Monday, November 25, 2013

A brief introduction to analyze Fixed Arches


Fixed arches are used as bridge arches or as tunnel roofs, where the span is short. Fixed arches restrict the horizontal, vertical or rotational movement at its two ends, so a fixed arch is an indeterminate structure.

As they have total of 6 supporting reactions, 3 at each end, they are indeterminate to a degree of 3. This can be  numerically justified as below:

Total numbers of support reactions:
Total numbers of equations of static equilibrium = r=3

So Indeterminacy,
E = R-r = 6-3 = 3
So, as to analyze a fixed arch we need a minimum of three extra equations, which can be derived from compatibility of the structure.

Here we will use the following compatibility conditions:
The displacement of the joint A with respect to joint B are zero in
(a) x- direction
(b) y-direction
(c) rotational movement (Angular)

These conditions can be used to form three equations.
Use the Castigliano's first theorem, which says that partial derivative of total strain energy of the structure with respect to the load gives us the displacement along the direction of the load.
So you can form the following three equations using this theorem:
(1) Partial derivative of total strain energy with respect to Ha is zero
(2) Partial derivative of total strain energy with respect to Ra is zero.
(3) Partial derivative of total strain energy with respect to Ma is zero.

These three equations along with the other three equations of static equilibrium makes total of six equations.
When they are solved simultaneously, you will get the six unknowns which are Ra, Ha, Ma, and Rb, Hb and Mb.

That is all for now, please write in to make this article better.

Sunday, November 24, 2013

Concept of Plane Strain

I will explain here the concept of the plane strain.
Plane Strain is the strain occurring in a body under the action of the external loads along a plane only. The loads can be three dimensional but the strain must be only in two dimensions.
As you know, when we apply an axial load, there is change in the lateral dimensions also, i.e. if we apply the load in the x-axis of a cubical element, as a result, along with the strain along the x-axis, it is accompanied by the change in the width and height of the element.

This change can be found out by multiplying the longitudinal strain with the Poisson's ratio. So, to get the plane strain we have to restrict the deformation of the element along the one of the two other lateral dimensions.

By restricting the strain along one of the three direction, we get the state of plane strain. Let P, is the axial load along the x-axis, which will result in the longitudinal strain,
ex = P/E

also, there will be lateral strain along the y-axis and z-axis
ey= -1/m(P/E)
and ez = -1/m(P/E)

To get the condition of plane strain suppose we want to restrict the strain along the z-axis, Then will will have to apply a load of -E.ez along the z-axis,
So we have to apply a load equal to 1/m(P) along the z-axis.

Net strain along the z-axis is 0. Only ey and ex will exit with further modification due application of the load along the z-axis.

Thank you!

(Reference: nptel- online video lectures)

Wednesday, November 20, 2013

Shear Center of beams.

Shear Centre is new topic introduced into the syllabus of the 3rd semester of the in Civil engineering by HPTU. So this was a topic was new to me, and I think it might be new to someone like me. So after studying it from the online video lectures offered by Prof. S.K. Maiti from IIT Bombay(nptel lectrures). I thought to form some online notes in writing by myself. So here is what I have prepared till now.

Shear Center is a point on the cross section of a beam at which when shear force is applied there is zero twisting moment in the cross section of the beam. If the load(Shear force) does not pass through the shear center then, the beam gets a twisting moment, so in order to avoid the twisting moment in the beams one must know the location of the shear center of a given cross section.
Consider a channel section subjected a shear force S. The shear stress near to the center of Gravity will be maximum and then its magnitude will decrease towards the ends of the web, similarly it will be maximum near to C.G. in the flange and will decrease when moving away from the C.G.
 To find out the total internal shear forces acting along the flange and the web we have to write use the equation of the shear stress at a point in the beam and then multiply it to a small element of the beam and then integrate it to the whole length of the element.
 Shear stress at a point is given by = S.Q/ Iz.b = S.A.y/ Iz.b
  where  S = Applied shear force
             A = Area under above the point under consideration
             y = Distance of the centroid of the area under consideration from the centroid of the whole cross section
   Iz = moment of inertia of the cross section about the z-axis.
   b= width of the cross section
Nos if you want to find the the value of V in the web, multiply the equation of the shear stress with the a small element of the area and then integrate it for the whole depth of the web.
Similarly for the flanges.
Once the V and F and are known, you have to calculate the amount of internal twisting moment about the center of the gravity of the section which in this case will be = 2F*d/2 + V*z2.
This internal moment has to be balanced by some external twisting moment. So external shear force S has to be applied in such a way that it create the equal twisting moment, but in opposite direction.
So S must be applied at some eccentricity from the centre of gravity such that
S*z1 = 2F*d/2+ V.z2
Here z1 tells us about the location of the shear center. So in same manner you can find out the internal twisting moment for any given section and then you will find out the location of the shear center.
I hope the article was of some help to you. Please leave a comment if you feel that there are some changes to be made.

Thank you!

Tuesday, November 19, 2013

Combined bending and shear and twisting stresses in beams

I will discuss here with you the way to analyse a beam for the amount of maximum normal and shear stresses developed at a point in a beam, which is subjected to combined stresses.

Beams subjected to combined stresses:
Beam subjected to combined normal, bending, shear and twisting stresses
I assume you are familiar with the normal stresses, shear stresses, bending stresses and shear stresses due to Twisting.
Also you might be familiar with the Principal stresses and maximum shear stresses at a point.
If you know all these terms then you have sufficient tools to analyze a beam who is subjected to combined stresses.
 Any point in a beam can be subjected a following stresses at the same time:
(1) Normal stress = Applied axial load/ Cross Sectional Area = P/A
(2) Bending stress = (M/I).y
(3) Shear Stress due to transverse load = (V.Q)/(I.b)
(4) Shear stress due to twist/Torque = (T/J).r
Combine all stresses and convert along the normal stress and
shear stress

If combination of two or more than two of the above stresses arises at a point in a beam, you have find the resultant of all of them along the normal and tangential direction along a face.

Generally normal loading is only in axial direction(x-axis) so the resultants will be only an axial load, and shear stress.

So when you have found the resultants you can easily apply the formula for the principal stress at an element when this is subjected to a normal stress accompanied by a shear stress.

(Reference: An online video lecture on youtube from IITK)

Sunday, November 17, 2013

Double Integration method and Macaulay's method(Structure Analysis)

How you doing?

  • Introduction:

Whenever a beam is subjected to transverse loading, or bending moment, it will deflect to a certain amount depending upon the rigidity of the beam and, the applied moment. Double integration method and Macaulay's methods are used to find out the slope and deflection at any section of a given statically determinate beam.

  • Moment Curvature Relationship and Differential equation of elastic curve:

Consider a beam shown in the figure which is subjected to a pure bending moment of 'M'. This is assumed that the beam has a circular shape when get deflected. Consider the section z-z, let M is the moment at the section. It has been studied in the flexure theory(Theory of simple bending)
M/I = f/y = E/R
1/R =M/EI -----(1)

Consider section at z-z, it can be well figured out that
1/R = -(d2Y/ dx2)----(2)
So from one and 2,
M/EI = -(d2Y/ dx2) ---(3)
equation(3) is known as the differential equation of the elastic curve of deflection of beam.

  • Double integration Method:

If we integrate the eqn(3) once we get dy/dx i.e. slope at any point x distant, and if we further integrate this once more, we get 'y', which is deflection at any point x distant.

After integrating the differential equation of elastic curve of deflection of beam, we get the deflection at the required point.

This must be noted that we have to put the equation of the Mx in the equation(3) and then we have to integrate the equation from o to x(required point). It may happen that the equation of the moment may vary from section to section of the beam, so you have to write different equation of moment for different sections, when the loading/shear force varies.

  • Macaulay's Method:

Macaulay's method is similar to double integration method, only modification is that, there is only one equation written for whole of the span for the bending moment, instead of writing different moment equation for different sections of the beam whenever shear force gets varied.

Reference: Strength of materials by R.K.Rajput

Thanks for your visit!

Friday, November 15, 2013

Approximate Analysis of Trusses/Frames

In this post I will introduce you with the approximate analysis methods for the indeterminate structures.
Introduction: In the approximate analysis we make some assumptions to make the structure determinate and so the analysis is based on those assumptions, so the final results are approximate. This method is different from the exact methods of structural analysis.
While designing any structure there are already a numbers of assumptions used even in the exact analysis, like the homogeneity of material, elasticity, the external loads are assumed but they can never be pre-determined 100% accurately. So talking in general even exact analysis of structure is approximate because there are a numbers of assumptions used.
Generally trusses are of similar geometry. One common truss used is as shown in the figure.

Suppose the external reactions are known, then structure is internally indeterminate to third degree. While designing with the exact analysis the geometry of the members must be known but generally geometry is known to us so we have to use the approximate method. For trusses we can use the following methods:
Method 1: Suppose the diagonal member is long/slender enough, so that they will buckle under the compressive loads. In that case we assume that one of the diagonal member will carry a tensile load, and other carries zero force, and we will design it accordingly.
Method 2: In case we use such members which can carry the compressive as well as tensile force effectively then we can assume that both the diagonal members will carry forces of equal magnitude but one will be tensile and another compressive to take up the shear V at the section as shown in the figure below.
As the two diagonal forces are equal, we can easily determine all the forces at the cut section.
Similarly we can cut a section at other panels of the truss and we can get the forces in the members desired.

Rigid Frame members:
Rigid frames can also be analyzed using the approximate analysis. Rigid frame are generally analyzed for the localized members such as the shown in the figure(a) below. Member AB is fixed to the rigid columns on both ends, so member has a redundancy of degree 3.
So in order to analyze this member locally using the approximate analysis we have to form three assumptions.
If we assume the two columns to be perfectly rigidly fixed to the member AB, then with the earlier experience of the exact analysis, we can say that the inflection points are located at a distance of 0.2L from either point, so the moment values are zero at these points.
Rigid member subjected to vertical load- approximate analysis
Second if the member AB has flexible ends, then beam AB can be analyzed as a simply supported beam.
In actual practice, joints A and B are neither perfectly rigid nor flexible, so we take the average of the two, so we will assume that the inflection points are located at distance of 0.1L from either (i.e. (0.2L+0)/2 = 0.1L).
Further for the vertical loads this can be assumed that the axial force in the member is zero. So up to length of 0.1L from either end beam will be analyzed at the cantilevers and middle part can be considered as the simply supported. Finally the required three assumptions as shown in figure(d) are:
(1) Moment is zero at a distance of 0.1L from left end,
(2) Moment is zero at a distance of 0.1L from right end,
(3) Axial force is zero in the member AB.
using these three assumptions beam can be easily analyzed to find out the bending moment and shear force diagram in the beam member AB.

Portal frames with lateral loads:
Portal frame with lateral loads(ends pinned)- approximate analysis 

Portal frames with lateral loads(ends fixed)- approximate analysis

(Reference:- Structure Analysis by R.C.Hibbler.. wonder fully explained in this book)

Wednesday, November 13, 2013

JNGEC-Strength of MaterialsI- Question Paper for internal session exams as per HPTU syllabus

Following are the questions asked as per syllabus of the subject CE-211, Civil Engineering, HPTU-Shimla for  courses for the internal session exams in the semester August - December 2013 at Jawahar Lal Engineering College of Engineering - Himachal Pradesh

3rd Sem Civil Engineering                                      Date:30-09-2013

 Time: 02 Hrs.                                                                       Course Code: CE-211
  Max Marks: 50                                                                  Course Title: Strength of Materials
Q.1 Explain the following:
(1) Body forces  (2) Contact forces (3) Stress tensor (4) Principal stresses (5) Stress invariants -(5*4)Marks.
Q.2  Briefly explain  the rectangular, cylindrical and polar co-ordinate system for stresses. – (5) Marks.

Q.3 An element in a stressed material has tensile stress of 500 MN/m2 and a compressive stress of 350 MN/m2 acting on two mutually perpendicular planes and equal shear stress of 100 MN/m2 on these planes. Find principal stresses, position of the principal planes and maximum shear stress, using:
(1) Transformation equations.    – (10 Marks)
(2) Mohr’s circle method.     ---(10 Marks).
Q.4 Write short notes on: (1) Differential equations static equilibrium (2) Octahedral stresses. –(5+5)Marks

3rd Sem Civil Engineering                                Date 01-11-2013

 Time: 02 Hrs.                                                                       Course Code: CE-211
  Max Marks: 50                                                                  Course Title: Strength of Materials
Q.1 Write short notes
(1) Transformation equations for strain
(2) Maximum Principal stress theory
(3) Maximum Strain Energy Theory
(4) Secant Formula
(5) Castigliano’s theorem
(1) A cast iron hollow column, of 30cm external diameter and 23cm internal diameter is used as a column 4m long, with both ends hinged. Determine the Rankine’s safe load with a factor of safety of 4. Take compressive strength = 567MN/m2 and  a= 1/1600                    -(15 Marks)

(2)  Using Castigliano’s theorem, determine the deflection of the free end of the cantilever beam shown in figure. Take EI = 4.9 MN.m2.                                                        ---(15 Marks).

JNGEC-Structural Analysis - II- Question Paper

Following are the questions asked as the HPU-Shimla syllabus for Engineering courses for the internal session exams in the semester August - December 2013 at Jawahar Lal Engineering College of Engineering - Himachal Pradesh

5th Semester –Civil Engineering              Date:30-09-2011
Time: 02 Hrs.                                                                       Course Code: CE-5001
  Max Marks: 50                                                                  Course Title: Structural Analysis-II
Q.1 A uniformly distributed load of 1kN per metre run, 6m long crosses a girder of 16m span. Construct the maximum S.F. and B.M. diagrams and calculate the values at section 3m, 5m and 8m from the left hand support.                                                                                   (20 Marks)

Q.2 Two concentrated rolling loads of 12 and 6kN, placed 4.5m apart, travel along a freely supported girder of 16m span. Sketch the graphs of maximum shearing force and maximum bending moment and indicate the position and magnitude of the greater value.                         (20 Marks)

Q.3 What is an influence line diagram? Two wheel loads of 16 and 8kN, at a fixed distance apart of 2m, cross a beam of 10m span. Draw the influence line diagram for bending moment and shear force for a given point 4m from the left abutment, and find out maximum bending moment and shear force at that point.                                                                               (2+ 8 Marks)

5th Semester –Civil Engineering                                    Date :01-11-2013
Time: 02 Hrs.                                                                       Course Code: CE-5001
Max Marks: 50                                                                  Course Title: Structural Analysis-II
Q.1  Give introduction:
            (1) Kani’s method
             (2) Matrix Method
                                                                    (5*2 Marks)

(1)  Draw the bending moment diagram and deflected shape of the frame shown using Kani’s Method.                                          (20 Marks)

(2)  Using force method, analyse the continuous beam shown in figure treating the support reaction at C as the redundant. Hence calculate the support reaction at B.  

(20 Marks)

Thursday, November 7, 2013

Castigliano's Theorem, Maxwell Betti's Reciprocal Theorem

How you doing?

If you are a Civil engineering student then these two theorems are important to understand. These both theorems are based on the energy principles.

 You can browse a numbers of results after you search them in the Google or you can find them in any books available on structural analysis.

Castigliano's first theorem and second theorem are used to determine the deflection and force respectively at the given points on a given structural system . 

Maxwell's reciprocal theorem is quite important because it allows you to skip a number of calculations when the relevant conditions arise. Let me explain a little more:

  • Castigliano's 1st Theorem: According to this theorem, partial derivative of the strain energy of a structural system with respect to a point load at the given coordinate gives us the amount of deflection along that coordinate.

  • Castigliano's second Theorem: According to his second theorem, partial derivative of strain energy with respect to deflection at the given coordinate gives the corresponding force.                                                                                     
To approach at the solution i.e. either force or deflection, first you have to write down the equation of strain energy of the system in terms of the force or deflection(to determine deflection and force respectively) and then you have to apply either first theorem or second theorem, according to the desired results. 

example:  application of Castigliano's theorem to solve for the Deflection and force in a Beam: 

  • Maxwell Betti's Reciprocal Theorem: According to this law the amount of work done by first load system due to displacements due to second load system, is equal to the amount of work by the second load system due to displacements due to first load system at their respective co-ordinates. 
In other words, if a load W is applied at A resulting a deflection "d" at B, then if same load W is applied at B, it will give the same deflection "d" at A.

This theorem can make you skip the calculations because if you know the deflection at B due to load at A, and you are asked about deflection at A due to load at B, you already know it, so why to do the calculations?
These theorems are really important to understand. I told same thing to my student in the class.

Note:  If you love the article please write a comment, or like it, and any suggestions or questions are most welcome.
Thank you.