Hi can you give ma a price for these questions, and let me know if you can do them in time. Also a good book to study is Understanding Structures Analysis, Materials, design by Dereck Seward 5th edition
BLAR12054 Assessment Item 2 - 2020.pdf Figure 2 (not to scale) Topic 2 - Page 1 Part 1: Topic 2 – The analysis of indeterminate structures Introduction This topic is an introduction to the analysis of indeterminate frames, carrying both vertical and horizontal loads using the ‘moment distribution method’, and from these results the development of a simplified ‘portal method’ for the approximate analysis of symmetrical, rigid framed, multi bay buildings This topic includes a description of the effect, and analysis, of sway in rigid frames. Learning outcome By successfully achieving the stated ‘enabling objectives’ for this topic, you should be able to describe the: the analysis of indeterminate structures by rigorous and approximate methods. Enabling objectives You may be assessed on the following objectives. These objectives should enable you to achieve the learning outcome stated above, descriptions of: the analysis of rigid frames for the ‘no sway’, and single sway mode situations the description of the methods of analysis for multi sway cases the development of a technique for the approximate analysis of symmetrical, multi bay rigid frames. What you will need Suggested study time Study guide Activities and exercises Total 12 hours Other resources Textbook Seward 2014 pp. 305-309 Reading 2-1 The Steel Construction Institute 1992 pp. 1036, 1080, 1081, 1083, 1085 Reading 2-2 Warner, Rangan, Hall, & Faulkes 1999 pp. 794–801 The resource material and a small collection of pins, drinking straws, cotton, cardboard and if possible a sheet of foam rubber, 25 mm thick by 300–450 mm square. Topic 2 - Page 2 The structural analysis of frames General All building frames must be stable under all loading conditions with their components possessing sufficient strength to resist the loads applied and sufficient stiffness to ensure that deflections are controlled. The stability is achieved by component configurations and the joint systems employed. The joint systems may be either of ‘pinned’ (jointed), as in analysis of statically determinate trusses (Topic 5, course BLAR11032 Structural Forms and Analysis (later referred to as SF), or ‘rigid’ (jointed). Many pin jointed frames, (but not all) are statically determinate, that is, they may be solved by the direct application of the three equations of equilibrium,: in the case of rigid jointed frames most (but not all) cannot be solved directly by the three equations of equilibrium, and are said to be statically indeterminate. Determinacy is described SF Topic 3 and, in general terms, in the textbook, pages 305-309, where many common frames and structures are described. A common example of a statically determinate rigid frame is the ‘three hinged arch’, described in Section 12.3 of the textbook, and students are directed to rework Examples 12–1 and 12–2 from the textbook, to ensure that they understand fully the analysis, and calculation of the bending moments, shear forces, and axial forces that occur in the examples shown. This topic ‘the analysis of indeterminate frames’, describes the: methods of analysis for indeterminate frames, with examples worked in using the moment distribution (or ‘successive approximation’) method the general introduction to the concept of ‘sway’ in rigid frames some standard solutions for the analysis of simple rigid frames an approximate method of solution for single story and multi-storey frames. The analysis of statically indeterminate frames Indeterminate structures are those which cannot be solved directly by the three equations of equilibrium. For example, in Reading 1-3, page 18, the frame ‘b’, has four external reactions, (two vertical and two horizontal) and thus cannot be solved by the three fundamental equations of equilibrium. Similarly, the frame ‘c’, on the same page, has six external reactions (two vertical, two horizontal and two moments) and similarly cannot be solved by the three equations of equilibrium. The frame, in the first case has one additional external force than the minimum required for equilibrium and is said to be ‘indeterminate to the first degree’, and second case is ‘indeterminate to the third degree’. In the first case, one external force needs to be determined prior to the application of the three equations of equilibrium, and the second case three external forces need to be determined before the three equations of equilibrium can be applied. Topic 2 - Page 3 Now, there are several traditional methods which can be used to solve these external forces, including the displacement method, the flexibility method and the moment distribution method. All, frankly, are tedious to apply from first principles; however, it is important that students should understand how one method, at least, is applied, in order to understand the ‘approximate (portal) method’ which forms a major thrust of this component. This course uses the ‘moment distribution method’ to analyse a simple rigid frame. The following sections show the application of the methodology to acontinuous beams and a non-sway simple rigid frame; students may care to observe that a ‘frame’, for bending analysis alone, is identical to a three-span beam, but with the end spans rotated through 90 degrees. Moment distribution method This section describes a method that is often used for calculating shear forces and bending moments in frames. However, the method described in this section should only be used when considering frames where sway is prevented. This means that the frame is constrained in some way, or the loads and frame are symmetrical, resulting in no significantly sideways deflection. Frames where sway is not prevented require additional calculations that are outside the scope of this course. In this course, we will only be using this method for calculating moments in continuous beams. This is just a particular case of a frame where sway is prevented. Moment distribution is a method of successive approximation. Initially, we fix all the joints of the frame. The moments at these joints are calculated from standard tables, assuming that each member is a fixed-end beam with the design loads applied. Any joints that are not fixed are then released. The moments are successively distributed across the members according to the relative stiffness of those members. Note that in this method the reactions are calculated after the bending moments. Example Find the reactions and bending moments for the points A, B and C on the following beam, assuming it does not change section (i.e. I is constant for the whole beam): Figure 2-1: Loads on indeterminate two span beam First, we assume that joint B is fixed (we do not fix a free end like C), and calculate the fixed-end moments: A B C 6 kN/m 2 kN/m 4 m 5 m Topic 2 - Page 4 The beam AB is built-in: ?? = − ??2 12 = −6?(4?)2 12 = − 8??? ?? = + ??2 12 = 6?(4?)2 12 = + 8??? The beam BC is a propped cantilever: ??? = − ??2 8 = 2?(5?)2 8 = −6.25??? The moment at C is zero. Now we need to find the relative stiffness of the beams AB and BC. The stiffness factor K is defined as: L I K for a built-in beam And L I K 4 3 for a propped cantilever Therefore I I K AB 25.0 4 And I I KBC 15.0 54 3 (Remember I is constant) Therefore, we can calculate the distribution factors: 625.0 40.0 25.0 15.025.0 25.0 .. II I FD AB and 375.0 40.0 15.0 15.025.0 15.0 .. II I FD BC Note that the distribution factors of all members meeting at a joint always add to 1.0. Of course, you can see that the moments on each side of the joint B are not equal. On the AB side, moment is + 8 kNm while on the BC side, moment is - 6.25 kNm. This is because we have assumed the joint B is fixed. Once we release the joint B, the moments will need to be distributed so that B ends up Topic 2 - Page 5 having an equal moment both sides of the support. When we distribute the unbalanced moment, one-half of the moment will carry over to the fixed-end A. Let’s see how it works: Table 2-1: Distribution table A B C Distribution Factors 0.625 0.375 Fixed End Moments -8.000 +8.000 -6.250 0 Distribution -1.094 -0.656 Carry Over -0.527 Final Moments -8.547 +6.906 -6.906 0 Now follow the steps: Enter the distribution factors and fixed-end moments into the table. Note that the fixed-end moment of 8 kNm is entered as -8 at the joint A and +8 at the joint B. Release the joint B. Find the required moment adjustments as follows: o Distribution AB = - 0.625 x (8 – 6.25) = -1.094 o Distribution BC = - 0.375 x (8 - 6.25) = -0.656 Figure 2-2: Loads on first span Now carry over the moment to the fixed-end A. o Carry over A = -1.094 2 = -0.547 All moments are now balance so we can add up the moments to get the final moments. Now that we have found the moments at A and B, we can find the reactions at A, B and C. All we need to do is look at the free-body diagram of section AB and section BC of the beam: 0BM − ?? ? 4? − 8.547 + 6 ?? ? ?(4?)2 2 + 6.906 = 0 RA = 11.6 kN Topic 2 - Page 6 0AM ??1 ? 4? − 8.547 − 6 ?? ? ?(4?)2 2 + 6.906 = 0 RB1 = 12.4 kN Figure 2-3: Loads on second span 0CM − ??2 ? 5? + 2 ?? ? ?(5?)2 2 − 6.906 = 0 RB2 = 3.6 kN Therefore