For typical steel tanks, the weight of the tank is much less than the weight of the liquid contained. As a result, a crescent-shaped strip of the base plate also rises from the foundation.
BACKGROUND
1.2 is the maximum holding force due to the weight of the liquid lying on the raised part of the base plate. The clamping force is equal to the weight of the liquid resting on the HE part of the base plate.
SCOPE AND ORGANIZATION
After studying the behavior of the shell in Chapter 3, the relationship between lift and restraining force from the axisymmetric analysis of Chapter 2 is used to define the properties of the equivalent nonlinear bed. The validity of this method of equivalent springs is verified in Chapter 5 by solving a coupled, non-axisymmetric problem for the base plate and shell using the finite-difference energy method.
AXISYMMETRIC UPLIFT PROBLEM
DEFINITION OF THE PROBLEM
The stresses and displacements due to the hydrostatic fluid load and an asymmetric upward force per unit length, P, applied to the top of the tank are to be determined.
AXISYMMETRIC SHELL PROBLEM
A [tsa characteristic length that determines the rate of decay of bending moments in the shell;.
GENERAL THEORY FOR BASE PLATE
A point on the center surface of the base plate is assumed to move from a point (r,O) in the original (empty, not lifted) configuration to a point (R,w) = (r + u,w) in the loaded (full and lifted) configuration. The radial and tangential moments are considered positive when they cause stress on the bottom of the base plate, and are given by
SOLUTION FOR MODERATE DEFLECTIONS
Now consider the solution of the initial value problem where N r and Qr are known at the point of contact. This scheme is applied iteratively, starting with the values of the state variables at the contact point given in Eq.
EXAMPLE PROBLEM
The restraining effect of the tank wall is represented by the radial diaphragm force at the edge. In contrast, for the linear theory, the bending moments increase as the square of the lifted width, and the shape of the bending moment diagram is close to parabolic.
SOLUTION FOR LARGE DEFLECTIONS
The results for large deviations are so close to those for moderate deviations that the difference could not be seen in a plot. This confirms that the deviations in this problem (involving rotations up to about 0.2 radians) are characterized as moderate, not large.
COMPARISON WITH EXPERIMENTAL RESULTS
In the experiment, hydrodynamic pressures were measured, which are of the order of half of the hydrostatic pressure. Therefore, the effect of hydrodynamic pressure alone is not considered a valid explanation for the drop in experimental radial membrane stress near the edge.
CONCLUSIONS
For sufficiently large uplift, buckling of the base plate occurs due to surrounding compressive forces. Base plate buckling is the most likely explanation for the difference between theoretical and experimental radial membrane strains shown in Figs.
ANALYSIS AND BEHAVIOR OF THE CYLINDRICAL SHELL
- DEFINITIONS
- COMMENTS ON THE SOLUTION FOR AN ANCHORED TANK
- COMMENTS ON SOLUTION FOR IMPOSED RADIAL DISPLACEMENTS AND ROTATIONS AT THE BASE
- COMMENTS ON THE SOLUTION FOR IMPOSED VERTICAL DISPLACEMENTS AT THE BASE
52 - . a) the solution for an anchored tank subjected to the loads experienced by the unanchored tank, and .. b) the solution for imposed displacements at the bottom of the tank, and no other applied loads. Due to the diaphragm action of the base plate, the horizontal displacements (radial and circumferential) are assumed to occur at the base. The rotation of the tank wall about the circumferential axis is assumed to be unrestrained and a vertical displacement U is imposed at the base.
Consider determining the vertical force that must be applied to the base to generate the displacements described by Eqs.
ANALYTICAL SOLUTION FOR A LIMITING CASE
Since there can be no moment applied at a point of contact, this means that the loads applied to the tank must be such that the tank is exactly at the tipping point. Let be the distributed loads due to the lateral pressure of the liquid and the weight of the tank wall. The simplest case occurs when F1 is such that the reservoir is at tipping point.
Under such conditions, the distribution of vertical stresses at the base for an unanchored tank is the same as for the anchored tank.
NUMERICAL SOLUTION FOR THE GENERAL CASE
For the asymmetric analysis of the bottom plate, it is assumed that the entire bottom plate consists of one continuous sheet of 0.09 inch thick aluminum. The distribution of vertical stresses at the base of the shell is shown in figure. The equation ignores the small stresses (about 20 psi) caused by the weight of the tank wall.
For the code analysis of the aluminum model, the. a) open top (b) close top Code Analysis:.
CLOSING REMARKS
These possibilities will be discussed further in the next chapter when the same experimental data are compared with the results of a more.
ASSUMPTIONS
The foundation is frictionless, except in certain locations close to the center where tank shifting can be prevented by horizontal Winkler springs.
FORMULATION
Once these three steps are accomplished, the rest of the finite element formulation follows standard procedures. The circle of Winkler springs under the bottom of the tank wall, with stiffness k (force per unit length per unit deflection). There is thus a limitation at node 1 due to the inner part of the bottom plate, and at node NN due to the cylindrical shell and roof.
The variation of the strain energy for such constraints can be written in the form
This means that making NH greater than 2NW is of no use under such circumstances. If no horizontal displacements are allowed while the vertical displacements are applied, the horizontal forces required to achieve this are also of Fourier order 2NW. Because the strain energy variation is not exactly integrated, the displacements do not necessarily converge from the bottom up.
Therefore, the displacements (or, strictly speaking, the work done by the applied loads) are a lower limit.
IMPLEMENTATION AND COMPUTATIONAL CONSIDERATIONS
TEST PROBLEMS
This means that the deflection mode displacements vary as cos 39 or sin 39 in the circumferential direction. For the NAAOAP analysis, the tangent stiffness matrix should become singular as the buckling pressure is approached. This introduces a small displacement (about 0.001 and in the transverse direction) with a circumferential variation similar to the bending mode.
Also, for the Fourier coefficient n = 3, the behavior is exactly what can be expected: The small displacement due to the non-symmetrical line loading is greatly amplified as the critical pressure is approached.
RESULTS
For this modified analysis, the non-linear portion of the base plate extends inward from the edge to the joint with the gasket. The radial spacing of the nodes in the non-linear part of the base plate is 0.05 in. Consequently, the shell stiffness matrix also changes due to nonlinear effects associated with hydrostatic pressure.
For simplicity, such changes in the shell stiffness matrix are not included in the analysis.
SUMMARY AND CLOSING REMARKS
It was found that for the long aluminum tank, the yield of the aluminum and the flexibility of the gasketed connection to the base plate have a strong influence on the lift. Some important effects may have been lost in the linearized formulation for the shell. Finally, the friction between the base plate and the foundation is not considered in the analysis.
This in turn can affect the membrane operation in the raised portion of the base plate.
THE PREUPLIFT METHOD
EXPERIMENTS
The vertical seam in the tank wall was patched and bound with 1/4" wide double-sided tape. Two types of tests were performed: i) The tank was filled with water to a depth of zero tilt, and the angle of tilt was increased in increments of approx. 3, and measure the maximum lift at each increment of 0. ii) The tilt angle was kept fixed, and the tank was slowly filled through the aluminum tube shown in Fig.
6.2 until the first signs of a buckle (much smaller than that shown in Figure 6.3) could be visually detected using light reflected from the tank wall.
ANALYSIS
This ensures that the entire perimeter of the tank wall is supported by the filler, even if there is a small defect. In the case of pre-uplift, the vertical uplift varies gradually around the perimeter, therefore the accuracy of the estimated method can be expected to be better for the case. Consider the problem of the tank whose bottom plate has been replaced by a ring of non-linear Winkler springs.
The pre-uplift can be easily taken into account by changing the force-deflection relationship of the Winkler springs.
UPLIFT
DISCUSSION OF RESULTS
The theoretical and experimental lift values obtained with and without pre-lift are shown in the figure. On the other hand, the stiffness of the bead of epoxy that bonds the base plate to the shell and the stiffness of the small base extension. the plate on the outside of the tank wall was neglected in the analysis. Perhaps one of the more important factors contributing to this difference is the reinforcing effect of the bead of epoxy on the rim.
This means that the damping action due to the axial stiffness of the epoxy bead will tend to decrease the lift for a given water level and angle of inclination.
WATER DEPTH (in.)
CLOSING REMARKS
Depending on the relative frequencies of the earthquake and the reservoir, this means that the prelifted reservoir may experience higher or lower lateral loading. As a result, some of the effectiveness of the pre-lift may be lost and there may be some risk of scraping at the end. This can be achieved by choosing a cross-section of the ring filler that matches the deformed shape of the base plate.
For example, the local hydrodynamic pressure mentioned in (iv) above can contribute to the formation of the "elephant foot bulge".
SUMMARY AND CLOSURE
This is achieved by the finite difference energy method, using an expansion of the displacements as a Fourier series in . A tangent stiffness matrix is obtained in which there is coupling between the different Fourier coefficients of the displacements. These discrepancies may be due to geometrically nonlinear effects in the shell, aluminum yield.
The following pages provide an overview of the derivation of such explicit expressions and report the final results.
