
weights of members acting over their length are calculated and transferred to the joints for
analysis. Some bending of the members due to their own weight or loads acting in between
the joints is generally neglected.
The centroidal lines of the members meet at the joint. By careful fabrication and
design, this can be reasonably achieved. If the lines are not concurrent, some moment due to
eccentricity is developed.
Two methods employed for the analysis of a pin-jointed frame and principle involved in
each case:
The basic approach to the analysis of a frame is the section method. We take a section
cutting a number of bars, and consider the equilibrium of either of the two parts so obtained.
On solution, the equilibrium equations so formulated can give us values of unknown bar
forces and reactions. Depending upon the method of taking a section, there are basically two
methods of analysing a frame, as shown in figs.
5
Section around a joint or method of joints:
In this case, as in fig, a section is taken around a joint, isolating the joint completely.
The important point to note is that the isolated joint is in equilibrium under the action of a set
of concurrent forces. Thus, there are two equilibrium equations for each joint, H=0 and
V=0, where H and V re the summations o components along two mutually perpendicular
directions.
In a stable, determinate frame, there are 2j equations available and the number of
members is only (2j-3). The three extra equations available can be used to calculate the three
unknown reactions or for checking.
Ritter’s method of section:
Here, the section is taken as shown in fig. the truss is separated into two parts by such
a section and each part is in equilibrium under the action of a general coplanar fore system.
There are three equilibrium equations. H=0, V=0, and M=0, available for such a force
system and three unknown forces can, therefore, be determined.
If the retains are calculated from the conditions of equilibrium for the frame as a
whole, then the advantage of Ritter’s method of section is that it enables us to determine the
force in any member by taking a section cutting that member. In the method of joints, it is
necessary to go from free ends as in a cantilever truss.
Before we discuss these methods in detail, we need to look at some simple procedures
to enable us to find forces in some members through visual inspection or to check the results.
Two methods of building a frame work:
Staring with a triangle of three members and three joints, the frame can be built up to
any extent by adding two members for every additional joint. This gives an internally stable
frame work, which can be supported suitably for external stability.
Starting from a firm foundation, two members can be made to form a joint. The frame
can be built up further as described. Note that the frame work is dependent upon its
attachment to the foundation for internal
stability.
6
Analyse the frame shown in fig. and find the forces in all the members.
Solution :
There are three members forces and three reactive components – RAH, RAV, and RCV, we
formulate two equations for each of the joints a, B and C and determine the six unknowns.
Joint A from the free body diagram shown in fig.
 H  0  , RAH  FAC  FAB 
 V=0  +,
RAV  FAB 
2
0
2.5
1.5
0
2.5
(i)
(ii)
Note that members AB and AC are assumed to be tension. The ratio of the length, the
vertical, and horizontal projections for member AB are 2.5, 1.5 and 2.0.
Joint B From the free body diagram in fig.
7
 H  0  , 80-FAB 
2
2
 FBC 
0
2.5
2.5
(iii)
 V=0  +,
1.5
1.5
 FBC 
0
2.5
2.5
(iv)
-40-FAB 
Joint C from the free body diagram in fig.
 H  0  , FAC  FBC 
2
0
2.5
(v)
 V=0  +,
1.5
0
2.5
(vi)
RCV  FBC 
These equations can be solved to evaluate the six unknowns.
FAB = -14.58 kN
(The member is under compression and not tension as assumed)
FBC = - 52.09 kN
(The member is in compression)
FAC = 41.67 kN
The member is in tension as assumed)
and
RCV = 31.25 kN,
RAH = -30 kn
RAV = 8.75 kN
Analyse the frame loaded as shown in fig. and determine the forces in all the members.
8
solution :
The reactions are determined from the equilibrium conditions of the frame as a whole. Let
RAH and RAV be the vertical reaction at B.
H = 0  +, RAH+30=0,
RAH = - 30kN
Acting towards the left.
 M  0 2A
+,30  2+30  3+60  3-RBV  6  0, RBV  55kN 
 V  0, RAV  55  30  60  0, R AV  35kN 
From visual inspection, considering joint C shown in fig. FBC = 60 kN (tensile) and FAC =
FCD, both tensile or both compressive. As shown in fig, the ratio of the length, horizontal and
vertical projections are 13 , 3 and 2 for AB and BD.
From the free bodies of the joints shown in fig. we observe the following.
Joint A
9
2
 V  0, R AV  FAB 
 H=0, -30-63.1
13  0
3
13
, FAB  35 
13
 63.1 kN (compressive)
2
 FAC  0, FAC  82.5 kN (tensile)
From joint C,
FCD = 82.5 kN (tensile)
Joint D
 V  0, 55+FDB 
2
13
 0, FDB  99.15 kN (Compressive)
The forces in all the members are known. As a check, the equilibrium of joint B can be
verified. H at joint B gives.
30  63.1
3
13
 99.15 
3
13
0
V at joint B is equal to -30-60 +63.1  2/(13)1/2 +99.15  s/(13)1/2, which is equal to zero.
The forces in all the members are shown in fig.
The direction of RAH is opposite to that assumed. All the member forces are shown in their
correct nature in fig.
It may be easier in many cases to find the reactions from the equilibrium conditions for the
whole frame. The member forces can be analysed joint by joint.
Analyse the truss shown in figure and determine the forces in all the members.
Solution:
The reactions are found from the equilibrium conditions of the frame as a whole H=0 gives
RAH =0
M=0 @ A
+gives
20  2 + 50  2+ 20  4 + 30  4 – RFV  6 = 0, RFV= 56.67 kN 
V = 0 gives
RAV + 56.67 – 20 – 50 – 30 = 0, RAV = 63.33 kN 
10
A visual inspection of the frame shows that
Force in member BC (considering joint C)=20 kN (tensile)
Force in member AC = Force in member CE
Considering joint D, the members BD and DF are collinear. There is a vertical load
of 30 kN at D. If the axes are selected as shown in fig, force in member DE= 30kN for
equilibrium in a direction perpendicular to BF. The force is member DE is compressive.
Members AB and BE are inclined at 45 to the horizontal, giving a ratio of 1, 1, 2 for h,v,
and l of these members. Members BD and DF have a ratio of 2,1 and 5 for h, v and l
Joint A V= 0 gives (fig)
63.33  FAB
1
2
0
FAB =-89.56 kN (Compressive ) H=0 gives
11
1
89.56 
2
 FAC  0,
FAC  63.33kN (tensile)
FCE =63.33 kN (tensile)
From joint C,
Joint F From the free body shown in fig , V=0 gives
56.67  FDF 

1
5
0
FDF  126.72kN (compressive)
 H  0 gives
126.72 

2
5
 FFE  0
FFE = 113.34 kN (tensile)
From joint D,
FBD =126.72 kN(compressive)
Joint E V=0 gives
-30-20 +FEB 
1
2
 FEB =70.71 kN (tensile)
=0
We have determined the forces in all the members. As a check, let us consider the
equilibrium of joint B. H at joint B.
89.56 
1
2
 70.71
1
2
 126.72 
2
5
0
V at joint B,
50  20  89.56 
1
2
 70.71
1
2
 126.72 
1
5
0
Figure shows the forces in all the members.
Analyse the Warren truss shown in fig, and find the forces in all the members.
Solution:
12
The reactions are determined as
RAH =30 kN 
M=0 @ A
+, 30  2.598 +60  3 +60  3 +60  6+30  12 – RBV  9 =0
RBV= 108.66 kN, RAV=41.34 kN
The forces in the diagonal members meeting at any of the joint B, D, F or H must be
equal and opposite to give V=0 as shown in Fig. The diagonal members are inclined at 60
to the horizontal giving an h,v, l ratio of 1, 3 , 2 from the free body diagrams in fig. we can
observe the following.
 V=0 gives
Joint J
FHJ 

3
 30  0
2
FHJ  34.64 kN (tensile)
H=0 gives
FJG = 17.32 kN (compressive)
FHG =34.64 kN (Compressive)
H= 0 at joint H gives
FHF = 34.64 kN (tensile)
13
Joint G V=0 gives
108.66  FGH 
3
3
 FGF 
0
2
2
3
3
 FGF 
0
2
2
FGF = 90.83 kN (compressive)
108.66  34.64 
FFE = 90.83 kN (tensile)
From joint F,
H =0 gives’
14
FGE 
90.83 34.64

 17.32  10.77kN(tensile) F
2
2
H=0
Joint F
90.83 90.83

 34.64  56.19 kN(compressive)
2
2
Joint E V=0 gives
FED
3
3
 90.83
 60  0
2
2
FED =21.54 kN(compressive)
H=0 gives
FEc 
21.54 90.83

 10.77  0,FEC  66.97(tensile)
2
2
Joint D V =0 gives
FDC =21.54 kN (tensile)
H=0 gives
21.54
21.54
 56.19 
2
2
=77.73 kN (compressive)
Joint C
V =0 gives
FDB =
FCB 

3
3
 21.54
 60  0
2
2
FCB =47.74 kN (tensile)
H=0
1
1
 21.54   66.97  0
2
2

FCB =53.87 kN (tensile)
FAB =47.74 kN (compressive)
The forces in the members are shown in fig.
FCB  47.74 
One two unknown member forces can be evaluated at a joint by the method of joints,
explain.
15
In the method of joints, section is taken around the joint to isolate it. Since the joint is
subjected to a concurrent, coplanar force system, two conditions of equilibrium, H=0 and
V=0, are available for the joint from which two unknown forces can be determined. The
commonly used graphical method is the graphic equivalent of the method of joints. We will
discuss a number of methods which use different techniques but the basic principle is the
joint equilibrium.
As mentioned earlier, there are 2j equations for a frame with j joints, but only (2j-3)
members. Three equations can be used for finding reactions or for checking. The reactions
can also be determined from the equilibrium conditions for the frame as a whole, i.e H=0,
V=0 for the loads and reactive forces. The following examples illustrates the conventional
method of joints.
Plane truss and Space truss
A plane truss is a two dimension truss structure composed of number of bars hinged
together to form a rigid framework, all the members are lie in one plane. Eg: Roof truss in
industries.
A space truss is a three dimension truss structure composed of number of bars hinged
together to form a rigid framework, the members are lie in different plane. Eg: Transmission
line towers, crane parts.
Methods used to analyze the plane & space frames


Analytical method.
1. Method of joints
2. Method of sections (Method of moments)
3. Tension coefficient method.
Graphical method.
Assumptions made in the analyze of a truss
1. In a frame or truss all the joints will be pin jointed.
2. All the loads will be acting at the joints only.
3. The self-weight of the members of the truss is neglected. Only the live load is
considered.
4. The frame is a perfect one.
Cantilever truss
If anyone of the member of the truss is fixed and the other end is free, it is called a
cantilever truss. There is no reaction force at the fixed end.
Simply supported truss
If the members of the truss are
supported by simple supports, then it is
16
called simply supported truss. Reaction forces are at the simply supported ends.
Hints to be followed while analyzing a cantilever truss using method of joints





There is no need to find the support reactions.
The analysis is to be started from the free end where there is a maximum of
two unknown forces, using the condition of equilibrium  Fx = 0, and  Fy =
0.
All the members are assumed to be tensile.
Consider tensile forces as positive and compressive as negative.
The force convention is, upward force assigns positive sign and downward
force assigns negative sign.
Hints to be followed while analyzing a simply supported truss using method of joints

The support reactions are determined first.

The analysis is to started from the free end where there is a maximum
of two unknown forces, using the condition of equilibrium  Fx = 0, and  Fy = 0
.
 All the members are assumed to be tensile.
 Consider tensile forces as positive and compressive as negative.
 The force convention is, upward force assigns positive sign and downward
force assigns negative sign.
Relation between the numbers of members and joints in a truss
n = 2j – 3, Where, n = number of members, j = number of joints. This relation is used
to find the type of the frames. Perfect frame is only solved by method of joints.
Primary and secondary stresses in the analysis of a truss
If the stresses are produced due to direct loads like tension, compression and torsion
then the stresses are called primary stress. If the stresses are produced due to expansion,
compression and temperature variation then the stresses are called secondary stress.
Statically indeterminate structure:
The simple equations are not sufficient to solve some problems. Such problems are
called statically indeterminate structures.
For solving statically indeterminate problems, the deformation characteristics of the
structure are also taken into account along with the statical equilibrium equations. Such
equation, which contain the deformation characteristics, are called compatibility equations.
Statically indeterminate structures.
If the forces on the members of a structure cannot be determined by using conditions
of equilibrium (Fx = 0 and Fy = 0 and M = 0), it is called statically indeterminate
structures.
17
Example: Fixed beam, continuous beam
Types of statically indeterminate structures:
1. Simple statically indeterminate structures
2. Indeterminate structures of equal lengths
3. Composite structures of equal length
Continuous beam:
A beam, which is supported on more than two supports, is called a continuous beam.
Such a beam, when loaded will deflect with convexity upwards, over the intermediate
supports and with concavity upwards over the mid of the spans. The intermediate supports of
a continuous beam are always subjected to some bending moment. The end supports, if
simply supported will not be subjected to any bending moment. But the end supports, if fixed,
will be subjected to fixing moments and the slope of the beam, at the fixed ends will be zero.
1. Beams of unsymmetrical sections
2. Beams of uniform strength
3. Flitched beams
A square bar of 20 mm side is held between two rigid plates and loaded by an axial
force P equal to 450 kN as shown in fig,
Find the
reactions
at the ends A and C and the extension of the portion AB. Take e=200 GPa.
Solution : Given : area of bar (A) = 20  20 = 400 mm2 ; Axial force (P) = 450 kN = 450 
103 N; Modulus of elasticity (E) = 200 GPa = 200  103 N/mm2 ; Length of AB (lAB) = 300
mm and length of BC (lBC) = 200 mm.
Reaction at the ends
Let
RA = reaction at A, and
RC = reaction at C.
Since the bar is held between the two rigid plates A and C, therefore, the upper portion will
be subjected to tension, while the lower
portion will be subjected to compression as
18
shown in fig.
Moreover, the increase of portion AB will be equal to the decrease of the portion BC.
We know that sum of both the reaction is equal to the axial force, i.e
RA + RC = 450  103
(1)
Increase in the portion AB,
R A l AB R A  300

AE
AE
And decrease in the portion BC,
l AB 
l AB 
R A l BC R C  200

AE
AE
(2)
Since the value lAB is equal to that of lBC therefore equating the equations (1)and (2),
R A  300 R C  200

AE
AE
RC 
R A  300
 1.5 R A
200
Now substituting the value of RC in equation (2)
R A  1.5 R A  450
RA 
and
450
 180 kN
2.5
or 2.5 R a  450
Ans.
RC=1.5 RA = 1.5  180 = 270 kN Ans.
Prove the clapeyron’s theorem of three moments.
It state, “if a beam has n supports, the end ones being fixed, then the same number of
equations required to determine the support moments may be obtained from the consecutive
pairs of spans i.e AB-BC, BC-CD, CD-DE and so on.”
Proof :
Consider a continuous beam ABC, fixed at A and C and supported at B as shown in fig.
Let
l1 = Span of the beams,
l1 = Moment of inertia of the beam in span AB,
l2, l2 = Corresponding value for the span BC,
MA, = Support moment at A,
MB = Support moment at B,
MC = Support moment at C,
19
Fig.
x= Bending moment at any section X, considering the beam between
two
support as simply supported and
x’ = fixing moment at any section X, of the beam,
We know that in the span AB, the bending moment at any section X at a distance x from A,
M x   x  'x

EI1
 M d2 y 
... 
 2
 EI dx 
d2 y
  x  'x
dx 2
Multiplying the above equation by x and integrating the same for the whole span AB i.e, from
0 to I1.
l
l
l
x.d2 y
EI
  x.x.dx   x. x '.dx
dx
0
0
0
4
 dy

EI1  x.
 y   a1 x1  a'1 x '1
 dx
0
or EI[l1iB  yB )  0(iA  y A )]  a1 x1  a'1 x '1
EI[l1 iB  yB ]  a1 x1  a'1 x '1
Since yB is equal to zero, therefore
EI1 l1 iB = a1 x1  a'1 x'1
a1 = Area of the  diagram in the span AB,
x1 = distance of centre of gravity of  diagram from A in the span AB,
20
a'1 x '1 = Corresponding values for the  diagram and
iB = Slope of the beam AB at B.
we know that the shape of the ’ diagram is trapezoidal, having end ordinates equal to M A
and MB as shown in fig. therefore splitting up this trapezium into two triangles,
l l  
l 2l 

a'1 .x '1   MA  1  1    MB  1  1 
2 3 
2 3 

 (MA  2MB )
l12
6
Substituting this value of a’1. x'1 in equation (2)
l12
EI1 l1 iB  a1 x1  (MA  2MB )
6
or EI1 iB 
E . iB 
a1 x1
l
 (MA  2MB ) 1
l1
6
 (3)
a1 x a
l
 (MA  2MB ) 1
I1 l1
6I1
Similarly, in the span BC, taking C as the origin and x positive to the left,
a1 x 2
l
 (MC  2MB ) 1
 (4)
I2 l1
6I2
a2 = area of the  diagram in the span BC,
EiB' =
Where,
'
x 2 = Distance of centre of gravity of  diagram from C in the span
BC,
a '2 = Area of the  diagram in the span BC,
'
x 2 = Distance of the centre of gravity of the  diagram from C in the
span BC and
i'B = Slope of the beam BC at B,
Since iB is equal to –iB, therefore E . iB is equal to – E . I’B
21
or
 a x2
a1 x1
l
l 
 (MA  2MB ) 1    2  (MC  2MB ) 2 
I1 l1
6I1
6I2 
 I2 l2
l
l
6a x1 6a x 2
(MA  2MB ) 1  (MC  2MB ) 2   1  2
I1
I2
I1 l1
I2 l2
MA

 6a x1 6a x 2 
l1
l
l
l
 2MB 1  MC 2  2MB 2    1  2 
I1
I1
I2
I2
I2 l2 
 I1 l1
 6a x1 6a x 2 
l 
l l 
l 
MA  1   2MB  1  2   MC  2     1  2 
I2 l2 
 l2 
 I1 I2 
 I2 
 I1 l1
Flexural Rigidity of Beams.
The product of young’s modulus (E) and moment of inertia (I) is called Flexural
Rigidity (EI) of Beams. The unit is N mm2.
Constant strength beam.
If the flexural Rigidity (EI) is constant over the uniform section, it is called Constant
strength beam.
Composite beam.
A structural member composed of two or more dissimilar materials jointed together to
acts as a unit. The resulting system is stronger than the sum of its parts. The composite
action can better utilize the properties of each c constituent material.
Example : Steel – Concrete composite beam, Steel-Wood beam.
Application of the theorem of three moments to a fixed beam:
Sometimes, a continuous beam is fixed at its one or both ends. If the beams is fixed at
the left end A, then an imaginary zero span is taken to the left of A and the three moments
theorem is applied as usual. Similarly, if the both beam is fixed at the right end, then an
imaginary zero span is taken after the right end support and the three moments theorem is
applied as usual.
Carry over factor:
Consider a beam Av fixed at A and simply supported at B, let a clockwise moment be
applied at the support B of the beam as shown in fig below.
22
Let
l = Span of the beam
 = Clockwise moment applied at B (i.e MB) and
MA = Fixing moment at A.
Since the beam is not subjected to any external loading, therefore the two reactions ®- must
be equal and opposite as shown in fig.
Fig.
Taking moments about a and equating the same.
R.l = MA + 
(1)
Now consider any section X, at a distance x from A. we know that the moment at X.
Mx = MA-R.x
Or
EI
d2 y
 MA  R.x
dx 2

d2 y 
...  M=EI 2 
dx 

Integrating the above equation,
EI
dy
Rx 2
 MA .x 
 C1
dx
2
Where C1 is the first constant of integration, we know that when x = 0, then
dy
=0. therefore
dx
C1=0.
Or
EI
dy
Rx 2
 MA .x 
dx
2
(2)
Integrating the above equation once again,
EI.y 
MA .x 2 Rx3

 C2
2
6
Where C2 is the second constant of integration. We know that when x=0, then y=0. therefore
C2=0.
23
MA .x 2 Rx3
EI.y 

2
6
Or
(3)
We also know that when x = l, then y=0. Therefore substituting these values in equation (3)
0
MA .l2 R.l3

2
6
Rl3 MA .l3

6
2

or R.l=3MA
Substituting the value in equation (1)
3MA  MA  
or MA 
 MB
y

2
2
MA 1

MB 2

It is thus obvious, that carry over factor is on-half in this case,
We see from equation (2) that
EI
dy
Rl2
 MA .x 
dx
2
Now for slope at B,
EI.ib  MA .l 
Rl2
2
...(
R.l=.MA )
3
 MA .l  MA .l
2
=-
MA .l
l

2
4


...  MA  
2

24
iB  
=
=
l
4EI
l
4EI
…(Minus sign means that the tangent at B makes
an angle with AB in the negative or anticlockwise
direction)
4EI.iB
l
Define the term ‘Carry over factor’, derive a relation for the stiffness factor for a beam
simply supported at it both ends.
Consider a beam AB simply supported at A and B. let a clockwise moment be applied at the
support B of the beam as shown in fig.
Let
l
=
of the beam, and
 = Clockwise moment at B.
Span
Since the beam is simply supported at A, therefore there will be no fixing moment at
A. moreover, as the beam is not subjected to any external loading, therefore the two reactions
must be equal and opposite as shown in fig.
Taking moments about A,
R.l = 
(1)
Now consider any section X, at a distance x from a, we know that the moment at X,
Mx  R.x

d2 y 
...  M=EI 2 
dx 

d2 y
EI 2  R.x
dx
Integration the above equation,
dy
Rx 2
EI

 C1
dx
2
 (2)
Where C1 is the first constant of integration. Integrating the above equation once again,
Rx3
EI.y  
 C1x +C2
6
25
Where C2 is the second constant of integration. We know that when x=0, then y=0. Therefore
C2=0.
Rx3
EI.y  
 C1x
6
Or
 (3)
We also know that when x=l, then y=0. therefore substituting these values in the above
equation,
Rl3
0
 C1l
6
C1 
Rl2 l

6
6
...(
R.l = )
Substituting this value of C1 in equation (2),
EI
dy
Rx 2 l
Rlx 2 l

 

dx
2
6
2l
6

x 2 l

2l
6
...(
R.l=)
Now for slope at B, substituting x=1 in the above equation,
EI.iB  

is  
=
=
l
3EI
l
3EI
l2 l
l l
l
   
2l 6
2 6
3
…(Minus sign means that the tangent at
B makes an angle with AB in the negative
or anticlockwise direction)
3EI.iB
i
Stiffness factor.
It is the moment required to rotate the end while acting on it through a unit rotation,
without translation of the far end being
(i) Simply supported is given by k = 3 EI / L
(ii) Fixed is given by k = 4 EI / L
Where, E = Young’s modulus of the beam material.
I = Moment of inertia of the beam
L = Beam’s span length.
26
It is the moment required to rotate the end, while acting on it, through a unit angle
without translation of the far end. We have seen that the moment on a beam having one end
fixed and the other freely supported,
4EI.iB
l

 Stiffness factor for such a beam (substituting iB = 1),
k1 
4EI
l
Similarly, we have that the moment on a beam having simply supported ends,

3EI.i B
l
 Stiffness factor for such a beam (substituting iB = 1),
k2 
3EI
l
Derive a relation for the stiffness factor for a beam fixed at the end simply supported at
the other.
Consider a beam AB fixed at A and simply supported at B. Let a clockwise moment be
applied at the support B of the beam as shown in fig.
Let
l = Span of the beam,
 = Clockwise moment applied at B (i.e. MB) and
MA = Fixing moment at A.
Since the beam is not
subjected to any external
loading, therefore the
two reactions ® must be
equal and opposite as
shown in fig.
27
Taking moments about A and equating the same,
R.l = MA + 
(1)
Now consider any section X, at a distance x from A. we know that the moment at X,
Mx = MA – R.x
Or
EI

d2 y 
...  M=EI 2 
dx 

d2 y
 MA  R.x
dx 2
Integrating the above equation,
EI
dy
Rx 2
 MA .x 
+C1
dx
2
Where C1 the first constant of integration. We know that when x=0, then
dy
 0 . Therefore
dx
C1 = 0.
Or
EI
dy
Rx 2
 MA .x 
dx
2
(2)
Integrating the above equation once again.
EI.y 
MA .x 2 Rx 2

2
6
Where C2 is the second constant of integration. We know that when x = 0, then y=0.
Therefore C2 = 0.
Or
EI.y 
MA .x 2 Rx3

2
6
 (3)
We also know that when x=l, then y=0. Therefore substituting these values in equation (3)
M .l2 R.l3
2 A .
2
6

Rl3 MA .l2

6
2
or
R/l = 3MA
 (4)
It is thus obvious, that carry over factor is one-half in this case,
We see from equation (2) that
28
dy
Rl2
 MA .x 
dx
2
Now for slope at B, substituting x = l in the above equation,
EI
EI.iB  MA .l 
Rl2
2
...(R.l=3MA )
3
 MA .l  MA .l
2


iB  
=
Ma .l
l

2
4
l
4EI
l
4EI
=


... MA  
2

…(Minus sign means that the tangent at
B makes an angle with AB in the negative
or anticlockwise direction)
4EI.iB
l
Distribution factor and importance in the moment distribution.
Sometimes, a moment is applied on a structural joint to produce rotation, without the
translation of its members. This moment is distributed among all the connecting member of
the joint in the proportion of their stiffness.
Consider four members OA, OB, OC and OD meeting at A. let the members OA and
OC be fixed at A and C, whereas the members OB and OD be hinged at B and d. let the joint
O be subjected to a moment  as shown in fig.
Fig.
29
Let
l1 = length of the moment OA,
I1 = Moment of inertia of the member OA,
E1 = Modulus of elasticity of the member OA,
l1, I2, E2 = Corresponding values for the member OA,
l3, I4, E3 = Corresponding values for the member OC,
l4, I4, E4 = Corresponding values for the member OD.
A little consideration will show that as a result of the moment , each member gets
rotated through some equal angle. Let this angle through which each member is rotated be .
We know that the stiffness of member OA,
4E I
k1  1 1
...( End A is fixed)
l1
3E I
Similarly,
k2  2 2
...( End B is hinged)
l2
4E3I3
and
k3 
...( End C is fixed)
l3
3E I
and
k4  4 4
...( End D is hinged)
l4
now total stiffness of all members,
k = k1+k2+k3+k4
and total moment applied at the joint,
 = k
 Moment on the member OA,
1 = k1
Similarly,
2  k 2 :

2  k 3
and  4  k 4
1 k1 k1


 k k
Similarly,
30

k
2 k 2

k
 ;  3  3 and 4  4

k

k

k
 1  k1

k
Similarly,
2  k 2


; 3  k 3
k
k
and  4  k 4

k
k1 k 2 k 3
k
,
,
and 4 are known as distribution factors for the members OA,
k k k
k
OB, OC and OD respectively. The moments 1, 2, 3, 4 are known as distributed moments.
The quantities
Moment distribution method. (Hardy Cross Method).
It is widely used for the analysis of indeterminate structures. In this method, all the
members of the structure are first assumed to be fixed in position and fixed end moments due
to external loads are obtained.
Distribution factor.
When several members meet at a joint and a moment is applied at the joint to produce
rotation without translation of the members, the moment is distributed among all the
members meeting at that joint proportionate to their stiffness.
Distribution factor = Relative stiffness / Sum of relative stiffness at the joint
If there is 3 members, Distribution factors =
k3
k1
k2
,
,
k1  k2  k3 k1  k2  k3 k1  k2  k3
Carry over moment and Carry over factor.
Carry over moment: It is defined as the moment induced at the fixed end of the beam
by the action of a moment applied at the other end, which is hinged. Carry over moment is
the same nature of the applied moment.
Carry over factor (C.O): A moment applied at the hinged end B “carries over” to the
fixed end A, a moment equal to half the amount of applied moment and of the same rotational
sense. C.O. = 0.5.
Five members OA, OB, OC, OD and OE meeting at O, are hinged at A and C and fixed
at B, d and E. the lengths of OA, OB, OC, OD and OE are 3m, 4m, 2m, and 5m and
their moments if inertia are 400mm4, 300mm4, 200mm4, 300mm4 and 250mm4
respectively. Determine the distribution factors for the members and the distributed
moments.
31
Solution :
Given Length OA=3m, Length OB = 4m, Length OC = 2m, Length OD = 3m; Moment of
inertia of OC = 200mm4 ; Moment of inertia of OD = 300 mm4 ; Moment of inertia of OE =
250 mm4 and moment on D = 4000 kN-m.
We know that stiffness factor for OA,
kA 
similalrly, k B 
3EI 3  E  400

 400E
I
4
...(
Member is hinged at A)
4EI 4  E  300

I
4
...(
Member is fixed at B)
Now complete the column for stiffness for all the members, keeping in mind whether
the member is hinged or fixed at the end. Now find out the distribution factor and distribution
moments for each member as shown in the above chart.
Now from the above chart, we find that the distribution factors for OA, OB, OC OD and OE
1 3
3 1
1
,
,
and
respectively .
are ,
4 16 16 4
8
The moment of 4000 kN-m applied at the joint O will be distributed among the member as
obtained from the above in the following table.
OA
Length
(m)
3
M.I
(mm4)
400
OB
4
300
OC
2
200
OD
3
300
OE
5
250
Member
Stiffness (k)
3E  400
 400
3
4E  300
 300
4
3E  200
 300
2
4E  300
 400
3
4E  250
 200
5
Distribution
factor
400E 1

1600E 4
300E
3

1600E 16
300E
3

1600E 16
400E 1

1600E 4
200E 1

1600E 8
Distributed
moments N-m
1
 4000  1000
4
3
 4000  750
16
3
 4000  750
16
1
 4000  1000
4
1
 4000  500
8
Thus distributed moments for OA, OB, OC, OD and OE are 1000, 750, 750, 1000 and 500 Nm respectively.
Statically determinate structures and statically indeterminate structures:
32
Sl. No
1.
2.
3.
Statically determinate structures
Statically indeterminate structures
Conditions of equilibrium are sufficient Conditions of equilibrium are insufficient
to analyze the structure
to analyze the structure.
Bending moment and shear force is
independent of material and cross
sectional area.
No stresses are caused due to
temperature change and lack of fit.
Bending moment and shear force is
dependent of material and independent of
cross sectional area.
Stresses are caused due to temperature
change and lack of fit.
Continuous beam:
A Continuous beam is one, which is supported on more than two supports. For usual
loading on the beam hogging (- ive) moments causing convexity upwards at the supports and
sagging (+ ive) moments causing concavity upwards occur at mid span.
Advantages of Continuous beam over simply supported beam:
1. The maximum bending moment in case of continuous beam is much less than in case
of simply supported beam of same span carrying same loads.
2. In case of continuous beam, the averaging bending moment is lesser and hence lighter
materials of construction can be used to resist the bending moment.
General form of Clapeyron’s three moment equations for the continuous beam:
A
A
ll
Ma 11  2 Mb 12  Mc 12  (
L2
B
C
6 A1 x1 6 A2 x2

)
11
12
Where,
Ma = Hogging bending moment at A
Mb = Hogging bending moment at B
Mc = Hogging bending moment C
l1 = length of span between supports A, B
l2 = length of span between supports B, C
x1 = CG of bending moment diagram from support A
x2 = CG of bending moment diagram from support C
A1 = Area of bending moment diagram between supports A, B
A2 = Area of bending moment diagram between supports B, C
Clapeyron’s three moment equations for the continuous beam with sinking at the
33
supports:
A
B
l1
l2
C
 6A x   6A x 
  
Ma 11  2 Mb 12  Mc 12   1 1    2 2  - 6EI  1  2 
 11   12 
 11 12 
Where,
Ma = Hogging bending moment at A
Mb = Hogging bending moment at B
Mc = Hogging bending moment C
l1 = length of span between supports A, B
l2 = length of span between supports B, C
x1 = CG of bending moment diagram from support A
x2 = CG of bending moment diagram from support C
A1 = Area of bending moment diagram between supports A, B
A2 = Area of bending moment diagram between supports B, C
δ1 = Sinking at support A with compare to sinking at support B
δ2 = Sinking at support C with compare to sinking at support B.
Clapeyron’s three moment equations for the fixed beam
A
B
l
 6 Ax 
Ma  2 Mb  

 12 
Where,
Ma = Hogging bending moment at A
Mb = Hogging bending moment at B
l = length of span between diagram from support A
x = CG of bending moment diagram from support A
A = Area of bending moment diagram between supports A, B
Clapeyron’s three moment equations for the continuous beam carrying UDL on both
the spans.
A
l1
B
34
l2
C
 6 A x   6 A x  w l3 w l3
Ma 11  2 Mb 12  Mc 12   1 1    2 2  = 1 1  2 2
4
4
 11   12 
Where,
Ma = Hogging bending moment at A
Mb = Hogging bending moment at B
Mc = Hogging bending moment C
l1 = length of span between supports A, B
l2 = length of span between supports B, C
Values of (6A1 x1 / l1), (6A2 x2 / l2) values for different type of loading.
Type of loading
UDL for entire span
Central point loading
Uneven point loading
6A1x1 / l1
Wl3 / 4
(3/8)Wl2
(wa / l)/(l2 –a2)
6A2x2 / l2
Wl3 / 4
(3/8)Wl2
(wb/l) / (l2-b2)
Procedure for analyzing the continuous beams with fixed ends using three moment
equations:
The three moment equations, for the fixed end of the beam, can be modified by
imagining a span of length 1 0 and moment of inertia, beyond the support the and applying
the theorem of three moments as usual.
Two conditions for the analysis of composite beam.
(i). Strain (Stress x E) in all the material are same (e 1 = e 2)
(e 1 = (Pl1 / A1 E1); e2 = (Pl2 / A2E2))
(ii). The total load = P1 + P2 + P3 (P1 = Stress x area)
Distribution factor for the given beam.
A
A
L
B
Join
Member
A
AB
BA
BC
CB
CD
Relative
stiffness
4EI / L
3EI / L
4EI / L
4EI / L
4EI / L
DC
4EI / L
B
C
D
L
C
L
D
Sum of Relative stiffness
Distribution factor
4EI / L
(4EI / L)/(4EI / L) = 1
(3EI / L) / (7EI / L) = 3/7
(4EI / L)/ (7EI / L) = 4/7
4EI / L) / (8EI / L) = 4/8
4EI / L) / (8EI / L) = 4/8
3EI / L + 4EI/L = 7EI / L
4EI/L + 4EI/L = 8EI / L
4EI / L
Distribution factor for the given beam.
35
(4EI / L)/(4EI / L) = 1
Figure
Join
Member
Relative stiffness
A
AB
BA
BC
CB
4E (3I) / L
4E (3I) / L
4EI / L
4EI / L
B
C
Sum of Relative
stiffness
12 EI / L
12EI / L + 4EI / L =
16EI / L
4EI / L
Distribution factor
(12 EI / L) / (12EI / L) = 1
(12EI / L)/(16EI / L) = ¾
(4EI / L) / (16EI / L) = ¼
(4EI / L) /(4EI / L) = 1
Distribution factor for the given beam
Join
Member
Relative stiffness
Sum of Relative
stiffness
Distribution factor
B
BA
BC
0 (no support)
(4EI / L)
(4EI / L)
0
(4EI / L) / (4EI / L) = 1
C
CB
CD
3EI / L
4EI / L
D
DC
4EI / L
3EI / L + 4EI/L =
7EI / L
4EI / L
UNIT – II
Strain Energy due to axial, bending and torsional loads
Castigliano’s theorems
36
(3EI / L) / (7EI / L) = 3/7
(4EI / L) / (7EI / L) = 4/7
(4EI / L) /(4EI / L) = 1
Maxwell's and Betis Reciprocal theorem
UNIT I load method
Application to beams, trusses, frames, rings, etc.
ENERGY METHODS
Strain Energy:
The strain energy of a member will be defined as the increase in energy associated
with the deformation of the member. The strain energy is equal to the work done by a slowly
37
increasing load applied to the member.
2. Define Strain energy density.
The strain-energy density of a material will be defined as the strain energy per unit
volume.
3. Define Modulus of toughness.
The area under the entire stress-strain diagram was defined as the modulus of
toughness and is a measure of the total energy that can be acquired by the material.
4. Define Modulus of resilience.
The area under the stress-strain curve from zero strain to the strain y at yield is
referred to as the modulus of resilience of the material and represents the energy per unit
volume that the material can absorb without yielding. We wrote
uy 
 2y
2E
5. Write the expression for strain energy under axial load.
If the rod is of uniform cross section of area A, the strain energy is
U 
L
0
p2
dx
2AE
6. Write the expression for strain energy due to bending.
For a beam subjected to transverse loads the strain energy associated with the normal
stresses is
2
L M
U 
dx
0 2EI
where M is the bending moment and EI the flexural rigidity of the beam.
7. Write the expression for strain energy due to shearing stresses.
The strain energy associated with shearing stresses, the strain-energy density for a
material in pure shear is
2xy
u
2G
where txy is the shearing stress and G the modulus of rigidity of the material.
8. Write the expression for strain
38 energy due to torsion.
For a shaft of length Land uniform cross section subjected at its ends to couples of
magnitude T the strain energy was found to be
T 2L
U
2GJ
Where J is the polar moment of inertia of the cross-sectional area of the shaft.
9. Explain strain energy for a general state of stress.
The strain energy of an elastic isotropic material under a general state of stress and
expressed the strain energy density at a given point in terms of the principal stresses a, b
and c at that point:
u
1
a2  b2  c2  2v  a b  bc  c a  
2E 
The strain-energy density at a given point was divided into two parts: u, associated
with a change in volume of the material at that point, and ud, associated with a distortion of
the material at the same point. We wrote u = u + ud, where
u 
1  2v
2
  a  b   c 
6E
ud 
1 
2
2
2
 a  b    b  c    c  a  

12G
and
10. Define Castigliano’s theorem.
In any beam or truss subjected to any load system, the deflection at any point r is
given by the partial differential coefficient of the total strain energy stored with respect to a
force Pr acting at the point r in the direction in which the deflection is desired.
Figure
39
Figure shows a structure AB carrying a load system P1, P2, P3 ….Pr, ….Pn.
Let the deflection at the point r be yr.
Let We = External work done be the given load system
Wi = Corresponding strain energy stored.
 We = Wi
Wi
yr = Lim
Pr 0 .
Pr
Wi
yr =
Pr
=Partial differential coefficient of the total strain energy stored with respect to Pr.
11. Define Maxwell’s reciprocal theorem.
In any beam of truss the deflection at any point D due to a load W at any other point C
is the same as the deflection at C due to the same load W applied at D.
Figure (i) shows a structure AB carrying a load W applied at any point C. Let the
deflection at C be c. Let the deflection at another point D be d.
Figure (ii) shows the same structure AB carrying the same load W at D. Let the
deflections at C and D be c and d respectively.
Figure
12. Give the relation between number of joints and the number of members in a perfect
frame.
Let there be n members and j joints in a perfect frame, Fig. (a)
Fig. (a)
40
Suppose we remove three members AB, BC and CA and the three joints A, B and C.
We are now left with (n – 3) members and (j – 3) joints.
Studying this remaining part of the frame (Fig. (b)), we find that the number of
members in such that, for each joint, there are two members.
Hence for the (j – 3) joints we have 2(j – 3) members.
Fig. (b)


n – 3 = 2 (j – 3)
n = 2j – 3
Hence for a stable frame the minimum number of members required = twice the
number of joints minus three.
13. Derive the expression for Strain Energy under Axial Loading.
Strain Energy under Axial Loading
When a rod is subjected to centric axial loading, the normal stresses x can be
assumed uniformly distributed in any given transverse section. Denoting by A the area of the
section located at a distance x from the end B of the rod and by P the internal force, we write
x = P/A.
Figure
U 
2
P
dV
2EA 2
or, setting dV = A dx,
U 
L
0
P2
dx
2AE
41
In the case of a rod of uniform cross section subjected at its ends to equal and
opposite forces of magnitude P.
Figure
U
P2L
2AE
14. A rod consists of two portions BC and CD of the same material and same length, but
of different cross sections. Determine the strain energy of the rod when it is subjected to
a centric axial load P, expressing the result in terms of P, L, E, the cross-sectional area
A of portion CD, and the ratio n of the two diameters.
Figure
1 
1 
P2  L  P2  L 
2
2 
 2   P L 1 1 
Un  


2 
2AE
2 n2 A E 4AE  n 


or
1  n2 P2L
Un 
2n2 2AE
We check that, for n = 1, we have
U1 
P2L
2AE
which is the expression given in equation for a rod of length L and uniform cross section of
area A. We also note that, for n > 1, we have Un < U1; for example, when n = 2, we have U2 =
5
 8  U1. Since the maximum stress occurs in portion CD of the rod and is equal to max =
 
P/A, it follows that, for a given allowable stress, increasing the diameter of portion BC of the
rod results in a decrease of the overall
energy-absorbing capacity of the rod.
42
Unnecessary changes in cross-sectional area should therefore be avoided in the design of
members that may be subjected to loadings, such as impact loadings, where the energyabsorbing capacity of the member is critical.
15. Derive the expression for strain energy in bending.
Strain Energy in Bending
Consider a beam AB subjected to a given loading and let M be the bending moment at
a distance x from end A. Neglecting for the time being the effect of shear, and taking into
account only the normal stresses x = My/I,
Figure

M2 y 2
U 
dV  
dV
2E
2EI2
2
x
Setting dV = dA dx, where dA represents an element of the cross-sectional area, and recalling
that M2/2EI2 is a function of x alone, we have
U 
L
0
M2
2EI2
  y dA  dx
2
Recalling that the integral within the parentheses represents the moment of inertia I of the
cross section about its neutral axis, we write
U 
L
0
M2
dx
2EI
16. Determine the strain energy of the prismatic cantilever beam AB taking into account
only the effect of the normal stresses.
43
Figure
The bending moment at a distance x from end A is M = - Px. Substituting this
expression we write
2 2
LP x
P2L3
U 
dx 
0 2EI
6EI
17. Derive expression for strain energy due to torsion.
Strain Energy in Torsion
Consider a shaft BC of length L subjected to one or several twisting couples.
Denoting by J the polar moment of inertia of the cross section located at a distance x from B
and by T the internal torque in that section, we recall that the shearing stresses in the section
are xy = T/J. Substituting for xy we have
U 
2xy
2G
dV  
T 2 2
dV
2GJ2
Setting dV = dA dx, where dA represents an element of the cross-sectional area, and
observing that T2/2GJ2 is a function of x alone, we write
U 
L
0
T2
2GJ2
   dA  dx
2
Recalling that the integral within the parentheses represents the polar moment of inertia J of
the cross section, we have
U 
L
0
T2
dx
2GJ
44
Figure
In the case of a shaft of uniform cross section subjected at its ends to equal and
opposite couples of magnitude T yields.
T 2L
U
2GJ
18. A circular shaft consists of two portions BC and CD of the same material and same
length, but of different cross sections. Determine the strain energy of the shaft when it is
subjected to a twisting couple T at end D, expressing the result in terms of T, L, G, the
polar moment of inertia J of the smaller cross section, and the ratio n of the two
diameters.
Figure
1 
1 
T2  L  T2  L 
2
 2    2   T L 1 1 
Un 
2GJ
4GJ  n4 
2G n4 J
 
or
Un 
1  n4 T 2L
2n4 2GJ
We check that, for n = 1, we have
T 2L
U1 
2GJ
Which is the expression given in equation for a shaft of length L and uniform cross section.
 17 
We also note that, for n > 1, we have Un < U1; for example, when n = 2, we have U2 =  
 32 
U1. Since the maximum shearing stress occurs in the portion CD of the shaft and is
proportional to the torque T, we note as we did earlier in the case of the axial loading of a rod
that, for a given allowable stress, increasing the diameter of portion BC of the shaft results in
a decrease of the overall energy-absorbing capacity of the shaft.
19. Find the deflection at the free end of a cantilever carrying a concentrated load at the
free end. Assume uniform flexural rigidity.
45
Solution:Figure shows a cantilever carrying a point load P at the free end A. The bending
moment at any section distant x from the free end is given by
M = - Px
 Strain energy stored by the cantilever
Wi  
l
M2dx
P2 x 2dx P2 l3


.
2EI
2EI
2EI
3
0
p2 l3
Wi 
6EI

 By the first theorem of Castiglione, the deflection in the line of action of the forceP,
3
Wi  2P  l
Pl3



P
6EI
3EI
20. Find the central deflection of a simply supported beam carrying a concentrated load
at mid span. Assume uniform flexural rigidity.
Solution:Figure shows a beam AB simply supported at A and B and carrying a central load P.
Each reaction
P

2
The bending moment at any section in AC, distant x from the end A is given by,
P
M x
2
Figure
 Strain energy stored by the beam
46
Wi  

Wi 



M2dx
2
2EI
2
3
1/2
p2 x 2 dx
4 2EI

0
23
p 1 l
pl
. . 
4EI 3 8 96EI
p2l3
96EI
The deflection in the line of action P is given by
Wi 2pl3


P 96EI
pl3

48EI
21. A simply supported beam carries a point load P eccentrically on the span. Find the
deflection under the load. Assume uniform flexural rigidity.
Solution:Figure shows a beam AB of span l which carries a load P at C.
Figure
Let
AC = a and BC = b.
Pb
l
.Pa
Reaction at B =
l
The strain energy stored by the beam AB
Reaction at A=
Wi = strain energy stored by AC
+ strain energy stored by BC
47
2
2
 Pb  dx
 Pa  dx
 
x
 
x
l  2EI 0  l  2EI
0
a

Since
b
p 2 b 2 a 3 p 2 a 2 b3 p 2 a 2 b 2


a  b
6EIl2
6EIl 2
6EIl 2
a+b=l
p2a 2 b2
Wi 
6EIl
 Deflection under load P is given by
Wi .  2P  a b
Pa 2 b 2


P
6EIl
3EIl
22. Define: Strain Energy
2
2

When an elastic body is under the action of external forces the body deforms and
work is done by these forces. If a strained, perfectly elastic body is allowed to recover slowly
to its unstrained state. It is capable of giving back all the work done by these external forces.
This work done in straining such a body may be regarded as energy stored in a body and is
called strain energy or resilience.
23. Define: Proof Resilience.
The maximum energy stored in the body within the elastic limit is called Proof
Resilience.
24. Write the formula to calculate the strain energy due to axial loads (tension).
U
P2
dx
2 AE
limit 0 to L
Where,
P = Applied tensile load.
L = Length of the member
A = Area of the members
E = Young’s modulus.
25. Write the formula to calculate the strain energy due to bending.
U
M2
dx limit 0 to L
2 EI
Where,
M = Bending moment due to applied loads.
E = Young’s modulus
48
I = Moment of inertia
26. Write the formula to calculate the strain energy due to torsion
T2
dx
limit 0 to L
2GJ
Where, T = Applied Torsion
G = Shear modulus or Modulus of rigidity
J = Polar moment of inertia
U
27. Write the formula to calculate the strain energy due to pure shear
T2
U  K
dx
limit 0 to L
2GA
Where, V = Shear load
G = Shear modulus or Modulus of rigidity
A = Area of cross section
K = Constant depends upon shape of cross section.
28. Write the down the formula to calculate the strain energy due to pure shear, if shear
stress is given.
U
 2V
2G
Where, τ = Shear stress
G = Shear modulus or Modulus of rigidity
V = Volume of the material.
29. Write the down the formula to calculate the strain energy, if the moment value is
given
M2 L
2 EI
Where, M = Bending moment
L = Length of the beam
E = Young’s modulus
I = Moment of inertia
U
30. Write the down the formula to calculate the strain energy, if the torsion moment
value is given.
T2L
U
2GJ
Where, T = Applied Torsion
49
L = Length of the beam
G = Shear modulus or Modulus of rigidity
J = Polar moment of inertia
31. Write down the formula to calculate the strain energy, if the applied tension load is
given.
P2 L
U
2 AE
Where, P = Applied tensile load.
L = Length of the member.
A = Area of the member
E = Young’s modulus.
32. Write the Castigliano’s first theorem.
In any beam or truss subjected to any load system, the deflection at any point is given
by the partial differential coefficient of the total strain energy stored with respect to force
acting at a point.
U

P
Where, δ = Deflection
U = Strain Energy stored
P = Load
33. What are the uses of Castigliano’s first theorem?
1. To determine the deflection of complicated structure.
2. To determine the deflection of curved beams, springs.
34. Define: Maxwell Reciprocal Theorem.
In any beam of truss the deflection at any point ‘A’ due to a load ‘W’ at any other
point ‘C’ is the same as the deflection at ‘C’ due to the same load ‘W’ applied at ‘A’.
35. Define: Unit load method.
The external load is removed and the unit load is applied at the point, where the
deflection or rotation is to found.
36. Give the procedure for unit load method.
50
1. Find the forces P1, P2, ………. in all the members due to external loads.
2. Remove the external loads and apply the unit vertical point load at the joint if the
vertical deflection is required and find the stress.
3. Apply the equation for vertical and horizontal deflection.
37. Compare the unit load method and Castigliano’s first theorem.
In the unit load method, one has to analyze the frame twice to find the load and
deflection. While in the latter method, only one analysis is needed.
38. Find the strain energy per unit volume, the shear stress for a material is given as 50
N/mm2. Take G = 80000 N/mm2.
U
2
per unit volume.
2G
= 50 2 / (2 x 80000).
= 0.015625 N / mm2. per unit volume.
39. Find the strain energy per unit volume, the tensile stress for a material is given as
150 N/mm2. Take E = 2 x 10 N/mm2.
f2
Per unit volume
2E
= (150)2 / (2 × (2x 10 2)
= 0.05625 N/mm 2 per unit volume.
U
40. Define: Modulus of resilience.
The proof resilience of a body per unit volume. (ie) The maximum energy stored in
the body within the elastic limit per unit volume.
41. Define Trussed Beam?
A beam strengthened by providing ties and struts is known as Trusted Beams.
UNIT – III
Columns with various end conditions
Euler’s Column
Rankine’s formula
Column with initial curvature
51
Eccentric loading
Southwell plot
Beam column, Short column, Long column, Stability of columns.
PART – A
1. Define crippling load.
The compression members which we come across do not fail entirely by crushing.
These members are considerably long in comparison with their lateral dimensions. Hence,
these members start bending, i.e buckling when the axial load reaches a certain critical value.
52
Once a member shows signs of buckling it will lead to the failure of the member. This load at
which the member just buckles is called buckling load or critical load or crippling load. The
buckling load is less than the crushing load. The value of the buckling load is low for the long
members and relatively high for short members. The value of the buckling load for a given
member depends upon the length of the member and the least lateral dimension.
2. Define effective length of a column.
The effective length of a given column with given end conditions is the length of a
equivalent column of the same material and section with hinged ends having the value of the
crippling load equal to that of the given column.
3. Write the relation between effective length and actual length for various end
conditions of column.
Actual length of column = l, effective length = L
Case 1 Both ends hinged
L=l
Case 2 One end fixed, One end free
L = 2l
Case 3 Both ends fixed
l
L
2
Case 4 One end fixed, one end hinged
l
L
2
4. What are the assumption made in Euler’s theory?
Euler’s formula for the cripping load is based on the following assumption :
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
The column I initially perfectly straight and is axially loaded.
The section of the column is uniform.
The column material is perfectly elastic, homogeneous and isotropic and obeys
Hooke’s law.
The length of the column is very large compared to the laternal dimensions.
The direct stress is very small compares with bending stress corresponding to the
buckling condition.
The self-weight of the column is ignorable.
The column will fail by buckling alone.
5. Define slenderness ratio of the column.
The ratio
L
effectiveness length

is called the slenderness ratio of the column.
K least radius of gyration
6. Write Rankine’s formula for colum.
53
Let P be the actual crippling load. Rankine stated his empirical formula
1 1 1


P Pc Pc
Where
Pc = Fc A = crushing load
Pe = Buckling load according to Euler’s formula
2EI
= Eulerian load =
L2
7. Define factor of safety for column.
This is the ratio of the critical load to the safe load on the column.
8. Define Column
A structural member, subjected to an axial compressive force is called a strut. A strut
may be horizontal, inclined or even vertical. But a vertical strut used in buildings or frames
is called a column and whose lateral dimensions are small as compared to its height.
9. What are the types of column failure?
1. Crushing failure:
The column will reach a stage, when it will be subjected to the ultimate
crushing stress, beyond this the column will fail by crushing. The load corresponding to the
crushing stress is called crushing load. This type of failure occurs in short column.
2. Buckling failure:
This kind of failure is due to lateral deflection of the column. The load at
which the column just buckles is called buckling load or crippling load or critical load. This
type of failure occurs in long column.
10. What is slenderness ration (buckling factor)? What is relevance in column?
It is the ratio of effective length of column to the least radium of gyration of the cross
sectional ends of the column.
Slenderness ration = 1 eff / r
1 eff = effective length of column
r = least radius of gyration
Slenderness ratio is used to differentiate the type of column. Strength of the column
depends upon the slenderness ratio, it is increased the compressive strength of the column
54
decrease as the tendency to buckle is increased.
11. What are the factors affect the strength column?
1. Slenderness ratio
Strength of the column depends upon the slenderness ratio, it is increased the
compressive strength of the column decrease as the tendency to buckle is increased.
2. End conditions: Strength of the column depends upon the end conditions also.
12. Differentiate short and long column
Short column
1. It is subjected to direct compressive
stresses only.
2. Failure occurs purely due to crushing
only.
3. Slenderness ratio is less than 80.
4. It’s length to least lateral dimension is
less than 8. (L / D < 8)
Long column
It is subjected to buckling stress only.
Failure occurs purely due to buckling
only.
Slenderness ratio is more than 120.
It’s length to least lateral dimension is
more than 30. (L / D > 30)
13. What are the assumptions followed in Euler’s equation?
1. The material of the column is homogeneous, isotropic and elastic.
2. The section of the column is uniform throughout.
3. The column is initially straight and load axially.
4. The effect of the direct axial stress is neglected.
5. The column fails by buckling only.
14. What are the limitations of Euler’s formula?
1. It is not valid for mild steel column. The slenderness ratio of mild steel column is
less than 80.
2. It does not take the direct stress. But in excess of load it can withstand under direct
compression only.
15. Write the Euler’s formula for different end conditions.
1. Both ends fixed.
PE 
2. Both ends hinged
PE 
 2 EI
(0.5 L)2
 2 EI
( L)2
3. One end fixed, other end hinged.
 2 EI
PE 
(0.7 L)2
L = Length of the column
55
4. One end fixed, other end free.
 2 EI
PE 
(2 L)2
16. Define: Equivalent length of the column.
The distance between adjacent points of inflection is called equivalent length of the
column. A point of inflection is found at every column end, that is free to rotate and every
point where there is a change of the axis. i.e, There is no moment in the inflection points.
(Or)
The equivalent length of the given column with given end conditions, is the length of
an equivalent column of the same material and cross section with hinged ends, and having the
value of the crippling load equal to that of the given column.
17. What are the uses of south well plot? (Column curve).
The relation between the buckling load and slenderness ratio of various column is
known as south well plot.
The south well plot is clearly shows the decreases in buckling load increases in
slenderness ratio.
It gives the exact value of slenderness ratio of column subjected to a particular
amount of buckling load.
18. Give Rankine’s formula and its advantages.
PR 
fC A
(1  a (1eff / r )2 )
Where , PR = Rankine’s critical load
fC = yield stress
A = cross sectional area
a = Rankine’s constant
leff = effective length
r = radius of gyration
In case of short column or strut, Euler’s load will be very large. Therefore, Euler’s
formula is not valid for short column. To avoid this limitation, Rankine’s formula is
applicable for both long and short column.
19. Write Euler’s formula for maximum stress for a initially bent column?
σ max = P / A + ( M max / Z)
Pa
=P/A+
(1-(P/PE ))Z
56
Where, P = axial load
A = cross section area
PE = Euler’s load
a = constant
Z = section modulus
20. Write Euler’s formula for maximum stress for a eccentrically loaded column?
σ max = P / A + ( M max / Z)
P Sec(1 eff/2) (P/EI)
=P/A+ e
(1  ( P / PE ))Z
Where, P = axial load
A = cross section area
PE = Euler’s load
e = eccentricity
Z = section modulus
EI = flexural rigidity
21. What is a beam column? Give examples.
Column having transverse load in addition to the axial compressive load are termed as
beam column.
Eg: Engine shaft, Wing of an aircraft.
22. Write the expressions for the maximum deflection developed in a beam column
carrying central point load with axial load, hinged at both ends.
 max = O /(1  ( P / PE ))
Where, δO = QL3 / 48EI
Q = central point load
P = axial load
PE = Euler’s load
23. Write the expressions for maximum bending moment and max. Stress developed in
a beam column carrying central point load hinged at both ends.
[1  0.18( P / PE )]
[1  ( P / PE )]
σ max = P / A + ( M max / Z)
M max = MO
Where,
MO = QL / 4
Q = central point load
P = axial load
PE = Euler’s load
Z = section modulus
57
24. Write the expressions for the maximum deflection developed in a beam column
carrying uniformly distributed load with axial load, hinged at both ends.
 max  O /(1  ( P / PE ))
Where,
δO = 5w L4 / 384 EI
w = uniformly distributed load / m run.
P = axial load
PE = Euler’s load.
25. Write the expressions for the maximum bending moment and max, stress developed
in a beam column carrying uniformly distributed load with axial load, hinged at both
ends.
[1  0.03( P / PE )]
[1  ( P / PE )]
σ max = P / A + ( M max / Z)
M max = MO
Where,
MO = wL2 / 8
w = uniformly distributed load / m run
P = axial load
PE = Euler’s load
Z = section modulus
26. Write the expressions for the deflection developed in a beam column carrying
several point loads at different distance with an axial load, hinged at both ends.
y  sin kx [ Q1 sin kc1  Q2 Sin kc2  Q3 sin kc3........ ]  ( x / PL)[Q1c1  Q2c2  Q3c3...... ]Pk sin kL
 sin k( L  x)[Q2 sin k(L-c2 )  Q3 sin k( L  c3 )]  (( L  x) / PL))[Q2 ( L  c2 )  Q3 ( L  c3 )]Pk sin kL
27. Write the general expressions for the maximum bending moment, if the deflection
curve equation is given.
BM = - EI (d2y / dx2)
PART – B
1. Derive the expression for crippling load when both ends of the column are hinged.
Case 1 : when both ends of the column are pinned or hinged.
The following fig. shows a column AB of the length l and uniform sectional area a, hinged at
both the ends A and B. let P be the crippling load at which the column has just buckled.
58
Consider any section at a distance x from the end B. let y be the deflection (lateral
displacement) at the section.
The bending moment at the section is given by,
El
d2 y
 Py
dx 2
 El

d2 y
 Py  0
dx 2
fig.
d2 y P
 y0
dx 2 El
The solution to the above differential equation is


P
P
y  C1 cos  x
 C2 sin  x

 EI 
 EI 




Where C1 and C2 are constants of integration.
At B, the deflection is zero.
 At
x = 0, y = 0

C1 = 0
A also, the deflection is zero.
i.e at
x = l, y = 0
 P

0 = C2 sin  l

 EI 
Since C1 = 0 we conclude that C2 cannot be zero.
This is because if both C1 and C2 are zero the column will not bend at all.

Hence sin  l


 l


P
0
EI 
P
  0, , 2,3,4
EI 
Considering the least practical value,
59
l
P

EI
 P=
2EI
l2
2. Derive the expression for crippling load when one end of a column is fixed and other
is free.
Case 2 : when one is fixed and the other is free.
The following fig. shows a column AB of length l whose lower end B is fixed, the
upper end A being free. Let due to the crippling load P the column just buckle. Let a be the
deflection at the top end.
El
d2 y
 P(a  y)
dx 2
Where y is the deflection at X.

d2 y
EI 2  Py  Pa
dx

d2 y P
Pa
 y
2
dx
EI
EI
The solution to the above differential equation is


P
P
y  C1 cos  x
 C2 sin  x
a

 EI 
 EI 




Where C1 and C2 are constants of integration.
At B, the deflection is zero.
 At

x = 0, y = 0
0 = C1 + 0
 C1 = -a
The slope at any section is given by


dy
P
P
P
P
 C1
sin  x
 C2
cos  x

 EI 
 EI 
dx
EI
EI




At B the slope is zero,
60
Fig.
At

dy
0
dx
P
0= C2
EI
C2 = 0
x = 0,
At A the deflection is a
AAt


x = l, y = a
 P
a = -a cos  l
  a
 EI 
 P
cos  l
0
 EI 


Considering the least practical value,
l
P 

EI 2
 P=
2EI
4l2
3. Derive the expression for crippling rod when both ends of the column are fixed.
Case 1 : when both ends of the column are fixed.
The following fig. shows a column AB of the length l whose ends A and are
both fixed. Obviously there will be a restraint moment say M0 at each end.
Let P be the crippling load.
Consider any section X distance x from the lower end B. The bending moment
at the section X, is given by.
61
El
d2 y
 M0  Py
dx 2
 El
d2 y
 Py  M0
dx 2

fig.
M
d2 y P
 y 0
2
dx
El
EI
The solution to the above differential equation is


P
P  M0
y  C1 cos  x
 C2 sin  x


 EI 
 EI  P




 (1)
Where C1 and C2 are constants of integration. The slope at any section is given by


dy
P
P
P
P
 C1
sin  x
 C2
cos  x

 EI 
 EI 
dx
EI
EI




At B, the deflection is zero.
 At

x = 0, y = 0
C1 = 0

0  C1 
M0
P
 C1  
M0
P
At B, the slope is zero,
At

dy
0
dx
P
0= C2
 C2  0
EI
x = 0,
At A, the deflection is zero
At
x = l, y = 0

0
 P  M0
M0
cos  l

 EI  P
P


62
 (2)

 P 
M0 
1  cos  l
   0
P 
 EI  

 P
cos  l
1
 EI 



l
P
 0, 2 4,6, ... ....
EI
Considering the first practical value,
l
P
 2
EI
 P=
42EI
l2
4. Derive the expression for crippling load when one end of the column is fixed and the
other end is hinged.
Case 4 : when one end of the column is fixed and other end is pinned or hinged.
The following fig. shows a column AB of the length l whose upper end A is hinged while its
lower end B is fixed.
Let P be the crippling load. Studying the nature of bending we realize that there
will be a restraint moment Mb at the lower fixed end.
The existence of the restraint moment therefore justifies the need for
horizontal force also at the top end a without
which no bending moment can occur at B.
hence the hinge at A must exert a horizontal force H at A.
Consider any section X at a distance x from the lower fixed end B.
The bending moment at the section is given by,
63
El
 El

d2 y
 Py  H(l  x)
dx 2
d2 y
 Py  H(l  x)
dx 2
fig.
d2 y P
H
 y  (l  x)
2
dx
El
EI
The solution to the above differential equation is


P
P H
y  C1 cos  x
 C2 sin  x
 (l  x)

 EI 
 EI  P




 (1)
Where C1 and C2 are constants of integration. The slope at any section is given by,


dy
P
P
P
P H
 C1
sin  x
 C2
cos  x


 EI 
 EI  P
dx
EI
EI




 (2)
At B, the deflection is zero.
 At

x = 0, y = 0
H
H
0 = C1 + l  C1   l
P
P
At B, the slope is zero,
At

dy
0
dx
P H
0= C2
EI P
x = 0,
 C2 
H EI
P P
At A, the deflection is zero
At
x = l, y = 0

0
Simplifying we get,
 P  P
tan l 
 l
 EI   EI 

 

 P   H EI   P 
H
lcos  l

sin l
 EI   P P   EI 
P

 
 

The solution to this equation is
64
l

P
 4.5 radians
EI
l2P

=(4.5)2  20.25
EI
20.25 EI
P
l2
Approximately 20.25 = 22

P
22 EI
l2
5. Calculate the safe compressive load on a hollow cast iron column (one end rigidly
fixed and the other hinged) of 150 mm external diameter and 100 mm internal diameter
and 10m in length. Use Euler’s formula with a factor of safety of 5 and E = 95 K N/mm2.
Solution :
D = 150 mm d = 100 mm
l
10
L

 5 2m  5 2  1000 mm
2
2
=5000
2 mm
I=

(150 4  100 4 )  1.994  107 mm4
64
P
safe load=
2EI 2  95  1.994  107

 373.92 kN
L2
(5000 2)2
373.92
 74.78 kN
5
6. A hollow alloy tube 5 metre long with external and internal diameters equal to 40 mm
and 25 mm respectively was found to extend by 6.4 mm under a tensile load of 60 kN.
Find the buckling load for the tube, when used as a column with both ends pinned. Also
find the safe compressive load for the tube, with a factor of safety of 4.
Solution :
Area of the section = A 

(402  252 )mm2  765.76 mm2
4
When subjected to tension
Extension  
Wl
AE
65
WL 60  103  5  103

 61213.6 N/mm2
A
765.76  6.4

I
(40 4  25 4 )  106489 mm 4
When used as a column
64

E
P
Buckling load
2Ei 2  61213.6  106489

 2573.43 N
L2
50002
Safe load, with a factor of safety of 4

5573.43
 643.36 N
4
7. A bar of length 4 meters when used as a simply supported beam and subjected to a
uniformly distributed load of 30 kN/m. Over the whole span deflects 1.5 mm at the
centre. Determine the crippling loads when it is used as a column with the following
conditions:
i)
Both ends pinjointed
ii)
One end fixed and the other hinged,
iii)
Both ends fixed
Solution :
Analysis as a beam
w = 30 kN/m = 30 N/mm
l = 4000 mm,  = 15 mm.
5 wl4

384 EI
5 wl4
5
30  (4000)4 2
EI 
.


  1013 Nmm2
384 
384
15
3

Analysis as a column
i)
When both ends are hinged
Crippling load
ii)
2EI
2
1
 2   1013 
= 412335 N=4112.335kN
2
l
3
(4000)2
When one end is fixed and the other end is hinged.
Crippling load
iii)
P
P
22EI
 2  4112.335 =8224.67 kN
l2
When both ends are fixed
66
Crippling load
P
42EI
 4  4112.335 =16449.34 kN
l2
8. Determine the ratio of the buckling strengths of two circular columns – one hollow
and the other solid. Both the columns are made of the same material and have the same
length, cross-sectional area and end conditions. The internal diameter of the hollow
column is half of its external diameter.
Solution :
D1 = External diameter of the hollow column
D2 = 0.5 d1 = internal diameter of the hollow column
D = Diameter of the solid column
Since both the columns have the same area
 2
D2
2
(D1  0.25D1 ) 
4
4
D12 4

D2 3
2EIh
L2
2E Is
Buckling load for the solid column = Ps 
L2
1
D14  D14
4
4
Ph Ih D1  D2
16
 

Ps Is
D2
D4
Buckling load for the hollow column= Ph=
2
15 D 4 15  4 
5
 . 14  .   
16 D
16  3 
3
9. Derive the expression for load when column subjected to eccentric loading.
Rankine’s Method: Consider a short column subjected to an eccentric load P. let e be the
eccentricity from the geometric axis. Let A be the sectional area of the member.
P P.e
P P.e
 Maximum compressive stress =Pmax  
yc  
yc
A
I
A AK 2
67
=

P=
P  ey c 
1 2 
A 
k 
Pmax A
ey
1  2c
k
Let f be the safe stress for the column material
Safe load for the column at the eccentricity e is given by
fA
P
 eyc 
1 k 2 


If the effect of buckling be also included, the safe eccentric load
fA
P
L2 
 ey c  
1

1





k 2  
k2 

Where l = effective length of the column
ii) Euler’s Method:
Consider a column AB of length l subjected to an eccentric load O at eccentricity e.
let the top of the column be free and the bottom of the column be fixed. Let y be the
deflection at any section X distant x from the fixed end B. let a be the deflection at A.
The bending moment at the section X is given by
EI
d2 y
 P(a  e  y)
dx 2
d2 y P
P(a  e)
 y
2
dx
EI
EI
fig.
The solution to the above differential equation is given by,
y  C1 cos x
P
P
 C2 sin x
 (a  e) ..(i)
EI
EI
The slope ay any section is given by
68
dy
P
P
P
P
 C1
sin x
 C2
cos x
dx
EI
EI
EI
EI
At B,
x = 0 and y = 0, and

0 = C1 + (a+e)
and
0 = C2
At A,
x = l and y = a

a = - (a+e) cos l


dy
0
dx
P
EI
P
 (a  e)
EI

P
a = (a+e) 1  cosl

EI 

P
(a+e) cos l
e
EI
a + e = e sec l
P
EI
The maximum bending moment for the column occurs at B and is equal to P (a+e)

Max.B.M. = M = Pe sec l
P
EI
Hence the maximum compressive stress for the column section at B
P
 
A
Pe sec l
P
EI
Z
If both ends of the column had been hinged, it can be shown that the maximum bending
moment
L P
 M  Pe sec
2 EI
10. A column of circular section made of cast iron 200 mm external diameter and 20
mm thick is used as a column 4 metres long. Both ends of the column are fixed. the
column carries a load of 150 kN at an eccentricity of 25 mm from the axis of the column.
Find the extreme stresses on the column section.
Find also the maximum eccentricity in order there may be no tension anywhere on the
section. Take E = 94000 N/mm2.
69
Solution :
Area of the column = A 

(2002  1602 )  11310 mm2
64
Moment of inertia of the section about a diameter

(20 4  16 4 )  4.637 10 7 mm 7
= I
64
 Section modulus
4.637  107
Z
 4.637  105 mm3
100
Effective length of the column  L 
l 4
  2 metres = 2000 mm
2 2
Maximum bending moment M  Pe sec
Let us determine the angle
L P
2 EI
L P
2 EI
L P
150  1000
 1000
 0.1855 radian
2 EI
94000  4.637  107
=1063=1038' say 1040'
sec 1040'=1.017
Maximum bending moment M = 150  1000  25  1.017 = 3813750 N mm
 Maximum compressive stress = Pmax 
=
P M

A Z
150  1000
3813750

 13.26  8.22
11310
4.637  105
= 21.48 N/mm2
If tension is just to be avoided corresponding to the maximum eccentricity.
70
P M

A Z
P

A
Pe sec
L P
2 EI
Z
150  1000 150  e  1.1017

11310
4.637  105
e
4.637  105
 40.3 mm
11310  1.017
11. Derive the expression for maximum compressive stress if column with initial
curvature.
The following fig. shows a column AB of length l with both its ends pinned. The column
has an initial curvature having a central deflection a’.
Let at a distance x from the end B the initial deflection be y’. For purpose of analysis
let us assume a sine curve for the initial deflection be y’. For purpose of analysis let us
assume a sine curve for the initial profile of the centre line of the column,
So that,
y'=a' sin
x
l

dy' a'
x

cos
dx
l
l

d2 y '
2a'
x


sin
2
2
dx
l
l
...(i)
When the loading on the column reaches the critical value P, the column will deflect
to the form ACB, so that the deflection at x changes from y’ to y. this happens due to the
bending moment Py.
d2 (y  y ')
 Py
dx 2

Ei

d2 (y  y ')
P
 y
2
dx
EI

d2 y P
d2 y '
2a'
x

y



sin
2
2
dx
EI
dx
l
l
..(ii)
71
Let the solution to the above differential equations be given by
y  Ca' sin
x
l
Where c is a constant of integration.

dy
x
x
 Ca' cos
dx
l
l
and
d2 y
2
x
 Ca' 2 sin
2
dx
l
l
Substituting the expressions for y and
 Ca'
d2 y
in equation (ii) we have,
dx 2
2
x P
x
 2a '
x
sin

Ca'
sin


sin
2
2
l
l EI
l
l
l
 2 P  2
 C 2    2
EI  l
l
C=
2
l2

P

2
l
EI
2

1
1

2
P
Pl
1
1 2
P0
 EI
Hence the equation to the deflected form of the column is given by
y
Pe

a'sin
Pe  P
l
...(iii)
The deflection will be a maximum at the mid-section C.
Let a be the central deflection
l

At
x= , y=a
2
a=
Pe
a'
Pe  P
Maximum B.M.=M=B.M. at the mid section =Pa
=
PPe
a'
Pe  P
72
Maximum compressive stress =Pmax  P0  Pb 
=
=
P M P My e
  
A Z A AK 2
PPe
y
P

a' c 2
A Pe  P AK

Pe
a' y 
Pe a' y e 
P
 2 c   P0 1 
. 2 
1 
A  Pe  P
K 
 Pe  P k 

Pe a' y c 
= P0 1 
. 2 
 p0  P K 
or rearranging,
 Pmax   P0  a' yc
 1 1    2

K
 P0
  Pe 
73
UNIT – IV
Maximum Stress theory
Maximum strain theory
Maximum shear stress theory
Distortion Theory
Maximum strain energy theory
Simple problems of shaft under combined loading.
74
Maximum stress theory:
This theory states that the failure of a material occurs when the maximum principal
(tensile) stress reaches the elastic limit stress  of the material in simple tension or the
maximum principal stress (that is, the maximum principal compressive stress) reaches the
elastic limit stress  in simple compression.
Maximum strain theory:
This theory states that the failure of a material occurs when the maximum principal
tensile strain in the material reaches the strain at the elastic limit in simple tension or when
the minimum principal strain (i.e. maximum principal compressive strain) reaches the elastic
limit strain in simple compression.
Maximum shear stress theory.
This theory states that the failure of a material occurs when the maximum shear stress
in the material reaches the maximum shear stress at the elastic limit in simple tension.
Distortion energy theory.
According to this theory, the elastic failure of a material occurs when the distortion
energy of the material reaches the distortion energy at the elastic limit in simple tension.
Maximum strain energy theory.
This theory states that the failure of a material occurs when the maximum strain
energy in the material reaches the maximum energy of the material t the elastic limit in
simple tension.
Maximum principal stress theory.
This theory states that the failure of a material occurs when the maximum principal
(tensile) stress reaches the elastic limit stress  of the material in simple tension or the
maximum principal stress (that is, the maximum principal compressive stress) reaches the
elastic limit stress  in simple compression.
In this theory, the maximum or the minimum principal stress is taken as the criterion
of failure. Minimum principal stress is actually the maximum compressive principal stress.
Mathematically, the failure condition is
1  
|3 ’
|3| means the numerical values of 3
(only if 1 is tensile)
(only if 3 is compressive)
75
To prevent failure, 1 < 
|3| < 
At the point of failure, 1 = 
|3| = 
(1)
 (2)
The above theory is in good agreement with experimental results of brittle materials such as
C.I
The above theory I contradicted in the following cases.
i)
In simple, tension test on mid steel sliding occurs approximately 45 to the axis of
the specimen, showing that the failure in this case is due to maximum shear stress
rather than the direct tensile stress.
ii)
A material even though weak in simple compression has been found to sustain
hydrostatic pressure far in excess of the elastic limit in simple compression.
In design problems  and ’ in equations (1) and (2) are replaced by the safe values of
stresses obtained by diving the elastic limit stress (or the limit of proportionally or yield point
stress or the ultimate stress) by a factor of safely. Let these safe stresses be t and c(1 =
safe tensile stress and c = safe compressive stress).
and
 For design purposes : 1 = t
|3|= c
Principal stresses in a M.S. body are +40 MN/m2 and -100 MN/m2, the third principal
stress being zero. Find the factor of safety based on the elastic limit if the criterion of
failure for the material is the maximum principal stress theory. Take the elastic limit
stress in simple tension as well as in simple compression to be equal to 210 MN/m2.
Solution : here 1 = 40 MN/m2, 2 = 0 and 3 = -100 MN/m2
Now
1 = t
Or
1 =

f.o.s
 210
 5.25
f.o.s= 
1
40

Also
|3 | c
or
|3 |
'
f.o.s

f.o.s=
210
 2.1
100
So the material, according to the maximum principal stress theory, will fail due to
compressive principal stress.
76
f.o.s = 2.1
Principal stresses in a c.I. body are + 35 MN/m3 and – 80 MN/m2, the third principal
stress being zero. Find the f.o.s based on the elastic limit if the criterion of failure is the
maximum principal stress theory. Take the elastic limit stress in simple tension as 70
MN/m2 and in simple compression as 500 MN/m2.
Solution :
Here 1 = 35 MN/m2, 2 = and 3 = -80 MN/m2
Now
1 = t
Or
1 =

f.o.s
 70
2
f.o.s= 
1 35

Also
|3 | c 

f.o.s=
'
f.o.s
'
500

 6.25
|3 | 80
 The material will fail due to tensile principal stress.
 f.o.s = 2
maximum principal strain theory:
This theory states that the failure of a material occurs when the principal tensile strain
in the material reaches the strain at the elastic limit in simple tension or when the minimum
principal strain (i.e. maximum principal compressive strain) reaches the elastic limit strain in
simple compression.
Now
1 = principal strain in the direction of principal stress1
1
=  1  (2  3 )
E
3 = Principal strain in the direction of the principal stress 3
1
3  (1  2 )
E
 According to the maximum principal strain theory, the conditions to cause failure are
77
1


E
and | 3 |
[ 1 must be positive]
'
E
[ 3 must be negative]
or
1

[3  (2  3 )] 
E
E
and
1
'
|  3  (1  2 ) |
E
E
or
[1  (2  3 )  
and
[3  (1  2 )   '
To prevent failure
[1  (2  3 )  
and
[3  (1  2 )   '
At the point of elastic failure
and
[1  (2  3 )  
 (1)
[3  (1  2 )   '
 (2)
In design ’ in equation (1) and (2) are replaced by safe stresses which can be designated by
t, and c respectively.
 For design purposes
[1  (2  3 )  
[3  (1  2 )  c
The above theory is not used much in practice as it does not fit well with the
experimental results except for brittle materials for biaxial tension-compression state of stress
for which it is sometimes recommended.
Maximum shear stress theory:
This theory states that the failure of a material occurs when the maximum shear stress in the
material reaches the maximum shear stress
at the elastic limit in simple tension.
78
Maximum shear stress =
1  3
2
Maximum shear stress at the elastic limit in simple tension =

1  3 
 to prevent failure
2
2
or
1  3   to prevent failure

2
At the point of failure
1 -3 = 
(1)
In actual design  in the above equation is replaced by the safe stress t.
This theory gives good arrangement with experimental results on ducts materials for
which  = ’ approximately. This theory is quite simple to apply as compared to the other
theories (next two theories applicable to ductile materials and gives quite safe results. This
theory is very commonly used in machine design for ductile materials.
The above theory is contradicted in the following cases.
i.
The theory does not give agreement with experiments for materials for which  is
quite different from ’. So it cannot be applied to brittle material such as C.I.
ii.
Under hydrostatic pressure or equal tensions in these direction, the maximum
shear stress is always equal to zero, meaning thereby that material will never fail
under hydrostatic pressure or equal tensions. This is physically impossible. Under
equal tensions in three principal directions, a brittle failure is expected and hence
the maximum principal stress theory must be more relevant in this case.
iii.
The theory does not give as close results as found by experiments on ductile
materials, However, it gives the safe results.
A M.S shaft 10 cm diameter is subjected to a maximum torque of 15 kNm and a
maximum bending moment of 10 kNm at a particular section. Find the f.o.s according
to the maximum shear stress theory if the elastic limit in simple tension is 240 MN/m2.
Solution :
79
b  maximum bending stress =
  maximum shear stress
=
M 10  1000 10 6  320


N / m2
3
Z

  10 


32  100 
T
15  1000 240  10 6


3
 3

  10 
d


16
16  100 
Principal stresses are given by
2


 106  320 
1
1 106  320
 240
2
6 
  b   b  4  
 
 10 
  4
2
2


 







2
106  320 
480   160  106

1  1  

(1  1.805)
 
2


 320  


 148  106 or -41 10 6 N/m2
According to the maximum shear stress theory,
1-3 = 
Here 1  148  106 , 2  0 and 3  41 106 N/m2


(148+41)  106  
=189  10 6 N/m2

240  106
f.o.s=
 1.27 Ans.
189  106
Maximum strain energy theory:
This theory states that the failure of a material occurs when the maximum strain energy in the
material reaches the maximum energy of the material at the elastic limit in simple tension.
Mathematically, at the point of failure
V 2
2
1  22  32  2(12  23  31 ) 
V, giving
2E
2E
12  22  32  2(12  23  31
 (1)
In actual design  in the above equation is replaced by the allowable stress obtained by
dividing  (or the ultimate stress or yield point stress) by a f.o.s.
The results of this theory are similar to the experimental results for ductile materials (i.e. the
materials which fail by general yielding)
for which  =  ’ approximately.
80
Also note that order of 1, 2 and 3 is immaterial here.
i. The theory does not apply to materials for which  is quite different from ’.

ii. Under hydrostatic pressure 1 = -p, 2 –p and 3 = -p, giving p 
when
3(1  2)
1, 2 and 3 are put in equation (1). It has been found that this result does not
agree with experiments. Actually much higher pressures can be applied than those

given by the equation p 
.
3(1  2)
iii. It does not give results exactly equal to the experimental results even for ductile
materials, even through the results are quite close to the experimental.
Distortion energy theory.
According to this theory, the elastic failure of a material occurs when the distortion energy of
the material reaches the distortion energy at the elastic limit in simple tension.

V
(1  2 )2  (2  3 )2  (3  1 )2 
12G
and shear strain energy due to the elastic limit stress 

V
V
(1  0)2  (0  0)2  (0  )2  
 2 2
12G
12G

1  ,  2  0 and 3  0 
Equating the two energies, we get
(1  2 )2  (2  3 )2  (3  1 )2
2
 (1)
or 2  12  22  32  (12  23  31 )
 (2)
2 
In actual design  in equations (1) and (2) is replaced by safe equivalent stress t, in simple
tension.
The above theory has been to give best results for ductile for ductile materials for which =’
approximately.
Note that order of 1, 2 and 3 is immaterial for this theory.
i.
The theory does not agree with the experimental results for the material for which
 is quite different from ’.
ii.
For hydrostatic pressure or tension, the above theory gives  = 0. This means that
the material will never fail under any hydrostatic pressure or tension and this is
obviously not correct. Actually
when three equal tensions are
81
applied in three principal directions, brittle fracture occurs and as such maximum
principal stress theory will give reliable results in this case.
A solid shaft transmits 1000 kW at 300 r.p.m maximum torque is 2 times the mean. The
shaft is subjected to a bending moment, which is 1.5 times the men torque. The shaft is
of ductile material for which the permissible tensile and shear stresses are 120 MPa and
60 MPa respectively. Determine the shaft diameter using suitable theory of failure. Give
justification of the theory used. Find result by use of max. principal stress theory also.
(AMIE Sec B, Winter 94).
Solution : It is problem of combined bending and torque.
Principal Stresses :
b 
32M
16T
, = 3
3
d
d
1
1  32M 32
 16
Principal Streses = b  b2  42    3  3 M2  T 2   3 (M  M2  T 2 )
 2  d
2
d
 d
1 


16
16
M  M2  T 2 , 2  0 and 3  3 (M  M2  T 2 )
3
d
d
Best theory is the distortion energy theory, which gives,
12  32  13  2t
2

 
or
 16 
2
2
 d3  M  M  T


or
 16 
2
2
2
2
2
 d3  2M  2M  2T  T   t


or
i.e,
Now
2

2

 16 
2
2
 d3  4M  3T


32
3
1  3 M2  T 2
d
4
2
 M  M2  T 2


2
 2t
 (1)
2NTmean
 1000  103
60
82

2
2
 16 
2
2
2
 d3   M  M  T




2
2
 16 
  3   2t
 d 
2  300  Tmean
 1000  103
60
or
100  103 100

kNm


i.e
Tmean 

T-Max.torque=
and
M=Max.B.M.=
200
kNm

150
kNm

Putting in equation (1) we get,
2
120  106 
d3 
2
32  150   200 

 103



3
d      
32  103
3
32
3
(150)2  (200)2 
1502  (200)2
6 2
2
3
120  10 
4
120  10
4
3
3
or d  6.191 10
 d=0.1836 m= 183.6 mm Ans.
Justification for the theory used: The theories generally used for ductile materials are (i)
maximum shear stress theory and (ii) distortion energy theory. Out of these theories,
distortion energy theory is best for ductile materials, as the experimental results for these
materials fit very well with this theory. Maximum shear stress theory gives results on safer
side, that is a little more materials is used than the actual required as given by distortion
energy theory.
Maximum shear stress theory results:
M2  T 2 
 3
d t
32
2
3
or
10
or
d3 
This gives
2
 3
 200   150 
6
        32 d  120  10

 

32  (200 / )2  (150 / )2
 103
6
  120  10
d3=6.755  10-3
83
and
d = 0.189 m = 189 mm Ans.
Since t = 2 (given), the equivalent torque criterion gives same results.
Maximum principal stress theory :

i.e
or

1
 3
M  M2  T 2 
d t
2
32
 3
M+ M2  T 2 
d t
16
2
2
150
150   200   3  3

6

 
      10  16 d  120  10

 


 



or
2d3
150+ (150)  (200)  3
 120  10 6
10  16
or
150+ 1502  2002 
or
d3  54.0379  10 4
2
2
2  120  103 d3
16

d=0.1755 m = 175.5 mm Ans.
15. What are the types of failures?
1. Brittle failure:
Failure of a material represents direct separation of particles from each other,
accompanied by considerable deformation.
2. Ductile failure:
Slipping of particles accompanied, by considerable plastic deformations.
Theories of failure:
1. Maximum Principal Stress Theory. (Rakine’s theory)
2. Maximum Principal Strain Theory. (St. Venant’s theory)
3. Maximum Shear Stress Theory. (Tresca’s theory or Guest’s theory)
4. Maximum Shear Strain Theory. (Von-Mises-Hencky theory or Distortion energy
theory)
5. Maximum Strain Energy Theory. (Beltrami Theory or Haigh’s theory)
Maximum Principal Stress Theory (Rakine’s theory)
According to this theory, the failure of the material is assumed to take place when the
value of the maximum Principal Stress (σ
1 = f y ).
84
Maximum Principal Strain Theory. (St. Venant’s Theory)
According to this theory, the failure of the material is assumed to take place when the
value of the maximum Principal Stain (e1) reaches a value to that of the elastic limit strain (fy
/ E) of the material.
e 1 = fy / E
e1  1/ E[ 1  (1/ m)( 2   3 )]  f y / E  [ 1  (1/ m)( 2   3 )]  f y
In 3D
In 2D, σ 3 = 0  e1  1/ E[ 1  (1/ m)( 2 )]  f y / E  [ 1  (1/ m)( 2 )] = fy
Maximum Shear Stress Theory. (Tresca’s theory)
According to this theory, the failure of the material is assumed to take place when the
maximum shear stress equal determined from the simples tensile test.
In 3D, ( 1   3 ) / 2  f y / 2  ( 1   3 )  f y
In 2D, ( 1   2 ) / 2  f y / 2  ( 1 )  f y
Maximum Shear Stress Theory. (Von-Mises-Hencky theory or Distortion energy
theory)
According to this theory, the failure of the material is assumed to take place when the
maximum shear strain exceeds the shear strain determined from the simple tensile test.
In 3D, shear strain energy
[( 1   2 )2  ( 2   3 )2  ( 3   1 )2 ]
due
to
distortion
U
=
(1/
12G)
Shear strain energy due to simple tension, U = f y2 / 6G
(1/ 12G) [( 1   2 )2  ( 2   3 )2  ( 3   1 )2 ] = f y2 / 6G
[( 1   2 )2  ( 2   3 )2  ( 3   1 )2 ] = 2 f y2
[( 1   2 )2  ( 2  0)2  (0   1 )2 ] = 2 f y2
In 2D,
Maximum Strain Energy Theory (Beltrami Theory)
According to this theory, the failure of the material is assumed to take place when the
maximum strain energy exceeds the strain energy determined from the simple tensile test.
In
3D,
strain
energy
due
2
2
2
[  1   2   3  (1/ m)( 1 2   2 2   2 2 ) ]
Strain
energy due
to
simple
85
to
deformation
tension, U = f y2 / 2 E
U
=
(1/2E)
(1/2E) [  12   22   32  (2 / m)( 1 2   2 2   2 2 ) ]= f y2 / 2 E
[  12   22   32  (2 / m)( 1 2   2 2   2 2 ) ]= f y2
[  12   22  (2 / m)( 1 2 )]  fy 2
Theories used for ductile failures
In 2D,
1. Maximum Principal Strain Theory. (St. Venant’s theory)
2. Maximum Shear Stress Theory. (Tresca’s theory or Guest’s theory)
3. Maximum Shear Strain Theory. (Von-Mises-Hencky theory or Distortion energy
theory)
Limitations of Maximum Principal Stress Theory (Rakine’s theory)
1. This theory disregards the effect of other principal stresses and effect of shearing
stresses on other planes through the element.
2. Material in tension test piece slips along 45 to the axis of the test piece, where normal
stress is neither maximum nor minimum, but the shear stress is maximum.
3. Failure is not a brittle, but it is a cleavage failure.
Limitations of Maximum Shear Stress Theory. (Tresca’s theory).
This theory does not give the accurate results for the state of stress of pure shear in
which the maximum amount of shear is developed (in torsion test).
Limitations of Maximum Shear Strain Theory.
Distortion energy theory)
(Von-Mises-Hencky theory or
It cannot be applied for the materials under hydrostatic pressure.
Limitations of Maximum Strain Energy Theory. (Beltrami Theory).
The theory does not apply to brittle materials for which elastic limit in tension and in
compression are quite different.
Failure theories and its relationship between tension and shear.
1. Maximum Principal Stress Theory. (Rakine’s theory) y = fy
2. Maximum Principal Strain Theory. (St. Venant’s theory) y = 0.8 fy
3. Maximum Shear Stress Theory. (Tresca’s theory or Guest’s theory) y =0.5 fy
4. Maximum Shear Strain Theory. (Von-Mises-Hencky theory or Distortion energy
theory)y = 0.577 fy
5. Maximum Strain Energy Theory. (Beltrami Theory or Haigh’s theory) y = 0.817 fy
Octahedral Stresses:
A plane, which is equally inclined
86
to the three axes of reference, is called
octahedral plane. The normal and shearing stress acting on this plane are called octahedral
stresses.
 oct  1/ 3  ( 1   2 )2  ( 2   3 )2  ( 3   1 )2
Plasticity ellipse:
The graphical surface of a Maximum Shear Strain Theory (Von-Mises-Hencky theory
or Distortion energy theory) is a straight circular cylinder. The equation in 2D is
 12   1 2   22  fy 2 which is called the Plasticity ellipse
UNIT V
Equilibrium and Compatibility conditions for elastic solids.
2D elasticity equations for plane stress,
plane strain and generalized plane strain
87
cases
Airy’s stress function.
Simple problems in plane stress / plane strain using Cartesian and polar coordinates.
Super position techniques.
Examples include (a) panels subjected to a Generalized plane strain Biaxial loading
(b) Uniform/Linearly varying edge loads on elastic half plane
(c) Thick cylindrical shells.
1. Write the differential equations of equilibrium for two-dimensional problems?
88
2. Write the condition of compatibility for elastic solids?
3. Define plane stress and plane strain.
4. Write
Airy’s stress function.
89
5. Write the components of stress and
strain.
90
6. Write the strain components by the method of superposition.
1. Explain the following (i) Notation for
forces and stresses (ii) Components of
91
stress (iii) Components of strain.
92
93
94
2. Explain the following (i) Plane stress and plane strain (ii) compatibility equations.
95
3. Explain (i) St.Venant’s Principle (ii) Airy’s stress function.
4.
96
Explain (i) Strain Rosette (ii) Compatibility equations.
97
AIRCRAFT STRUCTURES – I
PART – A
1. Give the relation between the number of members and the number of joints in a truss and
explain its significance.
2. A solid cylinder 100 cm long and 5 cm in diameter is subjected to a tensile force of 80
kN. One part this cylinder of length L1 is made of steel (E = 210 GPa) and the other part
of length L2 is made of aluminium (E = 70 GPa). Determine the length L1 and L2 so that
the two parts elongate to an equal amount.
3. Write down three moment equation in the general form.
4. Define and the S.I. units for (a) Stiffness (b) Flexural rigidity
5. A cantilever beam of length L is subjected to a tip load P, find the deflection at the tip
using Castigliano’s theorem.
6. State Reciprocal theorem. Give an example.
7. Find the slope at the support of a simply supported beam of length L and subjected to a
uniformly distributed load by unit load method.
8. Draw Euler’s curve for a column and explain critical slenderness ratio.
9. Give the Rankine’s formula and its significance.
10. A solid cube of steel (G = 80 GPa) is subjected to a shear of 56 MPa. Find the strain
energy per unit volume.
PART – B
11.
(a)
method.
Find the forces in the members of the truss shown in the figure by any one
Figure.
Or
(b)
Find the forces in the members of the landing gear tripod shown in the figure
98
12.
(a)
Find the support moments and draw bending moment diagram of the continuous
beam shown in Fig using three moment equation.
Or
(b) Find the support moments and draw bending moment diagram of the continuous
beam shown in the figure using moment distribution method.
13.
(a)
Find the horizontal reaction of the frame shown in the figure using strain energy
method.
(b) Determine the forces in the system shown in the figure, assuming the cross
section area of all bars equal and taking the force X in the diagonal AD as the
statically indeterminate quantity.
99
14.
(a)
The cross section of a hinged-hinged column 1 m long is Z with 3 cm x 2 mm
top and bottom flange and 5 cm x 2 mm middle web. Derive the formula used.
E = 70 GPa.
Or
(b) A beam-column made of steel simply supported at both ends is subjected to a
concentrated load of 1000 N at a distance 1 m from the right support and an
axial load of 1000 N. Find the deflection at mid-point and the maximum
deflection. Given: L = 4 m, b = 20 mm, E = 210 GPa, calculate the load the
column can carry. Derive the formula used.
15.
(a)
A simply supported beam column of length L is subjected to a axial load P and a
moment at M at one of the support. Find the slope at the supports.
Or
(b) A circular shaft of tensile yield strength 300 MPa is subjected to a combined
state of loading defined by a bending moment M = 15 kN-m and torque T = 20
kN-m. Calculate the diameter d which the bar must have in order to achieve a
factor of safety N = 3. Apply the following theories.
(i) Maximum shear stress theory
(ii) Maximum distortion energy theory
(iii)Octahedral shear stress theory.
100
AIRCRAFT STRUCTURES – I
PART – A
1. Where are the truss-type structures found in an aircraft?
2. Define the ‘carry-over’ factor used in the moment distribution method.
3. Transform the cross-section show in figure to a section made of aluminium alone. What
is the criterion used?
4. Write down the moment-curvature relationship for the section shown in figure.
5. Compute the strain energy stored in the bar indicated in the figure.
6. State Maxwell’s Reciprocal Theorem.
7. What is an ideal column?
8. The load carrying capacity of a column (increases/decreases/remains unchanged) because
of eccentricity?
9. State 2 failure theories applicable to brittle material.
10. What is the failure criterion according to the maximum shear stress theory?
PART – B
11.
(a) Determine the forces in the members of the truss indicated in figure.
101
Or
(b) Look at the 3-element truss configuration shown in figure E = 70 GPa for the
material used. The cross-section area of each member is 5 cm2. The length of
member (2) is 99.75 cm- this member is stretched so that a pin can be inserted at
B to connect the 3 members together. Determine the forces induced in the
members. Determine also the vertical and horizontal displacement of point B.
State and prove Clapeyron’s 3 – moment equation.
Or
(b) Determine all support reactions for the beam shown in figure Use Clapeyron’s
3-moment equation. Then, draw the shear force and bending moment diagrams.
EI is constant.
12.
(a)
13.
(a)
(b)
(i) State the prove Castigliano’s theorems.
(ii) Using energy methods, determines the slope at point B of the beam
shown in figure E = 210 GPa, I = 10- 4 m4.
Or
Determine all the support reactions of the beam shown in figure using energy
methods. Then draw the shear force and bending moment diagrams. E = 210
GPa, I = 10- 4 m4.
102
14.
(a)
(b)
15.
(a)
(i)
Derive and obtain the first 2 buckled shape and corresponding buckling
loads of a fixed-fixed column.
(ii) Explain Rankine’s hypothesis.
Or
Write notes on the following topics
(i) Inelastic column buckling.
(ii) The Southwell plot.
What is a beam-column? Where can a beam-column type of structure be
found in an aircraft? Explain the structural analysis of a beam-column type of
structure, with an example.
Or
(b)
(i) Explain the maximum distortion energy failure theory. Refer figure
Point A is a critical point located on the top surface of the lever arm.
Determine the maximum load P0 according to the maximum distortion energy
failure theory using a factor of safety of 1.5. The shaft is made of steel with a
yield stress value of 300 MPa.
(ii) Explain the maximum principal stress theory of failure.
AIRCRAFT STRUCTURES – I
PART – A
1. Differentiate between statically determinate and indeterminate trusses with examples.
2. What is equivalent rigidity of a
composite beam? Explain with an example.
103
3. Explain unit load method with an example.
4. Define carry-over factor in moment distribution method.
5. State Castigliano’s theorems.
6. State Reciprocal theorem.
7. Calculate the strain energy stored in a cantilever of length L, subjected to a tip load P.
8. Draw Euler’s curve for a column and explain critical slenderness ratio.
9. What is South well’s plot?
10. A solid cube of steel (G = 80 GPa) is subjected to a shear of 56 MPa. Find the strain
energy in per unit volume.
PART – B
11.
(a)
(b)
12.
(a)
Find the forces in the members of the truss shown in the figure by any one
method.
Or
The truss shown in figure is supported as cantilever at the joints A and H.
Find the forces in the members.
Find the support moments and draw bending moment diagram of the
continuous beam shown in
the figure and using three moment equation.
104
(b)
13.
(a)
(b)
14.
(a)
(b)
Or
Find the support moments and draw bending moment diagram of the
continuous beam shown in the figure and using moment distribution method.
A thin circular ring of radius R and bending rigidity EI is subjected to three
symmetric radial compressive loads lying in the plane of the ring structure.
Obtain the expression for the bending moment and plot its distribution.
Or
Calculate the vertical deflection of the point B and the horizontal movement of
D in the pin-jointed frame work shown in the figure. All members of the
frame work and linearly elastic and have cross sectional areas of 1800 mm2.
E = 200 GPa.
Find the critical load and stress for the column made of Steel, (E = 210 GPa)
shown in the figure, assuming both ends are pinned.
Or
A beam column made of steel simply supported at the both ends is subjected to
a concentrated load of 1000 N at a distance 1 m from the right support and an
axial load of 1000 N. Find the deflection at mid-point and the maximum
deflection. Given : L = 4 m, b = 20 mm, d = 40 nm, E = 210 GPa, calculate
the load the column can carry. Derive the formula used. b is the width of the
cross section and d is the
depth of the section.
105
15.
(a)
(b)
Explain the various theories of failure and their relative merits and demerits.
Or
A circular shaft of tensile yield strength of 350 MPa is subjected to a
combined state of loading defined by a bending moment M = 15 kN-m and
torque T = 10 kN-m. Calculate the diameter d which the bar must have in
order to achieve a factor of safety N = 2. Apply the following theories.
(i) Maximum shear stress theory.
(ii) Maximum distortion energy theory.
(iii) Octahedral shear stress theory.
AIRCRAFT STRUCTURES – I
PART - A
1.
2.
Explain the general criteria to determine whether a truss is statically determinate?
A 20 cm long steel tube 15 cm internal diameter and 1 cm thick is surrounded by a brass
tube of same length and thickness. The tubes carry an axial load of 150 kN. Estimate
the load carried by each. Es = 210 GPa, Eb = 100 GPa.
3.
Write down three moment equation in the general form.
4.
Define stiffness factor in moment distribution method.
5.
Explain unit load method with an example.
106
6.
State Reciprocal theorem.
7.
Differentiate between long and short column.
8.
What is Southwell’s plot?
9.
A solid cube of steel (E = 210 GPa) is subjected to a tension 200 MPa, find the strain
energy per unit volume.
10. Explain Octahedral shear stress theory.
PART – B
11.
(a)
(b)
12.
(a)
Find the axial loads in the members of the truss shown in the figure by any
one method.
Or
Find the axial loads in the members of the truss shown in the figure.
Find the support moments and draw bending moment diagram of the
continuous beam shown in the figure using three moment equation.
Or
107
13.
(b)
Find the support moments and draw bending moment diagram of the
continuous beam shown in the figure using moment distribution method.
(a)
In figure a vertical load P is supported by a vertical bar DB of length L and
cross-section A and by two equally inclined bars of length L and cross-section
area A1. Determine the forces in the bars and also the ration A1/A which will
make the forces in all the bars numerically equal.
(b)
14.
(a)
(b)
15.
(a)
(b)
Or
For the truss shown in the figure, determine the horizontal reaction at A and B.
(i) State and prove Castigliano’s theorem.
(ii) Find the defection at the point of load of a simply supported beam of
length L, subjected to a concentrated load P at a distance ‘a’ from the left
end and ‘b’ from the right end using unit load method.
Or
Find the buckling stress of a hinged-hinged column of length 100 cm and
having I-cross-section. The dimensions of the flange are 10 cm x 1 cm and
the web 12 cm x 1 cm. Derive the formula used. E = 70 GPa.
Derive an expression for the deflection curve of a beam-column simply
supported at both ends and subjected to a uniformly distributed load w and an
axial 1cap P.
Or
A circular shaft of tensile yield strength 300 MPa is subjected to a combined
state of loading defined by a bending moment M = 15 kN-m and Torque T =
15 kN-m. Calculate the
diameter d which the bar must have in order
108
to achieve a factor of safety N = 2. Apply the following theories.
(i) Maximum strain theory.
(ii) Maximum shear stress theory.
(iii) Maximum distortion energy theory.
AIRCRAFT STRUCTURES – I
PART – A
1. When is a structure said to be statically indeterminate?
2. What is the relationship between bending moment and shear force?
3. Sketch the bending moment diagram of a fixed-fixed beam of length L subject to a
concentrated moment Mo applied at x = L/2.
4. Define the carry-over factor.
5. Refer figure. What is the strain energy stored in the bar?
6. State Castigliano’s second theorem.
7. What is the ‘effective length’ of a column?
8. What is an ideal column?
9. Name 2 failure theories suitable for ductile materials.
10. Give example of brittle materials. Name a failure theory suitable for a brittle material.
PART – B
11.
Refer figure. Derive and obtain an expression for the mid-point slope and deflection
using moment-area theorems. Beam length is L and EI is constant.
12.
(a)
Refer figure. Beam length = 1.5 m. Cross-section is rectangular of size 2 x 0.5
cm. Material used is aluminium. Determine (i0 the reactions at the fixed end,
and (ii) the reaction and slope at the prop.
Or
109
(b)
Refer figure q = 20 kN / m. Determine all support reactions. Sketch the shear
force and bending moment diagrams. EI is constant.
Figure.
13.
14.
(a)
Refer figure. Derive and obtain an expression for the mid-point slope and
deflection using energy methods.
Or
(b)
Refer figure. Determine the axial force in the truss members. Determine also
the vertical and horizontal displacements of joint B. The material used has a
modulus of 70 GPa. The cross-section area of all the members is 3 sq. cm.
(a)
(i) Discuss the effect of initial imperfections in columns
(ii) Derive and obtain the Euler buckling load of a pinned-pinned column of
length L.
Or
(i) Explain the construction of a South well plot. What are its uses?
(ii) Describe how the buckling load of a given column can be estimated using
energy methods.
(b)
15.
(a)
(b)
(i) What are the typical materials used for aircraft construction? List the
criteria governing the selection of aircraft materials.
(ii) Explain the maximum principal stress failure theory.
(iii) Explain the maximum principal strain failure theory.
Or
(i) Explain the distortion energy failure theory.
(ii) A shaft is subjected to a maximum torque of 10 kNm and a bending
moment of 7.5 kNm at a particular section. The allowable stress in simple
tension is 160 MN/m2. Determine the shaft diameter using the distortion
energy failure and a factor of safety of 2.
110
Figure.
*****************
111