Introduction of Structural Analysis | Structural Analysis

Structural Analysis
INTRODUCTION
Assumptions
1. Material is homogeneous, isotropic and linearly elastic
2. Boundary conditions are ideal
3. Deflections are small
4. Loads are ideal
5. Structures are idealized as beams, plane frames, three-dimensional frames, etc.
Law of superposition It means the structure can be analysed for different loads separately and the
results be superposed to get the final results due to different load combinations.
Conditions of equilibrium In case of three-dimensional structure are
SFx = 0, SFy = 0, SFz = 0
SMx = 0, SMy = 0, SMz = 0
where x, y and z are Cartesian coordinate systems in space.
Compatibility conditions It means requirement of continuity at joints.
1. The members meeting at a joint will continue to meet at the same point even after deformation.
2. At rigid-joints, the angle between any two members remains the same even after deformation.
A system is called linear systems if its material has linear stress-strain relation and deflections are
small.
If the non-linearity is due to stress-strain relations not being linear, it is material non-linearity
problem.
If non-linearity is due to considerable change in the geometry, it is called geometric non-linearity
problem.
 INFLUENCE LINES
* The influence line diagram for a stress resultants is the one in which ordinate represents the value of
the stress resultant for the position of unit load at the corresponding abscissa.
* Influence line diagrams are used to determine the position of moving loads for the maximum stress
resultant and for the maximum value of the stress resultant.
ILD for Simply Supported Beams
Use of ILD:
1. In case of concentrated loads for
F = SWiyi and M = SWiy¢i
where, Wi are concentrated loads, y; are the ordinates to the ILD for shear force and y¢; are the

ordinates in the ILD for bending moment.
2. In case of udl smaller than the span,
SF = Intensity of udl × Area of ILD for SF under the load.
Similarly,
M = Intensity of load × Area of ILD for moment under the load.
* Maximum SFand BM due to moving loads.
1. Maximum SFand BM at point z from the end due to single concentrated load:
–ve Max SF = , when W is just to the left of the section.
+ve Max SF = , when the load is just to the right of the section.
Max BM = , when the load is on the section.
Absolute max. BM = , occurs at midspan when the load is at that section.
2. Max SFand BM at the given section when UDL is longer than the span
–ve Max SF = , when load is just to the left of section
+ve Max SF = , when load is just to the right of the section
Max MC = , when load covers the entire spam.
Absolute Max M = at midspan when load covers the entire span.
3. UDL smaller than the span
(i) Maximum –ve shear develops when the head of the load reaches the section.
(ii) Maximum positive shear develops when the tail of the udl reaches the section.
(iii) Moment at given section C will be maximum when the ordinates of ILD for MC at head and tail
of the udl are equal which means bending moment at a section is maximum when the load is so
placed that the section divides the load in the same ratio as it divides the span.
(iv) Absolute maximum moment occurs at midspan when the C.G. of udl coincides with midspan
point.
4. A train of concentrated loads
(i) Shear force has a peak value whenever a load is on the section. Highest value among these values
is to be selected to get maximum SF at a given section. It is to be noted that for maximum –ve SF,
most of the loads are to the right of the section, and for maximum +ve SF most of the loads are to
the right of the section.
(ii) Absolute maximum +ve SF occurs near support A when one of the loads is on A and maximum
+ve SF occurs at B when one of the loads is on support B.
(iii) At a given section maximum moment occurs when a concentrated load is on the section and when
that load moves from left to right, left side portion becomes heavier to lighter compared to right,
side portion.
(iv) Absolute maximum moment occurs under one of the loads when the resultant of all the loads and
the load under consideration are equidistant from the centre of the beam.
ILD for Bridge Trusses
Note:
1. To find ILD for the force in U1U2
, consider section 1–1, then consider moment about L2
.
2. To find ILD for the force in L2L3
, consider the section 2–2 and then moment about U2
.
3. To find ILD for inclined member U1L2
, consider the equilibrium of right hand side portion of
section 1–1 and when the load is in portion L2L6 consider the equilibrium of left hand portion.
When the load is between L1L2
it varies linearly.
Three-Hinged Arch
* It is a determinate structure. It may have circular or parabolic shape.
* In case of circular arch of radius R, span L and rise h,
= h(2R – h) or R =
x = – R sin q
y = R cos q – (R – h) = h – R (1 – cos q)
* In case of parabolic arch
y = (L – x), if springing point is origin

At a section if U is vertical shear, H is horizontal force, Q is radial shear and N is the normal thrust,
N = V sin q + H cos q
and q = V sin q – H cos q
where q is slope of tangent to horizontal.
* If support A and support B are below crown at distances h1 and h2
, in a parabolic 3-hinged arch,
Due to concentrated load, maximum moment in 3-hinged parabolic arch occurs when the load is it x
= 0.2113 L
and Mmax = 0.096225 WL.

* Due to moving udl, maximum +ve moment in 3-hinged parabolic arch occurs when the load is up to
0.234 L and its value is 0.01883 wL
2
.
6.3 CABLES AND SUSPENSION BRIDGES
* The cables may have either guided pulley support or roller pulley support.
In case of guided pulley support suspension cable continues as anchor cable also and hence tension in
the cable on either side of support is same.
In case of roller pulley support, the suspension and anchor cables are connected to a roller. Hence,
the horizontal components of tension in suspension cable and anchor cables are the same.
Since cable is flexible,
Mx = 0
and Hy = beam moment
where H = horizontal force developed at support.
* If the cable is subjected to udl and the supports are at the same level, horizontal thrust developed is
given by
H = , where l-span
* The length of cable is given by
L = l + , where h is maximum sag.
* If ends arc at different levels, say h1 and h2
from the crown point, due to ud

In case of guided pulley support,
Vertical load transmitted to tower = T (sin q + sin a)
Horizontal load transmitted to tower = T (cos q – cos a)
Max bending moment on the tower = T (cos q – cos a) h1
where q = Angle made by suspension cable with the horizontal
a = Angle made by anchor cable with the horizontal
T = Tension in the cable
and h1 = Height of tower
Suspension Bridge with Three-Hinged Stiffening Girder
Equivalent load in suspenders is given by We = Moment in stiffener.
where We equivalent udl is suspenders.
ILD for horizontal thrust H and bending moment at any point D are the same as those for 3-hinged
arches.
Suspension Cable with Two-Hinged Stiffening Girder
* Equivalent udl in suspenders is equal to the average load on the stiffening girder ILD:
Fig. 6.7 ILD for We
, H, V, MD and QD in suspension cable with two hinged stiffening girder
Degree of Static Indeterminancy
* Degree of static indeterminancy
= No. of unknowns – No. of independent static equilibrium equations
Consistant Deformation Method of Analysing Indeterminate Structures
1. Remove constraints to make the strucuture determinate.
2. Find the displacement of released structure in the direction of released constraints, due to given
loading.
3. Determine the expressions for deformation of released structure due to each of released reaction.
4. Impose consistency conditions to get the reactions.
Three-moment equation

Consistant Deformation Method of Analysing Indeterminate Structures
1. Remove constraints to make the strucuture determinate.
2. Find the displacement of released structure in the direction of released constraints, due to given
loading.
3. Determine the expressions for deformation of released structure due to each of released reaction.
4. Impose consistency conditions to get the reactions.
Three-moment equation
If ABC are three continuous supports, AB = L1 and BC = L2
,
MA =
where MA
, MB
, MC are moments at A, B and C respectively.
A1
, A2 area of moment diagram in AB and BC respectively.
a1 = Distance of C.G. of A1
from support A
a2 = Distance of C.G. of A2
from support C.
member in proportion to their stiffness. Distribution facto
Moment distribution procedure Find fixed end moments and distribution factors. Balance each
joint. Carry-over distributed moment to far ends. Stop the procedure when carry-over moments are
negligible.
Note: If the far end is simply supported/hinged it is to be balanced by carrying over 1/2 of it to
interior support. If stiffness of end member which has one end simply supported may be balanced
once for all and its stiffness modified as , then no moment is to be carried to the simply supported
end.
If there is sway in the frame
1. Carry out non-sway analysis first.
2. Due to sway only, end moments developed in the columns due to arbitrary sway are proportional to
the stiffness of columns. Assume arbitrary but proportional moments in columns. Carry out moment
distribution and find sway force. Find the ratio of horizontal force required to prevent sway in
non-sway analysis (S) and the sway force (S¢) in arbitrary sway cases. Then final moment = Non-
sway moment + arbitrary moment case
* In case of skew frames (Ref. Fig. 6.13)
D = horizontal displacement of BC
DAB =
DBC = – (D cot a + D cot b)
DCD = D cot b

Kani’s Method of Rotation Contribution
* The expression – is called rotation factor for member AB at joint A and the summation is over
various members meeting at A.
MAB = MFAB + 2 × Near and moment + far end moment
* Analysis of symmetric frame taking advantage of symmetry.
(a) Line of symmetry through columns: Treat joint at symmetric line as fixed support and
analysed half the fram



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