Core Concepts for the Civil PE Exam:

Structural Depth

Civil Morning Breadth and PE Structural Exam Practice Problems and Quick Reference Manual 

PE Exam - Structural Depth

  • PE Core Concepts PE Structural Exam Review & Quick Reference Guide designed to break down the specific information needed for the exam on every topic from the NCEES Syllabus

  • Comprehensive PE Civil Engineering Structural Practice Exam​.

  • 40 Civil Breadth practice problems with detailed solutions

  • 80 Structural Depth practice problems with detailed solutions

  • Breakdown of all NCEES listed codes including ACI, AISC, IBC, ASCE, Masonry design, NDS, AASHTO, OSHA, and PCI 

  • Available in Paperback for $34.95 or access all of our practice questions Online Only for $24.99 

Structural Depth Online Study Guide
Click the topics below to expand the core concepts. This material is included in the Paperback edition

ASCE

Live Load





Snow Loads


Snow loads are determined from Chapter 7. There are 3 types of snow loads to know: ground, flat roof, and design roof.

Pg = ground snow load. This is the snow load on the ground as per a specific geographic area.

Pf = flat roof snow load = Pg(0.7CeCtIS)

Ce = Exposure Factor from Table 7-2

Ct = Thermal Factor from Table 7-3

IS = Importance factor from 1-5.2

Ensure this is greater than the minimum = 20IS

The design roof snow load Ps = CsPf

Determine CS from Table 7-2c .




Snow Drift


Snow drift is additional load due to snow building up against a vertical wall from wind. The additional load is approximated by a triangular cross section of snow. Figure 7-8 depicts the necessary variables. To solve:

First determine the density of snow:

g = 0.13pg + 14 < 30 pcf

Then using the density you can determine the height of the roof snow hb = ps/g

hc = the vertical distance from the top of roof snow to upper roof = height to upper roof - hb

If hc < 0.2hb, Snow drift does not need to be applied

Then using figure 7-9 determine the drift height hd which will be the larger of:

For Leeward drift use lu = the length of the upper roof

For Windward use lu = the length of the lower roof and only use ¾ of the hd as determined from figure 7-9

Then calculate the width of the drift, w, for hd < hc w=4h, if hd > hc w = 4hd2/hc however w shall not be greater 8hc

The variables are better depicted in the diagram below:




Site Classification and Occupancy


  • From Table 20.3-1 the site classification can be determined
  • Table 1.5-2 for the risk category




Seismic Base Shear and Force Distribution


Equivalent Lateral Force Procedure Section 12.8

The seismic base shear by the equivalent force method V = CsW

CS = Seismic response coefficient = SDS/(R/Ie)

Determine Ie from table 1.5-2 using the risk category

R = response modification factor from table 12.2-1

The lateral seismic force at a given level shall be:

Cvx = vertical distribution factor

V = total design lateral force or shear at the base of the

structure (kip or kN)

wi and wx = the portion of the total effective seismic weight of the structure (W) located or assigned to Level i or x

hi and hx = the height (ft) from the base to Level i or x

k = an exponent related to the structure period as follows:

for structures having a period of 0.5 s or less,

k = 1 for structures having a period of 2.5 s or more,

k = 2for structures having a period between 0.5 and

2.5 s, k shall be 2 or shall be determined by linear

interpolation between 1 and 2




Effective Seismic Weight


Effective weight is the load which can be accounted for to offset horizontal seismic forces. This includes the dead load and any additional loading as outlined in section 12.7.2 such as:

  1. In areas used for storage, a minimum of 25 percent of the floor live load (floor live load in public garages and open parking structures need not be included).
  2. Where provision for partitions is required by Section 4.2.2 in the floor load design, the actual partition weight or a minimum weight of 10 psf (0.48 kN/m2) of floor area, whichever is greater.
  3. Total operating weight of permanent equipment.
  4. Where the flat roof snow load, Pf , exceeds 30 psf (1.44 kN/m2), 20 percent of the uniform design snow load, regardless of actual roof slope.




Site Coefficients and Spectural Response Factors


SDS = Design Spectural response acceleration parameter at short periods = 2/3SMS = 2/3FaSS

SMS = The risk targeted maximum considered earthquake ground motion acceleration parameter (MCER) Spectural response acceleration parameter at short periods.

Fa = Site coefficient defined in table 11.4-1

SS = Mapped (MCER) Spectural response acceleration parameter at short periods determined in accordance with section 11.4.1

SD1 = Design Spectural response acceleration parameter at a period of 1 s = 2/3SM1 = 2/3FvS1

SM1 = The risk targeted maximum considered earthquake ground motion acceleration parameter (MCER) Spectural response acceleration parameter at a period of 1 s.

Fv = Site coefficient defined in table 11.4-1

S1 = Mapped (MCER) Spectural response acceleration parameter at a period of 1 sec as determined in accordance with section 11.4.1

From Table 20.3-1 the site classification can be determined

Determine the Seismic design category based on short period response acceleration parameters from Table 11.6-1

Determine the Seismic design category based on 1-S period response acceleration parameters from Table 11.6-2





AASHTO

3D Statics


Statics in 3 dimensions introduces additional equations of equilibrium due to the third axis. Apply the same basic principles for the sum of the following forces:

FX = 0

FY = 0

FZ = 0

MX = 0

MY = 0

MZ = 0

  • First determine location of origin (0, 0, 0)
  • Determine X, Y, and Z component of all forces
  • Determine moment from each component about each axis
  • Moment about an axis is the perpendicular distance from a force component to that axis
  • Forces parallel to an axis has zero moment about that axis
  • Forces that run through an axis have zero moment about the axis




Moving Loads


  • Moving Loads are most often from Live Load due to traffic
  • Need to analyze position of load to cause the greatest stress
  • Shear in general is greatest when loads are at the support
  • Positive moment in general is greatest with the loads at midspan
  • Negative Moment is greatest with the load cloase to the support




Hinges


Hinges are supports at which there is a zero moment and only an axial and vertical force can be transferred

Hinges are best analyzed by taking free body diagrams to either side of the hinge




Cables


  • Cables carry load only in tension
  • Acts as axial two force tension members
  • Can be analyzed similarly to trusses use the method of joints

Consider the example below:

Determine support reactions by drawing free body diagram of entire system

Then you can take free body diagrams of individual points to determine axial tensions:





ACI

3D Statics


Statics in 3 dimensions introduces additional equations of equilibrium due to the third axis. Apply the same basic principles for the sum of the following forces:

FX = 0

FY = 0

FZ = 0

MX = 0

MY = 0

MZ = 0

  • First determine location of origin (0, 0, 0)
  • Determine X, Y, and Z component of all forces
  • Determine moment from each component about each axis
  • Moment about an axis is the perpendicular distance from a force component to that axis
  • Forces parallel to an axis has zero moment about that axis
  • Forces that run through an axis have zero moment about the axis




Moving Loads


  • Moving Loads are most often from Live Load due to traffic
  • Need to analyze position of load to cause the greatest stress
  • Shear in general is greatest when loads are at the support
  • Positive moment in general is greatest with the loads at midspan
  • Negative Moment is greatest with the load cloase to the support




Hinges


Hinges are supports at which there is a zero moment and only an axial and vertical force can be transferred

Hinges are best analyzed by taking free body diagrams to either side of the hinge




Cables


  • Cables carry load only in tension
  • Acts as axial two force tension members
  • Can be analyzed similarly to trusses use the method of joints

Consider the example below:

Determine support reactions by drawing free body diagram of entire system

Then you can take free body diagrams of individual points to determine axial tensions:





AISC

Prestressing Stresses


In addition to the stress from external forces, prestressed beams are subject to the stresses from the strands. There are 2 types of stress from the applied Prestressing force (P):

  1. Compression stress due to the strands P/A
  2. Bending due to the eccentricity of the strands Pe(c)/I

When calculating total stress, be aware of signs. The eccentric prestress force causes a negative moment which will offset any positive bending.




Prestressing Flexure





Shipping and Handling


  • Precast/prestressed members need to transported. This introduces stresses which may be different from the in-place design conditions.
  • PCI provides provisions for the handling of precast members to limit cracking.
  • The modulus of rupture of the section must be greater than applied stress due handling.
  • Modulus of Rupture, fr = 7.5sqrt(f'c)
  • PCI provides equations for the moments during lifting of typical pick point configurations. This can be found in Figure 8.3.1.
  • The situation in which lifting or transportation occurs requires an additional multiplier as outlined in Table 8.3.1.





NDS

Prestressing Stresses


In addition to the stress from external forces, prestressed beams are subject to the stresses from the strands. There are 2 types of stress from the applied Prestressing force (P):

  1. Compression stress due to the strands P/A
  2. Bending due to the eccentricity of the strands Pe(c)/I

When calculating total stress, be aware of signs. The eccentric prestress force causes a negative moment which will offset any positive bending.




Prestressing Flexure





Shipping and Handling


  • Precast/prestressed members need to transported. This introduces stresses which may be different from the in-place design conditions.
  • PCI provides provisions for the handling of precast members to limit cracking.
  • The modulus of rupture of the section must be greater than applied stress due handling.
  • Modulus of Rupture, fr = 7.5sqrt(f'c)
  • PCI provides equations for the moments during lifting of typical pick point configurations. This can be found in Figure 8.3.1.
  • The situation in which lifting or transportation occurs requires an additional multiplier as outlined in Table 8.3.1.





ACI 530 Masonry

3D Statics


Statics in 3 dimensions introduces additional equations of equilibrium due to the third axis. Apply the same basic principles for the sum of the following forces:

FX = 0

FY = 0

FZ = 0

MX = 0

MY = 0

MZ = 0

  • First determine location of origin (0, 0, 0)
  • Determine X, Y, and Z component of all forces
  • Determine moment from each component about each axis
  • Moment about an axis is the perpendicular distance from a force component to that axis
  • Forces parallel to an axis has zero moment about that axis
  • Forces that run through an axis have zero moment about the axis




Moving Loads


  • Moving Loads are most often from Live Load due to traffic
  • Need to analyze position of load to cause the greatest stress
  • Shear in general is greatest when loads are at the support
  • Positive moment in general is greatest with the loads at midspan
  • Negative Moment is greatest with the load cloase to the support




Hinges


Hinges are supports at which there is a zero moment and only an axial and vertical force can be transferred

Hinges are best analyzed by taking free body diagrams to either side of the hinge




Cables


  • Cables carry load only in tension
  • Acts as axial two force tension members
  • Can be analyzed similarly to trusses use the method of joints

Consider the example below:

Determine support reactions by drawing free body diagram of entire system

Then you can take free body diagrams of individual points to determine axial tensions:





PCI

Prestressing Stresses


In addition to the stress from external forces, prestressed beams are subject to the stresses from the strands. There are 2 types of stress from the applied Prestressing force (P):

  1. Compression stress due to the strands P/A
  2. Bending due to the eccentricity of the strands Pe(c)/I

When calculating total stress, be aware of signs. The eccentric prestress force causes a negative moment which will offset any positive bending.




Prestressing Flexure





Shipping and Handling


  • Precast/prestressed members need to transported. This introduces stresses which may be different from the in-place design conditions.
  • PCI provides provisions for the handling of precast members to limit cracking.
  • The modulus of rupture of the section must be greater than applied stress due handling.
  • Modulus of Rupture, fr = 7.5sqrt(f'c)
  • PCI provides equations for the moments during lifting of typical pick point configurations. This can be found in Figure 8.3.1.
  • The situation in which lifting or transportation occurs requires an additional multiplier as outlined in Table 8.3.1.





OSHA

Flexure


Flexural capacity of steel is determined by the length of lateral support. If a beam is fully laterally supported, the capacity is the plastic moment capacity:

The plastic moment is the yield strength times the plastic section modulus: MP = FyZ

The plastic section modulus is the summation of the areas in compression and tension multiplied by the distance from the center of gravity of each area to the plastic neutral axis.

For a symmetric W shape, Z = 2(AfYf +Aw/2Yw)

If a beam is not fully laterally supported, the capacity is a function of the unbraced length and will fail in lateral torsion buckling. For the purposes of this exam it is best to determine the capacity of a member from the design tables AISC 3-10. Find the intersection of the unbraced length and the applied load and then find the first dark line directly above this point to find the most efficient section.




Compression


The capacity of compression members are a function of the unbraced length. For non-doubly symmetric members there is a different capacity about each axis and it must be determined which is the controlling axis. To do this AISC Chapter 4 Tables provide the conversion factor to determine the equivalent unbraced length of the strong axis which is the ratio of the radii of gyration about each axis.

Then the design axial capacity can be determined from the appropriate table for each structural shape.




Tension


A steel tension member needs to be checked for 2 modes of failure. Yielding of the gross area and rupture of the effective net area

The gross area includes no holes

The net area includes the holes. The diameter of the hole is obtained by adding 1/8” to the diameter of the bolt




Fatigue


Fatigue is covered in the specifications Appendix 3 of the AISC. The stress range is determined by the following equation:

N is the fluctuations for the life of the structure

Cf = Found from table A3.1

FTH = Table A3.1




Welds





Block Shear


The main concept to understand for block shear is that a component of the capacity comes from tension which is the length of perimeter perpendicular to the load and the other comes in shear from the portion which is parallel to the tension. Then simply fill in the equation

Ubs = 1.0 if tension is uniform and 0.5 otherwise (Most often taken as 1.0)

Anv = net area in shear (in2)

Ant = net area in tension (in2)

Agv = gross area in shear (in2)

The diameter of the hole is 1/8” + diameter of the bolt




Bolt Strength


Bolted connections fail in shear or bearing.

For Shear:

Fn = Shear strength of bolt from table J3.2. N is for threads included, X for threads excluded

n = number of bolts

Also note the capacity is multiplied by the number of shear planes on the bolt

For Bearing:

Lc = clear edge distance measured from edge of hole to edge of connected material

Fu = ultimate strength of connected material

t = thickness of connected material

d = bolt diameter





IBC

Prestressing Stresses


In addition to the stress from external forces, prestressed beams are subject to the stresses from the strands. There are 2 types of stress from the applied Prestressing force (P):

  1. Compression stress due to the strands P/A
  2. Bending due to the eccentricity of the strands Pe(c)/I

When calculating total stress, be aware of signs. The eccentric prestress force causes a negative moment which will offset any positive bending.




Prestressing Flexure





Shipping and Handling


  • Precast/prestressed members need to transported. This introduces stresses which may be different from the in-place design conditions.
  • PCI provides provisions for the handling of precast members to limit cracking.
  • The modulus of rupture of the section must be greater than applied stress due handling.
  • Modulus of Rupture, fr = 7.5sqrt(f'c)
  • PCI provides equations for the moments during lifting of typical pick point configurations. This can be found in Figure 8.3.1.
  • The situation in which lifting or transportation occurs requires an additional multiplier as outlined in Table 8.3.1.





AWS

Weld Symbols and Types






Advanced Statics

3D Statics


Statics in 3 dimensions introduces additional equations of equilibrium due to the third axis. Apply the same basic principles for the sum of the following forces:

FX = 0

FY = 0

FZ = 0

MX = 0

MY = 0

MZ = 0

  • First determine location of origin (0, 0, 0)
  • Determine X, Y, and Z component of all forces
  • Determine moment from each component about each axis
  • Moment about an axis is the perpendicular distance from a force component to that axis
  • Forces parallel to an axis has zero moment about that axis
  • Forces that run through an axis have zero moment about the axis




Moving Loads


  • Moving Loads are most often from Live Load due to traffic
  • Need to analyze position of load to cause the greatest stress
  • Shear in general is greatest when loads are at the support
  • Positive moment in general is greatest with the loads at midspan
  • Negative Moment is greatest with the load cloase to the support




Hinges


Hinges are supports at which there is a zero moment and only an axial and vertical force can be transferred

Hinges are best analyzed by taking free body diagrams to either side of the hinge




Cables


  • Cables carry load only in tension
  • Acts as axial two force tension members
  • Can be analyzed similarly to trusses use the method of joints

Consider the example below:

Determine support reactions by drawing free body diagram of entire system

Then you can take free body diagrams of individual points to determine axial tensions:





Misc. Structural Topics

Flexure


Flexural capacity of steel is determined by the length of lateral support. If a beam is fully laterally supported, the capacity is the plastic moment capacity:

The plastic moment is the yield strength times the plastic section modulus: MP = FyZ

The plastic section modulus is the summation of the areas in compression and tension multiplied by the distance from the center of gravity of each area to the plastic neutral axis.

For a symmetric W shape, Z = 2(AfYf +Aw/2Yw)

If a beam is not fully laterally supported, the capacity is a function of the unbraced length and will fail in lateral torsion buckling. For the purposes of this exam it is best to determine the capacity of a member from the design tables AISC 3-10. Find the intersection of the unbraced length and the applied load and then find the first dark line directly above this point to find the most efficient section.




Compression


The capacity of compression members are a function of the unbraced length. For non-doubly symmetric members there is a different capacity about each axis and it must be determined which is the controlling axis. To do this AISC Chapter 4 Tables provide the conversion factor to determine the equivalent unbraced length of the strong axis which is the ratio of the radii of gyration about each axis.

Then the design axial capacity can be determined from the appropriate table for each structural shape.




Tension


A steel tension member needs to be checked for 2 modes of failure. Yielding of the gross area and rupture of the effective net area

The gross area includes no holes

The net area includes the holes. The diameter of the hole is obtained by adding 1/8” to the diameter of the bolt




Fatigue


Fatigue is covered in the specifications Appendix 3 of the AISC. The stress range is determined by the following equation:

N is the fluctuations for the life of the structure

Cf = Found from table A3.1

FTH = Table A3.1




Welds





Block Shear


The main concept to understand for block shear is that a component of the capacity comes from tension which is the length of perimeter perpendicular to the load and the other comes in shear from the portion which is parallel to the tension. Then simply fill in the equation

Ubs = 1.0 if tension is uniform and 0.5 otherwise (Most often taken as 1.0)

Anv = net area in shear (in2)

Ant = net area in tension (in2)

Agv = gross area in shear (in2)

The diameter of the hole is 1/8” + diameter of the bolt




Bolt Strength


Bolted connections fail in shear or bearing.

For Shear:

Fn = Shear strength of bolt from table J3.2. N is for threads included, X for threads excluded

n = number of bolts

Also note the capacity is multiplied by the number of shear planes on the bolt

For Bearing:

Lc = clear edge distance measured from edge of hole to edge of connected material

Fu = ultimate strength of connected material

t = thickness of connected material

d = bolt diameter





 
 
Civil PE Exam Structural Depth Sample Reference Manual
Civil PE Exam Structural Depth Sample Practice Problems
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