1. Lecture 5 January 31, 2006

2.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 2 In this Lecture Impulsive and convective base shear Critical direction of seismic loading

3.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

4.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 4 Base shear Seismic force in impulsive mode (impulsive base shear) V i = (A h ) i x impulsive weight Seismic force in convective mode (convective base shear) V c = (A h ) c x convective weight (A h ) i = impulsive base shear coefficient (A h ) c = convective base shear coefficient These are described in earlier lectures

5.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

6.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

7.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 7 Base shear : Ground supported tanks Hence, base shear in impulsive mode m i = mass of impulsive liquid m w = mass of container wall m t = mass of container roof g = acceleration due to gravity

8.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 8 Base shear : Ground supported tanks This is base shear at the bottom of wall Base shear at the bottom of base slab is : V i ’ = V i + (A h ) i x m b m b is mass of base slab Base shear at the bottom of base slab may be required to check safety against sliding

9.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 9 Base shear : Ground supported tanks Base shear in convective mode m c = mass of convective liquid

10.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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 = V i + V c

11.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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.

12.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 12 Bending moment: Ground supported tanks Next, we evaluate bending or overturning effects due to base shear Impulsive base shear comprises of three parts (A h ) i x m i g (A h ) i x m w g (A h ) i x m t g

13.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 13 Bending moment:Ground supported tanks m w acts at CG of wall m t acts at CG of roof m i acts at height h i from bottom of wall If base pressure effect is not included m i acts at h i * If base pressure effect is included Recall h i and h i * from Lecture 1

14.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 14 Bending moment:Ground supported tanks Bending moment at the bottom of wall Due to impulsive base shear h i = location of m i from bottom of wall h c = location of m c from bottom of wall h w = height of CG of wall h t = height of CG of roof Due to convective base shear

15.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 15 Bending moment:Ground supported tanks For bending moment at the bottom of wall, effect of base pressure is not included Hence, h i and h c are used

16.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 16 Bending moment:Ground supported tanks Bending moment at the bottom of wall Ground level (A h ) i m i hihi (A h ) i m w hwhw (A h ) i m t (A h ) c m c hchc htht

17.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 17 Bending moment:Ground supported tanks Total bending moment at the bottom of wall SRSS rule used to combine impulsive and convective responses

18.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 18 Overturning moment: Ground supported tanks Overturning moment This is at the bottom of base slab Hence, must include effect of base pressure h i * and h c * will be used Ground level

19.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 19 Overturning moment:Ground supported tanks Overturning moment in impulsive mode t b = thickness of base slab Overturning moment in convective mode

20.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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, t b is added to heights measured from top of the base slab Total overturning moment

21.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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. m i = 141.4 t; m c = 163.4 t m w = 65.3 t, m t =33.1 t, m b = 55.2 t, h i =1.5 m, h i * = 3.95 m, h c = 2.3 m, h c * = 3.63 m (A h ) i = 0.225, (A h ) c = 0.08 Roof slab 150 mm thick Base slab 250 mm thick 4 m 10 m Wall 200 mm thick

22.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 22 Example Solution: Impulsive base shear at the bottom of wall is V i = (A h ) i (m i + m w + m t ) 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 V c = (A h ) c m c g = 0.08 x 163.4 x 9.81 = 128.2 kN Total base shear at the bottom of wall is

23.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 23 Example For obtaining bending moment, we need height of CG of roof slab from bottom of wall, h t. h t = 4.0 + 0.075 = 4.075 m Impulsive bending moment at the bottom of wall is M i = (A h ) i (m i h i + m w h w + m t h t ) 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 M c = (A h ) c m c h c g = 0.08 x 163.4 x 2.3 x 9.81 = 295 kN-m Total bending moment at bottom of wall is

24.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 24 Example Now, we obtain base shear at the bottom of base slab Impulsive base shear at the bottom of base slab is V i = (A h ) i (m i + m w + m t + m b ) 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 V c = (A h ) c m c g = 0.08 x 163.4 x 9.81 = 128.2 kN Total base shear at the bottom of base slab is

25.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 25 Example Impulsive overturning moment at the bottom of base slab M i * = (A h ) i [m i (h i * + t b ) + m w (h w + t b ) + m t (h t +t b ) + m b t b /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 M c * = (A h ) c m c (h c * + t b ) 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%)

26.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 26 Base shear : Elevated tanks In elevated tanks, base shear at the bottom of staging is of interest M s is structural mass Base shear in impulsive mode Base shear in convective mode Total base shear

27.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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 (A h ) i x m i g A h ) i x m s g Convective base shear has only one part (A h ) c x m c g

28.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 28 Bending moment:Elevated tanks m i acts at h i * m c acts at h c * Bending moment at bottom of staging is being obtained Hence, effect of base pressure included and h i * and h c * are used

29.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 29 Bending moment:Elevated tanks Structural mass, m s comprises of mass of empty container and 1/3 rd mass of staging m s 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

30.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 30 Bending moment:Elevated tanks Bending moment at the bottom of staging h s = staging height Measured from top of footing to bottom of wall h cg = distance of CG of empty container from bottom of staging

31.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 31 Bending moment:Elevated tanks Bending moment at the bottom of staging Top of footing (A h ) i m s g hshs h cg hi*hi* hshs hc*hc* (A h ) i m i g (A h ) c m c g

32.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 32 Bending moment:Elevated tanks Total bending moment For shaft supported tanks, M* will be the design moment for shaft

33.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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 (A h ) i m i g + (A h ) i m s g Obtain forces in columns and braces Step 2: Analyze the frame for (A h ) c m c g Obtain forces in columns and braces Use SRSS rule to combine the member forces obtained in Step 1 and Step 2

34.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 34 Bending moment:Elevated tanks Approach 2: Apply horizontal force V at height h 1 such that V x h 1 = 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

35.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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 A is CG of empty container m i = 100t; m c = 180 t Mass of container = 160 t Mass of staging = 120 t h i * = 3 m, h c * = 4.2 m (A h ) i = 0.08, (A h ) c = 0.04 GL h s = 15 m 2.8 m

36.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 36 Example Structural mass, m s = mass of container + 1/3 rd mass of staging = 160 + 1/3 x 120 = 200 t Base shear in impulsive mode Base shear in convective mode = 78.5 + 157 = 235.5 kN

37.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 37 Example Total base shear Now, we proceed to obtain bending moment at the bottom staging Distance of CG of empty container from bottom of staging, h cg = 2.8 + 15 = 17.8 m

38.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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.

39.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 39 Example Base moment in convective mode Total base moment = 70.6 x 19.2 = 1356 kNm Note: 70.6 kN of force will act at 19.2 m from top of footing.

40.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

41.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 41 Example Approach 2: Total base shear, V = 245.8 kN will be applied at height h 1, such that V x h 1 = M* 245.8 x h 1 = 4420  h 1 = 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).

42.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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 1/3 rd staging mass shall be considered

43.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 43 Elevated tanks:Empty condition Lateral stiffness of staging, K s 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, S a /g will be more Usually, tank full condition is critical However, for tanks of low capacity, empty condition may become critical

44.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 44 Direction of seismic force 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 X Y

45.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

46.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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.

47.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

48.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

49.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

50.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 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

51.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 51 Direction of seismic force Critical directions for 4 - column staging Critical direction for shear force in brace Critical direction for axial force in column Bending Axis

52.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 52 Direction of seismic force For 6 – column and 8 – column staging, critical directions are given in Figure C-6 of the Guideline See next two slides More information available in Sameer and Jain (1994) Sameer, S. U., and Jain, S. K., 1994, “Lateral load analysis of frame staging for elevated water tanks”, Journal of Structural Engineering, ASCE, Vol.120, No.5, 1375-1393. (http://www.nicee.org/ecourse/Tank_ASCE.pdf)

53.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 53 Direction of seismic force Critical directions for 6 - column staging Critical direction for shear force and bending moment in columns Critical direction for shear force and bending moment in braces and axial force in columns Bending Axis

54.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 54 Direction of seismic force Critical directions for 8 - column staging Critical direction for shear force and bending moment in braces Critical direction for shear force, bending moment and axial force in columns Bending Axis

55.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 55 Direction of seismic force As an alternative to analysis in the critical directions, following two load combinations can be used 100 % + 30% rule Also used in IS 1893(Part 1) for buildings SRSS rule

56.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 56 Direction of seismic force 100%+30% rule implies following combinations EL x + 0.3 EL Y EL Y + 0.3 EL x EL x is response quantity when seismic loads are applied in X-direction EL Y is response quantity when seismic loads are applied in Y-direction

57.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 57 Direction of seismic force 100%+30% rule requires Analyze tank with seismic force in X-direction; obtain response quantity, EL X Response quantity means BM in column, SF in brace, etc. Analyze tank with seismic force in Y-direction; obtain response quantity, EL Y Combine response quantity as per 100%+30% rule Combination is on response quantity and not on seismic loads

58.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 58 Direction of seismic force Important to note that the earthquake directions are reversible Hence, in 100%+30% rule, there are total eight load combinations

59.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 59 Direction of seismic force SRSS rule implies following combination Note: EL x is response quantity when seismic loads are applied in X-direction EL Y is response quantity when seismic loads are applied in Y-direction Hence, analyze tank in two directions and use SRSS combination of response quantity

60.  Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 5/ Slide 60 At the end of Lecture 5 This completes seismic force evaluation on tanks There are two main steps Evaluation of impulsive and convective masses Evaluation of base shear coefficients for impulsive and convective modes SRSS rule is used to combine impulsive and convective responses Critical direction of seismic loading shall be properly ascertained Else, 100%+30% or SRSS rule be used