1. DESIGN OF TRUSS ROOF Chapter 7 University of Engineering & Technology, Taxila 1Prof Dr Z. A. Siddiqi

2. TABLE OF FORCES There are three important points to be considered while calculating the member forces: 1.Panel load multiplied with unit load forces gives the member forces by the principle of superposition. According to this principle, which is applicable for elastic structures, if a unit load is applied on a truss and the force in any member is calculated as F; then if we apply another unit load simultaneously at the same point, the force in the member will be 2F. 2Prof Dr Z. A. Siddiqi

3. Similarly, this member force will be 3F for three unit loads or P x F if P times unit loads are applied. 2.Effects of various types of loads are to be added while calculating member forces. Vertical and inclined loads on the truss cannot be added directly because of different lines of action of each. However, separate member forces due to vertical and inclined loads have the same direction (along the member longitudinal axis) and hence can be algebraically added together. 3Prof Dr Z. A. Siddiqi

4. 3.Probability of occurrence of various loads in a load combination and corresponding load factors (factor of safety) are also applied during these calculations. In case only dead, live and wind loads are acting on a truss, following combinations may be investigated: 1.1.2D+ 1.6L r + 0.65W (Wind effect is small and may be ignored especially suction is present throughout) 4Prof Dr Z. A. Siddiqi

5. 2.1.2D+ 0.5L r + 1.3W Wind towards the Right Wind towards the left 3. 0.9D + 1.3W Wind towards the Right Wind towards the Left For the roof design, the first (gravity) or second (wind) load combination is critical. It is to be noted that the downward wind load is to be considered in the second combination. 5Prof Dr Z. A. Siddiqi

6. The third combination is critical for reversal of forces and hence is evaluated for upward wind pressures. In case the wind load has two values, one downward and one upward, the downward value should be used for the second combination and the upward value should be used for the third combination. It would be unreasonable to include full wind and full snow (or live load) stresses together in a single combination. 6Prof Dr Z. A. Siddiqi

7. Similarly, when the wind is blowing at its full strength, the live load intensity may be reduced. The second combination reflects the condition when most severe windstorm is blowing and hence the live load intensity may be reduced to 0.5/ 1.6 or 0.31 times its maximum intensity, showing less probability of occurrence of full live load together with wind. The third combination represents an unoccupied building subjected to the heaviest wind. 7Prof Dr Z. A. Siddiqi

8. The negative sign with the wind forces only indicates that the combination is critical when the wind is producing member forces opposite in sense to that produced by the dead loads. The design forces may be calculated by entering the values into a Table of Forces as in Table 7.2. The first four columns of this table are directly taken from the unit load analysis of the truss while columns 5 to 9 are calculated using the first four columns and the algebraic values of the panel loads, already determined. 8Prof Dr Z. A. Siddiqi

9. 9 Member No. Unit gravity load member force Member force due to unit Wind load on hinge side Member force due to unit wind load on roller side (1.2P D +1.6P L ) x Col.2 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.4 + 1.3 P wL x Col.3 (0.9 P D ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 (0.9 P D ) x Col.2 +1.3 P ww x Col.4 +1.3 P wL x Col.3 Maximum factored tension (T u ) Maximum factored Compression (P u ) Remarks (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) Sample Table of Forces

10. After getting the values in these columns, the maximum factored tension and compression may be found out and entered in the next two columns. Usually tension is represented by positive sign and compression by a negative sign in columns 2 to 4 and hence maximum tension is defined as the maximum positive value and maximum compression as the minimum value (maximum negative answer) in columns 5 to 9. 10Prof Dr Z. A. Siddiqi

11. The remarks column is used to describe the type of the member for design such as pure tension member, pure compression member, member under reversal of stresses and zero force member. A computer spreadsheet may conveniently be used to construct such a table. The truss members may now be selected by using the procedure given in earlier chapters and connections may be designed by the methods outlined in the coming chapters. 11Prof Dr Z. A. Siddiqi

12. Unit Gravity Loads

13. 13Prof Dr Z. A. Siddiqi Member No. Unit gravity load member force Member force due to unit Wind load on hinge side Member force due to unit wind load on roller side (1.2P D +1.6P L ) x Col.2 Maximum factored tension (T u ) Maximum factored Compression (P u ) Remarks (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) Sample Table of Forces

14. Unit Wind Load on Hinge Side

15. Unit Wind Load on Roller Side

16. If Panel wind load is negative = Suction

17. P ww P ww = +ve (1.2P D +0.5P L ) X COL. 2 + 1.3 P WW X COL. 3

18. P ww P wl P ww = +ve (1.2P D +0.5P L ) X COL. 2 + 1.3 P WW X COL. 3 P wl = -ve

19. P ww P wl P ww = +ve P wl = -ve (1.2P D +0.5P L ) X COL. 2 + 1.3 P WW X COL. 3 + 1.3 P WL X COL. 4 Load Combination in Column No 6

20. P ww P wl P ww = +ve P wl = -ve (1.2P D +0.5P L ) X COL. 2 + 1.3 P WW X COL. 3 + 1.3 P WL X COL. 4

21. 21Prof Dr Z. A. Siddiqi Member No. Unit gravity load member force Member force due to unit Wind load on hinge side Member force due to unit wind load on roller side (1.2P D +1.6P L ) x Col.2 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 Maximum factored tension (T u ) Maximum factored Compression (P u ) Remarks (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) Sample Table of Forces

22. P ww = +ve P wl = -ve (1.2P D +0.5P L ) X COL. 2 + 1.3 P WW X COL. 4 + 1.3 P WL X COL. 3 P ww P wl Load Combination in Column No 7

23. 23Prof Dr Z. A. Siddiqi Member No. Unit gravity load member force Member force due to unit Wind load on hinge side Member force due to unit wind load on roller side (1.2P D +1.6P L ) x Col.2 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.4 + 1.3 P wL x Col.3 Maximum factored tension (T u ) Maximum factored Compression (P u ) Remarks (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) Sample Table of Forces

24. P ww P wl P ww = -ve P wl = -ve (0.9P D ) X COL. 2 + 1.3 P WW X COL. 3 + 1.3 P WL X COL. 4 Load Combination in Column No 8

25. 25Prof Dr Z. A. Siddiqi Member No. Unit gravity load member force Member force due to unit Wind load on hinge side Member force due to unit wind load on roller side (1.2P D +1.6P L ) x Col.2 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.4 + 1.3 P wL x Col.3 (0.9 P D ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 Maximum factored tension (T u ) Maximum factored Compression (P u ) Remarks (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) Sample Table of Forces

26. P ww = -ve P wl = -ve (0.9P D ) X COL. 2 + 1.3 P WW X COL. 4 + 1.3 P WL X COL. 3 P ww P wl Load Combination in Column No 9

27. 27Prof Dr Z. A. Siddiqi Member No. Unit gravity load member force Member force due to unit Wind load on hinge side Member force due to unit wind load on roller side (1.2P D +1.6P L ) x Col.2 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 (1.2 P D + 0.5 P L ) x Col.2 + 1.3 P ww x Col.4 + 1.3 P wL x Col.3 (0.9 P D ) x Col.2 + 1.3 P ww x Col.3 + 1.3 P wL x Col.4 (0.9 P D ) x Col.2 +1.3 P ww x Col.4 +1.3 P wL x Col.3 Maximum factored tension (T u ) Maximum factored Compression (P u ) Remarks (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12) Sample Table of Forces

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29. Assignment: Draw the Table of Forces for Your Data & find the Truss Member Forces. Time Allowed: 1 week