Tank structure | Fundamentals for Structural Engineering follow TCVN
The structure of water tanks and reinforced concrete swimming pools is present in most of the current civil construction designs. However, because they are secondary structures, people often skip the calculation step and follow the experience of conventional tank structures that have been constructed and exploited similarly. With tanks of special scale: large capacity, complex geological conditions, … requiring the design engineer to have basic knowledge of the work and calculation process of the tank structure. Including choosing the size, detailed design, choosing a reasonable foundation option (for underground tanks).
In general, the problem of designing a reinforced concrete tank structure is usually divided into small problems of the following components:
Tank cover structure: the problem of the floor beam structure subjected to bending due to the load acting on the tank cover.
Tank wall structure: the problem of the structure subjected to eccentric compression due to its own weight, the load transmitted from the tank cover, the horizontal pressure of the liquid contained in the tank and of the soil outside.
Tank bottom structure: the floor beam structure subjected to bending due to the pressure of the liquid and the load transmitted from the tank wall to the bottom of the tank. For underground tanks, the tank bottom floor is a floor on an elastic foundation or on a pile foundation. Calculations related to the foundation.
The following article presents a relatively basic way of calculating in practice the design of tank wall structures, which can be applied through the creation of spreadsheets.
1. Calculating tank walls according to intensity (TTGH1)
Due to hydrostatic pressure:
Standard load due to hydrostatic pressure:
$$q_w=H_w\gamma_w$$
The calculated load is obtained by multiplying by the reliability factor of nw=1.1. The calculation diagram and internal force in the wall are as follows:
Bending moment at the bottom of the tank (inside of the tank): $M_A=\frac{q_wH_w^2}{15}$
Maximum moment (outside of the tank): $M_{max}=\frac{q_wH_w^2}{33,6}$ at distance x from the bottom of the tank to the top $x=0,553H_w$
Due to the horizontal pressure of the soil on the outside of the tank:
Standard loads:
$$q_{e1}=K_a\gamma{h_o}+K_aq_s$$
$$q_{e2}=K_a\gamma(h_o+H)+K_aq_s$$
Ka: active pressure coefficient $K_a=tg^2\left(45^o-\frac{\varphi}{2}\right)$
Reliability coefficient of load due to horizontal pressure of soil: $n_e=1,15$
Calculation diagram of tank wall when bearing horizontal pressure load of soil
The internal force in the tank wall when calculating according to this case is the sum of the internal force determined according to diagram 1 and diagram 2. The internal force according to diagram 1 is similar to the case of calculating with water pressure above. The internal force according to diagram 2 is determined as follows:
Calculation of longitudinal reinforcement of tank wall
Calculation of bending structure design with width b=1m, determine the tensile reinforcement area As according to the internal bending moment M calculated above as for beams.
Calculation of shear resistance for tank wall
Shear forces caused by horizontal pressure are borne by tank wall concrete. According to formula (84) of reinforced concrete structure design standard TCVN 5574:2012, the condition for concrete to be able to withstand shear force, without calculating reinforcement is:
$$Q\leqslant{Q}_{bo}=\frac{\varphi}{c}$$
In which the limit $Q_{b3}\leqslant{Q}_{bo}\leqslant{2,5}R_{bt}bh_o$ with $Q_{b3}=\varphi_{b3}(1+\varphi_n)R_{bt}bh_o$
$\varphi_f$=0 because only the rectangular tank wall cross-section is considered
q1: The load is uniformly distributed causing internal shear force in the tank wall, taking the larger value from the two cases of horizontal pressure due to liquid in the tank or due to soil outside the tank.
Normally the shear force appearing in the tank $Q<Q_{bo}$, meaning that the concrete of the tank wall alone is capable of withstanding the shear. The transverse reinforcement is placed according to the structural conditions for the wall structure in the construction design.
2. Design of tank walls for Serviceability
Because the working conditions of the tank are relatively harsher than the indoor structure, the tank wall structure throughout its life cycle works in a liquid-containing environment, in contact with the outside ground. The calculation according to the second limit state of use conditions will focus on the control condition of cracks caused by loads not exceeding the limit values to ensure normal operation of the structure.
Specifically, for tanks, it is necessary to determine the crack resistance level and the allowable limit width value of cracks according to Table 1 of the Standard to ensure limited seepage for the tank structure and Table 2 to ensure safety for the reinforcement of the structure in the working conditions of the structure in an corrosive environment: outdoors or in the ground.
Due to the crack resistance level and the working conditions of the tank wall structure in which environment will determine the value of the tank’s limit crack width $a_{cr1}, a_{cr2}$.
The calculation procedure is similar to that for the design of flexural structures in the topic Cracking and deflection of concrete according to TCVN. However, the calculation formula is slightly different because the tank wall is an eccentrically compressed member.
The stress in the tensile reinforcement in the outermost layer σs is determined by the following formula for eccentrically compressed members:
$$\sigma_s=\frac{N(e_s-Z_b)}{A_sZ_b}$$
$e_s$: distance from the point of application of eccentric compressive force N to the center of gravity of the tensile reinforcement: $e_s=e_o+y_t-a$
$y_t$: distance from the axis of the member to the edge of the tensile force. With a rectangular cross-section $y_t=0,5h$
$e_o=\frac{M}{N}$: eccentricity of the longitudinal force
$h’_f=2a’$ with rectangular tank wall section: no wings in compression zone.
Also calculate 2 values: width $a_{cr}$ due to long-term effect of load (regular + long-term temporary) with φl>1, short-term width $a_{cr(1)}$ is determined by the sum of long-term $a_{cr}$ and the increase in crack width $Δa_{cr}$ due to the effect of short-term temporary load with coefficient $\varphi_l=1$
$$a_{cr(1)}=a_{cr}+\Delta{a}_{cr}$$
The value of $Δa_{cr}$ is calculated as the formula of $a_{cr}$ with $\sigma_s$ due to short-term temporary loads.
The permanent load is the self-weight of the structural layers. The long-term temporary load is the pressure of the liquid contained in the tank. The short-term temporary load is the earth pressure acting on the tank wall from the outside (because most of the working time of the tank is filled with water), in addition to the short-term part of the live load acting on the tank cover transmitted to the tank wall.
The calculation process according to serviceability limit state (LS2) can be applied to the design and construction of tank cover and tank bottom slabs (flexural components).
💞Calculation table of tank wall according to LS1 and LS2 here and here.
Reference:
TCVN 5574: 2018 “Concrete structure and reinforced concrete. Design code”
“Calculating the practice of reinforced concrete structure according to TCXDVN 356: 2005” Episode 2 – GS. Nguyen Dinh Cong – Construction Publishing House.
