TEXT BOOK:TEXT BOOK:R.C.Hibbeler, 2006, ”Structural Analysis”, Sixth Edition, 

llPrentice Hall. 

REFERENCES: K L t  C M  U  “F d t l   f St t l K.Leet, C.‐M. Uang,2002, “Fundamentals of Structural Analysis”, McGraw Hill. West, H.H., 1993, "Fundamentals of Structural Analysis", 2nd ed, John Wiley & Sons, Inc.Timoshenko, S.P., and Young, D.H., 19xx, "Theory of Structures", McGrawHill,Structures , McGrawHill,

COURSE EVALUATION:COURSE EVALUATION: The lectures will be given at times announced by The lectures will be given at times announced by the department. There will be additional problem solving tutorials taught by the teaching assistants  solving tutorials taught by the teaching assistants. There will be one mid‐term tests and a final 

i i  Th   ill b   l d b  examination. These will be supplemented by several homework assignments. Tentatively, the grading will be based on 25% for midterm, 60% for the final exam and 15% for the homework 5assignments. 

COURSE EVALUATION:COURSE EVALUATION: There will be 6 homeworks during the semester. There will be 6 homeworks during the semester. Homework grades will be given on an ACCEPTABLEand UNACCEPTABLE basis  Best five grades out of six and UNACCEPTABLE basis. Best five grades out of six will be included in the final grade; therefore there will be no late submission  be no late submission. Do not collaborate in solving assignments. Oth i           b i i   t d  ith Otherwise you may see your submission returned with UNACCEPTABLE grade assigned to it. 

INTRODUCTIONThis course is intended to develop the ability to l l d f d lcalculate deformation and analyze support 

reactions and internal forces for indeterminate structures (limited to structures with bar elements) using classical methods.elements) using classical methods.One may ask why, in this age of high‐powered 

t         l i l  th dcomputer programs, a course on classical methodsis needed. The software does all of the work for us, so isn't it sufficient to read the user’s guide to the software or to have a cursory understanding of y gstructural analysis?

INTRODUCTIONWhile there is no question that computer 

l bl l h h l lprograms are invaluable tools that help us solve complicated problems more efficiently, it is also p p ytrue that the software is only as good as the user’s level of experience and his/her knowledge of the level of experience and his/her knowledge of the software. A small error in the input or amisunderstanding of the limitations of the misunderstanding of the limitations of the software can result in completely meaningless 

t t   hi h  l d t     f  d i   ith output, which can lead to an unsafe design with potentially unacceptable consequences.

INTRODUCTION

Having a solid foundation in the fundamentals of analysis enables engineers to understand the behavior of structures and to recognize when output from acomputer program does not make sense.gWhichever analysis method is adopted during design, it must always be controlled by the designer, i.e. not a it must always be controlled by the designer, i.e. not a computer! This can only be the case if a designer has a highly developed knowledge and understanding of the highly developed knowledge and understanding of the concepts and principles involved in structural behaviorbehavior.

Deflection of Beams and FramesDeflection of Beams and Frames•When a structure is loaded its elements deformWhen a structure is loaded its elements deform•These deformations change the shape of the structure•Although deformations are generally small the designer•Although deformations are generally small, the designer has to be able to estimate their magnitude to make sure th d t d th li it i b th d i dthey do not exceed the limits given by the design code

•There are several methods to compute deflections in beams and frames:

– Double integration methodDouble integration method– Moment area method

Conjugate beam method– Conjugate beam method– Work/Energy methodsO h–Others

Importance of BeamImportance of Beam DeflectionsDeflectionsA designer should be able to determine deflections, i.eA designer should be able to determine deflections, i.e

I b ildi d ≤ L /300In building codes ymax ≤ Lbeam/300.

Analyzing statically indeterminate beams involve the use of various deformation relationships

Derivationofof

Beam’s Elastic Curve Differential Equation

Beam Bending Strain

To understand the bending stress in an arbitrary loaded beam, consider a small element cut from the b h i h di h l f h bbeam as shown in the diagram at the left. The beam type or actual loads does not effect the derivation of bending strain equation. Recall, the basic definitionbending strain equation. Recall, the basic definitionnormal strain is

ΔL/Lε = ΔL/L

Using the line segment, AB, the length before and after bending can be used to give

''

ABABBA −

=''

ε

The line length AB is the same for all locations before bending. However, the length A'B' becomes shorter above the neutral axis (for positive moment) and longer below The line AB and A'B' can be described using the radius of curvaturelonger below. The line AB and A B can be described using the radius of curvature, ρ, and the differential angle, dθ.

BENDING DEFORMATIONBENDING DEFORMATIONOF A STRAIGHT MEMBER

Assumptions:A) Plane section remains planeB) Length of longitudinal axis remains unchangedC) Plane section remains perpendicular to the

longitudinal axisFig. 6‐20

longitudinal axisD) In-plane distortion of section is negligible

Fig. 6‐21

Poisson effects that cause εy & εz can be neglected. 

Recall assumption:  any deformation of cross‐section within its own plane can be neglected.

POISSON’S RATIO EFFECTS

Ki tiKinematics

KinematicsKinematics

θρ dAB θρ dAB .=

θdBA )('' θρ dyBA ).( −=

θρθρ ddy)( −−θρ

θρθρεd

ddy.

.).(=

ρε y

−=ρ

This relationship gives the bending strain at any location as a function of the beam curvature and the distance from the neutral axis. However, this equation is of little use, and needs to be converted to stress. Also, radius of curvature is difficult to determine at a given b l ibeam location.

Beam Bending StressThe strain equation above can be converted to stress by using Hooke's law, σ = Eε, giving,

σ = Ey/ρ (1)σ = -Ey/ρ (1)

There is still the issue of not knowing the radius of curvature, ρ. If one thinks about g , ρit, the radius of curvature and the bending moment should be related. This relationship can be determined by summing the moment due to the normal stresses on an arbitrary beam cross section and equating it to the applied internal moment This is theit to the applied internal moment. This is the same as applying the moment equilibrium equation about the neutral axis (NA).

∑ = 0NAM

∫ Beam Section Cut∫ −= ).( dFyM

∫∫−= dAyM ..σ

For a positive moment, the top stresses will be in compression (negative stress) and thebottom stresses will be in tension (positive stress) and thus the negative sign in thebottom stresses will be in tension (positive stress) and thus the negative sign in theequation. This equation can be changed by using equation (1),

∫= dAyEM 2∫= dAyM .ρ

It is interesting to note that the integral is the area moment of inertia, I, or the second moment of the area. Many handbooks list the moment of inertia of common shapes .A review of moment of inertia is given below in the next sub-section. Using the areamoment of inertia gives g

M = E I / ρ But the radius of curvature, ρ, is still there. But equation (1),ρ = -Ey/σ , can be used again to eliminate ρ, giving,

E I /(-Ey/σ ) = M

Simplifying and rearranging gives,My

bending −=σThis equation gives the bending normal stress, and is also commonly called the flexure formula. The y term is the distance from the ne tral a is ( p is positi e)Ibending is the distance from the neutral axis (up is positive). The I term is the moment of inertia about the neutral axis.

Sk t hi D fl t d Sh f BSketching Deflected Shapes of BeamsThe deflected shape of a beam must be consistent withpThe restraints imposed by the supportThe curvature produced by the momentThe curvature produced by the moment

+

P iti   t b d  th  b  Positive moment bends the beam concave upward and negative moment bends the beam concave d ddownward.

+

Locating the Neutral Axis

If the cross section is symmetrical about the horizontal axis, then the neutral axis is halfway between the top and bottom. However, for non-symmetrical beam, such asa "T" cross section the neutral axis is not halfway between the top and bottom anda T cross section, the neutral axis is not halfway between the top and bottom, and needs to be determined before the bending stress equation can be used.

The neutral axis is located at the centroid (geometric center) of the cross sectionThe neutral axis is located at the centroid (geometric center) of the cross section. Recall from Statics, the centroid can be found using two methods. The first is by integration,

∫∫∫=

dA

dAyy

.

∫dAThe second, and more common method, is the method of parts The beam cross section is split into geometric shapes

Centroid for Arbitrary Shapeparts. The beam cross section is split into geometric shapes that are common (rectangle, triangle, circle, etc.). The centroid of basic shapes can be found in handbooks, eliminating the need for integration The centroid iseliminating the need for integration. The centroid is

∑= ii dAyy

.

∑=

idAy

Centroid Based on Sub-shapes

If there is a hole, then that area is considered to be negative, and the same equation can still be used As an example the diagram at the right would becan still be used. As an example, the diagram at the right would be

332211 yAyAyA −+

321

332211

AAAyAyAyAy

−++

=

Area Moment of InertiaSimilar to the centroid, the area moment of inertia can be found by either integration or by parts. The moment of inertia is also called the "second moment of the area" since that describes the integration equation,g q ,

∫= dAyI .2∫Moment of Inertia around Neutral Axis

using Integration

When using this with the bending stress equation, I is about the neutral axis and not using Integrationthe x-axis

A more common method to find the moment of inertia is by parts. Like finding the centroid( d b d fi ) h bj i li i ll b i h Th f i i(needs to be done first), the object is split into smaller basic shapes. The moment of inertia about the centroid of each part can be found in a handbook. Then the individual moment of inertia's are moved to the neutral axis using the parallel axis theorem. For a particular sub-shape, this gives

2iiiiNA yAII +=

where Ii is the moment of inertia about its own shape and

iiiiNA y−

INA-i is the moment of inertia about the object's neutral axis. All the moment of inertia terms can then be added together to give,g

INA = Σ INA-I

For the diagram at the rightt, the parts method gives, Moment of Inertia aroundthe Neutral Axis using Parts

I = (I1 + A1 y12) + (I2 + A2 y2

2) - (I3 + A3 y32)

Notice, for a hole, the moment of inertia is subtracted for , ,that shape.

Previously, beam stresses and strains were investigated andvarious equations were developed to predict bending andvarious equations were developed to predict bending andshear stresses. In addition to stresses, deflection and slopeare important and need to be calculated. This section (andthis chapter) will deal with various methods to calculateBeam Deflection  y  and Slope  dy/dx this chapter) will deal with various methods to calculatebeam deflections.

B D fl ti Diff ti l E ti

Beam Deflection, y, and Slope, dy/dx

Beam Deflection Differential EquationWhen a moment acts on a beam, the beam rotates anddeflects. The relationship between the radius of curvature, ρ,and the moment, M, at any given point on a beam wasdeveloped in the Bending Stress and Strain section as

EIM=ρ

1

This relationship was used to develop the bending stress equation but it can also be used toderive the deflection equation.Recall from calculus the radius of curvature for any point of a function y = f(x) isRecall from calculus, the radius of curvature for any point of a function, y = f(x), is

( )2123

dxdy

⎥⎦⎤

⎢⎣⎡ + ( )

2

2

dxyd

dx ⎥⎦⎢⎣=ρ

Kinematics of the cross section

•Plane section remains plane •No shear deformation

Pl i l lPlane section always normal to NA

Sign Conventiony

M > 0M > 0

0)('' >xy x0)( >xy

y

M 0

The deflection is measured from theoriginal neutral axis to the neutral axisof the deformed beam

x

M < 00)(" <xy

of the deformed beam.

The displacement y is defined as the deflection of the beam x0)( <xydeflection of the beam

1

dρdds

θθ=

(1) : 1 Curvaturedsdθ

ρκ ==∴

⎟⎞

⎜⎛ dydy (2) arctan,tan ⎟

⎠⎞

⎜⎝⎛==

dxdy

dxdy θθ

(3)t Sldy θθ (3) : tan Slopedxy θθ ≈=

θθθθ ⋅⋅⋅+++=2tan

53

θθθ

θθ

≈<<

+++=

tan,1 153

tan

ifdy (2)rpt. : tan Slopedxdy θ=

dsd

dsd

dd

dsdx

dxyd sec)(tan 22

2 θθθθθ

==

ddsdx

dxyd

dsd

1

(a) cos22

2

θ

θθ=∴

Curvaturedsdwhere : 1 , κ

ρθ

==

22

1 ⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

ddy

dds

⎠⎝⎠⎝ dxdx

1dx[ ] (b)

)/(11cos 2/12dxdyds

dx+

== θ

[ ] (4) )/(1

/12/32

22

dddxyd

dd

+===

θρ

κ(a) cos22

2

θθdsdx

dxyd

dsd

= [ ])/(1 /3dxdyds +ρdsdxds

dsdcurvature θ

ρ =≡ 1

curvature   :  the change in slope per unit lengthof distance along the curveof distance along the curve

dy/dx<<1 (dy/dx)2 ≈ 0

(5) 12

2

dxyd

≈ρ[ ] rpt. (4)

)/(1/1

2/32

22

dxdydxyd

dsd

+==

θρ dxρ

2 IE(6) : 2

2

vature Eq.Moment-CurEIM

dxyd= (..)rpt.

ρIEM =

(7) ,,'',' 2

2

etcdx

dMMdx

ydydxdyy ≡′≡≡

dxdxdx

(8):'' uationvature EqMoment CurMEIy = (8) : uation vature EqMoment-CurMEIy =

Solving the Deflection Differential Equation (Moment-Curvature Equation)Th diff ti l ti EI M i t f l b it lf b t d t b li d t bThe differential equation EIy = M is not useful by itself but needs to be applied to a beamwith specific boundary conditions. Generally, EI is constant and M is a function of the beamlength. Integrating the equation once gives,

( ) 1' . CdxxMEIy += ∫

Integration again gives,

( ) 21.. CxCdxxMEIy ++= ∫∫The integration constants, C1 and C2, are determined from the boundary conditions. For example, a pinned joint requires the deflection, y, equals 0. A fixed joint requires both the deflection, y, and slope, y´, equal 0. Each beam section must have at least two boundary conditions.section must have at least two boundary conditions. Each beam span must be integrated separately, just like when constructing a moment diagram. If the moment curve is discontinuous then a single equation cannot model theis discontinuous, then a single equation cannot model the deflection. Thus, each new support or load will start a new beam section that must be integrated (ie. Each Beam Section R i it O D fl ti E ti )Requires its Own Deflection Equation).Examples of beam sections are shown at the right.

Each Beam Section Requires its

Own Deflection Equation

Boundary conditionsBoundary conditionsF

A BC0=y 0=y0=Ay 0=By

FD0=y D0=Dy

0=Dθ D

PContinuity condition： +− = CCyy P

A BC

Smooth condition： +− = CC θθyyc

Boundary ConditionsD t i i th b d diti i llDetermining the boundary conditions is usually the most difficult part of solving the deflection differential equation. In particular, boundary q p , yconditions for multiple beam sections can be confusing.

For multiple beam sections, many times the boundary between the sections creates a yboundary. These type of conditions are also called "Continuity Conditions". For example, a point f b th d fl ti t b litforce on a beam causes the deflections to be split into two equations. However, the beam's deflection and slope will be continuous at the load plocation requiring y1 = y2 and y´1 = y´2. These conditions are needed to solve for the additional i t ti t tintegration constants.

The table at the right summarizes most common gboundary (and continuity) conditions for beam deflection and slope.

Boundary Conditions for Beam Sections

The Integration ProcedureThe Integration Procedure

Integrating once yields to slope dy/dx at any point in the beam. Integrating once yields to slope dy/dx at any point in the beam. 

Integrating twice yields to deflection y for any value of xIntegrating twice yields to deflection y for any value of x.

The bending moment M must be expressed as a function of the coordinate xbefore the integration

Differential equation is 2nd order the solution must contain two constantsDifferential equation is 2 order, the solution must contain two constantsof integration. They must be evaluated at known deflection and slopepoints (i.e. at a simple support deflection is zero, at a built in support bothslope and deflection are zero)

ExamplesLy p

xx

P

PxPLM +−=

MydEI2

PPL Mdx

ydEI =2

@ x = x PxPLd

ydEI +−=2

2

@ dx2

Integrating once 1

2

2cxPPLx

dxdyEI ++−=

2dx

@ x = 0 ( ) ( ) ( ) 020000 11

2

=⇒++−=⇒= ccPPLEIdxdy

Integrating twice 2

32

62cxPPLxEIy ++−=

( )0 3PL@ x = 0 ( ) ( ) ( ) 0

600

200 22

32 =⇒++−=⇒= ccPPLEIy

PL3

62

32 xPPLxEIy +−= @ x = L : y = ymax

PLPLLLPL 3332

EIPL3max =Δ

EIPLyPLLPLPLEIy3662

3

max

332

max −=⇒−=+−=

yWW kN per unit length

xx

( )22

xLWM −−=

yd 22 LWL M

dxydEI =2

@ x = x ( )22

2

2xLW

dydEI −−=2

2WL

2 2dx

( )3LWdIntegrating once ( )1

3

32cxLW

dxdyEI +

−=

@ x = 0 ( ) ( )63

02

003

11

3 WLccLWEIdxdy

−=⇒+−

=⇒=632dx

( )66

33 WLxLW

dxdyEI −−=∴

( ) 34 WLLWIntegrating twice

( )2

3

646cxWLxLWEIy +−

−−=

( )4@ x = 0 ( ) ( ) ( )

240

640

600

4

22

34 WLccWLLWEIy =⇒+−−

−=⇒=

( )24624

434 WLxWLxLWEIy +−−−= ( )

24624y

Max. occurs @ x = L

WLyWLWLLWEIy4444

−=⇒−=+−=EI

yEIy88246 maxmax −=⇒−=+−=

EIWL8

4

max =Δ

y x

x

L

2WL

2WL

xWWLM22xWxxWM −=

22 xWxWLydEI =222 Wx

dxEI −=

Integrating32

cxWxWLdyEI +−=Integrating 13222c

dxEI +=

Since the beam is symmetric 02

@ ==ddyLxy

2 dx

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

32

22L

WL

WLL 3WL( ) ⇒+⎠⎝−⎠⎝== 132

222

20

2@ cWWLEILx

241WLc −=

3WLWWLd2464

332 WLxWxWL

dxdyEI −−=∴

343 WLxWxWLIntegrating 2244634cxWLxWxWLEIy +−−=

( ) ( )43

@ x = 0 y = 0 ( ) ( ) ( ) ( ) 2

343

0244

063

04

0 cWLWWLEI +−−=⇒ 02 =⇒ c

xWLxWxWLEIy242412

343 −−=∴

242412

5 4WLMax. occurs @ x = L /2

3845

maxWLEIy −=

EIWL

3845 4

max =Δ

y x Px

PL0f

L/2

2P

2PL/2

0for2 LxxPydEI <<=

xPMLx22

0for =<<

20for

22 xxdx

EI <<=

Integrating2

cxPdyEI +=Integrating 122c

dxEI +=

Since the beam is symmetric 02

@ ==ddyLxy

2 dx

⎟⎠⎞

⎜⎝⎛

2

2L

PL 2PL( ) ⇒+⎠⎝== 122

20

2@ cPEILx

161PLc −=

2PLPd164

22 PLxP

dxdyEI −=∴

23 PLxPIntegrating 21634cxPLxPEIy +−=

( )3@ x = 0 y = 0 ( ) ( ) ( ) 2

23

0163

04

0 cPLPEI +−=⇒ 02 =⇒ c

xPLxPEIy1612

23 −=∴

1612

3PLMax. occurs @ x = L /2

48maxPLEIy −=

EIPL

48

3

max =Δ

B ’  El ti  C  Diff ti l E ti

Md 2

Beam’s Elastic Curve Differential Equation

EIM

dxyd=2

2

EIdxBut,

dMV dxdMV =

dxdVq = dxq

ExampleA mechanical assembly system moves sensitive electronicA mechanical assembly system moves sensitive electronic parts from one location to another using a small cantilever beam. The beam has two sections as shown in the diagram. Th l i ill l b l d h d dThe electronic parts will only be located on the extended section of the beam. The deflection of the beam tip is critical in the assembly process.What is known: A solid steel beam supports electronic parts over half of the beambeam. The parts have an average weight per area of 0.04 N/mm2. The steel stiffness, E, is 200 GPa. The two beam parts are rigidly connected Cantilever Beam Used to The two beam parts are rigidly connected. The beam is attached to the delivery mechanism and the connection can be assumed to be fixed.

Cantilever Beam Used to move Electronic Parts

Question :What is the deflection of the beam tip?Approach :Approach :•Modify the area load to a linear load. •Determine the moment of inertia and the moment equations for both beam sections.•Identify the boundary conditions. •Integrate the moment equations to find the deflection equations.

The deflection at the beam tip can be determined by first finding the moment equations and then integrating those g q g gequations. There are actually two moment equations, one for each half, since the load and beam structure is not continuous Four boundary conditions will benot continuous. Four boundary conditions will be needed, two for each beam section.

Moment Equation and DiagramTo determine beam deflections is to integrate the moment equation. This requires that the

t ti i k b f t ti th i t ti Thi b d b tti h

Beam Loading

moment equation is known before starting the integration. This can be done by cutting eachbeam section and developing a moment equation as a function of the beam location, x.For the cantilever beam, there are two sections, the first one is from point A to B and thesecond is from point B to C. Making a cut in the first sections and solving for the moment andshear at the cut surface gives,

V1 = -0.48x NM1 = -0.24x2 N-mm

The second cut in section 2 gives

V = 24 0 NV2 = -24.0 NM2 = -24 (x - 25) N-mm

= -24x + 600 N-mm

Moment Equation and DiagramTo determine beam deflections is to integrate the moment equation. This requires that the moment equation is known before starting the integration. This can be done by cuttingbefore starting the integration. This can be done by cutting each beam section and developing a moment equation as a function of the beam location, x.

For the cantilever beam, there are two sections, the first one is from point A to B and the second is from point B to C. M ki t i th fi t ti d l i f thMaking a cut in the first sections and solving for the moment and shear at the cut surface gives,

V1 = -0.48x NM1 = -0.24x2 N-mm

The second cut in section 2 gives

V2 = -24 0 NV2 24.0 NM2 = -24 (x - 25) N-mm

= -24x + 600 N-mm

The equations are plotted at the right. Moment and Shear Diagrams

B  P iBeam Properties

The moment of inertia for a rectangular cross section gives, 

12mm

3mm

I1 = 12(3)3/12 = 27.0 mm4

I2 = 16(5)3/12 = 166.7 mm4

The material stiffness, E, is given as

     G   / (   / )

5mm

E = 200 GPa = 200×109 N/m2 (1 m/1000 mm)2= 200,000 N/mm2 16mm

Integrating Moment Equations

The deflection of any beam can be found by integrating the basic moment differential equation, 

EIy´´ = M

However  each section must be integrated separately  Integrating section AB twice However, each section must be integrated separately. Integrating section AB twice gives, 

∫ +−= dxxy )60024.0()27)(000,200( '

13'6 08.0104.5 Cxy +−=×

4'6 0201045 CxCxy ++−=×

Recall, y is the deflection and y´ is the slope of the beam. The constants of integration, C1 and C2, must be determined from the boundary conditions (see below).

2102.0104.5 CxCxy ++=×

1 2, y ( )

Integrating the second beam section BC, gives

∫' ∫ +−= dxxy )60024()7.166)(000,200( '

32'6 600121033.33 Cxxy ++−=×

43236 30041033.33 CxCxxy +++−=×

Boundary Conditions

There are four constants of integrating that need to be defined. This requires four boundary conditions. 

The first two conditions are due to the fixed The first two conditions are due to the fixed joint at the right end. This requires both the deflections, v, and the slope, v´, to be zero. These are listed in the table at the right as These are listed in the table at the right as conditions 1) and 2).

A th  t   diti    b  id tifi d  t Another two conditions can be identified at the joint between beam sections 1 and 2.Since the beam is continuous, the beam deflection and slope on either side of the joint must be equal. This gives the third and fourth condition, as listed in the table. Four Boundary Conditions,

Determining Constants

With the four boundary conditions defined, four equations can now be constructed which will allow all four constants to be determined  determined. 

Generally, boundary conditions can be applied so that only one  i    i     i   i  H   i  constant is present in a given equation. However, sometimes 

two or three equations will need to be solved simultaneously. 

Boundary Condition 2)  v´2 = 0 at x = 100 mm33.33×106 y´ = ‐12x2 + 600x + C333.33×106 (0) = ‐12(100)2 + 600(100) + C333.33×10 (0)    12(100) + 600(100) + C3C3 = 60,000 N‐mm2

Boundary Condition 1)  v = 0 at x = 100 mmBoundary Condition 1)  v2 = 0 at x = 100 mm33.33×106 y = ‐4x3 + 300x2 + C3x + C433.33×106 (0) = ‐4(100)3 + 300(100)2

 6 ( )   C+ 60,000(100) + C4C4 = ‐5.0×106 N‐mm3

Four Boundary Conditions

Boundary Condition 4)  y´  y´ at x   50 mmBoundary Condition 4)  y 1 = y 2 at x = 50 mm

63

2

61

3

10333360012

104508.0

×++−

=×+ CxxCx

1033.33104.5 ××

6

2

61

3

1033.33000,60)5(600)50(12

104.5)50(08.0

×++−

=×

+− C

21 720,19 mmNC −=

Boundary Condition 3)  y1 = y2 at x = 50 mm

4323

214 300402.0 +++−

=++− CxCxxCxCx Four Boundary Conditions

66 1033.33104.5 ××

6

623

62

4 105)50(000,60)50(300)50(4)50(720,19)5(02.0 ×−++−=

++− C66 1033.33104.5 ××

32 000,145,1 mmNC −−=

Final Deflection EquationsFinal Deflection Equations

The final deflection equations for both beam sections qare

y = ‐3 704×10‐9x4 + 0 003652x ‐ 0 2120 mm y1 = ‐3.704×10 9x4 + 0.003652x ‐ 0.2120 mm applies for 0 ≤ x ≤ 50

  7 3   6 2   8    y2 = ‐1.2×10‐7x3 + 9.0×10‐6x2 + 0.0018x ‐ 0.15 mmapplies for 50 ≤ x ≤ 100 

The maximum deflection at the tip (x = 0) isyx=0 = ‐0.2120 mm

Final Deflection Curveyx 0

ExampleBeams commonly have distributed loads over only sections of the total beam, similar to the beam shown in the diagram. In this case, what is beam shown in the diagram. In this case, what is the deflection of the far right end (point C)?

Assume the modulus of elasticity  E  is equal to 10 Assume the modulus of elasticity, E, is equal to 10 GPa.

Partially Loaded Beam Solution

There are two basic ways to solve this problem using integration. First, the moment inboth beam sections, AB and BC, can be determined, and then integrated using twoboundary conditions for each spanboundary conditions for each span.

A second, and easier method, would be to find the rotation angle at B, and then just extrapolate the deflection from B to C  This is possible since there are no loads on the extrapolate the deflection from B to C. This is possible since there are no loads on the beam section BC and thus there will be no bending. The section BC will rotate, but it remain a straight line. Of course, the rotation at B still needs to be determined by i i   h     i  b  i  i   l     i  i d  f    i  integrating the moment equation, but it is only one equation instead of two equations needed in method 1. 

Due to symmetry, each of the two supports will carry half the load, giving,the load, giving,

Ay = By = 3(2)/2 = 3 kN

The moment equation for the first span, AB, is found by cutting the span at distance x from the left, and 

i    Thi   i  summing moments. This gives, 

M1 + 3x(x/2) ‐ 3x = 0M1 = 3x ‐ 1.5x2 kN‐m 

Now that the moment equation is known for the span, it q p ,can be integrated once to find the beam rotation, and a second time for beam deflection, 

''

Beam Support Reactions 

1''

1 MEIy =

∫ −= dxxxEIy )5.13( 2'1 dxCxxEIy )5051( 32 += ∫∫ dxxxEIy )5.13(1

132 5.05.1 Cxx +−=

dxCxxEIy )5.05.1( 11 +−= ∫21

43 125.05.0 CxCxx ++−=

The deflection, y1, for the span AB is know at x = 0 and x = 2 m. Using these two boundary conditions, givesthese two boundary conditions, gives

y1(x=0) = 0 = 0.5 (0)3 ‐ 0.125 (0)4 + C1 (0) + C2 ==> C2 = 0 

y1(x=2) = 0 = 0.5 (2)3 ‐ 0.125 (2)4 + C1 (2) + 0  ==> C1 = ‐1

Thi   i  th  b   t ti    This give the beam rotation as 

EIy' = (1.5x2 ‐ 0.5x3 ‐1) kN‐m2

The beam moment of inertia, I, is

I = 83(4)/12 = 170.7 cm4 = 170.7e‐8 m4

and EI = (10e9 N/m2)(170.7e‐8 m4 ) = 17.07 kN/m2

The beam rotation at B (x=2) is The beam rotation at B (x 2) is 

y'(x=2) = (6 ‐ 4 ‐1)/EI (1 kN m2)/(17 07 kn m2) = (1 kN‐m2)/(17.07 kn‐m2) 

= 0.05858 radians

The final deflection at C can be determined by noting that that beam rotation at B is The final deflection at C can be determined by noting that that beam rotation at B is also the slope at B. The final deflection is just the angle (or slope) times the distance, 

δC = θ d=0.05858 (2 m) = 0.1172 m 

Beam Deflection at Point C

2

0 1

-0,000366

Deformed shape, Comb: CASE1, Units: kN-mLi P 2 7 | E Silj k | @b i b | li bLinPro 2.7 | Enes Siljak | eness@bosnia.ba | www.line.co.ba

Integration of Load gEquation 

In the previous sections, Integration of the Moment Equation, was shown how todetermine the deflection if the moment equation is known This section willdetermine the deflection if the moment equation is known. This section willextend the integration method so that with additional boundary conditions, thedeflection can be found without first finding the moment equation.

Moment‐Shear‐Load RelationshipsWhen constructing moment‐shear diagrams  it was When constructing moment shear diagrams, it was noticed that there is a relationship between the moment and shear (and between the shear and the loading). That relationship can be derived by applying the basic equations relationship can be derived by applying the basic equations to a typical differential element from a loaded beam (shown at the right). First, summing the forces in the 

i l di i   ivertical direction givesΣFy = 0 V ‐ (V + dV) ‐ w(x) dx + 0.5 (dw) dx = 0 ( ) ( ) 5 ( )

Both dw and dx are small, and when multiplied together gives an extremely small term which can be ignored  gives an extremely small term which can be ignored. Assuming (dw)(dx) = 0, and simplifying gives,

( )dV Differential Element from Beamw(x)

dx−=

Next, summing moments about the right side (can be anywhere, but an edge is easier) and ignoring the 3rd order terms givesg g 3 g

ΣMright edge = 0 

‐M + (M + dM) ‐ V dx + [w(x) dx][0.5 dx] = 0 

Again, 2nd order terms such as dx2, are assumed extremely small and can be ignored. This gives

dM

N t   it l "V" i   h   d  t d fl ti

VdxdM

=

Note, capital  V  is shear and not deflection

Extending the Deflection Differential EquationExtending the Deflection Differential EquationRecall, the basic deflection differential equation (Moment‐Curvature Equation) was derived as Equation) was derived as 

MEIyydEI ''2

== MEIydx

EI 2

Thi    b   bi d  ith dM/d  V t   iThis can be combined with dM/dx = V to give

)(''' xVEIy = shear‐deflection equation

This equation assumes E and I are constant along the length of the beam section. They can be combined g ywith dV/dx = ‐w(x) to give

)('''' xwEIy = load‐deflection equation)(xwEIy load deflection equation

Th  th  d fl ti    b  d t i d di tl  Thus, the deflection can be determined directly from the load function, but it does require four integrations and four boundary conditions. 

h hWhere as using the moment‐curvature equation, only two integrations and two boundary conditions are needed, but the moment equation qmust first be determined.

w, M, V, Slope, and y Relationshipsand Sign Conventions 

Solving the Load‐Deflection Differential EquationThe differential equation EIv´´´´ = w(x) is not useful by itself but needs to be The differential equation EIv = w(x) is not useful by itself but needs to be applied to a beam with specific boundary conditions. It is assumed that EI is constant and w(x) is a function of the beam length. Note, the function, w(x), can be equal to 0  In fact  in most situations it does equal 0  be equal to 0. In fact, in most situations it does equal 0. Integrating the equation four times gives,

23∫∫ ∫∫ 432

2213

161)( CxCxCxCdxxwEIy ++++−= ∫∫∫∫

Th  i t ti   t t  C  C  C d C    The integration constants, C1, C2, C3 and C4, are determined from the boundary conditions. For example, a pinned joint at either end of a beam requires the deflection, v, equal 0 and the moment, M, equal 0.. A fixed joint requires both the deflection, v, and slope, v´, equal 0, but moment and p qshear are unknown. Each beam section must have at least four boundary conditions. Details about boundary conditions are given below. boundary conditions are given below. Each beam span must be integrated separately, just like when constructing a moment diagram. Thus, each new support or load will start a new beam each new support or load will start a new beam section that must be integrated. Examples of beam sections are shown at the right.

Each Beam Section Requires itsOwn Deflection Equation

Boundary ConditionsDetermining the boundary conditions (b.c) is usually the most difficult part of solving the g y ( ) y p gdeflection differential equation, especially when integrating four times. In particular, b.c for multiple beam sections can be confusing. The basic types of b.c are shown below. Those conditions that require two sections are sometimes yp qcalled continuity conditions instead of b.c. For example, a point force on a beam causes the deflections to be split into two equations. However, the beam's deflection and slope will be continuous at the load location requiring y1 = y2 and y´1 = y´2. Also, the shear difference will equal the applied point load at that location, and the moment will be equal in both beam sections at that point, M1 = M2. These conditions are needed to solve for the additional integration constants.

Typical Boundary Typical Boundary Conditions (y, ´, V, M) for Beam Sections

Using  Fourth Order Using  Fourth Order Load‐Deflection 

Equation

CASE STUDY SOLUTIONThe deflection at the tip of the equipment can be The deflection at the tip of the equipment can be determined by integrating the basic distributed load equation four times. There are actually four 

i    f   h h lf  i   h  l d  d b  equations, one for each half since the load and beam structure differ in each section. A total of eight boundary conditions will be needed to solve the four integration constants for both equations.

Beam Loading

Free Body DiagramFree‐Body DiagramTo help determine beam sections and boundary conditions, a free‐body diagram should be 

d h h b dconstructed. Each change in beam geometry and load requires a new beam section and deflection equation. For this cantilever beam, there will be qtwo sections, one from point A to B and a second section from point B to C.The uniform distributed load on the left part of the  Free‐Body DiagramThe uniform distributed load on the left part of the beam is 

w = (0.04 N/mm2)(12 mm) = 0.48 N/mm The actual values of the reactions do not need to 

y g

The actual values of the reactions do not need to be determined which is one of the advantages of this method.

Beam Properties

Th   t  f i ti   i  th   ti  f     t l     ti   iThe moment of inertia using the equation for a rectangular cross section gives,I1 = 12(3)3/12 = 27.0 mm4

I2 = 16(5)3/12 = 166.7 mm4

The material stiffness, E, is given asE = 200 GPa = 200×109 N/m2 (1 m/1000 mm)2/ ( / )= 200,000 N/mm2

Integrating the Load‐Deflection EquationsIntegrating the Load Deflection EquationsThe deflection of any beam can be found by integrating the basic load‐deflection differential equation, 

( )EIy´´´´ = ‐w(x)for each beam section.Section 1 (from point A to B)pThe load function w1(x) is the actual uniform distributed load of 0.48 N/mm

EIy1´´´´ = ‐0.48 N/mm EIy ´´´ = ‐0.48x + C => VEIy1   0.48x + C1 > V1EIy1´´ = ‐0.24x2 + C1 x + C2 => M1EIy1´ = ‐0.08x3 + C1 x2/2 + C2 x + C3EIy =  0 02x4 + C x3/6 + C x2/2 + C x + CEIy1 = ‐0.02x4 + C1 x3/6 + C2 x2/2 + C3 x + C4

Note that EIy´´´ is the shear and EIy´´ is the moment. This will be needed when applying the boundary conditions.

Section 2 (from point B to C)Section 2 is similar to section 1 except there is no uniform load. Thus, w2(x) is just 0  just 0. 

EIy2´´´´ = 0 N/mm EIy2´´´ = C5 => V2EI ´´  C    C  MEIy2´´ = C5 x + C6 => M2EIy2´ = C5 x2/2 + C6 x + C7EIy2 = C5 x3/6 + C6 x2/2 + C7 x + C8Boundary ConditionsBoundary Conditions

There are eight constants of integrating that need to be defined. This requires eight b.c, where the first two 

diti    d  t  th  fi d j i t  t th   i ht  d  conditions are due to the fixed joint at the right end. This requires both the deflections, y, and the slope, y´, to be zero. These are listed in the table at the right as conditions 1) and 2); and the next two conditions are due to continuity between beam sections 1 and 2. Since the beam is continuous, the beam deflection and slope on , peither side of the joint must be equal. Conditions 5 and 6 are from the free end which cannot have any shear or moment   And finally  similar to slope and deflection  moment.  And finally, similar to slope and deflection, the shear and moment need to be the same between beam sections 1 and 2. The shear is the the same since there is no point load at the joint  Likewise  the moment there is no point load at the joint. Likewise, the moment is the same since there is no applied point moment at the joint. 

Eight Boundary Conditions

Determining ConstantsWith the eight boundary conditions defined, eight equations can now be g y g qconstructed. Generally, boundary conditions can be applied so that only one constant is present in a given equation. However, sometimes two or three equations will need to be solved simultaneously. to be solved simultaneously. Boundary Condition 5) V1 = 0 at x = 0 mm

0 = ‐0.48(0) + C1C = 0 C1 = 0 

Boundary Condition 6) M1 = 0 at x = 0 mm0 = ‐0.24(0) + C1 (0) + C2C    C2 = 0 

Boundary Condition 7) V1 = V2 at x = 50 mm‐0.48x + C1 = C55‐0.48(50) + 0 = C5C5 = ‐24 N

Boundary Condition 8) M1 = M2 at x = 50 mmy ) 1 2 5‐0.24x2 + C1x + C2 = C5x + C6‐0.24(50)2 + 0 + 0 = 24(50) + C6‐600 = ‐1 200 + C6600 =  1,200 + C6C6 = 600 N‐mm 

Boundary Condition 2)  y´2 = 0 at x = 100 mm33 33 106 ´   24 2/2   600    C33.33×106 y´2 = ‐24x2/2 + 600x + C733.33×106 (0) = ‐12(100)2 + 600(100) + C7C7 = 60,000 N‐mm2

Boundary Condition 1)  v2 = 0 at x = 100 mm

33.33×106 y2 = C5x3/6 + C6x2/2 + C7x + C833.33×106 (0) = ‐4(100)3 + 300(100)2

+ 60,000(100) + C8+ 60,000(100) + C8C8 = ‐5.0×106 N‐mm3

Boundary Condition 4)  v´1 = v´2 at x = 50 mm21213080 +++++ CxCxCCxCxCx

67652

1

63212

1

1033.33104.508.0

×++

=×

+++− CxCxCCxCxCx

6

2

63

3

1033.33000,60)50(600)50(12

104.5)50(08.0

×++−

=×

+− C

272019 NC 23 720,19 mmNC −=

Boundary Condition 3)  v1 = v2 at x = 50 mm

687

262

1356

1

643

222

1316

14

1033.33104.502.0

×+++

=×

++++− CxCxCxCCxCxCxCx

6

623

64

4

1033.33105)50(000,60)50(300)50(12

104.5)50(720,1900)50(02.0

××−++−

=×

++++×− C

34 000,145,1 mmNC −−=

Final Deflection Equations

The final deflection equations for both beam sections aresections arey1 = ‐3.704×10‐9x4 + 0.003652x ‐ 0.2119 mm y2 = ‐1.2×10‐7x3 + 9.0×10‐6x2 + 0.0018x ‐ 0.15 mm 

The maximum deflection at the tip (x = 0) isyx=0 = ‐0.2119 mm

This is the same (within rounding error) of the previous solution. (yx=0 = ‐0.2120 mm) Final Deflection Curve

x 0

ExampleA beam is constructed where one end A beam is constructed where one end cannot deflect (pinned joint), but can rotate, and the other end cannot rotate, but can deflect  can deflect.. Determine the deflection equation by integrating from the loading function. Shelf Supports and LoadingSolutionSolutionStarting with the loading function, the deflection can be found by integrating it four ti  H  thi   ill  i  f   i  times. However, this will require four unique boundary conditions. Those four conditions are

1. θ(x=L) = y'(x=L) = 0 2. y(x=0) = 0 3. M(x=0) = y''(x=0) = 0 3 ( ) y ( )4. V(x=L) = y'''(x=L) = 0 

The loading is a constant distributed load, or 

Equivalent beam deflections using superposition principle 

gEI y'''' = ‐w 

Integration this gives the shear function, Integration this gives the shear function, V(x) = EIy''' = ‐wx + C1

The 4th boundary condition can be used to determine the integration constant, C1, V(x=L) = ‐wL + C1 = 01C1 = wL

Integrating again gives the moment equation,M(x) = EIy'' = ‐wx2/2 + xwL + CM(x)   EIy     wx /2 + xwL + C2

The 3rd boundary conditions gives, 0 = ‐w 02/2 + x 0 L + C2C  0 C2 = 0 

Thus, the final moment equation isM(x) = ‐wx2/2 + xwL

N   hi   i    b  i d    i   h   i   iNext, this equation can be integrated to give the rotation equation,θ(x) = EIy' = ‐wx3/6 + wLx2/2 + C3

Using boundary condition 1 gives, 0 = ‐wL3/6 + wL3/2 + C3C3 = ‐wL3/3 

The final rotation equation is,q ,θ(x) = EIy' = ‐wx3/6 + wLx2/2 ‐ wL3/3 

The deflection equation is just the integral of the rotation equation,EIy = ‐wx4/24 + wLx3/6 ‐ wL3x/3 + CEIy = ‐wx4/24 + wLx3/6 ‐ wL3x/3 + C4

Applying the last unused boundary conditions, number 2, gives, 0 = ‐ 0 + 0 ‐ 0 + C4 ==> C4 = 0

Th  fi l d fl ti   ti  i  The final deflection equation is y = ‐w (x 4/8 ‐ Lx3/2 + xL3) / (3EI)