NPTEL >> Courses >> Civil Engineering >> Advanced Structural Analysis (Video) >> Lecture - 1 Review of Basic Structural Analysis I
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 Question: consider a cantilever section twisting moment acts on which place? By: R.ohmprakaash     Date: 2012-01-11 Answer: The distribution of internal forces (bending moments, shear forces, twisting moments, axial forces) in the cantilever obviously depends on the loading (type, magnitude, position). Being statically determinate, this can be determined from static equilibrium. For example, when a concentrated torque T is applied at the free end, a twisting moment T will be generated uniformly at all sections along the span. By: Prof. Devdas Menon     Date: 2012-01-28

 Question: Is there going to be any videos on the lectures about Analysis of elastic instability and second-order effects??? Thank you By: TobiasNA     Date: 2012-01-12 Answer: Yes! This is covered in the last module (module 7) of this course. By: Prof. Devdas Menon     Date: 2012-01-28

 Question: sir..give us even more brief understandin of degree of static indeterminancy and kinematic indeterminancy..explain it with more number of examples By: anith     Date: 2012-02-07 Answer: Please go through the remaining videos on Review of Basic Structural Analysis I and II. Several examples will show up, and you will gain a better understanding. You may also look up the NPTEL course material on "Structural Analysis", and the textbook authored by me (published by Narosa), in which this topic is discussed in great detail. Please note that this understanding is a pre-requisite for this course on "Advanced Structural Analysis". By: Prof. Devdas Menon     Date: 2012-02-10

 Question: sir, are lectures giving a detailed insight into influence lines for determinate and indeterminate structures going to be included in the near future cause its a real necessity?Thank you By: debanjan banerjee     Date: 2012-03-04 Answer: This is a course on Advanced Structural Analysis, and in the review of basic structural analysis (first 18 lectures), influence lines is covered only briefly, in terms of application of Muller-Breslau's principle. For a 'detailed insight' into this topic, you may refer to my book, 'Structural Analysis' (Narosa). By: Prof. Devdas Menon     Date: 2012-03-09

 Question: Hello Prof. Devdas Menon, I enjoy and learn a lot from your lecture recordings. I am wondering if it is possible for you to upload your lecture slides to allow student to follow your lecture easily as the quality of the recording is not that great. By: Raymond Heng     Date: 2012-06-17 Answer: Thank you! i understand that all the videos (recorded professionally in a studio) are available on DVD which can be purchased at a nominal price from NPTEL. It will also be helpful if you get hold of the books referred to in the lectures. By: Prof. Devdas Menon     Date: 2012-06-24

 Question: sir i want know the difference between the nature of destabilizing and normal forces acting on beam (table 15 & 16 of IS 800:2007). could you please elaborate or give any reference By: dhiraj wakode     Date: 2012-06-20 Answer: This has reference to the estimation of 'effective length' in the structural design of steel beams. When the applied loading acts eccentric to the 'shear centre', there is a 'destabilizing' effect. To account for this, the effective length is to be enhanced by 20%. Kindly refer to clause 8.3.2 in IS 800:2007 for further explanation. By: Prof. Devdas Menon     Date: 2012-06-24

 Question: a beam ab length l is simply supported loaded as shown.moment of inertia I for left half and 2I for right half.calculate slope at ends and deflection at midspan and under the point loads By: SHILPA     Date: 2012-10-26 Answer: The magnitudes and positions of the point loads are not specified in your question. This problem can be easily solved by the conjugate beam method, as discussed in the video lecture and the textbook on Structural Analysis (where a similar example problem has been solved). By: Prof. Devdas Menon     Date: 2012-10-26

 Question: thank you sir for providing such fantastic lectures,the classroom environment(unlike many other NPTEL lectures) of these lectures really enhances the learning.I have got the book on structural analysis and was wondering if there could be answers to the back exercises provided anywhere?? By: vipul jain     Date: 2012-11-05 Answer: Thank you! Solutions to all the problems are available with us, and if you have any specific doubt you are welcome to write or email to me. However, the whole idea is to enable students to develop the self-confidence to be able to check their own solutions. There are many alternative ways of solving the same problem... By: Prof. Devdas Menon     Date: 2012-11-12

 Question: there are 2 portal frames ;- (3 member,same dimensions , same properties ).one has both fixed supports at column ends and the other has both hinged supports at column ends. From foundation point of view which portal frame is better ???? By: Digvijay     Date: 2012-11-08 Answer: In the case of fixed bases, there will be bending moments at the column bases, which have to be accounted for in the foundation design. In the case of hinged bases, there will be no bending moments transmitted to the foundations, whose design will be simpler (but higher moments will develop at the beam column junctions, compared to the fixed base case). Whether it is appropriate to model the base as hinged or fixed depends on the soil conditions and type of foundations! It would be inappropriate, for example to assume hinged supports when the foundation is resting on hard strata. But if the foundation is on relatively soft strata, even a small rotation at the base can release any assumed fixed end moment... By: Prof. Devdas Menon     Date: 2012-11-12

 Question: sir can we use finite element method for residential buildings By: Kumar Gaurav     Date: 2012-11-27 Answer: Yes, of course! It can be applied to any idealized structural system -- buildings, bridges, towers, ships, aircrafts, etc. The challenge lies in modelling the system (choosing appropriate elements, boundary conditions, material properties), in applying the loads and in interpreting the results. By: Prof. Devdas Menon     Date: 2012-12-22

 Question: 1Why is the most stable Element is triangular Element in FEM analysis. 2 Why we chose triangular element most of the time in our analysis? By: Kumar Gaurav     Date: 2013-03-03 Answer: You are referring to surface elements (used in plates), as applied in FEM, which is outside the scope of this course on Advanced Structural Analysis (restricted to skeletal elements). Triangulation is also an important concept in skeletal structures. The simplest stable configuration of axial elements is a triangle (pin jointed), and a simple truss is generated by beginning with a triangle and adding two elements for every additional joint... By: Prof. Devdas Menon     Date: 2013-03-15

 Question: brief description on three hinged girder By: Anoop Krishnan     Date: 2013-04-13 Answer: A three hinged arch or frame is typically one that is provided two hinged supports and an additional moment release (internal hinge) in the middle (typically crown of a symmetric arch). If a straight beam or girder is three hinged, it will become unstable! By: Prof. Devdas Menon     Date: 2013-04-26

 Question: Sir what actually the summation of moment of area = 0 signifies ? By: Soumitra     Date: 2013-06-21 Answer: Consider a plane area, such as a rectangle or T-shape. We want to find the centroidal axis of this area, which is also the 'neutral axis' of a prismatic beam (having this cross-section) when it is subject to pure bending. If the beam is subject to sagging, this means that the fibres above this axis are subject to compressive stress, and the ones below are subject to tensile stress. To satisfy equilibrium, the total compression must be equal to the total tension, and the moment of compressive forces acting with respect to this axis must be balanced by the moment of the tensile forces. The second (moment equilibrium) equation, integral of sigma*da = zero reduces to integral of y*da = zero, if we invoke the simple bending theory. Thus the neutral axis corresponds to the centroidal axis, which may be visualized as an axis passing through the 'centroid' of the area, such that the moments of area segments above the axis are balanced by moments of area segments below. By: Prof. Devdas Menon     Date: 2013-06-28

 Question: classification of partition based on materials By: nandu     Date: 2013-09-18 Answer: Your question relates to building materials, not structural analysis... By: Prof. Devdas Menon     Date: 2013-09-21

 Question: A portal frame has a beam and a column,the beam has its one end fixed and the column with its bottom end hinged.The beam column junction is an internal hinge carrying a vertical load on it.will there be any vertical reaction in the fixed end or the total load will be carried by column itself.Thanku By: vasishta saraj     Date: 2013-10-14 Answer: If the column is perfectly vertical (and relatively stiff and not slender), while the beam is perfectly horizontal, and the vertical load is applied exactly at the hinged beam-column junction, then the entire load will get transmitted as an axial compressive force in the column to the bottom hinged support below, and no load will be transmitted to the remote fixed end of the beam, according to 'first-order' structural analysis. [But if the axial deformation in the column is not negligible, then this will induce bending in the beam and a consequent vertical reaction at the fixed end; this requires an advanced 'second order' analysis...]. By: Prof. Devdas Menon     Date: 2013-10-20

 Question: Sir in your book on \"Reinforced Concrete Design\" , in all the examples of beams and slab why you haven\'t checked the design for development length? Is there any specific reason behind this? By: Archan Dave     Date: 2013-11-03 Answer: In the initial design examples you refer to, the focus is on designing, among other things, the flexural reinforcement requirements, with broad detailing rules (which satisfy development length requirements). At the end of chapter 5, the development length requirements associated with reinforcement curtailment are explained with examples (continued in the next chapter on shear). Development length is properly introduced in chapter 8 on bond, with examples. By: Prof. Devdas Menon     Date: 2013-11-09

 Question: In three span continuous beam rotation at B and C due to equivalent joint loads is exactly same rotation at joint B and C due to intermediate loading acting on three span continuous beam. WHY???? By: aditaya bansal     Date: 2013-11-09 Answer: In matrix analysis, we have a discrete set of coordinates, typically corresponding to the degrees of freedom at the various joints in a structure. So, when we have intermediate loads, they need to be converted to equivalent joint loads (acting at these coordinate locations). This equivalence is established by treating the original system as a superposition of (1) a system with all degrees of freedom restrained (all joint rotations arrested in the continuous beam - resulting in fixed end moments) and (2) a system with the rotations allowed to freely occur. Obviously, this requires the rotations in the second case (with equivalent joint loads) to be exactly equal to the rotations in the original beam with the intermediate loading. That's WHY!!!! By: Prof. Devdas Menon     Date: 2013-11-15

 Question: Resp.sir,While generating member stiffness matrix for truss element if we give unit transverse displacement, length of member increases as hypotenuse has more length than original element length. Hence there will be strains in the member.Consqly. there will be internalforces.But we take o in matrix By: Shriniwas Walunjkar.     Date: 2013-12-13 Answer: Yes, you are right. But we are dealing here with very small displacements, and the elongation associated with the hypotenuse is assumed to be negligible. It is only with such assumptions, we are able to simplify the analysis and operate in a linear framework. By: Prof. Devdas Menon     Date: 2013-12-13

 Question: how door hinges are related to hinged joints.what are the forces acting on the hinge when the door opens? By: venkat     Date: 2013-12-15 Answer: The hinge performs the basic kinematic requirement of facilitating the free rotation of the door with respect to a vertical axis through the hinge. It also performs a static requirement of transmitting forces through the hinge. As it provides a restraint against pulling out of the door from the door frame, when a force is applied in this direction on the door, a support reaction is transmitted to the door frame through the hinge. By: Prof. Devdas Menon     Date: 2013-12-30

 Question: I hAVE GONE THROUGH YOUR BOOK ON STRUCTURAL ANALYSIS THANK U SIR FOR SUCH A WONDERFUL BOOK . MY QUESTION IS IN A CANTILEVER BEAM THE MOMENT AT THE FREE END IS ZERO STIL WE USE THE DEPTH OF AT LEAST 150MM THERE WHY ? By: showkat ahmad     Date: 2014-02-26 Answer: In practice, a beam is so designed to ensure that it can resist all kinds of loads that can be reasonably expected. A concentrated load can always act at the free end, causing a shear force to develop at the free end, and so there has to be a minimum depth. This will also give resistance to possible concentrated moment acting in rare cases at the free end. Hence, it is always wise to provide minimum requirements. In another context, don't you carry a minimum amount of money in your wallet when you step outside your house? Why? By: Prof. Devdas Menon     Date: 2014-02-27

 Question: suppose the steel for tension comes out to be in fraction for a beam, usually we provide more steel than what we get in the formula, isn\'t the more steel making the beam over reinforced explain sir please By: showkat ahmad     Date: 2014-02-27 Answer: This question pertains to RC design, not Adv. Structural Analysis! Nevertheless, here is the answer. The section will be over-reinforced only if the steel provided exceeds that required for a 'balanced' section. Usually, the section is under-reinforced (x_u < x_u,lim), and providing slightly more tension steel will not make the section over-reinforced. Even, if the section is balanced, but designed as singly reinforced, the presence of compression reinforcement (hanger bars, usually ignored) will help in making it under-reinforced. If doubly reinforced, then it is desirable to be more liberal in providing compression reinforcement, thereby ensuring x_u < x_u,lim. Further, the presence of a flange in compression (as in a T-beam) will also help. By: Prof. Devdas Menon     Date: 2014-02-28

 Question: For the simply supported beam of length L, subjected to a uniformly distributed moment M kN-m per unit length what will be the bending moment at the mid-span of the beam .it was posted in gate.sir please explain By: showkat ahmad     Date: 2014-03-01 Answer: Under a uniformly distributed moment, 'm' per unit length (kNm/m), the total moment on the beam, 'mL' (say, acting clockwise), has to be balanced by vertical support reactions, equal to mL/L = m (kN), acting downward at the left support and upward at the right support. Taking any section at a distance x from the left support, we find that the shear force S(x) = -m (kN) and bending moment M(x) = -mx + mx = 0 (kNm). The bending moment is zero at all sections of the beam! Please also refer to Problem 5.11 of my book on this topic. By: Prof. Devdas Menon     Date: 2014-03-07
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