Degree of indeterminacy of trusses

The derivations of degree of indeterminacy of  static and kinematic type are derived in the following videos in two parts. Also various methods for analysis of structure is also discussed.
Click here for first part

Click here for second part

The stablity of simple, compound and complex trusses

The stablity of simple, compound and complex trusses. Click here to watch the video.

A good book to start learning the subject, Structural Analysis by Pearson https://www.amazon.in/dp/9332586144/ref=cm_sw_r_cp_apa_i_v0vNDbFVJC35N.

The demonstration of determination of structural stability by experimental method

The demonstration of determination of structural stability by experimental method. Click here to watch the video.

Supports of a structure

In this video you'll understand what are the various types of supports, and how they offers their support reactions in the their response to structural loads. Click here to watch the video.

Role of equilibrium equations in structural analysis

Click here to watch the video to understand the role of equilibrium equations for structural analysis.

Solved problem on slope deflection method

Solved problem on slope deflection method.

Click Here

Udemy Test Series

Four test series which are specially designed to improve and check understanding of important concepts of degree of indeterminacy of structures. Click Here to take this mind blowing challenge.

Kinematic approach for analysis 'Displacement Method'

Kinetic approach of analysis is also known as displacement method of analysis in which the reactive forces and moments are treated as dependent variable in terms of displacements and rotations of nodes of the structure to idealise the structure with the help of relationships derived based upon material and geometrical  properties. Since the deformation of the nodes of the structure functions as independent variable in the equations relating reactions with deformations, the concept of kinematic degree of indeterminacy helps to judge the difficulty of obtaining solutions from the equations, which further will help to complete free body diagram and latter on with the help of application of method of sections, the internal forces and moments will be evaluated. These internal forces and moments will induce stresses and strains which can solved by applications of solid mechanics. For analysis of statically indeterminate structures with the help of kinematic approach, redundant support reactions are expressed in terms of nodal rotations and deflections associated with stiffness coefficients; therefore this method is also popular as stiffness method. There are two popular methods which works on this approach:

1. Slope-Deflection method
2. Moment Distribution method

Method of Section (A method for evaluating internal forces and moments)

When a structural body is subjected to external load, it gets deform. Externally, the  structure remains static as the external load vectors gets counterbalanced by the support reaction vectors and maintain its state of equilibrium. The method of section is a very powerful method to understand that how does the external load vectors and support reaction vectors deforms the structure. The internal force can be axial force, shear force, bending moment and twisting moment.

General steps required to determine the variations of internal forces in the structural member are as follows:

1.       Determine the all of the support reactions.

2.       Draw the free body diagram.

3.       Establish the coordinate system in such a manner that entire structure occupies first quadrant.

4.       Cut a section on the structure at a variable distance from origin.

5.       Write expressions for internal loading in terms of variable distance with using sign conventions.

6.       Go back to step 4 and step 5 to cover entire structure for accounting variations of internal loading due to change in loading pattern.

7.       Plot the results graphically.

The knowledge of equations which gives relationships in between internal loading and their corresponding stresses gives facility to the designer to identify the variations of stresses in the structure. Since strength of the structure is the minimum value of stress up to which structure remains serviceable and hence the concept of method of section is very important to understand the behaviour of the structure.

Note: Reader should not be confused in method of section of finding internal forces with method of section of finding member forces for trusses.

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Last Edited on 2017-12-13

Static approach for analysis 'Force method'

If the static degree of indeterminacy (Ds) of the structure is less than the kinematic degree of indeterminacy (Dk) of the structure, the application of static approach will give much faster results than kinematic approach. The statically determinate structures (when Ds = 0) the process become much much easier as just by the application of equilibrium equations of forces and moments in the orthogonal coordinates will give the value of support reactions. Then these support reactions will be used to evaluate the variations of internal forces and moments within the structural body and that will be use to estimate the values of stresses and strains.

    The force method was originally developed by James Clerk Maxwell in 1864 and later refined by Otto Mohr and Heinrich Müller-Breslau.*

    The equation can be written in two forms. First for relating displacements with forces and second for relating rotations with moments.

∆ = f  * F

Ө = f  * M

where; 

∆ is displacement;

Ө is displacement;

 f is flexibility coefficient;

 F is force and;

M is moment. 
   To counteract the static degree of indeterminacy of the structure the addition equations are generated by substituting the values for displacements and rotations by splitting the structure in a determinate structure and same structure with the redundant reactions and then superimposing the results, in this way redundant support reactions workout. In next step the remaining support reactions are evaluated by using equilibrium equations of forces and moments. 

*page number 385, sixth edition, R. C. Hibbeler, Structural Analysis, Pearson publication. 

Betti's Law

Betti's theorem states: For a structure acted upon by two system of loads and corresponding displacements, the work done by the first system of loads moving through the displacements of the second system is equal to the work done by this second system of loads undergoing the displacements produced by the first load system.

https://en.wikipedia.org/wiki/Betti%27s_theorem

F1 * D2 = F2 * D1

                 D1= (F1/ F2) *D2

Example:

Solution:

    The differential equation also known as Eular-Bernaulli equation  which is used to determine the small deflections of beam is;

EI y'' = - M

    This equation relates elastic curvature of beam with bending moment. The sign convention for this equation is as follows;

    positive x axis is towards right and;

    positive y axis is towards downward

    

    So, using this equation one can write a equation for deflection by integrating twice which will have two integrating constants and can be evaluated by substituting known boundary conditions of slope and deflection. Further, when the equation of bending moment is written in multiple expressions for entire domain (beam span), the continuity equations of slope and deflections are used to evaluate the coefficients of integration.

    In this particular example, if we directly apply this method it will take larger time, since the equation of bending moment will have four equations which will get eight integration constants to get evaluate in order to get solved to write equation for deflection.

    So, lets do some engineer's hack to get rid of this riddle. 

    And the key trick is, Betti's law. 

Wow!!!

But,

How???

   Let's do it;

Step 1: Referencing points. 

Step 2: Defining first system of loading and displacement.

Step 3: Defining second system of loads and displacements. The choice for positioning second system of loads is not fixed and user can put them on those places where it's easier to solve the equations and get result; however the  positions of second system of displacements has to match the reference points.

Step 4: Now you can write equation for deflection quit easily.

                  M = - P ( L - x )

And using Euler-Bernaulli's equation for elastic curve 

EI y'' = - M

                   = P L - P x

                                    EI y' = P L x - ( P x^2 ) / 2 + C1

                          

                                                                EI y = ( P L x^2 ) / 2 - ( P x^3 ) / 6 + x C1 + C2

Using boundary conditions at x = 0; y' = 0 and at x = 0; y = 0, we'll obtain, C1 = 0 and C2 = 0,

So,

                                                                          EI y = ( P L x^2 ) / 2 - ( P x^3 ) / 6  

    DA = (11 /384 EI )  P L^3

  DB = ( 5 / 48 EI ) P L^3

       DC =  ( 27 / 128 EI ) P L^3

DD = ( 1 / 3 EI ) P L^3

Step 5: Using Betti's Law, and substituting F1, F2 and D2 and obtaining D1 as follows:

                            F1 = P + 2 P + 3 P + 4 P

          = 10 P

F2 = P

                               D2 = DA + DB + DC + DD

                                                 = ( 260 /384 EI )  P L^3

So;

                                                     D1 = ( 10 P / P ) ( 260 /384 EI )  P L^3

                                          = ( 2600 / 384 EI ) P L^3

    

Maxwell's Reciprocating Theorem

The conclusion of Maxwell's reciprocal theorem for elastic structures is that the deformation (rotation or deflection) of point A; when load is placed at point B; is equals to the deformation of point B when that load is placed at point A. 

    Example: Consider two cantilever beams with same flexural rigidity (EI) as shown below.

An alternative note on degree of indeterminacy for structure for better grasp of the important concept

Before explaining this concept again, I need few question to be answer, just few algebraic expressions to solve to get a unique answer. Here it is,

Q1. Determine the value of x from the following equation

5 x + 15 = 45

Q2. Determine the value of x and y from the following equation

2 x + y = 5

That's it. 

    You can confidently say that the value of x in question Q1 is 6 with a great confidence  but when it comes the Q2, you can not have a unique answer, one can say for its x = 2 and y = 1 while second one can say that its x = 2.5 and y = 1 and third one can say its x = 0 and y = 5 and so on. In other words, you obtain a unique answer for Q1 and infinite number of answers for Q2. 

    Now, let me modify the Q2 and name it as Q3. Here it is,

Q3. Determine the value of x and y from the following equation

2 x + y = 5

     x + y = 3  

Now, for Q3 you can say that the answer is x = 2 and y = 1 with the same confidence as for Q1.

The idea for obtaining degree of indeterminacy for structure is same. For example when you evaluates static degree of indeterminacy for a structure you make a judgement will you certainly obtain the values of support reactions by using equilibrium equations for provided loading conditions.

A special type of structure

Fact:

    You know for any beam, frame or arch if you add one loop for each of its constituting structural part you will end up with a structure which will have its static degree of indeterminacy equals to number of its support reactions. I mean for these type of structures you will get

                                                                      

Ds = r

Wow!!!

But

How???

Proof:

    Did you remember the equations I obtained by just adding the expressions for static and kinematic degree of indeterminacy, in case you forgot click here. 

Ds + Dk = f ( j + l - n)

    where; f is the number of degree of freedom;

                j is the number of joints;

                l is the number of loops; and

                n is number of structural parts 

Since we said that for these particular type of structures number of loops in them is equals to the number of its structural parts, we can say 

l = n

So the expressions for Ds + Dk will be reduced to,

Ds + Dk = f * j

And we know that Dk = f * j - r 

Now substituting its value we'll have,

Ds + ( f * j - r ) = f * j

So we'll end up with

Ds = r

Examples:

Which method will take less time, the Force Method or the Displacement Method?

   There are basically two approaches to to analyse the structure, namely static approach and kinematic approach. Static approach is based on force method of analysis; dealing with the flexibility coefficients on other hand in kinematic approach deals with stiffness coefficients and practiced as Displacement Method. 

    So now the question is that which method will save your more time while approaching varieties of problems manually? In other words, which approach is better to be followed between static approach and kinematic approach prior to evaluation of internal forces ( axial forces, shearing forces, bending moments, twisting moments) and internal deformations (axial strains, lateral strains, shearing strain, slope, deflections, twists)? And the answer is, which way you will face lesser calculations more technically the approach with lesser degree of indeterminacy,  this means that if you are getting static degree of indeterminacy lesser than kinematic degree of indeterminacy prefer static approach (Force Method), if you are getting kinematic degree of indeterminacy less than static degree of indeterminacy prefer kinematic approach ( Displacement Method) and you will end up with less number of equations to work out. You might have experience this behaviour when you'll solve any statically determinate structures (either static or kinematic).

Static Indeterminacy Problems (Trusses)

Problem Set:

Solution Set:

Problem Number	1	2	3	4	5	6	7	8
Degrees Of Freedom (f)	2	2	2	2	2	2	2	2
Number Of Joints or Nodes (j)	7	21	9	15	12	9	5	9
Number Of Parts Involved (n)	11	45	16	30	23	13	5	10
Number Of Reactions (r)	3	3	3	5	5	4	4	8
Static Degree Of Indeterminacy (Ds=(n+r)-f*j)	0 (Statically Determinate Structure)	6	1	5	4	-1 (Unstable structure)	-1 (Unstable structure	0 (Statically Determinate Structure)
Kinematic  Degree Of Indeterminacy (Dk= f*j-r)	11	39	15	25	19	14	6	10

Static Indeterminacy Problems (Arches)

Problem Set:

Solution Set:

Problem Number	1	2	3	4	5	6	7	8
Degrees Of Freedom (f)	3	3	3	3	3	3	3	3
Number Of Joints or Nodes (j)	2	3	4	2	3	4	4	6
Number Of Loops (l)	0	0	2	0	0	0	0	0
Number Of Parts Involved (n)	1	2	1	1	2	3	3	5
Number Of Reactions (r)	6	8	5	4	6	8	10	14
External Static Degree Of Indeterminacy (Dse= r–f*n)	3	2	2	1	0	-1 (Unstable Structure)	1	-1 (Unstable Structure)
Internal  Static Degree Of Indeterminacy (Dsi= f*l)	0	0	6	0	0	0	0	0
Static Degree Of Indeterminacy (Ds=Dse+Dsi)	3	2	8	1	0 (Statically Determinate Structure)	-1	1	-1
Kinematic  Degree Of Indeterminacy (Dk= f*j-r)	0 (Kinematic  Determinate Structure)	1	7	2	3	4	2	4