Core Concepts for the Civil PE Exam:

Morning Breadth

Civil Morning Breadth PE Review, Practice Problems and Quick Reference Manual

Ensure You're Prepared on Exam Day, With PE Core Concepts

PE Exam -Morning Breadth 

  • Core Concepts PE Review & Quick Reference guide breaking down the specific information needed for the PE Exam on every Morning Breadth topic from the NCEES Syllabus

  • 80 Civil Breadth practice problems with representative difficulty and detailed solutions

  • Available in Paperback for $29.95 or access all of our practice questions Online Only for $24.99 

Civil PE Morning Breadth Online Study Guide
Click the topics below to expand the core concepts. This material is included in the Paperback edition

PROJECT PLANNING

Lateral Earth Pressure





Soil Consolidation


Settlement is when the soil supporting the foundation consolidates which causes a decrease in volume and a drop in elevation. This causes the foundation to no longer be fully supported and will introduce additional stress. There are three phases of settlement:

  1. Immediate Settling or Elastic Settling: This settlement occurs immediately after the structure is built. The load from the structure causes instant consolidation of the soil. This is the main component of settlement in sandy soil conditions.

  1. Primary Consolidation: A more gradual consolidation which is due to water leaving the voids over time. This is mostly a factor only in clayey soils.

  1. Secondary Consolidation: Also occurs at a very gradual rate. This is due to the shifting and readjustment of soil grains. Most often this is the lowest magnitude of consolidation phases.




Effective Stress





Bearing Capacity


For shallow foundations, the soil below must be suitable to support the load transferred through the footing. Different types of soils have different bearing capacities. Sand is often a good foundation material. Sand undergoes some small immediate settlement and then stabilizes since it drains quickly. Clay generally is poor in bearing capacity. Clays do not drain quickly and will retain water for longer periods of time leading to long-term settlements. Most soils in reality are some combination of sands, clays, and silts which will behave somewhere in-between sand and clay. Exceeding the allowable bearing capacity of a soil will cause shear failure or excessive settlements. Bearing capacity is determined using the Terzaghi-Meyerhof equation:




Slope Stability






Means and Methods

Dead and Live Load


Dead Loads: Loads which are permanent in the final condition of the structure. Examples include self weight and additional permanent loads (such as pavement). Dead load factors are often lower than other types of loads. This is due to the higher level of reliability being able to predict the magnitude and character of these loads.

Live Loads: Loads which will or may change over time. In general live loads represent pedestrian or vehicle loads. The load factor for live loads is often much higher due to the unpredictability.

In LRFD different types of loads are factored to represent a safety factor based on the reliability of our ability to accurately predict certain loading conditions. If only Dead and Live loads are present, the likely load combination is:

1.2D + 1.6L




Trusses


Trusses are structural members used to span long distances. Trusses are built up by members which are only in axial tension or axial compression. They can be analyzed by the method of joints as illustrated below. Consider the example truss with nodes labeled. To design, the axial force in each member must be determined. If we wanted to find the force in member BD, first like a typical beam, the reactions at A and B can be found by summing forces. Then take a free body diagram of only the joint at A. This is shown below. In summing vertical reactions and since the reaction at A was found, the force in AC can be determined. Then there are only 2 horizontal forces and the force in AB can be found: Then take a free body diagram of Joint B as shown below. Since there are only two horizontal forces, the axial force in member BD can then be determined:

The remainder of the truss can be analyzed similarly.

Zero force members:

When determining how many 0-Force members a truss has, analyze each joint individually as a free body diagram and follow these guidelines:

  1. In a joint with 2 members and no external forces or supports, both members are 0-force
  2. In a joint with 2 members and external forces, If the force is parallel to one member and perpendicular to the other, then the member perpendicular to the force is a 0-force member.
  3. In a joint with 3 members and no external forces, if 2 members are parallel then the other is a 0-force member

All other members are non-zero.




Bending Stress


Bending Stress

Mc/I

M = Applied Moment

c = Distance from the Centroid of the Cross Section to the Desired Location of Stress

I = Moment of Inertia of the Cross Section




Shear Stress


Shear stress at any point along a beam is the shear at that point over the area.

t = V/A

V = Shear at the point of interest

A = Cross sectional area

There is also horizontal shear stress due to bending

Horizontal shear stress t = VQ/Ib

V = Applied Shear Force (kips)

Q = First Moment of the Desired Area = ay.

a = Cross Sectional Area from Point of Desired Shear Stress to Extreme Fiber (in2)

y = Distance from Centroid of Beam to Centroid of Area “a” (in)

I = Moment of Inertia of Beam (in3)

b = Width of Member (in)




Axial Stress


Axial Stress:

P/A

P = Applied Force

A = Cross Sectional Area




Deflection


Deflection is the degree to which an element is displaced under load. Common equations for the maximum deflection of beams can be found in the Beam Chart




Beams


The chart below shows reactions, moments, and max deflections for common beam types:

Shear and Moment Diagrams

Shear and moment diagrams are a graphical representation of the forces applied along the length of a beam. The following rules are used to develop shear diagrams:

-A concentrated force causes a jump In the shear diagram of equal magnitude

-A distributed load causes a line in the diagram with slope equal to the distributed load

-Forces up are positive and down is negative

This is depicted graphically below:

The following are rules for construction of a moment diagram:

-The moment at any point on the graph is equal to the area under the shear diagram up to this point

-An isolated moment causes a jump in the diagram of equal magnitude

-The shear at any point in the beam is equal to the slope of the same point on the moment diagram

-A distributed load will cause a parabolic moment diagram curve




Columns





Slabs


One way slabs:

Slabs are structural elements whose length and widths are large compared to the thickness. Slabs are often used as floors or as foundation elements.

Flexure:

Slabs must be analyzed by simplified methods due to the indeterminacy of a full analysis. The most common of which is to analyze as a 1- foot wide strip and treat the span length as a beam. Transverse reinforcement is necessary to control temperature and shrinkage.

Shear:

Shear in slabs is also determined by taking one foot wide sections to analyze as a beam. However often shear will not control.




Footings





Retaining Walls


Retaining walls are built to facilitate an immediate change in elevation. Some uses are to support roadways or a need for a wide, level area to be formed from a sloping existing grade. Retaining walls are designed to resist lateral loads from active earth pressure (see geotechnical section for computation of these loads) and surcharge loads which is any additional load imposed on the soil above, which when close enough will cause an additional pressure on the load due to the distribution of this load through soil. The stem of retaining walls can be analyzed as a cantilever beam extending vertically from the footing. The footing is composed of the toe which is the portion on the side of the lower elevation of soil and the heel which is portion on the side of the higher elevation of soil. Retaining walls are analyzed on a per foot width:

Moment at base of stem= Mstem=Rahya

Rah = Horizontal Active Earth Pressure per ft Width

ya = Eccentricity of Horizontal Active Earth Pressure

For shear, the critical section is a distance, d, from the base of the stem where d is the distance from the main flexural reinforcement (Heel side) to the extreme compression face (Toe side):

Vstem=Rah





Soil Mechanics

Lateral Earth Pressure





Soil Consolidation


Settlement is when the soil supporting the foundation consolidates which causes a decrease in volume and a drop in elevation. This causes the foundation to no longer be fully supported and will introduce additional stress. There are three phases of settlement:

  1. Immediate Settling or Elastic Settling: This settlement occurs immediately after the structure is built. The load from the structure causes instant consolidation of the soil. This is the main component of settlement in sandy soil conditions.

  1. Primary Consolidation: A more gradual consolidation which is due to water leaving the voids over time. This is mostly a factor only in clayey soils.

  1. Secondary Consolidation: Also occurs at a very gradual rate. This is due to the shifting and readjustment of soil grains. Most often this is the lowest magnitude of consolidation phases.




Effective Stress





Bearing Capacity


For shallow foundations, the soil below must be suitable to support the load transferred through the footing. Different types of soils have different bearing capacities. Sand is often a good foundation material. Sand undergoes some small immediate settlement and then stabilizes since it drains quickly. Clay generally is poor in bearing capacity. Clays do not drain quickly and will retain water for longer periods of time leading to long-term settlements. Most soils in reality are some combination of sands, clays, and silts which will behave somewhere in-between sand and clay. Exceeding the allowable bearing capacity of a soil will cause shear failure or excessive settlements. Bearing capacity is determined using the Terzaghi-Meyerhof equation:




Slope Stability






Structural

Dead and Live Load


Dead Loads: Loads which are permanent in the final condition of the structure. Examples include self weight and additional permanent loads (such as pavement). Dead load factors are often lower than other types of loads. This is due to the higher level of reliability being able to predict the magnitude and character of these loads.

Live Loads: Loads which will or may change over time. In general live loads represent pedestrian or vehicle loads. The load factor for live loads is often much higher due to the unpredictability.

In LRFD different types of loads are factored to represent a safety factor based on the reliability of our ability to accurately predict certain loading conditions. If only Dead and Live loads are present, the likely load combination is:

1.2D + 1.6L




Trusses


Trusses are structural members used to span long distances. Trusses are built up by members which are only in axial tension or axial compression. They can be analyzed by the method of joints as illustrated below. Consider the example truss with nodes labeled. To design, the axial force in each member must be determined. If we wanted to find the force in member BD, first like a typical beam, the reactions at A and B can be found by summing forces. Then take a free body diagram of only the joint at A. This is shown below. In summing vertical reactions and since the reaction at A was found, the force in AC can be determined. Then there are only 2 horizontal forces and the force in AB can be found: Then take a free body diagram of Joint B as shown below. Since there are only two horizontal forces, the axial force in member BD can then be determined:

The remainder of the truss can be analyzed similarly.

Zero force members:

When determining how many 0-Force members a truss has, analyze each joint individually as a free body diagram and follow these guidelines:

  1. In a joint with 2 members and no external forces or supports, both members are 0-force
  2. In a joint with 2 members and external forces, If the force is parallel to one member and perpendicular to the other, then the member perpendicular to the force is a 0-force member.
  3. In a joint with 3 members and no external forces, if 2 members are parallel then the other is a 0-force member

All other members are non-zero.




Bending Stress


Bending Stress

Mc/I

M = Applied Moment

c = Distance from the Centroid of the Cross Section to the Desired Location of Stress

I = Moment of Inertia of the Cross Section




Shear Stress


Shear stress at any point along a beam is the shear at that point over the area.

t = V/A

V = Shear at the point of interest

A = Cross sectional area

There is also horizontal shear stress due to bending

Horizontal shear stress t = VQ/Ib

V = Applied Shear Force (kips)

Q = First Moment of the Desired Area = ay.

a = Cross Sectional Area from Point of Desired Shear Stress to Extreme Fiber (in2)

y = Distance from Centroid of Beam to Centroid of Area “a” (in)

I = Moment of Inertia of Beam (in3)

b = Width of Member (in)




Axial Stress


Axial Stress:

P/A

P = Applied Force

A = Cross Sectional Area




Deflection


Deflection is the degree to which an element is displaced under load. Common equations for the maximum deflection of beams can be found in the Beam Chart




Beams


The chart below shows reactions, moments, and max deflections for common beam types:

Shear and Moment Diagrams

Shear and moment diagrams are a graphical representation of the forces applied along the length of a beam. The following rules are used to develop shear diagrams:

-A concentrated force causes a jump In the shear diagram of equal magnitude

-A distributed load causes a line in the diagram with slope equal to the distributed load

-Forces up are positive and down is negative

This is depicted graphically below:

The following are rules for construction of a moment diagram:

-The moment at any point on the graph is equal to the area under the shear diagram up to this point

-An isolated moment causes a jump in the diagram of equal magnitude

-The shear at any point in the beam is equal to the slope of the same point on the moment diagram

-A distributed load will cause a parabolic moment diagram curve




Columns





Slabs


One way slabs:

Slabs are structural elements whose length and widths are large compared to the thickness. Slabs are often used as floors or as foundation elements.

Flexure:

Slabs must be analyzed by simplified methods due to the indeterminacy of a full analysis. The most common of which is to analyze as a 1- foot wide strip and treat the span length as a beam. Transverse reinforcement is necessary to control temperature and shrinkage.

Shear:

Shear in slabs is also determined by taking one foot wide sections to analyze as a beam. However often shear will not control.




Footings





Retaining Walls


Retaining walls are built to facilitate an immediate change in elevation. Some uses are to support roadways or a need for a wide, level area to be formed from a sloping existing grade. Retaining walls are designed to resist lateral loads from active earth pressure (see geotechnical section for computation of these loads) and surcharge loads which is any additional load imposed on the soil above, which when close enough will cause an additional pressure on the load due to the distribution of this load through soil. The stem of retaining walls can be analyzed as a cantilever beam extending vertically from the footing. The footing is composed of the toe which is the portion on the side of the lower elevation of soil and the heel which is portion on the side of the higher elevation of soil. Retaining walls are analyzed on a per foot width:

Moment at base of stem= Mstem=Rahya

Rah = Horizontal Active Earth Pressure per ft Width

ya = Eccentricity of Horizontal Active Earth Pressure

For shear, the critical section is a distance, d, from the base of the stem where d is the distance from the main flexural reinforcement (Heel side) to the extreme compression face (Toe side):

Vstem=Rah





Hydraulics and Environmental

Lateral Earth Pressure





Soil Consolidation


Settlement is when the soil supporting the foundation consolidates which causes a decrease in volume and a drop in elevation. This causes the foundation to no longer be fully supported and will introduce additional stress. There are three phases of settlement:

  1. Immediate Settling or Elastic Settling: This settlement occurs immediately after the structure is built. The load from the structure causes instant consolidation of the soil. This is the main component of settlement in sandy soil conditions.

  1. Primary Consolidation: A more gradual consolidation which is due to water leaving the voids over time. This is mostly a factor only in clayey soils.

  1. Secondary Consolidation: Also occurs at a very gradual rate. This is due to the shifting and readjustment of soil grains. Most often this is the lowest magnitude of consolidation phases.




Effective Stress





Bearing Capacity


For shallow foundations, the soil below must be suitable to support the load transferred through the footing. Different types of soils have different bearing capacities. Sand is often a good foundation material. Sand undergoes some small immediate settlement and then stabilizes since it drains quickly. Clay generally is poor in bearing capacity. Clays do not drain quickly and will retain water for longer periods of time leading to long-term settlements. Most soils in reality are some combination of sands, clays, and silts which will behave somewhere in-between sand and clay. Exceeding the allowable bearing capacity of a soil will cause shear failure or excessive settlements. Bearing capacity is determined using the Terzaghi-Meyerhof equation:




Slope Stability






Transportation

Dead and Live Load


Dead Loads: Loads which are permanent in the final condition of the structure. Examples include self weight and additional permanent loads (such as pavement). Dead load factors are often lower than other types of loads. This is due to the higher level of reliability being able to predict the magnitude and character of these loads.

Live Loads: Loads which will or may change over time. In general live loads represent pedestrian or vehicle loads. The load factor for live loads is often much higher due to the unpredictability.

In LRFD different types of loads are factored to represent a safety factor based on the reliability of our ability to accurately predict certain loading conditions. If only Dead and Live loads are present, the likely load combination is:

1.2D + 1.6L




Trusses


Trusses are structural members used to span long distances. Trusses are built up by members which are only in axial tension or axial compression. They can be analyzed by the method of joints as illustrated below. Consider the example truss with nodes labeled. To design, the axial force in each member must be determined. If we wanted to find the force in member BD, first like a typical beam, the reactions at A and B can be found by summing forces. Then take a free body diagram of only the joint at A. This is shown below. In summing vertical reactions and since the reaction at A was found, the force in AC can be determined. Then there are only 2 horizontal forces and the force in AB can be found: Then take a free body diagram of Joint B as shown below. Since there are only two horizontal forces, the axial force in member BD can then be determined:

The remainder of the truss can be analyzed similarly.

Zero force members:

When determining how many 0-Force members a truss has, analyze each joint individually as a free body diagram and follow these guidelines:

  1. In a joint with 2 members and no external forces or supports, both members are 0-force
  2. In a joint with 2 members and external forces, If the force is parallel to one member and perpendicular to the other, then the member perpendicular to the force is a 0-force member.
  3. In a joint with 3 members and no external forces, if 2 members are parallel then the other is a 0-force member

All other members are non-zero.




Bending Stress


Bending Stress

Mc/I

M = Applied Moment

c = Distance from the Centroid of the Cross Section to the Desired Location of Stress

I = Moment of Inertia of the Cross Section




Shear Stress


Shear stress at any point along a beam is the shear at that point over the area.

t = V/A

V = Shear at the point of interest

A = Cross sectional area

There is also horizontal shear stress due to bending

Horizontal shear stress t = VQ/Ib

V = Applied Shear Force (kips)

Q = First Moment of the Desired Area = ay.

a = Cross Sectional Area from Point of Desired Shear Stress to Extreme Fiber (in2)

y = Distance from Centroid of Beam to Centroid of Area “a” (in)

I = Moment of Inertia of Beam (in3)

b = Width of Member (in)




Axial Stress


Axial Stress:

P/A

P = Applied Force

A = Cross Sectional Area




Deflection


Deflection is the degree to which an element is displaced under load. Common equations for the maximum deflection of beams can be found in the Beam Chart




Beams


The chart below shows reactions, moments, and max deflections for common beam types:

Shear and Moment Diagrams

Shear and moment diagrams are a graphical representation of the forces applied along the length of a beam. The following rules are used to develop shear diagrams:

-A concentrated force causes a jump In the shear diagram of equal magnitude

-A distributed load causes a line in the diagram with slope equal to the distributed load

-Forces up are positive and down is negative

This is depicted graphically below:

The following are rules for construction of a moment diagram:

-The moment at any point on the graph is equal to the area under the shear diagram up to this point

-An isolated moment causes a jump in the diagram of equal magnitude

-The shear at any point in the beam is equal to the slope of the same point on the moment diagram

-A distributed load will cause a parabolic moment diagram curve




Columns





Slabs


One way slabs:

Slabs are structural elements whose length and widths are large compared to the thickness. Slabs are often used as floors or as foundation elements.

Flexure:

Slabs must be analyzed by simplified methods due to the indeterminacy of a full analysis. The most common of which is to analyze as a 1- foot wide strip and treat the span length as a beam. Transverse reinforcement is necessary to control temperature and shrinkage.

Shear:

Shear in slabs is also determined by taking one foot wide sections to analyze as a beam. However often shear will not control.




Footings





Retaining Walls


Retaining walls are built to facilitate an immediate change in elevation. Some uses are to support roadways or a need for a wide, level area to be formed from a sloping existing grade. Retaining walls are designed to resist lateral loads from active earth pressure (see geotechnical section for computation of these loads) and surcharge loads which is any additional load imposed on the soil above, which when close enough will cause an additional pressure on the load due to the distribution of this load through soil. The stem of retaining walls can be analyzed as a cantilever beam extending vertically from the footing. The footing is composed of the toe which is the portion on the side of the lower elevation of soil and the heel which is portion on the side of the higher elevation of soil. Retaining walls are analyzed on a per foot width:

Moment at base of stem= Mstem=Rahya

Rah = Horizontal Active Earth Pressure per ft Width

ya = Eccentricity of Horizontal Active Earth Pressure

For shear, the critical section is a distance, d, from the base of the stem where d is the distance from the main flexural reinforcement (Heel side) to the extreme compression face (Toe side):

Vstem=Rah





Materials

Lateral Earth Pressure





Soil Consolidation


Settlement is when the soil supporting the foundation consolidates which causes a decrease in volume and a drop in elevation. This causes the foundation to no longer be fully supported and will introduce additional stress. There are three phases of settlement:

  1. Immediate Settling or Elastic Settling: This settlement occurs immediately after the structure is built. The load from the structure causes instant consolidation of the soil. This is the main component of settlement in sandy soil conditions.

  1. Primary Consolidation: A more gradual consolidation which is due to water leaving the voids over time. This is mostly a factor only in clayey soils.

  1. Secondary Consolidation: Also occurs at a very gradual rate. This is due to the shifting and readjustment of soil grains. Most often this is the lowest magnitude of consolidation phases.




Effective Stress





Bearing Capacity


For shallow foundations, the soil below must be suitable to support the load transferred through the footing. Different types of soils have different bearing capacities. Sand is often a good foundation material. Sand undergoes some small immediate settlement and then stabilizes since it drains quickly. Clay generally is poor in bearing capacity. Clays do not drain quickly and will retain water for longer periods of time leading to long-term settlements. Most soils in reality are some combination of sands, clays, and silts which will behave somewhere in-between sand and clay. Exceeding the allowable bearing capacity of a soil will cause shear failure or excessive settlements. Bearing capacity is determined using the Terzaghi-Meyerhof equation:




Slope Stability






Site Development

Lateral Earth Pressure





Soil Consolidation


Settlement is when the soil supporting the foundation consolidates which causes a decrease in volume and a drop in elevation. This causes the foundation to no longer be fully supported and will introduce additional stress. There are three phases of settlement:

  1. Immediate Settling or Elastic Settling: This settlement occurs immediately after the structure is built. The load from the structure causes instant consolidation of the soil. This is the main component of settlement in sandy soil conditions.

  1. Primary Consolidation: A more gradual consolidation which is due to water leaving the voids over time. This is mostly a factor only in clayey soils.

  1. Secondary Consolidation: Also occurs at a very gradual rate. This is due to the shifting and readjustment of soil grains. Most often this is the lowest magnitude of consolidation phases.




Effective Stress





Bearing Capacity


For shallow foundations, the soil below must be suitable to support the load transferred through the footing. Different types of soils have different bearing capacities. Sand is often a good foundation material. Sand undergoes some small immediate settlement and then stabilizes since it drains quickly. Clay generally is poor in bearing capacity. Clays do not drain quickly and will retain water for longer periods of time leading to long-term settlements. Most soils in reality are some combination of sands, clays, and silts which will behave somewhere in-between sand and clay. Exceeding the allowable bearing capacity of a soil will cause shear failure or excessive settlements. Bearing capacity is determined using the Terzaghi-Meyerhof equation:




Slope Stability






 
Civil PE Exam Morning Reference Manual Example
Civil PE Exam Morning Practice Problem Example
Contact Us

© 2023 by Sphere Construction. Proudly created with Wix.com