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Lecture 5

Lecture 5. January 31, 2006. In this Lecture. Impulsive and convective base shear Critical direction of seismic loading. Base shear. Previous lectures have covered Procedure to find impulsive and convective liquid masses This was done through a mechanical analog model

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Lecture 5

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  1. Lecture 5 January 31, 2006

  2. In this Lecture • Impulsive and convective base shear • Critical direction of seismic loading E-Course on Seismic Design of Tanks/ January 2006

  3. Base shear • Previous lectures have covered • Procedure to find impulsive and convective liquid masses • This was done through a mechanical analog model • Procedure to obtain base shear coefficients in impulsive and convective modes • This requires time period, damping, zone factor, importance factor and response reduction factor • Now, we proceed with seismic force or base shear calculations E-Course on Seismic Design of Tanks/ January 2006

  4. Base shear • Seismic force in impulsive mode (impulsive base shear) Vi = (Ah)i x impulsive weight • Seismic force in convective mode (convective base shear) Vc = (Ah)c x convective weight • (Ah)i = impulsive base shear coefficient • (Ah)c = convective base shear coefficient • These are described in earlier lectures E-Course on Seismic Design of Tanks/ January 2006

  5. Base shear • Now, we evaluate impulsive and convective weights • Or, impulsive and convective masses • Earlier we have obtained impulsive and convective liquid mass • Now, we consider structural mass also E-Course on Seismic Design of Tanks/ January 2006

  6. Base shear : Ground supported tanks • Impulsive liquid mass is rigidly attached to container wall • Hence, wall, roof and impulsive liquid vibrate together • In ground supported tanks, total impulsive mass comprises of • Mass of impulsive liquid • Mass of wall • Mass of roof E-Course on Seismic Design of Tanks/ January 2006

  7. Base shear : Ground supported tanks • Hence, base shear in impulsive mode • mi = mass of impulsive liquid • mw = mass of container wall • mt = mass of container roof • g = acceleration due to gravity E-Course on Seismic Design of Tanks/ January 2006

  8. Base shear : Ground supported tanks • This is base shear at the bottom of wall • Base shear at the bottom of base slab is : Vi’ = Vi + (Ah)i x mb • mb is mass of base slab • Base shear at the bottom of base slab may be required to check safety against sliding E-Course on Seismic Design of Tanks/ January 2006

  9. Base shear : Ground supported tanks • Base shear in convective mode • mc = mass of convective liquid E-Course on Seismic Design of Tanks/ January 2006

  10. Base shear : Ground supported tanks • Total base shear, V is obtained as: • Impulsive and convective base shear are combined using Square Root of Sum of Square (SRSS) rule • Except Eurocode 8, all international codes use SRSS rule • Eurocode 8 uses absolute summation rule • i.e, V = Vi + Vc E-Course on Seismic Design of Tanks/ January 2006

  11. Base shear : Ground supported tanks • In the latest NEHRP recommendations (FEMA 450), SRSS rule is suggested • Earlier version of NEHRP recommendations (FEMA 368) was using absolute summation rule • FEMA 450, 2003, “NEHRP recommended provisions for seismic regulations for new buildings and other structures”, Building Seismic Safety Council, National Institute of Building Sciences,, USA. • FEMA 368, 2000, “NEHRP recommended provisions for seismic regulations for new buildings and other structures”, Building Seismic Safety Council, National Institute of Building Sciences,, USA. E-Course on Seismic Design of Tanks/ January 2006

  12. Bending moment:Ground supported tanks • Next, we evaluate bending or overturning effects due to base shear • Impulsive base shear comprises of three parts • (Ah)i x mig • (Ah)i x mwg • (Ah)i x mtg E-Course on Seismic Design of Tanks/ January 2006

  13. Bending moment:Ground supported tanks • mw acts at CG of wall • mt acts at CG of roof • mi acts at height hi from bottom of wall • If base pressure effect is not included • mi acts at hi* • If base pressure effect is included • Recall hi and hi* from Lecture 1 E-Course on Seismic Design of Tanks/ January 2006

  14. Bending moment:Ground supported tanks • Bending moment at the bottom of wall • Due to impulsive base shear • Due to convective base shear • hi = location of mi from bottom of wall • hc = location of mc from bottom of wall • hw = height of CG of wall • ht = height of CG of roof E-Course on Seismic Design of Tanks/ January 2006

  15. Bending moment:Ground supported tanks • For bending moment at the bottom of wall, effect of base pressure is not included • Hence, hi and hc are used E-Course on Seismic Design of Tanks/ January 2006

  16. Bending moment:Ground supported tanks (Ah)imw (Ah)imi hw hi • Bending moment at the bottom of wall (Ah)imt (Ah)cmc hc ht Ground level E-Course on Seismic Design of Tanks/ January 2006

  17. Bending moment:Ground supported tanks • Total bending moment at the bottom of wall • SRSS rule used to combine impulsive and convective responses E-Course on Seismic Design of Tanks/ January 2006

  18. Overturning moment:Ground supported tanks Ground level • Overturning moment • This is at the bottom of base slab • Hence, must include effect of base pressure • hi* and hc* will be used E-Course on Seismic Design of Tanks/ January 2006

  19. Overturning moment:Ground supported tanks • Overturning moment in impulsive mode • Overturning moment in convective mode • tb = thickness of base slab E-Course on Seismic Design of Tanks/ January 2006

  20. Bending moment:Ground supported tanks • Overturning moment is at the bottom of base slab • Hence, lever arm is from bottom of base slab • Hence, base slab thickness, tb is added to heights measured from top of the base slab • Total overturning moment E-Course on Seismic Design of Tanks/ January 2006

  21. Example • Example: A ground-supported circular tank is shown below along with some relevant data. Find base shear and bending moment at the bottom of wall. Also find base shear and overturning moment at the bottom of base slab. mi = 141.4 t; mc = 163.4 t mw = 65.3 t, mt =33.1 t, mb = 55.2 t, hi =1.5 m, hi* = 3.95 m, hc = 2.3 m, hc* = 3.63 m (Ah)i = 0.225, (Ah)c = 0.08 Roof slab 150 mm thick 10 m Wall 200 mm thick 4 m Base slab 250 mm thick E-Course on Seismic Design of Tanks/ January 2006

  22. Example Solution: Impulsive base shear at the bottom of wall is Vi = (Ah)i (mi + mw + mt) g = 0.225 x (141.4 + 65.3 + 33.1) x 9.81 = 529.3 kN Convective base shear at the bottom of wall is Vc = (Ah)c mc g = 0.08 x 163.4 x 9.81 = 128.2 kN Total base shear at the bottom of wall is E-Course on Seismic Design of Tanks/ January 2006

  23. Example For obtaining bending moment, we need height of CG of roof slab from bottom of wall, ht. ht = 4.0 + 0.075 = 4.075 m Impulsive bending moment at the bottom of wall is Mi = (Ah)i (mihi + mwhw + mtht) g = 0.225 x (141.4 x 1.5 + 65.3 x 2.0 + 33.1 x 4.075) x 9.81 = 1054 kN-m Convective bending moment at the bottom of wall is Mc = (Ah)c mc hc g = 0.08 x 163.4 x 2.3 x 9.81 = 295 kN-m Total bending moment at bottom of wall is E-Course on Seismic Design of Tanks/ January 2006

  24. Example Now, we obtain base shear at the bottom of base slab Impulsive base shear at the bottom of base slab is Vi = (Ah)i (mi + mw + mt + mb) g = 0.225 x (141.4 + 65.3 + 33.1 + 55.2) x 9.81 = 651.1 kN Convective base shear at the bottom of base slab is Vc = (Ah)c mc g = 0.08 x 163.4 x 9.81 = 128.2 kN Total base shear at the bottom of base slab is E-Course on Seismic Design of Tanks/ January 2006

  25. Example Impulsive overturning moment at the bottom of base slab Mi* = (Ah)i [mi (hi* + tb) + mw(hw + tb) + mt(ht +tb) + mb tb/2]g = 0.225 x [141.4(3.95 + 0.25) + 65.3(2.0 + 0.25) + 33.1(4.075 + 0.25) + 55.2 x 0.25/2] x 9.81 = 1966 kN-m Convective overturning moment at the bottom of base slab Mc* = (Ah)c mc (hc* + tb) g = 0.08 x 163.4 x (3.63 + 0.25) x 9.81 = 498 kN-m Total overturning moment at bottom of base slab Notice that this value is substantially larger that the value at the bottom of wall (85%) E-Course on Seismic Design of Tanks/ January 2006

  26. Base shear : Elevated tanks • In elevated tanks, base shear at the bottom of staging is of interest • Ms is structural mass • Base shear in impulsive mode • Base shear in convective mode • Total base shear E-Course on Seismic Design of Tanks/ January 2006

  27. Bending moment:Elevated tanks • Bending moment at the bottom of staging • Bottom of staging refers to footing top • Impulsive base shear comprises of two parts • (Ah)i x mig • Ah)i x msg • Convective base shear has only one part • (Ah)c x mcg E-Course on Seismic Design of Tanks/ January 2006

  28. Bending moment:Elevated tanks • mi acts at hi* • mc acts at hc* • Bending moment at bottom of staging is being obtained • Hence, effect of base pressure included and hi* and hc* are used E-Course on Seismic Design of Tanks/ January 2006

  29. Bending moment:Elevated tanks • Structural mass, ms comprises of mass of empty container and 1/3rd mass of staging • ms is assumed to act at CG of empty container • CG of empty container shall be obtained by considering roof, wall, floor slab and floor beams E-Course on Seismic Design of Tanks/ January 2006

  30. Bending moment:Elevated tanks • Bending moment at the bottom of staging • hs = staging height • Measured from top of footing to bottom of wall • hcg = distance of CG of empty container from bottom of staging E-Course on Seismic Design of Tanks/ January 2006

  31. Bending moment:Elevated tanks • Bending moment at the bottom of staging (Ah)i mig (Ah)c mcg hi* hc* (Ah)i msg hcg hs hs Top of footing E-Course on Seismic Design of Tanks/ January 2006

  32. Bending moment:Elevated tanks • Total bending moment • For shaft supported tanks, M* will be the design moment for shaft E-Course on Seismic Design of Tanks/ January 2006

  33. Bending moment:Elevated tanks • For analysis of frame staging, two approaches are possible • Approach 1: Perform analysis in two steps • Step 1: • Analyze frame for (Ah)imig + (Ah)imsg • Obtain forces in columns and braces • Step 2: • Analyze the frame for (Ah)cmcg • Obtain forces in columns and braces • Use SRSS rule to combine the member forces obtained in Step 1 and Step 2 E-Course on Seismic Design of Tanks/ January 2006

  34. Bending moment:Elevated tanks • Approach 2: • Apply horizontal force V at height h1 such that • V x h1 =M* • V and M* are obtained using SRSS rule as described in slide nos. 26 and 32 • In this approach, analysis is done in single step • Simpler and faster than Approach 1 E-Course on Seismic Design of Tanks/ January 2006

  35. Example • Example: An elevated tank on frame staging is shown below along with some relevant data. Find base shear and bending moment at the bottom of staging. A is CG of empty container mi = 100t; mc = 180 t Mass of container = 160 t Mass of staging = 120 t hi* = 3 m, hc* = 4.2 m (Ah)i = 0.08, (Ah)c = 0.04 A 2.8 m hs = 15 m GL E-Course on Seismic Design of Tanks/ January 2006

  36. Example Structural mass, ms = mass of container + 1/3rd mass of staging = 160 + 1/3 x 120 = 200 t Base shear in impulsive mode = 78.5 + 157 = 235.5 kN Base shear in convective mode E-Course on Seismic Design of Tanks/ January 2006

  37. Example Now, we proceed to obtain bending moment at the bottom staging Distance of CG of empty container from bottom of staging, hcg = 2.8 + 15 = 17.8 m Total base shear E-Course on Seismic Design of Tanks/ January 2006

  38. Example Base moment in impulsive mode = 78.5 x 18 + 157 x 17.8 = 4207 kNm Note: 78.5 kN of force will act at 18.0m and 157 kN of force will act at 17.8 m from top of footing. E-Course on Seismic Design of Tanks/ January 2006

  39. Example Base moment in convective mode = 70.6 x 19.2 = 1356 kNm Note: 70.6 kN of force will act at 19.2 m from top of footing. Total base moment E-Course on Seismic Design of Tanks/ January 2006

  40. Example • Now, for staging analysis, seismic forces are to be applied at suitable heights • There are two approaches • Refer slide no 33 • Approach 1: • Step 1: Apply force of 78.5 kN at 18 m and 157 kN at 17.8 m from top of footing and analyze the frame • Step 2: Apply 70.6 kN at 19.2 m from top of footing and analyze the frame • Member forces (i.e., BM, SF etc. in columns and braces) of Steps 1 and 2 shall be combined using SRSS E-Course on Seismic Design of Tanks/ January 2006

  41. Example • Approach 2: • Total base shear, V = 245.8 kN will be applied at height h1, such that V x h1 = M* 245.8 x h1 = 4420  h1 = 17.98 m • Thus, apply force of 245.8 kN at 17.98 m from top of footing and get member forces (i.e., BM, SF in columns and braces). E-Course on Seismic Design of Tanks/ January 2006

  42. Elevated tanks:Empty condition • Elevated tanks shall be analysed for tank full as well as tank empty conditions • Design shall be done for the critical condition • In empty condition, no convective liquid mass • Hence, tank will be modeled using single degree of freedom system • Mass of empty container and 1/3rd staging mass shall be considered E-Course on Seismic Design of Tanks/ January 2006

  43. Elevated tanks:Empty condition • Lateral stiffness of staging, Ks will remain same in full and empty conditions • In full condition, mass is more • In empty condition mass is less • Hence, time period of empty tank will be less • Recall, T = • Hence, Sa/g will be more • Usually, tank full condition is critical • However, for tanks of low capacity, empty condition may become critical E-Course on Seismic Design of Tanks/ January 2006

  44. Direction of seismic force Y X • Let us a consider a vertical cantilever with rectangular cross section • Horizontal load P is applied • First in X-Direction • Then in Y-direction (see Figure below) • More deflection, when force in Y-direction • Hence, direction of lateral loading is important !! P P E-Course on Seismic Design of Tanks/ January 2006

  45. Direction of seismic force • On the other hand, if cantilever is circular • Direction is not of concern • Same deflection for any direction of loading • Hence, it is important to ascertain the most critical direction of lateral seismic force • Direction of force, which will produce maximum response is the most critical direction • In the rectangular cantilever problem, Y-direction is the most critical direction for deflection E-Course on Seismic Design of Tanks/ January 2006

  46. Direction of seismic force • For frame stagings consisting of columns and braces, IS 11682:1985 suggests that horizontal seismic loads shall be applied in the critical direction • IS 11682:1985, “Criteria for Design of RCC Staging for Overhead Water Tanks”, Bureau of Indian Standards, New Delhi Clause 7.1.1.2 Horizontal forces – Actual forces and moments resulting from horizontal forces may be calculated for critical direction and used in the design of the structures. Analysis may be done by any of the accepted methods including considering as space frame. Clause 7.2.2 Bending moments in horizontal braces due to horizontal loads shall be calculated when horizontal forces act in a critical direction. The moments in braces shall be the sum of moments in the upper and lower columns at the joint resolved in the direction of horizontal braces. E-Course on Seismic Design of Tanks/ January 2006

  47. Direction of seismic force • Section 4.8 of IITK-GSDMA Guidelines contains provisions on critical direction of seismic force for tanks • Ground-supported circular tanks need to be analyzed for only one direction of seismic loads • These are axisymmetric • Hence, analysis in any one direction is sufficient E-Course on Seismic Design of Tanks/ January 2006

  48. Direction of seismic force • Ground-supported rectangular tanks shall be analyzed for two directions • Parallel to length of the tank • Parallel to width of the tank Stresses in a particular wall shall be obtained for seismic loads perpendicular to that wall E-Course on Seismic Design of Tanks/ January 2006

  49. Direction of seismic force • RC circular shafts of elevated tanks are also axisymmetric • Hence, analysis in one direction is sufficient • If circular shaft supports rectangular container • Then, analysis in two directions will be necessary E-Course on Seismic Design of Tanks/ January 2006

  50. Direction of seismic force • For elevated tanks on frame staging • Critical direction of seismic loading for columns and braces shall be properly ascertained • Braces and columns may have different critical directions of loading • For example, in a 4 - column staging • Seismic loading along the length of the brace is critical for braces • Seismic loading in diagonal direction gives maximum axial force in columns • See next slide E-Course on Seismic Design of Tanks/ January 2006

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