Solution 1 Date of assignment: 18 January 2006 Date of solution: 20 January 2006 E-course on Seismic Design of Tanks
E-Course on Seismic Design of Tanks / January 2006 Part II: True / False Identify the following statements as True or False 1.1)Hydrodynamic pressure varies linearly with liquid height. False Hydrodynamic pressure has curvilinear variation along the height. In fact, Hydrostatic pressure varies linearly with height. 1.2)Impulsive mass is less than convective mass for short tanks. In short tanks, convective liquid or liquid undergoing sloshing motion is more. In tall tanks, impulsive liquid mass is more. E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part II: True / False 1.3)Net hydrodynamic force on the container wall is zero. False Rather, net hydrostatic force on wall is zero 1.4)Hydrodynamic pressure on base causes overturning moment on tank. True Refer slide no. 46 of Lecture 1 1.5)Convective mass acts at lower height than impulsive mass. Liquid in upper portion, undergoes convective or sloshing motion, hence, convective mass is always located at higher height than impulsive mass. Note the graphs for hi and hc in figure 2b and 3b of guidelines E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part II: True / False 1.6)With the inclusion of base pressure effect, overturning moment on tank reduces. False Hydrodynamic pressure on the base produces overturning moment in the same direction as that due to hydrodynamic pressure on wall. Hence, with the inclusion of base pressure effect, overturning moment increases. This can be seen from the fact that hi* is always greater than hi and hc* is always greater than hc. 1.7)Impulsive and convective masses depend only on height of liquid. Impulsive and convective masses depend on aspect ratio (h/D or h/L) rather than only on height, h. E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions 1.1)A circular tank has internal diameter of 12 m and water height of 5 m. Obtain mi, mc, Kc, hi, hc, hi* and hc*. Solution: Total volume of water = /4 x 122 x 5 = 565.5 m3 Total water mass, m = 565.5 x 1.0 = 565.5 t D = 12 m, h = 5 m h/D = 5/12 = 0.417 For this value of h/D, from Figures 2a and 2b of the Guideline, values of various parameters can be read. One can also use formulae given in Table C-1 of the Guideline. Here, these formulae will be used to obtain various parameters. E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions = 0.466 = 0.503 Since, h/D is less than 0.75, hi/h = 0.375 = 0.579 E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions = 0.947 = 0.878 E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions Thus, we get, mi/m = 0.466 mi = 0.466 x 565.5 = 263.5 t mc/m = 0.503 mc = 0.503 x 565.5 = 284.5 t hi/h = 0.375 hi = 0.375 x 5 = 1.88 m hc/h = 0.579 hc = 0.579 x 5 = 2.90 m hi*/h = 0.947 hi* = 0.947 x 5 = 4.735 m hc*/h = 0.878 hc* = 0.878 x 5 = 4.39 m Kch/mg = 0.694 Kc = 0.694mg/h = 0.694 x 565.5 x 9.81/5.0 = 770 kN/m E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions 1.2) A rectangular tank has internal plan dimension of 10 m x 16 m. Water height is 8 m. Obtain mi, mc, hi, hc, hi* and hc* in both the directions. Solution: Y 16 m X 10 m Plan view Total volume of water = 10 x16 x 8 = 1280 m3 Total mass of water, m = 1280 x 1.0 = 1280 t E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions First, consider base motion in X-direction: L = 16 m, h = 8 m h/L = 8/16 = 0.5 From Figure 3a, 3b of guidelines, for h/L = 0.5: mi/m = 0.54 mi = 0.54 x 1280 = 691.2 t mc/m = 0.48 mi = 0.48 x 1280 = 614.4 t hi/h = 0.375 hi = 0.375 x 8 = 3.0 m hc/h = 0.57 hc = 0.57 x 8 = 4.6 m hi*/h = 0.80 hi* = 0.8 x 8 = 6.4 m hc*/h = 0.87 hc* = 0.87 x 8 = 7.0 m E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions Now, consider base motion in Y-direction: L = 10 m, h = 8 m h/L = 8/10 = 0.8 From Figure 3a, 3b of guidelines, for h/L = 0.8: mi/m = 0.72 mi = 0.72 x 1280 = 921.6 t mc/m = 0.32 mi = 0.32 x 1280 = 409.6 t hi/h = 0.40 hi = 0.40 x 8 = 3.2 m hc/h = 0.65 hc = 0.65 x 8 = 5.2 m hi*/h = 0.58 hi* = 0.58 x 8 = 4.64 m hc*/h = 0.73 hc* = 0.73 x 8 = 5.84 m E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur
E-Course on Seismic Design of Tanks / January 2006 Part III: Solutions Thus, in x-direction, wherein, h/L = 0.5, impulsive and convective masses are almost same. Whereas, in Y-direction, for which h/L = 0.8, more liquid contributes to impulsive mass. Also note the difference in hi and hi* values ( or hc and hc*) for two cases. For shallow tanks, inclusion of base pressure significantly changes the height . E-Course on Seismic Design of Tanks / January 2006 Sudhir K. Jain, IIT Kanpur