This is a preview. Some pages have been omitted ... Solutions ..... 69 PM Buildings Solutions ..... 91 PM Bridges Solutions..... 111 ... These standards apply to the Vertical and Lateral

This is a preview. Some pages have been omitted. Copyrighted by NCEES. For permission to reuse, email [email protected].

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 [email protected] or to NCEES Exam Publications, PO Box 1686, Clemson, SC 29633. ISBN 978-1-932613-69-8 Printed in the United States of America October 2014 First Printing


iii

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 .............................................................................. 13

PM Buildings Practice Exam ............................................................... 49

PM Bridges Practice Exam ................................................................. 61

AM Solutions ....................................................................................... 69

PM Buildings Solutions ....................................................................... 91

PM Bridges Solutions........................................................................ 111 Lateral Forces ......................................................................................... 125

Exam Specifications .......................................................................... 127

AM Practice Exam ............................................................................ 133

PM Buildings Practice Exam ............................................................. 169

PM Bridges Practice Exam ............................................................... 179 AM Solutions ..................................................................................... 187

PM Buildings Solutions ..................................................................... 207

PM Bridges Solutions ........................................................................ 223


1

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.


2

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

Friday—Vertical Forces Saturday—Lateral Forces

AM Session (Breadth)

40 multiple-choice questions (101–140)

See p. 7 for details and specifications.

40 multiple-choice questions (101–140)


PM Session (Depth)

Choose EITHER: Buildings (601–604)

or Bridges (701–703)


Choose EITHER: Buildings (801–804)

or Bridges (901–903)


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.


3

STRUCTURAL ENGINEERING Design Standards1 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 DESIGN STANDARD TITLE

AASHTO AASHTO LRFD Bridge Design Specifications, 6th edition, 2012, American Association of State Highway & Transportation Officials, Washington, DC.

IBC International Building Code, 2012 edition (without supplements), International Code Council, Falls Church, VA.

ASCE 7 Minimum Design Loads for Buildings and Other Structures, 2010, American Society of Civil Engineers, Reston, VA.

ACI 318 Building Code Requirements for Structural Concrete, 2011, American Concrete Institute, Farmington Hills, MI.

AISC Steel Construction Manual, 14th edition, American Institute of Steel Construction, Inc., Chicago, IL.

AISC Seismic Design Manual, 2nd edition, American Institute of Steel Construction, Inc., Chicago, IL.

AISI 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.

TMS 402/602 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.


13

VERTICAL FORCES AM PRACTICE EXAM


Vertical Forces

Copyright 2014 by NCEES 14 GO ON TO THE NEXT PAGE

40'-0"

20'-0" 20'-0"

40'-0"120'-0"

SINGLE COLUMN

ABUTMENT ABUTMENT

TOP OF FOOTING

ELEVATION

SUPERSTRUCTURE

101. The figure shows an elevation view of a concrete highway bridge.

Design Code: AASHTO LRFD Bridge Design Specifications, 6th edition, 2012.

Design Data:

Concrete modulus of elasticity Ec = 3,605 ksi Column moment of inertia 636,000 in4 Coefficient of thermal expansion 6 10–6 in./in./°F Concrete shrinkage after 28 days sh = 0.0002 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


AM Practice Exam


HINGE

A B C

48'-0" 12'-0" 60'-0"

10'-0"15'-0"20'-0"

A

I = 200 I = 300 I = 100

B C D

109. The beam shown is subjected to a uniform load and a moving concentrated load.

Design Data: Uniform live load 1 klf Concentrated live load 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

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) 0.33 0.67 (B) 0.43 0.57 (C) 0.5 0.5 (D) 0.67 0.33


Vertical Forces


PLATE 3/4" × 14" × CONTINUOUS

L 4 × 4 × 1/4 × CONTINUOUS(TYP.)

BUILT-UP SECTION

116. 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) stop the construction of the project

(B) report the condition to the contractor

(C) report the condition to the owner

(D) 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, Fy = 36 ksi Assumptions:

K = 1.0 The unbraced lengths in the X and Y directions are 10 ft.

The critical buckling stress Fcr (ksi) is most nearly: (A) 22 (B) 26 (C) 35 (D) 48


Vertical Forces


2 × 12 ROOF JOISTS

4 × 12 LEDGER

8" CONCRETE MASONRY

WALL12'-0"

133. 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), 2011.

Design Data:

Hollow concrete masonry units mf 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 #5 @ 48-in. o.c. vertically at the centerline of the wall and #5 @ 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) 107,300 249,900


AM Practice Exam


2 × 12 ROOF JOISTS

4 × 12 LEDGER

8" CONCRETE MASONRY

WALL12'-0"

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), 2011.

Design Data:

Hollow concrete masonry units mf 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 #5 @ 48-in. o.c. vertically at the centerline of the wall and #5 @ 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) 2,335


49

VERTICAL FORCES PM BUILDINGS PRACTICE EXAM


Vertical Forces


601. 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, 2010. TMS 402/602: Building Code Requirements and Specifications for Masonry Structures (and

Related Commentaries), 2011. ACI 318: Building Code Requirements for Structural Concrete, 2011.

Design Data:

Masonry mf = 2,000 psi Grout cf = 2,000 psi Reinforcing ASTM A615, Grade 60 Masonry weight (per surface area) 84 psf Concrete cf = 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) 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 16-11 (ASD). Do not use load information given in Figure 601B for this requirement.

(b) 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).

(c) 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.


PM Buildings Practice Exam


1'-6"

1'-0"

1'-4" 1'-0"8"

FIGURE 601B

H

W SOIL

CL CL

P

SECOND FLOOR

2X STUD WALLSHEATHING

100 psf LL SURCHARGEP

8"

FINISHED GRADE

FIGURE 601A

#5 @ 32"

d = 5-1/4"

8" CMU

SOLID GROUTED

4'-0"

1'-0"

4'-0"

SLAB ON GRADE

STUD = CMU = FTG

FIGURE 601B FOR REQUIREMENT (b) ONLY

TOTAL HORIZONTAL FORCE DUE TO

SOIL & SURCHARGE LOADS, H = 600 plf

VERTICAL SOIL LOAD ON HEEL, WSOIL = 450 psf

TOTAL DL = 3,000 plf (NOT INCL. FTG. WT.)

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 CMU L

CL

4"

4"

EXPANSION MATERIAL

LOAD (P) AT C OF CMUL

SOIL

LEVEL

601. (Continued)


61

VERTICAL FORCES PM BRIDGES PRACTICE EXAM


Vertical Forces


701. 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, 2012.

Design Data: Concrete slab, cf 4 ksi Concrete prestressed girders, cf 6 ksi Density of concrete 0.150 kcf Ductility factor 1.0 Redundancy factor 1.0 Importance factor 1.0 Low-relaxation 1/2-in.-diameter strands, puf 270 ksi Area of 1/2-in.-diameter strands, sA 0.153 in2 per strand Reinforcing bars, yf 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, Ag 713 in2 Moment of inertia, Ig 392,638 in4 Neutral axis to bottom fiber 32.12 in. Neutral axis to top fiber 30.88 in. Section modulus (bottom), Sncb 12,224 in3

Section modulus (top) Snct 12,715 in3

Composite Section Properties: Cross-sectional area, Ag 1,117 in2 Moment of inertia, Ic 704,000 in4 Neutral axis to bottom fiber 44.64 in. Section modulus (bottom), Scb 15,767 in3

Section modulus (top) Sct 38,365 in3

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.


PM Bridges Practice Exam


48'-0"48'-0"24'-0"

120'-0"

CL2 STRANDS

6 STRANDS 12 STRANDS

2 STRANDS

TOTAL - 34 STRANDS

DRAPED - 8 STRANDS

N.A.

66"

(C. TO C. OF GIRDERS)

42"

30.88"

26"

32.12"

44.64"

63"

3"

6"#5 BARS

7-1/2" SLAB

N.A.

8 STRANDSDRAPED

(SHOWN @ MIDDLE

OF GIRDER)

8 STRANDS DRAPED(SHOWN @ ENDS OF GIRDER)

2 1/2"

GIRDER ELEVATIONNOT TO SCALE

COMPOSITE SECTIONNOT TO SCALE

GIRDER SECTIONNOT TO SCALE

3 SPACES

AT 2" = 6"

4 SPACES

AT 2" = 8"

FIGURE 701

701. (Continued) TABLE 701

Maximum Live Load Moment per Girder (Dynamic Allowance Included) at 10th Points from a Support

Location from support (ft) 0.0 0.1L 0.2L 0.3L 0.4L 0.5L

Maximum live load moment (ft-kips) 0 681 1,201 1,560 1,772 1,830

REQUIREMENTS: On the actual exam, any sketches necessary for these requirements must be neatly drawn in your solution pamphlet. (a) Verify the flexural adequacy for Strength I Limit State of the composite girder at midspan,

including checking for minimum reinforcement. Assume rectangular section behavior.

(b) Determine anchorage zone vertical web reinforcement for the girder, using #5 bars, and show them on a sketch.


69

VERTICAL FORCES AM SOLUTIONS


AM Solutions

71

20'-0"

120'-0"

101. Due to symmetry, the loads at each pier will be equal.

sh

3

3

2 4

3

L Lpier T2 2

120 120pier 0.0002( 12) 0.000006(40)( 12)2 2

0.32 in.

Ph 3EIFor fixed-pin condition, ; Solve for P3EI h

3(3,605 kips/in )(636,000 in )(0.32 in.)P 159 kips

(20 12 in.)

THE CORRECT ANSWER IS: (C)

102. w = (0.64 kips/ft)/lane AASHTO Art. 3.6.1.2 2

LL35

M 0.64 98ft-kips/lane8

IM = 1.33 AASHTO Table 3.6.2.1-1

LL+IM 1.75(1.33 360 ft-kips 98 ft-kips) 1,010 ft-kips


103. ASCE 7-10, Figure 7-9 Formula. 3 4

3 4

0.43 10 1.5

30, 100 ft (given)

0.43 100 30 10 1.5 3.52 ft

d u g

g u

d

h l p

p l

h



91

VERTICAL FORCES PM BUILDINGS SOLUTIONS


Vertical Forces

92

601. (a) CMU wall design

ASD solution: D + H + F + 0.75 L + 0.75(Lr or S or R) IBC Eq. 16-11

Vertical Loads: Given PD = 980 plf

PLR = 160 0.75 = 120 plf

PLF = 640 0.75 = 480 plf Maximum stress in wall occurs at bottom of wall: Wall weight = 84 psf (given) 4.67 ft = 392 plf Ptotal = 980 + 392 + 120 + 480 = 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 = 0.35 110 pcf = 38.5 psf/ft) (soil ht = 4 ft) = 154 psf

1 4 ftmoment at base (35 psf 4 ft) 2 ft surcharge 154 4 ft (active)

2 3

280 411

691 ft-lb/ft

Check compression:

k = 2.1 (fixed base and free top)

/ 12 7.625/ 12

2.20

r t

2.1(4.67 ft)(12 in/ft)54 99

2.20 k h r

Use Eq. 2-21(neglecting compressive stress in steel per Sec. 2.3.3.3): TMS 402, Sec. 2.3.4

2

m

2

(0.25 f ) 1140

2,000 psi (given)

7.675 in. 12 in/ft 91.5 in /ft

a n

m

n

hP A

r

f

A


PM Buildings Solutions

93

601. (Continued) 2

2total

54(0.25)(2,000)(91.5 in /ft) 1 38,944 lb/ft >

140

aP P

Check flexure:

2 2

m m

s

b m

2s

m

12 in.0.31 in 0.116 in

32 in.

E 900 f 1,800,000 TMS 402-11, Sec.1.8.2.2.1

E 29,000,000 TMS 402-11, Sec.1.8.2.1

F 0.45 f 900 psi TMS 402-11, Sec. 2.3.3.2.2

E n29 (0.116 in )(n 16.11 n

E 1.8 b d

st

st

A

A

2

16.11)

(12 in.)(5.25 in.)

0.0297

k 2 n ( n) n

0.216

kj 1 0.928

3

allow

2

2 2b

allow

M (steel) jd

in.-lb ft-lb(0.116 in /ft)(32 ksi)(0.928)(5.25 in.) 18,085 1,507

ft ft

F jkbd (900)(0.928)(0.216)(12)(5.25)M (CMU)

2 229,834 in.-lb/ft

2,486 ft-lb/ft

st sA F

Steel stress controls 1,507 ft-lb > 691 ft-lb OK Check combined stresses:

b 2 2

a

a b m

2M (2)(691 ft-lb 12 in./ft)f 250 psi

jkbd (0.928)(0.216)(12 in.)(5.25)

1,972 lb/ftf 22 psi

(12 in./ft)(7.625 in.)

per 2.3.4.2.2 f f 0.45 f

22 250 272 < 900

#5 @ 32 in., 8 in. CMU wall OK for loads


111

VERTICAL FORCES PM BRIDGES SOLUTIONS


Vertical Forces

112

701. (a) Determine flexural strength for Strength I Limit State at midspan. Check minimum steel.

Aps= (34 strands) (0.153 in2/strand) = 5.20 in2

ps pup

py

pu

pu

1

py pu

cf f 1 k Eq. 5.7.3.1.1-1d

fk 2 1.04 Eq. 5.7.3.1.1-2

f

f 270 ksi

0.85 Art. 5.7.2.2

f 0.9 f Table 5.4.4.1-1

Determine dp:

p

2 2

12 2.5 12 4.5 6 6.5 2 8.5 2 10.5x @ C = 4.74 in.L 34d @ C = 70.5 4.74 65.76 in.L

k 2 1.04 0.90 0.28

Assume rectangular section behavior.

270 ksic 5.20 in (270 ksi) / (0.85) 4 ksi 0.85 66 + 0.28 5.20 in

65.76

7.14 in

. 7.50 in. OK Eq. 5.7.3.1.1-4

ps pup

n ps ps p

2n

c 7.14f f 1 k 270 1 0.28 261.8 ksi Eq. 5.7.3.1.1-1

d 65.76

a 0.85 7.14 6.07 in. Art. 5.7.2.2

aM A f d Art. 5.7.3.2.3

2 Eq. 5.7.3.2.2-16.07 1

M 5.20 in 261.8 ksi 65.76 7,116 ft-kips2 12

1.0 Assume tension-controlled p/s concrete Art 5.5.4.2.1

Verify tension-controlled assumption.

S

S

E0.003

c d c0.003(d c)

Ec

0.003(65.76 in. 7.14 in.)0.025 > 0.005

7.14 in.Tension-controlled, 1.0

d

c

ES

0.003


PM Bridges Solutions

113

701. (Continued)

DC moment:

2120

1.3 kips / ft 2,340 ft-kips8

DW moment:

21200.26 kips / ft 468 ft-kips

8

LL+ IM = 1,830 ft-kips Strength I Tables 3.4.1-1 Mu = 1.25 DC + 1.50 DW + 1.75 (LL+IM) 3.4.1-2 = 1.25 (2,340 ft-kips) + 1.50 (468 ft-kips) + 1.75 (1,830 ft-kips) = 6,830 ft-kips < (1.0) (7,116 ft-kips) OK

Check minimum steel: Art. 5.7.3.3.2

n cr

ccr 3 1 r 2 cpe c dnc

nc

M M

SM f f S M 1

S

Eq. 5.7.3.3.2-1

r c

1

2

3

f 0.24 f

0.24 6 0.59 ksi

1.2

1.1

1.00

Art. 5.4.2.6

Sc = 15,767 in3 S @ bottom of composite section

Sncb = 12,224 in3 S @ bottom of precast section

Mdnc = 2,340 ft-kips Noncomposite DC moment

pe

prestress e cprestressf

A I

Prestress = Pjack – losses = (0.75)(270 ksi) – 40.5 ksi = 162 ksi Prestress force = (162 ksi)(5.202 in2) = 843 kips e = 32.12 – 4.74 = 27.38 in.


133

LATERAL FORCES AM PRACTICE EXAM


Lateral Forces


101. 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, 2010.

Design Data:

Site shear wave velocity for the top 100 ft, sv 1,100 ft/sec Mapped spectral response acceleration at 0.2-sec period, SS 1.00 Mapped spectral response acceleration at 1-sec period, S1 0.45 Building period, T 0.80 sec Long-period transition period, TL 8 sec

Assumption:

No clay, peat, or liquefiable soils

The design spectral response acceleration Sa is most nearly:

(A) 1.10 (B) 0.73 (C) 0.58 (D) 0.47


AM Practice Exam


30'-0"

60'-0"

412

15'-0"

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, 2010.

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, Kzt = 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) 40.0


AM Practice Exam


A

100'-0"

200'-0"

ROOF PLAN

100'-0"

5'-0"

30'-0"

30'-0"

50'-0"

CG

75 kips

50'-0"

CR

50'-0"

50'-0"

50'-0"

50'-0"

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, 2010. Design Data:

Roof 26 ft, A.F.F. Top of panel 26 ft, A.F.F. Roof DL 15 psf Panel DL 70 psf

Assumption:

sC = 0.15

The seismic chord force (kips) at Point A is most nearly: (A) 11.3 (B) 21.1 (C) 24.9 (D) 45.5

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, 2010.

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) 750


AM Practice Exam


4 SIDES

HSS 3 × 3 × 1/4

W16 × 36

HSS 6 × 6 × 5/16

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) 60 90 (B) 80 120 (C) 105 160 (D) 120 175


Lateral Forces


100'-0"

25'-0"

40'-0"RIDGE

2× TOP PLATE

2× TOP PLATE

SECTION B-BTOP PLATE SPLICE

AB B

W

131. 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:

CM = Ct = Cg = C = Ceg = Cdi = Ctn = 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) 25 31 (B) 30 37 (C) 39 48 (D) 47 58


169

LATERAL FORCES PM BUILDINGS PRACTICE EXAM


Lateral Forces


801. 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, 2010. TMS 402/602: Building Code Requirements and Specifications for Masonry Structures (and Related Commentaries), 2011.

Design Data:

Wind load 142 mph, Exposure C Kzt = 1.67 Seismic load SDS = 0.70 Masonry mf = 1,500 psi

Weight = 60 psf

fy = 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) 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).

(b) Determine the design wind pressure and seismic design force on the parapet. For wind, neglect corner zones. Consider interior zones only.

(c) 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.

(d) 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.


PM Buildings Practice Exam


16'-0"

4'-0"

FIGURE 801

lb/ftDL ~ 180

2"

lb/ftLL ~ 240

ROOF WOOD FRAMING

WITH WOOD SHEATHING

1'-4" WIDE × 10"UNREINFORCED FOOTING

CONCRETE SLAB-ON-GRADE

1'-4"

8" EXTERIOR PARTIALLY GROUTED CMU

801. (Continued)


179

LATERAL FORCES PM BRIDGES PRACTICE EXAM


Lateral Forces


901. 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, 2012.

Design Data: Concrete strength, cf 3.5 ksi Yield strength of reinforcement, yf 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.

Mu = 4,050 ft-kips Vu = 600 kips Pu = 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) For the Extreme Event I load combination, determine the vertical reinforcement required

for the column. Neglect any slenderness effect for the column.

(b) For the Extreme Event I load combination, design the required spiral reinforcement.

(c) Sketch an elevation view of the columns showing the size and spacing of the vertical and spiral reinforcements.


PM Bridges Practice Exam


1'-7" 4'-5" 4'-5" 1'-7"

CL CLCOL CL COLBOX GIRDER

4'-0"

5'-6"

20'-0"

TRANSVERSE SECTION OF PIER

11'-0" 11'-0"

FIGURE 901A

4'-0"

2" CLEAR

f 'c

f

= 3.5 ksi

= 60 ksiy

9,000

8,000

7,000

6,000

6,000

5,000

5,000

4,000

4,000

3,000

3,000

2,000

2,000

1,000

1,000

0

0−1,000

NOMINAL MOMENT STRENGTH, Mn, ft-kips

NO

MIN

AL

AX

IAL

ST

RE

NG

TH

, Pn,

kip

s

= 0.035= 0.030= 0.025= 0.020= 0.015

ρ

FIGURE 901B

901. (Continued)


187

LATERAL FORCES AM SOLUTIONS


AM Solutions

189

101. Not Site Class F or E since no clay, peat, or liquefaction ASCE 7-10, Section 20.3, and Table 20.3-1

sv 1,100 ft/sec Site Class D

a

MS a s

DS MS

v

M1 v 1

D1 M1

0 D1 DS

S D1 DS

F 1.1 Table11.4-1

S F S (1.1)(1.00) 1.10 Eq. 11.4-1

S 2/3 S 0.73 Eq. 11.4-3

F 1.55 Table11.4-2

S F S (1.55)(0.45) 0.70 Eq. 11.4-2

S 2/3 S 0.47 Eq. 11.4-4

T 0.2 S /S 0.13 Section 11.4.5

T S /S

D1S L a

0.64

S 0.47Building period > T and < T S 0.58 Eq.11.4-6

T 0.80


102. 2h h zt dq 0.00256 K K K V ASCE 7-10, Eq. 28.3-1

4 in. 30 ft/215 15

12h 17.5 ft2

h 17.5

zt

d

2z

K K 0.875 Table 28.3-1

K 1.0

K 0.85 Table 26.6-1

V 120 (Risk Category I - IBC Table 1604.5)

q 0.00256(0.875)(1.0)(0.85)(120) 27.4 psf

THE CORRECT ANSWER IS: (B)

103. (640)(192 ft)Force 61.44 kips Note: Load factor = 1.0 for wind for strength design.

2

Force for one panel 30.72 kips

30.72 kipsNumber of connectors 3.6; use 4 connectors

8.5 kips



207

LATERAL FORCES PM BUILDINGS SOLUTIONS


Lateral Forces

208

LOAD CASE A (WINDWARD)

p = p1 + p

2

= (65.9)(0.90) + (65.9)(1.8)

= 178 psf

POSITIVE

WALL

PRESSURE

NEGATIVE

WALL

PRESSURE

NEGATIVE

WALL

PRESSURE

LOAD CASE B (LEEWARD)

p = p3 + p

4

= (65.9)(0.90) + (65.9)(0.99)

= 125 psf

OR

POSITIVE

WALL

PRESSUREp

1p

3p

2p

4

ASCE 7 FIGURE 30.9-1

801. (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. 30.9 p = qp (GCp – GCpi) (Components and cladding elements of parapets) qp = 0.00256 KzKztKdV2 ASCE 7, Eq. (30.3-1) Kz = 0.90 Exposure C, z = 20 ft (top of parapet) ASCE 7, Table 30.3-1 Kzt = 1.67 (given) Kd = 0.85 ASCE 7, Table 26.6-1 V = 142 mph (given) qp = 0.00256 0.90 1.67 0.85 1422 = 65.9 psf GCpi = 0.00 (solid parapet, open building condition) ASCE 7, Table 26.11-1 GCp, h 60 ft ASCE 7, Fig. 30.4-1 Effective wind area ASCE 7, Sec. 26.2 = span length 4-ft height span 4 ft/3 = 16/3 = 5.3 ft2 < 10 ft2 Use 10 ft2 Zone 4 wall positive pressure GCp = +1.0 0.90 = +0.90 (Ref footnote 5 for reduction) Zone 4 wall negative pressure GCp = – 1.1 0.90 = –0.99 (Ref footnote 5 for reduction) Zone 2 roof negative pressure GCp = – 1.8 ASCE 7 Fig. 30.4-2A


PM Buildings Solutions

209

801. (Continued) Seismic design force on the parapet: ASCE 7, Sec. 13.3.1

p DS p

pP P

0.4 a S W zF 1 2

R / I h

ASCE 7, Eq. (13.3-1)

where pz 16'

I 1.0h 16' ASCE 7, Sec 13.1.3

p pa 2.5; R = 2.5 (parapets unbraced) ASCE 7, Table 13.5.1

p0.4 2.5 0.70 60

F 1 2 50 psf 2.5

1.0

Controlling design load on the parapet = 178 psf (windward pressure)

(c) Check if the existing #5 @ 48" o.c. vertical at centerline of wall is adequate for the parapet:

max

2

max

V 100 psf (1 ')(4 ') 400 lb / ft wall

100 psf (1')(4 ')M 800 ft-lb / ft wall

2

Check shear stress: Since no shear reinforcement is provided, assume one face shell resists shear force.

v3V 3(400)

f 40 psi2 A 2(1 1/ 4")(12")

Fv shall not exceed: TMS 402 Sec 2.2.5.2

m

v n

n

(a) 1.5 f 1.5 1,500 58 psi

(b) 120 psi

(c) 37 psi 0.45 N / A

37 0.45 (180 240) / A

An based on two face shells and one grouted cell per 48"

n

2n

(2)(1 1/ 4")(48") (6" 1" 1 1/ 4")(7 5 / 8" (2)1 1/ 4")A

4 ft

A 40.6 in / ft

37 0.45 (420) / 40.6 41.7 psi

v vF = 41.7 psi > f = 40 psi OK


223

LATERAL FORCES PM BRIDGES SOLUTIONS


Lateral Forces

224

901. (a) Vertical reinforcement required for the column

Design for moment and axial force Column requirement, Art. 5.10.11.4.1 h/Dc = 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. 5.7.4.3, may be ignored per problem statement. Vertical reinforcement, flexural strength, Art. 5.10.11.4.1b Design for Mu = 4,050 ft-kips and Pu max = 2,700 kips = 0.9 Mn = nominal moment = Mu/ = 4,050/0.9 = 4,500 ft-kips Pn = nominal axial force = Pu/ = 2,700/0.9 = 3,000 kips From the interaction diagram for Mn = 4,500 ft-kips and Pn = 3,000 kips = 0.025 Limits of vertical reinforcement, Art. 5.10.11.4.1a 0.01 < = 0.025 < 0.04 OK As = 0.025 (48)2/4 = 45.24 in2 Using 29 #11 bars with As = 29 1.56 = 45.24 in2 will satisfy the requirement.

(b) Design of spiral reinforcement

Shear stress, Art. 5.8.3 Resistance factor, Art. 1.3.2.1 = 1.0 where Vu = 600 kips B = 48 in. Dr = 48.0 – 2(2.00 +0.875) – 1.41 = 40.84 in. (assuming #7 spiral)

e48.0 40.84

d 37.0 in.2

Eq. C5.8.2.9-2

dv = 0.9 de = 33.3 in. Shear carried by concrete, Art. 5.8.3.3

minvUse 2, 45 if A provided and no tension in column per Art. 5.8.3.4.1 Check if concrete is effective in end regions per Art. 5.10.11.4.1c. Pu = 2,700 kips Pu > 0.10(3.5 ksi)(1,810 in2) = 634 kips Concrete is effective

cV 0.0316(2) 3.5(48 in.)(33.3 in) 189 kips Eq. 5.8.3.3-3


PM Bridges Solutions

225

901. (Continued) Spacing of spiral reinforcement, Eq. C5.8.3.3-1

s = pitch = v y v s A f d /V

where Av = 0.6 in2 (Try #7 spiral) fy = 60 ksi bv = 48 in. dv = 33.3 in.

s required600 189(1.0)

V 411 kips1.0

and (2 0.6) 60 33.3

s 5.83 in. Use 5.0 in.411

Check minimum transverse reinforcing, Art. 5.8.2.5

min2v

v cy

ys

48.0 5.0b sA 0.0316 f 0.0316 3.5 0.24 in OK

f 60

A f d 2 0.60 60 33.3V = = = 480 kips

5 5.0

v v

The nominal shear resistance Vn is the lesser of:

n c s

n v vc

V V V 189 480 669 kips Governs, OK Eq. 5.8.3.3-1

V 0.25 f b d 0.25(3.5)(48.0)(33.3) 1,399 kips Eq. 5.8.3.3-2

b

Check maximum spacing, Art. 5.10.6.2

6 d = 6(1.41) = 8.46 in.

6 in. governs, 5.0 in. OK

End regions, Art. 5.10.11.4.1c He = 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. 5.10.11.4.1d

s g c yh

s c y

2s s c c s c

2s

s c

c0.45(A /A 1)f /f 0.45 (1,810/1,521) 1 3.5/60 0.0050 Eq. 5.7.4.6-1

or 0.12 f f 0.12 3.5/60 0.007 whichever is larger Eq. 5.10.11.4.1d-1

4A ( D ) / s D 4A /sD

4A 4(0.6 in )s 7.8 in.

D 0.007(44 in.)


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