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2 Copyright 2014 by NCEES. All rights reserved. All NCEES sample questions and solutions are copyrighted under the laws of the United States. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means without the prior written permission of NCEES. Requests for permissions should be addressed in writing to permissions@ncees.org or to NCEES Exam Publications, PO Box 1686, Clemson, SC ISBN Printed in the United States of America October 2014 First Printing

3 CONTENTS Introduction to NCEES Exams... 1 About NCEES Updates on exam content and procedures Exam-day schedule Admission to the exam site Examinee Guide Scoring and reporting Staying connected Structural exam format Design Standards... 3 Vertical Forces... 5 Exam Specifications... 7 AM Practice Exam PM Buildings Practice Exam PM Bridges Practice Exam AM Solutions PM Buildings Solutions PM Bridges Solutions Lateral Forces Exam Specifications AM Practice Exam PM Buildings Practice Exam PM Bridges Practice Exam AM Solutions PM Buildings Solutions PM Bridges Solutions iii

4 About NCEES The National Council of Examiners for Engineering and Surveying (NCEES) is a nonprofit organization made up of engineering and surveying licensing boards from all U.S. states and territories. Since its founding in 1920, NCEES has been committed to advancing licensure for engineers and surveyors in order to protect the health, safety, and welfare of the American public. NCEES helps its member licensing boards carry out their duties to regulate the professions of engineering and surveying. It develops best-practice models for state licensure laws and regulations and promotes uniformity among the states. It develops and administers the exams used for engineering and surveying licensure throughout the country. It also provides services to help licensed engineers and surveyors practice their professions in other U.S. states and territories. Updates on exam content and procedures Visit us at ncees.org/exams for updates on everything exam-related, including specifications, exam-day policies, scoring, and corrections to published exam preparation materials. This is also where you will register for the exam and find additional steps you should follow in your state to be approved for the exam. Exam-day schedule Be sure to arrive at the exam site on time. Late-arriving examinees will not be allowed into the exam room once the proctor has begun to read the exam script. The report time for the exam will be printed on your Exam Authorization. Normally, you will be given 1 hour between morning and afternoon sessions. Admission to the exam site To be admitted to the exam, you must bring two items: (1) your Exam Authorization and (2) a current, signed, government-issued identification. Examinee Guide The NCEES Examinee Guide is the official guide to policies and procedures for all NCEES exams. All examinees are required to read this document before starting the exam registration process. You can download it at ncees.org/exams. It is your responsibility to make sure that you have the current version. NCEES exams are administered in either a computer-based format or a pencil-and-paper format. Each method of administration has specific rules. This guide describes the rules for each exam format. Refer to the appropriate section for your exam. Scoring and reporting NCEES typically releases exam results to its member licensing boards 8 10 weeks after the exam. Depending on your state, you will be notified of your exam result online through your MyNCEES account or via postal mail from your state licensing board. Detailed information on the scoring process can be found at ncees.org/exams. Staying connected To keep up to date with NCEES announcements, events, and activities, connect with us on your preferred social media network. 1

5 Structural exam format The 16-hour Structural exam is a breadth and depth exam offered in two components on successive days. The 8-hour Vertical Forces (Gravity/Other) and Incidental Lateral component is offered only on Friday. It focuses on gravity loads and includes minor lateral loads such as earth pressures. The 8-hour Lateral Forces (Wind/Earthquake) component is offered only on Saturday and focuses on wind/earthquake loads. Each component of the structural exam has a breadth (morning) module and a depth (afternoon) module. Breadth modules (AM session): These modules contain questions covering a comprehensive range of structural engineering topics. All questions are multiple-choice. Depth modules (PM session): In these modules examinees must choose either buildings or bridges problems. Examinees must work the same topic area on both components. That is, if buildings is the topic area chosen in the Vertical Forces component, then buildings must be the topic area chosen in the Lateral Forces component. All questions are constructed response (essay). Examinees must take the breadth module of each component and one of the two depth modules in each component. To pass the structural exam, you must receive acceptable results on both components. The components may be taken and passed in different exam administrations. Summary of Structural Exam Format AM Session (Breadth) PM Session (Depth) Friday Vertical Forces 40 multiple-choice questions ( ) See p. 7 for details and specifications. Choose EITHER: Buildings ( ) or Bridges ( ) See p. 11 for details and specifications. Saturday Lateral Forces 40 multiple-choice questions ( ) See p. 127 for details and specifications. Choose EITHER: Buildings ( ) or Bridges ( ) See p. 131 for details and specifications. NCEES will send a results notice to your licensing board each administration that you take a component. The notice will show the history of your performance on each component attempted. The results for each component will be listed as acceptable or unacceptable. After you have received an acceptable result on both components within a 5-year period, NCEES will notify your board that you have passed the Structural exam. 2

6 STRUCTURAL ENGINEERING Design Standards 1 These standards apply to the Vertical and Lateral components of the Structural Engineering exam. Changes to design standards are posted on ncees.org/exams. Effective Beginning with the April 2015 Examinations ABBREVIATION AASHTO IBC ASCE 7 ACI 318 AISC AISC AISI TMS 402/602 DESIGN STANDARD TITLE AASHTO LRFD Bridge Design Specifications, 6th edition, 2012, American Association of State Highway & Transportation Officials, Washington, DC. International Building Code, 2012 edition (without supplements), International Code Council, Falls Church, VA. Minimum Design Loads for Buildings and Other Structures, 2010, American Society of Civil Engineers, Reston, VA. Building Code Requirements for Structural Concrete, 2011, American Concrete Institute, Farmington Hills, MI. Steel Construction Manual, 14th edition, American Institute of Steel Construction, Inc., Chicago, IL. Seismic Design Manual, 2nd edition, American Institute of Steel Construction, Inc., Chicago, IL. North American Specification for the Design of Cold-Formed Steel Structural Members, 2007 edition with Supplement No. 2 (2010), American Iron and Steel Institute, Washington, DC. NDS National Design Specification for Wood Construction ASD/LRFD, 2012 edition & National Design Specification Supplement, Design Values for Wood Construction, 2012 edition, American Forest & Paper Association, Washington, DC. NDS Special Design Provisions for Wind and Seismic with Commentary, 2008 edition, American Forest & Paper Association, Washington, DC. PCI PCI Design Handbook: Precast and Prestressed Concrete, 7th edition, 2010, Precast/Prestressed Concrete Institute, Chicago, IL. Building Code Requirements and Specifications for Masonry Structures (and related commentaries), 2011; The Masonry Society, Boulder, CO; American Concrete Institute, Detroit, MI; and Structural Engineering Institute of the American Society of Civil Engineers, Reston, VA. Notes 1. Solutions to exam questions that reference a standard of practice are scored based on this list. Solutions based on other editions or standards will not receive credit. All questions use the US Customary System (USCS) of units. 3

7 VERTICAL FORCES AM PRACTICE EXAM 13

8 Vertical Forces This is a preview. Some pages have been omitted The figure shows an elevation view of a concrete highway bridge. Design Code: AASHTO LRFD Bridge Design Specifications, 6th edition, Design Data: Concrete modulus of elasticity E c = 3,605 ksi Column moment of inertia 636,000 in 4 Coefficient of thermal expansion in./in./ F Concrete shrinkage after 28 days sh = in./in. Assumptions: Columns are fixed at top of footing. Superstructure is pinned at the top of the columns. Bearings at the abutments are frictionless. Columns are very flexible compared to superstructure. The unfactored horizontal load (kips) at the top of each column due to shrinkage after 28 days and a temperature fall of 40F is most nearly: (A) 14 (B) 70 (C) 159 (D) 263 ABUTMENT ABUTMENT 40'-0" 120'-0" SUPERSTRUCTURE 40'-0" 20'-0" SINGLE COLUMN TOP OF FOOTING 20'-0" ELEVATION Copyright 2014 by NCEES 14 GO ON TO THE NEXT PAGE

9 AM Practice Exam 109. The beam shown is subjected to a uniform load and a moving concentrated load. Design Data: Uniform live load Concentrated live load 1 klf 10 kips Assumption: The uniform live load may extend the entire length of the beam from A to C. The maximum live load shear (kips) on the right of Support B is most nearly: (A) 32 (B) 36 (C) 40 (D) 46 HINGE A B C 48'-0" 12'-0" 60'-0" 110. The figure shows a line diagram for a continuous beam that is supported as shown. The distribution factors at Joint B for Spans BA and BC are most nearly: BA BC (A) (B) (C) (D) I = 200 I = 300 I = '-0" 15'-0" 10'-0" A B C D Copyright 2014 by NCEES 21 GO ON TO THE NEXT PAGE

10 Vertical Forces This is a preview. Some pages have been omitted You are working as a quality control representative for a contractor on a project. If you discover a potentially unsafe condition at the project site, the initial action you should take is to: (A) (B) (C) (D) stop the construction of the project report the condition to the contractor report the condition to the owner report the condition to OSHA 117. A built-up column section is shown. Design Code: AISC: Steel Construction Manual, 14th edition. Design Data: Steel ASTM A 36, F y = 36 ksi Assumptions: K = 1.0 The unbraced lengths in the X and Y directions are 10 ft. The critical buckling stress F cr (ksi) is most nearly: (A) 22 (B) 26 (C) 35 (D) 48 PLATE 3/4" 14" CONTINUOUS L 4 4 1/4 CONTINUOUS (TYP.) BUILT-UP SECTION Copyright 2014 by NCEES 26 GO ON TO THE NEXT PAGE

11 Vertical Forces This is a preview. Some pages have been omitted The figure shows an 8-in. reinforced concrete masonry wall. Design Code: TMS 402/602: Building Code Requirements and Specifications for Masonry Structures (and related commentaries), Design Data: Hollow concrete masonry units f m 1,500 psi with Type S mortar. Cells with reinforcement are grouted. Steel reinforcement ASTM A615 Grade 60 Assumptions: Seismic forces do not govern. The wall is reinforced with 48-in. o.c. vertically at the centerline of the wall and 32-in. o.c. horizontally. The design axial load (plf) for the masonry wall is most nearly: ASD LRFD (A) 14,200 32,400 (B) 25,500 59,400 (C) 39,300 91,500 (D) 107, , ROOF JOISTS 4 12 LEDGER 12'-0" 8" CONCRETE MASONRY WALL Copyright 2014 by NCEES 40 GO ON TO THE NEXT PAGE

12 AM Practice Exam 134. The figure shows an 8-in. reinforced concrete masonry wall. Design Code: TMS 402/602: Building Code Requirements and Specifications for Masonry Structures (and related commentaries), Design Data: Hollow concrete masonry units f m 1,500 psi with Type S mortar. Cells with reinforcement are grouted. Steel reinforcement ASTM A615 Grade 60 Assumptions: Allowable stress design provisions apply. The wall is reinforced with 48-in. o.c. vertically at the centerline of the wall and 32-in. o.c. horizontally. The maximum allowable moment (ft-lb/ft) on the masonry wall based on the maximum allowable masonry flexural stress is most nearly: (A) 665 (B) 1,070 (C) 1,465 (D) 2, ROOF JOISTS 4 12 LEDGER 12'-0" 8" CONCRETE MASONRY WALL Copyright 2014 by NCEES 41 GO ON TO THE NEXT PAGE

13 VERTICAL FORCES PM BUILDINGS PRACTICE EXAM 49

14 Vertical Forces This is a preview. Some pages have been omitted The first floor of a three-story apartment building is partially buried as shown in Figure 601A. Design Codes: IBC: International Building Code, 2012 edition (without supplements). ASCE 7: Minimum Design Loads for Buildings and Other Structures, TMS 402/602: Building Code Requirements and Specifications for Masonry Structures (and Related Commentaries), ACI 318: Building Code Requirements for Structural Concrete, Design Data: Masonry fm = 2,000 psi Grout fc = 2,000 psi Reinforcing ASTM A615, Grade 60 Masonry weight (per surface area) 84 psf Concrete fc = 3,000 psi Concrete density 150 pcf Soil weight 110 pcf Rankine coefficient of passive soil pressure 2.5 Rankine coefficient of active soil pressure 0.35 Allowable soil bearing pressure 3,000 psf Coefficient of static friction (concrete on soil) 0.28 Assumptions: Neglect wind and seismic loads. No hydrostatic pressure on wall. Wall is not backfilled until construction is complete. Neglect slab on grade and soil over toe side of footing. REQUIREMENTS: On the actual exam, any sketches necessary for these requirements must be neatly drawn in your solution pamphlet. (a) (b) (c) For the CMU wall design shown in Figure 601A, verify that the wall stem is adequate for vertical loads shown and provided soil information. Use IBC load combination equation 16-2 (LRFD) or (ASD). Do not use load information given in Figure 601B for this requirement. For the footing shown in Figure 601B, show by calculation whether the footing size subjected to the loads indicated is or is not adequate for bearing and stability. Do not use vertical load data provided for Requirement (a). Neatly sketch the connection of the masonry wall to the concrete footing. Show the required dimensions of the embedment and anchorage of the steel reinforcement to the footing. Copyright 2014 by NCEES 50 GO ON TO THE NEXT PAGE

15 PM Buildings Practice Exam 601. (Continued) FIGURE 601A FOR REQUIREMENT (a) ONLY CONCENTRIC LOAD (P) FROM ALL ELEMENTS TOTAL DL = 980 plf FLOOR LL = 640 plf ROOF LL = 160 plf C L STUD = C L CMU = C L FTG SECOND FLOOR SHEATHING 2X STUD WALL 100 psf LL SURCHARGE P 4'-0" FINISHED GRADE 8" 8" CMU SOLID GROUTED d = 5-1/4" 32" 4'-0" C CMU L P EXPANSION MATERIAL SLAB ON GRADE 1'-0" 4" 4" FIGURE 601A H 1'-6" W SOIL SOIL LEVEL FIGURE 601B FOR REQUIREMENT (b) ONLY 1'-0" LOAD (P) AT C OF CMU L TOTAL DL = 3,000 plf (NOT INCL. FTG. WT.) 1'-4" 8" FIGURE 601B 1'-0" TOTAL HORIZONTAL FORCE DUE TO SOIL & SURCHARGE LOADS, H = 600 plf VERTICAL SOIL LOAD ON HEEL, W SOIL = 450 psf Copyright 2014 by NCEES 51 GO ON TO THE NEXT PAGE

16 VERTICAL FORCES PM BRIDGES PRACTICE EXAM 61

17 Vertical Forces This is a preview. Some pages have been omitted Figure 701 shows the elevation and cross sections of an interior prestressed concrete girder for a 120-ft-long, simple single-span highway bridge. The girders spaced on 5'-6" centers are composite with a 7 1/2-in. concrete slab. Each girder is prestressed with 34 strands, 8 of which are draped as shown. All strands are 1/2-in. nominal-diameter low-relaxation strands and prestressed to their allowable limits. Live load moments and shears are shown in Table 701. Design Specification: AASHTO LRFD Bridge Design Specifications, 6th edition, Design Data: Concrete slab, fc 4 ksi Concrete prestressed girders, f c 6 ksi Density of concrete kcf Ductility factor 1.0 Redundancy factor 1.0 Importance factor 1.0 Low-relaxation 1/2-in.-diameter strands, f pu 270 ksi Area of 1/2-in.-diameter strands, A s in 2 per strand Reinforcing bars, f y 60 ksi Dead load of the girder and slab, DC 1.30 kips/ft per girder Superimposed dead load, DW 0.26 kips/ft per girder Prestressed Girder Properties: Cross-sectional area, A g 713 in 2 Moment of inertia, I g 392,638 in 4 Neutral axis to bottom fiber in. Neutral axis to top fiber in. Section modulus (bottom), S ncb 12,224 in 3 Section modulus (top) S nct 12,715 in 3 Composite Section Properties: Cross-sectional area, A g 1,117 in 2 Moment of inertia, I c 704,000 in 4 Neutral axis to bottom fiber in. Section modulus (bottom), S cb 15,767 in 3 Section modulus (top) S ct 38,365 in 3 Effective slab width 66 in. Assumptions: Total strand prestress loss is 40.5 ksi. For simplicity, the support centerlines are to be considered at the end faces of the girder. All strands are fully bonded. There is no nonprestressed tension or compression reinforcement. Section is in tension-controlled region. Copyright 2014 by NCEES 62 GO ON TO THE NEXT PAGE

18 701. (Continued) TABLE 701 Maximum Live Load Moment per Girder (Dynamic Allowance Included) at 10th Points from a Support PM Bridges Practice Exam Location from support (ft) L 0.2L 0.3L 0.4L 0.5L Maximum live load moment (ft-kips) ,201 1,560 1,772 1,830 REQUIREMENTS: This is a preview. Some pages have been omitted. On the actual exam, any sketches necessary for these requirements must be neatly drawn in your solution pamphlet. (a) (b) Verify the flexural adequacy for Strength I Limit State of the composite girder at midspan, including checking for minimum reinforcement. Assume rectangular section behavior. Determine anchorage zone vertical web reinforcement for the girder, using #5 bars, and show them on a sketch. 2 STRANDS C L 2 STRANDS TOTAL - 34 STRANDS DRAPED - 8 STRANDS 66" (C. TO C. OF GIRDERS) 6 STRANDS 12 STRANDS 48'-0" 24'-0" 48'-0" N.A. 120'-0" GIRDER ELEVATION NOT TO SCALE 7-1/2" SLAB 30.88" 42" 8 STRANDS DRAPED ENDS OF GIRDER) 3" 3 SPACES AT 2" = 6" 44.64" #5 BARS 63" 32.12" 6" 26" N.A. 8 STRANDS DRAPED 2 1/2" 4 SPACES AT 2" = 8" MIDDLE OF GIRDER) COMPOSITE SECTION NOT TO SCALE GIRDER SECTION NOT TO SCALE FIGURE 701 Copyright 2014 by NCEES 63 GO ON TO THE NEXT PAGE

19 VERTICAL FORCES AM SOLUTIONS 69

20 AM Solutions 101. Due to symmetry, the loads at each pier will be equal. pier L L sh T 2 2 pier ( ) (40)( ) in. Ph3 For fixed-pin condition, ; Solve for P 3EI 3EI h (3, 605 kips/in )(636, 000 in )(0.32 in.) P 159 kips (20 12 in.) 3 20'-0" 120'-0" THE CORRECT ANSWER IS: (C) 102. w = (0.64 kips/ft)/lane AASHTO Art MLL ft-kips/lane 8 IM = 1.33 AASHTO Table M 1.75( ft-kips 98 ft-kips) 1, 010 ft- kips LL+I THE CORRECT ANSWER IS: (C) 103. ASCE 7-10, Figure 7-9 Formula. h l p d u g p h d g 30, l 100 ft (given) u ft THE CORRECT ANSWER IS: (C) 71

21 VERTICAL FORCES PM BUILDINGS SOLUTIONS 91

22 Vertical Forces This is a preview. Some pages have been omitted (a) CMU wall design ASD solution: D + H + F L (L r or S or R) IBC Eq Vertical Loads: Given P D = 980 plf P LR = = 120 plf P LF = = 480 plf Maximum stress in wall occurs at bottom of wall: Wall weight = 84 psf (given) 4.67 ft = 392 plf P total = = 1,972 lb/ft Lateral loads: Surcharge: (surcharge load live = 100 psf) (active soil pressure coefficient = 0.35) = 35 psf Active soil pressure: (active pressure = pcf = 38.5 psf/ft) (soil ht = 4 ft) = 154 psf 1 4 ft moment at base (35 psf 4 ft) 2 ft surcharge 1544 ft (active) ft-lb/ft Check compression: k = 2.1 (fixed base and free top) r t/ / (4.67 ft)(12 in/ft) kh r Use Eq. 2-21(neglecting compressive stress in steel per Sec ): TMS 402, Sec h Pa (0.25 f m An) r f 2,000 psi (given) A m n in. 12 in/ft 91.5 in /ft 2 92

23 601. (Continued) Pa (0.25)(2,000)(91.5 in /ft) 1 38,944 lb/ft > P 140 b This is a preview. Some pages have been omitted. m 2 total PM Buildings Solutions Check flexure: Ast 2 12 in in in 32 in. Em 900 fm 1,800,000 TMS , Sec Es 29,000,000 TMS , Sec F 0.45 f 900 psi TMS , Sec Es 29 Ast n (0.116 in )( 16.11) n n Em 1.8 b d (12 in.)(5.25 in.) k 2n ( n) n k j M allow (steel) Ast Fs jd 2 in.-lb ft-lb (0.116 in /ft)(32 ksi)(0.928)(5.25 in.) 18,085 1,507 ft ft 2 2 Fb jkbd (900)(0.928)(0.216)(12)(5.25) M allow (CMU) ,834 in.-lb/ft 2,486 ft-lb/ft Steel stress controls 1,507 ft-lb > 691 ft-lb OK Check combined stresses: 2M (2)(691 ft-lb 12 in./ft) fb 250 psi 2 2 jkbd (0.928)(0.216)(12 in.)(5.25) 1,972 lb/ft fa 22psi (12 in./ft)(7.625 in.) per fa fb 0.45 fm < in., 8 in. CMU wall OK for loads 93

24 VERTICAL FORCES PM BRIDGES SOLUTIONS 111

25 Vertical Forces This is a preview. Some pages have been omitted (a) Determine flexural strength for Strength I Limit State at midspan. Check minimum steel. A ps = (34 strands) (0.153 in 2 /strand) = 5.20 in 2 c fps fpu 1 k d p Eq k fpy f pu Eq f 270 ksi pu 1 py 0.85 Art f 0.9 f Table pu Determine d p : C L = 4.74 in. 34 d C L = in. k Assume rectangular section behavior ksi c 5.20 in (270 ksi) / (0.85) 4 ksi in in in. OK Eq c 7.14 fps fpu 1 k ksi Eq d p a in. Art a Mn Apsfpsdp Art Eq Mn 5.20 in ksi ,116 ft-kips Assume tension-controlled p/s concrete Art Verify tension-controlled assumption. d c E S E S c d c 0.003(d c) ES c 0.003(65.76 in in.) > in. Tension-controlled,

26 PM Bridges Solutions 701. (Continued) DC moment: kips / ft 2,340 ft-kips 8 DW moment: kips / ft ft-kips LL+ IM = 1,830 ft-kips Strength I Tables M u = 1.25 DC DW (LL+IM) = 1.25 (2,340 ft-kips) (468 ft-kips) (1,830 ft-kips) = 6,830 ft-kips < (1.0) (7,116 ft-kips) OK Check minimum steel: Art M M n cr Sc Mcr 31fr 2fcpeSc Mdnc 1 S nc f 0.24 f r c ksi S c = 15,767 in 3 bottom of composite section S ncb = 12,224 in 3 bottom of precast section M dnc = 2,340 ft-kips Noncomposite DC moment prestress prestressec fpe A I Prestress = P jack losses = (0.75)(270 ksi) 40.5 ksi = 162 ksi Prestress force = (162 ksi)(5.202 in 2 ) = 843 kips e = = in. Eq Art

27 LATERAL FORCES AM PRACTICE EXAM 133

28 Lateral Forces This is a preview. Some pages have been omitted The following information is for a building that is located in a seismic zone. Design Code: ASCE 7: Minimum Design Loads for Buildings and Other Structures, Design Data: Site shear wave velocity for the top 100 ft, v s 1,100 ft/sec Mapped spectral response acceleration at 0.2-sec period, S S 1.00 Mapped spectral response acceleration at 1-sec period, S Building period, T 0.80 sec Long-period transition period, T L 8 sec Assumption: No clay, peat, or liquefiable soils The design spectral response acceleration S a is most nearly: (A) 1.10 (B) 0.73 (C) 0.58 (D) 0.47 Copyright 2014 by NCEES 134 GO ON TO THE NEXT PAGE

29 AM Practice Exam 102. The figure shows a diagram for an agricultural building. Design Codes: IBC: International Building Code, 2012 edition (without supplements). ASCE 7: Minimum Design Loads for Buildings and Other Structures, Design Data: Basic wind speed, V Risk Category I 120 mph Risk Category II 132 mph Risk Category III-IV 143 mph Assumptions: Topographic factor, K zt = 1.0 Building is located in flat open country. The wind velocity pressure (psf) at mean roof height for MWFRS is most nearly: (A) 21.9 (B) 27.4 (C) 33.2 (D) '-0" 60'-0" 30'-0" Copyright 2014 by NCEES 135 GO ON TO THE NEXT PAGE

30 AM Practice Exam 108. The figure shows the plan view of a tilt-up concrete wall warehouse roof. Design Code: ASCE 7: Minimum Design Loads for Buildings and Other Structures, Design Data: Roof Top of panel Roof DL Panel DL Assumption: C = 0.15 s 26 ft, A.F.F. 26 ft, A.F.F. 15 psf 70 psf The seismic chord force (kips) at Point A is most nearly: (A) 11.3 (B) 21.1 (C) 24.9 (D) '-0" A 200'-0" ROOF PLAN 100'-0" 109. The figure shows the plan for a rigid floor diaphragm. Design Codes: IBC: International Building Code, 2012 edition (without supplements). ASCE 7: Minimum Design Loads for Buildings and Other Structures, Design Data: There are shear walls on all four sides. The calculated story seismic shear is 75 kips. The calculated center of gravity is shown. The total torsion (ft-kips) to be distributed to the shear walls is most nearly: (A) 0 (B) 375 (C) 563 (D) '-0" 30'-0" 50'-0" CR 50'-0" 50'-0" 50'-0" 50'-0" CG 50'-0" 5'-0" 75 kips Copyright 2014 by NCEES 141 GO ON TO THE NEXT PAGE

31 AM Practice Exam 124. An office building is supported by special concentrically braced frames. Design Code: AISC: Seismic Design Manual, 2nd edition. Design Data: Seismic Design Category D A500 Grade B hollow structural section tubes. Assumption: Amplified seismic brace force = 175 kips. The required tensile strength of the bracing connection (kips) is most nearly: ASD LRFD (A) (B) (C) (D) W SIDES HSS 6 6 5/16 HSS 3 3 1/4 Copyright 2014 by NCEES 153 GO ON TO THE NEXT PAGE

32 Lateral Forces This is a preview. Some pages have been omitted The roof diaphragm of a single-story building is shown in the figure. Design Code: NDS: National Design Specification for Wood Construction ASD/LRFD, 2012 edition & National Design Specification Supplement, Design Values for Wood Construction, 2012 edition. Design Data: = 1.0 if LRFD method is used. Wind load, W = 333 plf Assumption: C M = C t = C g = C = C eg = C di = C tn = 1.0 The number of 10d common nails required on each side of the splice to connect the two 2 Spruce Pine-Fir top plates together for the diaphragm chord force at Point A is most nearly: ASD LRFD (A) (B) (C) (D) '-0" 25'-0" RIDGE B A B 40'-0" W 2 TOP PLATE 2 TOP PLATE SECTION B-B TOP PLATE SPLICE Copyright 2014 by NCEES 160 GO ON TO THE NEXT PAGE

33 LATERAL FORCES PM BUILDINGS PRACTICE EXAM 169

34 Lateral Forces This is a preview. Some pages have been omitted The preliminary design of a renovation of an existing single-story office building with a wood roof framing system and masonry exterior walls is shown in Figure 801. Design Codes: IBC: International Building Code, 2012 edition (without supplements). ASCE 7: Minimum Design Loads for Buildings and Other Structures, TMS 402/602: Building Code Requirements and Specifications for Masonry Structures (and Related Commentaries), Design Data: Wind load 142 mph, Exposure C K zt = 1.67 Seismic load S DS = 0.70 Masonry fm = 1,500 psi Weight = 60 psf f y = 60,000 psi 8" partially grouted CMU REQUIREMENTS: On the actual exam, any sketches necessary for these requirements must be neatly drawn in your solution pamphlet. (a) (b) (c) (d) During the preliminary design phase, the existing building is to be checked for compliance with the 2012 IBC wind and seismic loads. List three items to be checked at each of the following: exterior CMU wall, roof framing, and foundation system (list 9 items total). Determine the design wind pressure and seismic design force on the parapet. For wind, neglect corner zones. Consider interior zones only. For a horizontal service level wind pressure of 100 psf, check whether #5 at 48" o.c. vertical reinforcement at the centerline of the wall is adequate for the parapet. Check both the shear stress and flexural stress of the reinforced CMU parapet. The roof diaphragm requires attachment to the masonry wall for an out-of-plane anchorage force of 420 plf. Neatly sketch a complete wall anchorage connection at 48" o.c. Identify all required components but do not design. Copyright 2014 by NCEES 170 GO ON TO THE NEXT PAGE

35 PM Buildings Practice Exam 801. (Continued) 2" DL ~ 180 lb/ft LL ~ 240 lb/ft 4'-0" ROOF WOOD FRAMING WITH WOOD SHEATHING 16'-0" 8" EXTERIOR PARTIALLY GROUTED CMU CONCRETE SLAB-ON-GRADE 1'-4" 1'-4" WIDE 10" UNREINFORCED FOOTING FIGURE 801 Copyright 2014 by NCEES 171 GO ON TO THE NEXT PAGE

36 LATERAL FORCES PM BRIDGES PRACTICE EXAM 179

37 Lateral Forces This is a preview. Some pages have been omitted Figure 901A shows a transverse section of a bridge pier. Figure 901B is a column interaction diagram to be used for design of the column. The columns are fixed at the top and bottom in both directions. Design Specification: AASHTO LRFD Bridge Design Specifications, 6th edition, Design Data: Concrete strength, f c Yield strength of reinforcement, f y 3.5 ksi 60 ksi Factored loads for the Extreme Event I load combination for each column are shown below. Seismic forces have been reduced by the appropriate reduction (R) factor. M u = 4,050 ft-kips V u = 600 kips P u = 2,700 kips (compression) Seismic Zone = 4 REQUIREMENTS: On the actual exam, any sketches necessary for these requirements must be neatly drawn in your solution pamphlet. (a) (b) (c) For the Extreme Event I load combination, determine the vertical reinforcement required for the column. Neglect any slenderness effect for the column. For the Extreme Event I load combination, design the required spiral reinforcement. Sketch an elevation view of the columns showing the size and spacing of the vertical and spiral reinforcements. Copyright 2014 by NCEES 180 GO ON TO THE NEXT PAGE

38 PM Bridges Practice Exam 901. (Continued) 1'-7" 4'-5" 4'-5" 1'-7" 11'-0" 11'-0" C L COL C L BOX GIRDER C L COL 5'-6" 4'-0" 4'-0" 20'-0" 2" CLEAR TRANSVERSE SECTION OF PIER f ' c f y = 3.5 ksi = 60 ksi FIGURE 901A 9,000 8,000 NOMINAL AXIAL STRENGTH, P n, kips 7,000 6,000 5,000 4,000 3,000 2,000 1,000 0 = = = = = ρ 1, ,000 2,000 3,000 4,000 5,000 6,000 NOMINAL MOMENT STRENGTH, M n, ft-kips FIGURE 901B Copyright 2014 by NCEES 181 GO ON TO THE NEXT PAGE

39 LATERAL FORCES AM SOLUTIONS 187

40 AM Solutions 101. Not Site Class F or E since no clay, peat, or liquefaction ASCE 7-10, Section 20.3, and Table v 1,100 ft/sec Site Class D s F 1.1 Table a S F S (1.1)(1.00) 1.10 Eq MS DS v a s S 2/3 S 0.73 Eq M1 v 1 D1 MS F 1.55 Table S F S (1.55)(0.45) 0.70 Eq S 2/3 S 0.47 Eq M1 T 0.2 S /S 0.13 Section D1 DS T S /S S D1 DS 0.64 SD Building period > T S and < TL Sa 0.58 Eq T 0.80 THE CORRECT ANSWER IS: (C) qh KhKztKdV ASCE 7-10, Eq in. 30 ft/ h 17.5 ft 2 K K Table z h 17.5 K 1.0 zt K 0.85 Table d V 120 (Risk Category I - IBC Table ) q (0.875)(1.0)(0.85)(120) 27.4 psf THE CORRECT ANSWER IS: (B) (640)(192 ft) Force kips Note: Load factor = 1.0 for wind for strength design. 2 Force for one panel kips kips Number of connectors 3.6; use 4 connectors 8.5 kips THE CORRECT ANSWER IS: (C) 189

41 LATERAL FORCES PM BUILDINGS SOLUTIONS 207

42 Lateral Forces This is a preview. Some pages have been omitted (a) Items at exterior CMU wall: 1. Adequacy of CMU parapet for wind and seismic out-of-plane loads. 2. Adequacy of CMU wall for combined loading of vertical loads with wind and seismic out-of-plane loads. 3. Adequacy of CMU wall for wind and seismic in-plane loads (shear walls). Items at roof framing: 1. Out-of-plane anchorage for exterior CMU wall. 2. Adequacy of wood sheathing diaphragm for wind and seismic loads. 3. In-plane connection to exterior CMU wall for transfer of diaphragm forces. 4. Adequacy of struts and chords of diaphragm. 5. Uplift connection to exterior CMU wall for transfer of wind forces. Items at foundation system: 1. Adequate safety factor for overturning. 2. Adequate safety factor for sliding. 3. Adequate safety factor for uplift. 4. Adequacy of soil bearing pressure. 5. Adequacy of connection to CMU wall. (b) Design wind pressure on the parapet ASCE 7, Sec p = q p (GC p GC pi ) (Components and cladding elements of parapets) q p = K z K zt K d V 2 ASCE 7, Eq. (30.3-1) K z = 0.90 Exposure C, z = 20 ft (top of parapet) ASCE 7, Table K zt = 1.67 (given) K d = 0.85 ASCE 7, Table V = 142 mph (given) q p = = 65.9 psf GC pi = 0.00 (solid parapet, open building condition) ASCE 7, Table GC p, h 60 ft ASCE 7, Fig Effective wind area ASCE 7, Sec = span length 4-ft height span 4 ft/3 = 16/3 = 5.3 ft 2 < 10 ft 2 Use 10 ft 2 Zone 4 wall positive pressure GC p = = (Ref footnote 5 for reduction) Zone 4 wall negative pressure GC p = = 0.99 (Ref footnote 5 for reduction) Zone 2 roof negative pressure GC p = 1.8 ASCE 7 Fig A POSITIVE WALL PRESSURE ASCE 7 FIGURE NEGATIVE WALL PRESSURE NEGATIVE WALL PRESSURE POSITIVE WALL PRESSURE p 1 p 2 OR p 3 p 4 LOAD CASE A (WINDWARD) p = p 1 + p 2 = (65.9)(0.90) + (65.9)(1.8) = 178 psf 208 LOAD CASE B (LEEWARD) p = p 3 + p 4 = (65.9)(0.90) + (65.9)(0.99) = 125 psf

43 PM Buildings Solutions 801. (Continued) Seismic design force on the parapet: ASCE 7, Sec a p S DS Wp z Fp 12 R /I h P P ASCE 7, Eq. (13.3-1) where z 16' Ip 1.0 ASCE 7, Sec h 16' a 2.5; R = 2.5 (parapets unbraced) ASCE 7, Table p p F p 1250 psf Controlling design load on the parapet = 178 psf (windward pressure) (c) Check if the existing 48" o.c. vertical at centerline of wall is adequate for the parapet: V 100 psf (1')(4') 400 lb / ft wall max M max 100 psf (1')(4') ft-lb / ft wall Check shear stress: Since no shear reinforcement is provided, assume one face shell resists shear force. 3V 3(400) fv 40psi 2A 2(11/4")(12") F v shall not exceed: TMS 402 Sec (a) 1.5 f 1.5 1, psi m (b) 120 psi (c) 37 psi 0.45 N / A v n ( ) / An A n based on two face shells and one grouted cell per 48" A A n n (2)(1 1/ 4")(48") (6" 1" 1 1/ 4")(7 5 / 8" (2)1 1/ 4") 4ft in / ft (420) / psi F = 41.7 psi > f = 40 psi OK v v 209

44 LATERAL FORCES PM BRIDGES SOLUTIONS 223

45 Lateral Forces This is a preview. Some pages have been omitted (a) Vertical reinforcement required for the column Design for moment and axial force Column requirement, Art h/d c = ratio of clear height to maximum plan dimension of column = 20/4 = 5 > 2.5 The column qualifies to be designed as a column and not as a pier. Slenderness of the column, Art , may be ignored per problem statement. Vertical reinforcement, flexural strength, Art b Design for M u = 4,050 ft-kips and P u max = 2,700 kips = 0.9 M n = nominal moment = M u / = 4,050/0.9 = 4,500 ft-kips P n = nominal axial force = P u / = 2,700/0.9 = 3,000 kips From the interaction diagram for M n = 4,500 ft-kips and P n = 3,000 kips = Limits of vertical reinforcement, Art a 0.01 < = < 0.04 OK A s = (48) 2 /4 = in 2 Using 29 #11 bars with A s = = in 2 will satisfy the requirement. (b) Design of spiral reinforcement Shear stress, Art Resistance factor, Art = 1.0 where V u = 600 kips B = 48 in. D r = ( ) 1.41 = in. (assuming #7 spiral) de 37.0 in. Eq. C d v = 0.9 d e = 33.3 in. Shear carried by concrete, Art Use 2, 45 if Av min provided and no tension in column per Art Check if concrete is effective in end regions per Art c. P u = 2,700 kips P u > 0.10(3.5 ksi)(1,810 in 2 ) = 634 kips Concrete is effective Vc (2) 3.5(48 in.)(33.3 in) 189 kips Eq

46 PM Bridges Solutions 901. (Continued) Spacing of spiral reinforcement, Eq. C s = pitch = Avfyd v/v s where A v = 0.6 in 2 (Try #7 spiral) f y = 60 ksi b v = 48 in. d v = 33.3 in (1.0) Vsrequired 411 kips 1.0 and (2 0.6) s 5.83 in. Use 5.0 in. 411 Check minimum transverse reinforcing, Art bs v Av f min c in OK f 60 y Afd v y v V s = = = 480 kips The nominal shear resistance V n is the lesser of: Vn Vc Vs kips Governs, OK Eq Vn 0.25 fc bvdv 0.25(3.5)(48.0)(33.3) 1,399 kips Eq Check maximum spacing, Art d b = 6(1.41) = 8.46 in. 6 in. governs, 5.0 in. OK End regions, Art c H e = length of end region = column dimension = 48 in. or = 20 12/6 = 40 in. or = 18 in. Length of the end region = 48 in. Governs Check confinement, Art d s 0.45(A g/ac 1)f c /fyh 0.45 (1,810/1,521) 13.5/ Eq or 0.12 f f / whichever is larger Eq d-1 s c y 2 s 4A s( D c) / s Dc 4A s/sdc 4A 2 s 4(0.6 in ) s 7.8in. D 0.007(44 in.) s c 225

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