Seismic Principles

• Solution

  Best answer: (b)

   

  Explanation

  The Seismic Design Category for a building depends on what it is used for (the Risk Category) and the potential for ground shaking at the site (S_DS and S_D1). The Seismic Design Category is not influenced by irregularities.

   

  Step 1 - Determine the Risk Category

  Table 1604.5 in the California Building Code (CBC) can be used to identify the Risk Category. The Risk Category depends on what the building is going to be used for. From Table 1604.5, the Risk Category for a hospital with an emergency room would be IV.

   

  Step 2 - Identify the Seismic Design Category

  The procedures and tables for determining the Seismic Design Category are found in Section 1613.3.5 of the CBC. Based on the S_DS and S_D1 values that were given, and Risk Category IV, the building will be in Seismic Design Category D.

   

  The best answer is (b).

   

  Did you get this problem wrong?

  The book Seismic Principles is the ultimate resource to help you as you are preparing for and taking the real exam. There is an explanation of Seismic Design Categories and three fully-worked examples in Section 3.8 of Seismic Principles.

• Solution

  Best answer: (c)

   

  Explanation

  Minimum requirements for anchor bolts are found in Section 2308.3.1 of the California Building Code (CBC). The 5/8 in. diameter bolts are required because the Seismic Design Category is E. The plate washers are required because it is a shear wall in Seismic Design Category D or E. The 6 ft spacing is okay because the structure in not over two stories.

   

  Did you get this question wrong?

  The book Seismic Principles, is the ultimate resource to help you as you are preparing for and taking the real exam. There are great illustrations in Seismic Principles that show you the requirements, instead of just telling you. For example, pages 152 and 153 illustrate anchor bolt requirements to give you a better understanding of what the CBC is talking about.

• Solution

  Best answer: (b)

   

  Explanation

  Even though the sketches show both buildings deforming to the right, the expected displacements could occur either way. The critical location for the buildings to pound into each other would be at the roof elevation of the shorter building. Since it is unlikely that each building will experience its maximum deformation at the same point in a time, a square-root-sum-of-the-squares approach is used to calculate the appropriate spacing (see AISC 7-10 Section 12.12.3).

       [(3.7 in.)^2 + (3.5 in.)^2]^0.5 = 5.1 in.

  The best answer is (b).

   

  Did you get this question wrong?

  The book Seismic Principles, is the ultimate resource to help you as you are preparing for and taking the real exam. There are more than 200 examples in Seismic Principles that illustrate how to solve typical problems. Example 6.19 is a little harder than this problem because the inelastic displacement is not given and has to be calculated.

• Solution

  Best answer: (c)

   

  Explanation

  ASCE 7 puts an upper limit on the period that can be used when computing the base shear force (the seismic response coefficient).

   

  The four basic steps the check the upper limit are: 1) look up the parameters C_t and x, 2) compute T_a, 3) look up C_U, and 4) compare C_UT_a with the period determined from the computer analysis.

   

  Step 1 - Look up C_t and x

  An empirical procedure for estimating the period of a building requires values for C_t and x. Values for these parameters can be looked up in Table 12.8-2 of ASCE 7-10.

  Since the building has steel moment frames, C_t =0.028 and x=0.8.

   

  Step 2 - Compute T_a

  An approximate period, T_a, is calculated with the equation: T_a=C_th_n^x, where h_n is the height of the building.

     T_a = (0.028)*(200)^(0.8) = 1.94

   

  Step 3 - Look up C_U

  The upper limit when using a period determined from a computer model is equal to C_UT_a.

  For S_D1=0.6g, C_U will be 1.4

   

  Step 4 - Compare C_UT_a with the period determined from the computer analysis

     C_UT_a = (1.4)(1.94) = 2.7 sec.

  Since the period determined from the computer analysis (given as 3.0 seconds) exceeds this limit, the limiting value of 2.7 seconds is the maximum value that can be used for the period when determining the seismic response coefficient (the base shear).

   

  Choice (c) is the best answer.

   

  Did you get this question wrong?

  The book Seismic Principles, is the ultimate resource to help you as you are preparing for and taking the real exam. Versions of the ASCE 7 tables are provided in Seismic Principles, so that all the information you need for a problem (tables, equations, explanations, and examples) are all on one page. For example, look at Section 3.13 of Seismic Principles and see for yourself.

• Solution

  Best answer: (c)

   

  Explanation

  SCBF stands for special concentrically braced frame. The requirements for special concentrically braced frames are found in the AISC Seismic Provisions (AISC 341, Chapter F). K-bracing configurations (similar to image B) are specifically prohibited for SCBFs. The other bracing configurations shown are allowed, although additional requirements apply to cases A and C.

   

  Did you get this question wrong?

  The California Building Code (CBC) requires most steel buildings to constructed in accordance with AISC 341, but it does not reprint all of the requirements. Most people do not have a current copy of AISC 341 and are not familiar with this document. The book Seismic Principles, provides a summary of the requirements for various steel systems, making questions like this one easy. Take a look at Section 10.3 from Seismic Principles.

• Solution

  Best answer: (a)

   

  Did you get this question wrong?

  You are not alone. Very few engineering graduates have learned anything about diaphragm analysis. Unfortunately, several of the questions on the real exam will involve diaphragm analysis. The procedures for flexible and rigid diaphragm analysis are not covered in ASCE 7 or the CBC. You will need an additional reference in order to have the pertinent equations and examples for their use. Seismic Principles devotes an entire chapter to explaining flexible diaphragm analysis and has more than twenty-five fully-worked examples.

   

  Explanation

  Example 4.8 in Seismic Principles is identical the question above.

• Solution

  Best answer: (d)

   

  Explanation

  This question is about non-structural components. The requirements for non-structural components are found in Chapter 13 of ASCE 7. Many non-structural components are exempt and do not need to be specifically designed to resist seismic forces.

   

  Choices (a) and (b) are exempt, because the Seismic Design Category (SDC) is B, and all non-structural components are exempt in B except parapets supported by bearing or shear walls and tall bookshelves/cabinets.

   

  Choice (c) is exempt, because all mechanical and electrical components that weigh less than 20 lbs are exempt.

   

  Choice (d) is not exempt because it is a floor-supported cabinet that is over 6 ft tall. This non-structural component needs to be designed to withstand seismic forces without tipping over.

   

  Did you get this question wrong?

  Table 11.1 in Seismic Principles (including footnotes) explains which types of non-structural components are exempt.

• Solution

  Best answer: (b)

   

  Explanation

  The procedure for calculating diaphragm design forces is given in ASCE 7 Section 12.10.1. For this particular problem:

     F_px = (150 kips)*(475 kips)/(150 kips) = 150 kips

     F_px,min = 0.2*(1.7)*(1.0)*(475 kips) = 161 kips

     F_px,max = 0.4*(1.7)*(1.0)*(475 kips) = 322 kips

  Since the minimum value is greater than the value given by the first equation, it will govern. The best answer is (b).

   

  Did you get this question wrong?

  The book Seismic Principles, is the ultimate resource to help you as you are preparing for and taking the real exam. The formulas for computing diaphragm design forces can be confusing. Section 4.8 of Seismic Principles presents the diaphragm formulas and demonstrates how they are used for both floor and roof diaphragms.

• Solution

  Best answer: (d)

   

  Explanation

  Stiffness formulas for common walls, frames, and columns are given in Figure 2.7 of Seismic Principles. For a cantilever wall, the stiffness is a function of:

     A - the cross-sectional area of the wall

     I - the moment of inertia of the wall

     h - the height of the wall

     E - the modulus of elasticity

     G - the shear modulus of elasticity.

   

  Step 1 - Get values for A, I, h, and G (being careful with units)

  For the wall in question, the cross-sectional area is:

     A = (8 in.)(25 ft * 12 in./ft) = 2400 in.^2.

  For the wall in question, the moment of inertia is:

     I = (8 in.)(25 ft * 12 in./ft)^3/12 = 18,000,000 in.^4

  The height, modulus of elasticity, and shear modulus were given:

     h = (20 ft * 12 in./ft) = 240 in.

     E = 3500 ksi

     G = 1500 ksi

   

  Step 2 - Compute the stiffness

  Using the formula from Figure 2.7 in Seismic Principles:

     k = 1/{(240 in.)^3/(3*(3500 ksi)(18,000,000 in.^4)) +

                     1.2*(240 in.)/[(2400 in.^2)(1500 ksi)]} = 6530 k/in.

  The stiffness of the wall is 6530 k/in. so (d) is the best answer.

   

  Did you get this question wrong?

  You cannot solve problems like this in a reasonable amount of time unless you have a reference with the basic stiffness formulas. The book Seismic Principles, is the ultimate resource to help you as you are preparing for and taking the real exam. Figure 2.7 has stiffness formulas for basic structural elements. Example 2.5 in Seismic Principles illustrates the procedure for another wall, and provides more detailed discussion of the steps and how to be careful with units.

• Solution

  Best answer: (c)

   

  Explanation

  The concept is simple, but the explanation is a bit lengthy.  See Section 7.5 of Seismic Principles, particularly Example 7.8.

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