Design Basis Ground Motion

Abstract

Earthquake can cause large destruction in industrial structures.

Appendix 1: Development of New Attenuation Relationships

2.1.1 A.1 Introduction

As explained in Sect. 2.7, there are two approaches available for developing attenuation relationships . First one is regression from strong motion database, and the second one is from simulation. The steps involved in development of new attenuation relationships are given below:

• Step 1: Obtain acceleration time histories from either earthquake records or from simulation for various magnitudes and distances.

• Step 2: Evaluate response spectra corresponding to required damping from the available acceleration time histories.

• Step 3: For each frequency of the response spectrum , segregate the spectral acceleration data for all magnitudes and distances.

• Step 4: Select a suitable form for attenuation relationship.

• Step 5: Evaluate the coefficients of attenuation relationship for each frequency using the regression procedure explained in subsequent section.

• Step 6: Evaluate the standard deviation for each frequency

2.1.2 A.2 Details of Evaluation of Coefficients of Attenuation Relationship Using Regression

After segregating the spectral acceleration for all magnitudes and distances corresponding to each frequency and selection of the form, the coefficients of attenuation relationship are evaluated using regression as described below:

• Step 1: The first step in regression is to obtain a system of linear equations from the data and attenuation relationship form.

• Step 2: Solution of set of linear equations to obtain the coefficients of attenuation relationship.

• Step 3: Evaluate the standard deviation of the data.

This procedure is illustrated in the example given below:

Example A.1

From the data of PGA (g) for different magnitudes and distances given in Table 2.14, evaluate the coefficients of attenuation relationship given below. Also, estimate the standard deviation.

$$ \log \left( {\text{PGA}} \right) = C_{1} + C_{2} M + C_{3} M^{2} + C_{4} { \log }R $$

Table 2.14 Sample data of PGA for various magnitudes and distances

• Step 1:

• The first step in regression is to obtain a system of linear equations.

• $$ AX = B $$

• In which A, B, and X are given as follows:

• $$ \begin{aligned} A \, & = \left[ {\begin{array}{*{20}c} 1 & {M_{1} } & {M_{1}^{2} } & {\log R_{1} } \\ 1 & {M_{2} } & {M_{2}^{2} } & {\log R_{2} } \\ 1 & {M_{3} } & {M_{3}^{2} } & {\log R_{3} } \\ 1 & {M_{4} } & {M_{4}^{2} } & {\log R_{4} } \\ 1 & {M_{5} } & {M_{5}^{2} } & {\log R_{5} } \\ 1 & {M_{6} } & {M_{6}^{2} } & {\log R_{6} } \\ 1 & {M_{7} } & {M_{7}^{2} } & {\log R_{7} } \\ 1 & {M_{8} } & {M_{8}^{2} } & {\log R_{8} } \\ 1 & {M_{9} } & {M_{9}^{2} } & {\log R_{9} } \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}c} {1.0000} & {4.5000} & {20.2500} & {1.3010} \\ {1.0000} & {4.5000 } & {20.2500} & {1.9031} \\ {1.0000} & {4.5000 } & {20.2500} & {2.1761} \\ {1.0000} & {5.0000} & {25.0000} & {1.4771} \\ {1.0000} & {5.0000} & {25.0000} & {1.7404} \\ {1.0000} & {5.5000} & {30.2500} & {1.8451} \\ {1.0000} & {6.0000} & {36.0000} & {1.4771} \\ {1.0000} & {6.0000} & {36.0000} & {2.3010} \\ {1.0000} & {6.5000} & {42.2500} & {1.6021} \\ \end{array} } \right] \\ \end{aligned} $$
  $$ \begin{aligned} B & = \left[ {\begin{array}{*{20}l} {{ \log }\left( {\text{PGA}} \right)_{1} } \\ {{ \log }\left( {\text{PGA}} \right)_{2} } \\ {{ \log }\left( {\text{PGA}} \right)_{3} } \\ {{ \log }\left( {\text{PGA}} \right)_{4} } \\ {{ \log }\left( {\text{PGA}} \right)_{5} } \\ \end{array} } \right] \\ & = \left[ {\begin{array}{*{20}c} { - 2.8774} \\ { - 2.7122} \\ { - 2.6540} \\ { - 2.8223} \\ { - 2.7510} \\ { - 2.7256} \\ { - 2.8223} \\ { - 2.6297} \\ { - 2.7870} \\ \end{array} } \right] \\ X & = \left[ {\begin{array}{*{20}l} {C_{1} } \\ {C_{2} } \\ {\begin{array}{*{20}l} {C_{3} } \\ {C_{4} } \\ \end{array} } \\ \end{array} } \right] \\ \end{aligned} $$

• Step 2: Obtain the least square solution for the set of linear equations, by solving

• $X = A^{ - 1} B$ and X can be obtained as:

• $$ X = \left[ {\begin{array}{*{20}l} { - 3.2728} \\ { 0.0325} \\ {\begin{array}{*{20}c} { - 0.0029} \\ {0.2444} \\ \end{array} } \\ \end{array} } \right] $$

• Step 3: Evaluate the standard deviation of $\log \left( {\text{PGA}} \right)$, which is obtained as 0.0816.