Methods of structural analysis - broad classification

You might have read my earlier posts on the determinate and indeterminate structures. Determinate structures can be analysed easily with the conditions of the equilibrium but the indeterminate structure needs other methods of structural analysis. These methods are broadly classified into two categories:

(1) Force Methods/ flexibility methods/compatibility methods.

(2) Displacement Methods/stiffness methods/equilibrium methods.

In the force methods the forces are taken as the redundant and then writing the equations in terms of the forces for the displacements and then applying the compatibility equations on these equations, we can find out the various forces. There are numbers of methods which come under this category:

(1) Virtual work method/Unit load method

(2) column analogy method

(3) Elastic Center method

(4) Three moment theorem

(5) Castigliano's theorem of minimum strain energy

(6) Mohr-Maxwell Theorem

 In the displacement methods the equations are written for the forces and then the unknown displacements are found by applying the equilibrium conditions on these equations. These displacements are put back into the initial equations to find out the forces.  Various Displacement methods are listed below:

(1) Slope displacement method

(2) Moment Distribution Method

(3) Minimum potential energy method

There is a condition to chose among the two and the condition is if,

(a) Static indeterminacy(Ds) < Kinematic indeterminacy(Dk), then we should use the force methods

(b) Kinematic indeterminacy(Dk) > Static indeterminacy(Ds) , then we should use the displacement methods.

Some of these methods are dealt in details in my other posts. 

Thank you.

Elastic constants (E, G & K)

If you are studying the structural engineering then there are these three elastic constants which you have to use in various theories.
(1)Young's modulus of elasticity E
(2) Modulus of rigidity G
(3) Bulk modules  K
E is the elastic constant which is defined as the linear stress over linear strain.
G is defined as the shear stress over shear strain.
K Bulk modulus is defined as the stress over volumetric strain.
Poison's ratio is another constant which is defined as the lateral strain over linear strain.(n)
Numerically E= 3k(1-2n)
                 also E=2G(1+n)

Two-Hinged and Three-Hinged Arches

Introduction:

Arches are the structures, which look somewhat different from the columns and beam. They have the curved shape, of an arch, which can be circular or parabolic. In Civil Engineering, you have to study the analysis of the arches.

In engineering terms, there are three types of arches. They are listed below.

1. Two hinged arches
2. Three hinged arches.
3. Fixed Arches

  

Three hinged arches are the determinate structures, because there are four unknown support reactions, and again there are four numbers of equations of equilibrium, to get the values of these unknowns.

Three Hinged Arches:

See above in fig.2, there are three hinges in the arch, A, B and C. Generally there are three numbers of equilibrium equation, but the fourth equation is derived from the fact the algebraic sum of all the moments at the hing C is 0.  So there are four numbers of equilibrium equations, and we can determine all the four support reactions, Va, Vb, Ha, and Hb.

Three Hinged Arch Image

Two Hinged Arches:

In the fig.1 in the above figure there are two hinges A and B, and there are four support reactions. There are only three numbers of equations of static equilibrium. Hence, the two hinged arches are indeterminate to a degree of 1.

 If we have to find out all the four unknown reactions of the two hinged arch, then, we need one more equilibrium equation. So the indeterminacy of a two hinged arch is equal to 1. 

We can easily find out the Va and Vb, by taking algebraic sum of all the moments about A or B equal to 0.  To find out the horizontal reactions Ha and Hb, many books advise to use the Castigliano's first theorem.

The relative displacement of the either hinge with respect to other is zero, so the partial derivative of the strain energy of the beam with respect to the horizontal reaction will be zero.

 So first we have to find the equation of the strain energy of the whole beam, and then partially differentiate it w.r.t. to the horizontal reactions, and then equate it to zero. 

It becomes the fourth equation, and we can get the value of the horizontal reaction. Now as all the support reactions are found, we can easily plot the bending moment diagram, for the arch. 

Now at any cross of the arch the vertical and the horizontal forces can be resolved along two directions, one is tangent to the cross sectional surface of the arch and another is normal to the cross sectional surface of the arch. It gives rise to another two terms:

1. Radial Shear
2. Normal thrust. 

Radial Shear and Normal Thrust

Due to the curved shape of the arch,  unlike shear force and axial thrust in a straight  beam, the direction of the normal thrust and the radial shear change constantly from one end of the arch to the other.
To find out the magnitude of the radial shear and the normal thrust, first we have to find out the direction of these two.

To find out the direction, we simply find out the slope of the arch(angle w.r.t. horizontal), at that section which, can be found by writing the equation of the circle or the parabola in terms of y and x co-ordinates, depending upon the shape of the arch and  then differentiating the y w.r.t. x gives us the slope of the equation.

Putting the corresponding value of the x at the given cross section, gives us the slope at the given cross section. After finding the angle, it becomes easy to resolve the forces into two directions, of the radial shear and the normal thrust. In the figure shown below, you have to resolve the forces at the section along the R and N, to find out the Radial Shear and the Normal Thrust.

Rib Shortening: 

As we can see that there is a normal thrusting force, which applies the compressing action to the rib/arch. If the magnitude is high it may result in the change in the length of the arch, due to corresponding strain, which can be found out by the Hooke's law. Final result will be the shortening of the arch. This effect is known as the Rib shortening. 

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column analogy method

Hi,

Column analogy method is a method of structure analysis, used to analyse the indeterminate structures, specifically the fixed beams, arches etc. 

   

It is based on the analogy between the fixed end moments induced in a fixed beam to the pressure induced at the base of a column subjected to loading of the bending moment diagram of the basic determinate structure of the given fixed beam.

It means that first we have to make the indeterminate fixed beams determinate beams- mostly  simple supported beams, and then we have to find out the bending moment diagram for that structure. This Bending Moment diagram is used as the loading diagram on the top of a column having the cross section as that of the beam-  length equal to the span of the beam and width is kept as one.

 The pressures induced at the two ends of the column base are calculated and they are equal to the fixed end moments induced at the ends of the fixed beams.They are superimposed with the simple supported bending moments and that gives the final bending moment diagram.

Thanks!
Check these relevant books.

   

Moment distribution method - An approximate method

Moment distribution method of structure analysis is the method which was introduced after the slope deflection method by prof. Hardy Cross. This method uses the following properties of the beam for 

(1) one end hinged and far end fixed: The moment required to apply to produce the unit slope at hinged end of the beam when the far end is fixed is equal to 4EI/L. It is known as the stiffness of the beam.  At the same time half the moment applied at A is carried over to the far fixed end B, known as the carry over moment. 

(2) One end hinged and far end also hinged:  

When the far end is also hinged, the stiffness of the beam to produce the unit slope at end A by applying the moment at a is 3EI/L. The moment is not carried over to the far end because it is free.

Procedure to follow the Moment distribution analysis:

 All the joints are initially fixed  so that each span between joints behaves as the independent fixed beams.
Calculate the fixed end moments on each span after adding algebraically moments due to  the external loading, sinking of the supports. Now release the joints which are not fixed one at one time. There will be some unbalanced moment at joints (where two or more than two spans meet), and the ends with hinged or simple supports. balance the joints by distributing the moment at the joints among the member meeting at the joint proportional to their relative stiffness. The carry over moment to the far ends becomes the unbalanced moment for the next cycle of balancing.
Note: It is brief introduction of the method, you can write in if you need any further support in this method.

Slope deflection method of structural analysis

This is method of structural analysis which is given by the Prof. Maney in early of the 20th century (b/w 1910 to 1920). This is a displacement method of structural analysis, in which the end moment at the joints of the beams are expressed in the terms of the displacements and the fixed end moments. These expressions are known as the slope deflection equations.
Then the compatibility or the equilibrium equations are found for the structures, and these equilibrium equations are solved to get the unknown displacement at the joints. When the unknown displacements are found we can put these displacements in the slope deflection equations to find the final end moments. Once the final end moments are found, we can find the vertical reactions and the horizontal reactions at the various supports, to draw the shear force and the Bending moment diagrams.

A simple form of the slope deflection equation for a span AB of a continuous beam is written in the form as shown in the picture below.

In the beam shown in the figure, the kinematic indeterminacy is only two, i.e. angle at B and angle at C. So you need two equilibrium equations to get these values. Once you get the values, you can find the end moments by using the equations above.
The 2 equilibrium equations for the structure shown above can be created,on the basis of the fact that the sum of the left end moment and the right end moments at the joints B and C will be zero, because both of the joints are hinges, so they will not take any moment. So by equating the sum of the end moments at the two joints, and solving the two linear equations, simultaneously, I can get the unknown values of the slopes.
So now putting these values in the equations above, I can get the final end moments. 

Working stress method to design the concrete structures.

Working stress method of design is the old method which is replaced by the limit state design method now. It is still taught in the engineering studies, because it is important to understand the method, and some might also use this method.
In this method we take the design strength of the material as their yield strength. Further this strength is divided by a factor of safety to have a safer structure.
Generally the factor of safety for the concrete is almost equal to 3. So if you are using a M15 concrete then the design strength of the concrete in the bending compression for a RCC structure will be 5 N/sq.mm. M15 is not generally used for the construction of the RCC structures, generally M20 and M25 is used.
Design has two processes going side by side, one is analysis and other is the design. Structure is designed for the safety and economy at the same time, so we have to design the structural components by a number of trials until we get both economy and safety. 

Principle stresses and strains

There are generally two types of stresses, which we deal with in the structural mechanics or structural analysis of the Civil engineering. One is normal stress and another is shear stress. An element of a structure can be loaded with different combinations of the loading. There can be normal loading or shear loading on a face or the combination of both. These combination of loads results in the induction of the maximum stresses along some planes. There will be one particular plane of the element where the shear stress will be equal to zero but the normal stress will be maximum.
Principle Stress: 
This is the normal stress on a particular plane of a structural element subjected to external loads, where its value is maximum, but the value of the shear stress is zero.
Strain produced due to the principle stress is known as the principle strain. 

Strain energy stored in a member

When an elastic body is subjected to the external loading it may undergo the deformation or strain. If the material is strained withing its elastic limit, then according to the principle of conservation of energy, the work by the external load on the body should be equal to the strain energy that is stored in the body. Strain energy can also be called as the potential energy of the body.
If you remove the loading from the body, it will regain its original shape, so this energy is again converted to the mechanical energy.
Axial Loading on bar:
If a member having elasticity of "E" is subject to an axial loading of  stress "f",
then the amount of the strain energy stored in  body per unit of its length can be calculated to be equal to
U= f^2/2E
There are numbers of methods which uses the strain energy as the basis for the analysis of the given structures which can not be analysed by the general conditions of equilibrium, which are only 3 in numbers. Castigliano's theorems are much famous which use this energy principle also.
Bending loading on an elastic bar:
If the loading on the bar is a bending moment of value 'M', then the flexural stresses are induced in the cross section of the bar, which vary from the center of the cross section to the extreme fibers.
at any distance 'y' from the center, the value of stress is given by,
f = M.y/I
M is the applied moment, and I is the second moment of the area of the section about the central axis of the cross section.
Energy stored in any element of length, 'ds' and cross sectional area of 'da' is given by,
 (f^2.ds.da)/2E = (M.y/I)^2.ds.da/ 2E
Now the energy stored in the total cross section of the bar can be found by adding the energy stored in the infinite nos. of elements coming in the section of length ds.
now in the above equation, da.y^2 is the second moment of area. So summing up all the second moments of areas will give us the moment of Inertia of the section.
So the energy stored in the length of ds = M^2.ds/2EI
the total energy stored in the whole length of the bar can be found out by integrating the above equation with the limits of the length of the bar.

Indeterminacy (Determinate and indeterminate structures)

Hi,



There are the structures which can be easily analysed using the conditions of the equilibrium of the structure. These structures are easy to find their support reactions and the internal forces generated in the different members of these structures.

   

• Beams: 

The simply supported beam, beam with one end hinged one with roller support, perfect frames etc are the structures which are determinate. There are structures which have other support conditions, like if the beam has both ends fixed, or both ends hinged then the structure is statically indeterminate.
There are four reaction components if the beam is hinged at both the ends, and the conditions of the equilibrium are only three. So with these three conditions of equilibrium I can only find the 3 support reactions. So the structure is indeterminate.
if, R= total nos. of support reactions
    r = nos. of conditions of equilibrium.
then, the indeterminacy = R-r.
so in case of the fixed ends beam the indeterminacy is 1.

• Pinned frames:

In case of  frames, there are two kinds of indeterminacy. One is external and another is internal.
In case of the pinned frames, the external indeterminacy is R-r,
 but the internal indeterminacy is found by the formula:
I = m- (2j-r)
I = Internal indeterminacy
j= nos. of joints in the frame,
r= nos. of conditions of equilibrium.

• Rigid Frames:

In case rigid frames, the external indeterminacy is again R-r but the internal indeterminacy is equal to
I = 3a
a= nos. of the closed areas in the frame structure,
I = Internal indeterminacy.
Note: In any structure if a pin is introduced then it increases the numbers of equilibrium equations by 1. The equation is governed by the fact that the sum of all the moments is zero at the pin.

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Columns and strut - Euler and Rankine's formulae

Hi

Difference between a column and a strut:

We see the columns everywhere around us because they are very important component of structures. 

Column is the vertical member of a structure which generally takes the compression from the other components mainly, slabs. So the main function is to transfer the vertical load to the lower foundations of the structure. : 

   

There is other component which is known as the strut, it is again a compression member which takes up the compression or also they may be designed to take up the tension, such members are used in the roof trusses.

columns

The major structural difference: Columns have higher slenderness ratio so, due to more slenderness the columns fail  due to buckling in general and the struts fail due to crushing under the action of the compression.

You have to understand the concept of effective length, least radius of gyration and slenderness ratio first to calculate the Euler's buckling load.
Slenderness ratio:

Generally denoted by "r", the slenderness ratio is the ratio of the effective length of the component to the least radius of gyration.

r = L/k

Where L= effective length of a column;( Effective length is the length which participate in the buckling(which actually buckles)

   k = Least radius of gyration, which can be computed by the formula given as,
       k= (Imin/A)^(1/2) , Where A is the Area of cross section of the column

                                        and, I =  Least moment of inertia

     (Note: Least moment of inertia can be taken as the least of the moment of inertia of the cross section about both the axis)

Euler's Formula:

In 1757, mathematician Leonhard Euler created a formula for the buckling load for a column without considering the lateral loads.

Here F is the load under which a column will just start to buckle.

E is the Young's modulus of elasticity of the material of the column.

I is the least moment of Inertia.

K is the effective length factor.

L is the total length of the column
Effective length of a column depends upon the end support conditions of the columns:
a) For both ends pinned effective length is "L". So K = 1
b) For both ends fixed, this is "L/2", so K= 1/2
c) For one end Fixed and other free, this is 2L so, K = 2
d) For one end fixed and other hinge, this is L/1.414 so, K =1/1.414

If you understand the Euler's formula which is applicable to the columns which fail under buckling, then you can easily understand that this formula is not of much use to the struts which are likely to get failed due to crushing.
 In some columns you might have to consider both the factors. Considering the limitation of the Euler's Formula,  Rankine gave a formula, known as the
Rankine Gordon-Formula,
1/(critical load) = 1/(crushing load) + 1/(buckling load)
 = 1/Pc + 1/Pe (Pe = Euler's Load =F ; Pc = Crushing strength of the material of column)
Or,
       Pcr= Pc/[1+ a.(Le/k)^2]
Crushing load can be found by multiplying the crushing strength with the cross section area of the column. Buckling load can be found in the usual manner by Euler formula. Rankine formula is applicable to all types of columns, short as well as long columns.

Reference: