Machine Foundations

Structural Engineering Simplified

Chapter 2. Block Foundation

Chapter 2

2.1 Description of Block Foundation?2.2 Dynamic Impedance Models 2.3 Richart-Whitman Model 2.4 Veletsos Model 2.5 Veletsos Model Calculator 2.6 Examples

2.6 Examples

Figure 2.6.1 Machine Foundation of Example 2.6.1

Example 2.6.1

A centrifugal machine is supported on the rigid reinforced concrete block shown in Figure 2.6.1. The block rests directly on the surface of a deep, uniform granular deposit, so that no embedment is available ( $l = 0$ ). Using the constant halfspace parameters of the Novak–Beredugo model, calculate the stiffness and damping constants of the foundation for the vertical mode $v$ , the coupled horizontal–rocking mode $u - \psi$ in the X–Y plane, and the torsional mode $\eta$ . All constants are to be referred to the centre of gravity of the machine-plus-block system. Then adjust the constants to account for the soil material damping.

Table 2.6.1 Given Data for Example 2.6.1

Item	Quantity	SI	US Customary
Machine	Weight	88.96 kN	20,000 lb
Machine	Height of horizontal excitation	3.657 m	12 ft
Footing	Reinforced concrete unit weight $\gamma_c$	23.57 kN/m³	150 lb/ft³
Footing	Plan dimension $a$ (rocking direction)	3.048 m	10 ft
Footing	Plan dimension $b$	4.877 m	16 ft
Footing	Thickness $c$	2.44 m	8 ft
Footing	Embedment depth $l$	0	0
Footing	Height of the system C.G. above the base $y_c$	1.448 m	4.75 ft
Soil	Unit weight $\gamma$	15.714 kN/m³	100 lb/ft³
Soil	Mass density $\rho$	1602 kg/m³	3.105 slug/ft³
Soil	Shear wave velocity $V_s$	150 m/s	492.1 ft/s
Soil	Material damping $\tan \delta = 2 \beta_m$	0.10	0.10
Soil	Poisson's ratio $\nu$ (granular)	0.25	0.25
System	Total mass of machine + block $m$	9.60 × 10⁴ kg	6583 slug

The foundation rests on the surface of a homogeneous halfspace, so the embedment length is $l = 0$ and the embedment ratio $\delta = l / R$ is also zero. Every side-layer term $S_{i1}$ and $S_{i2}$ therefore drops out of the Novak–Beredugo equations, and only the halfspace parameters $C_{i1}$ and $\overline{C}_{i2}$ remain. The constants are calculated for the three vibration modes: the vertical mode, the coupled horizontal–rocking mode in the X–Y plane, and the torsional mode.

Step 1 — Equivalent radii of the rectangular base

The rectangular base is replaced by an equivalent circular disc. The translational radius is obtained from an equal contact area, while each rotational radius is obtained from an equal moment of inertia about the axis of rotation,

R_u = R_v = \sqrt{\frac{a b}{\pi}} = \sqrt{\frac{3.048 \times 4.877}{\pi}} = \sqrt{\frac{14.865}{\pi}}

R_u = R_v = 2.175 \space \text{m} = 7.13 \space \text{ft}

R_{\psi} = \sqrt[4]{\frac{b a^3}{3 \pi}} = \sqrt[4]{\frac{4.877 \times (3.048)^3}{3 \pi}} = \sqrt[4]{\frac{138.11}{9.425}}

R_{\psi} = \sqrt[4]{14.65} = 1.957 \space \text{m} = 6.42 \space \text{ft}

R_{\eta} = \sqrt[4]{\frac{a b (a^2 + b^2)}{6 \pi}} = \sqrt[4]{\frac{14.865 \times \big( (3.048)^2 + (4.877)^2 \big)}{6 \pi}}

R_{\eta} = \sqrt[4]{\frac{14.865 \times 33.07}{18.85}} = \sqrt[4]{26.08} = 2.260 \space \text{m} = 7.41 \space \text{ft}

Note that $a = 3.048 \space \text{m}$ is the plan dimension measured in the direction of rocking (parallel to the X axis), and that the aspect ratio $b/a = 1.6 < 2$ , so the equivalent-radius idealisation is acceptable.

Step 2 — Soil shear modulus and impedance coefficient

The shear wave velocity is related to the shear modulus of the soil by,

V_s = \sqrt{\frac{G}{\rho}} \quad \Rightarrow \quad G = \rho V_s^2

G = 1602 \times (150)^2 = 3.6045 \times 10^{7} \space \text{N/m}^2 = 7.528 \times 10^{5} \space \text{lb/ft}^2

All of the damping constants are proportional to the quantity $\sqrt{\rho G}$ , which is evaluated once and reused throughout,

\sqrt{\rho G} = \sqrt{1602 \times 3.6045 \times 10^{7}} = 2.403 \times 10^{5} \space \text{N.s/m}^3

Step 3 — Halfspace parameters

The soil is granular with $\nu = 0.25$ . Reading the halfspace columns of Table 2.4.1 (identical to Table 5.4.1.2 of ACI 351.3R-18) gives the following constant approximations, which are valid for a dimensionless frequency $a_o < 2.0$ ,

C_{v1} = 5.2 \qquad \overline{C}_{v2} = 5.0

C_{u1} = 4.7 \qquad \overline{C}_{u2} = 2.8

C_{\psi 1} = 3.3 \qquad \overline{C}_{\psi 2} = 0.5

C_{\eta 1} = 4.3 \qquad \overline{C}_{\eta 2} = 0.7

Step 4 — Vertical mode

The general expressions for an embedded footing are, for the stiffness and the damping constant respectively,

k_{vv} = G R_v \Big( C_{v1} + \frac{G_s}{G} \delta S_{v1} \Big)

c_{vv} = R_v^2 \sqrt{\rho G} \Big( \overline{C}_{v2} + \delta \sqrt{\frac{\rho_s G_s}{\rho G}} \overline{S}_{v2} \Big)

With $\delta = 0$ these reduce to the surface-footing form,

k_{vv} = G R_v C_{v1} = 3.6045 \times 10^{7} \times 2.175 \times 5.2

k_{vv} = 4.077 \times 10^{8} \space \text{N/m} = 2.794 \times 10^{7} \space \text{lb/ft}

c_{vv} = R_v^2 \sqrt{\rho G} \space \overline{C}_{v2} = (2.175)^2 \times 2.403 \times 10^{5} \times 5.0

c_{vv} = 5.685 \times 10^{6} \space \text{N.s/m} = 3.896 \times 10^{5} \space \text{lb.s/ft}

Step 5 — Coupled horizontal and rocking mode

Because the footing has a finite height, a horizontal translation of the base produces a moment about the centre of gravity, and a rotation produces a horizontal force. The motion is therefore coupled and four constants are required: $k_{uu}$ , $k_{\psi \psi}$ , the cross term $k_{u \psi} = k_{\psi u}$ , and their damping counterparts. All of them are referred to the centre of gravity, which lies at $y_c = 1.448 \space \text{m}$ above the base.

Horizontal translation

k_{uu} = G R_u C_{u1} = 3.6045 \times 10^{7} \times 2.175 \times 4.7

k_{uu} = 3.685 \times 10^{8} \space \text{N/m} = 2.525 \times 10^{7} \space \text{lb/ft}

c_{uu} = R_u^2 \sqrt{\rho G} \space \overline{C}_{u2} = (2.175)^2 \times 2.403 \times 10^{5} \times 2.8

c_{uu} = 3.184 \times 10^{6} \space \text{N.s/m} = 2.182 \times 10^{5} \space \text{lb.s/ft}

Rocking

The rocking stiffness about the centre of gravity is the sum of the true rocking resistance of the base and the moment generated by the base shear acting at the lever arm $y_c$ ,

k_{\psi \psi} = G \Big[ R_{\psi}^3 C_{\psi 1} + R_u^3 \Big( \frac{y_c}{R_u} \Big)^2 C_{u1} \Big]

k_{\psi \psi} = 3.6045 \times 10^{7} \Big[ (1.957)^3 \times 3.3 + (2.175)^3 \Big( \frac{1.448}{2.175} \Big)^2 \times 4.7 \Big]

k_{\psi \psi} = 3.6045 \times 10^{7} \big[ 24.71 + 21.44 \big] = 3.6045 \times 10^{7} \times 46.15

k_{\psi \psi} = 1.664 \times 10^{9} \space \text{N.m/rad} = 1.227 \times 10^{9} \space \text{lb.ft/rad}

c_{\psi \psi} = \sqrt{\rho G} \Big[ R_{\psi}^4 \overline{C}_{\psi 2} + R_u^4 \Big( \frac{y_c}{R_u} \Big)^2 \overline{C}_{u2} \Big]

c_{\psi \psi} = 2.403 \times 10^{5} \Big[ (1.957)^4 \times 0.5 + (2.175)^4 \Big( \frac{1.448}{2.175} \Big)^2 \times 2.8 \Big]

c_{\psi \psi} = 2.403 \times 10^{5} \big[ 7.33 + 27.78 \big] = 2.403 \times 10^{5} \times 35.11

c_{\psi \psi} = 8.436 \times 10^{6} \space \text{N.m.s/rad} = 6.222 \times 10^{6} \space \text{lb.ft.s/rad}

Cross (coupling) terms

The cross terms carry a negative sign under the sign convention in which a positive translation $u$ and a positive rotation $\psi$ are as shown in Figure 2.6.1,

k_{u \psi} = k_{\psi u} = - G R_u \space y_c \space C_{u1} = - 3.6045 \times 10^{7} \times 2.175 \times 1.448 \times 4.7

k_{u \psi} = - 5.336 \times 10^{8} \space \text{N/rad} = - 1.200 \times 10^{8} \space \text{lb/rad}

c_{u \psi} = c_{\psi u} = - \sqrt{\rho G} \space R_u^2 \space y_c \space \overline{C}_{u2} = - 2.403 \times 10^{5} \times (2.175)^2 \times 1.448 \times 2.8

c_{u \psi} = - 4.610 \times 10^{6} \space \text{N.s/rad} = - 1.036 \times 10^{6} \space \text{lb.s/rad}

Step 6 — Torsional mode

Torsion is uncoupled from the other degrees of freedom, so a single pair of constants is needed,

k_{\eta \eta} = G R_{\eta}^3 C_{\eta 1} = 3.6045 \times 10^{7} \times (2.260)^3 \times 4.3

k_{\eta \eta} = 3.6045 \times 10^{7} \times 11.54 \times 4.3 = 1.789 \times 10^{9} \space \text{N.m/rad}

k_{\eta \eta} = 1.319 \times 10^{9} \space \text{lb.ft/rad}

c_{\eta \eta} = R_{\eta}^4 \sqrt{\rho G} \space \overline{C}_{\eta 2} = (2.260)^4 \times 2.403 \times 10^{5} \times 0.7

c_{\eta \eta} = 4.388 \times 10^{6} \space \text{N.m.s/rad} = 3.236 \times 10^{6} \space \text{lb.ft.s/rad}

Table 2.6.2 Summary of the Stiffness and Damping Constants (material damping neglected)

Mode	Constant	SI	US Customary
Vertical	$k_{vv}$	$4.077 \times 10^{8}$ N/m	$2.794 \times 10^{7}$ lb/ft
Vertical	$c_{vv}$	$5.685 \times 10^{6}$ N.s/m	$3.896 \times 10^{5}$ lb.s/ft
Horizontal	$k_{uu}$	$3.685 \times 10^{8}$ N/m	$2.525 \times 10^{7}$ lb/ft
Horizontal	$c_{uu}$	$3.184 \times 10^{6}$ N.s/m	$2.182 \times 10^{5}$ lb.s/ft
Rocking	$k_{\psi \psi}$	$1.664 \times 10^{9}$ N.m/rad	$1.227 \times 10^{9}$ lb.ft/rad
Rocking	$c_{\psi \psi}$	$8.436 \times 10^{6}$ N.m.s/rad	$6.222 \times 10^{6}$ lb.ft.s/rad
Cross	$k_{u \psi}$	$-5.336 \times 10^{8}$ N/rad	$-1.200 \times 10^{8}$ lb/rad
Cross	$c_{u \psi}$	$-4.610 \times 10^{6}$ N.s/rad	$-1.036 \times 10^{6}$ lb.s/rad
Torsion	$k_{\eta \eta}$	$1.789 \times 10^{9}$ N.m/rad	$1.319 \times 10^{9}$ lb.ft/rad
Torsion	$c_{\eta \eta}$	$4.388 \times 10^{6}$ N.m.s/rad	$3.236 \times 10^{6}$ lb.ft.s/rad

Step 7 — Effect of the soil material damping

The constants above were derived for a perfectly elastic halfspace, so $c_i$ contains geometric (radiation) damping only. The soil of this example has a hysteretic material damping $\tan \delta = 2 \beta_m = 0.1$ , hence $\beta_m = 0.05$ . Multiplying the elastic impedance by the complex factor $(1 + 2 i \beta_m)$ gives the adjusted impedance,

k_i^*(adj) = (k_i + i \omega c_i)(1 + 2 i \beta_m) = \big( k_i - 2 \beta_m \omega c_i \big) + i \omega \Big( c_i + \frac{2 \beta_m k_i}{\omega} \Big)

Separating the real and the imaginary parts,

k_i(adj) = k_i - 2 \beta_m \omega c_i

c_i(adj) = c_i + \frac{2 \beta_m k_i}{\omega}

These expressions are frequency dependent, so a value of $\omega$ has to be selected. If the footing is checked at a known operating speed, that speed is substituted directly. If complete response curves are required, it is better to evaluate the adjustment at the natural frequency of the mode being examined. For the vertical mode of this footing,

\omega_v = \sqrt{\frac{k_{vv}}{m}} = \sqrt{\frac{4.077 \times 10^{8}}{9.60 \times 10^{4}}} = 65.2 \space \text{rad/s} \space \space (10.4 \space \text{Hz})

The corresponding dimensionless frequency confirms that the constant parameters of Step 3 were read in their valid range,

a_{ov} = \frac{\omega_v R_v}{V_s} = \frac{65.2 \times 2.175}{150} = 0.945 < 2.0 \quad \text{OK}

Substituting into the adjustment equations,

k_{vv}(adj) = 4.077 \times 10^{8} - 2 (0.05)(65.2)(5.685 \times 10^{6})

k_{vv}(adj) = 4.077 \times 10^{8} - 0.371 \times 10^{8} = 3.706 \times 10^{8} \space \text{N/m}

c_{vv}(adj) = 5.685 \times 10^{6} + \frac{2 (0.05)(4.077 \times 10^{8})}{65.2}

c_{vv}(adj) = 5.685 \times 10^{6} + 0.626 \times 10^{6} = 6.311 \times 10^{6} \space \text{N.s/m}

The material damping reduces the vertical stiffness by about 9 % and raises the vertical damping constant by about 11 %. The same procedure is applied to the remaining modes once their natural frequencies are known.

Step 8 — Equivalent damping ratio and code limits

It is often more convenient to express the result as a damping ratio. For the vertical mode, using the elastic (geometric) constants,

\beta_v = \frac{c_{vv}}{2 \sqrt{k_{vv} \space m}} = \frac{5.685 \times 10^{6}}{2 \sqrt{(4.077 \times 10^{8})(9.60 \times 10^{4})}} = 0.45

A geometric damping ratio of 45 % is very high. Both experiments and field measurements show that large foundations vibrating at small amplitudes develop less damping than the halfspace theory predicts, because reflected waves return energy to the footing. ACI 351.3R-18 (section 5.4.2) therefore recommends capping the calculated value, for example the EPRI limit of 50 % for vertical motion, the DIN 4024-2 limit of 25 % for rigid block foundations, or the 50 % reduction of the analytical value suggested by Novak.

Observations

Only the halfspace parameters survive when $l = 0$ . Embedding the same block would add the side-layer terms $S_{i1}$ and $S_{i2}$ , which increase every stiffness and, more importantly, increase the damping substantially. This is why guideline No. 8 for trial sizing recommends embedding the block whenever it is practical.

The rocking stiffness about the centre of gravity is dominated by neither term alone. The true rocking resistance of the base contributes $24.71 / 46.15 = 54 \%$ and the base shear acting at the lever arm $y_c$ contributes the remaining 46 %. Lowering the centre of gravity, by making the block thicker and the machine mount lower, reduces the coupling.

The cross terms $k_{u \psi}$ and $c_{u \psi}$ are negative and are not small compared with the diagonal terms. They cannot be ignored: the horizontal and rocking modes must be solved as a two degree of freedom coupled system, which is carried out in Chapter 4.

Material damping acts in both directions. It removes stiffness in proportion to $\omega c_i$ and adds damping in proportion to $k_i / \omega$ , so its relative importance grows as the operating frequency falls.

Figure 2.6.2 Embedded Machine Foundation of Example 2.6.2

Example 2.6.2

The block of Example 2.6.1 is enlarged and is now cast against undisturbed native soil so that half of its thickness is below grade, giving an embedment depth $l = 1.219 \space \text{m} \space (4 \space \text{ft})$ . The soil is heavier and has a higher Poisson's ratio. Calculate the stiffness and damping constants for the vertical mode $v$ , the coupled horizontal–rocking mode $u - \psi$ in the X–Y plane, and the torsional mode $\eta$ , all referred to the centre of gravity of the machine-plus-block system. Separate the base contribution from the side-layer contribution in every mode, and comment on what the embedment buys.

Table 2.6.3 Given Data for Example 2.6.2

Item	Quantity	SI	US Customary
Machine	Weight	88.96 kN	20,000 lb
Machine	Height of horizontal excitation	4.572 m	15 ft
Footing	Reinforced concrete unit weight $\gamma_c$	23.57 kN/m³	150 lb/ft³
Footing	Plan dimension $a$ (rocking direction)	4.572 m	15 ft
Footing	Plan dimension $b$	6.096 m	20 ft
Footing	Thickness $c$	2.438 m	8 ft
Footing	Embedment depth $l$	1.219 m	4 ft
Footing	Height of the system C.G. above the base $y_c$	1.524 m	5 ft
Soil	Unit weight $\gamma$	18.85 kN/m³	120 lb/ft³
Soil	Mass density $\rho$	1922 kg/m³	3.727 slug/ft³
Soil	Shear wave velocity $V_s$	150 m/s	492.1 ft/s
Soil	Material damping $\tan \delta = 2 \beta_m$	0.10	0.10
Soil	Poisson's ratio $\nu$	0.33	0.33
Side layer	Undisturbed native soil, $G_s = G$ , $\rho_s = \rho$	—	—
System	Total mass of machine + block $m$	1.724 × 10⁵ kg	11,800 slug

The embedment length is no longer zero, so every constant now carries two contributions: a base term generated by the soil reactions under the footing, governed by the halfspace parameters $C_{i1}$ and $\overline{C}_{i2}$ , and a side-layer term generated by the soil reacting against the buried faces of the block, governed by the parameters $S_{i1}$ and $\overline{S}_{i2}$ . Both are calculated below and kept separate so that the value of the embedment can be seen directly.

Choice of soil class. Table 2.4.1 tabulates the parameters for only two broad classes: granular, for which $\nu = 0.25$ is presumed, and cohesive, for which $\nu = 0.4$ is presumed. The value $\nu = 0.33$ given here falls between the two. This example adopts the cohesive column, which is the appropriate class for a soil able to stand against the buried faces of the block and supply the side-layer reactions. The choice is not cosmetic: reading the granular column instead would lower $k_{vv}$ by about 27 % and $k_{\psi \psi}$ by about 16 %. Where $\nu$ is known with confidence, the rigorous route is the Veletsos–Verbic closed form of section 2.4, whose coefficients are tabulated at exactly $\nu = 1/3$ ; it is implemented in the calculator of section 2.5.

Step 1 — Equivalent radii of the rectangular base

R_u = R_v = \sqrt{\frac{a b}{\pi}} = \sqrt{\frac{4.572 \times 6.096}{\pi}} = \sqrt{\frac{27.871}{\pi}} = 2.979 \space \text{m} = 9.77 \space \text{ft}

R_{\psi} = \sqrt[4]{\frac{b a^3}{3 \pi}} = \sqrt[4]{\frac{6.096 \times (4.572)^3}{3 \pi}} = \sqrt[4]{\frac{582.6}{9.425}}

R_{\psi} = \sqrt[4]{61.81} = 2.804 \space \text{m} = 9.20 \space \text{ft}

R_{\eta} = \sqrt[4]{\frac{a b (a^2 + b^2)}{6 \pi}} = \sqrt[4]{\frac{27.871 \times \big( (4.572)^2 + (6.096)^2 \big)}{6 \pi}}

R_{\eta} = \sqrt[4]{\frac{27.871 \times 58.06}{18.85}} = \sqrt[4]{85.85} = 3.044 \space \text{m} = 9.99 \space \text{ft}

The aspect ratio is $b/a = 1.33 < 2$ , so the equivalent-radius idealisation remains acceptable.

Step 2 — Soil properties

The mass density follows from the given unit weight,

\rho = \frac{\gamma}{g} = \frac{18 \space 850}{9.81} = 1922 \space \text{kg/m}^3 = 3.727 \space \text{slug/ft}^3

G = \rho V_s^2 = 1922 \times (150)^2 = 4.323 \times 10^{7} \space \text{N/m}^2 = 9.030 \times 10^{5} \space \text{lb/ft}^2

\sqrt{\rho G} = \sqrt{1922 \times 4.323 \times 10^{7}} = 2.882 \times 10^{5} \space \text{N.s/m}^3

The block is cast against undisturbed native soil, so the side layer has the same properties as the base soil,

\frac{G_s}{G} = 1 \qquad \text{and} \qquad \sqrt{\frac{\rho_s G_s}{\rho G}} = 1

Step 3 — Embedment ratios and side-layer parameters

The embedment ratio is referred to the equivalent radius of the mode being considered, $\delta_i = l / R_i$ ,

\delta_v = \delta_u = \frac{1.219}{2.979} = 0.409 \qquad \delta_{\psi} = \frac{1.219}{2.804} = 0.435 \qquad \delta_{\eta} = \frac{1.219}{3.044} = 0.401

Reading the cohesive rows of Table 2.4.1 for both the halfspace and the side layer,

C_{v1} = 7.5 \quad \overline{C}_{v2} = 6.8 \quad S_{v1} = 2.7 \quad \overline{S}_{v2} = 6.7

C_{u1} = 5.1 \quad \overline{C}_{u2} = 3.2 \quad S_{u1} = 4.1 \quad \overline{S}_{u2} = 10.6

C_{\psi 1} = 4.3 \quad \overline{C}_{\psi 2} = 0.7 \quad S_{\psi 1} = 2.5 \quad \overline{S}_{\psi 2} = 1.8

C_{\eta 1} = 4.3 \quad \overline{C}_{\eta 2} = 0.7 \quad S_{\eta 1} = 10.2 \quad \overline{S}_{\eta 2} = 5.4

Step 4 — Vertical mode

k_{vv} = G R_v \Big( C_{v1} + \frac{G_s}{G} \delta_v S_{v1} \Big) = G R_v C_{v1} + G_s \space l \space S_{v1}

k_{vv} = 4.323 \times 10^{7} \times 2.979 \times 7.5 + 4.323 \times 10^{7} \times 1.219 \times 2.7

k_{vv} = 9.658 \times 10^{8} + 1.423 \times 10^{8} = 1.108 \times 10^{9} \space \text{N/m} = 7.593 \times 10^{7} \space \text{lb/ft}

c_{vv} = R_v^2 \sqrt{\rho G} \Big( \overline{C}_{v2} + \delta_v \sqrt{\frac{\rho_s G_s}{\rho G}} \overline{S}_{v2} \Big) = \sqrt{\rho G} R_v^2 \overline{C}_{v2} + \sqrt{\rho_s G_s} \space R_v \space l \space \overline{S}_{v2}

c_{vv} = 2.882 \times 10^{5} \times (2.979)^2 \times 6.8 + 2.882 \times 10^{5} \times 2.979 \times 1.219 \times 6.7

c_{vv} = 1.739 \times 10^{7} + 7.013 \times 10^{6} = 2.440 \times 10^{7} \space \text{N.s/m} = 1.672 \times 10^{6} \space \text{lb.s/ft}

Step 5 — Coupled horizontal and rocking mode

Horizontal translation

k_{uu} = G R_u C_{u1} + G_s \space l \space S_{u1} = 4.323 \times 10^{7} \times 2.979 \times 5.1 + 4.323 \times 10^{7} \times 1.219 \times 4.1

k_{uu} = 6.568 \times 10^{8} + 2.161 \times 10^{8} = 8.729 \times 10^{8} \space \text{N/m} = 5.981 \times 10^{7} \space \text{lb/ft}

c_{uu} = \sqrt{\rho G} R_u^2 \overline{C}_{u2} + \sqrt{\rho_s G_s} \space R_u \space l \space \overline{S}_{u2}

c_{uu} = 2.882 \times 10^{5} \times 8.872 \times 3.2 + 2.882 \times 10^{5} \times 2.979 \times 1.219 \times 10.6

c_{uu} = 8.183 \times 10^{6} + 1.110 \times 10^{7} = 1.928 \times 10^{7} \space \text{N.s/m} = 1.321 \times 10^{6} \space \text{lb.s/ft}

Rocking

The complete expression for the rocking stiffness about the centre of gravity is,

k_{\psi \psi} = G R^3 \Big[ C_{\psi 1} + \Big( \frac{y_c}{R} \Big)^2 C_{u1} + \frac{G_s}{G} \delta S_{\psi 1} + \frac{G_s}{G} \delta \Big( \frac{\delta^2}{3} + \frac{y_c^2}{R^2} - \delta \frac{y_c}{R} \Big) S_{u1} \Big]

Substituting $\delta = l / R$ and expanding, each term takes the equivalent radius that belongs to it,

k_{\psi \psi} = G \big[ R_{\psi}^3 C_{\psi 1} + R_u \space y_c^2 \space C_{u1} \big] + G_s \big[ R_{\psi}^2 \space l \space S_{\psi 1} + I_l \space S_{u1} \big]

where $I_l$ is the second moment of the buried face about the centre of gravity, obtained from $\int_0^{l} (y_c - y)^2 dy$ ,

I_l = y_c^2 \space l - y_c \space l^2 + \frac{l^3}{3} = (1.524)^2 (1.219) - (1.524)(1.219)^2 + \frac{(1.219)^3}{3}

I_l = 2.831 - 2.265 + 0.604 = 1.170 \space \text{m}^3

The base contribution is,

4.323 \times 10^{7} \big[ (2.804)^3 \times 4.3 + 2.979 \times (1.524)^2 \times 5.1 \big]

= 4.323 \times 10^{7} \big[ 94.79 + 35.28 \big] = 5.624 \times 10^{9} \space \text{N.m/rad}

and the side-layer contribution is,

4.323 \times 10^{7} \big[ (2.804)^2 \times 1.219 \times 2.5 + 1.170 \times 4.1 \big]

= 4.323 \times 10^{7} \big[ 23.96 + 4.80 \big] = 1.243 \times 10^{9} \space \text{N.m/rad}

k_{\psi \psi} = 5.624 \times 10^{9} + 1.243 \times 10^{9} = 6.867 \times 10^{9} \space \text{N.m/rad} = 5.065 \times 10^{9} \space \text{lb.ft/rad}

The rocking damping constant follows the same pattern,

c_{\psi \psi} = \sqrt{\rho G} \big[ R_{\psi}^4 \overline{C}_{\psi 2} + R_u^2 \space y_c^2 \space \overline{C}_{u2} \big] + \sqrt{\rho_s G_s} \big[ R_{\psi}^3 \space l \space \overline{S}_{\psi 2} + I_l \space \overline{S}_{u2} \big]

= 2.882 \times 10^{5} \big[ 61.81 \times 0.7 + 8.872 \times 2.323 \times 3.2 \big]

+ \space 2.882 \times 10^{5} \big[ 22.05 \times 1.219 \times 1.8 + 1.170 \times 10.6 \big]

= 2.882 \times 10^{5} \big[ 43.27 + 65.94 \big] + 2.882 \times 10^{5} \big[ 48.37 + 12.41 \big]

c_{\psi \psi} = 3.148 \times 10^{7} + 1.752 \times 10^{7} = 4.900 \times 10^{7} \space \text{N.m.s/rad} = 3.614 \times 10^{7} \space \text{lb.ft.s/rad}

Cross (coupling) terms

The side-layer reaction acts over the depth $l$ , so its resultant sits at mid-depth and the lever arm about the centre of gravity is $(y_c - l/2)$ ,

k_{u \psi} = - \Big[ G R_u \space y_c \space C_{u1} + G_s \space l \space \Big( y_c - \frac{l}{2} \Big) S_{u1} \Big]

= - \big[ 4.323 \times 10^{7} \times 2.979 \times 1.524 \times 5.1 + 4.323 \times 10^{7} \times 1.219 \times 0.914 \times 4.1 \big]

k_{u \psi} = - \big[ 1.001 \times 10^{9} + 1.976 \times 10^{8} \big] = - 1.199 \times 10^{9} \space \text{N/rad} = - 2.694 \times 10^{8} \space \text{lb/rad}

c_{u \psi} = - \Big[ \sqrt{\rho G} R_u^2 \space y_c \space \overline{C}_{u2} + \sqrt{\rho_s G_s} \space R_u \space l \space \Big( y_c - \frac{l}{2} \Big) \overline{S}_{u2} \Big]

= - \big[ 2.882 \times 10^{5} \times 8.872 \times 1.524 \times 3.2 + 2.882 \times 10^{5} \times 2.979 \times 1.219 \times 0.914 \times 10.6 \big]

c_{u \psi} = - \big[ 1.247 \times 10^{7} + 1.014 \times 10^{7} \big] = - 2.262 \times 10^{7} \space \text{N.s/rad} = - 5.084 \times 10^{6} \space \text{lb.s/rad}

Step 6 — Torsional mode

k_{\eta \eta} = G R_{\eta}^3 \Big( C_{\eta 1} + \frac{G_s}{G} \delta_{\eta} S_{\eta 1} \Big) = G R_{\eta}^3 C_{\eta 1} + G_s R_{\eta}^2 \space l \space S_{\eta 1}

k_{\eta \eta} = 4.323 \times 10^{7} \times 28.20 \times 4.3 + 4.323 \times 10^{7} \times 9.266 \times 1.219 \times 10.2

k_{\eta \eta} = 5.243 \times 10^{9} + 4.982 \times 10^{9} = 1.022 \times 10^{10} \space \text{N.m/rad} = 7.542 \times 10^{9} \space \text{lb.ft/rad}

c_{\eta \eta} = \sqrt{\rho G} R_{\eta}^4 \overline{C}_{\eta 2} + \sqrt{\rho_s G_s} R_{\eta}^3 \space l \space \overline{S}_{\eta 2}

c_{\eta \eta} = 2.882 \times 10^{5} \times 85.85 \times 0.7 + 2.882 \times 10^{5} \times 28.20 \times 1.219 \times 5.4

c_{\eta \eta} = 1.732 \times 10^{7} + 5.352 \times 10^{7} = 7.084 \times 10^{7} \space \text{N.m.s/rad} = 5.225 \times 10^{7} \space \text{lb.ft.s/rad}

Table 2.6.4 Summary of the Stiffness and Damping Constants (material damping neglected)

Mode	Constant	SI	US Customary	Side layer share
Vertical	$k_{vv}$	$1.108 \times 10^{9}$ N/m	$7.593 \times 10^{7}$ lb/ft	13 %
Vertical	$c_{vv}$	$2.440 \times 10^{7}$ N.s/m	$1.672 \times 10^{6}$ lb.s/ft	29 %
Horizontal	$k_{uu}$	$8.729 \times 10^{8}$ N/m	$5.981 \times 10^{7}$ lb/ft	25 %
Horizontal	$c_{uu}$	$1.928 \times 10^{7}$ N.s/m	$1.321 \times 10^{6}$ lb.s/ft	58 %
Rocking	$k_{\psi \psi}$	$6.867 \times 10^{9}$ N.m/rad	$5.065 \times 10^{9}$ lb.ft/rad	18 %
Rocking	$c_{\psi \psi}$	$4.900 \times 10^{7}$ N.m.s/rad	$3.614 \times 10^{7}$ lb.ft.s/rad	36 %
Cross	$k_{u \psi}$	$-1.199 \times 10^{9}$ N/rad	$-2.694 \times 10^{8}$ lb/rad	16 %
Cross	$c_{u \psi}$	$-2.262 \times 10^{7}$ N.s/rad	$-5.084 \times 10^{6}$ lb.s/rad	45 %
Torsion	$k_{\eta \eta}$	$1.022 \times 10^{10}$ N.m/rad	$7.542 \times 10^{9}$ lb.ft/rad	49 %
Torsion	$c_{\eta \eta}$	$7.084 \times 10^{7}$ N.m.s/rad	$5.225 \times 10^{7}$ lb.ft.s/rad	76 %

Step 7 — Effect of the soil material damping

The material damping is unchanged at $\tan \delta = 2 \beta_m = 0.1$ , so $\beta_m = 0.05$ , and the adjustment of ACI 351.3R-18 Eq. (5.4.4b) and (5.4.4c) applies as before. Evaluating at the vertical natural frequency,

\omega_v = \sqrt{\frac{k_{vv}}{m}} = \sqrt{\frac{1.108 \times 10^{9}}{1.724 \times 10^{5}}} = 80.2 \space \text{rad/s} \space \space (12.8 \space \text{Hz})

k_{vv}(adj) = k_{vv} - 2 \beta_m \omega c_{vv} = 1.108 \times 10^{9} - 2 (0.05)(80.2)(2.440 \times 10^{7})

k_{vv}(adj) = 1.108 \times 10^{9} - 1.957 \times 10^{8} = 9.12 \times 10^{8} \space \text{N/m}

c_{vv}(adj) = c_{vv} + \frac{2 \beta_m k_{vv}}{\omega} = 2.440 \times 10^{7} + \frac{2 (0.05)(1.108 \times 10^{9})}{80.2}

c_{vv}(adj) = 2.440 \times 10^{7} + 1.382 \times 10^{6} = 2.578 \times 10^{7} \space \text{N.s/m}

The dimensionless frequency should now be checked against the range over which the constant parameters were tabulated,

a_{ov} = \frac{\omega_v R_v}{V_s} = \frac{80.2 \times 2.979}{150} = 1.59

The halfspace parameters $C_{i1}$ and $\overline{C}_{i2}$ are valid for $a_o < 2.0$ , so they are still in range. The side-layer parameters $S_{i1}$ and $\overline{S}_{i2}$ , however, are tabulated only for $0.5 < a_o < 1.5$ , and $a_{ov} = 1.59$ has just passed that limit. Since the side layer supplies 29 % of the vertical damping and 76 % of the torsional damping in this footing, the constant-parameter approximation is being asked to do more work than it comfortably can. For a foundation of this size, the frequency-dependent expressions of section 2.4 and the calculator of section 2.5 should be used to confirm the result.

Step 8 — Equivalent damping ratio and code limits

\beta_v = \frac{c_{vv}}{2 \sqrt{k_{vv} \space m}} = \frac{2.440 \times 10^{7}}{2 \sqrt{(1.108 \times 10^{9})(1.724 \times 10^{5})}} = \frac{2.440 \times 10^{7}}{2 (1.382 \times 10^{7})} = 0.88

A geometric damping ratio of 88 % is far beyond anything that should be carried into a response calculation. It is roughly double the 45 % obtained for the surface footing of Example 2.6.1, the increase coming partly from the embedment and partly from the larger base area. Either way the result must be capped: ACI 351.3R-18 (section 5.4.2) cites the EPRI limit of 50 % for vertical motion and the DIN 4024-2 limit of 25 % for rigid block foundations. Taking the DIN limit,

\beta_{v} \space \text{(design)} = 0.25 \quad \Rightarrow \quad c_{vv} \space \text{(design)} = 2 (0.25) \sqrt{k_{vv} \space m} = 6.91 \times 10^{6} \space \text{N.s/m}

which is less than one third of the analytical value. This is the single most important practical lesson of the embedded case: the theory predicts damping generously, and the code limit, not the halfspace formula, usually governs the design.

Observations

Embedding the block over only half of its thickness adds 13 % to the vertical stiffness but 29 % to the vertical damping, and 25 % to the horizontal stiffness but 58 % to the horizontal damping. Embedment buys damping far more efficiently than it buys stiffness. This is exactly why the trial-sizing guidelines recommend embedding the block whenever site conditions allow.

Torsion benefits the most. The side layer contributes 49 % of $k_{\eta \eta}$ and 76 % of $c_{\eta \eta}$ , because the side reactions act at the full lever arm $R_{\eta}$ from the axis of twist while the base reactions are spread over the contact area.

The rocking stiffness rose from $1.664 \times 10^{9}$ in Example 2.6.1 to $6.867 \times 10^{9}$ N.m/rad, a factor of about 4.1. Most of that comes from the larger plan dimension, since $R_{\psi}^3$ scales roughly with $a^{9/4}$ , and only 18 % comes from the embedment.

A practical caution: the side-layer terms assume full, permanent contact between the buried faces and the soil. Backfill that shrinks away from the block, or a gap opened by the vibration itself, removes that contribution entirely. Where the contact cannot be relied on, the safe check is to verify the foundation twice, once with the embedment and once as a surface footing.