structural analysis1-76 course for phd.pdf

ADVANCED STRUCTURALANALYSIS
By:
Prof. Dr. Jahangir Bakhteri
B.S. Civil Engineering, Kabul University
M.E. & PhD Structural Engineering, IIT Roorkee, India
Professional Body Memberships & Affiliations:
F.ASCE, CMSEI, P.E.(Malaysia),
MIE & CEng(India) MISET, MSAAE,
Architect and the Engineer
An architect usually deals with a design problem in a comprehensive
manner, and stresses an overall, rather than an elemental approach to
design thinking.
An engineer thinks in the reverse manner, starting with details, and
without sufficient regards for overall picture.

ADVANCED STRUCTURALANALYSIS Dr. Jahangir
Structure
A structure can be defined as a body capable of resisting applied forces
without exceeding an acceptable limit of deformation of one part
relative to the other. Over the centuries different structural form evolved
for different types of structures. For example, a building structure has
essentially beams and columns to resist the applied loads. A bridge
structure can be constructed from beams, arches, cables and so on. All
structures are an assembly of basic elements.
Idealization of Structures
Line Elements
Fig.1(a) shows a bar which can resist a compressive or tensile force along
its length and under tensile force it will extend by ∆.
Fig.1(b) shows a beam which is supported at its ends and can resist the
loads acting normal to its axis and exhibits bending.
Fig.1(c) shows a shaft which can resist twisting moment / torque and under
the action of loads the shaft twists.

Continuum Elements Dr. Jahangir
Fig. 2(a) shows a slab or plate which resists loads applied normal to
the plate and because of its two-dimensional nature its behavior is
more complex.
Fig. 2(b) shows a wall which, like a bar resists compressive and
tensile loads applied to it but behave as two dimensional. The above
elements are called continuum elements because their length and
breath are much larger than their thicknesses.
Fig. 2(c) shows a typical structure assembled from two-dimensional
elements which can be described as a three-dimensional.
Fig. 2(d) indicates the cross-section of a dam structure which can
best to be described as three-dimensional model.

Types of Joints Dr. Jahangir
Structures are assembled by joining elements at elements intersections.
In steel structures, the joints are joined by welding or bolting.
In reinforced concrete structures, the joint is made monolithic by proper
reinforcement systems. Fig. 3 represents some typical joints connecting beams and
columns
Fig. 3 Joints at beam-column junction

Types of Supports Dr. Jahangir
Ideal types of “support joints” are classified as roller support, pinned or hinged
support, and fixed support.
(a) Roller support
Fig. 4(a) indicates schematic conventional representations of a roller support and the
force acting on it.
Conventional representation of support Nature of forces transmitted
Fig. 4(a) Schematic representation of roller support

(b) Hinged or Pinned support Dr. Jahangir
A hinged support resists both horizontal and vertical movements but allows
unrestricted rotational movement i.e. it support the forces acting in horizontal and
vertical directions as shown in Fig. 4(b).
Fig. 4(b) Schematic representation of pinned or hinged support

(c) Fixed support Dr. Jahangir
A fixed support is achieved by connecting a member to a heavy foundation which
resists horizontal, vertical and rotational movements. Therefore, at a fixed support
the forces acting are a vertical force, a horizontal force and a couple as shown in
the Fig. 4(c).
Fig. 4(c) Schematic representation of a fixed support

(d) Ball and socket support Dr. Jahangir
A ball and socket support can transmit forces in three-dimension namely Rx, Ry,
and Rz as shown in the Fig. 4(d)
Fig. 4(d) Ball and socket support

(f) Rigid support in space Dr. Jahangir
A rigid support in space resists forces and moments in x, y and z directions i.e.
Rx, Ry, Rz, Mx, My and Mz as shown in the Fig. 4(e).
Fig. 4(e) Rigid support in space

Principle of Superposition Dr. Jahangir
The main reason for assuming linear behavior of structures or linear load-deformation
relationship is that it allows the use of the principle of superposition. The principle
states that the displacements resulting from each of a member forces may be added to
obtain the displacements resulting from the sum of the forces.
As an example consider the cantilever beam shown in Fig. 5. The deflections caused
by the three separate loads are shown in Fig. 5(a). The same final deflections would
result if all the three loads are applied together as shown in Fig.5 (b)

Sign Convention in Structural Analysis Dr. Jahangir
An essential part of structural analysis is the adoption of an appropriate sign convention for
representing forces and displacements.
For a global X, Y and Z coordinates system as shown in Fig, 5 the positive direction of the forces
coincides with the direction of global coordinate axes and the moments follow the right-hand screw
rule.
(a) Axial force: An axial force is considered
positive when it produces tension in the
member. A compressive force is therefore
negative.
(b) Shear force: Shear force which tends to
shear the member as shown in Fig. (a)
is positive. The positive shear force forms
a clockwise couple on a segment
Fig. 5 Sign Convention System

Dr. Jahangir
(c) Bending Moment: The moment which produces compressive stresses in the top
fiber or tension in the bottom fiber is positive as shown in Fig.(b).
In the joint convention, the moment that tends to rotate
the joint clockwise or the member end anti-clockwise
is denoted positive as shown in Fig.(c).
(d) Twist: The twist moment is considered positive
as shown in Fig.(d). The positive twist
corresponds to the right-hand screw rule.

Equilibrium Dr. Jahangir
The fundamental requirement of equilibrium is concerned with the guarantee that a
structure, or any of its parts, will not move.
The laws governing the motion of bodies, published by Isaac Newton in 1687 are
called Newton’s Laws. The particular cases of these laws governing equilibrium, i.e.
the lack of motion are of basic importance in structural theory and the structures are
designed based on them.
The conditions of equilibrium are best established with reference to coordinate axes
X, Y and Z. Thus for static equilibrium, the algebraic sum of all the forces along
coordinate axes must be zero, or mathematically,
ΣFx = 0 ΣMx = 0
ΣFy = 0 ΣMy = 0
ΣFz = 0 ΣMz = 0

Loads and Forces in Structures Dr. Jahangir
Loads acting in structures are:
(a) Dead Loads: Dead loads include the weight of all permanent components of the
structure and any other immovable loads that are constant in magnitude and
permanently attached to the structure.
(b) Imposed Loads: Imposed loads are loads and forces that act on a structure by
character of use of the structure to the nature of use, activities due to people,
machinery installation, external forces etc. These loads are:
i) Live Loads
ii) Wind Loads
iii) Earthquake Loads
iv) Snow and Rain Loads
v) Soil and Hydrostatic Forces
vi) Erection Loads
vii) Other Forces; such as impact, vibration, temperature effects, shrinkage, creep
and forces due to the settlement of foundations.

The Essential Properties of Structural Materials Dr. Jahangir
There are various types of materials used in the construction
of structures such as: stone and masonry, wood, steel,
aluminum, reinforced and pre-stressed concrete, fibers, plastic,
etc. They all have in common certain essential properties that
suit them to withstand loads.
Whether the loads act on a structure permanently,
intermittently, or only briefly, the deformation of the structure
must not increase indefinitely, and should disappear after the
action of loads end.

Elastic Materials Dr. Jahangir
A material whose deformation vanishes with the removal of the loads is
called elastic material. Beam shown in Fig.6 has elastic behavior. Most
of the structural materials are elastic to certain extent (elastic range).
Fig. 6 Elastic behavior

Linearly Elastic Materials Dr. Jahangir
When the deformation of a material is directly proportional to the applied
loads, the material is called linear elastic as shown in Fig.7. Thus, if a
beam made from linear elastic material deflects one inch under a vertical
load of 10 kips, it will deflect 2 inches under 20 kips loads.
Fig. 7 Linearly elastic behavior

Plastic Materials Dr. Jahangir
When the total deformation in a material does not vanish by removing
the loads, the material is said to behave plastically as shown in Fig.8.
The load at which a material starts behaving in plastic manner is called
yield load or yield point.
Fig. 8 Plastic behavior

Brittle Materials Dr. Jahangir
The materials that are linearly elastic up to the failure ( Fig.9) such as
glass, some plastics and fibers, are highly unsuitable for structural
purposes. They cannot give warning of approaching failure and often
shatter under impact.
Fig. 9 Failure in brittle materials

Note: Fig. 10 indicates different ranges of the stress- strain curve for
mild steel in tension
Dr. Jahangir

Displacement – Geometric Method Dr. Jahangir
A necessary part of analysis of structure is the evaluation of deflections. Deflections
are evaluated not only to check that they do not exceed the design limitations, but
also to use them in the analysis statically indeterminate structures.
Let us consider a propped cantilever beam under the action of distributed load w as
shown in the figure. The deflected shape is shown in broken line. The deflected
shape of the beam is shown by a smooth and
continuous curve.
Considering a continuous beam with a point
load in a span, the deflected shape is shown
in the figure below. At supports A and B there is no deflections but may have slopes
i.e. rotations. At point C the fixed end, zero slope and zero deflection. At inflection
point the curvature changes sign and moment will be zero.

Dr. Jahangir
Considering a portal frame shown in the figure below subjected to symmetrical
loading w. The axial deformations are considered negligible compared to bending
deformation. Therefore, points B and C will remain in the same position. However,
connection at B and C do rotate by the same amount.

Dr. Jahangir
Considering a portal frame subjected to lateral force as shown in the figure below.
Neglecting axial deformations, the lateral displacement of point B and C must be
same.

Dr. Jahangir
Let us consider a segment of an elastic line deflected by a moment as shown in
Fig.I(a). The moment producing the deflected shape is shown in Fig.I(b) in the form
of moment area diagram ordinates divided by flexural rigidity EI of the beam. This
diagram is known as M/EI diagram or curvature diagram.
From S.O.M., the differential equation
of such an elastic curve is
d2y/dx2 = M/EI….(I)
The slope of the elastic line can be
expressed as θ = dy/dx ….(II)
Therefore, Eq.(I) can be written
as dθ/dx = M/EI …..(III)
or dθ = Mdx/EI ……(IV)

Dr. Jahangir
Considering a differential length of the beam between point B and C as shown in
the figure, the change of slope is denoted by dθ, that is, the angle between the
tangents drawn from B and C.
From Eq.(IV) the value of change in slope between points B and C is equal to slope
between A and C is equal to the area of M/EI diagram between those points.
The change of slope between A and C may be obtained by integrating Eq.(IV)
dθ = θC – θA = ΔθCA = Mdx/EI ……(V)
From the results of Eq.(V) the first moment-area theorem related to the change of
slope can be stated as follows;
The change of slope between two points on an elastic line is equal to the area of the
M/EI diagram between those points.
As can be seen from the elastic line, the slope at A is negative. From A to C the
slope increases which is shown in M/EI diagram. Using consistent dimensions for
M, E and I and the length of the member, using Eq.(V) the value for θ can be
obtained which will be in radians.

Dr. Jahangir
Let us consider moment-area concept to evaluate deflection.
Referring to the Fig.I(a) the vertical deviation of the two tangents drawn from B and
C can be obtained from the product of angle between the tangents and distance to the
reference line under consideration.
Therefore, at point A which is located at a distance x from the differential element
BC, the vertical deviation dt between B and C is
dt = (x) dθ ……… VI
Considering small deflection the deviation of the tangent at A from the tangent at C
denoted by tAC can be obtained using the following equation;
tAC = (x) dθ ……VII
Substituting dθ from Eq.IV
tAC = Mxdx/EI ……….VIII
From Eq.VIII the theorem related to the deflection can be stated as follows;
The tangential deviation of A from a tangent to the elastic curve at C is equal to the static
moment of the area of the M/EI diagram between A and C taken about A.
The is deviation is obtained on a vertical line passing through A. If the area of M/EI diagram
between A and C is denoted by A1 Eq.VIII can be written as tAC = A1x1……..IX
where x1 represents the horizontal distance of the centroid of the area from point A.

Example-1 Dr. Jahangir
A cantilever beam having stepped moment of inertia is subjected to a 4kN load at the free
end as shown in the figure below. Calculate the
slope and deflection at free end of the beam.
E = 204x106kN/m2 (204000MPa)
The slope at B can be taken as the change
of slope from A to B since the slope at A is zero.
The slope at B is denoted by θB as shown in
figure (b). Since the slope of the elastic curve
at A is zero, the deviation of point B on the
elastic curve from the tangent drawn at A
represents directly the deflection of beam.
This deflection at point B is denoted by tBA.
For calculating θB and tBA first draw the
bending moment diagram. The M/EI diagram
can be obtained by dividing M diagram
by EI as shown in figure (d). The slope at B
can be determined as follows: θA = 0

Dr. Jahangir
θB = θB – θA = ΔθBA = (M/EI) dx
The area under the M/EI diagram can be subdivided into triangles and rectangle.
θB = A1 + A2 + A3
Hence from the figure θB = [ (1) (-6.54)/2 + (1) (-2.45) + (1) (-2.45)/2] x10-3
θB = - 0.0069 radians
The deflection at B is determined by taking the static moment of A1, A2 and A3 about B
tBA = A1x1 + A2x2 + A3 x3
where, x1, x2 and x3 are the horizontal distances from B to the centroids of the
respective areas.
From the figure, tBA = [ (-3.27)(2/3) + (-2.45)(1.5) + (-1.225)(5/3)] x 10-3
= -7.895 x 10-3 m
or tBA = -7.9mm
The negative sign indicates that point B is below the tangent drawn through point A, i.e.
the deflection is downward.

Dr. Jahangir
Example-2
A simply supported beam is loaded as shown
in Fig.(a). Determine the location and
magnitude of maximum deflection in
the beam. EI is constant, E = 204 x 106 kN/m2
and I = 50 x 10-6m4 (50 x106mm4). The elastic
curve and moment diagram for the beam are
shown in Fig.(b) & (c). If the max. deflection
occurs at D which is a distance x from right
support. The max. deflection occurs where
the tangent to the elastic curve is horizontal.
From Fig.(c) the max. deflection occurs in
region CB. First determine slope θB at
support B. Then find x for which the value
of slope changes by θB. From Fig.(b), θB can be
obtained by dividing deviation tAB by the distance
between point A and B.
The M/EI diagram is shown in Fig.(c). To evaluate
tAB , take the static moment of the M/EI diagram area about point A.

Thus, Dr. Jahangir
tAB = A1x1 = 1/ EI [ (1/2) (6)(40)(8/3)] = 320.4/EI m
and θB = tAB/6 = 53.4/EI radians
The desired value for x is, therefore, the length in which the area M/EI diagram
changes by this value of θB. That is,
1/2(x)(40/EI)(x/4) = 53.4/EI
This yields to x = 3.27m
The magnitude of max. deflection is found by evaluating tangential deviation tBD
which is equal to the static moment or the portion of M/EI diagram between D and
B about B. Therefore,
tBD = (1/2)(3.27)(40/EI)(3.27/4)(2/3)(3.27) = 116.55/EI m
(point B is above the tangent drawn through point D)
Substituting for EI,
Δmax = -116.55/(204x50) = -11.43x10-3m
Δmax = -11.43mm

Dr. Jahangir
The following method is based on a
mathematical correspondence that exists
between moment vs load function and
deflection vs M/EI function in a beam.
If the deflected shape of the beam is
described by a function y(x),
the following general relationships exist.
y = deflection of the elastic curve
dy/dx = θ = slope of the elastic curve
dθ/ dx = d2y/dx2 = Mx/EI
d3y/dx3 = dMx/EIdx = Vx/EI
d4y/dx4 = dVx/EIdx = wx/EI
Sign convention is the same as before except
the load is positive when applied upwards.
In the beam shown in Fig.(A) the uniform
load acting upwards is positive.
Fig.(B), (C), (D) and (E) show the shear,
moment, slope and deflection quantities. The moment function can be obtained
by successively integrating twice the load function. Similarly, the deflection
function can be obtained by successively integrating twice the M/EI function.

Dr. Jahangir
These relationships lead to the conjugate beam concept for evaluating deflections
and slopes. If the M/EI diagram for a given beam and loading is considered to be
the loading on an imaginary beam known as conjugate beam, the following two
principles of conjugate beam can be stated;
Theorem 1: The shear at any point on a conjugate beam is equal (in sign and value)
to the slope at the corresponding point on the real beam.
Theorem 2: The moment at any point on the conjugate beam is equal (in sign and
value) to the deflection at the corresponding point on the real beam.
The supports of the conjugate beam are such that the shear and moment that are
obtained in conjugate beam are consistent with the slope and deflections in the real
beam.

Example 1 Dr. Jahangir
Determine the deflection at center point C and slope at ends A and B of the beam shown in
Fig.(A) by conjugate beam method. Assume E = 205 x 106 kN/m2 (205000 MPa) and I =
80 x10-6m4 (80 x 106mm4)
The conjugate beam loaded with M/EI diagram
of the real beam is shown in Fig.(B). Because
the real beam moment is positive, the M/EI
loading on the conjugate beam is also shown
to be upwards in the positive direction.
The end supports are such that as to generate
only shearing force but no moment.
The arrangement gives slopes at the end
supports but no deflection. The slope at A in
the real beam is determined by evaluating
the shear at A’ in the conjugate beam. The shear
at A’ is equal to the support reaction V’A and
is equal to half the loading on the conjugate beam.

Therefore, V’A + A1 + A2 + A3 = 0 Dr. Jahangir
V’A = -1/EI [1/2(2)50 + 2(25) +1/2(2)(25)] = -125/EI
That is, θA = -125/(205 x80) = -0.0076 radians (clockwise)
The shear is negative in its sense and, therefore, the slope at A in the real beam is
also negative. The slope θB = at support B is again equal to shear at B’ in the
conjugate beam. Due to symmetry of the beam and loading V’A = V’B, but the shear
is positive to the left of B. Therefore, the slope at B is
θB = V’B = + 0.0076 radians (anticlockwise)
The deflection at C in the real beam is determined by evaluating M’C in the
conjugate beam. The value of M’C is obtained by summing up of moment at point
C’ on the conjugate beam. That is,
M’C = (-125/EI) (4) + 50/EI (2+2/3) + 50/EI(1) + 25/EI (2/3) = -300/EI
The moment in the conjugate beam at C’ is negative. The deflection at C in the real
beam is also negative, therefore, that is, the deflection is downwards and its value is
ΔC = -300/(205x80) m or ΔC = 18.29mm

Example 2 Dr. Jahangir
Consider the beam and its loading shown in Fig.(1) below. Using conjugate beam
method, find the slope at A and deflection at point C. Assume EI is constant.
The M/EI load on conjugate beam is shown
in Fig.(2). The M/EI loading is drawn by parts
for the convenience of areas and centroids by
taking the effects of two concentrated loads
separately. For finding the possible slope and
deflection in the real beam, the conjugate
beam in Fig.(2) is supported on the roller
at A’ hinged at B’ and fixed at C’ The pin
at B’ results in zero moment, which is consistent
with zero deflection at B in the real beam.
The slopes at A and B in the real beam are
determined by evaluating the shear at A’ and B’
in the conjugate beam. The shear forces can
be evaluated by considering free-body diagram
of conjugate beam shown in Fig.(3).

Dr. Jahangir
Assuming V’A and V’B as acting upwards and summing up moment about B’ will
give,V’A (6) + 1/2(6) (45/EI)(3) + 1/2 (6)(-40/EI)(6/3) = 0
V’A = -27.5/EI,
θA = -27.5/EI (clockwise)
Again equating all transverse forces on segment A’B’ to zero we have,
ΣFv = 0 (on conjugate beam)
V’A + V’B + 1/2(6)(45/EI) + (1/2)(6)(-40/EI) = 0
V’B = 12.5/EI
θB = 12.5/EI (anti-clockwise)
M’C can be obtained from B’C’ segment on the conjugate beam.
M’C = -V’B (2) + (1/2) (2) (-40/EI) (4/3) = 1/EI[-25.0-53.33] = -78.33/EI
The deflection in the real beam at C is
ΔC = M’C = -78.33/EI (downward)

Deformation of Elements Dr. Jahangir
Consider the bar subjected to a distributed axial load q(x) as shown in the figure
Let the axial force at two sections dx apart to be F and (F + (dF/dx) dx) as shown. If
q(x) is distributed axial force per unit length, then the equilibrium equation will be
– F + [F + (dF/dx) dx] + q(x)dx = 0
or, (dF/dx) + q(x) = 0 ……..(I)
The differential equation is in terms of forces and independent of material properties.

Dr. Jahangir
For the determination of displacement, the axial force F to be expressed in terms of
axial displacement u(x). The axial strain ϵ = du / dx.
From Hook’s law, the stress σx is σx = Eϵx, where E is Young modulus.
The axial force F = Aσx, where A = cross-sectional area
Therefore, F = AE(du/dx)
Substituting for F in terms of displacement u, the differential equation (I) becomes
d/dx(AEdu/dx) + q(x) = 0 ………….(II)
The axial rigidity AE is constant, the differential equation is simplified to
AE(d2u/ dx2) + q(x) = 0 …………….(III)
Equation (III) is the fundamental differential equation for axially loaded bar with
constant properties along the axis.

Exact analysis shows that if taper is less than 200 to the horizontal, the error in
assuming uniform distribution of σx and zero shear stress is negligible.
Example -1
Calculate the axial displacement of the bar shown in the figure.
Dr. Jahangir

Soln: q = 0, therefore, equation (III) will be d2u / dx2 = 0 Dr. Jahangir
Integrating successively, du / dx = C1
u = C1 x + C2
Boundary condition are, u = 0 at x = 0 because of fixity
F = F1 at x = L
Since F = AE (du / dx), therefore, du / dx = F1 / AE at x = L
Using boundary conditions and solving for C1 and C2
C2 = 0, C1 = F1 /AE
Therefore, u = F1(x) / (AE), du / dx = F1 /AE
F = AE (du/dx) = F1, The axial displacement Δ at the tip is given by u when x = L,
Therefore, Δ = F1L/(AE).
The results can be summarized as follows:
In an axially loaded bar of constant cross-sectional fixed at one end subjected to an
axial force the other end

(a) The displacement varies linearly i.e. u = F1x/AE Dr. Jahangir
(b) The strain and therefore the axial force is constant i.e. F = F1
(c) The axial extension is given by Δ = F1L/AE. The axial stiffness of the bar defined
as the force F1 required for unit extension Δ is therefore given by AE/L. AE is
often called the axial rigidity of the cross-section.
Example – 2
Calculate the axial displacement in the bar subjected to a uniformly distributed axial
load of constant intensity q per unit length as shown in the figure.

From the differential equation AE d2u / dx2 + q(x) = 0 Dr. Jahangir
d2u / dx2 = -q / AE, where q is constant
Integrating successively with respect to du / dx = (-q/AE) x + C1
u = -qx2 / (2AE) + C1 x + C2
The boundary conditions are, u = 0 at x = 0 because of fixity and
F = (AEdu) / dx = 0 at x= L, because the end x = L is free from force. Therefore at x = L,
(AEdu)/dx = 0 or du/dx = 0.
Solving for C1 and C2, C2 = 0 and C1 = qL / (AE)
Substituting for C1 and C2 in the expression for u,
u = qx(2L – x) / (2AE), Therefore, F = AE du /dx = q(L – x)
The extension Δ for the bar at free end is given by u at x = L,
therefore, Δ = qL2 / (2AE)
If the total applied force is F1, then F1 = qL.
Therefore, Δ = 0.5F1L / (AE). The results indicate that:

Dr. Jahangir
(a) The displacement of the bar varies parabolically. It is zero at the fixed end and qL2/(2AE)
at the free end
(b) The strain and therefore the axial force vary linearly. It is a maximum at fixed end and zero
at the free end
(c) The total axial extension Δ = 0.5F1L / (AE) is only half the extension that would have
occurred if a concentrated force equal to F1 had been applied at the tip
Assignment
Calculate the forces at the ends of a prismatic bar fixed at the ends and carrying uniformly
distributed axial load q as shown in the figure.

Stiffness Matrix for a Bar Element Dr. Jahangir
For the axially loaded bar shown in the figure below establish the relationship
between the axial forces and the corresponding axial displacements at the end of the
bar. Assume q is constant.
If the axial force and displacement
at x = 0 and x = L are respectively
(Fa1, Δa1) and (Fa2, Δa2).
From the differential equations,
AE (d2u/dx2) + q = 0, F = AE (du/dx),
du / dx = (-q/AE)x + C1,
u = (-qx2)/2AE + C1 x + C2
The boundary conditions are u = Δa1 at x = 0 and u = Δa2 at x = L ,
therefore, C2 = Δa1,
At x = L, Δa2 = (- qL2) / 2AE + C1 L + Δa1,
C1= (Δa2 - Δa1) / L +qL / (2AE)

u = (1 - x/L) Δa1 + (x / L) Δa2 + qx (L - x) / (2AE) Dr. Jahangir
F = AE (du/dx) = (AE / L) [Δa2 - Δa1] + q(L - 2x) / 2
At x = 0, Fa1 = – F = (-AE / L) ( Δa2 - Δa1 ) – 0.5qL …….(I)
Note that Fa1 is a compressive force.
At x = L, Fa2 = F = (AE / L)( Δa2 - Δa1) – 0.5qL …………(II)
Eqns. (I) and (II) can be written in matrix form as follows
If Δa1 = Δa2 = 0, then

Dr. Jahangir
The forces which result at the ends due to the load on the bar when the ends are
prevented from displacing are called fixed end forces and the corresponding force
vector as fixed end force vector
Similarly, if q = 0 then
The above equation is a relationship between the forces at the ends and the
corresponding displacements at the ends. Since the relationship in a spring between
the force and the corresponding extension is known as the stiffness of the spring, by
analogy we have,
{Force vector} = [Stiffness matrix] {Displacement vector}

Dr. Jahangir
Where, [Stiffness matrix] =
{Displacement vector} =
{Force vector} =
Therefore the relationship between the forces and displacements can be written as,
{Force vector}= [Stiffness matrix] {Displacement vector}+ {Fixed end force vector}

In general therefore, for a bar element Dr. Jahangir
Where Ffa1 and Ffa2 are the fixed end forces at the end 1 and 2 respectively due to axial loads
applied to the bar along its length. Or they can be due to temperature change as well.

Simple Elastic Spring: Dr. Jahangir
Assume the spring shown in the figure is part of a structure
Force vector is
and displacement vector is
Stiffness Matrix for Single Elastic Spring:

The stiffness matrix for the spring is therefore of 2 X 2 order and the relationship is:
=
Case - 1: Force F1 applied at end A, end B is fixed
Force-displacement at node 1 is related by F1a = Ku1
Equilibrium of forces acting on the spring
is F1a + F2a = 0, F2a = -F1a = -Ku1
Case - 2: Force F2 applied at end B, end A is fixed
F2b = Ku2 = -F1b
Combination of Case - 1 and Case – 2:
Using superposition principle, the
total force acting @ node 1 will be
F1 = F1a + F1b = Ku1 – Ku2 …….(a)
Total force @ node 2: F2 = F2a + F2b
or F2 = -Ku1 + Ku2 ……….(b)

Dr. Jahangir
Writing in matrix form,
= ,
Important property of Ke is that it is a symmetric matrix.

Home Work Problem Dr. Jahangir
A four spring system shown in Fig.(a) below is subjected to axial loads P1 and P2
(a) Determine the displacements Δ1 and Δ2 and spring forces, if they are
connected through a weightless rigid body Q.
(b) If K1= 50kN/m, K2 = 75kN/m, K3 = 50kN/m, K4 = 60kN/m, P1 = 20kN, and
P2 = 40kN.
What will be the amount of Δ1 , Δ2, F1, F2, F3, F4 the spring forces.
Soln.:
(a) Since the springs are interconnected through a rigid body Q, the
compatibility condition requires that spring 1, 4, and 3 must elongate by Δ2
each. The free-body diagram of the springs are shown in Fig(b), (c), (d) & (e).
Force displacement relations give,
F1 = K1Δ2, F2 = K2 Δ1, F3 = K3 Δ2, F4 = K4 (Δ2 – Δ1) ……(I)
Joint equilibrium equations give
F2 - F4 = P1 ……...(II)
and F1 + F4 + F3 = P2 ……(III)

structural analysis1-76 course for phd.pdf

There are six relations and six unknowns F1, F2, F3, F4, Δ2, and Δ1.
Substituting the values for F1 to F4 in equations (II) and (III) give
K2 Δ1 - K4 (Δ2 – Δ1) = P1
And K1Δ 2 + K4 (Δ2 – Δ1) + K3 Δ2 = P2
or ……..(IV)
Eqn. IV gives the values of Δ1 and Δ2. The spring force can be obtained using Eqn. (I)
Eqn. (IV) gives

Bending Stresses
Consider a portion of a beam subjected to bending moment M only. The beam
bends in an arc form the radius of curvature being R and center of curvature O as
shown in the figure below.

Dr. Jahangir
From the figure, the top fibers are in compression and bottom fibers in tension and
the surface in between being neutral. The intersection of neutral surface with the
cross-section is called neutral axis (N.A.).
The longitudinal strain in a fiber at a distance y from the neutral axis maybe given by
ϵ = [(R + y) θ – Rθ] / (Rθ) …………..(I)
The normal stress σ = E ϵ = (Ey) / R ……….(II)
Therefore, σ / y = E / R ………………… (III)
Since the beam is subjected to pure bending, net normal force on any cross-section
must vanish,
Therefore, Aʃ σdA = 0, where, dA is an elemental area at distance y from N. A.
Therefore, Aʃ (Ey) dA / R = 0, E / R A ʃ ydA = 0
or Aʃ ydA = 0
But Aʃ ydA is the static moment of area which is zero only about a centriodal axis.
Hence N. A. coincides with centroidal horizontal axis. This gives an information
whereform y is to be measured. Secondly, the moment due to be bending stresses
about the N. A. must be equal to the applied B. M.

Hence M = Aʃ σdA.y = Aʃ (EydA.y) / R = E/R Aʃ y2 dA
or M = (EI) / R …………………..(IV)
where, I is the second moment of area about N. A., also called moment of inertia of the
cross-section.
Thus M / I = E / R …………………(V)
Combining eqns III and V M / I = σ / y = E / R ……….(VI)
Eqn. VI is known as Bending Equation or Flexure Formula.
The most commonly used form of bending stress eqn. is
σ = My / I ………………….(VII)
This shows that bending stress varies linearly over the depth of the beam as shown in
Fig.(c) above.
σmax = Mymax / I …………………(VIII)
where, ymax may have in general different values in tension and compression.

Section Modulus
Section modulus is a property of the section which is defined as
Z = I / ymax …………………..(IX)
If the N.A. is not at mid-depth the section modulus may be different in tension and
compression zones.
Maximum Moment of Resistance:
It is the max moment of resistance that can be resisted by the cross-section of a
beam. If σperm be the permissible bending stress in a beam material then,
Moment of resistance MR = (I / ymax) σperm = Z (σperm) …………….(X)
Sectional properties of some of the common beam sections are given in the
following table.

Example #1
A rectangular section beam 200mm deep having width to depth ratio of 2/3 is simply
supported over a span of 4m. If the bending stress is to be limited to 140MN/m2, what
distributed load over the span can the beam carry?
q = ?
Mmax = qL2 / 8 = q(4)2 / 8 = 2qkNm = 2q x 103 Nm
Z = bd2/6 = [(2/3) d x d2] / 6 = d3/ 9 = (200)3/ 9 mm3
= [(8x 106)/9] x 10-9 = [(8x10-3) / 9] m3
Mmax = (σmax ) Z
Or 2q x 103 = [140 x 106 (8 x 10-3)] / 9
q = (140x8) / (2x9) = 62.22 kN/m
q = 62.22 kN /m

Example #2
A T-section beam of dimensions shown in the figure below is used as a cantilever of
span 1m. If it carries a 5 kN load at the free end what additional uld can be applied on
it if bending stress is not exceed 120 MN/m2. Neglect the self weight of the beam.
M = 5 x 1 + qL2/2 = (0.5q + 5) kNm
y- = (ΣAiyi) / ΣAi = [100x20x10 +80x20 (20+80/2)] / (20x100+80x20)
y- = 32.22mm from the top, ymax = 100 -32.22 = 67.78mm
I = [ Ixx + Ad2]flange + [Ixx + Ad2]stem
= [(100x203)/12 + (20x100)(22.22)2] + [(20x803)/12 + 20x80(27.78)2]
= 3.142x106mm4
I = 3.142x106mm4
MR = (I/ ymax) σallow
(0.5q+5.0)x106 = (3.142x106x120)/(67.78)
q = 1.125 kN/m

Example problem
The I-section beam shown in the figure below is simply supported over a span of
6m and carries concentrated load of 30kN at distance 2m from left end. Find the
bending stresses at top and bottom of the section at the load point and the moment
of resistance of the section if allowable bending stress is 120 N/mm2.

Example
The cross-section of an unequal-flange beam is shown in the figure below. The
beam is so supported and loaded that the upper portion is in compression and
lower portion in tension, the permissible stress being
260N/mm2 and 180N/mm2 respectively. Find the
moment of resistance and draw the stress
diagram for this moment.
Soln.: Let N.A. lie at a distance y- from
the top fiber of the beam section.
y- = (ΣAy) / (ΣA)
= {(150x50x25)+[(40x200) (50+100)]
+ [(250x80) (250+40)]}/(150x50+40x200+250x80)
y- = 202.46mm from top,
Therefore, yc = 202.46mm
and yt = (50+200+80) – 202.46 = 127.54mm

Ie = I1 + A1d1
2 +I2 + A2d2
2 +I3A3+d3
2
= (150)(50)3/12 + 50x150(177.46)2
+ (40)(200)3/12+ 40x200(52.46)2
+ (250)(80)3/12+ 80x250(87.54)2
= 1562500+236190387+26666666.67+22016412.8+10666666.67+153265032
Ie = 4.5036x108mm4
M1 = (Ie x σt) / yt = (4.5036x108 mm4 x180N/mm2) / 127.54mm = 6.356x108 Nmm
M2 = (Ie x σc) /yc = (4.5036x108mm4 x 260N/mm2) / 202.46mm = 5.783x108 Nmm
The smaller of the two values i.e. 5.783x108Nmm is the moment of resistance of the
section. At this moment, σcmax = 260N/mm2 and σtmax = 163.78N/mm2

Strain Energy Method (Displacement – Energy Method)
There is a relationship between strain energy and deformation in a structure. This
relationship can be used to determine the slopes and deflections in a statically
determinate as well as indeterminate structure. The strain energy method is one of
the most powerful tools for the analysis of any structure. This method is applicable
to linear elastic or non-linear elastic deformable systems.
In mechanics, energy is defined as the capacity to do work, and work is the product
of the force and the distance it move along its direction. The internal work is stored
in the body as the internal elastic energy of deformation or elastic strain energy.
Both the stain energy and complementary energy methods are special cases of the
principle of virtual work. The energy theorems may be grouped as follows;
(1) Strain energy theorems:
(a) Castigliano’s first theorem
(b) Principle of virtual displacement or unit displacement method
(c) Principle of stationary potential energy

(2) Complementary energy theorems:
(a) Castigliano’s second theorem
(b) Principle of virtual forces or unit load method
(c) Principle of stationary complementary energy
The theorems in the first group are used for the stiffness method or displacement
method of analysis, while the theorems in the second group are used for flexibility
method or force method of analysis.
Work and Complementary Work
Force is equal to mass times acceleration according to the Newton’s law of motion.
In structural mechanics, it is defined as the stress times area. When the point of
application of a force moves, then a work is said to be done. The work is equal to
force times the displacement in the direction of the applied force.
Consider a structure acted upon a load P0 producing a displacement Δ0. A typical
load displacement characteristic is shown in Fig.(a) for the a linear structure when
the load is applied gradually.

The work done by a load P in causing a small displacement dΔ is shown by the shaded
strip.
dW = PdΔ ………………..(I)
The work done by P0 is represented by the area
OAB which is equal to the strain energy stored in
the structure. The total work is found by summing
all the increments of work represented by the similar
strips as the displacement increases gradually from
zero to the final value Δ0. This is equal to the area
under the force-deformation curve:
W = 0ʃ Δ0PdΔ = U ……………………(II)
The work done may also be obtained by considering
the horizontal strip shown in the Fig.(a).
dW* = ΔdP ……………………………(III)
The term W* represents complementary work. The load is increased by an amount dP.
The total work done as the load increases from zero to its final value P0 is obtained by
summing all the increments of work represented by the similar strips.

This is equal to the area above the force-deformation curve.
W* = 0 ʃP0 ΔdP = U* ………………….(IV)
For the linear elastic system as shown in Fig.(a)
W = W* and U = U*
Following Hook’s law the integral representing the areas under load-deformation
curves are reduced to areas of the triangles.
For non-linear elastic system as shown in
Fig.(b) the work and complementary work
are not equal. Therefore, it is necessary to
clearly distinguish them. The area OAC
represents the complementary work or
complementary strain energy. The
complementary strain energy is useful
in computing deflections in a non-linear
elastic structure.

Strain Energy
Axial Strain Energy
Consider a member of length L subjected to an axial load P. Let the area of cross-section of
the member to be A at a distance x from one end as shown in Fig.(c)
Stress in member σ = P/A
Strain in member ϵ = σ/E = P/AE
Elongation in length dx = (Pdx)/AE
The total elongation in the member will
be ΔL = 0ʃL (Pdx)/AE
If the force P was increases gradually from
zero to P, average force causing elongation
ΔL will be (0+P)/2 = P/2
Therefore, work done
U = Force x Displacement or, dU = (P/2) x ΔL ………….(IV)

Or, U = (P/2)0ʃL L(Pdx/AE) = 0ʃL (P2dx)/(2AE) ………………(V)
If area of the cross-section of the member is uniform
Total strain energy U = P2L/(2AE) ……………………………..(VI)
The strain energy may be tensile or compressive.
Shear Strain Energy
Consider a small element of a structure dx, dy and dz subjected to a shear force V as
shown in Fig.(d). It causes a shear strain equal θ and a lateral displacement equal to
‘a’ at top fiber. If the she ar force is applied gradually,
Average shear force = (0+V)/2 = V/2
Or, V = τdydz and a = dxθ
Shear stress / shear strain = modulus of rigidity
Or, τ / θ = G or θ = τ / G,
therefore, a = (τ / G) dx

Work done U = Force x displacement
dU = (V / 2) a = (V/2) (τ/G) dx = (1/2) (τ2 /G) dxdydz ………….( VIII)
Alternatively τ = V/ Ae
where,
Ae = effective area of cross-section in shear
Therefore, Eqn (VII) becomes
dU = (1/2) (V2/AeG)dx
U = 0ʃL [V2/(2AeG)] dx

Flexural Strain Energy
Consider a small element of length ds subjected to a uniform bending moment M
which produces an angle dθ in this length ds as shown in the figure. If moment is
applied gradually,
Average bending moment = (0+M) / 2 = M / 2
Work done dU = (M / 2) dθ but dθ = (Mds) / EI by moment- area theorem
Therefore, dU = (M2ds) / 2EI
Total strain energy U = ʃ (M2 ds) / 2EI

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