civil engineering license exam...

Civi l Engineering License Exam Review

Structural - Session 1

Old Dominion University Brian Crowder [email protected]

Structural Engineering

Exam Technical ContentThe following structural engineering subjects are covered in the Civil Engineering License Exam:

Structural Engineering Content in the Breadth (Morning) Exam (20%)

Loadings: Dead & Live Loads, Construction Loads

Analysis: Determinate Analysis

Mechanics of Materials: Flexure, shear, tension & compression, deflection, shear and moment diagrams, combined stresses

Materials: plain/reinforced concrete, structural/light gage/reinforcing steel

Member Design: beams, slabs, footings

Breadth Exam Structural Engineering Content

Old Dominion University - Virginia Applied Technology and Professional Development Center

Depth Exam Subject (% points) Depth Exam Detailed Topics

Structural (87.5%) Loadings (Dead, Live, Construction, Moving, Wind, EQ, Snow, load path / combinations, impact, etc.) - 12.5%

Analysis (Determinate, Indeterminate) - 12.5%

Mechanics of Materials (Flexure, shear, torsion, tension & compression, combined stresses, deflection, shear diagrams, moment diagrams, progressive collapse?, torsion, buckling, fatigue, thermal deformation) - 12.5%

Materials ( Plain/Reinf. Concrete, Pre-stressed/post-tensioned concrete, structural/light gage/reinforcing steel, timber, masonry, composite construction) - 25%

Member Design (Beams, slabs, columns, footings, trusses, braces, frames, connections, shear walls, bearing walls, diaphragms) -12.5%

Design Criteria (IBC, ACI 318/530, PCI, AISC, NDS, AASHTO, ASCE7, AWS) - 12.5%

Other Topics (12.5%) Engineering Properties of soils and materials (index properties)

Soil Mechanics Analysis (Expansive soils)

Shallow Foundations (mat and raft foundations)

Deep Foundations (Axial capacity, lateral capacity and deflection, and behavior of single pile/drilled shafts and groups, settlement)

Engineering Economics (Value engineering and costing)

Material QC and Production (welding and bolting testing)

Temporary Structures (formwork, falsework and scaffolding, shoring and reshoring, concrete maturity and early strength evaluation, bracing, anchorage)

Worker Health, Safety and Environmental (OSHA regulations, safety management)

Structural Engineering Depth Exam Content


ReferencesThe following references have been approved for use on the Principles and Practice exam effective with the October 2006 examinations.

Abbreviation Design Standard Title

AASHTO AASHTO LRFD Bridge Design Specifications, 5th Edition, 2010, American Association of State Highway & Transportation Officials, Washington, DC

ACI318 Building Code Requirements for Structural Concrete, 2008, American Concrete Institute, Farmington Hills, MI

ACI530/ASCE 5/TMS402 & ACI 530.1 / ASCE 6/ TMS 602

Building Code Requirements for Masonry Structures, 2008; & Specification for Masonry Structures, 2008, American Concrete Institute, Detroit, MI; Structural Engineering Institute of the American Society of Civil Engineers, Reston, VA; and The Masonry Society, Boulder, CO (Use ASD except for walls with out of plane loads)

AISC Steel Construction Manual, 13th Edition, American Institute of Steel Construction, Inc., Chicago, IL (Can use ASD or LRFD using 13th edition)

NDS National Design Specification for Wood Construction, 2005 ASD/LRFD Edition & National Design Specification Supplement 2005, American Forest & Paper Association, Washington, DC (use ASD only on the exam)

PCI PCI Design Handbook, Sixth Edition, 2004, Precast/Prestressed Concrete Institute, Chicago, IL

ASCE7 Minimum Design Loads for Buildings and Other Structures, 2005, American Society of Civil Engineers, New York, NY

IBC International Building Code, 2009 Edition (without supplements), International Code Council, Falls Church, VA

Technical Areas for Structural Engineering Session 1In this session will first review technical areas that overlap between geotechnical and structural engineering. These areas can be the subject of questions in either the breadth or depth portion of the Civil Engineering principles and practices exam. The following subjects will be reviewed:

• Strength of Materials - Shear and Moment Diagrams

• Strength of Materials - Stresses, Properties of Areas

• Loadings

• Shallow Foundations

• Retaining Walls


In order to cover the later two topics, we will need to review basic flexural and shear design of reinforced concrete. So, first we’ll take a look at loadings and mechanics of materials, and then we’ll move into reviewing some basic concepts in reinforced concrete.

Strength of Materials - Shear and Moment DiagramsLindeburg covers Strength of Materials in Chapter 44. It would be wise to review and become familiar with the topics covered in this chapter as they are fundamental and clearly a large part of the structural engineering content within the breadth portion of the Civil Engineering exam. While the basics of compressive stress, tensile stress, and stress concentrations can be reviewed within the text, it is more important to work example problems concerning shear and moment diagrams to ensure competence and the ability to quickly answer these types of questions in the exam. Lindeburg reviews shear and moment diagrams in Section 12 of Chapter 44. The shear and moment diagram can be found using simple free body diagrams (FBD). The solutions are expedited once you are familiar with typical characteristics of the diagrams such as:

• The magnitude of the shear diagram at any point is equal to the slope of the moment line at that point. For this reason, where the shear is equal to zero, the moment diagram is at a peak and potential maximum value.

• The shear diagram is straight and sloping over uniformly distributed loads

• The shear diagram is straight and horizontal between concentrated loads with no other applied loads in between

• The shear diagram is a vertical line and it is undefined at points of concentrated loads or reactions

• The moment diagram is zero at free ends or at frictionless hinges.

The best way to re-familiarize yourself with these diagrams is by working several problems.


Problem S-1

Source: Practice Problems for the Civil Engineering PE Exam, 12th Edition

Problem S-1AThe maximum moment is most nearly?

(A) 150 ft-lbf

(B) 250 ft-lbf

(C) 390 ft-lbf

(D) 830 ft-lbf

Problem S-1BThe maximum shear is most nearly?

(A) 80 lbf

(B) 120 lbf

(C) 150 lbf

(D) 190 lbf


Solution S-1The typical first step is to solve for the support reactions. The vertical reaction at Point A is denoted by “L”. The vertical reaction at Point C is denoted by “R”. The reactions are most quickly determined by summing moments about an opposite support. For example, taking the sum of moments and enforcing equilibrium gives:

∑M = R * 12’ - 100 lb * 2 ft - 80 lb * 14ft - (20 lb/ft * 14 ft) * 7 ft. = 0 therefore, R = 273.3 lb

∑F(y) = L + R - 100 lb - 80 lb - (20 lb/ft * 14ft) = 0 therefore, L = 186.7 lb

Using these reactions, the shear diagram can be built by examining FBDs from the left or right end of the beam. For example, the shear at the left end is equal to the reaction, L, or 186.7 lb. The shear will then decrease linearly due to the uniform load at a slope of -20 lb/ft. So just prior to the concentrated load at x = 2ft., the shear will be 186.7lb - 20 lb (2 ft.) = 146.7 lb. Using the knowledge that the concentrated load will create a vertical line for the shear diagram, the concentrated load will reduce the shear from 146.7 lb to 46.7lb. Then shear will then continue to reduce at 20 lb/ft until it reaches 0 lb. at x = 4.3 ft. The maximum moment will occur at a location where the shear is equal to zero.

The maximum moment is the area of the shear diagram up until the point of zero shear, or:

Mmax = 0.5*(186.7-146.7 lb)( 2 ft) + 146.7 lb * 2 ft + 0.5*46.7*(4.3 ft - 2 ft) = 387.1 ft-lb ANS C

As seen on the shear diagram, the maximum shear is 186.7 lb. or most nearly 190 lb. ANS D

-200-150-100-50

050

100150200

0 2.0 4.0 6.0 8.0 10.0 12.0 14.0

Shear Diagram


Problem S-2

Which of the moment diagrams shown best represents the loading on the beam shown in the figure?

Source: NCEES Civil: Construction Sample Questions and Solutions, 2011

Solution S-2

This problem is solved without calculation, based on the typical characteristics and relationships of shear and moment diagrams. If you can visualize the shear diagram, then you can derive the shape of the moment diagram.

Since there are no applied vertical loads past point B, we know the shear is zero from B to the end of the beam. We know the shear diagram will be sloped in the region of the uniform load. Between the fixed end and the uniform load, the shear will be constant since there are no other applied vertical loads. Therefore the diagram will look like the following:


From this shear diagram, we can visualize the moment diagram. As indicated in the summary of shear and moment diagram characteristics, the moment will be linear in regions of constant shear. So we know there will be a fixed end moment and the moment will have a linear slope until point A. In regions of linear shear, the moment will be parabolic (moment is integral of shear diagram). Then in a region where the shear is zero, the moment diagram will be constant (slope of zero). This is can also be observed by cutting a section in the last region and drawing the free body diagram. The moment will be equal and opposite to the applied concentrated moment throughout the region since there are no other applied loads. Therefore the answer is (A).

Problem S-3 The point load (kips) placed at the centerline of a 30 ft. beam that produces the same maximum shear in the beam as a uniform load of 1 kip / ft is most nearly:

(A) 7.5

(B) 15

(C) 30

(D) 60


Shear Diagram


Solution S-3The shear and moment diagrams and formulae for calculating the maximum values for many typical loading conditions can be found in Appendix 44.A. “Case 6” on page A-90 shows the maximum shear Vconc = P /2. “Case 9” on page A-91, shows the maximum shear Vuniform = wL/2.

For w = 1 kip / ft, Vuniform = [ ( 1 kip/ft ) * 30 ft ] / 2 = 15 kips

Setting Vconc = 15 kips = P /2, gives P = 15 kips x 2 = 30 kips. ANS (C)

Strength of Materials - Stresses and Properties of AreasStrength of Materials is covered in Chapter 44 of Lindeburg. In order to calculate the stresses in cross sections you also need to Chapter 42 on Properties of Areas. Likely problems will ask you to solve for the axial or flexural stresses due to an applied loading or you will be tested on your general understanding of the concepts and the formulae.

Problem S-4The beam sections shown are fabricated from 1/2-in. x 6-in. steel plates. Which of the following cross sections will provide the greatest flexural rigidity about the x-axis.



Solution S-4Lindeburg discusses stiffness and rigidity in Section 5 of Chapter 44. Flexural rigidity is the reciprocal of deflection in members that are in bending. While you would need to know the loading condition to find the exact flexural rigidity, the term flexural rigidity also refers to the product of the Elastic Modulus (E) and the Moment of Inertia (I) - EI. Since the material is the same for all cross sections, the cross section with the highest Moment of Inertia (I), will have the greatest flexural rigidity.

The moment of inertia of an area is discussed in Section 5 of Chapter 42. The most convenient way to solve this problem is by considering the Parallel Axis Theorem (Section 6, Chapter 42). The cross section with the greatest area, furtherest away from the x-axis will have the highest moment of inertia. Reviewing the choices provided, it is clear that cross section (D) will have the highest moment of inertia, and therefore flexural rigidity. ANS (D)

(Note: ANS (C) is the same section, with the same area above and below the x-axis. However, when you consider the distance between the centroidal axis and the centroid of each area, the distance is smaller for each area making up section (C). Ans (C) is the moment about the “weak” axis of the cross section shown in Ans (D).

Problem S-5Consider two beams with equal cross-sections, made of the same material, having the same support conditions, and each loaded with equal uniform load per length. One beam is twice as long as the other. The maximum bending stress in the longer beam is larger by a factor of: (Source: NCEES Civil: Construction Sample Questions and Solutions, 2011)

(A) 1.25

(B) 2

(C) 3

(D) 4

Solution S-5As shown in Lindeburg Chapter 44, Section 14, the bending stress = M x c / I, where M = moment, c = distance from the neutral axis to the extreme fiber, and I = centroidal moment of inertia. The only quantity that varies in this case is the bending moment due to the length. Examining Appendix 44.A, shows that the maximum bending moment varies by the square of the length (L2), regardless of the support condition for a uniformly loaded beam (such as Mmax = wL2/8 for a simply supported beam).

For the short beam, the moment will be proportional to L2.

For the longer beam, the moment will be proportional to ( 2 x L )2 , or 4L2. Therefore the answer is (D).


LoadingsIn the NCEES design standards list there are two primary documents defining loadings. These are ASCE 7-05 Minimum Design Loads for Buildings and other Structures and IBC 2009 International Building Code. Even if you aren’t taking the structural engineering depth exam, you should be familiar with where and how to obtain reference dead and live loads. It only requires a small commitment of time to gain familiarity with the references. If you learn about the documents, you may find this results in some easy to earn points on the exam.

The tables below highlight the location of some important basic information in ASCE7.

Topic ASCE 7-02

Dead Loads Section C3, Pg. 245

Minimum Live Loads Section 4, Pg. 9

WInd Loads Section 6, Pg. 23

Seismic Loading Section 9, Pg. 95

Snow Load Section 7, Pg. 77

Load Combinations Section 2, Pg. 5

It would reasonable to find an exam problem that examines your knowledge of where information is located in the documents and the proper application. In addition, it is reasonable in the breadth portion of the exam to expect you to be able to determine the basic application of loads and the resulting forces on structural components (with basic determinate analysis).

Let’s take a look at a couple of problems to reinforce some of these concepts.


Problem S-6

Given the building cross section above (symmetric) and the following additional information, answer the following questions. (Source: NCEES Principles and Practice of Engineering - Civil Engineering - Sample Questions and Solutions and Brian Crowder)

Additional Information:

Roof Joists are spaced 24” o.c.

Roof sheathing is 3/4” plywood. Roofing is built-up membrane roof. Roof slope = 0.5” / ft.

Building is symmetric about roof beam shown. Building Length = 100 ft. Girder Span = 15 ft.

Problem S-6A:The approximate dead load from the roof sheathing and roof joists is most nearly:

(A) 4 psf

(B) 5 psf

(C) 6 psf

(D) 8 psf


Problem S-6B:The minimum design roof live load is most nearly:

(A) 15 psf

(B) 20 psf

(C) 25 psf

(D) 30 psf

Problem S-6C: Given that the total roof dead load is 10 psf, the specified roof live load is 25 psf, and that the central roof beam simply spans 15 ft., the design moment for the roof beam is most nearly? (Assume no snow load or other superimposed loading.)

(A) 11,800

(B) 22,500

(C) 23,600

(D) 24,200

Problem S-6D:Given the following:

Roof dead load = 15 psf

Non-reducible roof snow load = 40 psf

Average wall dead load = 54 psf

Design wind (pressure or suction) = 20 psf

The total axial load, P, in pounds per linear foot at the midheight of the wall is most nearly?

(A) 60

(B) 980

(C) 1310

(D) 1970


Problem S-6E:Given the loads in S-1D, the maximum design moment (ASD) in ft-lb per linear foot of wall at mid-height of the wall is most nearly?

(A) 200

(B) 395

(C) 435

(D) 560

Solution S-6A:From ASCE7-02 Section C3, Table C3-1, Page 246, you can find typical values of the dead load of various common construction materials. From this table, it is found that the dead load from the sheathing is 3 psf per inch of thickness. Therefore, a 3/4” plywood sheathing would weigh approx. (0.75) x ( 3 psf ) = 2.25 psf. From ASCE7 Table C3-2 you can find timber weighs approximately 35 pcf. 2x12 -> A = 1.5” x 11.25” = 16.875 in2 = 0.118 ft2 . Therefore 2x12 weighs 0.118 ft2 x 35 pcf = 4.1 lb/ft. Since they are spaced 2’ o.c., the area weight is 4.1 plf / 2 ft. = 2.05 psf. That gives a total of 2.25 psf + 2.05 psf, or 4.3 psf of dead load from the sheathing and roof joists. ANS (A)

Solution S-6B:From ASCE Section 4.9.1, you find that the minimum roof live load, Lr = 20 psf (R1) (R2) where 12 psf ≤ Lr ≤ 20 psf. Support area for joist = 2’ x 24’ = 48 sq. ft. Support area for girder = 24’ x 15’ = 360 sq. ft. You would use the smallest value, Therefore A = 48 sq. ft.. For an area this small R1 = 1.0. F = number of inches of rise per ft for a sloped roof. Therefore F = 0.5. For F≤4, R2 = 1.0. Therefore, Lr = 20 * 1.0 * 1.0 = 20 psf. Therefore, the appropriate minimum roof live load is 20 psf. ANS(B)

Solution S-6C:The total load is 35 psf. The tributary width for the roof beam is 24 ft. Since the beam has several closely spaced concentrated loads, the moment on the beam will be approximately the same as a uniformly distributed load. Therefore, the uniform load on the beam is equal to w = 35 psf * 24 ft. = 840 plf. Determinate beam tables can be found in Appendix 44.A of Lindeburg. On page A-81 the case of a simply supported beam with uniform load is covered.

M(max) = w * l2 / 8 = 840 plf * ( 15 ft )2 / 8 = 23,625 ft lb ANS(C)

Solution S-6D:P = 6 ft * 54 psf ( wt. of wall ) + 12 ft. ( 15 psf + 40 psf ) ( load from roof tributary area) = 984 plf ANS(B)


Solution S-6E:There are two components of load creating moment on the wall - (1) out of plane wind load and (2) eccentricity of vertical load. The moment will be calculated for each component of load and on a per foot of wall basis.

(1) out of plane wind load - w = 20 psf x 1 ft width of wall = 20 plf

M(wind) = w * l2 / 8 = 20 plf * ( 12 ft )2 / 8 = 360 ft lb / ft (max at mid-height)

(2) axial load component - w(roof, dead) = 12 ft * ( 15 psf ) = 180 plf.

eccentricity = e = 7.625”/2 + 3.5” = 7.31 inches

Maxial,dead (at roof level) = w * e = 180 plf * (7.31 in. / 12 in/ft ) = 109.65 ft-lb / ft

Maxial,dead (at mid-height) = 0.5 * 109.65 ft-lb / ft = 54.82 ft-lb / ft

(3) axial load component w(roof, snow) = 12 ft * ( 40 ) = 480 plf.

eccentricity = e = 7.625”/2 + 3.5” = 7.31 inches

Maxial,dead(at roof level) = w * e = 480 plf * (7.31 in. / 12 in/ft ) = 292.4 ft-lb / ft

Maxial,dead(at mid-height) = 0.5 * 292.4 ft-lb /ft = 146.2 ft-lb / ft

Load combinations for ASD from ASCE7 that include dead, snow, and wind are:

D + S = (54.82 + 146.2 ) ft-lb/ft = 201 ft-lb/ft

D + 0.75S + 0.75W = (54.82 + 0.75*146.2 + 0.75*360) ft-lb/ft = 434.5 ft-lb / ft

0.6D + W = (0.6*54.82 + 360) ft-lb/ft = 392.9 ft-lb/ft

So max case is D+0.75S+0.75W = 434.5 ft-lb/ft , ANS(C)


3 in.

Reinforced Concrete ReviewBefore we can continue on to review the structural design of footings or retaining walls, we will first need to do a quick review of flexural and shear design of reinforced concrete.

Unfortunately with the time available for review we can not review the design of reinforced concrete entirely. However, I would encourage everyone to consider the design of reinforced concrete flexural members and footings to be part of their preparation even for the breadth portion of the exam. Understanding the basic equations and their use may garner a few extra points with a limited investment of review time.

I suggest that instead of reviewing your college text on reinforced concrete and subsequently taking that into the exam, you consider a different reference. “Notes on ACI 318-02 Building Code Requirements for Structural Concrete - with Design Applications” is published by the Portland Cement Association (PCA). With a copy of ACI 318-02 and this PCA reference you will be able to cover a large number of potential concrete design related questions on the exam. You will find it easier and quicker to use the PCA reference compared to a typical textbook as the text is organized by application. So you can quickly turn to example problems of designing spread footings, design of beams, etc.

Important Changes in ACI 318-02With the 2002 edition of ACI 318 there are a few important changes to be aware of in case you originally studied reinforced concrete design using an earlier version of the code and you have not had a chance to practice with the newer code. The following table summarizes the changes. We’ll take a quick look at these subjects to help you prepare for the exam.

Topic Change

Minimum reinforcement of flexural element

Previous codes provided the minimum reinforcement ratio, ρmin. As of the 2002 code, the actual minimum area of steel, As,min, is specified and the equations have been updated to directly consider the compressive strength of the concrete.

Maximum reinforcement of flexural element

The 2002 code now defines reinforcement limits in terms of net tensile strain, εt, instead of the balanced ratio ρ/ρbal previously used. So you’ll have to be more familiar with determining the strain distribution in the cross section and some of the design aids to know where you stand. The minimum εt for a flexural element is now 0.004. This would be compared to the previous code limit of ρ < 0.75 ρbal which provided εt = 0.0036.

Phi factor for flexure In previous codes, resistance factor for flexure was 0.90. In the 2002 code, the value of the factor is dependent on the strain in the tension steel. If the strain is equal to or greater than 0.005 the section is said to be tension controlled and ϕ = 0.90. If the strain is less than 0.005 the section is in a transition region and the factor must be determined.


Topic Change

Load factors

ACI318 - 9.2

The basic load combinations in previous ACI codes were different than those used for other materials. In the 2002 code, ACI 318 adopted the load combinations used in ASCE7. This has the result of lowering the ultimate loads slightly in some situations. For example, the typical dead load and live load combination was previously 1.4D + 1.7L. This combination is now 1.2D + 1.6L. So make sure you include a review of the appropriate load combinations in your review. (Note: there is an Appendix to ACI318-02 that allows use of the older load combinations)

Flexural DesignLet’s first recall the basic derivation of the stress distribution in flexure for a reinforced concrete cross section.


(Source: Notes on ACI 318-02 Building Code Requirements for Structural Concrete - with Design Applications)

The force equilibrium is shown for a tension controlled section, i.e. the reinforcing steel yields prior to the compression concrete reaching its crushing strain. Lindeburg reviews basic reinforced concrete beam design in Chapter 50.

It is convenient to make some additional substitutions and simplifications to the equation for the nominal moment in order to allow the use of some design aids.


further, Rn is defined as Rn = Mn / (b d2 f’c)

The PCA reference provides a graph and table as design aids taking advantage of these relationships. These tools can be used for design or analysis. Since the exam format is multiple choice and that most problems are single part problems, it is likely that the problems will prescribe all but one quantity about a concrete section and the problem will be largely an analysis type problem. Therefore, these design aids may enable you to quickly answer some problems related to the flexural design of reinforced concrete. A copy of these design aids is provided for your information.


3”

Let’s look at how you can use these design aids for a some simple beam problem.


Problem S-7Given the following beam cross section, f’c = 4000 psi, and fy = 60 ksi, the design moment strength of the cross section is most nearly? (Ignore the longitudinal hanger bars in the compression region.) (Source: PCA “Notes on ACI 318-02 Building Code Requirements for Structural Concrete, with Design Applications”)

(A) 135 ft kips

(B) 140 ft kips

(C) 122 ft kips

(D) 120 ft kips

First we’ll look at the long form calculation, then we’ll examine how to shorten the calculation using the design aids.


Now let’s look at how to shorten this solution using the design aids.

First calculate ω = ρ * ( fy / f’c ) = ( 2.37 sq. in. / 10” x 13.5” ) * ( 60,000 psi / 4000 psi ) = 0.2633

( ρ = 0.0176 )

Now using Table 7-1, go down the left column to 0.26 to determine the appropriate row. Then find the 0.003 column and go down to that row. This gives Mn / ( f’c * b * d2 ) = 0.2222 . Solving for Mn gives Mn = 1619.83 kip-in.= 135 ft.-kips

Using Table 6-1, it is found that for f’c = 4000 psi , ωt = 0.2709 and ρt = 0.01806 (these are the values when the strain in the tension steel is equal to 0.005, the limit for a tension controlled section. Since ρ < ρt the section is tension controlled and ϕ = 0.90 .

Therefore, the design moment strength ϕMn = 0.90 x 1619.83 kip-in. = 1457.8 kip-in. = 121.5 ft-kips

ANS( C) using both procedures.

Shear DesignShear design of beams is discussed starting on page 50-24 of Lindeburg. We’ll briefly review this material during the lecture. There have been some refinements to the shear design equations compared to previous ACI codes, but in many cases the more complex equations are optional and the more familiar equations remain. Important things to understand is determining the shear capacity contribution of the concrete and reinforcing steel and determining when shear reinforcing is not required. In spread footing design, the footing thickness is typically selected such that no shear reinforcing is required (see page 50-25 in Lindeburg). In slabs, it is difficult to place shear reinforcing. So slabs are typically sized to avoid the need for stirrups. However, the limit of when no shear reinforcement is required in a slab is different than beams due to the width of slab sections (See Page 51-3 of Lindeburg). Knowing these relationships may also help you make an educated guess on a seemingly more difficult problem.

DeflectionWe don’t have time to review the detailed calculation of short or long-term deflections of concrete sections. However, a couple of things you should be aware of is the serviceability guidance for the depth / thickness of beams and slabs. On Page 50-18, Lindeburg discusses the minimum beam depths required to avoid explicit deflection calculations. On page 51-2 Lindeburg discusses the minimum slab thicknesses to avoid deflection calculations if the slab doesn’t support deflection sensitive components. A question based on understanding this basic requirement would make a good exam problem.


Problem S-8

Problem S-8AFor a reinforced concrete beam simply spanning 20 feet, normal weight concrete, and reinforcing yield strength of 60 ksi, the minimum thickness of the beam for serviceability is most nearly?

(A) 20 in.

(B) 15 in.

(C) 16 in.

(D) 11 in.

Problem S-8BFor a reinforced concrete slab simply spanning 20 feet, normal weight concrete, and reinforcing yield strength of 60 ksi, the minimum thickness of the slab for serviceability is most nearly?

(A) 20 in.

(B) 15 in.

(C) 16 in.

(D) 12 in.

Solution S-8A:Using Lindeburg Table 50.4 - for a simply supported beam, the thickness, h, should be 1/16 of the span length. Therefore, h = 20/16 ft. x 12 in./ft. = 15 in. ANS(B)

Solution S-8B:Using Lindeburg Table 51.1 - for a simply supported slab, the thickness, t, should be 1/20 of the span length. Therefore t = 20/20 ft. x 12 in./ft. = 12 in. ANS(D)

Moment CoefficientsAs discussed in Lindeburg Chapter 47, page 47-19, under certain conditions, ACI permits the use of approximate moment coefficients for continuous beams and one-way slabs. While these may seem a little complicated at first glance, they are fairly straightforward. Also it is more likely that determinate analysis or something like these approximate methods will be encountered in a short multiple choice question than a full indeterminate analysis such as moment distribution. So knowing how to use these moment coefficients may allow you to quickly focus in on the right choice in a multiple choice question with a low burden as far as review goes.


T-BeamsReinforced concrete T-beam systems are common in practice and they are also common in exam problems. There is only a limited amount of additional information to review in order to add T-beams to your repertoire if you are already up to speed on flexural design. Lindeburg discusses T-beams on Page 50-19. The first thing to note is how to determine the effective flange width per ACI code (Note: for a bridge problem you would use AASHTO criteria to determine the flange width). When the neutral axis located within the flange of the T-beam the only difference in calculating the capacity compared to a rectangular beam is you use the effective flange width for the beam width, b. If the neutral axis is below the flange, then you have to adjust to take into account the T-shape of the compression region (see Example 50.8 in Lindeburg). As a minimum know how to find the effective flange width and how do a quick check to see if the neutral axis is in the flange. We’ll reinforce this with an example problem.

Problem S-9

Given problem statement to the right, a slab thickness of 3”, and a span of 20 ft. the design moment strength of this design per repeating unit is most nearly?

(A) 164 ft kips

(B) 148 ft kips

(C) 152 ft kips

(D) 127 ft kips

(Source: 101 Solved Civil Engineering Problems, 3rd Edition )

Solution S-9:Effective width, b,eff = minimum of:

1/4 span = 5 ft. = 60 inches

Stem width + 16*t = 10” + 16*3” = 58”

or, beam spacing = 34”

Therefore, b,eff = 34”

T = AsFy = 60 ksi * 3 sq.in. = 180 kips

a = (AsFy) / (0.85 * f’c * beff ) = 180 kips / (0.85 * 3 ksi * 34 in.) = 2.07 in.

c = a / β1 = 2.07” / 0.85 = 2.44” < 3 inches so N.A. is in flange.


AASHTO T-Beam Flange Guidelines:Interior Girder beff = minimum of:•1/4 effective span •2(6 x tslab) + greater of web thickness or 1/2 the width of the top flange of the girder;•or, c. to c. girdersExterior Girder beff = minimum of:•1/8 span •(6 x tslab) + greater of 1/2 web thickness or 1/4 the width of the top flange of the girder;• or, width of the overhang

ρ = As / beffd = 3 sq. in. / 34” x 12” = .0074

ω = ρ * fy / f’c = 0.0074 * 60 ksi / 3 ksi = 0.1471

From Table 6-1, ρt = 0.01355, therefore section is tension controlled.

From Table 7-1, Mn / (f’c b d2 ) = 0.1342 therefore Mn = 1971.13 kip-in. = 164.3 kip-ft.

or you could use Mn = T * ( d - a/2) = 180 kips * ( 12 - 2.07/2) = 1973.7 kip-in. = 164.5 kip-ft.

Since section is tension controlled, ϕ=0.90. So ϕMn = 0.90 x 164.5 kip-ft = 148 ft kips

ANS (B)

Shallow FoundationsIn Geotechnical Engineering Session 1 we reviewed the general bearing capacity and vertical pressure distribution. In Geotechnical Engineering Session 2 we reviewed calculating the bearing capacity. In this session we will take a closer look at the structural design of the foundation. Lindeburg covers shallow foundations in Chapter 36. The structural design of a reinforced concrete footing is covered in Chapter 55. Let’s examine an example problem that will review the basic steps of completing the structural design of the footing.

Problem S-10A 16 in. thick, 10 ft. square footing on grade supports a 12 inch square column as shown. The net allowable soil pressure is 3000 psf. (Source: Practice Problems for the Civil Engineering PE Exam, 8th Edition).


Problem S-10AIf the column is not subjected to a moment, the maximum service load that the footing can transmit safely to the soil is most nearly?

(A) 200 kips

(B) 250 kips

(C) 300 kips

(D) 350 kips

Problem S-10BIf the column has a total axial service load (dead plus live) of 200 kips, the maximum moment about the centroid of the footing that the footing can support is most nearly?

(A) 140 ft-kips

(B) 170 ft-kips

(C) 190 ft-kips

(D) 220 ft-kips

Problem S-10CThe area of the critical section resisting punching shear is most nearly?

(A) 680 sq. in.

(B) 960 sq. in.

(C) 1200 sq. in.

(D) 1400 sq. in.

For problems S-0D through S-10F, assume the column is subjected to a factored axial force of 340 kips and no moment.

Problem S-10DThe punching shear stress at the critical section is most nearly?

(A) 200 psi

(B) 250 psi


(C) 280 psi

(D) 320 psi

Problem S-10EThe one-way shear stress at the critical section is most nearly:

(A) 60 psi

(B) 80 psi

(C) 100 psi

(D) 120 psi

Problem S-10FThe design moment at the critical section is most nearly:

(A) 280 ft-kips

(B) 300 ft-kips

(C) 340 ft-kips

(D) 440 ft-kips

In Problems S-10G through S-10J, assume the loading on the footing consists of a factored load of 340 kips plus a moment of 150 ft-kips.

Problem S-10GThe maximum punching shear stress at the critical section is most nearly:

(A) 220 psi

(B) 250 psi

(C) 360 psi

(D) 410 psi

Problem S-10HThe maximum one-way shear stress at the critical section is most nearly?


(A) 86 psi

(B) 97 psi

(C) 100 psi

(D) 120 psi

Problem S-10IThe design moment at the critical section is most nearly?

(A) 350 ft-kips

(B) 380 ft-kips

(C) 410 ft-kips

(D) 510 ft-kips

Problem S-10JThe number of No. 6 bars needed as reinforcement for the moment of problem S-5I is most nearly, if the depth to the reinforcing from the top of the footing is 12 in. and f’c = 3000 psi?

(A) 12

(B) 14

(C) 19

(D) 24

Problem S-10KThe one way shear stress capacity of the footing is most nearly? f’c = 3000 psi , fy = 60 ksi

(A) 110

(B) 82

(C) 219

(D) 164

Problem S-10LThe punching shear stress capacity of the footing is most nearly? f’c = 3000 psi, fy = 60 ksi


(A) 110

(B) 82

(C) 219

(D) 164

Solution S-10A:At this point, the only question relates to the bearing capacity. I.e. based on the net allowable bearing capacity how much axial load can a footing of this size carry. P = 3 ksf * 10 ft. * 10 ft. = 300 kips ANS(C)

Solution S-10B:Similar to Problem S-10A, this question is only assessing the capacity of the footing to carry axial load and moment given the net allowable bearing pressure. The total pressure = P/A + Mc/I

Using Eqn. 55.2, If = (1/12) ( LB3) = (1/12)(10 ft.)4 = 833.3 ft4

Use Eqn. 55.1 disregarding correction for burial -

qa = (Ps / Af) + ϒc * h + ϒs ( H - h ) + ( Ms ( B/2 ) ) / If

3 ksf = (200 kips / 100 sq. ft. ) + ( Ms * ( 10 ft / 2 ) ) / 833.3 ft4

Ms - 166.7 ft.-kips ANS (B)

Solution S-10C:Leaving room for bars and cover, assume:

d ≈ h - 4 in. = 16” - 4” = 12 in.

The critical section for punching shear is located a distance d/2 away from the face of the column. Therefore the length of each leg of the critical section for a square column is equal to the column width + 2 (d/2).

b1 = b2 = column size + 2 ( d/2) = 12 in. + 2 ( 12 in. / 2) = 24 inches

Ap = ( 4 ) (24 in.) ( 12 in. ) = 1152 sq. in. ANS (C)

Solution S-10D:Use Eq. 55.14, to find the resultant force over acting over the area b1xb2. This area doesn’t contribute to the shearing force at the critical plane.

R = Pu * b1 * b2 / Af = (340 kips ) * ( 24 in. x 24 in.) / ( 120 in. x 120 in. ) = 13.6 kips

Side Note:P/A ± Mc/I A = BL, I = LB3/12 , c = B/2Mc/I = M(B/2) / (LB3/12) = M(6) / LB2

M = P*e, therefore(P*e) (6) / ((B*L)(B)) = P/A ( 6e/B) and substituting...P/A ± Mc/I = P/A (1 ± (6e/B))


vu = ( Pu - R ) / Ap = ( 340 kips - 13.6 kips ) / 1152 sq. in. = 0.283 ksi = 283 psi ANS (C)

Solution S-10E:The distance from the critical sections to the edge is e = (ftg. width - column width)/2 - d = (10 ft - 1 ft) / 2 - 1 ft. = 3.5 ft.

qu = 340 kips / 100 sq. ft. = 3.4 ksf.

Use eqn. 55.11 for one-way shear -

vu = ( ( 3.4 ksf )( 3.5 ft ) ( 1000 lb / kip)) / ( 1 ft * 144 sq. in. / sq. ft.) = 82.64 psi ANS(B)

Solution S-10F:The length of the critical section for flexure is 4.5 ft. (face of column to edge of footing).

Using Eqn. 55.23 - Mu = qu * L * l2 / 2 = ( 340 kips / ( 100 sq. ft. ) ) * ( 10 ft. ) * ( 4.5 )2 / 2

(This is simply the moment of a cantilever with uniformly distributed load where qu * L is the load. The use of this equation only applies when there is no applied moment to the footing.)

Mu = 344.3 ft. kips ANS (C)

Solution S-5G:From Eqn. 55.16,

J = ( (12 in.) * ( 24 in ) 3 ) /6 x ( 1 + (12 in. / 24 in.)2 + (3) * ( 24 in. / 24 in.) )

J = 117,504 in4

Determine the fraction of the moment assumed to contribute to shear stress, ϒv

ϒv = 1 - ( (1) / ( 1 + ( 2/3 ) * ( √(24 in.)/(24 in.) ) = 0.4

Using Eqn. 55.13,

vu = ( 340 kips - 13.6 kips) / ( 1152 sq. in.) + ( 0.4)(150 ft-kips)(12 in./ft.)(0.5)(24 in.) / 117,504 in4

vu = 0.283 ksi + 0.074 ksi

vu = 0.356 ksi = 356 psi ANS (C)


Solution S-10H:Determine the pressure distribution. Use Eq. 55.1.

q = ( 340 kips ) / ( 100 sf. ) ± (( 150 ft-lb ) ( 10 ft. / 2)) / 833.4 ft4

q = 3.4 ksf ± 0.9 ksf = ( 4.3 ksf , 2.5 ksf )

At the critical section,

q = 4.3 ksf - ( 3.5 ft / 10 ft. ) ( 4.3 ksf - 2.5 ksf) = 3.67 ksf

Q = ( 4.3 ksf + 3.67 ksf ) ( 3.5 ft / 2 ) * ( 10 ft. ) = 139.5 kips

vu = V/A = ((139.5 kips ) ( 1000 lb / kip )) / (( 10 ft ) ( 1 ft ) ( 144 sq.in. / sq. ft. )) = 96.86 psi ANS(B)

Solution S-10I:From the free-body diagram,

Side Note:P/A ± Mc/I A = BL, I = LB3/12 , c = B/2Mc/I = M(B/2) / (LB3/12) = M(6) / LB2

M = P*e, therefore(P*e) (6) / ((B*L)(B)) = P/A ( 6e/B) and substituting...P/A ± Mc/I = P/A (1 ± (6e/B))


A = ( 3.49 ksf ) ( 10 ft ) ( 4.5 ft.) = 157.05 kips

B = (0.81 ksf ) ( 10 ft ) ( 4.5 ft.) ( 0.5 ) = 18.23 kips

Mu = ( 157.05 kips ) ( 4.5 ft. / 2 ) + (18.23 kips ) ( 2/3 ) ( 4.5 ft. ) = 408.05 ft-kips ANS (C)

Solution S-10J:Assume section will be tension controlled. Therefore, ϕ=0.90.

Mu / ( ϕ f’c b d2 ) = 408.05 ft.kips * 12 in. / 1 ft. / ( 0.90 * 3 ksi * 120 in. * 144 sq. in.) = 0.1050

From Table 7-1, ω = 0.1125 --> ρ = ω * f’c / fy = 0.1125 * 3 ksi / 60 ksi = .0056

As = ρ* b * d = 0.0056 * 120” * 12” = 8.1 sq. in. --> #6 = 0.44 sq. in. --> 18.4 bars ANS (C)

From Table 6-1, ρt = 0.01355 Since ρ = 0.0056 < 0.01355 section is tension controlled as assumed.

Solution S-10K:ϕ = 0.75 (new to ACI318-02)

vc = 2 * √f’c = 2 * √3000 psi = 109.5 psi

ϕ vc = 0.75 * 109.5 psi = 82.1 psi ANS (B)

Solution S-10L:ϕ = 0.75 (new to ACI318-02)


vc = ( 2 + y ) * √f’c (Eqn. 55.17)

y = min ( 2, 4/βc, 40d/bo) = min ( 2, 4, 8.9) =2 (Eqn. 55.18 (a consolidated version of several ACI 318 Eqns))

vc = ( 4 * √3000 psi ) = 219.1 psi

ϕ vc = 0.75 * 219.1 psi = 164.3 psi ANS (D)

Note: Solution to Parts K and L suggest that for the higher load of 340 kips, a thicker footing is required since vu one-

way = 96.9 psi and vu two-way = 356 psi both exceed the capacities just calculated.

Problem S-11A square, reinforced concrete footing is installed so that the footing bearing surface is 5 ft. below the soil level, at a point where the allowable soil pressure is 3500 psf. Other than the soil above the footing, there is no surcharge. The soil unit weight is 100 pcf. The footing is located at the corner of a building and is loaded through a concentric 14 in. square column. The column transmits a 125,000 lb service dead load and a 175,000 lb service live load to the footing. The dead load includes the column’s weight but does not include the footing’s weight. The compressive strength for all of the concrete used is 3000 psi. Assume a footing thickness of 24 inches. Answer the following questions. (Source: Civil Engineering Sample Examination, 5th Edition)

Problem S-11AThe minimum footing size is most nearly?

(A) 9 ft x 9 ft

(B) 10 ft x 10 ft

(C) 11 ft x 11 ft

(D) 12 ft x 12 ft

Problem S-11BThe critical (plan) area contributing to two-way punching shear is most nearly?

(A) 60 sq. ft. or less

(B) 70 sq. ft.

(C) 80 sq. ft.

(D) 90 sq. ft. or more


Problem S-11CThe nominal concrete shear strength resisting punching shear is most nearly

(A) 350,000 lb

(B) 450,000 lb

(C) 550,000 lb

(D) 650,000 lb

Problem S-11DThe ultimate two-way punching shear is most nearly?

(A) 350,000 lb

(B) 400,000 lb

(C) 550,000 lb

(D) 650,000 lb

Problem S-11EThe critical area for beam-action shear forces is most nearly?

(A) 20 sq.ft.

(B) 30 sq.ft.

(C) 40 sq. ft.

(D) 50 sq. ft.

Solution S-11A:The weight of the footing must be added to the dead load. The allowable soil pressure is given for the 5 ft depth. The allowable soil pressure does not require a factor of safety. The net allowable soil pressure is

q(net) = pa - tϒconc = 3500 psf - 2 ft. * 150 pcf = 3200 psf

The applied load is not factored in calculating the footing size. The approximate area required is:

A = ( Pd + Pl ) / q(net) = (125000 lb + 175000) / 3200 psf = 93.75 sq. ft.

Since the footing is square, sides of 10 ft would work. ANS (B)


Solution S-11B:Typically you would expect two layers of tension steel, one in each direction, plus 3 inches of cover. Assume the tension bars are approx. 1” in diameter ( #8 ) .

d = t - dia #8 - (dia #8 / 2 ) - cover = 24 in. - 1 in. - 1/2 in. - 3 in. = 19.5 in. say 19 inches.

The critical section for two-way punching shear starts at a distance d/2 from the column face.

d/2 = 9.5 inches

The length of each side of the inner periphery of the two-way punching shear area is

l0 = 9.5 in. + 14 in. (column) + 9.5 in. = 33 in.

The critical perimeter length is

b0 = 4 * l0 = (4) (33 in.) = 132 inches.

The critical area for shear force lies outside the critical perimeter. The critical area is

A critical = bw2-l0 2 = [( 120 sq. in.) 2 - (33 sq. in.) 2 ]/ ( 12 in/ft )2 = 92.44 sq. ft. ANS (D)

Solution S-11C:The nominal concrete shear strength is given by Lindeburg Eq. 55.17.

vc = ( 2 + y ) * √f’c (Eqn. 55.17)

y = min ( 2, 4/βc, 40d/bo) = min ( 2, 4, 5.8) =2 (Eqn. 55.18)

vc = ( 4 * √3000 psi ) = 219.1 psi

Vc = vc * bo * d = 219.1 psi * 132 in. * 19 in. = 549,500 lb. ANS (C)

Solution S-11D:The footing area is, A = B x B = 100 sq. ft.

The ultimate load carried by the footing is

Pu = 1.2 PD + 1.6 PL


Pu = 1.2 ( 125000 lb ) + 1.6 ( 175000 lb ) = 150,000 lb + 280,000 lb. = 430,000 lb.

The factored soil pressure acting on the footing is

pu = Pu / A = 430,000 lb. / 100 sq. ft. = 4300 psf.

The ultimate two-way punching shear is

Vu = pu * A critical = 4300 psf * 92.44 sf. = 397,492 lb. ANS (B)

Solution S-11E:The critical area for beam action is located a distance d from the column face.

The length of the critical section is

l = bw - bw /2 - bc / 2 - d

= 120” - 120”/2 - 14”/2 -19”

= 34 in.

The critical area for beam action shear force is

A critical = bw * l = (120 in ) ( 34 in. ) / ( 12 in/ft )2

A critical = 28.3 sq. ft.

ANS (B)

Retaining WallsIn Geotechnical Session 1 and 2 we examined the lateral pressures applied to retaining walls and the stability analysis of retaining walls. Now we will add the structural design of a reinforced concrete wall. It is important to note that the same flexural and shear design techniques reviewed previously can be used to solve this type of problem.


Problem S-12

The retaining wall shown supports a 400 psf surcharge in addition to the active backfill loading. 3000 psi concrete and 60,000 psi reinforcing steel are used. The specific weights of the backfill and the concrete are 100 pcf and 150 pcf, respectively. The active earth pressure coefficient is 0.5. Rear face of wall is vertical. Answer the following questions. (Source: Practice Problems for the Civil Engineering PE Exam, 8th Edition).

Problem S-12A:The factor of safety against overturning is most nearly?

(A) 2.0

(B) 2.1

(C) 1.5

(D) 1.6

Problem S-12B:The minimum theoretical heel depth is most nearly?

(A) 28 in.

(B) 30 in.

(C) 36 in.

(D) 24 in.

Problem S-12C:Using a heel depth of 30 in. and No. 8 reinforcing bars, the required bar spacing in the heel is most nearly?

(A) 12 in.

(B) 10 in.

(C) 9 in.

(D) 8 in.


Solution S-12A:H = 18.25 ft. + 1.75 ft. = 20 ft.

The soil pressure resultant is:

Ra = 1/2 ( 0.5) ( 100 pcf ) ( 20 ft )2 = 10,000 lb / ft of wall

Ra acts at 20 ft / 3 = 6.67 ft. up from the base.

The surcharge loading is

Rq,h = (0.5) ( 400 psf ) * 20 ft = 4000 lb.

Rq,h acts at 20 ft. / 2 = 10 ft. up from base.

Split the wall, fill, and surcharge into areas to calculate the forces and moments. First find the forces and distances from the heel.

Element length or area (ft or sq. ft)

q or ϒ (lb / sq. ft. )

F or W (lb) r (dist from heel) , (ft)

1 8 400 3200 4

2 146 100 14,600 4

3 22.8 150 3420 ≈ 8.5

4 24.5 150 3675 7

TOTAL 24,895 lb

Now take the moments about the toe, where “x” is the distance from the resultant force to the toe.

Element F ( lb ) x = 14 - r moment ( ft-lb )

1 3200 10 32,000

2 14,600 10 146,000

3 3420 5.5 18,810

4 3675 7 25,725

TOTAL 222,535 ft.lb.

∑Ra * ya = (10,000 lb ) ( 6.67 ft. ) + ( 4000 lb. ) ( 10 ft. ) = 106,700 ft-lb. (moment of lateral forces about toe

F.S. (overturning ) = 222,535 ft.lb. / 106,700 ft.lb. = 2.09 ANS (B)


Solution S-12B:heel weight = ( 8 ft. ) ( 1.75 ft. ) ( 150 pcf. ) = 2100 lb / ft of width (use toe thickness as initial heel thickness est.)

Vu = (1.2 ) ( 2100 lb/ft + 14,600 lb/ft ) + ( 1.6 ) ( 3200 lb/ft) = 20,040 lb/ft + 5,120 lb/ft = 25,160 lb/ft

The nominal shear strength of concrete is

ϕvn = ϕvc = (2) ( 0.75 ) ( √(3000 psi ) = 82.2 psi

The required heel thickness, without requiring shear reinforcement and neglecting the soil pressure beneath the footing, is

(25,160 lb / ft ) / ( ( 82.2 psi * 12 in/ft )) = 25.5 in. Round to 26”

Add at least 3” of cover and h = 29 in. ANS (B)

Solution S-12C:With a heel thickness of 30”, the heel weight becomes

(8 ft.) ( 30 in. / 12 in/ft ) ( 1 ) ( 150 pcf ) = 3000 lb/ft

Take moments about the stem face.

Element F ( lb ) distance ( ft. ) M ( ft-lb / ft. of wall)

1 3200 4 12,800

2 14,600 4 58,400

heel 3000 4 12,000

The ultimate moment on the heel at the stem face is given by Eqn. 54.10b

Mu = 1.2 * ( 58,000 ft-lb + 12,000 ft-lb ) + 1.6 * ( 12,800 ft-lb ) = 104,480 ft-lb / ft. of wall

Use Eqn. 54.13 in Lindeburg or design aids presented in PCA reference.

Mn = Mu / ϕ = 104,480 ft-lb / ft. of wall / 0.90 (assume) = 116,089 ft-lb / ft. of wall

for b = 12 in. d = 26 in. ---> Mn / f’c bd2 = 0.0572

Using Table 7-1, ω = .059 = ρ * fy / f’c ----> ρ = 0.059 * 3000 psi / 60000 psi = .0030

As = ρ * b * d = 0.92 sq. in. / ft. of wall

As,min found using Eqn. 50.16, As,min = 3 * √ f’c * b * d / fy ≥ 200 * b * d / fy

As,min = 3 * √3000 * 12 * 26 / 60,000 = 0.85 sq. in. ≥ 200 * 12” * 26” / 60000 psi = 1.04 sq. in. / ft.

Therefore, As, min. governs at 1.04 sq. in. So spacing provided, s, can be found from -> 0.79 sq. in (#8) / s = 1. 04 sq. in.

s = 0.75 ft. = 9 inches. So #8 at 9” ANS (C)


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