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ME31B: CHAPTER FOUR

ME31B: CHAPTER FOUR. DESIGN OF STRUCTURAL MEMBERS. DESIGN OF MEMBERS IN DIRECT STRESS:. Structural members under direct stress are mainly ties, cables, and short columns. Solution Concluded. Design of Short Columns. Example.

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ME31B: CHAPTER FOUR

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  1. ME31B: CHAPTER FOUR DESIGN OF STRUCTURAL MEMBERS

  2. DESIGN OF MEMBERS IN DIRECT STRESS: • Structural members under direct stress are mainly ties, cables, and short columns.

  3. Solution Concluded

  4. Design of Short Columns

  5. Example • A square concrete column(pier) which is 0.5 m high is made of a nominal concrete mix of 1:2:4, with a permissible direct stress of 5.3 N/mm2. What is the required cross-sectional area if the column is required to carry an axial load of 300 kN? • Solution: • Area = Force/stress = 300 x 103 N = 56600 mm2 5.3 N/mm2 • ie. the column should be minimum 240 mm square. 

  6. Note • Maximum compressive stress (fc) occurs in the section where the bending moment is maximum. In the design of simple beams, section modulus (Z) should be selected such that fc does not exceed the allowable value. Allowable working stress values can be found in building codes or Engineering handbooks. • For safe bending, fw > f = Mmax/Z where fw is the allowable bending stress; f is the actual stress and Mmax is the maximum bending moment.

  7. b)Deflection of Beams • Excessive deflections cause cracking of plaster in ceiling and can lead to jamming of doors and windows. • Most building codes limit the amount of allowable deflection as a • proportion of the member's length ie. 1/180 , 1/240, or 1/360 of the length.

  8. 4.2.1 Steps in the Design of Simple beams a) Calculate loading on the beam • b) Calculate the bending moment, shear forces, etc. • c) For the maximum bending moment (Mmax), provide a suitable section: • Z = Mmax/fw , fw is allowable stress (from tables). • d) Check for shear stress, deflection and buckling of web if necessary.

  9. Example • Consider a floor where beams are spaced at 1200 mm and have a span of 4000 mm. The beams are seasoned cypress with the following properties: fw = 8 N/mm2 , E = 8400 N/mm 2 , density = 500kg/m3 . Loading on floor and including floor is 2.5 kN/m2 . Allowable deflection is L/240. Design the beam. Allowable shear stress is 0.7 N/mm2

  10. 4.2.2 Bending Moment Caused By Askew Loads • When the resulting bending moment on a beam is not about one of the axis, the moment need be resolved into components acting about the main axis. • The stresses are then calculated separately relative to each axis and the total stress is found by adding the stresses caused by the components of the moment.

  11. Roof Truss Showing Purlin

  12. Example • Design a timber purlin, which will span rafters 2.4 m on centre. The angle of the roof slope is 30 ° and the purlin will support a vertical dead load of 250 N/m and a wind load of 200 N/m acting normal to the roof. The allowable bending stress(fw ) for the timber used is 8 N/mm 2 . The timber density is 600 kg/m3

  13. Forces in the Purlin

  14. Solution • Assume a purlin cross sectional size of 50 x 125 mm • i) Find an estimated self load • w = 0.05 m x 0.125 m x 600 kg/m3 x 9.81 = 37 N/m • Total dead load = 250 + 37 = 287 N/m • ii) Find the components of the loads relative to the main axes • wx = 200 N/m + 287 N/m cos 30 ° = 448.5 N/m • wy = 287 N/m sin 30 ° = 143.5 N/m

  15. Solution Contd. • iii) Calculate the BM about each axis for a udl. The purlin is assumed to be a simple beam • Mmax = w L 2 /8 • Mmax x = wx L2= 448.5 x 2.42 8 8 • = 323 x 103 N mm • Mmax y = wy L2 = 143.5 x 2.42 • 8 8 • = 103 x 10 3 N mm

  16. Solution Contd. • iv) The actual stress in the timber must be less than the allowable stress. • f = Mmax x + Mmax y < fw Zx Zy • v) Try the assumed purlin size of 50 x 125 mm • Zx = b d2 = 50 x 1252 = 130 x 10 3 mm3 6 6

  17. Solution Concluded

  18. 4.2.3 Design Using the Universal Steel Beams

  19. Example • A steel beam used as a lintel over a door opening is required to span 4.5 m between centres of simple supports. The beam will carry a 220 mm thick and 3.2 m high brick wall, weighing 20 kN/m3 . Allowable bending stress is 165 N/mm2. Assume allowable shear stress of 100 N/mm2, E is 2 x 105 N/mm2. Assume self weight of the beam as 1.5 kN.

  20. With Z = 221 cm3 and I = 2070 cm4 , From SDM, choose a UB 254 x 102 x 22 with Z = 225 .4 cm3 and I = 2863 cm4Check for shear stress: Shear stress = Qmax D. tFrom Table 4.4. Q max = W /2 = 32.43 kNShear stress = 32.43 x 103 N = 22.01 N/mm2254 mm x 5.8 mmSince 22.01 N/mm2 < the allowable shear stress of 100 N/mm2 , ie. Beam is very satisfactory.USE UB 254 x 102 x 22

  21. Composite Beams

  22. 4.3 DESIGN OF COLUMNS • Columns are compression members but the manner in which they tend to fail and the amount of load which causes failure depend on: • i) The material of which the column is made eg. a steel column can carry a greater load than timber column of similar cross-sectional size. • ii) The shape of the cross-section of the column. A column having high c/s area compared to the height is likely to fail by crushing rather than by buckling.

  23. Columns Contd. • iii) The end conditions of the column. • To account for buckling of slender columns, the allowable compressive strength is reduced by a factor k , which depends on the slenderness ratio and the material used. • Pbw = k . cw . A where Pbw is the allowable load wrt buckling; • k is the reduction factor which depends on the slenderness ratio and • A is the cross-sectional area of the column.

  24. 4.3.2 Slenderness Ratio • Slenderness ratio can be defined as: • = k L = l • r r  • Where is slenderness ratio; • k is effective length factor whose value depends on how the ends of the column are fixed; • L is the length of the column; r is the radius of gyration (r = I/A and l is the effective length of the columns (k. l)

  25. 4.3.3 Types of End Conditions of a Column: • Columns can either be • (a) fixed in position nor direction (the weakest condition); • (b) fixed in position but not in direction (pinned); • (c) fixed in direction but not in position • (d) fixed in position and in direction

  26. 4.3.4 Design of Axially Loaded Timber Columns • Timber columns are designed with the following formulae: • = k L and Pbw = k . cw. A r • NB: In some building codes, a value of slenderness ratio in case of sawn timber is taken as l/b instead of l/r

  27. Example • Design a timber column which is 3 m long with a compressive load of 15 kN. Allowable compressive stress ( cw) for the timber is 5.2 N/mm2, kp is 1.00

  28. Note • Actual load/allowable load = • 15 kN/16.9 kN = 0.89 • This ratio is all right and shows that the section is economical. • A ratio of 0.85 to 1.00 is acceptable.

  29. 4.3.5 Design of Axially Loaded Steel Columns • The allowable loads for steel with respect to buckling can be calculated in the same manner as for timber. • The relationship between the slenderness ratio and the reduction factor ( k ) is slightly different (see Table 4.6).

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