Simulescu Student Workbook VOL1.pdf

TECHNICAL UNIVERSITY OF CIVIL ENGINEERING

Mechanics of Materials Student Workbook

-Volume I-

Eng. Ion S. Simulescu, MPh, PhD, PE Associate Professor

and

Eng. Cristian Ghindea

Assistant Professor

Bucharest 2004

- I -

PREFACE The present textbook is the first volume from a series of textbooks, titled Mechanics of Materials Student Workbook, intended to familiarize the students enrolled in the first semester of their sophomore year at the Technical University of Bucharest, School of Civil Engineering, with the practical application of the theoretical concepts developed during the weekly lectures. This textbook has in fact a complementary role to the more theoretically orientated textbook published under the name of Lectures in Mechanics of Materials, volume I. A number of four chapters are covered in the textbook. Briefly, the organization is as follows: Chapter 1 - Stress-Strain Diagram and Material Properties; Chapter 2 - Geometrical Characteristics of the Beam Cross-Section; Chapter 3 - Equilibrium of the Plane Linear Member and Chapter 4 - Axial Deformation. Each chapter starts with a theoretical section, named Theoretical Background, where the most important theoretical aspects are succinctly discussed. This section is followed by the Solved Problems section which contains a number of representative solved problems. Finally, the last section is the Proposed Problems section where a relatively large number of problems are proposed to the student for private exercise. With the intent to increase the student appetite towards using the modern capability of the numerical computer, the problems contained in the Solved Problems sections are solved using the MATHCAD software capabilities in parallel to the more classical method of the manual calculation. The first two chapters have a number of appendices attached. These appendices contain important engineering data necessary in solving some of the proposed problems. It is our pleasure to acknowledge the help that we received during the preparation of this textbook from our younger colleague Eng. George Vezeanu. Finally, the authors express their sincere gratitude to Prof. Dr. Eng. Dan Cretu, the Chairman of the Strength of Material Department, for his encouragements and support in the realization of this textbook. Dr. Eng Ion S. Simulescu Eng. Cristian Ghindea Bucharest, Romania. November, 2004.

- II -

TABLE OF CONTENTS Chapter 1 Stress-Strain Diagrams and Material Properties 1

1.1 Theoretical Background 1.1.1 Tension Static Test 1.1.2 Material Behavior 1.1.3 Linear Elasticity, Hook’s Law and Poisson’s Ratio

1.2 Solved Problems 1.3 Proposed Problems Appendix 1.1 Modulus of Elasticity and Poisson’s Ratio Appendix 1.2 Yield and Ultimate Stress

Chapter 2 Geometrical Characteristics of the Beam Cross-Section 40

2.1 Theoretical Background 2.2 Solved Problems 2.3 Proposed Problems Appendix 2.1 Geometrical Characteristics of Plane Areas Appendix 2.2 Properties of the Romanian Rolled Shaped Sections Appendix 2.3 Properties of the U.S.A Rolled Shaped Sections

Chapter 3 Equilibrium of the Plane Linear Members 118

3.1 Theoretical Background 3.1.1 Type of Loads, Supports and Reactions 3.1.2 Cross-Sectional Internal Resultants 3.1.3 Types of Statically Determinate Beams 3.1.4 Method 1 - Calculation of the Internal Resultants Using Method of

Sections 3.1.5 Method 2 - Differential Relations between Loads and Cross-

Section Internal Resultants 3.2 Solved Problems 3.3 Proposed Problems

- III -

Chapter 4 Axial Deformation 154

4.1 Theoretical Background 4.1.1 Basic Theory of Axial Deformation 4.1.2 Uniform-Axial Deformation 4.1.3 Nonuniform-Axial Deformation

4.2 Solved Problems 4.3 Proposed Problems

References 201

- 1 -

CHAPTER 1 Stress-Strain Diagrams and Material Properties

1.1. Theoretical Background 1.1.1 Tension Static Test Theoretical Stress-Strain Diagram for Structural Steel The test is conducted by subjecting a specimen made of structural steel to a monotonically increasing loading. The specimen is schematically depicted in Figure 1.1.1 During the test a series of pairs (P*,d*) are collected and tabulated. Consequently, the normal stress * and the corresponding axial strain * are calculated as:

0

**

AP

(1.1)

0

*

0

0*

*

LL

LLL

(1.2)

where 4

* 20

0

dA is the initial area of the specimen and *L is the current

elongation.

Figure 1.1.1

The normal stress * and axial strain * calculated using the formulae (1.1) and (1.2) are called engineering stress and engineering stain, respectively. If the normal stress and the normal strain are calculated using the value of the measurements at the particular moment the normal stress and the corresponding axial strain are called true stress and true strain, respectively. They are calculated as:

true

true AP *

* (1.3)

- 2 -

)1ln()1ln()ln( *

00

**

LL

LL

true (1.4)

Figure 1.1.2

The typical tensile stress-strain diagram for structural steel behavior is shown in Figure 1.1.2. This stress-strain diagram plotted in the plan is characterized by a number of defining points:

Point A (PLPL) - where PL is the proportional limit; Point B (YY) - where Y is the yielding point; Point C (EYEY) - where EY is the end stress of the perfect plastic region; Point D (UU) - where U is the ultimate stress; Point E (FF) - where F is the fracture stress; Point E’ (FF’) - where F is the true fracture stress.

The true stress-strain diagram is plotted with a dashed line above the engineering stress-strain diagram. For some materials (i.e. aluminum), which do not have after the proportional limit point a perfect plasticity region, the yield point is not easily identify and, consequently, it is determined using a method called the offset method. A straight line parallel to the initial linear part of the stress-strain diagram and passing through

002.0 is drawn. The construction is shown in Figure 1.1.3. The point A is located at the intersection between the stress-strain diagram and the parallel line. The stress corresponding to point A is called offset yield stress and is used instead of the yield stress.

- 3 -

Figure 1.1.3

1.1.2 Material Behavior Non-Linear Elastic Behavior

Figure 1.1.4

The non-linear elastic behavior is characterized by a one-to-one correspondence between the stress and the strain. During loading or unloading for a given value of always corresponds the same value of . Mathematically this can be expressed as:

)( f (1.5)

- 4 -

Non-Linear Elasto-Plastic Behavior

Figure 1.1.5

The elasto-plastic behavior is characterized by a different behavior during loading and unloading phases. During the unloading phase the material behaves elastically. Consequently, even when the load is completely removed a residual strain remains.

Figure 1.1.6 An idealized elasto-plastic behavior typical for structural steel is the Prandthl’s curve shown in Figure 1.1.6. This curve represents a material with an elastic-perfect plastic behavior.

- 5 -

Ductile and Brittle Materials A material is ductile if can undergo large plastic strain before fracture. In contrast, a material which fails at small strain in classified as brittle. The difference between the behavior of a ductile and brittle material is schematically pictured in Figure 1.1.7.

Figure 1.1.7 The ductility of a material in tension is characterized by its elongation and reduction of the area at the cross-section where the failure occurs. The percentage elongation is defined as:

100*_0

0

LLL

elongationpercentage failure (1.6)

where 0L and failureL are the original and failure gage lengths, respectively.

The percentage reduction in area is obtained as:

100*__0

0

AAA

areareductionpercentage failure (1.7)

where 0A and failureA are the original and failure areas, respectively.

1.1.3 Linear Elasticity, Hook’s Law and Poisson’s Ratio A bar is loaded in tension, as shown in Figure 1.1.8.b, the axial elongation is accompanied by lateral contraction.

- 6 -

Figure 1.1.8

The structural steel axially loaded under the proportionality limit PL behaves linearly elastic. Mathematically, the relation between the stress and strain is expressed by the Hook’s Law: *E if PL (1.8) where E is the modulus of elasticity. The lateral strain lateral is proportional with the longitudinal strain . The relation is:

*lateral (1.9)

where is the Poisson’s ratio. Consequently, the elastic behavior of a material is characterized by two material constants: the modulus of elasticity, E, and the Poisson’s ratio, . 1.2 Solved Problems Problem 1.2.1- Strain Measurements A mechanical extensometer uses the lever principle to magnify the elongation of a test specimen enough to make the elongation (or contraction) readable. The extensometer shown in Figure 1.2.1 is held against the test specimen by a spring that forces two sharp points against the specimen at A and B. The pointer AD pivots about a pin at C, so that distance between the contact points at A and B is exactly La = 15 cm (the gage length) of this extensometer when the pointer points to the origin, O, on the scale. In a particular test, the extensometer arm points "precisely" at point O when the load P is zero. Later in the test, the 25.5 cm long pointer points a distance d = 0.30 cm below point O. What is the current extensional strain in the test specimen at this reading? A. General Observations

The body of the extensometer is considered rigid in comparison to the specimen subjected to deformation. Consequently, the distance BC remains unaffected by the deformation of the specimen. This finding implies that during the elongation only the point A can move from A to A*.

- 7 -

Figure 1.2.1

B. Calculations The distance AA* represents the current elongation *L . Using the geometrical ratio:

4.2554.2

* dL

the current elongation *L is obtained:

cm 03.04.25

54.230.054.2

4.25*

dL

The strain is calculated as:

002.024.15

03.0

0

*

LL

Problem 1.2.2- Stress-Strain Diagram The tension specimen, shown in Figure 1.2.2.a, with an initial diameter d0=12 mm and a gage length L0 = 50 mm is used to obtain the load-elongation data contained in Table 1.2.1. Using the test data plot the stress-strain diagram and then calculate the following: (a) the proportional limit, (b) modulus of elasticity, (c) yield stress at 0.2% offset, (d) ultimate stress, (e) percent elongation and (g) percentage area.

Figure 1.2.2.a

- 8 -

Table 1.2.2 Load-Elongation Test Data

Force (kN)

Elongation (mm)

Force (kN)

Elongation (mm)

0.0 0.000 55.603 1.524 23.131 0.127 56.492 1.778 41.813 0.229 57.382 2.032 43.148 0.254 57.827 2.286 44.482 0.330 58.272 2.540 47.151 0.508 58.717 2.794 50.265 0.762 58.717 3.048 52.489 1.016 57.827 3.302 54.268 1.270 56.048 3.505

A. General Observations The initial measurements of the diameter and gage length of the specimen are:

d0 12 mm

L0 50 mm

Consequently the size of the original area is obtained as:

A0

d02

4

4

12* 2A0 113.097mm

2

B. Calculations B.1 Calculation of the Stress-Strain Diagram The stress and strain corresponding to each measurement step are calculated using the collected values contained in Table 1.2.2. There are a number of eighteen (18) measured steps readn . The measured values of the elongation L

and applied force P

are collected into two separate vectors:

- 9 -

Force Elongation

P

0.000

23.131

41.813

43.148

44.482

47.151

50.265

52.489

54.268

55.603

56.492

57.382

57.827

58.272

58.717

58.717

57.827

56.048

103 N

L

0.000

0.127

0.229

0.254

0.330

0.508

0.762

1.016

1.270

1.524

1.778

2.032

2.286

2.540

2.794

3.048

3.302

3.505

mm

The stress and strain values corresponding to each one of the measured steps are calculated using the formulae below: i 1 nread

i 1Li 1

L0

engineering strain

i 1Pi 1

A0

engineering stress

The obtained values are show in a tabular format below.

- 10 -

Strain Stress

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

0

2.54·10 -3

4.58·10 -3

5.08·10 -3

6.6·10 -3

0.01

0.015

0.02

0.025

0.03

0.036

0.041

0.046

0.051

0.056

0.061

0.066

0.07

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

0

2.045·10 8

3.697·10 8

3.815·10 8

3.933·10 8

4.169·10 8

4.444·10 8

4.641·10 8

4.798·10 8

4.916·10 8

4.995·10 8

5.074·10 8

5.113·10 8

5.152·10 8

5.192·10 8

5.192·10 8

5.113·10 8

4.956·10 8

Pa

The minimum and maximum values obtained are:

min 0

min 0

max 17 07.0

max 15 Pa810*192.5

The graphical representation of the stress-strain diagram is shown in Figure 1.2.2.b. A qualitative analysis of the stress-strain diagram indicated an elasto-plastic behavior and consequently, a ductile behavior.

- 11 -

0 0.008 0.016 0.024 0.032 0.04 0.048 0.056 0.064 0.072 0.080

6 107

1.2108

1.8108

2.4108

3 108

3.6108

4.2108

4.8108

5.4108

6 108

min

max

i

min max

i

Figure 1.2.2.b

B.2 Calculation of the modulus of elasticity To obtain the value of the modulus of elasticity, E, representative for the elastic behavior of the material, the ratio of the stress and strain increased corresponding to each measured step is calculated: k 1 nread 1

k k k 1 stress increased

k k k 1 strain increased

Ekk

k

the ratio stress-strain

The numerical values of the ratio stress-strain obtained are tabulated below and plotted in Figure 1.2.2.c. It should be remarked that despite the fact that theoretically the modulus of elasticity is constant in the elastic range, due to the measurement errors a small variation is obtained.

- 12 -

E

0

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

0

8.052·10 10

8.097·10 10

2.361·10 10

7.76·10 9

6.629·10 9

5.42·10 9

3.871·10 9

3.096·10 9

2.324·10 9

1.547·10 9

1.549·10 9

7.745·10 8

7.745·10 8

7.745·10 8

0

-1.549·10 9

-3.874·10 9

Pa

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 172 1010

1.2510105 109

2.51091 1010

1.7510102.51010

3.2510104 1010

4.7510105.51010

6.2510107 1010

7.7510108.51010

9.2510101 1011

Ek

k

Figure 1.2.2.c

The elastic range is represented by the “almost constant” variation and in this case ends after the third measurement point. The theoretical value of the modulus of elasticity is obtained by averaging the calculated values of the measurement steps pertinent to the elastic behavior: nn 2

- 13 -

Eaverage1

nn

k

Eknn

2

10*097.810*052.8 1010

Eaverage 8.075 1010 Pa

B.3 Calculation of the 0.2% Offset Stress The construction is shown in Figure 1.1.3. The line anchored at the offset strain value and parallel to the linear portion of the stress-strain diagram is constructed below

using two description points: )0,( offset and ),( maxmax

averageoffset E

offset 0.002

j 0 1 strain stress

line

offset

max

Eaverageoffset

line

0

max

line2 10

3

8.43 103

line

0

5.192 108

Pa

The graphical construction is shown in Figure 1.2.2.d. The value of the stress offset

corresponding to 002.0offset is obtained by reading the stress scale as:

Paoffset

810*9.3

This value can be considered as the yielding stress.

- 14 -

0 0.008 0.016 0.024 0.032 0.04 0.048 0.056 0.064 0.072 0.080

6 107

1.2108

1.8108

2.4108

3 108

3.6108

4.2108

4.8108

5.4108

6 108

min

max

i

linej

min max

i linej

Figure 1.2.2.d

B.4 Calculation of the Ultimate Stress Analyzing the stress-strain diagram the value of the ultimate stress is obtained as:

U 15 U 5.192 108 Pa

B.5 Calculation of the Percentage Elongation The percentage elongation is calculated at failure: elongation_failure L17

pelongationelongation_failure

L0100 100*

50

505.3pelongation 7.01 %

B.5 Calculation of the Percentage of Area Reduction The calculation of the area at failure is based on the assumption that volume remained constant during the deformation:

- 15 -

Lfailure L0 elongation_failur 505.350 Lfailure 53.505mm

Afailure

A0 L0

Lfailure

505.53

50*097.113Afailure 105.689mm

2

The percentage reduction of the area is calculated as:

pA

Afailure A0

A0100

097.113

097.113689.105pA 6.551 %

Problem 1.2.3 Two tension specimens with initial identical dimensions, diameter d0=12 mm and gage length L0 = 50 mm, are made of structural materials A and B, respectively. They are tested in tension until the failure is reached. The test data obtained is shown in Table 1.2.3. Conduct the following tasks: (a) calculate the percent elongation and the percent of reduction in the area at failure (b) draw to scale the idealized stress-strain diagram pertinent to both materials; (c) classify the material as either brittle or ductile and explain the judgment.

Table 1.2.3 Tensile Test Data

Data Material A B

1. Gage at failure 73.66 mm 56.39 mm 2. Diameter at failure 6.68 mm 11.96 mm 2. Modulus of Elasticity 6.9x1010 Pa 7.2x1010Pa 3. Yield Stress 3.5x107 Pa 5.0x108Pa 4. Ultimate Stress 8.9x107Pa 5.7x108Pa 5. Failure Stress 1.25 510 Pa 4.14 510 Pa 6. Ultimate Strain 0.85 of ultimate 0.85 of ultimate

A. General Observations The initial measurements of the specimen dimensions (gage length and diameter) are: d0 12mm L0 50mm The initial area of the specimen is:

A0 d0 2

4

4

12 2A0 113.097mm

2

- 16 -

B. Calculations B.1 Calculation of the percentage elongation The percentage elongations corresponding to both materials are:

pea

Lfa L0

L0100

100

50

5066.73pea 47.32 0

0 material “A”

peb

Lfb L0

L0100

100

50

5039.56peb 12.78 0

0

material “B”

B.2 Calculation of the percentage reduction of the area The diameters and areas at failure, corresponding to material “A” and “B”, respectively, are: dfa 6.68 mm

dfb 11.96 mm

The areas at failure are:

Afa

dfa2

4

4

68.6* 2Afa 35.046mm

2

Afb

dfb2

4

4

96.11* 2Afb 112.345mm

2

The percentage of the area reduction is obtained as:

parea_a

Afa A0

A0100

097.113

097.113046.35parea_a 69.012 0

0 material

“A”

parea_b

Afb A0

A0100

097.113

097.113345.112parea_b 0.666 0

0

material

“B” B.3 Schematic plot of stress-strain relations The strain corresponding to the yielding point is:

- 17 -

ya

ya

Ea

10

7

10*9.6

10*5.3ya 5.072 10

4 material “A”

yb

yb

Eb

10

8

10*2.7

10*5.5yb 6.944 10

3 material “B”

The strain at failure is calculated as:

fa

Lfa L0

L0

50

5066.73 fa 0.473 material “A”

fb

Lfb L0

L0

50

5039.56 fb 0.128

material “B”

The strain corresponding to the ultimate stress: ua 0.85 fa 473.0*85.0

ub 0.85 fb 128.0*85.0

The representative points of the stress-strain curves are:

material “A” material “B”

a

0

ya

ua

fa

a

0

ya

ua

fa

b

0

yb

ub

fb

b

0

yb

ub

fb

The plot of the two stress-strain diagrams is shown in Figure 1.2.3.

- 18 -

0 0.1 0.2 0.3 0.4 0.50

1.2108

2.4108

3.6108

4.8108

6 108

strain

stre

ss

a

b

a b

Figure 1.2.3

B.4 Classification of the materials The percentage of elongation previously calculated for the two materials is: pea 47.32 0

0 material “A”

peb 12.78 0

0

material “B”

The ductility ratios, other ductility indicators, are calculated as:

a

ua

ya a 792.948

material “A”

b

ub

yb b 15.643

material “B”

It can be concluded that both materials show ductile behavior. Obviously the material “A” is more ductile than “B”.

- 19 -

Problem1.2.4- Mechanical Properties of Materials A tensile specimen of a certain alloy has an initial diameter of 13 mm and a gage length of 200 mm. Under a load P = 20 kN, the specimen reaches its proportional limit and is elongated by 3 mm. At this load the diameter is reduced by 0.064 mm. Calculate the following material properties: (a) the proportional limit, (b) PL the modulus of elasticity, E, and (c) the Poisson's ratio,

Figure 1.2.4

A. General Observations The initial measurements of the specimen dimensions (gage length and diameter) are: L0 200mm

d0 13 mm

The original area of the specimen is:

A0

d02

4

4

13* 2A0 132.732mm

2

At the application of the axial load kNP 20 the proportional limit, defined by the

stress PL and the strain PL , is attained. B. Calculations B.1 Calculation of proportional limit corresponding stress and strain The strain PL is calculated using the measured elongation L as:

PLL

L0

mmmm

200

3PL 0.015

The stress PL is obtained as:

PLP

A0

mmN

732.132

20000PL 1.507 10

8 Pa

- 20 -

B.2 The modulus of elasticity The modulus of elasticity is calculated:

EPL

PL

015.0

10*507.1 8 PaE 1.005 10

10 Pa

B.3 The Poisson’s Ratio The Poisson’s Ratio represents the ratio between the transversal strain transvPL _ and

the longitudinal strain PL . Consequently, the transversal strain transvPL _ is obtained

employing the reduction of the diameter d :

PL_transd

d0

mmmm

13

064.0PL_trans 4.923 10

3

The Poisson’s Ratio is calculated:

PL_trans

PL

015.0

10*923.4 3

0.328

Problem1.2.5 A wire of length L0 = 2.50 m and diameter d0= 1.6 mm is stretched by tensile forces P = 1250 N. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation:

*3001

*10*24.1 5

03.00 where is nondimensional and has MPa

units. Conduct the following tasks: (a) construct a stress-strain diagram for the material, (b) determine the elongation of the wire due to the forces P, (c) if the forces are removed, what is the permanent strain of the bar considering an average elastic modulus Eaverage=7.086x1010 Pa and (d) if the forces are applied again, what is the proportional limit? A. General Observations The initial measurements of the specimen dimensions (gage length and diameter) are: L0 2.50 m

d0 1.6 mm

The original area of the specimen is:

- 21 -

A0

d02

4

4

6.1* 2A0 4.021 mm

2

The strain is limited to the value: limit 0.03 B. Calculations B.1 Plot the stress-strain diagram A number of thirteen (13) points are considered nn 13 and consequently, the step

increased of the strain is calculated as limit

nn 1( )

12

03.0

The points representing the strain and stress diagram are obtained:

i 0 nn 1 i i i124000 i

1 300 i10

6 Pa

The stress-strain diagram values are first tabulated and then plotted in Figure 1.2.5.

0

0

1

2

3

4

5

6

7

8

9

10

11

12

0

2.5·10 -3

5·10 -3

7.5·10 -3

0.01

0.013

0.015

0.018

0.02

0.023

0.025

0.028

0.03

0

0

1

2

3

4

5

6

7

8

9

10

11

12

0

1.771·10 8

2.48·10 8

2.862·10 8

3.1·10 8

3.263·10 8

3.382·10 8

3.472·10 8

3.543·10 8

3.6·10 8

3.647·10 8

3.686·10 8

3.72·10 8

Pa

- 22 -

0 0.003 0.006 0.009 0.012 0.015 0.018 0.021 0.024 0.027 0.030

4 107

8 107

1.2108

1.6108

2 108

2.4108

2.8108

3.2108

3.6108

4 108

3.72 108

0

i

lineari

0.030 i lineari

Figure 1.2.5

B.2 Calculation of the stress and strain corresponding to load P=1250 N The stress is obtained as:

PP

A0

021.4

1250P 3.108 10

8 Pa

The corresponding strain is calculated using the expression:

P

P

1.24 1011 Pa 3 10

2 P

8211

8

10*108.3*10*310*24.1

10*108.3

P 0.01

B.3 Calculation of the remnant strain after unloading The remnant strain is obtained by constructing the unloading line which is anchored at point ),( PP and has a slop of PaEaverage

1010*086.7 .

The remnant strain is obtained as:

rem P

P

Eaverage

10

8

10*086.7

10*108.301.0 rem 5.724 10

3

- 23 -

The unloading line is constructed using two points described as:

linear

rem

P

linear

0

P

B.4 The loading proportional limit The loading follows the same linear behavior described by the unloading and the new proportional limit is ),( PP , the point where the unloading begun.

PP

A0 P 3.108 10

8 Pa

1.3 Proposed Problems Problem 1.3.1 A "pencil" laser extensometer, like the mechanical lever extensometer in Prob.1.2.1, measures elongation, from which extensional strain can be computed, by multiplying the elongation. In Figure 1.3.1 the laser extensometer is being used to measure strain in a reinforced concrete column. The target is set up across the room from the test specimen so that the distance from the fulcrum, C, of the laser to the reference point O on the target is dOC = 5m. Also, the target is set so that the laser beam points directly at point O on the target when the extensometer points are exactly Lo = 150 mm apart on the specimen, and the cross section at B does not move vertically. At a particular value of (compressive) load P, the laser points upward by an angle that is indicated on the target to be = 0.0030 rad. Determine the extensional strain in the concrete column at this load value.

Figure 1.3.1 Problem 1.3.2 A tensile test is conducted on a flat-bar steel specimen having the dimensions shown in Figure 1.3.2. Using the experimental load-elongation data, shown in Table 1.3.2, collected during the test conduct the following tasks: (a) plot a curve of engineering stress, , versus engineering strain, ; (b) determine the modulus of elasticity of

- 24 -

this material; (c) use the 0.2%-offset method to determine the yield strength, YS , of

this material. Table 1.3.2 Tension Test Data

Force (kN)

Elongation (mm)

Force (kN)

Elongation (mm)

0.000 0.000 26.467 0.127 5.338 0.020 27.801 0.152 10.676 0.041 28.913 0.191 16.014 0.061 29.581 0.254 21.351 0.081 30.470 0.317 25.355 0.102 30.693 0.381

Figure 1.3.2

Problem 1.3.3 A standard ASTM tension specimen, shown in Figure 1.3.3, with an original diameter d0=13 mm and a gage length L0 = 50 mm is used to obtain the load-elongation data contained in Table 1.3.3. Conduct the following tasks: (a) plot a curve of engineering stress, , versus engineering strain, ; (b) determine the modulus of elasticity of this material; (c) use the 0.2%-offset method to determine the yield strength, YS , of

this material.

Table 1.3.3 Tension Test Data

Force (kN)

Elongation(mm)

Force (kN)

Elongation (mm)

0.000 0.000 42.258 0.305 8.452 0.051 44.482 0.368 16.903 0.102 46.262 0.457 25.355 0.152 47.374 0.610 33.806 0.203 48.930 0.762 40.034 0.254 49.153 0.914

- 25 -

Figure 1.3.3

Problem 1.3.4 The tension specimen, shown in Figure 1.3.3, with an initial diameter d0=13 mm and a gage length L0 = 50 mm is used to obtain the load-elongation given in Table 1.3.4 Conduct the following tasks: (a) plot a curve of engineering stress, , versus engineering strain, , (b) determine the modulus of elasticity of this material, (c) use the 0.2%-offset method to determine the yield strength, YS , of this material.

Table 1.3.4 Test Data

Force (kN) Elongation (mm) P(kN) ∆L (mm)

0.0 0.000 27.5 1.68 9.3 0.050 28.4 2.0 14.9 0.200 28.6 2.33 17.7 0.325 28.9 2.68 22.4 0.675 28.4 3.00 25.2 1.000 27.50 3.33 26.6 1.330 26.10 3.68

Problem 1.3.5 A specimen of a methacrylate plastic shown in Figure 1.3.5 is tested in tension at room temperature, producing the stress-strain data listed in the accompanying Table 1.3.5. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity, the yield stress at 0.2% offset and establish if the material is brittle or ductile.

Figure 1.3.5

- 26 -

Table 1.3.5 Test Data

Stress (MPa) Strain Stress (MPa) Strain

0.0 0.000 44.0 0.0184 8.0 0.0032 48.2 0.0209 17.5 0.0073 53.9 0.0260 25.6 0.0111 58.1 0.0331 31.1 0.0129 62.0 0.0429 39.8 0.0163 62.1 fracture

Problem 1.3.6 The data shown in the accompanying Table 1.3.6 were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13 mm and a gage length of 50 mm as shown in Figure 1.3.6. At fracture, the elongation between the gage marks was 3.0 mm and the minimum diameter was 10.7 mm. Plot the conventional stress-strain curve for the steel. Determine the following: (a) the proportional limit, (b) modulus of elasticity, (c) yield stress at 0.1% offset, (d) ultimate stress, (e) percent elongation, and (f) percent reduction in area.

Figure 1.3.6

Table 1.3.6 Test Data Force (kN) Elongation (mm) Force (kN) Elongation

(mm) 0.000 0.00000 61.385 0.16000 4.448 0.00508 62.275 0.22900 8.896 0.01500 64.054 0.25900 26.689 0.04800 67.613 0.33000 44.482 0.08400 74.730 0.58400 53.379 0.09900 81.847 0.85300 57.382 0.10900 88.964 1.28800 59.606 0.11900 99.640 2.81400 60.496 0.13700 100.530 fracture

- 27 -

Problem 1.3.7 A tensile test is performed on an aluminum specimen that is 13 mm in diameter using a gage length of 50 mm, as shown in Fig. 1.3.7. When the load is increased by an amount P = 8 kN, the distance between gage marks increases by an amount L = 0.0430 mm. Calculate: (a) the value of the modulus of elasticity, E, for this specimen,

(b) If the proportional limit stress for this specimen is PL= 280 MPa, what is the distance between gage marks at this value of stress?

Figure 1.3.7

Problem 1.3.8 A short brass cylinder ( mm15d0 , L0 = 25.5mm) is compressed between two

perfectly smooth, rigid plates by an axial force P = 22.73 kN. (a) If the measured shortening of the cylinder, due to this force is 0.02667 mm, what is the brass specimen modulus of elasticity E? (b) If the increase in diameter due to the load P is 0.00533 mm, what is the value of Poisson's ratio ?

Figure 1.3.8 Problem 1.3.9 A tensile force of 500 kN is applied to a uniform segment of a titanium-alloy bar. The cross section is a 50 mm x 50 mm square, and the length of the segment being tested is 200 mm. Using titanium-alloy data from Appendix 1, determine: (a) the change in the cross-sectional dimension of the bar, and (b) the change in volume of the 200 mm segment being tested.

- 28 -

Problem 1.3.10 A cylindrical rod with an initial diameter of 8 mm is made of 6061-T6 aluminum alloy. When a tensile force P is applied to the rod, its diameter decreases by 0.0101 mm. Using the appropriate aluminum-alloy data from Appendix 1, determine (a) the magnitude of the load P, and (b) the elongation over a 200 mm length of the rod. Problem 1.3.11 Under a compressive load of P = 110 kN, the length of the concrete cylinder in Figure 1.3.11 is reduced from 305 mm to 304.924 mm, and the diameter is increased from 150 mm to 150.008 mm. Determine the value of the modulus of elasticity, E, and the value of Poisson's ratio, . Assume linearly elastic deformation.

Figure 1.3.11 Problem 1.3.12 The cylindrical rod in Figure 1.3.12 is made of annealed (soft) copper with modulus of elasticity E = 8101718.1 2kN/m and Poisson's ratio = 0.33, and it has an initial diameter, d0, of 51 mm. For compressive loads less than a "critical load" Pcr, a ring with inside diameter d = 51.005 mm is free to slide along the cylindrical rod. What is the value of the critical load Pcr?

Figure 1.3.12

Problem 1.3.13 A steel pipe column of length L0 = 3.65 m, outer diameter d0 = 102 mm, and wall thickness t0 = 13 mm is subjected to an axial compressive load P = 570 kN as shown

- 29 -

in Figure 1.3.13. If the steel has a modulus of elasticity E = 100 GPa and Poisson's ratio = 0.30, determine: (a) the change, L, in the length of the column, and (b) the change, t in the wall thickness.

Figure 1.3.13 Problem 1.3.14 A rectangular aluminum bar (w0 = 25 mm; t0, = 13 mm) is subjected to a tensile load P by pins at A and B (Figure 1.3.14). Strain gages measure the following strains in the longitudinal, x, and transversal, y, directions: x = 566 , and y = -187 . (a) What is the value of Poisson's ratio for this specimen? (b) If the load P that produces these values of x and y P = 27.5 kN, what is the modulus of elasticity, E, for this specimen? (c) What is the change in volume, V, of a segment of bar that is initially 50 mm long?

Figure 1.3.14

Problem 1.3.15 Three different materials, designated A, B, and C, are tested in tension using test specimens having diameters of 12 mm and gage lengths of 50 mm. At failure, the distances between the gage marks are found to be 54.5, 63.2, and 69.4 mm, respectively. Also, at the failure cross sections the diameters are found to be 11.46, 9.48, and 6.06 mm, respectively. Determine the percent elongation and percent

- 30 -

reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.

Figure 1.3.15

Problem 1.3.16 A bar made of structural steel having the stress-strain diagram shown in Figure 1.3.16 has a length of 1.525 m. The yield stress of the steel is 280 MPa and the slope of the initial linear part of the stress-strain curve, modulus of elasticity, is 210 GPa. The bar is loaded axially until it elongates 5.334 mm, and then the load is removed. How does the final length of the bar compare with its original length?

Figure 1.3.16 Problem 1.3.17 A bar of length 0.8 m is made of a structural steel having the stress-strain diagram shown in the Figure 1.3.17. The yield stress of the steel is 250 MPa and the slope of the initial linear part of the stress-strain curve (modulus of elasticity) is 200 GPa. The bar is loaded axially until it elongates 2.5 mm, and then the load is removed. How does the final length of the bar compare with its original length?

- 31 -

Figure 1.3.17

Problem 1.3.18 An aluminum bar has length L = 40.5 cm and diameter d = 18 mm. The stress-strain curve for the aluminum alloy is shown in Figure 1.3.18. The initial straight-line part of the curve has a slope, the modulus of elasticity, of 28 kN/m100.6893 . The bar is loaded by a tensile force P = 57 kN and then unloaded. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit?

Figure 1.3.18 Problem 1.3.19 A circular bar of magnesium alloy is 750 mm long. The stress-strain diagram for the material is shown in the Figure 1.3.19. The bar is loaded in tension to an elongation of 4.5 mm, and then the load is removed. (a) What is the permanent set of the bar? (b) If the bar is reloaded, what is the proportional limit?

- 32 -

Figure 1.3.19

Problem 1.3.20 A round bar of length L = 2.5 m and diameter d = 10 mm is stretched by tensile a force P = 60 kN. The bar is made of an aluminum alloy for which the stress-strain relationship may be described mathematically by the following equation:

]*10*338.5

31[*

700009

22

where has units of megapascals (MPa) and is nondimensional. Conduct the following calculations: (a) construct a stress-strain diagram for the material, (b) determine the elongation of the bar due to the force P, (c) if the forces are removed, what is the permanent strain of the bar and (d) if the forces are applied again, what is the proportional limit? Problem 1.3.21 A high-strength steel bar used in a large crane has diameter d = 57 mm as shown in Figure 1.3.21 is compressed by axial forces. The steel has modulus of elasticity E = 28 kN/m101.999 and Poisson's ratio = 0.30. Because of clearance requirements, the diameter of the bar is limited to 57.025 mm. What is the largest compressive load Pmax that is permitted?

Figure 1.3.21

- 33 -

Problem 1.3.22 The round bar, shown in Figure 1.3.22 has the initial diameter of 12 mm diameter and is made of aluminum alloy 6061-T6. When the bar is stretched by axial force P, its diameter decreases by 0.012 mm. Find the magnitude of the load P. (Obtain the material properties from Appendix 1).

Figure 1.3.22

Problem 1.3.23 A nylon bar having diameter d1 = 70 mm is placed inside a steel tube having inner diameter d2, =70.25 mm as shown in Figure 1.3.23. The nylon cylinder is then compressed by an axial force P. At what value of the force P will the space between the nylon bar and the steel tube be closed, assuming that the nylon has the modulus of elasticity E = 26 kN/m103.102 and the Poisson’s ratio = 0.4?

Figure 1.3.23

Problem 1.3.24 A prismatic bar of circular cross section is loaded by a tensile force P as shown in Figure 1.3.24. The bar has an initial length L0 = 3.0 m and diameter d0 = 30 mm. The bar is made of aluminum alloy 2014-T6 with modulus of elasticity E = 73 GPa and Poisson's ratio = 0.333. (a) If the bar elongates by 7.0 mm, what is the decrease in diameter d0? (b)What is the magnitude of the load P?

- 34 -

Figure 1.3.24

Problem 1.3.25 A bar of monel metal with an initial length L0 = 0.38 m and a diameter mm8d0 is

loaded axially by a tensile force P = 12 kN. Using the data in Appendix 1.1, determine the increase in length of the bar and the percent decrease in its cross-sectional area. Problem 1.3.26 A high-strength steel wire with an initial diameter of d0= 3 mm stretches 37.1 mm when a 15-meter length of it is stretched by a force of 3.5 kN. (a) What is the modulus of elasticity, E, of the steel? (b) If the diameter of the wire decreases by 0.0022 mm, what is Poisson's ratio? Problem 1.3.27 A hollow bronze cylinder, shown in Figure 1.3.27, is compressed by a force P. The cylinder has inner diameter d1 = 47 mm, outer diameter d2 = 55 mm, and modulus of elasticity 110320E Mpa. When the force P increases from zero to 35 kN, the outer diameter of the cylinder increases by 0.0432 mm. Determine: (a) the increase in the inner diameter, (b) wall thickness and (c) the Poisson's ratio for the bronze.

Figure 1.3.27

- 35 -

Problem 1.3.28 A plate of length L, width b, and thickness t is subjected to a uniform tensile stress applied at its ends as shown in Figure 1.3.28. The material has a modulus of elasticity E and Poisson's ratio . Before the stress is applied, the slope of the diagonal line OA is b/L. What is: (a) the slope when the stress is acting; (b) the increase in area of the front face of the plate; (c) the decrease in cross-sectional area?

Figure 1.3.28 Problem 1.3.29 An axially loaded member having before loading a squared cross-section area of 3cm x 3cm and a length of 180 cm becomes 0.001 cm wider and 0.07 cm shorter after loading. Determine the Poisson’s ratio. Problem 1.3.30 At the proportional limit, the 205 mm gage length of a 12.5 mm diameter alloy bar has elongated 0.3 mm and the diameter has been reduced by 0.0064 mm. The total axial load carried was 22 KN. Determine the following properties of this material: (a) the modulus of elasticity; (b) the Poisson's ratio and (c) the proportional limit. Problem 1.3.31 A 455 kN axial load is slowly applied to a 2.50 m long rectangular bar. The bar cross-section is 2.5 cm wide and 10.5 cm deep. When loaded, the 10.5 cm side of the cross-section measures 10.445 cm and the length has increased by 0.2286 cm. Determine Poisson's ratio and Young's modulus for the material. Problem 1.3.32 In a 0.65 cm diameter steel tie rod 3.2 m long, there is an axial tensile stress of 1.38 N/m2. Poisson's ratio for this steel is 0.25. How much has the rod elongated, and how much has its diameter been altered? Problem 1.3.33 A 70 mm by 150 mm rectangular alloy bar elongates 0.003 cm. The member has an original length of 1.55 m and is loaded with an axial load of 44.5 kN. Considering that

- 36 -

the proportional limit of the material is 2.4*105 kN/m2, calculate the following: (a) the axial stress in the bar, (b) the modulus of elasticity of this material, (c) if Poisson's ratio for the material is 0.25 what will be the total change in each lateral dimension? Problem 1.3.34 A steel rod characterized by a 38 mm diameter solid circular cross-section and a length of 6 m elongates 12 mm under an axial load of 235 kN. The rod diameter decreased 0.025 mm during the loading. Determine the following properties of the material: (a) the Poisson's ratio, (b) the modulus of elasticity and (c) the modulus of rigidity. Problem 1.3.35 A steel and an aluminum bar are coupled together end to end and loaded axially at the extreme ends. Both bars are 50 mm in diameter; the steel bar is 1.55 m long, and the aluminum bar is 1.25 m long. When the load is applied, it is found that the steel bar elongates 0.102 mm in a gage length of 205 mm. Poisson's ratio for this steel is 1/4, and the modulus of elasticity of the aluminum is 69 GPa. Determine: (a) the load, (b) the total change in length measured between the bar ends and (c) the change in the diameter of the bar.

- 40 -

CHAPTER 2 Geometrical Characteristics of the Beam Cross-Section

2.1. Theoretical Background The geometrical characteristics of the beam cross-section are as follows: Area (Figure 2.1.1)

A

dAA (2.1)

Figure 2.1.1

First Moments or Static Moments (Figure 2.1.1)

A

y dAzS (2.2)

A

z dAyS (2.3)

Position of the Centroid located at point ),( CC zyC (Figure 2.1.2)

A

dAy

ASy Az

C

(2.4)

A

dAz

AS

z AyC

(2.5)

Figure 2.1.2

Second Moments or Moments of Inertia (Figure 2.1.3)

- 41 -

dAzIA

y 2 (2.6)

dAyIA

z 2 (2.7)

Figure 2.1.3

Polar Moment of Inertia (Figure 2.1.3)

yz

AApo

II

dAzydAI

)( 222 (2.8)

Product of Inertia (Figure 2.1.3)

A

yz dAzyI (2.9)

Parallel-Axis Theorems for Moment of Inertia (Figure 2.1.4)

CyCy IAzI 2 (2.10)

CzCz IAyI 2 (2.11)

CIAI C 2 (2.12)

Figure 2.1.4 Moment of Inertia about Inclined Axes (see Figure 2.1.5)

)2sin()2cos(22

'

yzzyzy

y IIIII

I

(2.13)

)2sin()2cos(22

'

yzzyzy

z IIIII

I

(2.14)

Figure 2.1.5

- 42 -

)*2cos()2sin(2

''

yzzy

zy III

I

(2.15)

Principal Moments of Inertia

22

max1 4

)(

2 yzzyzy

p IIIII

II

(2.16)

22

min2 4

)(

2 yzzyzy

p IIIII

II

(2.17) Note: For the calculation of the principal moments of inertia related to centroidal

axes the moments of inertia yzzy III ,, contained in equations (2.16), (2.17)

should be replaced by CCC yzzy III ,, , respectively.

Principal Directions of Inertia

The value of the angle corresponding to the principal directions is obtained using the equation:

zy

yz

III

2

)2tan( 0 (2.18)

The two solutions of equation (2.18), angles 01 and 02 , are related as:

20102

(2.19)

The angle, 01 or 02 , corresponding to the maximum moment of inertia max1 II p , is

the angle which verifies the relation (2.20):

0tan 0

yzI

(2.20)

Maximum Product Moment of Inertia The maximum value of the product of inertia is obtained for an angle of rotation

4

measured in the counter-clockwise direction from the position of the principal

axes and has the following expression:

- 43 -

22

21

max 4

)(

2'' yz

zyppzy I

IIIII

(2.21)

The corresponding moments of inertia are:

2221

''

zyppzy

IIIIII

(2.22)

Mohr’s Circle Representation of the Moments of Inertia (Figure 2.1.6)

Figure 2.1.6

Practically the Mohr’s circle is constructed using the following steps:

(a) The coordinates system O is drawn as shown in Figure 2.1.6. The

horizontal axis O represents the moments of inertia, while the vertical axis O represents the product of inertia (note that the positive axis is drawn

upwards). The drawing should be done roughly to scale. The representation considers that the following conditions are met: zy II , 0yI , 0zI and

0yzI ;

(b) Using the calculated values of the moments of inertia yI and zI and the product

of inertia yzI two points noted as Y and Z are placed on the drawing. The line

YZ intersects the horizontal axis in point C which represents the center of the Mohr’s circle;

(c) The distance CY is the radius of the circle. Using the radius CY and the position of the center C the Mohr’s circle is constructed. The intersection points,

1P and 2P , between the circle and the horizontal axis represent the maximum and the minimum moments of inertia;

- 44 -

(d) The absolute value of the )2tan( 0 can be calculated from the graph;

(e) The angle of the principal direction 1 is the angle measured in the counter-

clockwise direction between lines CY and CP1. To obtain the position of the two principal directions corresponding to the cross-section system Oyz an additional point Z’, the reflection of the point Z in reference to axis , has to be constructed. The lines Z’P1 and Z’P2 represent the principal direction 1 (associated with the maximum moment of inertia) and 2 (associated with the minimum moment of inertia), respectively. The two directions can then be transcribed on the cross-section sketch.

Discussion regarding the correlation between the calculated principal directions

and the Mohr’s circle representation Four cases can be identified. They are as follows:

(a) if zy II and 0yzI then:

0)2tan( 0

180290 0 is located in the second quadrant 9045 01 is located in first quadrant and 0)tan( 01

180135 02 is located in second quadrant and 0)tan( 02

conclusion: the angle 02 verifies the relation (2.20) and represents the angle of the

principal direction. The graphical representation of the principal directions in both planes, Oyz and O , is shown in Figure 2.1.7.

(a) (b)

P

P

I

I

I

Dir 1

I,( ) I( ,I )YZ'

Dir 2

= 0

20 C

+z

-z

180°

+y

90°

45°

315°125°

135°

0°

-y

270°

Dir 1

Dir 2

)IZ( I,-I-

I

Figure 2.1.7

(b) if zy II and 0yzI then:

0)2tan( 0

9020 0 is located in the first quadrant

- 45 -

450 01 is located in first quadrant and 0)tan( 01 13590 02 is located in second quadrant and 0)tan( 02



Z' I,

I

P

Dir 1

(Y ,II )

(a) (b)

270°z-315°125°

-

180°y

135°

90°z+

45°

+

0°y

Dir 1

Dir 2

P

Dir 2

-I

0

I IZ -,(I

I

)

C 2

)I(

Figure 2.1.8

(c) if zy II and 0yzI then:

0)2tan( 0


13590 02 is located in second quadrant and 0)tan( 02



Dir 2

Dir 1

0

I-

P

I

Dir 2

2CI

),(Z I -I

I

(Y ,I I

P

)

Dir 1

Z'(I I, )

(a) (b)

270°z-315°125°

-

180°y

135°

90°z+

45°

+

0°y

Figure 2.1.9

- 46 -

(d) if zy II and 0yzI then:

0)2tan( 0


180135 02 is located in second quadrant than 0)tan( 02



Dir 2

Dir 1Dir 1

(

Dir 2

0

-I

P

I

Y = 0

I

,I I )

C

Z I(

2

I

I-,

I

)

P

Z' I( ,

)

(a) (b)

270°z-315°125°

-

180°y

135°

90°z+

45°

+

0°y

Figure 2.1.10

Radii of Gyration The radii of gyration relative to the original coordinate system yz0 are calculated as:

AI

r yy (2.23)

AIr z

z (2.24)

For the case of the principal moments of inertia, the corresponding radii of gyration are:

AI

r pp

11 (2.25)

AI

r pp

22 (2.26)

- 47 -

dx

dy

)(1 xfy

2 )y f (x

y

xa b

0

Mathematical Observations Concerning the Evaluation of the Surface Integrals The equations (2.1) through (2.9) require, for the practical cases, the evaluation of the surface integral. These equations can be replaced by the following general equation:

dAzyfIA

),( (2.27)

If the function ),( zyf is continuous on the rectangular domain D in plane Oyz , based on Fubini’s theorem the surface integral is expressed as a double integral:

D

dzdyzyfI **),( (2.28)

The domain D , as shown in Figure 2.1.11, is bounded above and below by the curves )(1 xfy and )(2 xfy , respectively, and by ax to the left and bx to the right.

In practice, the double integral is calculated using the method of iterated integrals as:

b

a

xf

xf

dxdyyxfI)(

)(

2

1

*)*),(( (2.29)

Figure 2.1.11

In general the domain D is not of rectangular shape and the double integral (2.28) is easier to be evaluated using the natural coordinate system Ouv of the curves describing the boundary. Considering the transformation:

),( vux (2.30)

),( vuz (2.31) the double integral can be expressed as:

T

dvduvuDzyDvuvufI **),(

),(*)),(),,(( (2.31)

where ),(

),(

vuDzyD

is the Jacobian of the transformation.

- 48 -

uz

vy

vz

uy

vz

uz

vy

uy

vuDzyD

**),(

),( (2.32)

and T is the domain mapped from D by the transformations (2.30) and (2.31). The same method of iterated integrals is used to evaluate the double integral (2.31). The advantage of using the variables pertinent to the natural coordinate system is that the integral separates into two independent integral with constant limits, facilitating the integration process.

2.2. Solved Problems 2.2.1 Cross-Sections with Analytical Described Boundary The geometrical characteristics of the cross-section are obtained by direct integration of the equations (2.1) through (2.26). Problem 2.2.1.1 Rectangular cross-section (Figure 2.2.1)

z'

y'0

C y

z

h= 1

0 m

h2

h2

b 2 b 2

b= 4 m

ydy

zdz

dA=dy dz.

Figure 2.2.1 Rectangular Cross-Section

A. General Observations A.1 The coordinate system used is the centroidal coordinate system Cyz . A.2 The rectangle is characterized by a double symmetry against the axes of the

centroidal coordinate systemCyz . Consequently, the axes of the centroidal coordinate system identify with the principal directions. This conclusion is verified.

- 49 -

A.3 For numerical application of the formulae the following global dimensions describing the rectangular shape are used:

mb 4 mh 10 B. Calculations B.1 Step 1 - calculation of the cross-sectional area

A b h( )b

2

b2

zh

2

h2

y1

d

d b h

A b h( ) substitute b 4 m h 10 m 40 m2 B.2 Step 2 - calculation of the cross-sectional static moments Note: The static moments are calculated in reference to the centroidal coordinate

system.

Szc b h( )h

2

h

2

zb

2

b

2

yy

d

d 0

Syc b h( )b

2

b

2

yh

2

h

2

zz

d

d 0

B.3 Step 3 - calculation of the cross-sectional moments of inertia

Iyc b h( )b

2

b

2

yh

2

h

2

zz2

d

d1

12b h3

Iyc b h( ) substitute b 4 m h 10 m1000

3m4 float 4 333.3 m4

Izc b h( )b

2

b2

yh

2

h2

zy2

d

d1

12b3 h

Izc b h( ) substitute b 4 m h 10 m160

3m4 float 4 53.33 m4

Iyzc b h( )b

2

b2

yh

2

h2

zy z

d

d 0

- 50 -

B.4 Step 4 - calculation of the cross-sectional radii of gyration

ry b h( )Iyc b h( )

A b h( )

1

63 h2

1

2

simplify

assume b 0 h 01

63 h

ry b h( ) substitute b 4 m h 10 m5

33 m float 4 2.887 m

rz b h( )Izc b h( )

A b h( )

1

63 b2

1

2

simplify

assume b 0 h 01

63 b

rz b h( ) substitute b 4 m h 10 m2

33 m float 4 1.155 m

Note: For a rectangular cross-section the product of inertia calculated about the

centroidal coordinate system is zero 0yzcI and consequently, the centroidal

coordinate system axes coincide with the principal axes of inertia. If the moments of inertia about axes of a cartesian coordinate system Oyz parallel with the centroidal coordinate system axes and passing through the lower corner of the rectangle are required, the parallel-axis theorem for moments of inertia must be employed in the calculation:

Iy b h( ) Iyc b h( ) A b h( )h2

2

1

3b h3

Iy b h( ) substitute b 4 m h 10 m4000

3m4 float 4 1333. m4

Iz b h( ) Izc b h( ) A b h( )b2

2

1

3b3 h

Iz b h( ) substitute b 4 m h 10 m640

3m4 float 4 213.3 m4

Iyz b h( ) Iyzc b h( ) A b h( )h2

b2

1

4b2 h2

Iyz b h( ) substitute b 4 m h 10 m 400 m4

Problem 2.2.1.2 Tubular Cross-Section (Figure 2.2.2) A. General Observations A.1 The coordinate system used is the centroidal coordinate system Cyz . A.2 The circle is characterized by a double symmetry against the axes of the


- 51 -

A.3 For numerical application: mRa 2 mRb 5

C y

z

dA=

d

d d d

R = 2 mR =

5 m

b

a

y

z

Figure 2.2.2 Tubular Cross-Section

B. Calculations The parametric representation of a circle is: y cos z sin The Jacobian of the transformation is calculated as:

D y d

d

z d

d

y d

d

z d

d

simplifytrig

B.1 Step 1 - calculation of the cross-sectional area

A Ra

Rb

0

2

D

d

d collect Rb2 Ra

2

A substitute Ra 2 m Rb 5 m

float 465.98 m2

B.2 Step 2 - calculation of the cross-sectional moments of inertia.

Iyc Ra

Rb

0

2

z 2D

d

dcollect

simplify1

4Rb

4 Ra4

- 52 -

Iyc substitute Ra 2 m Rb 5 m

float 4478.5 m4

Izc Ra

Rb

0

2

y 2D

d

dcollect

simplify1

4Rb

4 Ra4

Izc substitute Ra 2 m Rb 5 m

float 4478.5 m4

Iyzc Ra

Rb

0

2

y z D

d

dcollect

simplify0

Ic Ra

Rb

0

2

2D

d

dcollect

simplify1

2Rb

4 Ra4

Ic substitute Ra 2 m Rb 5 m

float 4956.7 m4


ryc Iyc A

simplify

assume Ra 0 Rb 01

2Ra

2 Rb2

1

2

ryc

substitute Ra 2 m Rb 5 m

assume m 0

simplify

float 4

2.693 m

rzc Izc A

simplify

assume Ra 0 Rb 01

2Ra

2 Rb2

1

2

rzc

substitute Ra 2 m Rb 5 m

simplify

assume m 0

float 4

2.693 m

Note: For a circular cross-section the product of inertia calculated about the centroidal coordinate system is zero 0yzcI and consequently, the centroidal

coordinate system axes coincide with the principal axes of inertia. This conclusion is misleading. In fact, one can easily observe that the

0

0)2tan( 0 and consequently, the principal directions are undetermined. It

is concluded that any pair of orthogonal diameters can identify themselves with the principal directions.

- 53 -

The geometrical characteristics of the solid circular cross-section, where 0aR ,

are calculated in a similar manner by following the steps described above. The only difference is the changing of the limits corresponding to variable appearing in the double integrals or by directly replacing them in above calculated expressions. Consequently, the following geometrical characteristics are obtained:

A substitute Ra 0 m Rb 5 m

float 478.55 m2

Ic substitute Ra 0 m Rb 5 m

float 4981.9 m4

Iyc substitute Ra 0 m Rb 5 m

float 4491.1 m4

Izc substitute Ra 0 m Rb 5 m

float 4491.1 m4

r yc

substitute R a 0 m R b 5 m

simplify

assume m 0

float 4

2.500 m

r zc

substitute R a 0 m R b 5 m

assume m 0

simplify

float 4

2.500 m

Problem 2.2.1.3 Elliptical Cross-Section (Figure 2.2.3)

a = 5 ma = 5 m

b =

10

mb

= 1

0 m

C

z

y

Figure 2.2.3

- 54 -

A. General Observations A.1 The coordinate system used is the centroidal coordinate system Cyz . A.2 The ellipse is characterized by a double symmetry against the axes of the


A.3 For numerical application:

ma 5 mb 10 B. Calculations The parametric representation of an ellipse is: y a cos assume a 0 a cos z b sin assume b 0 b sin

The Jacobian of the transformation is calculated as:

D y d

d

z d

d

y d

d

z d

d

simplifytrig a b

B.1 Step 1 - calculation of the cross-sectional area

A 0

1

0

2

D

d

d collect a b

A substitute a 5 m b 10 m

float 4157.1 m2

B.2 Step 2 - calculation of the cross-sectional moments of inertia.

Iy 0

1

0

2

z 2D

d

dcollect

simplify1

4 b3 a

Iy substitute a 5 m b 10 m

float 43928. m4

Iz 0

1

0

2

y 2D

d

dcollect

simplify1

4 a3 b

- 55 -

Iz substitute a 5 m b 10 m

float 4981.9 m4

Iyz 0

1

0

2

y z D

d

dcollect

simplify0

I 0

1

0

2

y 2z 2 D

d

dcollect

simplify1

4 a b a2 b2

I substitute a 5 m b 10 m

float 44911. m4


ry Iy A

simplify

assume a 0 b 01

2b

ry substitute a 5 m b 10 m

float 2

assume m 0

5. m

rz Iz A

simplify

assume a 0 b 01

2a

rz substitute a 5 m b 10 m

float 2

assume m 0

2.5 m

Note: For an elliptical cross-section the product of inertia calculated about the

centroidal coordinate system is zero 0yzI and consequently, the centroidal

coordinate system axes coincide with the principal axes of inertia.

a = 5 m

b =

10

m

C y

z

z

y

0 y

z

Figure 2.2.4

- 56 -

The geometrical characteristics of a quarter elliptical cross-sections, shown in Figure 2.2.4, are calculated in a similar manner with these corresponding to a solid elliptical cross-section. The only notable difference is the limits corresponding to variable appearing in the double integrals. The following geometrical characteristics are obtained:

Aq 0

1

0

2

D

d

d collect 1

4 a b

Aq

substitute a 5 m b 10 m

float 439.28 m2

Sz_q 0

1

0

2

y D

d

dcollect

simplify1

3a2 b

Sz_q substitute a 5 m b 10 m

float 483.33 m3

Sy_q 0

1

0

2

z D

d

dcollect

simplify1

3b2 a

Sy_q substitute a 5 m b 10 m

float 4166.7 m3

yc_q Sz_q Aq

4

3

a

yc_q substitute a 5 m b 10 m

float 42.122 m

zc_q Sy_q Aq

4

3

b

zc_q substitute a 5 m b 10 m

float 44.243 m

Iy_q 0

1

0

2

z 2D

d

dcollect

simplify1

16 b3 a

Iy_q substitute a 5 m b 10 m

float 4981.9 m4

Iz_q 0

1

0

2

y 2D

d

dcollect

simplify1

16 a3 b

- 57 -

Iz_q substitute a 5 m b 10 m

float 4245.5 m4

Iyc_q Iy_q Aq zc_q 2 simplify1

144a b3

9 2 64

Iyc_q substitute a 5 m b 10 m

float 4274.6 m4

Izc_q Iz_q Aq yc_q 2 simplify1

144a3 b

9 2 64

Izc_q substitute a 5 m b 10 m

float 468.66 m4

ryc_q Iyc_q

A simplify

assume a 0 b 01

12 9 2 64 1

2

b

ryc_q substitute a 5 m b 10 m

float 2

assume m 0

1.3 m

rzc_q Izc_q

A simplify

assume a 0 b 01

12 9 2 64 1

2

a

rzc_q substitute a 5 m b 10 m

float 2

assume m 0

.61 m

2.2.2. Composite Cross-Sections Problem 2.2.2.1 L Shaped Cross-Section (Figure 2.2.5.a) A. General Observations A1. The "L" shaped area, shown in Figure 2.2.5.a, is subdivided into two

rectangular areas called SHAPE1 and SHAPE2 (Figure 2.2.5.b). The geometrical characteristics of the two rectangular areas are calculated using the formulae previously obtained. The decomposition of the original area into those two particular rectangular areas and the usage of the coordinate system pictured are arbitrary and any other alternative can be employed.

nshapes 2

number of rectangular areas considered

A2. The Cartesian orthogonal coordinate system 0y

0z

0 is used as the original

reference coordinate system.

- 58 -

z

y

z

y

z

y

Figure 2.2.5

A.3. For numerical application:

a 200 mm b 100 mm tv 20 mm th 20 mm

B. Calculations B.1 Step 1 - collecting data pertinent to individual rectangular areas: Data pertinent to area SHAPE1 Note: The local coordinate system 0

1y

1z

1 is used.

az

1a th

20200 az1

180 mm

ay1

tv ay120 mm

dimensions

z01

az1

2th 20

2

180z01

110 mm

centroid O1

y01

ay1

2

2

20y01

10 mm

A1 az1

ay1

20*180 A1 3.6 103 mm

2 area

Iy1

az1

3 ay1

12

12

20*1803

Iy1

9.72 106 mm

4

moments of inertia

Iz1

az1

ay1

3

12

12

20*180 3

Iz11.2 10

5 mm4

Iyz10 mm

4 product of inertia

- 59 -

Data pertinent to area SHAPE2 Note: The local coordinate system 0

2y

2z

2 is used.

az

2th az2

20 mm

ay2

b ay2100 mm

dimensions

z02

az2

2

2

20z02

10 mm

y02

ay2

2

2

100y02

50 mm centroid O2

A2 az2

ay2

100*20 A2 2 103 mm

2

area

Iy2

az2

3 ay2

12

12

100*203

Iy2

6.667 104 mm

4

moments of inertia

Iz2

az2

ay2

3

12

12

100*20 3

Iz2

1.667 106 mm

4

Iyz20 mm

4

product of inertia

B.2 Step 2 - calculation of the cross-section area

Atotal1

nshapes

n

An

20003600 Atotal 5.6 10

3 mm2

total area

B.3 Step 3 - calculation of the cross-section centroid C Note: The global coordinate system 0y

0z

0 is used.

Sz1

nshapes

n

An y0n

)50*2000()10*3600( Sz 1.36 10

5 mm3

Sy1

nshapes

n

An z0n

)10*2000()110*3600( Sy 4.16 10

5 mm3

yc

Sz

Atotal

3

5

10*6.5

10*36.1

yc 24.286mm

position of the centroid C

zc

Sy

Atotal

3

5

10*6.5

10*44.3 zc 74.286 mm

B.4 Step 4 - calculation of the cross-section moments of inertia, polar moment and product of inertia

- 60 -

Note: The centroidal coordinate system Cyczc is used.

Iyc1

nshapes

n

IynAn z0n

zc 2

])286.47(10*10*210*[6.667

])286.47(110*10*3.610*9.72[234

236

Iyc 2.264 107 mm

4

Izc1

nshapes

n

IznAn y0n

yc 2

])286.42(50*10*2104*[1.667

])286.42(10*10*3.610*2.1[23

235

Izc 3.844 106 mm

4

Ipolar Iyc Izc 67 10*844.310*264.2 Ipolar 2.649 10

7 mm4

Iyzc1

nshapes

n

IyznAn y0n

yc z0nzc

)]286.47(10*)286.42(50*10*[2)]286.47(110*)286.42(10*10*3.6[ 33

Iyzc 5.143 106 mm

4

B.5 Step 5 - calculation of Principal Moments and Principal Directions of Inertia Note: The centroidal coordinate system Cyczc is used.

Ip1

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

2626767

)10*142.5(4

)10*844.310*264.2(

2

10*844.310*264.2

Imax Ip1

Imax 2.396 107 mm

4

Ip2

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

2626767

)10*142.5(4

)10*844.310*264.2(

2

10*844.310*264.2

Imin Ip2

Imin 2.529 106 mm

4

0_112

atan2 Iyzc

Iyc Izc

]

10*844.310*264.2

)10*142.5(*2[tan

2

167

61

0_1 14.342 deg 0255.0)342.14tan(

- 61 -

0_2 0_12

90342.14 0_2 104.342deg

0911.3)342.104tan(

angle corresponding to maximum principal moment of inertia

01 14.342 deg

angle corresponding to minimum principal moment of inertia 02 104.342deg

z

y

y

z

y

z

Dir 1

Dir 2

Figure 2.2.6

B.6 Step 6 - calculation of the cross-section radii of gyration Note: Two sets of radii of gyrations are calculated: first set is related to the

centroidal moments of inertia, while the second involves the principal moments of inertia.

ryc

Iyc

Atotal

3

7

10*6.5

10*264.2ryc 63.589mm

radii of gyration

rzc

Izc

Atotal

3

6

10*6.5

10*844.3rzc 26.199mm

rp1

Ip1

Atotal

3

7

10*6.5

10*396.2rp1 65.409mm

01 0_1

tan 0_1 Iyzc

0if

0_2

tan 0_2 Iyzc

0if

02 0_1 01 0_2if

0_2 01 0_1if

- 62 -

rp2

Ip2

Atotal

3

6

10*6.5

10*529.2rp2 21.251 mm

B.7 Step 7 - construction of the Mohr's circle (Figure 2.2.7)

point Y ( Iyc 2.264 107 mm

4 , Iyzc 5.143 106 mm

4 )

point Z ( Izc 3.844 106 mm

4 , Iyzc 5.143 106 mm

4 )

point Z’ ( Izc 3.844 106 mm

4 , Iyzc 5.143 106 mm

4 )

P P

I

I

IDir 1

I,-(Z )

I( , I )YZ'

Dir 2

= 0

20 C

Figure 2.2.7

B.8 Step 8 - variation of the centroidal moments of inertia (Figure 2.2.8) We consider that the axes rotates with an angle i between 0 to

i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc

2cos 2 i Iyzc sin 2 i

Izri

Iyc Izc

2

Iyc Izc


Iyzri

Iyc Izc

2sin 2 i Iyzc cos 2 i

- 63 -

0 30 60 90 120 150 180

2 107

7.5 106

5 106

1.75 107

3 107

0

Iyri

Izri

Iyzri

180

180

i

Figure 2.2.8

Problem 2.2.2.2 C Shaped Cross-Section (Figure 2.2.9.a)

y

z

y

z

y

zy

z

Figure 2.2.9 A. General Observations A1. The "C" shaped area, shown in Figure 2.2.9.a, is subdivided into three

rectangular areas called SHAPE1, SHAPE2 and SHAPE3 (Figure 2.2.9.b). The geometrical characteristics of the rectangular area are expressed using the previously obtained formulae. nshapes 3

number of rectangular areas considered

- 64 -

A2. The cartesian orthogonal coordinate system 0y0z0 is used as the original

reference coordinate system. A3. For numerical application:

a 200 mm b1 80 mm

b2 100 mm

tv 10 mm

th1 10 mm

th2 20 mm


1y

1z

1 is used.

az

1th1 az1

10 mm

ay1

b1 tv ay1

70 mm

dimensions

z01

ath12

2

10200 z01

195 mm

centroid 01

y01

ay1

2tv 10

2

70y01

45 mm

A1 az1

ay1

70*10 A1 700 mm2

area

Iy1

az1

3 ay1

12

12

70*103

Iy15.833 10

3 mm4

moments of inertia

Iz1

az1

ay1

3

12

12

70*10 3

Iz12.858 10

5 mm4

Iyz10 mm

4

product of inertia


2y

2z

2 is used.

az

2a az2

200 mm

ay2

tv ay210 mm

dimensions

z02

az2

2

2

200

z02

100 mm

y02

ay2

2

2

10y02

5 mm

centroid O2

A2 az2

ay2

10*200 A2 2 103 mm

2

area

- 65 -

Iy2

az2

3 ay2

12

12

10*2003

Iy26.667 10

6 mm4

moments of inertia

Iz2

az2

ay2

3

12

12

10*200 3

Iz21.667 10

4 mm4

Iyz20 mm

4

product of inertia

Data pertinent to area SHAPE3 Note: The local coordinate system O

3y

3z

3 is used.

az

3th2 az3

20mm

ay3

b2 tv ay3

90 mm

dimensions

z03

az3

2

2

20z03

10mm

y03

ay3

2tv 10

2

90y03

55mm

centroid O3

A3 az3

ay3

90*20 A3 1.8 103 mm

2

area

Iy3

az3

3 ay3

12

12

90*203

Iy36 10

4 mm4

moments of inertia

Iz3

az3

ay3

3

12

12

90*20 3

Iz31.215 10

6 mm4

Iyz30 mm

4

product of inertia


Atotal1

nshapes

n

An

33 10*8.110*2700 Atotal 4.5 103 mm

2 total area

B.3 Step 3 - calculation of the cross-section centroid C Note: the general coordinate system 0y

0z

0 is used.

Sz1

nshapes

n

An y0n

)55*10*8.1()5*10*2()45*700( 33

Sz 1.405 10

5 mm3

- 66 -

Sy1

nshapes

n

An z0n

)10*10*8.1()100*10*2()195*700( 33

Sy 3.545 10

5 mm3

yc

Sz

Atotal

3

5

10*5.4

10*405.1yc 31.222mm

centroid C

zc

Sy

Atotal

3

5

10*5.4

10*545.3zc 78.778 mm

B.4 Step 4 - calculation of the cross-section moments of inertia, polar moment and

product of inertia Note: The centroidal coordinate system Cyczc is used.

Iyc1

nshapes

n

IynAn z0n

zc 2

])778.87(10*10*8.110*[6

])778.87(100*10*210*[6.667

])778.87(195*00710*833.5[

234

236

23

Iyc 2.56 107 mm

4

Izc1

nshapes

n

IznAn y0n

yc 2

])222.31(55*10*8.110*[1.215

])222.31(5*10*210*[1.667

])222.31(45*00710*858.2[

236

234

25

Izc 4.043 106 mm

4

Ipolar Iyc Izc 67 10*043.410*56.2 Ipolar 2.965 107 mm

4

Iyzc1

nshapes

n

IyznAn y0n

yc z0nzc

)]778.87(10*)222.31(55*10*8.1[

)]778.87(100*)222.31(5*10*[2)]778.87(195*)222.31(45*007[3

3

Iyzc 2.936 106 mm

4

B.5. Step 5 - calculation of the cross-section principal moments of inertia Note: The centroidal coordinate system Cyczc is used.

- 67 -

Ip1

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

2626767

)10*936.2(4

)10*043.410*56.2(

2

10*043.410*56.2

Imax Ip1

Imax 2.6 107 mm

4

Ip2

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

2626767

)10*936.2(4

)10*043.410*56.2(

2

10*043.410*56.2

Imin Ip2

Imin 3.651 106 mm

4

0_112

atan2 Iyzc

Iyc Izc

]

10*043.410*56.2

)10*936.2(*2[tan

2

167

61

0_1 7.617 deg

01337.0)617.7tan(

0_2 0_12

90617.7 0_2 97.617 deg

0478.7)617.97tan(


01 7.617 deg

angle corresponding to minimum principal moment of inertia 02 97.617 deg

B.6 Step 6 - calculation of the cross-section radii of gyration Note: Two sets of radii of gyrations are calculated. The first set is related to the

centroidal moments of inertia, while the second set involves the principal moments of inertia.

ryc

Iyc

Atotal

3

7

10*5.4

10*56.2ryc 75.43mm

radii of gyration

rzc

Izc

Atotal

3

6

10*5.4

10*043.4rzc 29.975mm

01 0_1

tan 0_1 Iyzc

0if

0_2

tan 0_2 Iyzc

0if

02 0_1 01 0_2if

0_2 01 0_1if

- 68 -

rp1

Ip1

Atotal

3

7

10*5.4

10*6.2rp1 76.006mm

rp2

Ip2

Atotal

3

6

10*5.4

10*651.3rp2 28.483mm

z

y

y

z

z

y

Dir 1

Dir 2

Figure 2.2.10



4 , Iyzc 2.936 106 mm

4 )


4 , Iyzc 2.936 106 mm

4 )


4 , Iyzc 2.936 106 mm

4 )

(Y ,I I

I

= 0)

-,I I )Z(

Z'

PP

20Dir 1

Dir 2

I C

Figure 2.2.11

- 69 -


i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc


Izri

Iyc Izc

2

Iyc Izc


Iyzri

Iyc Izc


0 30 60 90 120 150 180

2 107

7.5 106

5 106

1.75 107

3 107

0

Iyri

Izri

Iyzri

0 180

180

i

Figure 2.2.12 Problem 2.2.2.3 Z-Shaped Cross-Section (Figure 2.2.13.a) A. General Observations A1. The "Z" shaped area, shown in Figure 2.2.13.a, is subdivided into three

rectangular areas called SHAPE1, SHAPE2 and SHAPE3 (Figure 2.2.13.b). The geometrical characteristics of the rectangular are calculated using the previously obtained formulae. nshapes 3

number of rectangular areas

A2. The Cartesian orthogonal coordinate system 0y

0z

0 is used as the original


- 70 -

yz

yz

yz

y

z

Figure 2.2.13

A3. For numerical application:

a 400 mm b1 100 mm

b2 150 mm

tv 10 mm

th1 20 mm

th2 30 mm


1y

1z

1 is used.

az1

th1 az120 mm

ay1

b1 ay1100mm

dimensions

z01a

th1

2

2

20400 z01

390mm centroid O1

y01

ay1

2tv

10

2

100y01

40 mm

A1 az1ay1 100*20 A1 2 10

3 mm2 area

Iy1

az1 3 ay1

12

12

100*203

Iy16.667 10

4 mm4

Iz1

az1ay1 3

12

12

100*20 3

Iz11.667 10

6 mm4

moments of inertia

Iyz10 mm

4

product of inertia

- 71 -


2y

2z

2 is used.

az2

a th1 th2 3020400 az2350mm

dimensions

ay2

tv ay210mm

z02

az2

2th2 30

2

350z02

205mm

y02

ay2

2

2

10y02

5 mm

centroid O

2

A2 az2ay2 10*350 A2 3.5 10

3 mm2

area

Iy2

az2 3 ay2

12

12

10*3503

Iy23.573 10

7 mm4

moments of inertia

Iz2

az2ay2 3

12

12

10*350 3

Iz22.917 10

4 mm4

Iyz20 mm

4

product of inertia

Data pertinent to area SHAPE3 Note: The local coordinate system O

3y

3z

3 is used.

az3

th2 az330 mm

ay3

b2 ay3150mm

dimensions

z03

az3

2

2

30z03

15 mm

y03

ay3

2

2

150

y03

75 mm centroid O

3

A3 az3ay3 150*30 A3 4.5 10

3 mm2

area

Iy3

az3 3 ay3

12

12

150*303

Iy33.375 10

5 mm4

Iz3

az3ay3 3

12

12

150*30 3

Iz38.437 10

6 mm4

moments of inertia

Iyz30 mm

4

product of inertia


Atotal1

nshapes

n

An

333 10*5.410*5.310*2 Atotal 1 104 mm

2

- 72 -

B.3 Step 3 - calculation of the cross-section centroid C Note: The local general coordinate system 0y

0z

0 is used.

Sz1

nshapes

n

An y0n

)75*10*5.4()5*10*5.3())40(*10*2( 333

Sz 2.75 10

5 mm3

Sy

1

nshapes

n

An z0n

)15*10*5.4()205*10*5.3()390*10*2( 333

Sy 1.565 106 mm

3

yc

Sz

Atotal

4

5

10

10*75.2yc 27.5 mm

centroid C

zc

Sy

Atotal

4

6

10

10*565.1zc 156.5 mm

B.4 Step 4 - calculation of the cross-section moments of inertia, polar moment and

product of inertia Note: The centroidal coordinate system Cyczc is used.

Iyc1

nshapes

n

IynAn z0n

zc 2

])5.156(15*10*5.410*[3.375

])5.156(205*10*5.310*[3.573

])5.156(390*10*210*667.6[

235

237

234

Iyc 2.435 108 mm

4

Izc1

nshapes

n

IznAn y0n

yc 2

])5.27(75*10*5.410*[8.437

])5.27(5*10*5.310*[2.917

])5.27(-40*10*210*667.1[

236

234

236

Izc 3.117 107 mm

4


4

Iyzc1

nshapes

n

IyznAn y0n

yc z0nzc

)]5.156(15*)5.27(75*10*5.4[

)]5.156(205*)5.27(5*10*5.[3)]5.156(390*)5.27(-40*10*2[3

33

- 73 -

Iyzc 6.559 107 mm

4

B.5 Step 5 - calculation of the cross-section principal moments of inertia Note: The centroidal coordinate system Cyczc located is used.

Ip1

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

2727878

)10*559.6(4

)10*117.310*435.2(

2

10*117.310*435.2

Imax Ip1

Imax 2.621 108 mm

4

Ip2

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

2727878

)10*559.6(4

)10*117.310*435.2(

2

10*117.310*435.2

Imin Ip2

Imin 1.255 107 mm

4

0_112

atan2 Iyzc

Iyc Izc

]

10*117.310*435.2

)10*559.6(*2[tan

2

178

71

0_1 15.853 deg

0283.0)853.15tan(

0_2 0_12

90853.15 0_2 105.853 deg

0521.3)853.105tan(


01 15.853 deg


B.6 Step 6 - calculation of the cross-section radii of gyration Note: Two sets of radii of gyrations are calculated. The first set is related to the centroidal moments of inertia, while the second set involves the principal moments of inertia.

ryc

Iyc

Atotal

4

8

10

10*435.2ryc 156.048mm

01 0_1

tan 0_1 Iyzc

0if

0_2

tan 0_2 Iyzc

0if

02 0_1 01 0_2if

0_2 01 0_1if

- 74 -

rzc

Izc

Atotal

4

7

10

10*117.3rzc 55.831 mm

rp1

Ip1

Atotal

4

8

10

10*621.2rp1 161.906 mm

rp2

Ip2

Atotal

4

7

10

10*255.1rp2 35.42 mm

y

y

z

y

z

z

Dir 2

Dir 1

Figure 2.2.14

B.7 Step 7.- construction of the Mohr's circle (Figure 2.2.15)


4 , Iyzc 6.559 107 mm

4 )


4 , Iyzc 6.559 107 mm

4 )


4 , Iyzc 6.559 107 mm

4 )

0

Z'

(

I

Z -I , I

C

)

2

Y I( I,= 0

)

IP

PDir 1

Dir 2

P

Figure 2.2.15

- 75 -


i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc


Izri

Iyc Izc

2

Iyc Izc


Iyzri

Iyc Izc


0 30 60 90 120 150 180

2 108

7.5 107

5 107

1.75 108

3 108

0

Iyri

Izri

Iyzri

180

180

i

Figure 2.2.16

Problem 2.2.2.4 Composite cross-section using plane areas (Figure 2.2.17.a) A. General Observations A1. The cross-sectional area, shown in Figure 2.2.17.a, is subdivided into one rectangular areas called SHAPE1, from which it will be deducted the following areas: one triangular area called SHAPE2 and two circular areas called SHAPE3 and SHAPE4 (Figure 2.2.17.b). The geometrical characteristics of these areas can be easily expressed using the formulae from Appendix 2.1. nshapes 4

A2. The cartesian orthogonal coordinate system 0y0z0 is used as the original


- 76 -

y

z

zy

y

z

y

z

y

z

Figure 2.2.17

A.3. The following dimensions are used in the calculation: a 400 mm b 300 mm c 130 mm f 170 mm d1 20 mm

dy1 80 mm

dz1 80 mm

d2 20 mm

dy2 80 mm

dz2 80 mm

B. Calculations B.1 Step 1 - collecting data pertinent to each one of the individual rectangular areas Data pertinent to rectangular area SHAPE1 Note: the local coordinate system 01y1z1 located at the centroid 01 of the area is used

az1

b mm 300za1

ay1

a mm 004a1y

global dimensions

z01

az1

2

2

300

z0

1150mm

centroid 01

y01

ay1

2

2

400

y0

1200mm

A1 az1

ay1

400*300 A1 1.2 105 mm

2

area

Iy1

az1

3 ay1

12

12

400*3003

Iy19 10

8 mm4

moments of inertia

Iz1

az1

ay1

3

12

12

400*300 3

Iz11.6 10

9 mm4

- 77 -

Iyz10 mm

4

product of inertia

Data pertinent to triangular area SHAPE2 Note: - the local coordinate system 02y2z2 is located at the centroid 02 of the area;

az

2b f 170300 az2

130 mm

ay2

a c 130400 ay2270 mm

dimensions

z02

baz

2

3

3

130300 z02

256.667 mm

y02

aay

2

3

3

270400 y02

310 mm

centroid O2

A21

2az

2 ay

2 270*130*

2

1A2 1.755 10

4 mm2

area

Iy2

az2

3 ay2

36

36

270*1303

Iy21.648 10

7 mm4

moments of inertia

Iz2

az2

ay2

3

36

36

270*130 3

Iz27.108 10

7 mm4

Iyz2

az2 2 ay2 2

72

72

270*130 22

Iyz21.711 10

7 mm4

product of inertia

Data pertinent to circular area SHAPE3 Note: the local coordinate system 0

3y

3z

3 located at the centroid 03 of the area is used

r1

d12

mm 01r2

201

dimension

z03

dz1 z0380 mm

centroid O3

y0

3dy1

y03

80 mm

A3 r12 210*π A3 314.159 mm

2

area

Iy3

r14

4

4

10*π 4

Iy37.854 10

3 mm4

moments of inertia

Iy3

r14

4

4

10*π 4

Iz37.854 10

3 mm4

- 78 -

Iyz30 mm

4

product of inertia

Data pertinent to circular area SHAPE4 Note: the local coordinate system 0

4y

4z

4 located at the centroid 04 of the area is used

r2

d22

mm 012

r2

20

dimension

z04

dz2 z0480 mm

centroid O4

y0

4a dy2

80400 y04

320 mm

A4 r22 210*π A4 314.159 mm

2

area

Iy4

r24

4

4

10*π 4

Iy47.854 10

3 mm4 moments of inertia

Iz4

r24

4

4

10*π 4

Iz47.854 10

3 mm4

Iyz40 mm

4

product of inertia


Atotal A1

2

nshapes

n

An

159.314159.31410*755.110*2.1 45

Atotal 1.018 105 mm

2 total area

B.3 Step 3 - calculation of the cross-section centroid C Note: the local general coordinate system 0y0z0 located at point 0 of the cross-section

is used

Sz A1 y01

2

nshapes

n

An y0n

320*159.31480*159.314310*10*755.1200*10*2.1 45

Sz 1.843 10

7 mm3

Sy A1 z01

2

nshapes

n

An z0n

80*159.31480*159.314667.256*10*755.1150*10*2.1 45

- 79 -

Sy 1.345 107 mm

3

static moments

yc

Sz

Atotal

5

7

10*018.1

10*843.1yc 181.04 mm centroid C

zc

Sy

Atotal

5

7

10*018.1

10*345.1zc 132.047mm

B.4 Step 4 - calculation of the cross-section moments of inertia Note: - the local coordinate system Cyczc located at the centroid C of the area is used;

moments of inertia

Iyc Iy1

A1 z01zc 2

2

nshapes

n

Iyn 2

nshapes

n

An z0nzc 2

2047.13280*159.3142047.13280*159.3142047.132667.256*410*755.1

310*854.7310*854.7710*648.12047.132150*510*2.1810*9

Iyc 6.479 10

8 mm4

Izc Iz1A1 y01

yc 2

2

nshapes

n

Izn 2

nshapes

n

An y0nyc 2

204.181320*159.314204.18180*159.314204.181310*410*755.1

310*854.7310*854.7710*108.7204.181200*510*2.1910*6.1

Izc 1.271 109 mm

4


4

polar moment of inertia product of inertia

Iyzc Iyz1A1 y01

yc z01zc

2

nshapes

n

Iyzn 2

nshapes

n

An y0nyc z0n

zc

047.13280*04.181320*159.314

047.13280*04.18180*159.314047.132667.256*04.181310*10*755.1

00710*171.1047.132150*04.181200*510*2.10

4

Iyzc 2.235 10

8 mm4

- 80 -

01 0_1

tan 0_1 Iyzc

0if

0_2

tan 0_2 Iyzc

0if

02 0_1 01 0_2if

0_2 01 0_1if

B.5 Step 5 - calculation of the cross-section principal moments of inertia Note: - the local coordinate system Cyczc located at the centroid C of the area is used;

Ip1

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

28

29898

10*235.24

10*271.110*479.6

2

10*271.110*479.6

Imax Ip1

Imax 1.343 109 mm

4

Ip2

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

28

29898

10*235.24

10*271.110*479.6

2

10*271.110*479.6

Imin Ip2

Imin 5.761 108 mm

4

0_112

atan2 Iyzc

Iyc Izc

98

81

10*271.110*479.6

10*235.2*2tan

2

1

0_1 17.828 deg 032.0)828.17tan(

0_2 0_12

908.19 0_2 72.172 deg

011.3)172.72tan(

angle corresponding to maximum principal moment of inertia 01 72.172 deg

angle corresponding to minimum principal moment of inertia

02 17.828 deg

- 81 -

y

z z

y

y

z

Dir 1

Dir 2

Figure 2.2.18

B.6 Step 6 - calculation of the cross-section radii of gyration Note: - two sets of radii of gyrations are calculated. The first set is related to the centroidal moments of inertia, while the second set involves the principal moments of inertia.

ryc

Iyc

Atotal

5

8

10*018.1

10*479.6 ryc 79.771mm

radii of gyration

rzc

Izc

Atotal

5

9

10*018.1

10*271.1 rzc 111.721 mm

rp1

Ip1

Atotal

5

9

10*018.1

10*343.1rp1 114.837 mm

rp2

Ip2

Atotal

5

8

10*018.1

10*761.5rp2 75.217 mm


point X ( Iyc 6.479 108 mm

4 , Iyzc 2.235 108 mm

4 )

point Y ( Izc 1.271 109 mm

4 , Iyzc 2.235 108 mm

4 )


4 , Iyzc 2.235 108 mm

4 )

- 82 -

Dir 2

C0

Z'

P

I

Z )( -, II

2

Y I,(I

P

I

)

= 0

Dir 1

Figure 2.2.19


i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc


Izri

Iyc Izc

2

Iyc Izc


Iyzri

Iyc Izc


0 30 60 90 120 150 180

5 108

5 108

1 109

1.5 109

0

Iyri

Izri

Iyzri

180

180

i

Figure 2.2.20

- 83 -

Problem 2.2.2.5 Cross-section made from plates and rolled shapes (Figure 2.2.21.a)

25130 xPl

36I

y

z

h

b

y

z

z

y

Figure 2.2.21

A. General Observations A1. The cross-section, shown in Figure 2.2.21.a, is composed from two structural shapes, an I36 and a plate 130mmx25mm, called SHAPE1 and SHAPE2 (Figure 2.2.21.b). The geometrical characteristics of the I36 are obtained from Appendix 2.2, Table 2.2.1, while those pertinent to the plate, which is a rectangular shape, are calculated. A.2. The vertical axis of the cross-section is a symmetry axis and consequently, the axes of the centroidal coordinate system identify with the principal directions. A3. The cartesian orthogonal coordinate system 0y0z0 is used as the original

reference coordinate system. B. Calculations B.1 Step 1 - collecting data pertinent to each one of the individual rectangular areas Data pertinent to I36 (area SHAPE1) Note: the local coordinate system 0

1y

1z

1 located at the centroid 0

1 of the area is used

h1 360 mm depth

b1 143 mm flange width

z01

0 mm y010 mm

centroid 01

A1 97.1 cm2

area

Iy119610 cm

4

Iz1818 cm

4

moments of inertia

- 84 -

Iyz10 mm

4

product moment of inertia

Data pertinent to plate 130mm x 25mm (area SHAPE2) Note: - the local coordinate system 0

2y

2z

2 is located at the centroid 02 of the area;

az2

25 mm

ay2130 mm

global dimensions

z02

h1

2

az2

2

2

25

2

360z02

192.5 mm

centroid 02

y02

0

A2 az2ay2

130*25

A2 3.25 10

3 mm2

area

Iy2

az2 3 ay2

12

12

130*253

Iy21.693 10

5 mm4 moments of inertia

Iz2

az2ay2 3

12

12

130*25 3

Iz24.577 10

6 mm4

Iyz20 mm

4



Atotal1

nshapes

n

An

32 10*25.310*1.97 Atotal 1.296 104 mm

2

total area

B.3 Step 3 - calculation of the cross-section centroid C Note: the local general coordinate system 0y

0z

0 located at point o of the cross-section

is used

Sz1

nshapes

n

An y0n

0*10*25.30*10*1.97 32 Sz 0mm3

Sy1

nshapes

n

An z0n

5.192*10*25.30*10*1.97 32 Sy 6.256 105 mm

3

static moments

- 85 -

yc

Sz

Atotal

410*296.1

0yc 0mm

position of the centroid

zc

Sy

Atotal

4

5

10*296.1

10*256.6

zc 48.274 mm


moments of inertia

Iyc1

nshapes

n

Iyn 1

nshapes

n

An z0nzc 2

232254 274.485.192*10*25.3274.480*10*1.9710*693.110*19610

Iyc 2.865 108 mm

4

Izc1

nshapes

n

Izn 1

nshapes

n

An y0nyc 2

0*10*25.30*10*1.9710*577.410*818 3264 Izc 1.276 10

7 mm4


4

polar moment of inertia

Iyzc1

nshapes

n

Iyzn 1

nshapes

n

An y0nyc z0n

zc

Iyzc 0 mm4

product of inertia

B.5 Step 5 - calculation of the cross-section principal moments of inertia

Note: - the product of inertia Iyzc 0 mm4 and consequently, the principal axes are

the centroidal axes. B.6 Step 6 - calculation of the cross-section radii of gyration

ryc

Iyc

Atotal

4

8

10*296.1

10*865.2ryc 148.683mm radii of gyration

- 86 -

rzc

Izc

Atotal

4

7

10*296.1

10*276.1rzc 31.374 mm

Dir 1

Dir 2

y

z

y

z

z

Figure 2.2.22 B.8 Step 8 - variation of the centroidal moments of inertia (Figure 2.2.23)

0 30 60 90 120 150 180

2 108

7.5 107

5 107

1.75 108

3 108

0

Iyri

Izri

Iyzri

180

180

i

Figure 2.2.23 We consider that the axes rotates with an angle i between 0 to

i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc


- 87 -

Izri

Iyc Izc

2

Iyc Izc


Iyzri

Iyc Izc



U 20

U 30

y

zh

b

h

b

y

z

z

y

Figure 2.2.24

A. General Observations A1. The cross-section, shown in Figure 2.2.24.a, is composed from two structural shapes, a U30 and a U20, called SHAPE1 and SHAPE2 (Figure 2.2.24.b). The geometrical characteristics of both U shapes are obtained from Appendix 2.2, Table 2.2.2 A2. The cartesian orthogonal coordinate system 0y

0z

0 is used as the original reference

coordinate system. nshapes 2

number of areas considered in the calculation

B. Calculations B.1 Step 1 - collecting data pertinent to each one of the individual areas Data pertinent to U30 (area SHAPE1) Note: the local coordinate system 01y1z1 located at the centroid 01 of the area

h1 300 mm total depth


ey1 2.70 cm

local eccentricity of the centroid O1

- 88 -

A1 58.8 cm2 area

Iy18030 cm

4

Iz1495 cm

4

moments of inertia

Iyz10 mm

4

product of inertia

z010 mm

y01

0 mm

centroid O1

Data pertinent to U20 (area SHAPE2) Note: the local coordinate system 02y2z2 located at the centroid 02 of the area



ez22.01 cm

local eccentricity of the centroid O2

A2 32.2 cm2 area

Iy2148 cm

4

Iz21910 cm

4

moments of inertia

Iyz20 mm

4

product of inertia

z02

h1

2ez2

1.20

2

300z02

129.9 mm centroid O2

y02ey1

h2

2

2

2000.27 y02

127mm


Atotal1

nshapes

n

An

22 10*2.3210*8.58 Atotal 9.1 103 mm

2

total area

B.3 Step 3 - calculation of the cross-section centroid C Note: the local general coordinate system 0y0z0 located at point 0 of the cross-section

is used

Sz1

nshapes

n

An y0n

127*10*2.320*10*8.58 22 Sz 4.089 105 mm

3

Sy1

nshapes

n

An z0n

9.129*10*2.320*10*8.58 22

Sy 4.183 105 mm

3

- 89 -

yc

Sz

Atotal

3

5

10*1.9

10*089.4yc 44.938mm

centroid C

zc

Sy

Atotal

3

5

10*1.9

10*183.4zc 45.965 mm


Iyc1

nshapes

n

Iyn 1

nshapes

n

An z0nzc 2

965.459.129*10*2.32965.450*10*8.5810*14810*8030 2244

Iyc 1.169 108 mm

4

Izc

1

nshapes

n

Izn 1

nshapes

n

An y0nyc 2

938.44127*10*2.32938.440*10*8.5810*191010*495 2244

Izc 5.761 107 mm

4


4

Iyzc1

nshapes

n

Iyzn 1

nshapes

n

An y0nyc z0n

zc

965.459.129*938.44127*10*2.32965.450*938.440*10*8.58 22

Iyzc 3.432 107 mm

4

B.5 Step 5 - calculation of the cross-section principal moments of inertia Note: - the local coordinate system cyczc located at the centroid C of the area is used;

Ip1

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

27

27878

10*432.32

10*761.510*169.1

2

10*761.510*169.1

Imax Ip1

Imax 1.326 108 mm

4

- 90 -

01 0_1

tan 0_1 Iyzc

0if

0_2

tan 0_2 Iyzc

0if

02 0_1 01 0_2if

0_2 01 0_1if

Ip2

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

27

27878

10*432.32

10*761.510*169.1

2

10*761.510*169.1

Imin Ip2

Imin 4.19 107 mm

4

0_112

atan2 Iyzc

Iyc Izc

78

71

10*761.510*169.1

10*432.3*2tan

2

1

0_1 24.594 deg 046.0)954.24tan(

0_2 0_1

2

90594.24 0_2 114.594 deg

018.2)594.114tan(



z

y

y

z

z

y

Dir 1

Dir 2

Figure 2.2.25


- 91 -

ryc

Iyc

Atotal

3

8

10*1.9

10*169.1 ryc 113.335mm

radii of gyration

rzc

Izc

Atotal

3

7

10*1.9

10*761.5 rzc 79.565mm

rp1

Ip1

Atotal

3

8

10*1.9

10*326.1 rp1 120.712mm

radii of gyration

rp2

Ip2

Atotal

3

7

10*1.9

10*19.4 rp2 67.854mm



4 , Iyzc 3.432 107 mm

4 )


4 , Iyzc 3.432 107 mm

4 )


4 , Iyzc 3.432 107 mm

4 )

Dir 2

2

-

P

Z'

0 I

(Z I , I )

C

Y I(

I

P

I, )

= 0

Dir 1

Figure 2.2.26


i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc


Izri

Iyc Izc

2

Iyc Izc


- 92 -

Iyzri

Iyc Izc


0 30 60 90 120 150 180

5 107

5 107

1 108

1.5 108

0

Iyri

Izri

Iyzri

180

180

i

Figure 2.2.27


U 30

L x80x1080

150 x 20Pl

y

z

y

z

y

z

y

z

h

b

Figure 2.2.28

A. General Observations A1. The cross-section, shown in Figure 2.2.28.a, is composed from three structural shapes, an U30 a, a plate 150mm x 25mm and an L80x80x10, called SHAPE1, SHAPE2 and SHAPE3, respectively (Figure 2.2.28.b). The geometrical characteristics of both U and L shapes are obtained from Appendix 2.2, Table 2.2.2 and Table 2.2.3, while the geometrical characteristics of the plate are calculated as for a rectangular area. A2. The cartesian orthogonal coordinate system 0y

0z

0 is used as the original reference

coordinate system. nshapes 3

number of rectangular areas considered in the calculation

- 93 -

B. Calculations B.1 Step 1 - collecting data pertinent to each one of the individual rectangular areas Data pertinent to U30 (area SHAPE1) Note: the local coordinate system 01y1z1 located at the centroid 01 of the area



ez1 2.70 cm

local eccentricity of the centroid 01

A1 58.8 cm2 area

Iy1495 cm

4

Iz18030 cm

4

moments of inertia

Iyz10 mm

4


z010 mm

y01

0 mm

centroid 01

Data pertinent to plate 150mm x 20mm (area SHAPE2) Note: the local coordinate system 02y2z2 located at the centroid 02 of the area

az2150 mm

ay220 mm

dimensions

A2 az2ay2

20*150

A2 3 10

3 mm2

area

Iy2

az2 3 ay2

12

12

20*1503

Iy2

5.625 106 mm

4

moments of inertia

Iz2

az2ay2 3

12

12

20*150 3

Iz21 10

5 mm4

Iyz20 mm

4

product of inertia

z02

az2

2ez1

0.27

2

150z02

48 mm

centroid 02

y02

h1

2

ay2

2

2

20

2

300y02

160 mm

Data pertinent to L80x80x10 (area SHAPE3) Note: the local coordinate system 03y3z3 located at the centroid 03 of the area

az380 mm

ay380 mm

flange dimensions

- 94 -

ez3 2.34 cm local eccentricity of the centroid 03

ey3 2.34 cm

A3 15.10 cm2 A3 1.51 10

3 mm2 area

Iy387.5 cm

4 Iy38.75 10

5 mm4

moments of inertia

Iz387.5 cm

4 Iz38.75 10

5 mm4

The product of inertia for the rotated position of the shape is calculated using additional data from Appendix 2.2, Table 2.2.3:

Iu3139 cm

4

Iv336.3 cm

4

Iuv30 cm

4

45 deg (as shown in Figure 2.2.29)

Figure 2.2.29

Iyz3

Iu3Iv3

2sin 2 Iuv3

cos 2

0)45*2sin(*2

3.36139

Iyz35.135 10

5 mm4

NOTE: The product of inertia was calculated using the formula (2.13) in which we consider =45˚ z03

ez1 ez3 4.230.27 z0350.4 mm centroid O3

y03

h1

2ey3

4.23

2

300y03

126.6 mm


Atotal1

nshapes

n

An

333 10*51.110*310*88.5 Atotal 1.039 104 mm

2

total area B.3 Step 3 - calculation of the cross-section centroid C Note: the local general coordinate system 0y0z0 located at point o of the cross-section

is used

Sz1

nshapes

n

An y0n

)6.126(*10*51.1160*10*30*10*88.5 333

Sz 2.888 10

5 mm3

Sy1

nshapes

n

An z0n

)4.50(*10*51.148*10*30*10*88.5 333

y

yv

u

=+45°

- 95 -

Sy 6.79 104 mm

3

yc

Sz

Atotal

4

5

10*039.1

10*888.2yc 27.799mm

centroid C

zc

Sy

Atotal

4

4

10*039.1

10*79.6zc 6.535 mm

B.4 Step 4 - calculation of the cross-section moments of inertia Note: - the local coordinate system cyczc located at the centroid C of the area is used;

Iyc1

nshapes

n

Iyn 1

nshapes

n

An z0nzc 2

23

2322464

535.64.50*10*51.1

535.648*10*3535.60*10*8.5810*5.8710*625.510*495

Iyc 2.175 107 mm

4

Izc1

nshapes

n

Izn 1

nshapes

n

An y0nyc 2

23

2322454

8.276.126*10*51.1

8.27160*10*38.270*10*8.5810*5.8710*110*8030

Izc 1.742 108 mm

4


4

Iyzc1

nshapes

n

Iyzn 1

nshapes

n

An y0nyc z0n

zc

535.64.50*8.276.126*10*51.1

8.2748*8.27160*10*3535.60*8.270*10*8.5810*135.5003

325

Iyzc 2.845 107 mm

4

B.5 Step 5 - calculation of the cross-section principal moments of inertia Note: - the local coordinate system Cyczc located at the centroid C of the area is used;

Ip1

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

27

28787

10*845.22

10*742.110*175.2

2

10*742.110*175.2

- 96 -

01 0_1

tan 0_1 Iyzc

0if

0_2

tan 0_2 Iyzc

0if

02 0_1 01 0_2if

0_2 01 0_1if

Imax Ip1

Imax 1.658 108 mm

4

Ip2

Iyc Izc

2

Iyc Izc

2

2

Iyzc2

27

28787

10*875.22

10*742.110*175.2

2

10*742.110*175.2

Imin Ip2

Imin 1.614 107 mm

4

0_112

atan2 Iyzc

Iyc Izc

87

71

10*742.110*175.2

10*027.3*2tan

2

1

0_1 11.171 deg 019.0)171.11tan(

0_2 0_12

90171.11 0_2 101.171 deg

006.5)171.101tan(


angle corresponding to minimum principal moment of inertia

02 11.171 deg

Dir 2

Dir 1

y

z z

y

zy

Figure 2.2.30


- 97 -

ryc

Iyc

Atotal

4

7

10*039.1

10*175.2 ryc 45.757mm

radii of gyration

rzc

Izc

Atotal

4

8

10*039.1

10*742.1 rzc 129.502mm

rp1

Ip1

Atotal

4

8

10*039.1

10*8.1 rp1 131.636mm

radii of gyration

rp2

Ip2

Atotal

4

7

10*039.1

10*596.1 rp2 39.198mm



4 , Iyzc 2.845 107 mm

4 )


4 , Iyzc 2.845 107 mm

4 )


4 , Iyzc 2.845 107 mm

4 )

Dir 2

2

-

P

Z'

0I

(Z I , I )

C

Y I(

I P

I, )

= 0 Dir 1

Figure 2.2.31


i 0 24 i24

i( )

Iyri

Iyc Izc

2

Iyc Izc


- 98 -

Izri

Iyc Izc

2

Iyc Izc


Iyzri

Iyc Izc


0 30 60 90 120 150 180

1 108

2.5 107

5 107

1.25 108

2 108

0

Iyri

Izri

Iyzri

180

180

i

Figure 2.2.32 2.3 Proposed Problems Calculate the following geometrical characteristics:

(a) area (b) position of the centroid (c) centroidal moments and product of inertia (d) centroidal principal moments and directions (e) radii of gyration

and draw:

(a) the Mohr’s circle (b) the variation of the centroidal moments of inertia and product of inertia

considering a rotation of the coordinate system with an angle varying from 0 to 2 for all the cross-sections shown in Figures 2.3.1 to 2.3.33

Note: Dimensions are in mm unless noted

- 99 -

40 c

m

20 cm b=60 cm

h=30

cm

parabola

Figure 2.3.1 Figure 2.3.2

500 mm

250

mm

20 mm 20 mm

10 m

m24

0 m

m

10 c

m50

cm

10 c

m

30 cm

5 cm

20 cm


400

280

15

1525

015

400

36010515

30

120

30

30

30

180


20

200

200 100mm

200

mm

400mm

Ø100

110

110

100

mm


- 100 -

150 mm 150 mm

50 50

50

r = 10cm

b = 30 cm

semi-circle


100

400400

200

200

200

460

3030

260

30 30


400

5010050

200

90°

200

150

400

100

200


a=20 cm

2a=20 cm

2a

a


- 101 -

100 100 100

100

100

100

100

50

50

50200

100


6 in x1/2 in

W 18 x 71

C 10 x 30

270 x10

U 24

Pl

270Pl 10x

Pl


10

10

L 80x80x62L 90x90x10

U 18240 x20Pl


- 102 -

240

1260

012

12

L 80x80x8

300 x 20

500 x 10

300 x20

200

Pl

Pl

Pl


240 x16

400 x 10

150 x16

U 30

L x50 x965

Pl

Pl

Pl


U 20 U 20

I 18

I 18

U 22

L x70 x870

180 x20Pl


- 103 -

NOTE: For cross-sections shown in Figures 2.3.29 and 2.3.30, net geometrical characteristics are required. (Hint: subtract the bolts holes)

450 x15

600 x 10

450 x15

L 90 x90 x11

60

60

200

M16

60

60

60

60

450 x12

L x90x1090

L x80x1080

M20

Pl

Pl

Pl

Pl


- 118 -

CHAPTER 3 Equilibrium of Plane Linear Members

3.1 Theoretical Background 3.1.1 Type of Loads, Supports and Reactions The external forces (loads) applied to plane linear beams are described in the

vertical plane of the beam and are classified as:

a. distributed forces acting on a particular segment or the entire length of the beam;

b. concentrated forces acting in a particular point of the beam;

c. distributed moments acting on a particular segment or the entire length of

the beam; d. concentrated moments (couples) acting in a particular point of the beam.

The movement of a point located on the left axis of the beam resulting from the application of forces in the Oxy plane is completely determined by three generalized components: two in-plane translations (displacements) and a rotation. The two displacements are chosen, for convenience, in the directions ox and oy of the coordinates system attached to the beam, while the rotation is described by the angular motion about an axis parallel to the oz axis located at the point of interest (normal to the plane of the definition plane). The forces and moments resulting from the constraints induced by the existence of supports, accordingly with the Newton’s First Law, are called reactions. By constraining the free movement of the material point three types of supports are created:

a. the roller support prevents the displacement in the direction normal to the rolling plane and is replaced in the free-body diagram by a corresponding reaction represented by a concentrated force as shown for case1 of Table 3.1. A similar case is that of a supporting cable illustrated in case 2 of Table 3.1;

b. the pined (clamped) support impedes both translational displacement in

the beam description plane and is replaced in the free-body diagram by a reaction, commonly represented by two orthogonal concentrated force components as depicted in case 3 of Table 3.1;

c. the fixed support completely prevents all translational displacement and

rotation in the beam definition plane and is replaced in the free-body

- 119 -

diagram by a set of reactions commonly represented by two orthogonal concentrated force components and a concentrated moment as depicted in case 4 of Table 3.1.

Table 3.1 Plan Supports and Reactions

The drawing of the beam containing the exterior loads and the reaction forces obtained by replacing the supports with their corresponding reaction forces is called the free-body diagram of the beam. If the number of unknown reactions is equal to the number of independent equilibrium equations the beam is a statically determinate beam. If the number of unknown reactions is larger than three the beam is called statically indeterminate beam. Only the statically determinate beams are treated in this chapter. 3.1.2 Cross-Sectional Internal Resultants In general there are six (6) internal resultants, )(xF , )(xVy , )(xVz , )(xT ,

)(xM y and )(xM z , are present in the cross-section. They are illustrated in Figure 3.1.a.

(a) (b)

Figure 3.1

(a) Three-dimensional case and (b) Loads in Oxy plane

- 120 -

If the loads are acting in the vertical planeOxy , only three internal resultants, )(xF ,

)(xVy and )(xMz have values different than zero. In order to simplify the notation the subscript indices are dropped and the internal resultants are noted as: )(xF ,

)(xV and )(xM . This is the case illustrated in Figure 3.1.b. 3.1.3 Types of Statically Determinate Beams In the technical language employed by structural engineers, beams are identified by the manner in which they are supported. Types of beams studied in the present chapter are:

a. cantilevered beam, shown in Figure 3.2, has only one end, point C, constrained against movement in both, horizontal and vertical, directions and against the rotation around an axis parallel to the oz axis

b. simply supported beam, shown in Figure 3.3, has one end, point A, constrained only in the vertical direction by a roller support, while the other end, point B, is constrained in both vertical and horizontal directions by a pined support or clamped support;;

c. beam with overhang, shown in Figure 3.4, is a simply supported type beam which has cantilevers located at one or both ends

Figure 3.2 Cantilevered Beam

Figure 3.3 Simply Supported Beam

- 121 -

Figure 3.4 Beam with Overhang

3.1.4 Method 1 - Calculation of the Internal Resultants Using Method of Sections The calculation is conducted as: Step 1. The beam is first conceptually freed of constraints and then, the corresponding reactions, as described in the previous section. At this point, the schematic diagram containing only the externally applied forces and reaction forces is identified. This way the free-body diagram is constructed; Step 2. Using this diagram the equations of equilibrium (3.1) of the entire beam are written as:

0 xF 0 yF 0O

zM (3.1)

Note: Only the number of equilibrium equations is fixed at three, but the way to write these equations is of personal preference. Step 3. The linear algebraic system is solved and the three unknown reaction forces are calculated. Step 4. After the reactions have been calculated, the beam is conceptually sectioned, as by a cutting-plane coincidental with a cross-section located at distance Cx from the origin point. This separation operation is called the method of sections. To maintain intact the equilibrium of the two separated parts, the corresponding internal resultants are introduced and two free-body diagrams are obtained. Writing the equilibrium of one of the two previously obtained free-body diagrams the internal resultants are obtained.

0 xF 0 yF 0C

zM (3.2)

Note: the sectioning of the beam is repeated as many times it is necessary to calculate a meaningful set of values for the internal resultants and define their variation along the length of the beam.

- 122 -

Step 5. Using the sign convention described below the graphs representing the internal resultants are plotted. The sign convention for the cross-section internal resultants, illustrated in Figure 3.5, is stated as:

1. the positive shear force,V , acts in the negative direction of the oy axis at the face x of the cross-section;

2. the positive bending moment, M , makes the face y of the beam

concave.

Figure 3.5 Sign Convention for Internal Resultants

Note: the variation of the internal resultant plotted in-between the cross-sections calculated is unknown and for practical reason is considered linear. For this reason, in order to obtain a variation closed to the real one, the number of cross-sections use in the calculation has to be judiciously chosen or the variation judged using other method. This method is not used for practical applications. 3.1.5 Method 2 - Differential Relations between Loads and Cross-Section Internal

Resultants The free-body diagram concept will be extended to write the equilibrium of the elementary volume around a point of the beam. The free-body diagram of the elementary volume is shown in Figure 3.6.

Figure 3.6 Equilibrium of Elementary Beam Volume

- 123 -

The external forces acting on the elementary volume are considered to be uniformly distributed, their variation being small enough to be neglected over the short length

x quantity. These are represented by two distributed forces, )(xpn and )(xpt , and a distributed moment )(xm , all acting in the positive sense relative to the respective axes. The sign convention for the external loads is: 1. the positive tangential distributed and concentrated loads on the beam

longitudinal axis ox act in the positive direction of ox axis; 2. the positive normal distributed and concentrated loads on the beam

longitudinal axis ox act in the positive direction of oy axis; 3. the positive distributed and concentrated bending moments act in the positive

direction of the oz axis according to the right-hand rule. The differential relations between external loads and cross-sectional internal resultants are:

)(xpdxdF

t (3.3)

)(xpdxdV

n (3.4)

)()( xmxVdxdM

(3.5)

If the differential equations (3.3), (3.4) and (3.5) are integrated and the initial boundary conditions at 0x are enforced, the following expressions are obtained:

dxxpFxF t *)()( 0 (3.6)

dxxpVxV n *)()( 0 (3.7)

dxxmdxxVMxM *)(*)()( 0 (3.8)

Note: From equations (3.6) through (3.8) important practical conclusions are

derived:

(a) the loading functions )(xpn , )(xpt and )(xm must be continuous functions on finite-intervals for the integration to be possible;

- 124 -

(b) the cross-sections where concentrated forces and moments are acting represent discontinuity points. Those cases are treated as limiting cases of the expressions obtained above;

(c) the integrals in equations (3.6) and (3.7) represent the area contained under the

loading curve over the length of the integration interval. The shear force related integral in equation (3.8) represents the area contained under the shear force curve over the length of the integration interval;

(d) values representing the boundary conditions of the cross-section internal

resultants at the beginning of the integration interval must be known in order to evaluate the cross-sectional internal resultants on that interval.

The integrals contained in the equations (3.6) through (3.8) can be evaluated if the load variation is established. Table 3.2 contains the most simple, but frequently used variations. Table 3.2 Variation of the Shear Force and Bending Moment

Case Load Distribution Shear Force Variation

Bending Moment Variation

1 No load constant linear 2 Constant linear Parabola 2nd Order 3 Linear Parabola 2nd Order Parabola 3rd Order

The following steps are required to construct de cross-sectional internal resultants graphs using the differential equations (3.4) trough (3.6): Step 1, 2 and 3 are identical to these used in the first method. Step 4. The beam length is divided in continuity intervals following the variation of the load. The discontinuity points, where the concentrated moments and forces are located, are also identified. Step 5. Starting with the first interval from the left end of the beam, one integrates the differential equations (3.4) through (3.6) to the particular variation of the load characterizing that particular interval of continuity. This way the internal resultants and their variation are obtained. The operation is repeated for all established continuity intervals. Step 6. The calculated values and the knowledge of their variation provide enough information for the plotting of the internal resultant graphical representation (diagrams).

- 125 -

3.2 Solved Problems Problem 3.2.1- Cantilevered Beam Calculate and draw the shear force and moment diagram for the plane beam loaded as shown in Figure 3.2.1.a.

Figure 3.2.1.a

A. General Observations A.1. A number of four (4) cross-sections define the variation of the loading, the continuity intervals, along the entire length of the beam. They are located at the following positions: x1 0 m x2 1 m x3 2 m x4 4 m measured from the fixed end of the cantilever. A.2 The existence of the reactions at point 1, the fixed end of the cantilever, identifies that point as a point of discontinuity. Also, in a similar situation, a discontinuity point, is the point where the concentrated force P is located. A.3. The external system of forces acting in the vertical plane of the beam, is composed from a concentrated force P 4000 N and an uniformly distributed

force q 3000Nm

.

B. Calculations B.1. Construction of the free-body diagram (Figure 3.2.1.b) The fixed support placed at the left end of the beam is replaced with two concentrated reaction forces, one horizontal, 1F , and one vertical, 1V , and one concentrated moment,

1M . The reaction forces and moment are illustrated in Figure 3.2.1.b and together with the external loading system represent the free-body diagram corresponding to the cantilevered beam.

- 126 -

12 3 4

V1F

1M P = =q

1x = x 2= x 3= x 4=

1

Figure 3.2.1.b

B.2 Calculation of the reaction forces and moments A number of three reactions are required to be calculated and an equal number of equations of equilibrium are available. It can be concluded that the system is statically determinate. First equilibrium equation (all forces projected on the horizontal direction -x):

0 xF NF *01 no axial force

Second equilibrium equation (all forces projected on the vertical direction -y):

0 yF

V1 P q x4 x3 0 solve V1 10000 N

02)(4*341V V1 10000 N

Third equilibrium equation (all moment in point 1 created by the forces

rotating about the normal out-of-plane axis –z ):

01

zM

M1 P x2 x1 q x4 x3 x3 x1x4 x3

2

0 solve M1 22000 N m

02

342*2)(4*30001*4000M1 M1 22000 N m

- 127 -

To verify the validity of the reactions calculation the equation of equilibrium of the moments around point 2 is used:

0

2 zM

0)](2

[*)(*)*( 2334

341211

xxxxxxqxxVM

000)]12(2

24[*)24(*3000)01(*1000022000

The free-body diagram, shown in Figure 3.2.1.b, is now complete. B.2. Calculation of the Shear Force and Bending Moment Diagrams Discontinuity Point 1 mxx *01 The location of the reaction force 1V and moment 1M at point 1 represent a discontinuity, a “jump”, in both shear force and bending moment diagrams. Consequently, the shear force and the bending moment in the section located in the right vicinity of section 1 are obtained as: V1left 0 N

shear force at the left of the point

M1left 0 N m

bending moment at the left of the point

V1right V1left V1 solve V1right 10000 N

shear force at the right of the point

1000001rightV V1right 10000N

M1right M1left M1 solve M1right 22000 N m


2200001rightM M1right 22000 N m

Continuity Interval 1_right to 2_left ]1,0[],0[ 12 xxx The initial values for this continuity interval are:

V0 V1right NV *100000

M0 M1right mNM **220000

- 128 -

The distributed loadings, forces and moments, acting in this interval are:

pn x( ) 0Nm

zero distributed load

md x( ) 0N m

m

zero distributed moments

The variation of the shear force is obtained using the formula (3.7) as:

V x( ) V00

xm

pn

d 10000 N

the shear force is constant in this interval

The value of the shear force at the end point of the interval is obtained: V2left V x( ) substitute x 1 m 10000.0 N

V2left 10000 N

The variation of the bending moment inside of this interval is calculated using the formula (3.8):

M x( ) M0 xV x( )

d xmd x( )

d 22000 N m 10000. N x

the bending moment has a linear variation. The value of the bending moment at the right end of this interval is calculated as: M2left M x( ) substitute x 1 m 12000. N m

M2left 12000 N m

Discontinuity Point 2 mxx *12 The location of the concentrated force

P

at the section 2 represents a discontinuity, a

“jump”, in the shear force diagram. Consequently, the shear force and the bending moment in the section located in the right vicinity of section 2 are obtained as: V2right V2left P V2right 6000 N

M2right M2left M2right 12000 N m

- 129 -

Continuity Interval 2_right to 3 ]1,0[],0[ 23 xxx

The initial values for this continuity interval are then calculated as:

V0 V2right V0 6 103 N

M0 M2right M0 1.2 104 N m

The distributed loadings, forces and moments, acting in this interval are:

pn x( ) 0Nm

no distributed load

md x( ) 0N m

m

no distributed moments The variation of the shear force is obtained using the formula (3.7) as:

V x( ) V0 xpn x( )

d float 2 6.0 103 N

the shear force is constant in this

interval. The value of the shear force at the end point of the interval is obtained: V3 V x( ) substitute x 1 m 6000.0 N

V3 6000 N


M x( ) M0 xV x( )

d xmd x( )

d 12000 N m 6000. N x

the bending moment has a linear variation. The value of the bending moment at the right end of this interval is calculated as: M3 M x( ) substitute x 1 m 6000. N m

M3 6000 N m

Continuity Interval 3 to 4 ]2,0[]3,0[ 4 xxx

- 130 -

The initial values for this continuity interval are:

V0 V3 V0 6 103 N

M0 M3 M0 6 103 N m

The distributed loadings, forces and moments, acting in this interval are: pn x( ) q

the

distributed load

q acting in the negative direction

md x( ) 0N m

m

no distributed moments


V x( ) V0 xpn x( )

d float 2 6.0 103 N 3.0 103Nm

x

the shear force varies linearly on this interval.

The value of the shear force at the end point of the interval is obtained: V4 V x( ) substitute x 2 m 0


M x( ) M0 xV x( )

d xmd x( )

d 6000 N m 6000. N x 1500.Nm

x2

the bending moment has a parabolic variation.

The value of the bending moment at the right end of this interval is calculated as: M4 M x( ) substitute x 2 m 0

Using the shear force and bending moment values calculated above and

knowing the variations of these functions along the beam length, able us to draw the correct diagrams. They are illustrated in Figure 3.2.1.c.

- 131 -

q ==PM 1

F 1V

4321

1

=

==

+

Figure 3.2.1.c

Problem 3.2.2 Simple-supported beam Calculate and draw the shear force and moment diagram for the plane beam loaded as shown in Figure 3.2.2.a.

Figure 3.2.2.a A. General Observations A.1. A number of four (4) cross-sections define the continuity intervals of the loading along the entire length of the beam. They are located at the following positions:

- 132 -

x1 0 m

x2 1 m

x3 2.5 m

x4 3.5 m

measured from pinned support of the beam. A.2 The discontinuity points are those where concentrated forces or reaction forces, (refers to forces or moments) act. For the simply supported beam studied they are identified as points 1,2 and 4. A.3. The external system of forces acting in the vertical plane of the beam, is

composed from an uniformly distributed force q1 24000Nm

, a concentrated force

P 36000 N and a linearly distributed forcemNxq )(2 .

B. Calculations B.1. Construction of the free-body diagram (Figure 3.2.2.b) The pinned support placed at the left end of the beam is replaced with two concentrated reaction forces, one horizontal, 1F , and one vertical, 1V . The roller support located at the left end of the beam is replaced by a force perpendicular on the rolling support, in this case vertical on the beam longitudinal axis. The reaction forces are illustrated in Figure 3.2.2.b and together with the external loading system represent the free-body diagram corresponding to the simply supported beam.

12 3

4

V1F

P = q1

1 2 3 4

1

2q

V4

Figure 3.2.2.b

B.2 Calculation of the reaction forces A number of three reactions are required to be calculated and an equal number of equations of equilibrium are available. It can be concluded that the system is statically determinate. First equilibrium equation (all forces projected on the horizontal direction -x):

0xF NF *01 no axial force

- 133 -

Second equilibrium equation (all moment in point 4 created by the forces rotating about the normal out-of-plane axis –z ):

0

4 zM

V1 x4 x1 P x4 x2

q1 x3 x2 x3 x2

2x4 x3

q112

x4 x3 23

x4 x3

0solve V1

float 24.6 104 N

0))5.25.3(*32*)5.25.3(*24000*

21(

))5.25.3(2

15.2(*)15.2(*24000()15.3(*36000)05.3(*1V

V1 4.6 104 N



01

zM

V4 x4 x1 P x2 x1

q1 x3 x2 x3 x2

2x2 x1

12

q1 x4 x3 13

x4 x3 x3 x1

0solve V4

float 23.8 104 N

0))5.25.3(*32*)5.25.3(*24000*

21(

))5.25.3(2

15.2(*)15.2(*24000()15.3(*36000)05.3(*1V

V4 3.8 104 N

To verify the validity of the reactions calculation the equilibrium equation of forces projected on the vertical axis y is used:

0 yF 0)(**21)*( 43412311 VxxqxxqPV

000)5.25.3(*24000*21)15.2(*240003600046000

- 134 -

The free-body diagram, shown in Figure 3.2.1.b, is now complete. B.2. Calculation of the Shear Force and Bending Moment Diagrams Discontinuity Point 1 mxx *01 The location of the reaction force 1V at point 1 represents a discontinuity, a “jump”, in the shear force diagram. The shear force and the bending moment in the section located in the right vicinity of section 1 are obtained as: V1left 0 N

shear force at the left of the point

M1left 0 N m


V1right V1left V1 V1right 4.6 104 N


M1right M1left M1right 0N m

bending moment at the right of the point

Continuity Interval 1_right to 2_left ]1,0[],0[ 12 xxx The initial values for this continuity interval are:

V0 V1right V0 4.6 104 N

M0 M1right M0 0N m

pn x( ) 0Nm

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 2 4.6 104 N

the shear force is constant in this

interval. The value of the shear force at the end point of the interval is obtained: V2left V x( ) substitute x 1 m 46000.0 N

V2left 46000 N

- 135 -


M x( ) M0 xV x( )

d xmd x( )

d 46000. N x

the bending moment has a linear variation. The value of the bending moment at the right end of this interval is calculated as: M2left M x( ) substitute x 1 m 46000. N m

M2left 46000 N m

Discontinuity Point 2 mxx *12 The location of the concentrated force P at point 2 represents a discontinuity, a “jump”, in the shear force diagram. The shear force and the bending moment in the section located in the right vicinity of section 1 are obtained as:

V2right V2left P V2right 1 104 N


M2right M2left M2right 4.6 104 N m

bending moment at the right of the point

Continuity Interval 2_right to 3 ]5.1,0[],0[ 23 xxx The initial values for this continuity interval are:

V0 V2right V0 1 104 N

M0 M2right M0 4.6 104 N m

pn x( ) q1

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 2 1.0 104 N 2.4 104Nm

x

The shear force is linear in this interval. The value of the shear force at the end point of the interval is obtained:

- 136 -

V3 V x( ) substitute x 1.5 m 26000.00 N V3 26000 N


M x( ) M0 xV x( )

d xmd x( )

d 46000 N m 10000. N x 12000.Nm

x2

The bending moment has a parabolic variation. The value of the bending moment at the right end of this interval is calculated as: M3 M x( ) substitute x 1.5 m 34000.00 N m M3 34000 N m

It is observed that the shear force changes it sign on this interval, and consequently, taking a zero value. The section where the shear force is zero represents a local maximum for this interval. The position of the section where the bending moment has a local maximum is calculated as:

V2right q1 xmax 0solve xmax

float 3.417 m the left side of the equation

represents the expression of the shear force on the interval. The position of the cross-section is calculated: xmax 0.417 m

The expression of the maximum bending moment is obtained by substituting the value

of maxx into the bending moment expression:

Mmax M2right q1 xmaxxmax

2 Mmax 4.809 104 N m

Continuity Interval 3 to 4 ]1,0[],0[ 34 xxx The initial values for this continuity interval are:

V0 V3 V0 2.6 104 N

M0 M3 M0 3.4 104 N m

pn x( ) q1 1x

l34

where

l34 x4 x3

is the length of the interval

- 137 -

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 2 2.6 104 N 2.4 104Nm

x .50x2

m

The shear force is parabolic in this interval. The value of the shear force at the end point of the interval is obtained: V4left V x( ) substitute x 1 m 38000.000 N V4left 38000 N


M x( ) M0 xV x( )

d float 5 34000. N m 26000. N x 24000.Nm

.50000 x2 .16667x3

m

The bending moment has a parabolic variation. The value of the bending moment at the right end of this interval is calculated as:

M4left M x( ) substitute x 1 m 8.000 10-2 N m

Note: The moment

leftM 4 should be zero, but because the numerical error induced by

fixing the number of decimals at five (5), a very small value is obtained.

Discontinuity Point 4 V4right V4left V4 V4right 0N

M4right M4left M4right 0 N m


knowing the variations of these functions along the beam length, the shear force and bending moment diagrams are draw as illustrated in Figure 3.2.2.c.

- 138 -

4V

q2

1

1q=P

F 1 V432

1

0

= ==

Figure 3.2.2.c

Problem 3.2.3 Simple-supported beam with overhang Calculate and draw the shear force and moment diagram for the plane beam loaded as shown in Figure 3.2.3.a.

Figure 3.2.3.a A. General Observations A.1. A number of six (6) cross-sections define the continuity intervals of the loading along the entire length of the beam. They are located at the following positions: x1 0 m

x2 4 m

x3 6 m

x4 9 m

x5 12 m

and

- 139 -

x6 14 m

measured from the left end of the beam.

A.2 The discontinuity points are those where concentrated forces or reaction forces, (refers to forces or moments) act. For the studied they are identified as points 2, 4, 5 and 6. A.3. The external system of forces acting in the vertical plane of the beam, is

composed from an uniformly distributed force q 3000Nm

, a concentrated force

P 8000 N and a concentrated moment Mext 24000 N m .

B. Calculations B.1. Construction of the free-body diagram (Figure 3.2.3.b) The pinned support placed at section 2 is replaced with two concentrated reaction forces, one horizontal, 2F , and one vertical, 2V . The roller support located at section 5 of the beam is replaced by a force perpendicular on the rolling support, in this case vertical, 5V , on the beam longitudinal axis. The reaction forces are illustrated in Figure 3.2.3.b and together with the external loading system represent the free-body diagram.

12 3 5

V2F

P =q

1 2 3 4

2 V5

Mext

5 6

46

=

Figure 3.2.3.b

B.2 Calculation of the reaction forces A number of three reactions are required to be calculated and an equal number of equations of equilibrium are available. It can be concluded that the system is statically determinate. First equilibrium equation (all forces projected on the horizontal direction -x):

0xF NF *01 no axial force

Second equilibrium equation (all moment in point 5 created by the forces


- 140 -

05

zM

q x2 x1 x2 x1

2x5 x2

V2 x5 x2 q x3 x2 x3 x2

2x5 x3

P x5 x4 Mext

0 solve V2 20250 N

024000)912(*8000

)]612(2

46[*)46(*3000)412(*)]412(2

02[*)02(*3000 2

V

V2 20250 N



02

zM

q x2 x1 x2 x1

2 q x3 x2

x3 x2 2

P x4 x2 V5 x5 x2 Mext

0 solve V5 5750 N

024000

)412(*)49(*8000]2

46[*)46(*3000]2

02[*)02(*3000 5

V

V5 5750 N

To verify the validity of the reactions calculation the equilibrium equation of

forces projected on the vertical axis y is used:

0 yF

0)(* 5213 VPVxxq 005750800020250)06(*3000 The free-body diagram, shown in Figure 3.2.1.b, is now complete. B.2. Calculation of the Shear Force and Bending Moment Diagrams Interval 1-2_left The initial values for this continuity interval are: V0 0 N

- 141 -

M0 0 N m

pn x( ) q

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 2 3.0 103 Nm

x

The shear force is linear in this interval. The value of the shear force at the end point of the interval is obtained: V2left V x( ) substitute x 4 m 12000.0 N V2left 12000 N


M x( ) M0 xV x( )

d xmd x( )

d 1500.Nm

x2

The bending moment has a parabolic variation. The value of the bending moment at the right end of this interval is calculated as: M2left M x( ) substitute x 4 m 24000. N m

M2left 24000 N m

Discontinuity Point 2 V2right V2left V2 solve V2right 8250 N

V2right 8250 N

M2right M2leftsolve M2right 24000 N m

M2right 24000 N m

Interval 2right-3 The initial values for this continuity interval are: V0 V2right

M0 M2right

pn x( ) q

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 3 8.25 103 N 3.00 103Nm

x

- 142 -

The shear force is linear in this interval. The value of the shear force at the end point of the interval is obtained: V3 V x( ) substitute x 2 m 2250.00 N

V3 2250 N


M x( ) M0 xV x( )

d xmd x( )

d 24000 N m 8250. N x 1500.Nm

x2

The bending moment has a parabolic variation. The value of the bending moment at the right end of this interval is calculated as: M3 M x( ) substitute x 2 m 13500. N m

M3 13500 N m

Interval 3-4left The initial values for this continuity interval are:

V0 V3

V0 2.25 103 N

M0 M3

M0 1.35 104 N m

pn x( ) 0Nm

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 3 2.25 103 N

The shear force is constant in this interval. The value of the shear force at the end point of the interval is obtained: V4left V x( ) substitute x 3 m 2250.00 N

V4left 2250 N


M x( ) M0 xV x( )

d xmd x( )

d 13500 N m 2250. N x

The bending moment has a linear variation. The value of the bending moment at the right end of this interval is calculated as: M4left M x( ) substitute x 3 m 6750. N m

M4left 6750 N m

- 143 -

Discontinuity Point 4 V4right V4left P solve V4right 5750 N

V4right 5750 N


M4right 6750 N m

Interval 4right-5left The initial conditions are:

V0 V4right

V0 5.75 103 N

M0 M4right

M0 6.75 103 N m

pn x( ) 0Nm

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 3 5.75 103 N

The shear force is constant in this interval. The value of the shear force at the end point of the interval is obtained: V5left V x( ) substitute x 3 m 5750.00 N

V5left 5750 N


M x( ) M0 xV x( )

d xmd x( )

d 6750 N m 5750. N x

The bending moment has a linear variation. The value of the bending moment at the right end of this interval is calculated as: M5left M x( ) substitute x 3 m 24000. N m

M5left 24000 N m

Discontinuity Point 5 V5right V5left V5 solve V5right 0

V5right 0 N


M5right 24000 N m

- 144 -

Interval 5right to 6left The initial values are: V0 V5right

V0 0N

M0 M5right

M0 2.4 104 N m

pn x( ) 0Nm

md x( ) 0N m

m


V x( ) V0 xpn x( )

d float 3 0

The shear force is zero in this interval. The value of the shear force at the end point of the interval is obtained: V6left V x( ) substitute x 2 m 0

V6left 0 N


M x( ) M0 xV x( )

d xmd x( )

d 24000 N m

The bending moment has a constant variation. The value of the bending moment at the right end of this interval is calculated as:

M6left M x( ) substitute x 2 m 24000 N m

M6left 24000 N m

Discontinuity Point 6 V6rigt V6left

V6rigt 0N

M6right M6left Mext

M6right 0N m


knowing the variations of these functions along the beam length, the shear force and bending moment diagrams are draw as illustrated in Figure 3.2.3.c.

- 145 -

=6

4

extM

5V2

q =P

F 2 V532

1

Figure 3.2.3.c

Problem 3.2.4 Using the shear force diagram shown in Figure 3.2.4.a, determine: (a) the free body diagram of the beam, (b) the corresponding bending moment diagram and (c) the possible support positions.

Figure 3.2.4.a

A. General Observations A.1. Analyzing the variation of the shear force diagram illustrated in Figure 3.2.4.a, it is observed that three (3) intervals of continuity are found. They are described by four (4) sections located at distances:

x1 0 m x2 3m x3 4.5 m x4 6 m measured from the left end of the beam.

- 146 -

A.2 The shear diagram has three “jumps” indicating the existence of concentrated forces at these locations. They are positioned at point 1, 3 and 4 as illustrated in Figure 3.2.4.b.

1 2 3 4

(a)1.leftV

V1.right

V2

V3.left

V3.rightV4.left

V4.right

Figure 3.2.4.b

B. Calculations B.1 Free-Body Diagram Following the reasoning described in the paragraph A.2 can be concluded the following: (a) in point 1 acts a concentrated vertical force kNP 181 . The force is positive orientated; (b) in point 3 operates a negative orientated vertical force kNP 183 and (c) in point 4 a positive orientated vertical force kNP 184 is found. Note: These conclusions can be made also by interpreting the equation (3.7). The construction of the free-body diagram is made in the following manner: - a positive orientated concentrated force kNP 181 is placed in 1; - the variation of the shear force in the interval 1_right to 2 is linear indicating

the existence of a constant loading in this interval (see Table 3.2). Because the value of the shear force decreases towards point 2, the uniform distributed force is negative orientated. The magnitude of the distributed force is calculated by writing the value of the shear force in point 2:

03*180)(* 1212 qxxqVV right kNq 6

- the shear force has zero magnitude in the interval 2 to 3_left, indicating that there is not any distributed load in this interval;

- a negative concentrated force kNP 183 acts in point 2; - the shear force is constant in the interval 3_right to 4, and consequently, no

distributed force acts in this interval;

- 147 -

- a concentrated positive orientated force kNP 184 is attached to point 4. The free-body diagram is illustrated in Figure 3.2.4.c

1P

q 3P =

P4

Figure 3.2.4.c

B.2 Calculation of the Bending Moment Diagram The moment diagram can be calculated using the formula (3.8) or by using the observation that the integral geometrically represents the area located under the function. This time the second method is used.

mkN 01 M

mkN 273*18*2102_1_12 shearAMM

mkN 270273_2_23 shearAMM mkN 05.1*18274_3_34 shearAMM

The bending moment diagram is shown in Figure 3.2.4.d.

2MM3

Figure 3.2.4.d

B.3 Discussion about the possible support configuration Two vertical supports are necessary to be placed, for the beam to be statically determinate, where the concentrated forces are located. The concentrated forces are future candidates to be reaction forces. Two configurations illustrated in Figures 3.2.4.e and 3.2.4.f are yielding similar shear force and bending moment diagrams. It can be concluded that problem has two possible configurations.

- 148 -

Figure 3.2.4.e

Figure 3.2.4.f

It is the configuration shown in Figure 3.2.4.g a real configuration?

Figure 3.2.4.g

Hint: To answer calculate the reaction forces, construct the free-body diagram and

compare with the free-body diagram shown in Figure 3.2.4.c.

- 149 -

3.3 Proposed Problems Problem 3.3.1 – 3.3.36 Conduct the following tasks: (a) Draw the free-body diagram, (b) calculate the reaction forces and moments, (c) calculate and draw the shear force and bending moment using method 2.





- 150 -





- 151 -






- 152 -






- 153 -

Problems 3.3.37 and 3.3.38 Calculate the bending moment diagram and discussed the possible free-body diagram related to the shear force diagrams shown in Figures 3.3.37 and 3.3.38.


- 154 -

CHAPTER 4 Axial Deformation 4.1 Theoretical Background 4.1.1 Basic Theory of Axial Deformation Definition 4.1 A plane linear member, when subjected to exterior loads and/or change of temperature, undergoes an axial deformation if after the deformation:

(a) the axis of the member remains straight; (b) the cross-sections remain plane, perpendicular to the longitudinal axis

of the beam and do not rotate about the same longitudinal axis after the deformation.

Equilibrium Equation

Figure 4.1 Equilibrium of an Elementary Beam Volume

The differential relation between the exterior load and the axial internal resultant is obtained from the equilibrium of the elementary volume:

)()( xpdx

xdFt (4.1)

where )( xp t is the distributed loading parallel to the beam longitudinal axis and

)(xF is the axial stress resultant .

Integrating equation (4.1) the stress resultant force )(xF in a particular cross-section is calculated as:

x

t dpFxF00 *)()( (4.2)

- 155 -

where )0(0 xFF is the value of the axial force at the origin of the integration interval. Strain-Displacement Equation

Figure 4.1 Geometrical Aspects of the Axial Deformation (a) Undeformed Member and (b) Deformed Member

The extensional strain )(xy along the longitudinal axis of the beam is obtained using

the notation shown in Figure 4.1:

dxdu

xxuxxu

xxxx xxx

])()([lim)(lim)( 0

*

0 (4.3)

The rest of the generalized strain tensor components, extensional and shear strains, are:

)(*)( xx xy (4.4)

)(*)( xx xz (4.5)

0 yzxzxy (4.6) The expression of the generalized strain tensor T is:

x

x

x

zzyzx

yzyyx

xzxyx

T

*000*000

(4.7)

The displacement )(xu is obtained by integration the differential equation (4.3):

- 156 -

dxxuxux

x *)()(00 (4.8)

where )0(0 xuu is the displacement at the beginning at the integration interval. The elongation is calculated using equation (4.8) as:

L

x dxxuLue0

*)()0()( (4.9)

where L is the total length of the bar. Constitutive Equation The constitutive equation reflects the relation between the stress and the strain. If the linear elastic material behavior is considered, applying the Hook’s Law, the relation between the normal stress ),,( zyxx and extension strain )(xx is written:

)(*),,(),,( xzyxEzyx xx (4.10) where E is the modulus of elasticity and is obtained performing tensile tests. Considering the assumption that the cross-section of the bar is a small surface, the variation of the modulus of elasticity ),,( zyxE is negligible on this surface, and consequently, the constitutive equation (4.10) is expressed as: )(*)()( xxEx xx (4.11) Note: equation (4.11) implies that the normal stress )(xx varies only along the length of the member, but has a constant value on the entire cross-section. The representation of the normal stress )(xx is shown in Figure 4.2.

Figure 4.2 Cross-Section Normal Stress Distribution

- 157 -

Note: equation (4.11) implies that the normal stress )(xx varies only along the length of the member, but has a constant value on the entire cross-section. The rest of the stress tensor T components are zero:

0)()( xx zy (4.12)

0 yzxzxy (4.13) Consequently, the generalized stress tensor is:

00000000x

zzyzx

yzyyx

xzxyx

T

(4.14)

Cross-Section Stress (Internal) Resultants The following cross-sectional stress (internal) resultants are obtained using the stress distribution expressed by equation (4.11) through (4.13): )(*)( *)( *)()( xAxdAxdAxxF x

Ax

Ax (4.15)

yx

Ax

Axy SxdAzxdAxzxM *)( **)( *)(*)( (4.16)

zx

Ax

Axz SxdAyxdAxyxM *)( **)(*)(*)( (4.17)

The relations between the normal stress )(xx and the cross-section resultants )(xF ,

)(xM y and )(xM z are derived using the notation shown in Figure 4.3.

Figure 4.3 Normal Stress and Stress Resultants

(a) Normal Stress and (b) Stress Resultants

- 158 -

If the axes oy and oz of the coordinate system intersect such that the x axis passes through the cross-section centroid, the static moments yS and zS are zero and the axial force )(xF remains the only non-zero stress resultant:

)(*)( *)( *)()( xAxdAxdAxxF xA

xA

x (4.18)

0)( xM y (4.19)

0)( xM z (4.20)

Then, from equation (4.18) the normal stress x is calculated as:

)()()(

xAxFxx (4.21)

Note: It is concluded that a beam made from a linear elastic material undergoes an axial deformation if the axial force passes through the cross-section centroid.

Thermal Effects on Axial Deformation The thermal strain was introduced as:

TT * (4.22) where is the thermal expansion coefficient and T is the change in the member temperature. The total elongation strain is the sum of the elongation strain induced by the exterior load action and thermal effects and is expressed as:

)(*)()(*)(

)()(*)()()(

)( xTxxAxE

xFxTxxExx x

x

(4.23)

Then, the total elongation of the member is written as:

LLL

x dxxTxdxxAxE

xFdxxe000

*)(*)(*)(*)(

)(*)( (4.24)

4.1.2 Uniform-Axial Deformation Definition 4.2 The uniform axial-deformation element is a linear member characterized by:

(a) a constant area along the entire length of the member;

- 159 -

(b) is made of a homogeneous elastic material;

(c) is subjected to a constant axial force .F Uniform-axial deformed members (shown in Figure 4.4)

. Figure 4.4 Member Exhibiting Uniform Axial-Deformation

Transcribing the requirements of the definition 4.2 the following expressions are obtained:

0)( AxA (4.25)

0)( ExE (4.26)

0)( FxF (4.27) Note: Equation (4.27) implies the absence of the distributed load )(xpt . Rewriting equations (4.21), (4.23) and (4.24) previously obtained for the case of the member with uniform axial-deformation the following equations are obtained:

00

0

)()()(

AF

xAxFxx for axial stress (4.28)

000

0

*)()(

)(

AE

FxEx

x xx for elongation strain (4.29)

TLLTLAELF

TLLdxxTxdxxeLL

x

*****

**

****)(*)(*)(

000

0

000

for total elongation (4.30)

- 160 -

Axial flexibility and stiffness coefficients

LAE

k 00 * the axial stiffness coefficient. (4.31)

00 *1

AEL

kf axial flexibility coefficient (4.32)

Substituting the equation (4.32) into the total elongation expression (4.30), the total elongation of a bar under uniform-axial deformation is recast as:

TLFfTLAELFe *****

**

000

0 (4.33)

4.1.3 Nonuniform-Axial Deformation If any one of the assumptions contained in definition 4.2 is violated the axial deformation is called nonuniform-axial deformation. The most common cases of nonuniform-axial deformation are: Member with non-homogeneous cross-section; Member with variable Cross-Section; Member loaded along its length. The formulae described in the section 4.1.1 have to be adapted function of the situation. 4.1.4 Verification of the Members Subjected to Axial Deformation The design formula used is the relationship between maximum normal stress max and the allowable normal stress all : all max (4.34) The formula (4.34) was used for a long period of time in a procedure known as the allowable-stress design. Due to the simplicity of application, this method is still commonly used in United States for the design of steel structures. The allowable normal stress all is defined by limiting the value of the normal stress in the axially deformed member. Dividing the yield stress Y pertinent to the material subjected to axial deformation by a safety factor SF the allowable axial stress is calculated:

SF

Yall

(4.35)

- 161 -

The safety factor is greater than one, usually taking values between 2 and 3. The yield stress for different materials is found in Appendix 1.2. 4.2 Solved Problems Problem 4.2.1 A solid brass rod AB and a solid aluminum rod BC are connected together by a rigid coupler of negligible length at B as shown in Figure 4.2.1.a. The diameters and the modulus of elasticity of the two segments are d1 = 65 mm, d2 = 50 mm, E1 = 105 GPa and E1 = 69 GPa, respectively. The system is loaded by two concentrated loads,

kNP B 180 and kNP C 45 , acting along the centroidal line of the system at point B and C , respectively . Calculate the axial stress existing in the two rods and the displacement at point B and C. Verify the rod segments.

Figure 4.2.1.a

A. General Observations Two axial forces are acting on the rod: PB 180 103 N

and PC 45 103 N The areas of the two rod segments are:

A1 d1

2

4

465* 2

A1 3.318 103 mm2

A2 d2

2

4

450* 2

A2 1.963 103 mm2

B. Calculations B.1 Free-Body Diagram The constraint located at point A is replaced by a horizontal reaction AH as shown in Figure 4.2.1.b.

- 162 -

Figure 4.2.1.b

B.2 Reaction Calculation The equation of equilibrium used is the projection of all axial forces on the horizontal axis x : 0X HA PB PC 0 solve HA 135000 N

HA 135000 N

The equilibrium equation contains only one unknown reaction force, AH , and consequently, the system is statically determinate. B.3 The Axial Force Diagram The axial force diagram is drawn in Figure 4.2.1.c.

HA BPCP

Figure 4.2.1.c The axial force on the interval AB is a constant tension force F1 135 103 N ,

while on the interval BC is a constant compression force F2 45 103 N . It can

be concluded that both segments of the rod are uniform-axial deformed members. B.3 Stress and Strain Calculation The stress and strain in the interval AB , where the rod is made of solid brass are obtained as:

- 163 -

1F1A1

23

5

10*318.310*35.1

mN

1 4.068 107 Pa

tension stress

11E1

9

7

10*10510*068.4

1 3.875 10 4 elongation strain

In a similar manner is calculated the stress and strain pertinent to the interval BC representing the aluminum made rod.

2F2A2

23

5

10*963.110*45.0

mN

2 2.292 107 Pa compression stress

22E2

9

7

10*6910*292.2


B.4 Flexibility Coefficients The flexibility coefficients are calculated as:

f1L1

E1 A1 2311 10*318.3*10*05.1

5.1mPa

mf1 4.305 10 9

mN

f2L2

E2 A2 2311 10*963.1*10*69.0

0.1mPa

mf2 7.381 10 9

mN

B.5 Calculation of the axial displacements The displacement Bu of the point B , the right end of the brass segment, is calculated as:

uB u0 f1 F1 NNmm *10*35.1**10*305.4*0 59 uB 5.812 10 4 m

where the displacement at the origin of the interval

u0 uA and uA 0 m , because

the point is constraint against the horizontal movement. The calculation of the displacement Cu implies the knowledge of the displacement of the origin point of the interval BC . u0 uB mu 4

0 10*812.5

The displacement in point C is calculated as:

- 164 -

uC u0 f2 F2 NNmm *10*45.0**10*381.7*10*81.5 594

uC 2.49 10 4 m

B.6 Verification of the Rod The verification of the rod segments is conducted as: for the brass segment

brass

brassy

SF_

brassallowable_1

PaPa 77 10*75.1310*068.4 ok

where brassallowable_ is the brass allowable stress, brassy _ is the brass yield stress which can be found in Appendix 1.2, and brassSF is the safety factor . For this calculation

Y_brass 275 106 Pa and SFbrass 2.0 .

for the aluminum segment

umalu

umaluy

SF min

min_aluminumallowable_2

PaPa 77 10*5.2010*292.2 ok

where aluminumsallowable_ is the brass allowable stress, umsaluy min_ is the aluminum yield stress which can be found in Appendix 1.2, and umaluSF min is the safety factor . For

this calculation Y_aluminum 410 106 Pa and SFaluminum 2.0 .

Problem 4.2.2 Two uniform, linearly elastic members are held together at point B and the resulting two-segment rod is attached to rigid supports at ends A and C. A single external load P = 4000 kN is applied at joint B. Member (1) has a length 2L1 m, diameter d1 = 120 mm and is made of steel with a modulus of elasticity 1E = 200 GPa. Member (2) has a length 2L = 1.8 m, diameter d2 = 150 mm and is made of an aluminum alloy with a modulus of elasticity 2E =75 GPa. Conduct the following tasks: (a) verify the axial stress in both members and (b) calculate the axial displacement at point B.

Figure 4.2.2.a

- 165 -

A. General Observations The areas of the two rod segments are:

A1 d1

2

4

4120* 2

A 1 1.131 104 mm2

A2 d2

2

4

4150* 2

A2 1.767 104 mm2

B. Calculations B.1 Free-Body Diagram The constraint located at point A is replaced by a horizontal reaction AH as shown in Figure 4.2.2.b.

Figure 4.2.2.b

B.2 Reaction Calculation The equation of equilibrium used is the projection of all axial forces on the horizontal axis x : HA PB HC 0 solve HA PB HC

The equation contains two (2) unknown reaction forces

AH and CH . Consequently,

the system is a statically indeterminate. An additional equation is necessary. This is equation is geometrical in nature and represents the fact that the total elongation of the beam is zero.

0 AC uue Considering that the calculation of the displacement starts at point A, uA 0 m

and

the axial displacement at the end B of the interval AB , representing the steel rod, is

calculated as:

- 166 -

uB uAL1

A1 E1F1

substitute F1 HA

substitute uA 0 muB

L1

A1 E1 HA

At the left end of the BC interval the axial displacement is expressed as:

uC uBL2

A2 E2F2

substitute uBL1

A1 E1HA

substitute F2 HCuC

L1

A1 E1 HAL2

A2 E2 HC

This way the second equation is obtained as: uC 0

and by substituting the flexibility coefficients

1f and

2f into the expression of the

axial displacement Cu the following algebraic system is obtained:

HA PB HC 0

f1 HA f2 HC 0

The solutions, representing the two reactions, AH

and

CH , are found:

Find HA HC collect PB

PBf2

f1 f2

f1PB

f1 f2

Substituting the numerical data the reaction forces are calculated as:

f1L1

A1 E1 Pam

m1122 10*2*10*131.1

0.2f1 8.842 10 10

mN

f2L2

A2 E2 Pam

m1122 10*75.0*10*767.1

8.1f2 1.358 10 9

mN

HAf2

f1 f2PB

6

910

9

10*4*10*358.110*842.8

10*358.1HA 2.423 106 N

- 167 -

HCf1

f1 f2PB

6

910

10

10*4*10*358.110*842.8

10*842.8HC 1.577 106 N

B.3 The Axial Force Diagram The axial force diagram is drawn in Figure 4.2.2.c.

HA BPCH

Figure 4.2.2.c

The axial force on the interval AB is a constant tension force F1 2.423 106 N ,

while on the interval BC is a constant compression force F2 1.577 106 N . It can

be concluded that both segments of the rod are uniform-axial deformed members. B.3 Stress and Strain Calculation The stress and strain in the interval AB , where the rod is made of solid steel are obtained as:

1F1A1

22

6

10*131.110*423.2

mN

1 2.142 108 Pa

tension stress

11E1

9

8

10*20010*142.2


In a similar manner is calculated the stress and strain pertinent to the interval BC representing the aluminum made rod.

2F2A2

22

6

10*767.110*577.1

mN

2 8.926 107 Pa compression stress

22E2

9

7

10*7510*926.8


B.4 Calculation of the axial displacement The displacement Bu of the point B , the right end of the steel segment, is calculated:

- 168 -

uB u0 f1 F1 NNmm *10*423.2**10*842.8*0 610

uB 2.142 10 3 m

where the displacement at the origin of the interval u0 uA and uA 0 m , because

the point is constraint against the horizontal movement. B.5 Verification of the Rod The verification of the rod segments is conducted as: for the steel segment

steel

steely

SF_

steelallowable_1

PaPa 88 10*667.210*142.2 ok

where brassallowable_ is the brass allowable stress, brassy _ is the brass yield stress which can be found in Appendix 1.2, and is steelSF the safety factor . For this calculation

Y_steel 400 106 Pa and SFsteel 1.5 .

for the aluminum segment

umalu

umaluy

SF min

min_aluminumallowable_2

PaPa 87 10*1.110*926.8 ok

where aluminumsallowable_ is the brass allowable stress, umsaluy min_ is the aluminum yield stress which can be found in Appendix 1.2, and SF is the safety factor . For this

calculation Y_aluminum 275 106 Pa and SFaluminum 2.5 .

Problem 4.2.3 A rigid beam AB, shown in Figure 4.2.3.a, is supported by two vertical rods made of steel with a modulus of elasticity E=200 GPa. The support rod located at the end A has a diameter d1=25 mm. The weight of the beam AB is negligible and is loaded at point C with a concentrated force P = 60 kN. Calculated: (a) the diameter, d2, of the hanger located at the end B, considering that the relation between the vertical displacement at the ends of the beam AB uu *2 , (b) under same condition the vertical displacement in node C, (c) if the hanger located at end B of the rigid beam has a diameter d2=20 mm, what should be the position of the concentrated load P for relation AB uu to be true, and (d) the axial stresses in the hangers considering the conditions stipulated in the previous question.

- 169 -

A. General Observations A.1 The steel rod (1) has a length L1 3 m

and

diameterd1 25 mm , while for

the steel rod (2) only the length is known L2 2 m . Both rods are made of steel with

a modulus of elasticity of steel Esteel 200 109 Pa .

Figure 4.2.3.a

A.2 The vertical concentrated force P 60 103 N is located at point C at a distance of a 1 m from the left end. The total distance in-between the rigid beam supports is L 3 m , while the distance from the application point C to point B is b 2m . B. Calculations B.1 Free-Body Diagram The rigid beam AB is supported at ends A and B by two steel rods which are playing the supporting role for the beam. Sectioning the rods and replacing them by two corresponding axial forces AV and BV , the free-body diagram of the system is obtained as shown in Figure 4.2.3.b. B.2 Reactions Calculation Two moment equations are written:

BM 0

VA L P b 0 solve VA PbL

mmN

32*10*60 3 VA 4 104 N

- 170 -

A

M 0

VB l P a 0 solve VB Pal

mmN

31*10*60 3 VB 2 104 N

Figure 4.2.3.b

The two equilibrium equations are containing two unknowns, the forces

AV and BV , and consequently, the system is statically determinate. The verification of the reaction forces is done using the equilibrium equation of the projection of the forces on the vertical direction: 0Y 0 BA VPV 00010*210*610*4 444 B.3 Calculation required by question (a) The axial forces in the rods are tension type forces:

F1 VA F1 4 104 N

F2 VB F2 2 104 N

Because the axial forces in the rods and they have constant areas, both rods are uniform-axial deformed members. The flexibility coefficients are obtained as:

f1L1

A1 Esteel

- 171 -

f2L2

A2 Esteel

The geometrical condition required is:

2 uA uB 0substitute uA f1 F1

substitute uB f2 F22 f1 F1 f2 F2 0 solve f2 2 f1

F1F2

L2

d22

4

2L1

d12

4

P

bL

PaL

simplify

solve d2

12 L1 b 2 L1 b L2 a

12

d1

12 L1 b 2 L1 b L2 a

12

d1

Because the rod diameter is always a positive value, from the two above solutions only the positive value is a valid solution. Consequently, the diameter of the rod (2) is calculated as:

d21

2 L1 b 2 L1 b L2 a 12

d1 1

1

2 ***2

1 dba

LL

d2 10.21 mm

B.4 Calculations required by question (b) The vertical displacement Cu is composed from two components: one elastic and equal with the vertical displacement Au and other rigid induced by the rigid rotation of the beam in the vertical plane. The elastic vertical displacement Au is calculated:

uA f1 F1 NNm 48 10*4*10*056.3 uA 1.222 10 3 m

where f1

L1A1 Esteel

Pamm

1124 10*2*10*909.43

f1 3.056 10 8mN

is

the flexibility coefficient of the rod (1). The vertical displacement Cu is obtained:

uC uAuB uA

La

310*222.110*222.1

33 uC 1.63 10 3 m

- 172 -

B.5 Calculations required by question (c) The geometrical condition imposed is written as:

uA uB 0substitute uA f1 F1

substitute uB f2 F2f1 F1 f2 F2 0

f1 F1 f2 F2 0

substitute F1 PL a

L

substitute F2 PaL

f1 P

L a( )L

f2 PaL

0 solve af1

f1 f2 L

The flexibility coefficient of the rod (2) is calculated considering that the rod diameter is d2 20 mm :

A2 d2

2

4

420* 2

A2 314.159 mm2

f2L2

A2 Esteel Pam

m1124 10*2*10*14159.3

2f2 3.183 10 8

mN

The position of the concentrated force measured from point A is obtained by substituting the previously calculated flexibility coefficients:

af1

f1 f2 L

m3*10*183.310*056.3

10*056.388

8

a 1.469m

B.6 Calculations required by the question (d) The axial forces in the hangers are:

F1 PL a

L

mmmN

3469.13*10*6 4 F1 3.061 104 N

for rod (1)

F2 PaL

m

mN3469.1*10*6 4 F2 2.939 104 N

for rod (2)

The axial stresses are obtained:

1F1A1

24-

4

10*4.90910*061.3

mN

1 6.236 107 Pa tension stress in rod (1)

- 173 -

2F2A2

24-

4

10*3.14210*939.2

mN

2 9.354 107 Pa tension stress in rod (2)

B.7 Verification of the rods Considering that the steel yielding stress and the safety coefficient employed are Y_steel 250 106 Pa

and SFsteel 2.5 , respectively, the allowable steel stress

is calculated as:

allwable_steelY_steelSFsteel

5.2

10*5.2 8 Paallwable_steel 1 108 Pa

Consequently,

PaPa steelallowable8

_7

1 1010*236.6 the rod (1) is ok

PaPa steelallowable8

_7

2 1010*354.9 the rod (2) is ok Problem 4.2.4 The system shown in Figure 4.2.4.a is composed of a rigid beam AD, pinned into the wall at point A, and two unequal linear elastic rods, BE and CF, made of steel with a modulus of elasticity 11

21 10*2 EE Pa . The steel rods lengths and areas are 0.21 L m and 3.12 L m , and, 0.61 A 2cm and 0.32 A 2cm , respectively. The

system is loaded with by vertical force P acting at point D. The steel rods and force P locations are determined by the following distances measured from point A:

5.1a m , 5.2b m and 0.4c m , respectively. What is the allowable force allP supported by the system if the allowable normal stress for steel is 810*1.2all Pa .

Figure 4.2.4.a

- 174 -

A. General Observations The system illustrated in Figure 4.2.4.a is geometrical defined by the data contained in the text. B. Calculations B.1 Free-Body Diagram The rigid beam AD is supported at the ends B and C by two steel rods, which are playing the supporting role for the beam, together with the pinned support located at point A . Sectioning the rods and replacing them by two corresponding axial forces 1F and 2F , and substituting the constraints introduced by the pinned support by its corresponding reaction forces, AH and AV , the free-body diagram of the system is obtained and illustrated in Figure 4.2.4.b.

Figure 4.2.4.b B.2 Reactions Calculation The following equilibrium equations are written:

0 xF 0AH no axial force

0 yF 021 PFFVA

0A

zM 0*** 21 cPbFaF

The last two equilibrium equations contain three unknown reaction forces 1F , 2F and AV . The system is statically indeterminate. In order to find these unknown quantities one additional equation is necessary. This equation is obtained from the deformation compatibility condition schematically described in Figure 4.2.4.c.

- 175 -

Because the beam AD is rigid, purely geometric relations between the rod elongations, 2e and 2e , and the rotation angle are written as:

*1 ae

*2 be

Figure 4.2.4.c

Using the elongation expressions the forces in the rods are calculated as:

****11

11 aka

fF *10*9*5.1*10*6 77 Nm

mN

****12

22 bkb

fF *10*923.6*5.2*10*615.4 77 Nm

mN

where the stiffness coefficients, 1k and 2k , are calculated from the geometrical and material properties characteristics of the rods as:

k1E1 A1

L1

mPa

0.210*6*10*2 411

k1 6 107Nm

k2E2 A2

L2

mPa

3.110*3*10*2 411

k2 4.615 107Nm

Substituting the expressions of the rod forces into the last two equilibrium equations the three original unknowns, 1F , 2F and AV , are replaced with two unknowns, AV and . The equilibrium equations are then written as: Given VA k1 a k2 b P 0

k1 a2 k2 b2 P c 0

Solving the algebraic system, the two unknowns are found as:

- 176 -

Find VA float 4

1.069 P

1.055 10-8P

Pa m2

The axial forces in the rods are calculated:

F1 k1 a

substitute 1.055 10 8P

Pa m2

float 4F1 .9495 P

F2 k2 b

substitute 1.055 10 8P

Pa m2

float 5F2 1.1199 P

To find the force P producing an allowable stress all in the rods the following equations are written:

A1 all F1

substitute F1 0.9495 P

solve P

float 4

1.327 105 Pa m2

P1_all 1.327 105 N

for rod (1)

A2 all F2

substitute F2 1.1199 P

solve P

float 4

5.626 104 Pa m2

P2 5.626 104 N

for rod (2)

The allowable force P for the entire system is:

Pall min P1 P2 )10*626.5,10*327.1min( 45 NN Pall 5.626 104 N

The vertical displacements at point B and C are then calculated:

uB1k1

P1_allsubstitute P1_all 1.327 105 N

float 3uB

2.21 10-3 Pa m( )

N

uB 2.21 103 m

- 177 -

uC1k1

P2_allsubstitute P2_all 5.626 104 N

float 3uC

9.38 10-4 Pa m( )

N

uC 9.38 10 4 m

Problem 4.2.5 An axial load P is applied to a tapered circular rod of length L as shown in Figure 4.2.5.a. The variation of the rod radius along its length is expressed as:

)1

1(*)( 0

Lxrxr

where )0(0 rr is the radius at the left end cross-section. Symbolically express: (a) the axial stress )(x , (b) the elongation strain )(x and (c) the total elongation e . Apply the above obtain expressions for the case when where P = 9 kN, L = 2.50 m, r0 = 0.50 m and the rod is made of an aluminum alloy for which E = 69 GPa.

Figure 4.2.5.a

A. General Observations The tapered aluminum rod is subjected to a non-uniform axial deformation, because the area is not constant along the length of the beam. The variation of the area along its axis is expressed as:

20

220

2 )(*)(**)(*)(xL

LAxL

LrxrxA

where 0A is the area representing the cross-section at the left end of the rod (x=0).

- 178 -

B. Calculations B.1 Free-Body Diagram The free-body diagram is illustrated in Figure 4.2.5.b.

Figure 4.2.5.b

B.2 Reaction Calculation The constraint located at the left end of the beam is replaced by a horizontal concentrated reaction force AH . The reaction force AH is calculated using the equilibrium equation involving the horizontal projection of all forces. 0X 0 PH A PH A The system is statically determinate. B.3 Expression of the Axial Stress The axial force in the rod is constant and represents a tension force:

PxF )( The general expression of the axial stress in the rod is obtained as:

20

20

)1(*)(*)()(

)()(Lx

xLLA

PxA

PxAxFx

where 0 is the axial stress in the cross-section located at the left end of the rod. B.4 Expression of the Axial Strain

20

20

)1(*)1(*

)()()(

Lx

ELx

xExx

where 0 is the axial stress in the cross-section located at the left end of the rod.

- 179 -

B.5 Expression of the Rod Total Elongation The total elongation is obtained as:

e

0

L

x0 1xL

2

d exp 1( )73

0 L

B.6 Numerical Application The following data is substituted in the above obtained expressions:

P 9 103 N r0 0.5 m

L 2.5 m E 69 109 Pa

The reaction force is:

HA P

HA 9 103 N

The area at the left end cross-section:

A0 r02

A0 0.785 m2

The axial stress and elongation strain at the left end cross-section are:

0P

A0

0 1.146 104 Pa

0

0E

0 1.661 10 7

The variation of the axial stress and elongation strain are calculated in a number of twenty-one sections: i 0 20

xiL20

i

i 0 1xi

L

2

i 0 1xi

L

2

The maximum values are calculated in the right end cross-section (x = L):

max 20 max 4.584 104 Pa

max 20

max 6.643 10 7

The total elongation of the rod is:

emax73

0 L

emax 9.688 10 7 m

- 180 -

The variation of the axial stress is illustrated in Figure 4.2.5.c.

0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.51 104

1.4104

1.8104

2.2104

2.6104

3 104

3.4104

3.8104

4.2104

4.6104

5 104

x

Figure 4.2.5.c

Problem 4.2.6 A magnesium-alloy rod (Emag = 45 GPa) of diameter dm = 30 mm is encased by a brass tube (Ebra = 100 GPa) with outer diameter db. Both bars have an equal length L = 500 mm. An axial load P = 40 kN is applied to the resulting bimetallic rod. Assuming that that the magnesium rod and the brass tube are securely bonded to each other calculate: (a) the outer diameter db of the tube, if three fourths of the load P is carried by the magnesium rod and one fourth by the brass tube and (b) the total elongation of the bimetallic rod. The bimetallic rod is illustrated in Figure 4.2.6. A. General Observations A.1 It is assumed, due to the bondage between the materials, that the deformation at the right end cross-section of the bimetallic rod is equal for both materials. A.2 The reaction force located at the left end of the rod is not necessary to be calculated, because the distribution of the external force P in-between the magnesium and brass cross-section is established.

Fmag14

P

Fmag 1 104 N

for magnesium

Fbra34

P

Fbra 3 104 N

for brass

- 181 -

Note: It can be remarked that the bimetallic rod it is statically determined system, but because the modulus of elasticity is not constant for the entire length of the rod, the rod is subjected to a nonuniform-axial deformation.

Figure 4.2.6 B. Calculations B.1 Calculation of the brass exterior diameter bd The equality of the deformation at the right end of the bimetallic rod is:

endbraendmag uu __ Using the flexibility coefficients magf and braf , corresponding to magnesium and brass rods the displacements equality is written as:

brabramagmag FfFf **

where magmag

mag AELf*

and brabra

bra AELf*

.

Substituting the forces and the flexibility coefficients into the above equation, the expression of the brass area is obtained:

LEmag Amag

14

PL

Ebra Abra34

P solve Abra 3 EmagAmagEbra

The area of the magnesium rod is calculated as:

- 182 -

Amag dm

2

4

430* 2

Amag 706.858mm2

and consequently, the area of the brass rod is obtained:

Abra 3EmagEbra

Amag

239

9

10*706858.0*10*10010*45*3 m

PaPa

Abra 954.259mm2

The exterior diameter of the brass tube is calculated from the following equation:

Abra db

2 dm2

4

solve db

float 5

45.990 mm

45.990 mm

db 45.990 mm

because only the positive value makes sense.

B.2 Calculation of total elongation The total elongation is equal to the displacement of the either material calculated at the right end of the rod. The calculation is conducted using the data pertinent for magnesium and is also verified using the brass data. e umag_end

umag_end fmag Fmag

fmagL

Emag Amag 239 10*706858.0*10*45

5.0mPa

mfmag 1.572 10 8

mN

umag_end 1.572 10 4 m

e 1.572 10 4 m The verification using brass:

fbraL

Ebra Abra 239 10*954259.0*10*100

5.0mPa

m

fbra 5.24 10 9

mN

ubra_end fbra Fbra

ubra_end 1.572 10 4 m

- 183 -

Note: It can be remarked that the deformation of both materials is 1.572*10-4 m and, this way, the geometrical condition imposed is also verified numerically. Problem 4.2.7 A uniform circular cylinder of diameter d and length L is made of a material with modulus of elasticity E. It is fixed to a rigid wall at end A and subjected to a distributed external axial loading of magnitude p(x) per unit length, as shown in Figure 4.2.7.a. The axial stress, )(x , varies linearly with x as shown in Figure 4.2.7.b. Determine: (a) the expression for the distributed loading, p(x) and (b) the expression for the axial displacement, u(x), of the cross section.

Figure 4.2.7

A. General Observations A.1 The member is subjected to a nonuniform-axial deformation and it is statically determinate system. A.2 The axial stress has the following expression (see Figure 4.2.7):

x( ) 0 1xL

B. Calculations The horizontal reaction force

AH is calculated as:

HA 0 A

The axial force pertinent to a particular cross-section is obtained from the equilibrium as:

F x( ) HA0

x

p

d 0 A0

x

p

d

- 184 -

The axial stress is expressed:

x( )F x( )

A

HA0

x

p

d

A0

0

x

p

d

A

Comparing the above expression with the expression given by the problem the following equation is written:

0

x

p

d

A0

xL

and after the algebraic manipulation the integral becomes:

0

x

p

d 0 AxL

Consequently,

p x( )0 A

L It can be concluded that the linear variation of the axial stress is induced by a constant axially applied load. Problem 4.2.8 Two rods are stress-free when welded together at point B and welded to two rigid walls at points A andC . The geometrical and material characteristics of the two rods are illustrated in Figure 4.2.8.a. Subsequently, rod (1) and rod (2) are heated by an amount 1T and 2T , respectively from their initial installation temperature. Determine an expression for the axial forces induced in each rod by the change in temperatures.

Figure 4.2.8.a

- 185 -

A. General Observations A.1 The rods have the same modulus of elasticity E , but different expansion coefficients 1 and 2 . A.2 It can be remarked the absence of any external load and the existence of a change in temperature B. Calculations B.1 Free-Body Diagram The free-body diagram is illustrated in Figure 4.2.8.b.

Figure 4.2.8.b B.2 Reaction Calculation The constraints located at the left and right end points, A and C , of the member are replaced by two horizontal reaction forces, AH and CH , respectively. The following equilibrium equation is written: 0X 0 CA HH CA HH . The system is statically indeterminate and in order to calculate the reaction forces an additional equation is necessary. This equation is the fact that the total elongation e is zero:

0 AC uue The expression of the displacements Bu and Cu are:

11111 *** TLFfuu AB

222111221122222 ********* TLTLFfFfuTLFfuu ABC

- 186 -

where

11

11 * AE

Lf and 22

22 * AE

Lf are the flexibility coefficients

The axial forces in the rods are:

AHxF )(1 and AHxF )(2 The elongation is calculated as:

22211121

2221112211

*****)(******

TLTLHffTLTLFfFfuue

A

AC

From the condition imposing that the total elongation to be zero, the reaction force

AH is obtained:

21

222111 ****ff

TLTLH A

Consequently, the reaction force CH is:

21

222111 ****ff

TLTLH C

and the forces ion the rods are:

21

22211121

****)()(

ffTLTLxFxF

4.3 Proposed Problems Problem 4.3.2 The three-part axially loaded member, shown in Figure 4.3.2, consists of a tubular segment (1) with outer diameter 32mmdo1 and inner diameter 22mmdi1 , a solid circular rod segment (2) with diameter 32mmd2 and a third solid circular rod segment (3) with diameter 22mmd3 .

Figure 4.3.2

- 187 -

All three applied loads shown are acting along the centroidal axis of the members. Considering that the rigid couplers have a negligible length determine: (a) the axial stresses in each one of the three respective segments, (b) the displacements in points B, C and D if all segments have equal lengths L1 = L2 = L3 = 0.50 m and the modulus of elasticity of the material is 21000MPaE .

Problem 4.3.3 The diameter of the central one-third of a 50 mm diameter steel rod is reduced to 20 mm, forming a three-segment rod, as shown in Figure 4.3.3. For the loading shown, determine the displacements of the rod points B, C and D, respectively. The rod is made of a material which has the modulus of elasticity E = 200 GPa.

Figure 4.3.3

Problem 4.3.4 A column in a two-story building is fabricated from square structural steel tubing having a modulus of elasticity E = 210 GPa. The cross-sectional dimensions of the two segments are shown in Figure 4.3.4. Two axial loads acting along the centroidal axis of the column are applied to the column at levels A and B. Calculate: (a) the axial stress both segments of the column and (b) the total shortening of the column length

Figure 4.3.4

- 188 -

Problem 4.3.5 A three-segment stepped aluminum-alloy column is subjected to the vertical axial loads shown in Figure 4.3.5 The cross-sectional areas of the segments are A1 =3870 mm2, A2 =5810 mm2 and A3 =9035 mm2, respectively. The material modulus of elasticity is E = 69 GPa. Calculate: (a) the axial stresses in all three segments and (b) the vertical displacement of the column at nodes A, B and C under the given loading system.

Figure 4.3.5

Problem 4.3.6 A uniform rod is subjected to three axial loads acting as shown in Figure 4.3.6 and is made of a material with a modulus of elasticity E = 70 GPa. What is the minimum allowable diameter of the cylindrical rod if the displacement at the right end D and the maximum axial stress in the rod can not excide 5 mm and 80 MPa, respectively?

Figure 4.3.6

- 189 -

Problem 4.3.7 A 1.2 in diameter aluminum-alloy hangar, illustrated in Figure 4.3.7, is supported by a steel pipe with an inside diameter of id =75 mm. The moduli of elasticity for the steel pipe and hanger are Esteel = 310210 MPa and Ealuminum = 31069 MPa, respectively. Determine the thickness of the steel pipe if the maximum axial displacement at the node C is 2.5 mm.

Figure 4.3.7

Problem 4.3.8 A 3.60 m rigid beam AB that weighs 0.80 kN supports an air conditioner that weighs Wair=4.45 kN. The beam is supported by hanger rods (1) and (2) located at its ends as shown in Figure 4.3.8.

Figure 4.3.8

- 190 -

Conduct the following calculations: (a) if the diameter of rod (1) is 95 mm what is the stress in the rod? (b) if the stress in rod (2) is to be the same as the stress in rod (1), what should the diameter of rod (2)? (c) what are the downward displacements at the ends of the rigid beam if the rods length are L1 = L2 = 1.80 m and they are made of steel with a modulus of elasticity E =210 103 MPa? Problem 4.3.9 A hanger rod CD is attached to a rigid beam AB. The beam is supported at its ends by two hanger rods. Assuming that all tree hangers are identical and are made from the same material, calculate: (a) the axial stress in all tree hangers, (b) the vertical displacement of the points A, B, C and D.

Figure 4.3.9

Problem 4.3.10 A commercial sign of weight W is supported by a structural system comprised from a rigid beam AB and a wire CD, as shown in Figure 4.3.10. The rigid beam has a negligible weight, while the wire has length L, cross-sectional area A, and modulus of elasticity E. Assuming that the attachment pin D is directly located above pin A, and when there is no load acting on the beam, the beam is in a perfect horizontal position calculate the following: (a) the axial stress in the wire CD when the sign is attached at points B and C of the beam, (b) the vertical displacement in point C of the beam.

- 191 -

Figure 4.3.10

Problem 4.3.11 The inclined rod AB, shown in Figure 4.3.12, is pinned to a fixed support at A and is pinned at the end B to a block that is forced to move only horizontally when the load P is applied. Determine: (a) an expression for the axial stress in the inclined rod as a function of P, L, E, A, and the angle , (b) an expression for the horizontal displacement at the end B.

Figure 4.3.11

Problem 4.3.12 A bimetallic bar is made by bonding together two homogeneous rectangular bars, each having a width b, length L and moduli of elasticity of the bars are E1 and E2, respectively. An axial force P is applied to the ends of the bimetallic bar at location (y= yp, z = 0) such that the bar undergoes an axial deformation only. Assuming the following data: L = 1.5 m, b = 50 mm, t1 = 25 mm, t2 = 15 mm, E1 = 70 GPa, E2 = 210 GPa and P = 48 kN. Calculate: (a) the normal stress in each material, (b) the value of yp, (c) the elongation of the bar.

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Figure 4.3.12

Problem 4.3.13 A bimetallic bar, shown in Figure 4.3.13, undergoes an axial deformation. The bar has the following geometrical and material characteristics: L = 2.55 m, b = 50 mm and h1 = h2 = 1.50 mm and E1=210 GPa. Calculate: (a) the modulus of elasticity E2 if the load P is applied at 10 mm, and (b) the total elongation of the bar for a load of P = 9 kN.

Figure 4.3.13 Problem 4.3.14 A steel pipe is filled with concrete, and the resulting column is subjected to a compressive load P = 360 kN. The pipe has an outer diameter of 325 mm and an inside diameter of 305 mm. The elastic moduli of the steel and concrete are: Estel = 210 GPa and Econc = 25 GPa. Determine: (a) the stress in the steel and the stress in the concrete due to this loading, (b) the shortening of the column if its initial length is L = 3.70 m, (Ignore- radial expansion of the concrete and steel due to Poisson's ratio effect.)

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Figure 4.3.14

Problem 4.3.15 A homogenous rod of length L and elasticity modulus E is a conical frustum with diameter d(x) that varies linearly from d0 at one end to 2*d0 at the other end, with d0 << L. An axial load P is applied to the rod, as shown in Figure 4.3.15. Determine analytical expressions for: (a) the stress distribution, )(x , on an arbitrary cross section and (b) the elongation of the rod, )(xe .

Figure 4.3.15

Problem 4.3.16 A uniform circular cylinder of diameter d and length L is made of a material with modulus of elasticity E and specific weight . It hangs from a rigid ceiling as shown in Figure 4.3.16. Determine: (a) the expression of the axial stress )(x , (b) the expression of the strain )(x , (c) the expression of the

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displacement )(xu and (d) the compression force necessary to be applied in order to return the bar at its initial length.

Figure 4.3.16

Problem 4.3.17 A steel pipe with outer diameter do = 50 mm, and inner diameter di = 38 mm and a solid aluminum-alloy rod of diameter d = 19 mm form a three-segment system that undergoes axial deformation due to a single external load PC = 55 Kn acting on a collar at point C, as shown in Figure 4.3.17. Considering the following data: L1 = L2 = 0.75m, L3 = 1.20 m, E1 = 210 GPa and E2 = E3 = 69 GPa, calculate: (a) the axial stresses induced in all three segments and (b) determine the axial displacement at points B and C.

Figure 4.3.17

Problem 4.3.18 A three-segment rod is attached to rigid supports at ends A and D and is subjected to equal and opposite external loads P at nodes B and C, as shown in Figure 4.3.18. The rod is homogeneous and linearly elastic, with modulus of elasticity E. Assuming A1 =

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A3 = A and A2 = 2A, L1 = 2L and L2 = L3 = L, calculate: (a) the axial stresses in all tree segments and (b) the horizontal displacements at nodes B and C, respectively.

Figure 4.3.18

Problem 4.3.19 A rigid beam AD, supported by a pin at its end point D and attached by the two vertical steel rods at points A and C, is loaded by a vertical load P at point B. Neglecting the weight of the beam and assuming that the support rods are stress-free when P = 0, calculate: (a) the forces F1 and F2 in the support rods after load P is applied, (b) the deformation of the supporting and (c) the expressions previously obtained if A1 = A2 = 500 2mm , L1 = 1m, L2 = 2m , E1 = E2 = 210 GPa, P = 50 kN, a = 0.50 m and b = 1.5 m.

Figure 4.3.19

Problem 4.3.20 A rigid beam AD, shown in Figure 4.3.20, is supported by a smooth pin at B and by two vertical rods attached to the beam at points and C. Neglecting the weight of the beam and assuming that the rods are stress-free when P = 0. Considering that A1 = 650 mm2, A2 = 325 mm2, L1 = L2 = 1.25 m, a = 0.60 m, b = 1.25m, E1 = E2 = 69 GPa

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and P = 22.70 kN, determine: (a) the axial forces in the support rods, (b) determine the axial stress in each support and (c) calculate the elongation of the support rods.

Figure 4.3.20

Problem 4.3.21 A rigid beam AD, shown in Figure 4.3.21, is supported by three identical vertical rods that are attached to the beam at points A, C, and D and loaded in node B with a vertical concentrated load P. Assuming that A = 650 mm2, L = 1.50 m, a = 0.50 m, b = 1.00 m, c = 1.50 m, E = 210 GPa and P = 45.5 kN, calculate: (a) the axial forces in the support rods and (b) the vertical displacements at nodes A, B, C and D.

Figure 4.3.21

Problem 4.3.22 The rod composed from two segments as shown in Figure 4.3.22, is attached to rigid walls at A and C. Determine expressions for the stresses pertinent to both segments resulting from a uniform temperature increase T of the entire rod. Numerical

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application: A1 = 1000 mm2, A2 = 1500 mm2, L1 = 2 m, L2 = 1.5 m, E1 = 210 Gpa, E2 = 120 GPa, and and T = 30 C .

Figure 4.3.22

Problem 4.3.23 A three-segment rod shown in Figure 4.3.23 is rigidly attached to walls at points A and D. Subsequently, the middle segment is heated by an amount, while the segments (1) and (3) are kept at their original temperature. Using the notation shown in Figure 4.3.23 and considering the system free at stress at the installation, calculate the stresses in the segments and the axial displacements at points B and C.

Figure 4.3.23

Problem 4.3.24 A steel pipe with outer diameter do, and inner diameter di and a solid aluminum-alloy rod of diameter d form a three-segment system as shown in Figure 4.3.24. The system is considered stress free when is welded to the rigid supports at points A and D. The installation temperature FT oninstallati

060 is recorded. Subsequently, the aluminum rod is cooled by 100°F ( FTT 0

32 100 ), while the steel pipe is held at the initial temperature )0( 0

1 FT . Assuming do = 50 mm, di = 38 mm, d = 19 mm, L1 = 1.25 m, L2 = L3= 0.75 m, E1 = E3 = 210 GPa, E2 = 69 GPa, and2

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Figure 4.3.24

Problem 4.3.25 The mechanical system shown in Figure 4.3.25 is composed from two identical steel rods (A = 40 mm2, E = 200 GPa, ) and a "rigid" beam AC. The beam is supported by a smooth pin at B. Assuming the two rods stress-free after installation, determine: (a) the axial stresses induced in rods if their temperature is decreased by 50°C and (b) the small angle through which the beam AC would rotate due to this temperature change.

Figure 4.3.25

Problem 4.3.26 The steel rod of diameter 20 mm is held without any initial stresses between two rigid walls as illustrated in Figure 4.3.26. Determine the temperature drop T at which the stress in the rod reaches 200 MPa. Use for the steel E=200 GPa and

Figure 4.3.26

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Problem 4.3.27 The bar AB, shown in Figure 4.3.27, is held between rigid supports and heated nonuniformly in such a manner that the temperature increase T at distance x from end A is given by the expression T (x) = T1*(x/L)2, where T1 is the increase in temperature at end B of the bar. Obtain a formula for the compressive stress in the bar. (Assume that bar has a length L and is made of a material with modulus of elasticity E and coefficient of thermal expansion .)

Figure 4.3.27

Problem 4.3.28 The copper bar AB of length 1.00 m is placed in position shown in figure 4.3.28 at room temperature. A gap of 0.2 mm exists between the end A of the bar and a rigid restraint. Calculate the axial compressive stress in the bar if the temperature is raised 90°F. Use for copper the following material constants: E=110 GPa and .

Figure 4.3.28

Problem 4.3.29 Three identical springs, 10 in. apart, are attached to a horizontal rigid bar at points A, B, and C, as illustrated in Figure 4.3.29. Three vertical loads of magnitudes 138 N, 45 N and 31 N act at points A, B, and C, respectively. Calculate the angle of rotation (degrees) of the rigid bar if the spring stiffness k is 14 N/m.

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Figure 4.3.34

Problem 4.3.30 The rigid bar ABCD, shown in Figure 4.3.30, is pinned at point B and supported by springs at A and D. The springs at A and D have stiffnesses k1 = 11.20 kN/m and k2 = 36.75 kN/m, respectively. The dimensions a, b, and c are 0.45 m, 0.90 m, and 0.35 m, respectively. A load P acts at point C. Determine the maximum permissible load Pmax if the angle of rotation of the bar due to the action of the load P is limited to 2°.

Figure 4.3.30

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REFERENCES

1. Craig, Roy R., Mechanics of Materials, Second Edition, John Willey & Sons, New York, 2000.

2. Gere, James M., Mechanics of Materials, Fifth Edition, Brooks/Cole,

Pacific Grove, CA, 2001.

3. Higdon, A., Ohlsen, E.H., and Stiles, W.B., Mechanics of Materials, Third Edition, John Willey & Sons, New York, 1962.

4. Popov, E.P., Introduction to Mechanics of Solids, Prentice-Hall Inc.,

Englewood Cliffs, New Jersy, 1968.

5. Mazilu P., Posea N. and Iordachescu E., Probleme de Rezistenta Materialelor, Vol I, Editura Tehnica Bucuresti, 1969.

6. Hartog, J.P.D., Strength of Materials, Dover Publications Inc., New York,

1961.

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APPENDIX 1.1 Modulus of Elasticity and Poisson’s Ratio

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APPENDIX 1.2 Yield and Ultimate Stress

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APPENDIX 2.1 Geometrical Characteristics of Plane Areas

B-B axis respect to with inertia ofmoment axes z andy theoforigin therespect to with inertia ofmoment polar

axes z andy therespect to with inertia ofproduct

lyrespective axes, z andy therespect to with inertia of momentsC centroid todistances,

area

BB

zyP

yz

zy

IIII

I,IIzy

A

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APPENDIX 2.2 Properties of the Romanian Rolled Shaped Sections

TABLE 2.2.1 Wide Flange Sections (I SHAPES)

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TABLE 2.2.2 Channel Sections (U SHAPES)

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TABLE 2.2.3 Angle Sections with Equal Legs (L SHAPES)

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TABLE 2.2.3 Angle Sections with Equal Legs (L SHAPES) - continuation

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TABLE 2.2.4 Angle Sections with Unequal Legs (L SHAPES)

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APPENDIX 2.3 Properties of American Rolled Shaped Sections .

TABLE 2.3.1 Wide flange sections (W SHAPES)

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TABLE 2.3.1 Wide flange sections (W SHAPES) - continuation

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TABLE 2.3.2 Channel Sections (C SHAPES)

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TABLE 2.3.3 Angle Sections with Equal Legs (L SHAPES)

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