Engineering science lesson 6 2

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  • can you suggest me a suitable universal column to withstand a load of 2MN having a length of 3m and be pinned at both ends ?
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Engineering science lesson 6 2

  1. 1. Chapter 1- Static engineering systems 1.2 Beams and columns 1.2.1 elastic section modulus for beams 1.2.2 standard section tables for rolled steel beams 1.2.3 selection of standard sections (eg slenderness ratio for compression members, standard section and allowable stress tables for rolled steel columns, selection of standard sections) 1
  2. 2. Standard sections• Standard section handbooks give values of section modulus for different section beams• Such data and guidance on the design of structures are given in British Standards and relevant ones for design using beams are BS 4 Part I:2005 (structural steel sections) and BS 5950 Part I:2000 (structural use of steelwork in building, code of practice for design, rolled and welded sections)• Example type of data sheet is illustrated on the next slide 2
  3. 3. • In selecting beams required section modulus can be obtained by dividing the maximum bending moment by allowable stress in the material, allowing for a suitable factor of safety 3
  4. 4. • If the allowable stresses are the same for both tension and compression, then a cross-sectional shape which has its neutral axis at the mid height of the beam selected• To minimize the weight of a beam, abeam can be selected that has not only has the required sectional modulus but also the smallest cross-sectional area and the smallest mass per unit length• The critical buckling stress depends on the material concerned and how a slender a column is. The measure of slenderness used in the slenderness ratio. where the effective length is essentially the length between points at which the member bows about. (eg: both ends fixed effective length is 0.70L, One end fixed and other end pinned 0.85L, vertical cantilever 2L) 4
  5. 5. Radius of gyration 5
  6. 6. 6From BS 5950 Part I:2000
  7. 7. • The maximum slenderness ratio for steel columns carrying dead and imposed loads is limited to 180; as table and graph indicates, when the slenderness ratio approaches this value the compressive strength drops to a quite a low value.• If in the design of structure, a larger slenderness ratio is indicated and it indicates the need to use a larger section size in order to decrease the slenderness ratio• The general procedure for designing axially loaded column is – Determine the effective length of the required column – Select a trial section – Using the radius of gyration value for the trial section, calculate the slenderness ratio – If the slenderness ratio is greater than 180, try a large cross section trial section – Using the slenderness ratio, obtain the compression strength from the tables – Compare the compression strength with that which is likely to be required in practice. If the compression strength is adequate then the trial section is suitable. If the compression strength is much greater than is required then it would be more economical to choose a section with a smaller radius of gyration. 7
  8. 8. 8
  9. 9. Standard Rolled Steel Shapes 16”W eF n e id l g a Te e A e i a Sa d r m c n t n ad r Ca nl hn e Age nl W x6 1 2 6 W 1 x3 T 31 S0 6 2x6 C2 2 1x5 Lx x / 5 3 12 W 16 x 26 TYPE Weight per Nominal Depth Linear Foot 9
  10. 10. Steps to design beams for bending1. For a given loading determine draw the Bending Moment Diagram (BMD) and determine Mmax.2. Specify the type of material to be used and determine the allowable bending stress σall based on the ultimate stress σ ult and a selected factor of safety FS (e.g.: σ all = σ ult/ FS).3. Determine the required cross section modulus S using equation (1) that will ensure that σ max is NOT more than σ all. S = Mmax / σ all … (1)4. For standard cross section (including I-shaped, S-cross sections, Wide-flange cross sections), some design tables often tabulate the values of section modulus for each shape. The design will then only requires selecting a suitable economical shape (e.g.: WF-shape or I-shape) such that the selected cross section will ensure that Sprovided ≥ Srequired … (2)4. Then if inequality (2) above is satisfied, the beam stresses will also satisfy the following design requirements σ max ≤ σ all … (3) As the design process is often iterative (repetitive), the above steps might have to be repeated till the requirements of inequalities 2 and 3 above are satisfied and the cross section is economical to construct with minimum cost 10 (very often it is the cross section with minimum weight).
  11. 11. Example• Select a standard rolled steel I-section for the simply supported beam shown in Figure. A factor of safety of 6 is to apply and the ultimate tensile strength of the material is 500 MPa. The selected section must have the least possible weight. The weight of the beam itself may be neglected when calculating the maximum bending moment 11
  12. 12. • Finding support reactions: The beam is symmetrically loaded and so the support reactions R A and RC will be equal. That is, RA = RC = 50/2 RA=RC=25 kN• Finding maximum bending moment, M which will be at the centre of the beam: M = RA X 2 = 25 X 103 X 2 M =50X 103Nm• Finding maximum allowable stress, σ:• Finding elastic section modulus, s: s s 12
  13. 13. • Multiply by 106 to convert section modulus to cm 3: s s• Finding section with next higher value and least mass per meter from standard section table: s 13

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