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Strength of MaterialsWORK-BOOK

RM Nkgoeng

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Copyright c 2013 Mashilo Nkgoeng

SEL F-PUBLISHED

WWW.MASHILONKGOENG.CO.ZA

Licensed under the Creative Commons Attribution-Non-commercial 3.0 Unported License (the

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ANY KIND, either express or implied. See the License for the specific language governing

permissions and limitations under the License.

First printing, December 2013

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Contents

1 Deflection of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1 What is a Beam? 5

1.1.1 Beam terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.2 Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.3 Assumption of Classical Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.4 Beam Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.5 Support Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.6 Stresses, strains and bending moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Notation 7

1.3 Second Order Method for Beam Deflections 7

1.4 Double Integration Using Bracket Functions 7

1.5 Examples 9

1.6 Exercises 16

2 Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.1 Introduction 19

2.1.1 Point Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1.2 Uniformly Distributed Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Exercise 22

2.3 Examples 23

2.4 Exercises 23

3 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.1 Introduction 25

3.2 Strain Energy of Bars 26

3.3 Castigliano’s Theorem 26

3.4 Structures 27

3.5 Castigliano’s theorem applied to Curved Beams 27

3.6 Castigliano’s theorem applied to Beams 27

3.6.1 Cantilever beam with a Point Load at the free end . . . . . . . . . . . . . . . . . . . . . . . 27

3.6.2 S/S beam with a Point Load at mid-point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.7 Examples 29

3.8 Exercises 34

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4 Unsymmetrical Bending of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.1 Symmetric Member in Pure Bending 39

4.2 Unsymmetrical Bending 40

4.3 Alternative procedure for stress determination 42

4.4 Deflection 44

4.5 Notation 44

4.6 xxx 44

4.6.1 Point Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.6.2 Uniformly Distributed Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.7 Examples 44

4.8 Exercises 47

5 Inelastic Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.1 Plastic Bending of Rectangular Beams 49

5.2 Plastic Bending of Symmetrical (I-Section) Beam 52

5.3 Partially plastic Bending of Unsymmetrical Sections 53

5.4 Limit Analysis-Bending 56

5.4.1 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.4.2 The principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.5 Solid Shaft 65

5.6 Hollow shaft 675.7 Exercises 69

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What is a Beam?

Beam terminology

Mathematical Models

Assumption of Classical Beam Theory

Beam Loading

Support Conditions

Stresses, strains and bending moments

Notation

Second Order Method for Beam Deflections

Double Integration Using Bracket

Functions

Examples

Exercises 1 — Deflection of Beams

Introduction

This lecture notes starts the presentation of methods for computed lateral deflections of plane

beams undergoing symmetric bending. The workbook just summarises the topic, but the

textbook 1 covers in detail this topic in chapter 9 and 10. We assume that as a student you are

familiar with the following:

1. integration of Ordinary Differential Equations

2. Statics of plane beams under symmetric bending. This is also covered in Chapter 5 and 6of the prescribed textbook.

1.1 What is a Beam?

Beams are the most common type of structural component, particularly in Civil and Mechanical

Engineering. A beam is a bar-like structural member whose primary function is to support

transverse loading and carry it to the supports. By bar-like, we mean that one of the dimensions

is considered larger than the other two. This dimension is called the longitudinal dimension

or beam axis. The intersection of planes normal to the longitudinal dimension with the beam

member are called cross sections. A longitudinal plane is one that passes through the beam axis.

A beam resist transverse loads mainly through bending action. Bending produces compressivelongitudinal stresses in one side of the beam and tensile stresses in the other. the two regions

are separated by a neutral surface of zero stress. The combination of tensile and compressive

stresses produces an internal bending moment. This moment is the primary mechanism that

transports loads to the supports. This is illustrated by Fig. #

1.1.1 Beam terminology

General Beam –it is a bar like member designed to resist a combination of loading actions such

as biaxial bending, transverse shears, axial stretching or compression, possibly torsion. If

the internal axial force is compressive, the beam has also to be designed to resist buckling.

If the beam is subject primarily to bending and axial forces, it is called a beam-column. If

it is subjected primarily to bending forces, it is called simply a beam. A beam is straight if its longitudinal axis is straight. It is prismatic if its cross section is constant.

Spatial Beam –it supports transverse loads that can act on arbitrary directions along the cross

section.

Plane Beam –it resists primarily transverse loading on a preferred longitudinal plane.

1.1.2 Mathematical Models

One-dimensional mathematical models of structural beams are constructed on the basis of beam

theories. Since beams are actually three-dimensional bodies, all models necessarily involve

some form of approximation to the underlying physics. The simplest and best known model for

straight, prismatic beams are based on the Bernoulli-Euler theory as well as the Timoshenko

1Mechanics of Materials by Gere and Goodno

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6 Deflection of Beams

beam theory. The B-E theory is the one that is taught in SOM3602 and is the only one that we

will be dealing with in SOM401M. The Timoshenko model incorporates a first order kinematic

correction for transverse shear effects. This model assumes additional importance in dynamics

and vibration.

1.1.3 Assumption of Classical Beam Theory

The classical beam theory for plane beams rests on the following assumptions:

1. Planar symmetry: The longitudinal axis is straight and the cross section of the beam has

a longitudinal plane of symmetry. The resultant of the transverse loads acting on each

section lies on that plane. The support conditions are also symmetric about this plane.

2. Cross section variation: The cross section is either constant or varies smoothly.

3. Normality: Plane sections originally normal to the longitudinal axis of the beam remain

plane and normal to the deformed longitudinal axis upon bending.

4. Strain energy: The internal strain energy of the member accounts only for bending

moment deformations. All other contributions, notably transverse shear and axial force,are ignored.

5. Linearisation: Transverse deflections, rotations and deformations are considered so small

that the assumptions of infinitesimal deformations apply.

6. Material model: The material is assumed to be elastic and isotropic. Heterogeneous

beams fabricated with several isotropic materials, such as reinforced concrete, are not

excluded.

1.1.4 Beam Loading

The transverse force per unit length that acts on the beam in the y+ direction is denoted by

f ( x) as shown in Figure #Point loads and moments acting on isolated beam sections can berepresented with Discontinuity Functions (DF’s).

1.1.5 Support Conditions

Support conditions for beams exhibit far more variety than for bar members. Two cases are often

encountered in engineering practice: simple support and cantilever support. These are shown

in Figs. # and # respectively. Beams often appear as components of skeletal structures called

frameworks, in which case the support conditions are of more complex type. Easily solved using

finite element methods.

1.1.6 Stresses, strains and bending moments

The Bernoulli-Euler model assumes that the internal energy of beam member is entirely due to

bending strains and stresses. Bending produces axial stresses σ x which is just abbreviated σ and

axial strains ε x which is just ε . The strains can be linked to the displacements by differentiating

the axial displacement u( x):

ε = ∂ u

∂ x= − y d

2v

dx2 = − yv′′ = − yκ (1.1)

where κ denotes the deformed beam axis curvature. The bending stress is linked to e through

one-dimensional Hooke’s law

σ = Ee = − Ey d 2vdx2

= − Eyκ (1.2)

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1.2 Notation 7

The most important stress resultant in classical beam theory is the bending moment M x, which is

defined as the cross sectional integral

M x = A −

yσ dA = E d 2v

dx2

A

y2dA = EI xxκ (1.3)

The bending moment is considered positive if it compresses the upper portion: y > 0. the product EI xx is called the bending rigidity of the beam with respect to flexure about the x axis.

1.2 Notation

Quantity Symbol

Generic load for ODE Work f ( x)Transverse Shear Force V ( x)

Bending Moment M ( x)Slope of deflection curve

dv( x)dx

= v′( x)Deflection Curve v( x)

Table 1.1: Notation

1.3 Second Order Method for Beam Deflections

The second-order method to find beam deflections gets its name from the order of the ODE to be

integrated: EI xxv′′( x) = M ( x) is a second order ODE. The procedure can be broken down into

the following steps:

1. Find the bending moment M ( x) = d 2v( x)

dx2 directly. i.e. by cutting the beam at distance x

and taking moments about x.

2. Integrate M ( x) once to get the slope v′( x) = dv( x)dx

3. Integrate the slope v′( x) once to get the deflection v( x)4. If there are no continuity conditions, the above steps will produce two integration constants

C 1 and C 2. Apply kinematics boundary conditions to determine the values of the integration

constants. If there are continuity conditions, more than two constants of integration may

appear and are solved the same way, i.e. by using boundary conditions.

5. Substitute the constants of integration into the deflection function to get v( x)6. Evaluate v( x) at points of interest on the beam.

1.4 Double Integration Using Bracket Functions

When the loads on a beam do not conform to standard cases, the solution for slope and deflection

may be found from first principles. Macaulay developed a method for making the integration

simpler. The basic equation governing the slope and deflection of beams is:

EI d 2 y

dx2 = M : Where M is a function of x (1.4)

When a beam has a variety of loads it is difficult to apply this theory because some loads may

be within the limits of x during the derivation but not during the solution of a particular point.

Macaluay’s method makes it possible to do the integration necessary by placing all the termscontaining x within a square bracket and integrating the bracket, not x. During evaluation, any

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bracket with a negative value is ignored because a negative means that the load it refers to is

not within the limit of x. The general method of solution is conducted as follows. Refer to Fig.

#. In a real example, the loads and reactions would have numerical values but for the sake of

demonstrating the general method we will use algebraic symbols. This has only point loads.

1. Write down the bending moment equation placing x on the extreme right hand end of thebeam so that it contains all the loads. Write all terms containing x in a square bracket.

EI d 2 y

dx2 = M = R A[ x]− P1[ x − a] − P2[ x − b]− P3[ x −c]

2. Integrate once treating the square bracket as the variable.

EI dy

dx= R A

[ x]2

2 − P1 [ x −a]

2

2 − P2 [ x − b]

2

2 −P3 [ x −c]

2

2 +C 1

3. Integrate again using the same rules.

EIy = R A[ x]3

6 − P1 [ x −a]

3

6 − P2 [ x − b]

3

6 −P3 [ x − c]

3

6 +C 1 x +C 2

4. Use boundary conditions to solve constants C 1 and C 2.

5. Solve slope and deflection by putting in appropriate value of x. IGNORE any brackets

containing negative values.

Evaluating the constants of integration that arise in the double-integration method can become

very involved if more than two beam segments must be analysed. We can simplify the calculations

by expressing the bending moment in terms of discontinuity functions, also known as Macaulay

bracket functions. Discontinuity functions enable us to write a single expression for the bending

moment that is valid for the entire length of the beam, even if the loading is discontinuous. By

integrating a single, continuous expression for the bending moment, we obtain equations for

slopes and deflections that are also continuous everywhere. As an illustration let us consider a

simply supported beam with three segments as shown in Fig. 1.1. I have already determined the

reactions. All we need to do is to segment it nicely and take moments about X − X as follows:

M xx = 480 x − 500 x − 2− 4502 x − 32 Nm (1.5)

Note that a bracket function is zero by definition if the expression in the brackets–namely ( x−a)

A B

q0 = 450 N /m

500 N

480 N 920 N

X

X

5m

2m 1m

x

Figure 1.1: Macaulay’s Method

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1.5 Examples 9

is negative; otherwise, it is evaluated as written. A bracket function can be integrated by the

same rule as an ordinary function–namely,

x

−a

n =

x −an+1

n + 1

+C (1.6)

This is called the global bending moment equation and it can be integrated to obtain the slope

and the deflection equations for the entire beam. The two constants of integration as mentioned

before can be computed from the boundary conditions. When you use this method ensure that

you include every load on the beam and leave the last reaction if the beam is simply supported.

If the beam is a cantilever, you can obtain the global bending moment equation by starting from

the free end or by starting from the fixed end. If you choose to start from the fixed end, calculate

the reaction first so that you can be able to include it in the global bending moment equation. At

the end of the day, you have the choice to use any method that you feel comfortable with. We

have not included other methods, simply because in practice we have found that we hardly use

the other methods in solving deflection of beam problems.

1.5 Examples

Example 1.1 Determine the equation for the deflection curve as well as the slope in Fig. 1.2.

q N /m

L

Figure 1.2: Cantilever Beam with UDL

Solution 1.1 Still to be done

Example 1.2 Determine the equation for the deflection curve as well as the slope in Fig. 5.25.

P

L

Figure 1.3: Cantilever Beam with UDL


Example 1.3 Determine the equation for the deflection curve as well as the slope in Fig.1.4 .

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A B

P

L

L/2

Figure 1.4: SS Beam with Point Load


Example 1.4 For the cantilever beam, Fig. 1.5 under triangular distributed loading or variably

distributed loading, determine the equation for the deflection curve as well as the slope.

q0 X

X

A

C

DB

L

x

Figure 1.5: Cantilever with VDL

Solution 1.4 We will use the same figure, Fig. 1.5 as the free-body diagram. Let the height

CD = q and use the law of triangles to get the expression of q in terms of the maximum heightor intensity q0, i.e.

q

x=

q0

L∴ q =

q0 x

L

The moment about X − X is as follows:

EIv′′( x) = −12

q × x × x3

= −q0 x3

6 L

We need to integrate the above twice to get two constants of integration

EIv′( x) = −q0 x4

24 L+C 1

EI v( x) = − q0 x5

120 L+C 1 x +C 2

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1.5 Examples 11

The kinematic boundary conditions are v′( L) = 0 and v( L) = 0

EIv′( L) = 0 = −q0 L4

24 L+C 1; ∴C 1 =

q0 L3

24

EI v( L) = 0 = − q0 L5

120 L+ q0 L

4

24 +C 2; ∴C 2 = −q0 L

4

30

The equation of interest are:

v( x) = 1

EI

− q0 x

5

120 L+

q0 L3 x

24 − q0 L

4

30

(1.7)

and

v′( x) = 1

EI

−q0 x

4

24 L+

q0 L3

24

(1.8)

Example 1.5 For a given beam shown below:

1. derive the equation of the elastic curve2. determine the point of maximum deflection

3. determine ymaxBefore one could jump in and provide a solution, it is advisable to do the following:

• Develop an expression for M ( x) and derive a differential equation for the elastic curve.• Integrate the differential equation twice and apply boundary conditions to obtain the elastic

curve equation

• Locate a point of zero slope or point of maximum deflection• Evaluate corresponding maximum deflection

P

L a

A B C

Figure 1.6: SS Beam with an Overhang

Solution 1.5 The reactions are found to be R A = Pa L

and R B = P

1 + a L

. Taking moments about

X − X we get

M ( x) = −Pax L

(0

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The boundary conditions are simply: x = 0 : y = 0 and x = L : y = 0. The first boundary conditionleads to C 2 = 0 and the second boundary condition leads to C 1 =

PaL6

. The resulting equations

are therefore

EI

dy

dx = −Pax2

2 L +

PaL

6 (1.9)

and

EIy = −Pax3

6 L+

PaL

6 x (1.10)

Determining the position of maximum deflection is relatively easy, set y′( x) = 0

dy

dx= 0 =

PaL

6 EI

1 − 3

xm L

2

∴ xm = L√

3= 0.577 L

ymax = PaL2

6 EI

0.577 − 0.5773

= 0.0642PaL2

6 EI

Example 1.6 For the uniform beam shown below,

1. Determine the reaction at A

2. Derive the equation for the elastic curve

3. Determine the slope at A (Note that the beam is statically indeterminate to the first degree)

A B

L

Figure 1.7: LHS SS and RHS Fixed

Solution 1.6 The solution will include the following:

• develop differential equation for the elastic curve. (this will be functionally dependent onreaction A)

• Integrate twice and apply boundary conditions to solve for the reaction at A and to obtainthe equation for the elastic curve.

• Evaluate slope at ATaking moments about D, i.e. Σ M D = 0

M = R A x − 12

w0 x

2

L

x

3

M = R A x − w0 x3

6 L= EI d

2

ydx2

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1.5 Examples 13

We need to integrate E I d 2 y

dx2 twice to get the equation for the elastic curve.

EI d 2 y

dx2 = R A x − w0 x

3

6 L

EI dydx

= R A x2

2 − w0 x4

24 L+C 1

EIy = R A x

3

6 − w0 x

5

120 L+C 1 x +C 2

The boundary conditions are as follows:

x = 0 : y = 0 C 2 = 0

x = L : dy

dx= 0 C 1 =

w0 L3

24 − R A L

2

2

x = L : dy

dx

= 0 C 1 = w0 L

3

120 − R A L

2

6

From the above we find that the reaction at A is

R A = w0 L

10

The resulting two equations were:

y = w0

120 EI

− x5 + 2 L2 x3 − L4 x (1.11)θ =

w0

120 EI −5 x4 + 6 L2 x2 − L4

(1.12)

Example 1.7 A cantilever beam is 4m long with a flexural stiffness of 20 MN m2. It has a point

load of 1kN at the free end and a u.d.l of 300 N /m along its entire length. Calculate the slope anddeflection at the free end.

P

q N /m

L

Figure 1.8: Cantilever Beam with UDL and Point Load

Solution 1.7 This problem can be solved by using various method. The first one that will be

used is Theory of Superposition for Combined loads and the second one is double integration

method from first principles.For the point load only we will use the following equations to

determine the slope as well as the deflection:

1. Superposition Method for Combined Loads

y = −PL33 EI

(1.13)

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dy

dx=

PL2

2 EI (1.14)

y = − 1000

×43

3 ×20 ×106 = −1.06mmdy

dx=

1000 × 422× 20 × 106 = 400 × 10

−6

For the U.D.L we will use the following equations:

y = − qL4

48 EI (1.15)

dy

dx=

qL3

6 EI (1.16)

y = − 300 × 44

48 × 20 × 106 = −0.48mmdy

dx=

300 × 436× 20 × 106 = 160 × 10

−6 rad

The total deflection and slope are y = −1.54mm and dydx

= 560 ×10−6 rad2. Double Integration Method from 1st Principles

Example 1.8 The beam shown in Fig. 1.9 is 7m long with E I = 200 MN m2. Determine theslope and deflection in the middle of the beam.

A B

30kN 40kN

7m

2m 4.5m X

X

x

Figure 1.9: SS Beam with Point Load loads

Solution 1.8 The first thing that we need to solve is the reaction on either side of interest and

this is done by taking moments about either A or B. If we take moments about B then our solution

will be:

0 = 7 R A − 30 × 5 −40 ×2.5 R A = 35.71kN

The bending moment equation is as follows:

M xx = R A[ x]− 30[ x − 2]− 40[ x − 4.5] (1.17)

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1.5 Examples 15

EI d 2 y

dx2 = 35.71[ x] −30[ x − 2] −40[ x − 4.5]

EI dy

dx= 35.71

[ x]2

2 −30 [ x − 2]

2

2 − 40 [ x −4.5]

2

2 +C 1

EIy = 35.71 [ x]3

6 −30 [ x − 2]3

6 − 40 [ x −4.5]3

6 +C 1 x +C 2

The boundary conditions are at x = 0 we have y = 0 and this leads to C 2 = 0. At x = L we have y = 0 and C 1 needs to be worked out

0 = 35.71[7]3

6 − 30 [7 − 2]

3

6 −40 [7 −4.5]

3

6 +C 1(7)

C 1 = −187.4We are now able to write the required expressions as

EI dy

dx

= 35.71[ x]2

2 −30

[ x − 2]2

2 −40

[ x −4.5]2

2 −187.4

EIy = 35.71[ x]3

6 −30 [ x − 2]

3

6 − 40 [ x −4.5]

3

6 − 187.4 x

From the above, it becomes easy to determine the deflection and rotation at the distance x = 3.5m.

Example 1.9 The beam shown in Fig. 1.10 is 6m long with E I = 300 MN m2. Determine theslope at the left hand end and the deflection at the middle of the beam.

A B

30kN

2kN /m

7m

2m X

X

x

Figure 1.10: SS Beam with UDL and Point Load

Solution 1.9 The reaction is determine by first taking moments about the right hand support.

0 = 6 R A − 30 × 4 −2 × 62/2 R A = 26kN

M xx = R A[ x] −30[ x − 2] − qx2

2

EIy′′ = 26[ x]− 30[ x −2]− 2[ x]2

2

EI y′ = 26[ x]2

2 − 30[ x − 2]

2

2 − 2[ x]

3

6 +C 1

EIy = 26[ x]3

6 − 30[ x − 2]

3

6 − 2[ x]

4

24 +C 1 x +C 2

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1.6 Exercises

Exercise 1.1 The simply supported beam ABC carries a distributed load of maximum

intensity q0 over its span of length L. Determine the maximum displacement of the beam.

A B

q0

L

L/2

Exercise 1.2 The intensity of the distributed load on the cantilever beam varies linearly from

zero to q0. Derive the equation of the elastic curve as well as the slope.

q0

L

L/2

Exercise 1.3 The intensity of the distributed load on the simply supported beam varies

linearly from zero to q0.

a. Derive the equation of the elastic curve

b. Find the location of the maximum deflection

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1.6 Exercises 17

A B

q0

L

L/2

Exercise 1.4 Determine the maximum displacement of the simply supported beam due to

the distributed loading shown below. (Hint: Utilize symmetry and analyse the right of the

beam only.)

Exercise 1.5 Determine the deflection at the free end for the cantilever beam below with a

variably distributed loaded. E I = 10 MN m2

Exercise 1.6 A 203mm × 133mm × 25kg/m I-section (parallel flange) is used as a can-tilever with a span of 6m. It carries a point load of 6kN at 2m from the fixed end and auniformly distributed load of 2kN /m from the free end to a point 1m from the fixed end. Thecantilever is propped at a point 2m from the free end, such that the load in the prop is 9.7kN .Calculate the deflection at the free end and under the 6kN load. Neglect the mass of the beam.

Exercise 1.7 A cantilever 3m long carries a point load of 22kN at 1m from the fixed end as

well as a uniformly distributed load of 12kN /m from the free end to a point 1m from the freeend. If the deflection at the free end is limited to 16mm, calculate the minimum necessary

value of I for the cross-section, neglecting the mass of the beam. A prop is now introduced

1m from the free end to reduce the deflection at the free end by half. What is the magnitude

of the load in the prop.

Exercise 1.8 Calculate the deflection 2m from the left-hand end and the slope at the left-hand

support of the beam shown below. E I = 10 MN m2

Exercise 1.9 Calculate the slope and deflection 1m from the left-hand support of the beam

shown below. E I = 10 MN m2

Exercise 1.10 Calculate the slope and deflection 1m from the left-hand support of the beam

shown below. E I = 10 MN m2

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Exercise 1.11 A beam AB of constant section, depth 400mm and I max = 250 × 10−6m4, ishinged at A and simply supported on a non-yielding support at C. The beam is subjected to

the given loading as shown in the figure. Determine:

a. The vertical deflection of B

b. The slope of the tangent to the bent centre line at C

E = 80GPa

Exercise 1.12 #

Exercise 1.13 #

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Introduction

Point Load

Uniformly Distributed Load

Exercise

Examples

Exercises

2 — Continuous Beams

2.1 Introduction

Definition 2.1.1 — Continuous Beams. -are beams that are supported on more than two

supports. These beams are statically indeterminate.

We will not dwell much on the derivation of the Three Moment Theorem also known as Clapey-

ron’s theorem. The following equations is what you will be using when dealing with continuous

beams.

M A L1 − 2 M B( L1 + L2)− M C L2 = 6

A1 x1

L1+

A2 x2

L2

(2.1)

A highway bridge shown in Fig. 2.1 is a clear example of a continuous beam

Figure 2.1: A continuous beam acting as a highway bridge

2.1.1 Point Load

Let us take a look at the span AB in Fig. 2.2 and tackle the two triangles ∆ACD and ∆CDB.

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20 Continuous Beams

A B

P

L

a b

• •

A

C

D B

x = a3

x = a + 2b3

M A B =

P a b L

Figure 2.2: Span with Point Load

6 Ax L

= 6 L

12 × a× Pab

L

× 2a

3 +

12 ×b × Pab

L

a + b

3

6 Ax

L=

6

L

Pa3b

3 L+

Pab2

2 L

a +

b

3

6 Ax

L=

6

L

Pa3b

3 L+

Pa2b2

2 L+

Pab3

6 L

6 Ax

L=

Pab2

6 L2 (2a + b)(a + b) BUT a + b = L

= Pab

L(2a + b) BUT b = L − a

= Pa L

( L − a)(2a + L − a) = Pa L

( L2 − a2)

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2.1 Introduction 21

2.1.2 Uniformly Distributed Load

We take a look at Fig. #. The centroid of the parabola is at the centre of the semi-circle and it is

x = L/2.

6 Ax L

= 6 L

× 2 L3 × qL2

8 × L

2

6 Ax

L=

qL3

4

Example 2.1 The uniform beam shown in fig. 2.3 carries the loads as indicated. Determine

the B.M at B and hence draw the S.F. and the B.M diagrams for the beam.

A B C

30 kN /m

20kN

2m 2m

0.5m

Figure 2.3:


Example 2.2 A beam ABCDE is continuous over four supports and carries the loads as shownin fig 2.4. Determine the values of the fixing moment at each support and hence draw the S.F.

and B.M. diagrams for the beam.

10 kN /m

20kN

10 kN /m

A B C D E

5m 4m

2m

5m 2m

Figure 2.4:


Example 2.3 A beam ABCDE is continuous over four supports ( A, B, C and D) and fixed at

one the other end E . Span AB carries a point of 10kN at 0.5m from A, Span CD carries a pointload 20kN at 0.5m from C. Span BC carries a udl of 30kN /m while span DE carries a 20kN /mudl. Determine the values of the fixing moment at each support and hence draw the S.F. and B.M.

diagrams for the beam.


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22 Continuous Beams

Example 2.4 A uniform continuous beam ABC is built-in at support C and simply supported

at A and B. AB is loaded with a udl of 20kN /m magnitude and BC has a point load of 10kN located at 1m from point B. Distance for AB = 2m and BC = 2m. Determine the reactions at thesupports and draw the bending moment diagrams as well as the shear force diagrams.


Example 2.5 Still to be done


2.2 Exercise

Exercise 2.1 A uniform continuous beam is built-in at one end.

a. Determine the moments as well as reactions forces.

b. Sketch the shear force and bending moment (True moment and correcting moment)

diagrams.

c. Determine the position of the point of contra-flexure nearest to the free end.

Exercise 2.2 For the continuous beam shown below, calculate the reactions at the supports.

I xx AB = 20×10−6m4 = I xx BC and I xxCD = 40×10−6m4. Also, calculate the moments and sketchshear force as well as bending moment diagrams.

Exercise 2.3 A continuous beam ABCD is simply supported over three spans AB = 1m, BC = 2m and CD = 2m. The first span carries a central load of 20kN and the third span auniformly distributed load of 30kN /m. The central span remains unloaded. Calculate the

bending moments at B and C; draw S.F. and B.M. diagrams. The supports remain at the same

level when the beam is loaded.

Exercise 2.4 Calculate the magnitude of the reactions at the supports of the continuous beam

shown below as well as the true bending moment and and correcting moment diagrams.

Exercise 2.5 Calculate the magnitude of the reactions at the supports of the continuous

beam shown below as well as the true bending moment and and correcting moment diagrams.

20 kN

2kN /m

4m2m 4m 1m 2m

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2.3 Examples 23

2.3 Examples

Example 2.6 #

Solution 2.6 #

Example 2.7 #

Solution 2.7 #

Example 2.8 #

Solution 2.8 #

Example 2.9 #

Solution 2.9 #

2.4 Exercises

Exercise 2.6 #

Exercise 2.7 A beam is continuous over four supports and are loaded as shown in Fig. ??.

1. Calculate the magnitude of the forces in the supports

2. Draw the shear force and bending moment diagrams.

20kN 30kN

10 kN /m

A B

1m

4m 4m 3m 1m

Exercise 2.8 A beam is continuous over four supports and are loaded as shown in Fig. ??.

1. Calculate the magnitude of the forces in the supports

2. Draw the shear force and bending moment diagrams.

8 kN /m 5 kN /m

30kN 10kN

A B C D

4m 3m 1m 4m 1m

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24 Continuous Beams

Exercise 2.9 #

Exercise 2.10 #

Exercise 2.11 #

Exercise 2.12 #

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Introduction

Strain Energy of Bars

Castigliano’s Theorem

Structures

Castigliano’s theorem applied to Curved

Beams

Castigliano’s theorem applied to Beams

Cantilever beam with a Point Load at the free

end

S/S beam with a Point Load at mid-point

Examples

Exercises

3 — Energy Methods

3.1 Introduction

The energy stored within a material when work has been done is called the strain energy. Energy

is normally defined as the capacity to do work and it may exist in many forms such as mechanical,

thermal, nuclear, electrical, etc. The potential energy of a body is the form of energy which

is stored by virtue of the work which has previously been done on that body. Strain energy

is a particular form of potential energy (PE) which is stored within materials which has been

subjected to strain, i.e. to some change in dimension.

Definition 3.1.1 Strain energy is defined as the energy which is stored within a material

when work has been done on the material.

Strain Energy U = Work Done (3.1)

When an axial force P is applied gradually to an elastic body that is rigidly fixed (no

displacement, rotation permitted), the force does work as the body deforms. We can see this

clearly in Fig. 3.1

L δ

P

Figure 3.1: Elastic Bar

P

δ O

B

C

Area=U

Figure 3.2: Load vs Displacement

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26 Energy Methods

This work can be calculated from U =

δ

0 P d δ , where δ is the work absorbing displacement

of the application of the load.

U = 1

2Pδ (3.2)

where U is the area under the force-displacement diagram. The work of several loads, i.e. P1,

P2, P3,..., Pn acting on an elastic body is independent of the order in which the loads are applied.

The work is thus

U = 1

2∑Piδ i (3.3)

where the same U is actually the energy (strain) stored in an elastic body. The unshaded area

above the line ’OB’ is called the complimentary energy, a quantity which is utilised in some

advanced energy methods of solution, [dro]. [dro]

3.2 Strain Energy of BarsLet us consider a bar of constant cross-sectional area A, length L and Young’s modulus E as

shown in Fig. #. If the axial load P is applied gradually to result in displacement δ , the strain

energy of the bar is

U = 1

2Pδ =

P2 L

2 AE (3.4)

This can be better expressed as follows:

U = 1

2

L0

P2

2 AE dx (3.5)

3.3 Castigliano’s Theorem

Theorem 3.3.1 — Castigliano’s First Theorem. Castigliano’s theorem states that if an

elastic body is in equilibrium under the external loads P1, P2, P3,..., Pn then

δ i = ∂ U

∂ Pi(3.6)

where δ i is the displacement associated with load Pi and U is the strain energy of the body.

The other way of writing Castigliano’s first theorem is as follows:

Theorem 3.3.2 — Castigliano’s First Theorem. If the total strain energy expressed in termsof the external loads is partially differentiated with respect to one of the loads the result is the

deflection of the point of application of that load and in the direction of that load. Deflection

in direction of Pi will be

δ Pi = ∂ U

∂ Pi(3.7)

In applications where bending provided practially all of the strain energy,

δ Pi =

A

M

EI

∂ M

∂ PidA (3.8)

Castigliano’s theorem for angular movements states:

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3.4 Structures 27

Theorem 3.3.3 — Castigliano’s theorem for angular movements. If the total strain energy

expressed in terms of the external moments be partially differentiated with respect to one of

the moments, the result is the angular deflection in radians of the point of application of that

moment and in its direction

θ =

A

M

EI

∂ M

∂ M idA (3.9)

where M i is the actual or dummy moment at the point where θ is required.

3.4 Structures

Displacement in the direction of the applied load is found using the following equation:

δ =

n

∑i=1

F i L

i∂ F

i Ai E i∂ P

(3.10)

There would be plenty of examples for trusses/structures in the example section

Statically Indeterminate/determinate

Pin jointed structures can either be statically determinate or indeterminate. The challenge is

when the structure is indeterminate. The following steps should be followed when approaching a

pin jointed structure:

• Count the number of joints, members and reactions. m + r − 2 j = 1. If 2 j < m + r , thestructure is indeterminate. There is one redundant member on the structure.

• The number of redundant members is equal to the degree of indeterminacy of the structure.Release the redundant members to render the structure statically determinate.

• Calculate the forces in the statically determinate structure, subjected to any external loadsplus the redundant reaction forces, using methods of joints of sections.

• Use Castigliano’s method to calculate the required deflections and also slopes.

3.5 Castigliano’s theorem applied to Curved Beams

The theorem can be applied to all types of beams, cantilever with point load or udl, simply

supported beam with point load or udl, etc. I will leave you to derive the rest of the beams. You

can confirm your answer by using first principles.

3.6 Castigliano’s theorem applied to Beams

3.6.1 Cantilever beam with a Point Load at the free end

When given any type of beam, follow this procedure:

1. Cut an elemental strip of width d x

2. Measure distance x from the free-end or the fixed-end. If it is from the fixed end, make

sure that you calculate the reaction first

3. Take moments about X − X line, i.e. distance from free-end to where the strip starts, M xx4. Determine partial derivative of moment with respect to the load applied, i.e. ∂ M xx/∂ P

For a cantilever given, Fig. 3.6.1, this is what you do:

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28 Energy Methods

P

L

xdx

Figure 3.3: Cantilever Beam with a Point Load

M xx = Px; ∂ M xx

∂ P= x

δ BP = 1

EI

L0

M xx∂ M xx

∂ Pdx

δ BP = 1

EI

L0

Px · xdx

δ BP = Px3

3 EI

L

0

= PL3

3 EI

3.6.2 S/S beam with a Point Load at mid-point

The simply supported beam, Fig. 3.6.2 is tackled a bit differently from the cantilever. In this

example the beam has constant cross-sectional area A, length L and Young’s modulus E In this

A

C B

P

Pb L Pa L

L

a b

x dx zdz

Figure 3.4: Simply Supported Beam with a Point Load

problem all we need to do is to take moments about X − X and Z − Z as follows:

M AC = Pb

L x;

∂ M AC ∂ P

= bx

L

M CB = Pa L

z; ∂ M CB∂ P

= az L

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3.7 Examples 29

The deflection is thus

δ = 1

EI

a0

Pb

L x × bx

Ldx +

1

EI

b0

Pa

L z× az

Ldz

δ = Pb2

L2 EI

x3

3

a0

+ Pa2

L2 EI

z3

3

b0

δ = Pb2a3

3 L2 EI +

Pa2b3

3 L2 EI

δ = Pb2a2(a + L)

3 L2 EI

But since a = L/2 = b then

δ = PL3

48 EI

3.7 Examples

Example 3.1 Calculate the vertical displacement at point B on the pin-jointed structure shown

in Fig. #. The cross-sectional area of both members is 2000mm2 and E = 200GPa

Solution 3.1 We start by determining what the length of AB and BC is, i.e. L AB = 2m and L BC = 2m. The only joint that we will deal with is Joint ’B’.

R The structure is statically determinate. j = 3, m = 2 and r = 4. 2 j = m + r

Σ F ↑ = 0 = F AB sin30◦ − F BC sin30◦ − QF BC = F BA − 2Q . . .Eq. 1

Σ F → = 0 = R −F AB cos30◦ −F BC cos30◦F BC = 1.155 R −F BA . . .Eq. 2

F AB = Q + 0.577 R with ∂ F BA

∂ Q= 1;

∂ F BA∂ R

= 0.577

F BC = 0.577 R

−Q with

∂ F BC

∂ Q

=

−1;

∂ F BC

∂ R

= 0.577

Member Length (m) Load ∂ F ∂ Q

∂ F ∂ R

FL AE

∂ F ∂ Q

FL AE

∂ F ∂ R

AB 2m Q + 0.577 R 1 0.577 50 ×10−6 28 ×10−6BC 2m −Q + 0.577 R -1 0.577 50 ×10−6 −28 × 10−6

100µ m 0

Table 3.1: Table caption

Example 3.2 A Plate 5mm thick and 30mm wide is bent into the shape shown below.

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30 Energy Methods

Solution 3.2 Taking moments about z − z

M zz = Fr sin θ ; ∂ M zz

∂ P

= r sin θ

δ = 1

EI

3π 2

0(Fr sin θ )(r sin θ )rd θ

δ = Fr 3

EI

θ

2 − sin2θ

4

3π 2

0

δ = 3π Fr 3

4 EI =

(200)(0.2)3(3π )

4 × 62.5 = 60.3mm

Example 3.3 The structure shown below is made from a pipe with inner and outer diameters

of 80mm and 100mm respectively. Calculate the resultant deflection at A due to bending. E = 200GPa.

Solution 3.3 #

Example 3.4 Determine the vertical deflection of point A on the bent cantilever as shown

below, Fig. 3.5, when loaded at A with a vertical load of 25 N . The cantilever is built-in at B and

EI is constant throughout and is equal to 450 Nm2. What would be the horizontal deflection at

point A?

P = 25 N

r =

1 2 5 m

m

200mm

A

B

Figure 3.5: Castigliano-Semicircular

Solution 3.4 We now look at Fig. 3.6 and we let R become a dummy load.

M xx = Px , ∂ M xx

∂ P= x

M zz = (0.2 + r sin θ )P + Rr (1 − cos θ ), ∂ M zz∂ P

= (0.2 + r sin θ ), ∂ M zz

∂ R= r (1 −cos θ )

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3.7 Examples 31

P = 25 N

R = 0 N

r =

1 2 5 m m

200mmr sin θ

r c o s θ X

X

x

A

B

Figure 3.6: Castigliano-Semicircular

The deflections are calculated as follows:

δ P = 1

EI

0.20

Px · xdx + 1 EI

π

0(0.2 + r sin θ )P(0.2 + r sin θ )rd θ

δ P =

Px3

3 EI

0.20

+ P

EI (0.04r + 0.4r 2 sin θ + r 3 sin2 θ )d θ

δ P = 0.0667

EI +

P

EI

0.04r θ −0.4r 2 cos θ + r 3

θ

2 − sin2θ

4

π 0

δ P = 0.0667

EI + P

EI (0.0157 + 0.0125 + 0.003068)

δ P = 0.8484

EI

δ R = 1

EI

π

0(0.2 + r sin θ )Pr (1 − cos θ )rd θ

δ R = Pr 2

EI

π

0(0.2 −0.2cos θ + r sin θ − r sin θ cos θ )d θ

δ R = Pr 2

EI 0.2θ −0.2sin θ − r cos θ +

r cos2 θ

2π

0

δ R = Pr 2

EI

0.2π − r (−2) + r

2(0)

δ R =

0.3431

EI

Example 3.5 The steel truss (Fig. 3.7) supports the load P = 30kN . Determine the horizontaland vertical displacements of joint E. Use E = 200GPa. The cross-sectional area for all membersis 500mm2.

Solution 3.5 Referring to Fig. 3.8 We are given A = 500mm2, P = 30kN and E = 200GPa. Wewill use a JOINT method starting with Joint E, D and C. Let us introduce a dummy load R at

point E.Joint E

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32 Energy Methods

• •

• ••

P

2m

2m

2 m

A

B

C

D E

Figure 3.7: Castigliano Truss Structure

• •

• •• R( Dummy)

P

2m

2m

2 m

A

B

C

D E

Figure 3.8: Solution Castigliano Truss Structure

ΣF v = 0 = P + 0.707F EC

∴ F EC = −1.414P∂ F EC

∂ P= −1.414

ΣF H = F ED + 0.707F EC − R∴ F ED = P + R

∂ F ED∂ P

= 1 ∂ F ED∂ R

= 1

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3.7 Examples 33

Joint D

ΣF V = 0 = F DC

ΣF H = F DE −F DB + R∴ F DB = P + R∂ F DB

∂ P= 1

∂ F DB∂ R

= 1

Joint C

ΣF v = 0.707F CE + 0.707F CB

ΣF v = −P + 0.707F CB∴ F CB = 1.414P

∂ F CB∂ P

= 1.414

ΣF H = 0 = 0.707F CE − 0.707F CB − F CAΣF H = −P − P − F CA

∴ F CA = 2P

∂ F CA∂ P

= 2

Member L(mm) Load ∂ F ∂ P

∂ F ∂ R

FL AE

∂ F ∂ P

FL AE

∂ F ∂ R

AC 2000 2P 2 0 1.2 0

BC 2828 1.414P 1.414 0 1.696 0BD 2000 P+R 1 1 0.6 0.6

CD 2000 0 0 0 0 0

DE 2000 P+R 1 1 0.6 0.6

CE 2828 -1.414P -1.414 0 -1.696 0

δ P = 2.4mm δ R = 1.2mm

Example 3.6 For the simply supported beam (Fig. 3.9) loaded with a uniformly distributed

load, determine the maximum deflection in the middle of the beam. Let the udl be q N /m.

Solution 3.6 We can solve this problem by concentrating on half the length of the beam. Since

we need to calculate the maximum deflection which occurs in the middle, we need to place adummy load at the middle of the beam. We then need to cut the beam at distance x from the left

hand support and take moments about X − X section as follows:

R A = qL

2 +

P

2 = R B

M xx = R A x − qx2

2 =

qLx

2 +

Px

2 − qx

2

2∂ M xx

∂ P=

x

2

Since deflection is determined using the following

δ P = 1

EI

A

M ∂ M ∂ P

dA (3.11)

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34 Energy Methods

A B

q N /m

L

L/2

Figure 3.9: SS Beam with UDL

δ P = 2

EI

L/20

qLx

2 +

Px

2 − qx

2

2

x

2 dx

= 1

EI

L/20

qLx2

2 +

Px2

2 − qx

3

2

dx

= 1

EI qLx3

6 +

Px3

6 − qx

4

8 L/2

0

= 1

EI

qL

L2

36

+ P

L2

36

− q

L2

48

= 1

EI

PL3

48 +

qL4

48 − qL

4

128

= 1

EI

PL3

48 +

5qL4

384

3.8 Exercises

Exercise 3.1 Calculate the vertical displacement as well as the horizontal displacement at

point B on the pin jointed structure shown in the figure below. The cross-sectional area of

both members is 2000 mm2 and E = 200GPa.

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3.8 Exercises 35

•

•

•

10 kN

C

B

A

2 m

2 m

2 m

Exercise 3.2 Calculate the magnitude of the force R on the pin-jointed structure, shown inthe figure below, if the vertical deflection at node E is zero. The cross-sectional area of all the

members is the same.

•

•

• •

•• • R

AB

C

FE

D

10 kN 3m 3m

4 m

Exercise 3.3 Calculate the resultant displacement at point E on the pin-jointed structure

shown in the figure below. The cross-sectional area for all members is 200mm2 and E =200GPa.

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36 Energy Methods

•

•

••

•

•

60◦

60◦60◦

90◦

20 kN

2 m

Exercise 3.4 Calculate the vertical displacement at point D and the horizontal displacement

at C on the pin-jointed structure shown in the figure below. The cross-sectional area for all

members is 1200mm2 and E = 200GPa.

•

•

•

3 m

•20 kN

A CD

B

4m 4m

Exercise 3.5 Calculate the resultant deflection at point A on the pin-jointed structure shown

in the figure below.

•

• • •

10 kN

2m

3 m

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3.8 Exercises 37

Exercise 3.6 Calculate the vertical deflection at point B on the pin-jointed structure shown

below. The cross-sectional are of the members in tension is 30mm2 and for those in compres-

sion is 200mm2. E = 200GPa

•

•

••

0 .6 m

0 .

8 m

0 .

5 m

10 kN

1 m

Exercise 3.7 Calculate the resultant deflection at point D on the pin-jointed structure shown

below. The cross-sectional area of the members in tension is 1000mm2 and for those in

compression is 2000mm2. E = 200GPa

•

•

•

•

8 kN

0 .

6 m

1m

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Symmetric Member in Pure Bending

Unsymmetrical Bending

Alternative procedure for stress

determination

Deflection

Notation

xxx

Point Load

Uniformly Distributed Load

Examples

Exercises

4 — Unsymmetrical Bending of Beams

Introduction

The most common type of structural member is a beam. I actual structures beams can be found

in an infinite variety of:

• sizes• shapes• orientations

Definition 4.0.1 — Beam. A beam may be defined as a member whose length is relativelylarge in comparison with its thickness and depth, and which is loaded with transverse loads

that produce significant bending effects as oppose to twisting or axial effects.

Beams are generally classified according to their geometry and the manner in which they are

supported. Geometrical classification includes such features as the shape of the cross section

whether the beam is straight, curved, tapered or has constant cross-section.

Beams can also be classified according to the manner in which they are supported. Some

types that occur in ordinary practice are shown in the figures below, Fig. 4.1 and 4.2. There are

many other types not shown.

Figure 4.1: Cantilever Beam

4.1 Symmetric Member in Pure Bending

• Internal forces in any cross section are equivalent to a couple. The moment of the coupleis the section bending moment .

• From statics, a couple M consist of two equal and opposite forces.• The sum of the components of the forces in any direction is zero.• The moment is the same about any axis perpendicular to the plane of the couple and zero

about any axis contained in the plane.

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40 Unsymmetrical Bending of Beams

Figure 4.2: Continuous Beam-Bridge Support

• These requirements may be applied to the sums of the components and moments of thestatically indeterminate elementary internal forces

4.2 Unsymmetrical Bending

Simple bending theory applies when bending takes place about an axis which is perpendicular

to a plane of symmetry. If such an axis is drawn through the centroid of a section, and another

mutually perpendicular to it also through the centroid, then these axes are principal axes. Thus a

plane of symmetry is automatically a principal axis. Second moments of area of a cross-sectionabout its principal axes are found to be maximum and minimum values, while the product second

moment of area,

xy dA, is found to be zero. All plane sections, whether they have an axis of

symmetry or not, have two perpendicular axes about which the product second moment of area

is zero. Principal axes are thus defined as the axes about which the product second moment of

area is zero. Simple bending can then be taken as bending which takes place about a principal

axis, moments are being applied in a plane parallel to one such axis.

In general, however, moments are applied about a convenient axis in the cross-section; the

plane containing the applied moment may not then be parallel to a principal axis. Such cases are

termed unsymmetrical bending1.

The most simple type of unsymmetrical bending problem is that of skew loading of the

sections containing at least one axis of symmetry as shown in Fig. #. This axis and the axisperpendicular to it are then principal axes and the term skew loading implies load applied at

some angle to these principal axes. The method of solution in this case is to resolve the applied

moment M A into M uu and M vv

How to approach an Unsymmetrical Bending problem:

1. Determine the position of the centroid (if not already known)

2. Calculate values of I xx, I yy and I xy3. Calculate angle θ p using

tan 2θ = 2 I xy

I xx

− I yy

(4.1)

1Mechanics of Materials, EJ Hearn

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4.2 Unsymmetrical Bending 41

4. Calculate principal second moments of area

I 11,22 = I xx + I yy

2 ±

I xx − I yy2

2

+ I 2 xy (4.2)

5. Calculate the moment M and resolve it into components of M uu = M cos θ and M vv =

M sin θ

6. Calculate combined bending stress

σ = M uu × v

I uu+

M vv ×u I vv

(4.3)

Figure 4.3: Asymmetrical Bending

7. Find position of neutral axis on cross-section (through centroid): σ = 08. Identify points on cross-section which are greatest distance from the neutral axis(one

tensile and one compressive) and determine x, y coordinates of each point.9. Use the following two equations to determine the distances u and v which are both positive

in the quadrant UGV.

u = x cos θ + y sin θ (4.4)

v = y cos θ − x cos θ (4.5)

10. Determine the maximum bending stress using Eqn. 4.3

Moment applied along the principal axis

The conditions that one should consider when working with this type of a problem are:

• Stress distribution acting over entire cross-sectional area to be a zero force resultant.• Resultant internal moment about y- axis to be zero.• Resultant internal moment about z- axis to be equal to M .

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• Express the three conditions mathematically by considering forces acting on differentialelement dA located at (0, y, z). Force is dF = σ dA, therefore

F R = F y; 0 = A σ dA( M R) y = σ M y; 0 =

A

zσ dA

( M R) z = σ M z; 0 =

A

− yσ dA

Figure 4.4: Asymmetrical Bending

• If material has linear-elastic behaviour, then we can substitute σ = − yc

σ max into 0 =

A − yσ dA and after integrating, we get

A yz dA = 0• The resultant general normal stress at any point on the cross section is

σ = M z y

I zz+

M y z

I yy(4.6)

BUT in out notation we will use x instead of z. You will not be penalized if you decide to

use what the book says you must use. The direction is important, so make sure that z is

pointing the right direction for a positive/ negative sense. M z = M cos θ and M y = M sin θ

Orientation of neutral axis• Angle α of the neutral axis can be determined by applying the stress equation with σ = 0,

since normal stress acts on neutral axis. The resulting equation is thus:

tan α = I zz

I yytan θ (4.7)

• For unsymmetrical bending, the angle θ defining direction of moment M is not equal toangle α , angle defining inclination of neutral axis unless I zz = I yy

4.3 Alternative procedure for stress determination

I prefer this method over the other methods.Let us consider any unsymmetrical section as shown in Fig. 4.5. The assumption that we make

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4.3 Alternative procedure for stress determination 43

Figure 4.5: Alternative Procedure

as we start this is that the stress at any point on the unsymmetrical section us given by

σ = Px + Qy (4.8)

where P and Q are constants; in other words it is assumed that bending takes place about the X

and Y axes at the same time, stresses resulting from each effect being proportional to the distance

from the respective axis of bending. Let there be a tensile stress σ on the element of area d A.

Then the force F acting on the element is F = σ dA. The moment of this force about the X axisis then σ dAy

M xx =

σ dAy

=

(Px + Qy) ydA =

PxydA +

Qy2dA

but we know from statics that

I xx =

y2dA

I yy =

x2dA

I xy =

xydA

therefore substituting the above in the moment equation, we get

M xx = PI xy + QI xx (4.9)

In a similar manner the moments about the Y axis is

M yy = −

σ dAx

M yy = −

(Px + Qy) xdA = −

PxydA −

Qy2dA

∴ M yy = −PI yy − QI xy (4.10)Since the stresses resulting from bending are zero on the neutral axis, the equation of the neutral

axis is derived by setting the stress to zero, i.e.

0 = Px + Qy

y x

= − PQ

= tan α N . A

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4.4 Deflection

The deflections of unsymmetrical sections in the directions of the principal axes may always be

determined by application of the standard deflection formulae, i.e.

δ = FL3

3 EI (4.11)

and this we know because it is the maximum deflection of a cantilever with a point load F at the

free end. The vertical deflection is determined as follows:

δ v = (F yy) L

3

3 EI xx(4.12)

and the horizontal deflection

δ h = (F xx) L3

3 EI yy

(4.13)

4.5 Notation

Quantity Symbol

Generic load for ODE Work f ( x)Transverse Shear Force V ( x)Bending Moment M ( x)

Slope of deflection curve dv( x)

dx = v′( x)

Deflection Curve v( x)

Table 4.1: Notation

4.6 xxx

#

4.6.1 Point Load

#

4.6.2 Uniformly Distributed Load

4.7 Examples

Example 4.1 A z-section shown in Fig.# is subjected to bending moment of M = 20kNm. Theprincipal axes y and z are oriented as shown such that they represent the maximum and minimum

principal moments of inertia, I yy = 900 × 10−6mm4 and I zz = 7540 × 10−6mm4 respectively.Determine the normal stress at point P and orientation of the neutral axis.

Solution 4.1 #

Example 4.2 Due to load misalignment, the bending moment acting on the channel sections

is inclined at an angle of 3◦ with respect to the y axis. If the allowable flexural stress for thisbeam is σ al = 150 MPa, what is the maximum moment, M max that may be applied.

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4.7 Examples 45

Solution 4.2 We start the solution by determining the components of the moment

M yy = − M sin 3◦ M xx = M cos 3

◦

The normal stresses due to the moment components are

σ zz1 = M yy × x

I yy= −( M sin 3

◦)× x I yy

σ zz2 = M xx × y

I xx=

( M cos 3◦)× y I xx

The combines stress is

σ zz = σ zz1 + σ zz2

= −( M sin 3◦)

× x

I yy +

( M cos 3◦)×

x

I xx

A1 = A3 = 500mm2

A2 = 800mm2

AT = 1800mm2

I xx = 2

10 × 503

12 + 500(8.9)2

+

80 ×103

12 + 800(11.1)2

= 0.3927 × 106mm4

I yy = 250 × 103

12 + 500(45)2+

10 × 80312

= 2.46 × 106mm4

−100 × 106 = −( M sin 3◦)× (−50)

2.46 × 106 + ( M cos 3◦)× (−33.9)

0.3927 × 106 M = 1.17 MN m

NB: I am not quite happy with the answer.

Example 4.3 A rectangular-section beam 80 mm × 50 mm is arranged as a cantilever 1.3mlong and loaded at its free end with a point load of 5kN inclined at an angle of 60◦ to thehorizontal axis as shown in Fig. #. Determine the position and magnitude of the greatest tensile

stress in the section. What would be the vertical deflection at the end? E = 210GPa

Solution 4.3 The moments of inertia are simple to calculate and are as follows:

I xx = 50 × 803

12 = 2.133 × 106mm4

I yy = 80 × 503

12 = 0.833 × 106mm4

M xx = 5000 ×1300cos30◦ = 5629 ×103 Nmm M yy = −5000 × 1300sin30◦ = −3250 × 103 Nmm

Using the general method of determining stress at any point, i.e.

σ = M xx y

I xx± M yy x

I yy

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Y

X •

C B

AD P

50mm

8 0 m m

Figure 4.6: Rectangular section

We will determine the stresses at points A(25,40), B(25,−40), C(−25,−40) and D(−25,40)

σ A = M xx y

I xx± M yy x

I yy=

(5629)(40)(1000)

2.133 × 106 − (−3250)(25)(1000)

0.833 × 106 = 203.1 MPa

σ B = M xx y

I xx± M yy x

I yy=

(5629)(−40)(1000)2.133 × 106 −

(−3250)(25)(1000)0.833 × 106 = −8.021 MPa

Using the alternative method, we must determine the constants P and Q. Since the section is

symmetric about both axes, we know that I xy = 0.

M xx = 5629 ×106 = 2.133Q ∴ Q = 2639× 106 M yy = −3250 × 106 = −0.833Q ∴ P = 3901.56 × 106 × 106

The stresses at various points are

σ A = 3901.56(25) + 2639(40) = 203.1 MPa

σ B = 3901.56(25) + 2639(−40) = 8.021 MPaσ C = 3901.56(−25) + 2639(−40) = −203.1 MPaσ D = 3901.56(−25) + 2639(40) = −8.021 MPa

Example 4.4 A cantilever if length 1.2m and of the cross-section shown in Fig. 4.7 carries avertical load of 10kN at its outer end, the line of action being parallel with the longer leg and

arranged to pass through the shear centre of the section (i.e. there is no twisting of the section).

Working from first principles, find the stress set up in the section at points A, B and C, given that

the centroid is located as shown. Determine also the angle of inclination of the neutral axis α NA.

Given: I xx = 4 × 10−6m4 and I yy = 1.08 ×10−6m4

Solution 4.4 #

Example 4.5 #

Solution 4.5 #

Example 4.6

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4.8 Exercises 47

Figure 4.7: Unequal Leg subjected to vertical load

Solution 4.6 #

Example 4.7 #

Solution 4.7 #

Example 4.8 #

Solution 4.8 #

4.8 Exercises

Exercise 4.1 #

Exercise 4.2 #

Exercise 4.3 #

Exercise 4.4 A T-section shown below has two loads acting on it. It is supported as a

cantilever of length 3m and P = 500 N is acting at 2m from the fixed end and P2 = 707 N .

Exercise 4.5 A beam of 3m length has a cross-section as shown below and is subjected to a

constamt bending moment of 600Nm about the X-axis. Calculate:1. the maximum stress induced in the section

2. the magnitude and the direction of the maximum deflection of the beam.

The following is given I xx = 363.05 × 10−9m4, I yy = 49.72 × 10−9m4, x = 9.45mm, y =27.22mm and E = 200GPa.

Exercise 4.6 #

Exercise 4.7 #

Exercise 4.8 #

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Exercise 4.9 #

Exercise 4.10 #

Exercise 4.11 #

Exercise 4.12 #

Exercise 4.13 #

Exercise 4.14 #

Exercise 4.15 #

Exercise 4.16 #

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Plastic Bending of Rectangular Beams

Plastic Bending of Symmetrical (I-Section)

Beam

Partially plastic Bending of Unsymmetrical

Sections

Limit Analysis-Bending

Bending

The principle of Virtual Work

Solid Shaft

Hollow shaft

Exercises

5 — Inelastic Bending

Introduction

When the design of components is based upon the elastic theory, i.e. the simple bending or

torsion theory, the dimensions of the components are arranged in such a way that the maximum

stresses which are likely to result do not exceed the allowable working stress. This is obtained

by taking the yield stress and dividing it by the applicable safety factor.

Under normal service conditions, we want to present yielding because the resulting permanent

deformation is generally undesirable. However, permanent deformation does not necessarily leadto catastrophic failure; it may only make the structure or component undesirable and considered

unsafe or unfit for further use. At the outer fibres yield stress may have been exceeded but some

portion of the component may be found to be still elastic and capable of carrying the load. The

strength of a component will normally be much greater than that assumed on the basis of initial

yielding at any position. To take advantage of the inherent additional strength, a different design

procedure is used which is often referred to as plastic limit design.

Definition 5.0.1 — Inelastic Bending. Inelastic materials are materials which follow

Hooke’s law up to the yield stress σ Y and then yield plastically under constant stress (see Fig.

5.1).

The figure below, Fig. 5.1, assumes material behaviour which:

1. Ignores the presence of upper and lower yields and suggests only a single yield point

2. takes the yield stress in tension and compression to be equal

3. When a plastic hinge has developed at one point, the moment of resistance at that point

remains constant until collapse of the whole structure takes place due to the formation of

the required number of plastic hinges at other points.

4. transverse sections of beams in bending remain in plane throughout the loading process,

i.e. strain is proportional to distance from the neutral axis.

It is now possible on the basis of assumption (4) to determine the moment which must be applied

to produce:

• maximum or limiting elastic condition in the beam material with yielding just initiated at

the outer fibres.

• yielding to a specific depth.• yielding across the complete section, i.e. fully plastic state or plastic hinge. Depending

on the support and loading conditions, one or more plastic hinges may be required before

complete collapse of the beam or structure occurs and the load required to produce this

situation is called the collapse load.

5.1 Plastic Bending of Rectangular Beams

Let us consider a cantilever beam loaded at the tip with a point load P which is large enough

to cause yielding in the shaded area. At section a-a, the stresses on the outer fibres have justreached yield stress, but the distribution is elastic as shown in Fig. 5.3. Applying the flexure

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50 Inelastic Bending

ε

σ

O ε Y

σ Y

ε Y

σ Y

Figure 5.1: Idealized Stress-Strain Diagram

Figure 5.2: Cantilever Beam subjected to load P

formula

M max = σ maxS

= σ maxbh2

6

we find that the magnitude of the bending moment at this section is

M Y = bh2

6 σ Y (5.1)

This moment is called yield moment because it is the moment responsible for yielding.

At section b-b, the cross section is elastic over the depth of 2 yi but plastic outside this depth as

shown in Fig. 5.4 The stress is constant at σ Y over the plastic portion and varies linearly over

the elastic region. The bending moment carried by this elastic region is given by the following

formula

M pp = I i

yi(5.2)

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5.1 Plastic Bending of Rectangular Beams 51

Figure 5.3: Rectangular section-Elastic

Figure 5.4: Rectangular section Partially Plastic

where I i is the moment of inertia of the elastic region of the cross section about the neutral axis.

For the plastic region, which is symmetrical about the neutral axis, the bending moment is

M pp = moment of elastic portion + total moment of the plastic region

= b(2 yi)

2σ Y 6

+ 2

σ Y b

h

2 − yi

1

2

h

2 − yi

+ yi

= 2σ Y by

2i

2 +

σ Y bh2

4 −σ Y by2i

M pp = σ Y b

12

3h2 −4 y2i

This moment is referred to as partial plastic moment. Instead of the stress at the outside increasing

due to an increase in loading, more and more of the section reaches the yield stress.

At section c-c, the beam is fully plastic. The stress is constant at σ Y over the tensile and

compressive portion of the cross section. The bending moment that causes this stress distribution

is called the fully plastic moment M f p. When the loading has been continues until the stress

distribution is as shown in Fig. 5.5, the beam will collapse. We note that yi = 0, all that remainsis h/2. The fully plastic moment is determined as follows:

M f p = 2σ Y yA

= 2σ Y

bh

2

h

4

= bh2

4

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Figure 5.5: Rectangular section Fully plastic

It is worth noticing that M f p = 2/3 M Y and this is valid for beams of rectangular cross-section.This ratio is also called the shape factor. Other shape factors for various cross sections are shown

in the table below, Tab. 5.1 For a rectangular section, this shape factor means that the beam can

Cross Section M f p/ M Y Solid Rectangle 1.5Solid Circle 1.7Thin-walled Circular Tube 1.27Thin-walled Wide-flange Beam 1.1

Table 5.1: Shape factors of different sections

carry 50% additional moment to that which is required to produce initial yielding at the edge of

the beam section before a fully plastic hinge is formed.

5.2 Plastic Bending of Symmetrical (I-Section) Beam

Figure 5.6: I-section Elastic and Fully Plastic

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The elastic moment or yield moment M Y is determined as follows

M Y = I

yσ Y

I =

BH 3

12 − 2

b2 h

3

12

= BH

3

12 − bh

3

12

y = H

2

M Y = σ Y

BH 3

12 − bh

3

12

2

H

= σ Y 6 H

BH 3 − bh3

The fully plastic moment is found to be:

M f p =

σ Y

4 BH

2

−bh2

(5.3)

The value of the shape factor is 1.18 for the I-beam indicating that only an 18% increase in

strength capacity using plastic design procedures.

5.3 Partially plastic Bending of Unsymmetrical Sections

Let us consider a T-section beam shown below in Fig. 5.7 and 5.8. Whilst stresses remain within

the elastic limit the position of the neutral axis can be obtained by taking moment of are about

the neutral axis.

Figure 5.7: T section-Elastic

It does not matter what state this section is in, i.e. elastic, partially plastic or fully plastic,

equilibrium of forces must always be maintained. At any section the tensile forces on one side of

the neutral axis must equal the compressive forces on the other side of the neutral axis.

Summation of stresses ×area above the N.A = Summation of stresses ×area below the N.AIn the fully plastic condition, the stresses σ Y will be equal throughout the section, the equation

then becomes:

Σ Aabove = Σ Abelow = Atotal

2

For partially plastic condition we will have

F 1 + F 2 = F 3 + F 4

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Figure 5.8: T section Fully Plastic

The sum of the moments of these forces about the N.A yields that partially plastic moment M pp .

This is best explained with an example, Eg. 5.1.

Example 5.1 Determine the shape factor of a T-section of dimensions 100mm × 170mm ×20mm as shown in Fig. 5.9

Figure 5.9: T Section Elastic and Fully Plastic

Solution 5.1 With reference to Fig. 5.9,

y = Σ Ai yi

Σ Ai=

100 ×20 × 160 + 20 × 150 × 75100 × 20 + 20 ×150 = 109mm

I NA =

100 × 203

12 + 2000(160 − 109)2

+

20 × 1503

12 + 3000(75 −109)2

= 14.362 × 106mm4

Yielding will start at the bottom of the cross-section when bending moment reaches M Y , i.e.

M Y = σ Y I

y=

14.362 ×106109

σ Y

= 131.761 × 103σ Y mm3

When the section becomes fully plastic the N.A is positioned such that the area below NA=half

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the total area. We now must locate the plastic neutral axis, i.e. y p above the base

20 × y p = 100 ×20 + 20(150 − y p)20 × y p = 2000 + 3000 −20 × y p40 × y p = 5000

y p = 125mm

The fully plastic moment is then obtained by considering the moments of forces on convenient

rectangular parts of the section, each being subjected to a uniform stress σ Y

M f p = σ Y (100 ×20)(45 − 10) + σ Y (45 − 20)(20) × 12

(45 − 20) + σ (125 ×20)( 1252

)

= 70000σ Y + 6250σ Y + 156250σ Y

= 232.5 × 103σ Y mm3

∴ f = 232.5 × 103131.761 ×103

= 1.765

Example 5.2 A cantilever is to be constructed from a T-section beam of Example 5.1 and

is designed to carry a udl over its entire length of 2m. Determine the maximum udl that the

cantilever beam can carry if yielding is permitted over the lower part of the web to a depth of

25mm. The yield stress of the material is σ Y = 225 MPa.

Figure 5.10: T-section Fully Plastic

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Solution 5.2

σ Y y

= σ Y i

145 − y ∴ σ Y i = σ Y

y(145 − y)

F 1 = σ

Y (20

×25) = 500σ

Y

F 2 = σ Y

2 (20 × y) = 10 yσ Y i.e. Average stress for the triangle

F 3 = σ Y

2

125 − y

y

×20(125 − y) = 10σ Y

y(125625 − 250 y + y2)

F 4 = σ Y

2

125 − y

y

+

145 − y

y

(100 × 20) = 1000σ Y

270 − 2 y

y

F 1 + F 2 = F 3 + F 4

500σ Y + 10 yσ Y = 10σ Y

y(125625 −250 y + y2) + 1000σ Y

270 −2 y

y

y = 85.25mm

We are now able to calculate the values for the forces and we find them to be F 1 = 112.5kN ,F 2 = 191.8kN , F 3 = 41.7kN and F 4 = 262.6kN . The moment of resistance M R that the beamcarry can now be obtained by taking moments about the neutral axis.

M R = F 1( y + 12.5) + F 2 × 23

y + F 32

3(125 − y) + F 4 [(125 − y) + 10]

= 112.5(85.25 + 12.5) + 191.8 × 2 ×85.25

3 +

41.7 × 2 × (125 − 85.25)3

+ 262.6 [(125 − 85.25 + 10)]= 10996.875 + 10901.34 + 1105.05 + 12065

= 36068.265 Nm

The maximum bending moment present on the beam will occur at the fixed end and it will be

calculated using the following formula

M max = qL2

2

= 18.034kN /m

5.4 Limit Analysis-Bending

Limit analysis is a method of determining the loading that causes a statically indeterminate

structure to collapse. This method applies only to ductile materials, which in this simplifieddiscussion are assumed to be elastic, perfectly plastic. The method is straightforward, consisting

of two steps. The first step is a kinematic study of the structure to determine which parts must

become fully plastic to permit the structure as a whole to undergo large deformations. The

second step is an equilibrium analysis to determine the external loading that creates these fully

plastic parts. We will be presenting only bending in the form of examples.

5.4.1 Bending

Revisiting Fig. 5.2, as the load P is increased, section c-c at the fixed end goes through elastic and

partially plastic states until it becomes fully plastic, whereas the rest of the beam remains elastic.

The fully plastic section is called a plastic hinge because it allows the beam to rotate about the

support without an increase in the bending moment. The bending moment at the plastic hinge iscalled the limiting moment M P. Once the plastic hinge has formed the beam will collapse.

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5.4 Limit Analysis-Bending 57

The collapse mechanism of a beam depends on the supports. Each extra support constrain

requires an additional plastic hinge in the collapse mechanism. A plastic hinge on the beam is

shown by a solid circle. A simply supported beam requires only one plastic hinge whereas a

beam with built-in ends will require three plastic hinges, that is if there is only one point load

acting on the beam. See Figs. 5.11, 5.12 and 5.13.

A B

W L/3 2 L/3

• y

Figure 5.11: Simply Supported Beam

W L/3 2 L/3

•

•

y

Figure 5.12: Simply Supported and Fixed

W L/3 2 L/3

•

•

• y

Figure 5.13: Fixed on both ends

In general, plastic hinges form where the bending moment is a maximum, which excludes

built-in supports and sections with zero shear force. The location is usually obvious for beams

subjected to concentrated loads. With statically indeterminate beams carrying distributed loads,the task tends to become difficult. Sometimes there is more than one collapse mechanism in

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which case we must compute the collapse load for each mechanism and choose the smallest

collapse load as the actual limiting load.

5.4.2 The principle of Virtual Work

Virtual Displacement

Definition 5.4.1 Virtual Work: states that for a structure that is in equilibrium and that is

given a small virtual displacement, the sum of the

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