Basic Properties of Section

Note: The source of the technical material in this volume is the Professional Engineering Development Program (PEDP) of Engineering Services.

Warning: The material contained in this document was developed for Saudi Aramco and is intended for the exclusive use of Saudi Aramco’s employees. Any material contained in this document which is not already in the public domain may not be copied, reproduced, sold, given, or disclosed to third parties, or otherwise used in whole, or in part, without the written permission of the Vice President, Engineering Services, Saudi Aramco.

Chapter : Civil Engineering For additional information on this subject, contact File Reference: CSE 106.01 PEDD Coordinator on 862-1026

Engineering Encyclopedia Saudi Aramco Desktop Standards

BASIC PROPERTIES OF SECTION

Engineering Encyclopedia Civil Engineering: Basic Properties of Section

Basic Strength of Materials

Saudi Aramco Desktop Standards i

Section Page

OBJECTIVES.................................................................................................................. 1

TERMINAL OBJECTIVE ...................................................................................... 1

ENABLING OBJECTIVES .................................................................................... 1

INFORMATION ............................................................................................................... 3

INTRODUCTION.................................................................................................. 3 MODULE INTRODUCTION.................................................................................. 4 SECTION PROPERTIES ..................................................................................... 5

Types of Sections ...................................................................................... 5 Uniform Sections............................................................................. 5

Non-uniform Sections...................................................................... 6

Simple Solid Sections ..................................................................... 7

Compound Sections........................................................................ 7

Composite Sections ........................................................................ 8

Structural Steel Shapes .................................................................. 9

Thin-Wall Shapes............................................................................ 9

Common Section Properties .................................................................... 10

Area, A (in2, mm2) ........................................................................ 11

Moment of Area, Q (in2, mm2)...................................................... 11

Centroid ........................................................................................ 12

Moment of Inertia, I (in4, mm4) = ∑Aixi2, ∑Aiyi

2 ............................. 13

Section Modulus, S (in3, mm3) = I/c ............................................. 14

Torsional Constant, R (in4, mm4) ................................................. 15

Radius of Gyration, r (in, mm) ....................................................... 15

Listed Properties - Standard Sections ..................................................... 16 Properties...................................................................................... 16

Standard Shapes .......................................................................... 16

Example 1: Calculating Section Properties for Standard Shapes ........... 18 Calculating Section Properties for Nonstandard Shapes ......................... 20

Compound Sections...................................................................... 20



Saudi Aramco Desktop Standards ii

Example 2: Properties of Compound Sections ....................................... 21 Composite Sections................................................................................. 22 Example 3: Calculating Properties of Composite Sections ..................... 23 Example 4: Calculating Properties of Reinforced Concrete Section ....... 24

SUMMARY ......................................................................................................... 25

WORK AIDS.................................................................................................................. 26

GLOSSARY .................................................................................................................. 68

PRACTICE PROBLEMS ............................................................................................... 70

List of Figures

Figure 1. Uniform Sections............................................................................................. 5

Figure 2. Non-uniform Sections...................................................................................... 6

Figure 3. Simple Solid Sections ..................................................................................... 7

Figure 4. Compound Sections........................................................................................ 7

Figure 5. Composite Sections ........................................................................................ 8

Figure 6. Structural Steel Shapes................................................................................... 9

Figure 7. Thin-Walled Shapes...................................................................................... 10

Figure 8a. Rectangular Beam ...................................................................................... 18

Figure 8b. Elliptical Sewer Pipe.................................................................................... 18

Figure 9. Structural Member Consisting of Two Back- to-Back Angles ........................ 19

Figure 11. Plate Girder ................................................................................................. 21

Figure 10. Bridge Pier .................................................................................................. 21

Figure 11. Plate Girder ................................................................................................. 21

Figure 12. Composite Sections .................................................................................... 23

Figure 13. Reinforced Concrete Section ....................................................................... 24

Figure 14. Concrete Beam ............................................................................................ 44

Figure 15. Elliptical Pipe............................................................................................... 45

Figure 16. Square Tube ............................................................................................... 47



Saudi Aramco Desktop Standards iii

Figure 17. Steel Girder ................................................................................................. 49

Figure 18. Back-to-Back Angles................................................................................... 50

Figure 19. Single Angle ................................................................................................ 51

Figure 20. Bridge Pier Section ..................................................................................... 53

Figure 21. Three-Section Procedure ............................................................................ 53

Figure 22. Bridge Pier Section ..................................................................................... 56

Figure 23. Steel I-Section............................................................................................. 58

Figure 24. I-section Parts ............................................................................................. 58

Figure 25. Location of Centroid .................................................................................... 59

Figure 26. Timber Beam Reinforced by Steel Plate and Steel Channel ....................... 61

Figure 27. Transformed Compound Section ................................................................ 62

Figure 28. Timber Beam Reinforced by Steel Plate and Steel Channel ....................... 64

Figure 29. Cross Section of a Concrete Beam Reinforced with Steel Bars .................. 65

Figure 30. Transformed Section................................................................................... 66

List of Tables

Table 1. Form Used to Record Properties of Compound Sections............................... 52

Table 2. Calculating Properties of Compound Sections ............................................... 57

Table 3. Calculation of Centroid and Moment of Inertia (Work Aid 15) ........................ 60

Table 4. Calculation of Centroid and Moment of Inertia (Work Aid 16) ........................ 64

Table 5. Summary of the calculation around x-axis...................................................... 73

Table 6. Summary of the calculation about x-axis ........................................................ 77

Table 7. Summary of the calculation around x-axis...................................................... 79



Saudi Aramco Desktop Standards 1

OBJECTIVES

TERMINAL OBJECTIVE

Upon completion of this module, the participant will be able to compute basic properties of sections to solve simple Saudi Aramco structural engineering problems.

ENABLING OBJECTIVES

In order to meet the terminal objective, the participant will be able to:

• Define and calculate geometric properties of sections and structural shapes.

Note: Definitions of words in italics are contained in the Glossary




This Page Intentionally Blank




INFORMATION

INTRODUCTION

As a Saudi Aramco civil/mechanical engineer you will need to use the principles of basic strength of materials in the design of components for structures and buildings. These components are constructed from various types of materials such as concrete, steel and timber. They can be of various shapes and sizes. In analysis and design, you will use information about the properties of materials of construction and their dimensions to ensure that the components are properly proportioned for the loads imposed on them. For example, you can use basic strength of materials concepts to determine if the floor in a building is strong enough to support the weights stored on it and whether it will deflect or sag excessively.

The Basic Strength of Materials course, CSE 106, consists of three modules:

• CSE 106.01 - Basic Properties of Sections.

• CSE 106.02 - Basic Structural Loads and Stresses.

• CSE 106.03 - Analysis of Stresses in Structural Components.

The information covered in these modules provide you with the data, principles, and procedures used in the more application-oriented courses such as CSE 104, Analysis and Design of Wood Structures; CSE 108, Analysis and Design of Reinforced Concrete Structures; CSE 109, Analysis and Design of Steel Structures; and CSE 110, Civil Engineering Aspects of Tanks, Vessels, and Piping. Many of the examples and exercises covered in CSE 106 relate to the topics addressed in those courses.




MODULE INTRODUCTION

The first module of the Basic Strength of Material course focuses on the properties of sections you will normally encounter in structural or mechanical engineering problems. In this module you will identify and use the various types of sections, section properties, computation of section properties and formulas and tables for sectional properties.




SECTION PROPERTIES

The behavior of a structural member subjected to loads is a function of the geometric properties of its cross-section as well as of the material properties. The cross-section is the section perpendicular to the longitudinal axis of the member. Structural members are usually prismatic, the cross-section being uniform. In many structural applications, the designer’s task is to select or proportion the most efficient section for the given applied loads and properties of the material being used. Section efficiency usually relates to least area, weight, or cost.

Types of Sections

Uniform Sections

Uniform - (prismatic) Sections - the same cross-section along the axis of the member.

Figure 1. Uniform Sections

y

x

A

B

A

B Section B-B

Section A-Ay

x




Non-uniform Sections

Non-uniform - (variable) Sections - one or more dimension change along the axis of member. In Figure 2 two examples of non-uniform members often seen in structural/mechanical applications are shown.

• Steel beam with cover plates.

• Tapered cantilever concrete beam.

Figure 2. Non-uniform Sections




Simple Solid Sections

Simple solid sections are usually square, rectangular, circular, elliptical, etc., as shown in Figure 3. Solid sections are easy to fabricate but they may not be efficient for some types of members, loads, or materials.

Figure 3. Simple Solid Sections

Compound Sections

Compound Sections - consists of combinations of two or more simple solid elements as shown in Figure 4 below.

Figure 4. Compound Sections

Compound sections are usually used for increased structural efficiency or for their architectural appearance.

a) Bridge Pier b) Wood Beam c) Precast Concrete Beam




Composite Sections

Composite Sections - consists of sections of two or more types of material. Composite sections are used to increase structural efficiency and to reduce cost. Examples are given in Figure 5 below.

• Reinforced Concrete

• Structural Steel - Concrete

• Structural Steel - Timber

Figure 5. Composite Sections

Concrete-FilledPipe Column or Pile

RC - SteelColumn

RC - SteelT – Beam

Beam Bar Joist




Structural Steel Shapes

Structural Steel Shapes - compound sections rolled and fabricated for structural efficiency as shown in Figure 6 below.

Wide Flange Channel Tee Angle (Equal & Unequal Legs)Wide Flange Channel Tee Angle (Equal & Unequal Legs)

Figure 6. Structural Steel Shapes

Thin-Wall Shapes

Thin-Wall Shapes are cold-formed steel or aluminum sections with wall thicknesses that are small compared to the overall dimensions of the sections. Examples of thin-walled shapes are shown in Figure 7.




Figure 7. Thin-Walled Shapes

Thin-walled shapes are often used for lightweight prefabricated (pre-engineered) construction, building wall and roof cladding, decking for concrete slab, and for sheet piling.

Common Section Properties

Sections have certain geometric properties depending on the shape and size of the structural member’s cross-section. These are covered in this section.

The geometric properties of a section most commonly used in structural analysis are:

• Area (A)

• Static moment of area (Q)

• Centroid (CG)




• Moment of inertia (I)

• Section modulus (S)

• Torsional constant (R)

• Radius of gyration ( r)

The uses, units, formulas, or data sources for these section properties for selected structural shapes are as follows:

Area, A (in2, mm2)

• Rectangular section: A = bd

• Circular section: A = 4d 2π

• Triangular section: A = 2bd

Some uses for area in strength of materials are:

• To calculate stress σ = AP

• To calculate axial deformation δ = AEPL

• To calculate axial stiffness K = P/δ =L

AE

E : Modulus of Elasticity Moment of Area, Q (in2, mm2)

The moment of area is used to locate the centroid of a compound or composite section. Moment of area is the product of an area and the distance from the reference point.

b

d

b

d

dd

b

d

b

d

PP PP




Q = ∑Aiyi

Q = A1y1 + A2y2

The moment of areas about the centroid )y,x( of a section is equal to zero.

Q = ∑Aixi = ∑Aiyi = 0

The moment of area is also used to compute the shear stress in a beam,

τ = Ib

VQ

V = shear force

I = moment of inertia = 12bh3

Centroid

The centroid of an area is the geometric center of a section. The centroidal coordinates )y,x( are determined by the first moment equations:

,A

xAA

Qxi

iix

∑∑

∑== ,

AyA

AQ

yi

iiy

∑∑

∑== (3)

A1

A y 2 Ref. Line

2

y 1




21

2211

AAxAxAx

++

=

21

2211

AAyAyAy

++

=

The longitudinal axis of an axial member is assumed to be at the centroid of the section. The neutral axis of a beam passes through the centroid of the beam section for the case of pure bending.

Moment of Inertia, I (in4, mm4) = ∑Aixi

2, ∑Aiyi

2 Moment of inertia is the second moment of area about a specified axis. That is, the moment of inertia is the sum of the product of area times the square of distance to the reference axis. The moment of inertia is a measure of the resistance of the section to rotation about the axis. It is calculated about the centroidal axes of a section by using the formulas:

Ix = ∑Aiyi2 with respect to x-axis

Iy = ∑Aixi2 with respect to y-axis




These values are for moment of inertia about the neutral axis of the sections. For moment of inertia about any other axis parallel to y axis, the following formula applies:

yI = Iy + Ax2

Where Iy is the moment of inertia about the centroidal axis, A is the area, and x is the distance of the centroid of the section from the new axis. This relationship is used in calculating the moment of inertia of a compound section.

Expressions for moment of inertia of simple sections are:

• Rectangular Section: Ix = 12bd3

• Circular Section: Ix or Iy = 4r 4π or

64d4π

• Triangular Section: Ix = 36bd3

The moment of inertia is used for computing bending stress, stiffness, and deflection in flexural members.

• Bending stress: σ = I

Mc

• Deflection: ∆ = EIPL3α

• Stiffness: K = 3LλEI

where α and λ are constants that depend on the boundary conditions of the beam.

Section Modulus, S (in3, mm3) = I/c

The section modulus is the moment of inertia, I, divided by the distance, c, from the neutral axis of a beam to the fiber farthest from the neutral axis. The section modulus provides the resistance to the bending of a beam. The expressions for section modulus of simple sections are:

Rectangular Section: Sx = 6

bd2




Circular Section: Sx or Sy = 4r 3π

Triangular Section: Sx = 24bd2

(top), 12bd2

(bottom)

The section modulus is used to calculate bending stress, σ:

SMσ =

Torsional Constant, R (in4, mm4)

The torsional constant (R) is the resistance to twisting due to torsion. For axisymmetric sections (e.g. circular sections), R is equal to the polar moment of inertia, Iz or J, which is the sum of moments of inertia about the x and y axes of the section. That is,

R = Iz = J = Ix + Iy

Circular Section: R = 2r 4π or

32d4π

For other sections, R may be significantly less than Iz, due to the warping effects of the section when it is loaded in torsion.

In such cases R is determined experimentally, e.g. for a rectangular section of thickness t and width b

R = αbt3

where α is a value that depends on the aspect ratio b/t ≥ 1.0 of the section, the torsional constant is used to calculate torsional stiffness and stress.

Radius of Gyration, r (in, mm)

A length called the radius of gyration (r) is a stability parameter that measures the resistance to buckling, and is defined by the following formulas:




r = AI

ry = AIy rx =

AIx

Rectangular Section: r = 12d

d = dimensions in buckling direction

Circular Section: r = 4d d = diameter

It is used to calculate slenderness ratio, L/r of a column.

Listed Properties - Standard Sections

The formulas for calculating section properties are provided in standard engineering handbooks. Some of these formulas are listed in Work Aid 3. These tables include the following:

Properties

A = Area

I = Moment of Inertia

S = Section Modulus

R = Torsional Resistance

r = Radius of Gyration

Standard Shapes

• Simple Solids (Work Aid 3).

- Rectangle, Triangle, Circle, and Ellipse.

• Hollow (Thick-Wall) Shapes (Work Aid 3).




- Rectangle, Circle, and Ellipse.

• Thin-Wall (Line) Sections: t< d/10 (Work Aid 4).

• Structural Steel Rolled Shapes.

- Designated in the form Ad x w or Lb x d x t.

Where A represents a letter W, S, or C defined as follows:

W = Wide-flange shape (Work Aids 5 & 6, pages 1 & 2)

S = Standard I-beam shape (Work Aids 5 & 6, page 3)

C = Channel shape (Work Aids 5 and 6, page 4)

And

L = Angle with equal or unequal legs (Work Aids 5 & 6, pages 5 & 6)

w = Weight per unit length

b = Width of angle

d = Depth of section

t = Thickness of angle legs

The actual values are listed for the section properties for the structural steel rolled shapes. Values for both the U.S. customary units (Work Aid 5) and SI units (Work Aid 6) are given.




Example 1: Calculating Section Properties for Standard Shapes

Calculate A, I, S, and r for the following sections about x-axis:

A. Concrete beam with nominal dimensions, 4 x 10 in. as shown in Figure 8a.

Figure 8a. Rectangular Beam

B. Elliptical sewer pipe having the dimensions shown in Figure 8b.

Figure 8b. Elliptical Sewer Pipe

10 ft

11 ft

6 ft5 ftx

4 in

10 in x-axis

5 in




C. Square, thin-wall steel tube, mean side 10 in. wide and 0.25 in. thick.

D. Steel girder, W610 x 155 (SI units).

E. Structural member consisting of two back-to-back angles, L8 x 6 x 1, (U.S. customary units) as shown in Figure 9.

8 in

1 in6 in6 in

1 in1 in

xx

y

Figure 9. Structural Member Consisting of Two Back- to-Back Angles

F. Same as in E above, but determine values about axis of symmetry.

x

x




Solution: Example 1

See Work Aids 8 to 12.

Calculating Section Properties for Nonstandard Shapes

For the nonstandard sections, such as the compound and composite sections discussed above, formulas or tabulated values for the section properties are usually not available. The engineer however, can calculate the values using the procedure outlined below.

Compound Sections

The steps involved in calculating the geometric properties of compound sections are as follows:

1. Divide the compound section into a number of simple sections and number them 1, 2, 3, etc.

2. Calculate the area, Ai, and moment of inertia, Ii, of each simple section.

3. Compute the moment of the areas, Aixi or Aiyi, about the reference axis for the compound section.

4. Locate the centroid of the compound section by the following relationship to calculate x or y:

,AxA

xi

ii

∑∑=

∑∑=

i

ii

AyA

y

5 Calculate the moment of inertia about the axis passing through the centroid using the relationship:

( ) ( )∑ ∑ +=+= 2iiiy

2iiix xAII or yAII

6. Calculate the other section properties for the compound section using the standard formulas.




You can use Work Aid 13 as a convenient format for doing these calculations. The use of this Work Aid is illustrated in Example 3, Work Aids 14 and 15.

Example 2: Properties of Compound Sections

Determine the area, moment of inertia, section modulus, and radius of gyration of the sections shown in Figures 10 and 11.

Figure 10. Bridge Pier Figure 11. Plate Girder

Solution: Example 2

See Work Aids 14 and 15.




Composite Sections

When a structural member consists of two or more types of material, the properties of the composite section can be determined by calculating the values for the transformed section. The transformed section is the equivalent homogeneous section based on the relative elastic modulus of the materials of the composite section. The relative elastic modulus is expressed in terms of modular ratio, n where:

Modular ratio, n = o

i

EE

where: Ei is the elastic modulus of the given material. Eo is the elastic modulus of the reference material.

To calculate the properties of a composite section follow these steps:

1. Identify the materials of the section and determine the elastic modulus Ei of each material.

2. Select one material as the reference material for the transformed section. Usually, the material with the lowest E value is used as the reference and denotes its E = Eo.

3. Determine the modular ratio, ni, for each of the other materials in the section. ni = Ei/Eo.

4. Replace each section of the other materials with an equivalent section transformed by multiplying the dimension parallel to the axis of the composite section by the appropriate modular ratio, ni. Then draw the transformed section.

5. From this point, follow the steps used to calculate the properties of the resulting compound section with the help of Work Aid 13.




Example 3: Calculating Properties of Composite Sections

An 8 x 12-in. timber beam (exact size) is reinforced by adding a 7-7/8 x 1/2-in. steel plate at the top and a C 7 x 9.8 steel channel at the bottom as shown in Figure 12. Determine the section properties, A, I, S, and r relative to the horizontal axis for the composite section.

Figure 12. Composite Sections

Solution: Example 3

See Work Aid 16.

8 in

12 in

C7 x 9.8 (Steel)

Timber Beam

7-7/8 in x 1/2 - in Steel Plate

x

y

Timber E = 1.8 x 106 psiSteel E = 29 x 106 psi

8 in

12 in

C7 x 9.8 (Steel)

Timber Beam

7-7/8 in x 1/2 - in Steel Plate

x

y

Timber E = 1.8 x 106 psiSteel E = 29 x 106 psi




Example 4: Calculating Properties of Reinforced Concrete Section

A 10 x 20-in. concrete beam is reinforced with four 1-in. diameter steel bars as shown in Figure 13. If the effect of the concrete below the neutral axis through the centroid of the section is ignored, determine the section modulus for calculating stresses in the concrete (top) and in the steel (bottom).

Figure 13. Reinforced Concrete Section

Solution: Example 4

See Work Aid 17.

Concrete

4- 1-in. Diam. Steel Bars

2 in

4 in

10 in




SUMMARY

This module has reviewed the purpose and significance of section properties in the analysis and design of engineering structures. For various materials, there are certain principles and data required for safe design. These principles and data requirements will be covered in more depth in the specific CSE courses on steel, concrete, and timber. Based on the information in this module, you should now be able to apply basic section properties in structural analysis and design.

You should be able to define the various section properties involved in structural analysis and design, identify the types of sections, and review the formulas for calculating these properties. You should be able to use these formulas and calculation procedures to determine the properties of compound and composite sections that relate to reinforced concrete, structural steel, and timber designs.




WORK AIDS

Work Aid 1: Properties of Materials

(U.S. Customary Units) Ultimate Strength Yield Strength3

Material

Specific Weight, lb/in.2

Tension,ksi

Compression, 2ksi

Shear, ksi

Tension,ksi

Shear,ksi

Modulus of Elasticity 104 psi

Modulus of Rigidity, 106 psi

Coefficientof ThermalExpansion,10-6/°F

Ductility,Percent Elongationin 2 in.

1 Properties of metals vary widely as a result of variations in composition, heat treatment, and mechanical working.

2 For ductile metals the compression strength is generally assumed to be equal to the tension strength.

3 Offset of 0.2 percent. 4 Timber properties are for loading parallel to the grain.




Work Aid 2: Properties of Materials

(SI Units) Ultimate Strength Yield Strength3

Material

Specific Weight, lb/in.2

Tension,ksi

Compression, 2ksi

Shear, ksi

Tension,ksi

Shear,ksi

Modulus of Elasticity 104 psi

Modulus of Rigidity, 106 psi

Coefficientof ThermalExpansion,10-6/°F

Ductility,Percent Elongationin 2 in.

1 Properties of metals very widely as a result of variations in composition, heat treatment, and mechanical working.

2 For ductile metals the compression strength is generally assumed to be equal to the tension strength.

3 Offset of 0.2 percent. 4 Timber properties are for loading parallel to the grain.




Work Aid 3: Properties of Standard Sections

Section Shape Area

A Moment of

Inertia I

Section Modulus

S

Radius of Gyration

r

Rectangle Centroid axis

b

d

bd 12bd3

6

bd2

12d

Rectangle Base axis

b

d

bd 3

bd3

3

bd2

3d

Triangle Centroid axis

b

d 3

d

2

bd

36bd3

24bd2

18d

Triangle Base axis

b

d

2

bd

12bd3

12bd2

6

d

Circle Centroid axis

d

4πd2

64πd4

32πd3

4d

Ellipse Centroid axis

a

b

πab 4

bπa3

4

bπa2

2a

Hollow square Centroid axis

a

b

22 ba − 12

ba 44 − 6a

ba 44 − 12

ba 22 +

Hollow circle Centroid axis

D d

( )22 dD

4π

−

( )44 dD64π

− D

dD32π 44 −

4dD 22 +

Hollow ellipse Centroid axis

d

b

a c

π(ab - cd) ( )dcba4π 33 − ( )

4adcbaπ 33 −

cdabdcba

21 33

−

−




Work Aid 4: Properties of Thin-Walled Sections (Cold Formed Steel or Aluminum)

Section Axis

Area A

Moment of Inertia

I

Section Modulus

S

Torsion Constant

R

Radius of Gyration

r

d

b

2t (b + d) ( )d3b6

td2

+ ( )d3b3

td+

db

d2tb 22

+ ( )

( )db12

d3bd2

+

+

t (2b + d) ( )d6b12

td2

+ ( )d6b6

td+ ( )d2b

3

t3

+ ( )( )d2b12

d6bd2

+

+

b

b

t (2b + d) 6

tb3

3

tb3

( )d2b3

t3

+ ( )d2b6

b3

+

d

πtd

3td8

π 2td

4

π 3td

4

π 0.3535d

d

b

x

t (b + d) ( )( )db12

d)4btd3

+

+ ( )d4b

6

td :top + ( )db

3

t3

+ ( )( )2

3

db12

d4bd

+

+

x d2 (b+d)

( )( )d2b6

d4btd :bottom

2

+

+

d

b

t (b + d)

12

tb3

6

tb2

( )db3

t3

+ ( )db12

b3

+

d

b x

t (b + d) ( )( )db12

d4btd3

+

+ ( )d4b

6

td :top + ( )db

3

t3

+ ─

x d2 (b+d)

( )( )d2b6

d4btd :bottom

2

+

+

d

b

t (2b + d) ( )d6b12

td2

+ ( )d6b6

td+ ( )d2b

3

t3

+ ( )( )db12

d6bd2

+

+

d

b x

t (2b + d) ( )( )d2b3

2dbtd3

+

+ ( )d4b

6

td :top + ( )d2b

3

t3

+

( )( )2

+

+

3b b 2d

3 2b d




Section Axis

Area A

Moment of Inertia

I

Section Modulus

S

Torsion Constant

R

Radius of Gyration

r x b2

(2b+d ( )

( )d2b6

d4btd :bottom

2

+

+




Work Aid 5: Properties of Rolled-Steel Shapes (U.S. Customary Units) (Page 1 of 6)

W Shapes(Wide-flange Shapes)

Properties of Rolled-steel Shapes(U.S. Customary Units)

bf

tw

tf

d X

Y

X

Y

Flange

Axis X-X Axis Y-Y

Designation† Area

A, in.2 Depth d, in.

Widthbf, in.

Thick-ness tf, in.

Web Thick-ness tw, in. Ix, in.4 Sx, in.3 rx, in. Iy, in.4 Sy, in.3 ry, in.

† A wide-flange shape is designated by the letter W followed by the nominal depth in inches and the weight in pounds per foot.




(Page 2 of 6)



bf

tw

tf

d X

Y

X

Y

Flange Axis X-X Axis Y-Y

Designation† Area


Width bf, in.

Thick-ness tf, in.


† A wide-flange shape is designated by the letter W followed by the nominal depth in inches and the weight in pounds per foot.




(Page 3 of 6)

S Shapes(American Standard Shapes


bf

tw

tf

d X

Y

X

Y


Designation† Area


Widthbf, in.

Thick-ness tf, in.


† An American Standard Beam is designated by the letter S followed by the nominal depth in inches and the weight in pounds per foot.




(Page 4 of 6)

C Shapes(American Standard Channels)


bf

tw

tf

dX

Y

X

Y

xx


Designation† Area


Width bf, in.

Thick-ness tf, in.

Web Thick-ness tw, in. lx, in.4 Sx, in.3 rx, in. ly, in.4 Sy, in.3 ry, in. ,x in.

† An American Standard Channel is designated by the letter C followed by the nominal depth in inches and the weight in pounds per foot.




(Page 5 of 6)

AnglesEqual Legs


x

yX

Y

X

Y

Z

Z

Axis X-X and Axis Y=Y

Size and Thickness, In. Weight per Foot, lb/ft Area, in.2 Ly, in.4 Sy, in.3 ry, in. x, in.

Axis Z-Z

r, in.




(Page 6 of 6)

AnglesUnequal Legs


x

yX

Y

X

Y

Z

Zα

Axis X-X Axis Y=Y Axis Z-Z Size and Thickness,

In.

Weight per Foot,

lb/ft Area, in.2 lx, in.4 Sx, in.3 rx, in. y, in. ly, in.4 Sy, in.3 ry, in. x, in. rx, in. tan α




Work Aid 6: Properties of Rolled-Steel Shapes (SI Units) (Page 1 of 6)

bf

tw

tf

d X

Y

X

YProperties of Rolled Steel Shapes

(SI Units)


Flange

Axis X-X Axis Y-Y

Designation† Area

A, mm2 Depth d, mm

Widthbf, mm

Thick-ness

tf, mm

Web Thick-ness

tw, mm lx

106 mm4Sx

103 mm3 rx

mm ly

106 mm Sy

103 mm3ry

mm

† A wide-flange shape is designated by the letter W followed by the nominal depth in millimeters and the mass in kilograms per meter.




(Page 2 of 6)

bf

tw

tf

d X

Y

X

YProperties of Rolled Steel Shapes

(SI Units)


Flange

Axis X-X Axis Y-Y

Designation† Area

A, mm2 Depth d, mm

Width bf, mm

Thick-ness

tf, mm

Web Thick-ness

tw, mm lx 106 mm4

Sx 103 mm3

rx mm

ly 106 mm

Sy 103 mm3

ry mm

† A wide-flange shape is designated by the letter W followed by the nominal depth in millimeters and the mass in kilograms per meter.




(Page 3 of 6)

S Shapes(American Standard Shapes

Properties of Rolled-steel Shapes(SI Units)

bf

tw

tf

d X

Y

X

Y

Flange

Axis X-X Axis Y-Y

Designation† Area

A, mm2 Depth d, mm

Width bf, mm

Thick-ness

tf, mm

Web Thick- ness

tw, mm lx

106 mm4Sx

103 mm3 rx

mm ly

106 mm4 Sy

103 mm3 ry

mm

† An American Standard Beam is designated by the letter S followed by the nominal depth in millimeters and the mass in kilograms per meter.




(Page 4 of 6)

C Shapes(American Standard Channels)


bf

tw

tf

dX

Y

X

Y

xx


Designation† Area

A, mm2 Depth d, mm

Width bf, mm

Thick-ness

tf, mm

Web Thick-ness

tw, mm lx 106 mm4

Sx 103 mm3

rx mm

ly 106 mm

Sy 103 mm3

ry mm

,x mm

† An American Standard Channel is designated by thee letter C followed by the nominal depth in millimeters and the mass in kilograms per meter.




(Page 5 of 6)

AnglesEqual Legs


x

y X

Y

X

Y

Z

Z

Axis X-X and Axis Y=Y

Size and Thickness, mm Weight per Meter, kg/m Area, mm2

I 106 mm

S 103 mm 3

r

mm x or y mm

Axis Z-Z

r mm




(Page 6 of 6)

AnglesUnequal Legs


x

yX

Y

X

Y

Z

Zα

Axis X-X Axis Y=Y Axis Z-Z

Size and Thickness, In.

Mass per Meter kg/m

Area, mm2

Ix 106mm4

Sx 103mm3

rx mm

y mm

Iy 106mm4

Sy 103mm3

ry mm

x mm

rx mm tan α




Work Aid 7: Properties of Safe Axial Loads for Structural Lumber




Work Aid 8: Calculating Section Properties for Standard Shapes Concrete Beam

Calculate A, I, S, and r for the following section:

Procedure:

A. Concrete beam

Dimensions:

b = 4 in.

d = 10 in.

Section Properties:

Using Formulas from Work Aid 3

A = (4)(10) = 40 in2

( )( )12104

12bdI

33

x == = 333.33 in4

( )( )6104

6bdS

22

== = 66.67 in3

b

d NA

3.4610

12dr == = 2.89 in Figure 14. Concrete Beam

40333.33

AIr x

x == = 2.89 in




Work Aid 9: Calculating Section Properties for Standard Shapes Elliptical Pipe

(Page 1 of 2)

Calculate A, I, S, and r for the following section about x-axis:

1. Elliptical sewer pipe having the dimensions shown in Figure 15.

5 ft. 6 ft.

10 ft.

11 ft.

xx

Figure 15. Elliptical Pipe

Elliptical pipe (hollow ellipse) - Work Aid 3.

Dimensions - From Figure 15.

a = half outside major axis = 211 = 5.5 ft

b = half outside minor axis = 26 = 3.0 ft

c = half inside major axis = 2

10 = 5.0 ft

d = half inside minor axis = 25 = 2.5 ft




(Page 2 of 2)

Section properties, Work Aid 3:

A = π (ab – cd) = 3.142 [(5.5)(3.0) – (5.0)(2.5)] = 12.57 ft2

Ix = ( )( ) ( )( ) ⎥⎦⎤

⎢⎣⎡ −=⎟

⎠⎞⎜

⎝⎛ − 32.55.033.05.5

43.1423cd3ab

4π = 55.27 ft4 *

* Note change in centroidal axis, that is, minor and major axes are interchanged.

Sx = 3.0

55.27bIx = = 18.42 ft3

rx = 12.5755.27

AIx = = 2.10 ft




Work Aid 10: Calculating Section Properties for Standard Shapes Square Tube

(Page 1 of 2)


Procedure:

1. Square tube

Square, thin-wall steel tube, mean side 10 in wide and 0.25 in thick.

A. Use formula for rectangular tube from Work Aid 4.

Dimensions

b = d = 10 in

t = 0.25 in Figure 16. Square Tube

Section properties:

A = 2t(b + d) = 4dt = 4(0.25)(10) = 10 in2

Ix = ( ) ( )( )332

100.2564td

64d3b

6td

==+ = 166.67 in4

Sx = ( )( )22

100.2568

68td

d2I

== = 33.33 in3

rx = 2.4510

6d

AI

== = 4.08 in




(Page 2 of 2)

B. Use formula for Hollow Square from Work Aid 3.

Dimensions:

a = 10 + 0.25 = 10.25 in

b = 10 – 0.25 = 9.75 in

Section Properties:

A = a2 – b2 = (10.75)2 – (9.75)2 = 10 in2

Ix = ( ) ( )12

9.7510.2512

ba 4444 −=

− = 166.77 in4

Sx = ( )10.25166.772

a2I

= = 32.54 in3

rx = AI =

10.0166.77 = 4.08 in

The results based on the two approaches are close. However, the second method is more accurate and should be used for sections with thick walls. The accuracy of the first approach is reduced as the wall thickness increases.

In general:

If t < 10d then use Work Aid 4 (Thin-Walled Sections) and

If t > 10d use Work Aid 3 (Thick-Walled Sections).




Work Aid 11: Calculating Section Properties for Standard Shapes Steel Girder


Procedure:

1. Steel girder, W610 x 155 (SI units)

Section Designation

W = Wide flange

d = Nominal depth = 610mm

w = Unit weight = 155kg/m

x x

Figure 17. Steel Girder

Section Properties from Work Aid 6, page 1.

Identify designated section, W610 x 155, and read across to value in the appropriate column. Note that for beams and girders section properties are related to the strong axis (X-X) of the section:

A = 19,700 mm2

Ix = 1290 x 106 mm4 (Axis X-X)

Sx = 4220 x 103 mm3 (Axis X-X)

rx = 256 mm

ry = 73.9 mm




Work Aid 12: Calculating Section Properties for Standard Shapes Back-To-Back Angels

(Page 1 of 2)


Procedures:

1. Back-to-Back Angles (U.S. customary unit):

Section Designation Two L8 x 6 x 1

L = Angle section

d = Depth of each angle = 8 in

b = Width of each angle = 6 in

t = Thickness of angle legs = 1 in

Figure 18. Back-to-Back Angles

Section Properties: Use Work Aid 5.

Note that the location of X-X axis for the back-to-back arrangement of the two angles is the same as for a single angle. However, the Y-Y axis relocates to the axis of symmetry which passes between the two angles.

– Identify designation L8 x 6 x 1 in Work Aid 5, page 6.

– Obtain section properties for a single angle by reading across to value in the appropriate column.

– Modify (multiply by 2) value from table to obtain value for two angles:

A = 2(13.0) = 26.0 in2

Ix = 2(80.8) = 161.6 in4

Sx = 2(15.1) = 30.2 in3

rx = AIx =

26.0161.6 = 2.49 in




(Page 2 of 2)

Procedures:

2. Back-to-back Angles in Figure 19. Properties about y'-y' axis (axis of symmetry).

Properties of single Angle, L8 x 6 x 1, from Work Aid 5.

– A = 13.0 in2, for single angle.

– Obtain values of I and x for a single angle about centroid y-y axis:

Iyo = 38.8 in4

x = 1.65 in

x

Figure 19. Single Angle

– Transpose I to axis y'-y' in Figure 19 using formula:

Iy’ = Iyo + A x 2

Iy’ = 38.8 + 13.0 (1.65)2 = 74.19 in4

– Modify (multiply by 2) for two angles

A = 2 (13) = 26.0 in2

Iy’ = 2 (74.19) = 148.4 in4

Sy’ = 6.00

148.4/cIy' = = 24.73 in3

ry’ = AIy' =

26.0148.4 = 2.39 in




Work Aid 13: Calculating Properties of Compound Sections Form

Table 1. Form Used to Record Properties of Compound Sections

Item Shape/ Size

Area, Ai

Io xi or yi Aixi or Aiyi iy or ix

2ixiA

or 2

iyiA

2ixiAoI +

or 2

iyiAoI +

Total

∑∑=

i

ii

AxA

x ___

or IX = ( )∑ + 2iio xAI _____________________

∑∑=

i

ii

AyA

y ___

IY = ( )∑ + 2iio yAI ____________________




Work Aid 14: Calculating Properties of Compound Sections - Bridge Pier

(Page 1 of 5)

Determine the area, moment of inertia, section modulus, and radius of gyration of the section shown in Figure 20.

4 in.

20 in.

10 in.

10 in.

40 in.

20 in. 10 in. 10 in.

x

Figure 20. Bridge Pier Section

Procedure:

4.24 in.

30 in. 10 in.

10 in. 6.667 in.

(3)

(1)

(2)

Section 3: Half Circle – Solid

(I about base of circle)

Section 2: Square – Solid

2a y,

12aI ,aA o

42 ===

Section 1: Triangle – Solid

32h y,

36bhI ,

2bhA o

3

===

Figure 21. Three-Section Procedure

248 4

2 8 9 3or rA I r yπ π

π π⎛ ⎞= = − =⎜ ⎟⎝ ⎠




(Page 2 of 5)

Procedure:

• Use form in Work Aid 13, (see Table 2 for completed form).

• Divide compound section into three simple shapes and mark 1, 2, and 3.

• Select reference axis through point O.

• Select formula and calculate Ai Ii and yi for each simple shape.

Section 1: Triangle, b = 20 in. h = 10 in

A1 = 2

bh = ( )21020 = 100 in2

I1 = 36bh3

= ( )361020 3

= 555.56 in4

y1 = 32h = ( )

3102 = 6.667 in

Section 2: Square, a = 20 in.

A2 = a2 = (20)2 = 400 in2

I2 = 12a4

= 12204

= 13333.33 in4

y2 = 2a10 + =

22010 + = 20 in

Section 3: Half circle, radius, r = 10 in

A3 = 2πr2

= ( )2

103.142 2

= 157.1 in2

I3 = 2

3

4

3π4rA

8πr

⎟⎠⎞

⎜⎝⎛− = 1098 in4

y3 = 3π4r30 + = ( )

( )3.142310430 + = 34.24 in

Enter values on the form in Work Aid 13 (see Table 2 for completed form) and complete the calculations.




(Page 3 of 5)

Procedure:

Calculate y at the bottom of the form to locate the centroid of the compound section:

y = 21.38 in

This means that the centroid is 21.38 in above O. By symmetry, the centroid also lies on the y-axis.

Determine ŷi, the distance of the local centroid of each section from the centroid of the compound section using the relationship:

yyy ii −=

Therefore, 1y = 6.667 – 21.38 = –14.71 in

2y = 20 – 21.38 = –1.38 in

3y = 34.24 - 21.38 = 12.86 in

Complete the last two columns in the form and obtain the totals.

Determine appropriate values for section properties:

• Area A = ΣAi = 2in657

• Moment of Inertia, Ix = ( )∑ + 2iioxi yAI = 63,351.7 in4

Alternatively, Ix = ( )∑ + 2iioxi yAI = ( )∑ + 2

iio yAI

= 14,986.89 + 48,364.8 = 63,352 in4




(Page 4 of 5)

Procedure:

Use the appropriate formulas to calculate the other section properties.

• Section Modulus, S = cI

40 in.

20 in.

C

C

CG+

+

–

Figure 22. Bridge Pier Section

- Top of section:

c+ = h - y

= 40 - 21.38 = 18.62 in

S+ = 18.6263,352 = 3402 in3

- Bottom of section:

c– = y = 21.38 in

S– = 21,3863,352 = 2963 in3

• Radius of Gyration, r = AI

r = 657

63,352 = 9.82 in




(Page 5 of 5)

Procedure:

Table 2. Calculating Properties of Compound Sections

Item Shape/ Size Area, Ai Io yi Aiyi iy

2iyiA 2

iyiAoI +

Units in2 in4 in. in3 in. in4 in4

1 Triangle 100 555.56 6.667 666.7 -14.71 21,638.4 22,194.0

2 Square 400 13,333.33 20 8000 -1.38 761.8 14,095.1

3 Half- Circle 157 1098 34.24 5376 12.86 25,964.6 27,062.6

Total 657 14.986.89 14,042.7 48,364.8 63,351.7

y = in.21.37in657

in14,042.7A

yA2

3

i

ii ==∑∑ Ix = ( )∑ =+ 42

iioxi in63,351.7yAI




Work Aid 15: Calculating Properties of Compound Sections - Plate Girder

(Page 1 of 3)

Determine the area, moment of inertia, section modulus, and radius of gyration of the section shown in Figure 23.

8 in.

3/4 in.

3/8 in. 25 3/4 in.

1 in.

10 in.

1

y12

3

y3

Rectangle, 8 x 0.75 in.

Rectangle, 0.375 x 24 in.

x Reference Axis

Rectangle, 10 x 1.0 in.

x

Figure 23. Steel I-Section Figure 24. I-section Parts

Procedures: • Divide section into three rectangles as shown and tabulate on the form from Work

Aid 13.

• Choose reference axis at the center of rectangle (2).

• Calculate Ai and Ii for each rectangle about its local axis by substituting the appropriate values for bi and hi in the relationship:

Ai = bihi

Ii = 12hb 3

ii

and enter values on the form (see Table 3).




(Page 2 of 3)

• Determine yi, the distance from the reference axis to the local centroid of each rectangle. Enter values on the form. Compute values of Aiyi and totals.

• • Locate centroid (CG) of compound section by calculating y (bottom of the form), and calculate the distance Iy from the local centroid of each rectangle to CG of the overall section.

Procedure: • Enter values of Iy on the form and complete calculations.

• Obtain from the form the values for A and I, and compute the other section properties required:

A = ΣAi = 25.0 in2

I = Σ (Io + Ai y i2) = 2811.43 in 4

• Section modulus:

S = I/c+ or I/c–

S+ = 14.78

2811.43 = 190.22 in 3

S– = 10.97

2811.43 = 256.28 in 3

C- = 10.97 in.

C+ = 14.78 in.

2.03CG

+ O

Figure 25. Location of Centroid




(Page 3 of 3)

• Radius of Gyration:

r = I/A

= 25.0

2811.43 = 10.60 in

Procedure:

Table 3. Calculation of Centroid and Moment of Inertia (Work Aid 15)


2iyiA 2

iyiAoI +

Units in2 in4 in in3 in in4 in4

1 Rectangle

8(0.75) = 6.0

128 (0.75) 3

=0.28 12 + 2

0.75

=12.37574.25 12.375 + 2.03

=14.405 1245.02 1245.3

2 Rectangle

0.375(24) = 9.0

120.375 (24)3

= 432 0 0 2.03 37.09 469.09

3 Rectangle

10(1.0)

= 10 1210 (1)3 = 0.83

-12- 21.0

= -12.5 -125 -12.5 + 2.03 =

-10.47 1096.21 1097.04

Total 25.0 433.11 -50.75 2378.32 2811.43

y = in.2.0325.050.75-

AyA

i

ii −==∑∑ I = ( )∑ + 2

iio yAI = 2811.43 in4




Work Aid 16: Calculating Properties of Composite Sections

(Page 1 of 4)

An 8 x 12-in. timber beam (exact size) is reinforced by adding a 7-7/8 x 1/2-in. steel plate at the top and a C 7 x 9.8 steel channel at the bottom as shown in Figure 26. Determine the section properties, A, I, S, and r relative to the horizontal axis for the composite section.

7-7/8 in x 1/2 in PL (Steel)

12 in Timber Beam

C7 X 9.8 (Steel)

8 in

Figure 26. Timber Beam Reinforced by Steel Plate and Steel Channel

Procedure:

Step 1: Identify materials and obtain Elastic Modulus from Work Aid 1.

Material Elastic Modulus

Timber 1.8 x 106 psi

Steel 29 x 106 psi

Steps 2 & 3: Select reference material and calculate modular ratio, n, for the other material.

Reference material: Timber

Modular ratio for steel: n = timber

steel

EE

= psi10 x1.8psi10 x29

6

6

= 16




(Page 2 of 4)

Procedure:

Step 4: Transform the steel sections and draw the transformed sections for the composite section. Multiply the horizontal steel dimensions by n = 16.

16 x 7.875 = 126 in (Not to Scale)

x 1

2

3

x

0.5 in

2.09 in

0.54 in

6 in

6 in

Figure 27. Transformed Compound Section

Step 5: Calculate the section properties for the transformed compound section (Figure 27) using Work Aid 13. (For details see Table 4).

a. The transformed section has three simple sections as indicated. Select the horizontal axis through the centroid of rectangle 2 as the reference axis for the sections.

b. Use the standard formulas to calculate A and I for the two rectangular sections.

A1 = b • d = 126(0.5) = 63 in2

I1 = 12bd3

= 12

126(0.5)3

= 1.31 in4

The distance from the reference axis to the centroid of rectangle (1) is:

y1 = 2

0.52

12+ = 6.25 in

A2 = 8 x 12 = 96 in2




I2 = 12

8(12)3

= 1152 in4 (Page 3 of 4)

y2 = 0

Procedure:

For the transformed channel Section 3, the values are obtained by multiplying the appropriate values from Work Aid 5 by n = 16.

Channel (3):

From Work Aid 5 for C7 x 9.8 channel: A3 = 2.87 in2 I3 = 0.968 in4 about weak axis

Therefore, the transformed values are: A3 = 2.87 x 16 = 45.92 in2 I3 = 0.968 x 16 = 15.49 in4

The distance from the back of the channel to its centroid is 0.54. Therefore,

y3 = –6 – 0.54 = –6.54 in.

c. Enter the values in Work Aid 13 and complete calculations, for Aiyi and totals.

d. Determine location of centroid (CG) by computing y and calculate the distance (yi) from the CG to the local axis of the individual sections.

e. Complete calculations for the last two columns and compute totals.

f. Obtain totals for area A and moment of inertia I. A = ΣAi = 204.92 in2

I = Σ(Io + Aiyi2) = 5551.21 in4

g. Compute S and r




(Page 4 of 4)

Figure 28. Timber Beam Reinforced by Steel Plate and Steel Channel

Table 4. Calculation of Centroid and Moment of Inertia (Work Aid 16)


2iyiA 2

iyiAoI +

Units in2 in4 in. in3 in. in4 in4

1 Rectangle 126 x 0.5 63 1.31 6.25 393.75 5.794 2114.94 2116.25

2 Rectangle 8 x 12 96 1152.0 0 0 -0.456 19.96 1171.96

3 Channel (C7 x 9.8) in 45.92 15.49 -6.54 -300.32 -6.996 2247.51 2263.00

Total 204.92 1168.80 93.43 4382.41 5551.21

y = ∑∑

i

ii

AyA

= 204.9293.43 = 0.456 in I = ( )∑ + 2

iio yAI = 5551.21 in4

S+ = +cI =

6.0445551.21 = 918.47 in3

S- = −cI =

8.5465551.21 = 649.57 in3

r = AI =

204.925551.21 = 5.20 in.




Work Aid 17: Calculating Properties of Reinforced Concrete Sections

(Page 1 of 3)

A 10 x 20-in. concrete beam is reinforced with four 1-in. diameter steel bars as shown in Figure 29. If the effect of the concrete below the neutral axis through the centroid of the section is ignored, determine the section modulus for calculating stresses in the concrete (top) and in the steel (bottom).

Concrete

(4) 1-in Diam. Steel Bars

2 in

10 in

20 in

Figure 29. Cross Section of a Concrete Beam Reinforced with Steel Bars

Procedure:

Step 1 Identify section materials, obtain their elastic modulus, and calculate modular ratio, n.

Material Elastic Modulus

Concrete 3 x 106 psi Steel 30 x 106 psi

Modular ratio, n = C

S

EE =

psi3x10psi30x10

6

6

= 10

Step 2. Determine the transformed steel area, At = nAS.

( ) ( )2t 14π410A = = 31.42 in2




(Page 2 of 3)

Step 3. Define/draw transformed section. This consists of the concrete above the neutral axis and the steel below the neutral axis. The concrete below the neutral axis is to be ignored.

Procedure:

If the neutral axis is located a distance, x, below the top of the concrete section, the transformed section appears as follows:

2 in 2 in

d = 18 in

d-x

At = nAs = 31.42 in2

Neutral Axisx/2

b = 10 in

x

CG

Figure 30. Transformed Section

Step 4. Determine the location of neutral axis by calculating y from the relationship:

ΣAixi = 0 about neutral axis

( ) ( )xdnAxbx s −−⎟⎠⎞

⎜⎝⎛

2 = 0

dnAxnA2

bxss

2

−+ = 0

( )1831.4231.42x2

10x2

−+ = 0

x2 + 6.284x – 113.1 = 0




(Page 3 of 3)

x = 2a

4acbb 2 −±− = ( ) ( )

( )12113.11146.2846.284 2 ++−

= 7.95 in.

Step 5. Calculate section properties.

Moment of Inertia, I (about N.A)

Iconcrete = Ic = 3

bx3

Isteel = Is = ( )2s xdnA −

I = Ic + Is = 2s

3

x)(dnA3

bx−+

= ( ) ( )23

7.951831.423

7.9510−+ = 4848.4 in4

Section Modulus, S

- Concrete (top), Sc = xI =

7.954848.4 = 609.9 in3

- Steel (bottom), Ss = x-d

I = 7.95-18

4848.4 = 482.4 in3




GLOSSARY

Area Usually refers to area of cross section which is used in calculating stress, axial deformation, axial stiffness.

Axial Compression Condition which causes compression or shortening of a structural member.

Axial Tension Condition which results in extension or stretch of a structural member.

Centroid Geometric center of a cross section area.

Elastic Modulus Ratio of stress to strain of the material in the elastic range. Also known as modulus of elasticity or Young’s modulus.

Flexure Condition which results when a structural member bends or flexes when it is subjected to an internal moment.

Instability Condition which occurs when a slender structural component subjected to compressive load buckles and fails suddenly before the yield strength of the material is reached.

Mechanical Properties Properties determined by testing standard samples of the material in a laboratory that relate to how the a material behaves when loads are applied like compressive and tensile strength and yield stress.

Moment of Area Unit used to locate the geometric center (centroid) area of a composite section or neutral axis of a beam, and to compute the shear stress in a beam.

Moment of Inertia Second moment of area about a specified axis, the sum of the product area times the square of the distance to the axis.

Physical Properties Properties determine by testing standard samples of the material in a laboratory other than mechanical, that affect the behavior of a structure or structural member.

Poisson’s Ratio Absolute value of the ratio of transverse to axial strain that relates the lateral and axial deformations of a material when it is loaded along the longitudinal axis.

Shear Force or stress that causes a structural member to distort so that angles are changed.




Shear Modulus Ratio of shear stress to shear strain used to calculate

shear and torsional stiffness and deformations in structural analysis. Also known as modulus of rigidity.

Specific Weight Weight per unit volume. Often used to determine loads due to the self-weight of materials.

Strain A dimensionless measure of deformation of a material when it is subjected to a load. Change in length per unit length.

Stress The measure of an internal resistance due to an external force measured as force per unit area (e.g. MPa or ksi).




PRACTICE PROBLEMS

Practice Problem 1

a) Find area, location of centroid ( )yx, . b) Find first moment of areas about x and y axis c) Find the second moment of areas about x and y axis d) Find xI and yI about the centroidal axis e) Find the radius of gyration xr and yr .

Solution

a)

( ) ( )

1 2 3 4

2 250 501 100 40 100 140 50 502 2 42000 10075 2500 1963.5 16538.5

A A A A A

mm

π π

= + + +

⎛ ⎞ ⎛ ⎞× ×= × × + × − + × +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠= + + + =




( ) ( )4 50 4 501002000 100 140 50 3927 100 50 50 25 1963.5 503 3 316538.5

39.89

i iA xx

A

mm

π π

=

⎛ ⎞ ⎛ ⎞× ×⎛ ⎞ ⎛ ⎞× + × × − × − + × × + × +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠=

=

∑

( )( ) ( ) ( ) 4 50402000 190 100 140 120 3927 120 50 50 25 1963.5 503 316538.5

104.87

i iA yy

A

mm

π

=

⎛ ⎞×⎛ ⎞× + + × × − × + × × + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=

=

∑

b)

3

3

16538.5 39.89 659720.77

16538.5 104.87 1734392.50

x

y

Q Ax mm

Q Ay mm

= = × =

= = × =

c)

xAxAxAxAx IIIII 4321 +++=

( )3 2

6 41

3 4 22 6 4

2

36 4

3

24 26 4

4

100 40 100 40 40190 82.87 10336 2100 140 50 50100 *140 120 165.46 10

12 8 2

50 50 2.08 10350 50 4 5050 2.85 10

16 4 3

A x

A x

A x

A x

I mm

I mm

I mm

I mm

π π

π ππ

× ×= + + = ×

⎛ ⎞ ⎛ ⎞× × ×= − + − = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

×= = ×

× × ×⎛ ⎞= + × − = ×⎜ ⎟⎝ ⎠

( ) 6 6 482.87 165.46 2.08 2.85 10 253.26 10xI mm⇒ = + + + × = ×




yAyAyAyAy IIIII 4321 +++=

( )3 2

6 41

23 4 2

2

6 4

36 4

3

24 26 4

4

40 100 100 40 100 3.33 10336 2

140 100 50 50 4 501003 8 2 3

19.84 1050 50 2.08 10

350 50 4 5050 11.19 10

16 4 3

A y

A y

A y

A y

I mm

I

mm

I mm

I mm

π ππ

π ππ

× ×= + = ×

⎛ ⎞⎛ ⎞× × × ×⎛ ⎞= − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠= ×

×= = ×

× × ×⎛ ⎞= + × + = ×⎜ ⎟⎝ ⎠

( ) 6 6 43.33 19.84 2.08 11.19 10 36.44 10yI mm⇒ = + + + × = ×

d)

2 6 2 6 4

2 6 2 6 4

253.26 10 16538.5 104.87 71.37 10

36.44 10 16538.5 39.89 10.12 10

x x

y y

I I Ay mm

I I Ax mm

= − = × − × = ×

= − = × − × = ×

e)

6

6

71.37 10 65.6916538.5

10.12 10 24.7416538.5

xx

yy

Ir mmA

Ir mm

A

×= = =

×= = =




Table 5. Summary of the calculation around x-axis

Item Shape/ Size Area, Ai Io*106 yi Aiyi*103 iy

2iyiA *106 2

iyiAoI + * 106

Units mm2 mm4 mm. mm3 mm. mm4 mm4

1 Triangle 100 x 40 2000 0.178 203.3 406.6 98.4 19.37 19.548

2 Rectangle – half circle 10075 20.412 120.0 1209.0 15.1 2.30 22.712

3 Squire 50 x 50 2500 0.521 25.0 62.5 -79.9 15.96 16.481

4 Quarter Circle R=50 1963.5 1.227 28.8 56.5 -76.1 11.37 12.597

Total 16538.5 22.34 1734.6 49.00 71.338

y = ∑∑

i

ii

AyA

= 31734.6 10

16538.5× =104.9 mm I = ( )∑ + 2

iio yAI = 71.338×106 mm4

Note:

The value of moment of inertia in the Table is slightly different than on page 76 due to rounding off of digits.




Practice Problem 2

a) Find area, location of centroid ( )yx, . b) Find first moment of areas about x and y axis c) Find the second moment of areas about x and y axis d) Find xI and yI about the centroidal axis e) Find the radius of gyration xr and yr .

Solution

a)

( ) ( ) ( )1 2 3

100 20 40 20 80 202000 800 1600 4400.0

A A A A

mm

= + +

= × + × + ×

= + + =




( ) ( ) ( )2000 70 800 40 1600 104400

42.73

i iA yy

A

mm

=

× + × + ×=

=

∑

( ) ( ) ( )2000 50 800 10 1600 404400

39.09

i iA xx

A

mm

=

× + × + ×=

=

∑

b)

3

3

4400 42.73 188000

4400 39.09 172000

x

y

Q Ay mm

Q Ax mm

= = × =

= = × =

c)

1 2 3x A x A x A xI I I I= + +

( )

( )

32 6 4

1

32 6 4

2

36 4

3

100 20 100 20 70 9.87 1012

20 40 40 20 40 1.39 1012

80 20 0.21 103

A x

A x

A x

I mm

I mm

I mm

×= + × = ×

×= + × = ×

×= = ×

( ) 6 6 49.87 1.39 0.21 10 11.47 10xI mm⇒ = + + × = ×




1 2 3y A y A y A yI I I I= + +

36 4

1

36 4 6 4

2

36 4

3

20 100 6.67 103

40 20 0.107 10 0.11 103

20 80 3.41 103

A y

A y

A y

I mm

I mm mm

I mm

×= = ×

×= = × ≈ ×

×= = ×

( ) 6 6 46.67 0.11 3.41 10 10.19 10yI mm⇒ = + + × = ×

d)

2 6 2 6 4

2 6 2 6 4

11.47 10 4400 42.73 3.44 10

10.19 10 4400 39.09 3.47 10

x x

y y

I I Ay mm

I I Ax mm

= − = × − × = ×

= − = × − × = ×

e)

6

6

3.44 10 27.964400

3.47 10 28.084400

xx

yy

Ir mmA

Ir mm

A

×= = =

×= = =




Table 6. Summary of the calculation about x-axis

Item Shape/ Size Area, Ai Io*103 yi Aiyi*103 iy

2iyiA *103 2

iyiAoI + *103

Units mm2 mm4 mm. mm3 mm. mm4 mm4

1 Rectangle 100 x 20 2000 66.667 70.0 140.0 27.27 1487.3 1553.967

2 Rectangle 20 x 40 800 106.667 40.0 32.0 -2.73 5.962 112.629

3 Rectangle 80 x 20 1600 53.333 10.0 16.0 -32.73 1714.00 1767.333

Total 4400 226.667 188.0 3207.267 3433.929

y = ∑∑

i

ii

AyA

= 3188.0 *10

4400=42.73 mm I = ( )∑ + 2

iio yAI = 3.43×106 mm4




Practice Problem 3

A 7618×W section is reinforced with a plate inin 112 × at the top of the section. Use work aid 5 to compute the following for the composite section: a) Moment of inertia xI b) Radius of gyration xr c) Section Modulus at top and bottom

Solution

For 7618×W section:

2 422.3 18.2 1330xA in d in I in= = =

For the composite section:

22.3 18.2 / 2 12 18.7 12.4622.3 12

i i

i

A yy in

A× + ×

= = =+

∑∑




a)

sec .x W PlateI I I−= +

( )

( )

2 4sec.

32 4

4

1330 22.3 12.46 9.1 1581.76

12 1 12 18.7 12.46 468.2512

1581.76 468.25 2050.0

x W

x Plate

x

I in

I in

I in

− −

−

= + − =

×= + − =

⇒ = + =

b)

2050 59.77 7.7322.3 12

xx

Ir in inA

= = = =+

c)

3

3

2050 304.1519.2 12.46

2050 164.5312.46

xtop

top

xbottom

bottom

IS inc

IS inc

= = =−

= = =

Table 7. Summary of the calculation around x-axis


2iyiA 2

iyiAoI +

Units in2 in4 in in3 in in4 in4

1 Rectangle 12 x 1 12.0 0.125 18.7 224.4 6.24 467.25 467.375

2 W 18 x 76 22.3 1330.0 9.1 202.9 -3.36 251.76 1581.760

Total 34.3 1330.13 427.3 719.01 2049.135

y = ∑∑

i

ii

AyA

= 427.334.3

=12.46 in I = ( )∑ + 2iio yAI = 2049.135 in4

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Basic Properties of Section