ewahyuni-mekbanenglish colour.pdf

7/25/2019 ewahyuni-MekbanEnglish colour.pdf

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Dr. Endah Wahyuni 1

MEKANIKA BAHAN

Mechanics of Materials)

3 CREDITS

Statically Determinate

Mechanics11

Lecturers:Lecturers:

UntilUntil EETSTS

EndahEndah WahyuniWahyuni, ST (ITS),, ST (ITS), MScMSc (UMIST), PhD ((UMIST), PhD (UoMUoM))[email protected]@gmail.com @end222@end222

--

Prof. Ir. Priyo Suprobo, MS, PhDProf. Ir. Priyo Suprobo, MS, PhD

22


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Dr. Endah Wahyuni 2

BILINGUAL CLASSBILINGUAL CLASS

Module in English, Class in Indonesian; orModule in English, Class in Indonesian; or

v ce versa.v ce versa.

Delivery of contents in 2 languagesDelivery of contents in 2 languages

(Indonesian & English).(Indonesian & English).

Technical terms in EnglishTechnical terms in English

33

MaterialsMaterials Books:Books:

1.1. E.P. Popov, 1978, Mechanics of MaterialsE.P. Popov, 1978, Mechanics of Materials

.. , ,, ,MaterialsMaterials

3.3. R.C. Hibbeler, 1997, Mechanics of MaterialsR.C. Hibbeler, 1997, Mechanics of Materials

4.4. Any related books, with topic: Mechanics ofAny related books, with topic: Mechanics ofMaterialMaterial

..

http://personal.its.ac.id/dataPersonal.php?userid=http://personal.its.ac.id/dataPersonal.php?userid=ewahyuniewahyuni

http://www.structuralconcepts.orghttp://www.structuralconcepts.org44


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E.P. Popov, 1978, Mechanics ofE.P. Popov, 1978, Mechanics of

Materials, 2Materials, 2ndnd editionedition

55

GereGere & Timoshenko& Timoshenko, 2008, Mechanics, 2008, Mechanics

of Materials, 7of Materials, 7thth editionedition

66


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R.C. Hibbeler,R.C. Hibbeler, 20102010, Mechanics of, Mechanics ofMaterialsMaterials, 8, 8thth editionedition

77

Other books: Mechanics of MaterialOther books: Mechanics of Material

88


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Dr. Endah Wahyuni 5

Learning MethodsLearning Methods ClassClass

Students areStudents are requiredrequired to read the courseto read the coursematerial to be provided in the existing classmaterial to be provided in the existing classscheduleschedule

ResponsivenessResponsiveness

Exercises in class with guidanceExercises in class with guidance

QuizQuiz

nn--c ass exam a any g ven mec ass exam a any g ven me Home workHome work

Students do the work to be doneStudents do the work to be done at homeat home withwiththe responsibility, not only collects the dutythe responsibility, not only collects the duty..

99

EvaluationsEvaluationsUTS (30%) UAS (30%)

Quiz1 (10%) Quiz2 (10%)

PR1 (10%) PR2 (10%)

*Prosentase bisa diubah sesuai yang menguntungkan mahasiswa

1010


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Notes:Notes: 20 minutes late20 minutes late, not permitted to enter the class, not permitted to enter the class..

Disturbing classDisturbing class go outgo out

Home work is collected before the class startingHome work is collected before the class starting

Keep the spirit on!Keep the spirit on!

1111

ContentsContentsMetode Pembel aj aran Bobo t Ni l ai

dan Evaluasi %

1 1 Ketepatan penjelaskan Kuliah lihat UTS

tentang tegangan, rergangan, modulus

elastisitas serta modulus geser

No

Dapat menjelaskan tentang tegangan, a. pendahuluan

regangan, modulus elastisitas serta modulus b. pengertian tegangan, regangan

geser c. pengertian modulus elastisitas

Minggu ke Kompetensi Indikator Kompetensi Materi Pembelajaran

2 2 & 3 Ketepatan perhitungan tegangan pada Kuliah lihat UTS

balok yang menerima beban lentur murni Responsi

PR 1 2

3 4 & 5 Ketepatan perhitungan tegangan geser Kuliah lihat UTS

pada balok akibat beban lentur Responsi

PR 2 2

4 6 Ketepatan perhitungan tegangan dan Kuliah lihat UTSDapat menghitung tegangan dan regangan a. pengertian torsi

penampang. lentur

c. shear center

d. geser pada profil berdinding tipis

Dapat menghitung tegangan geser pada balok a. hubungan momen dan gaya

yang disebabkan oleh beban lentur, lintang

pada balok-balok dengan berbagai bentuk b. tegangan geser akibat beban

baik semasih pada kondisi elastis maupun non elastis

sesudah mencapai kondisi non elastis

pada sebuah balok akibat beban lentur murni b. lentur muni pada balok dengan

baik pada balok dengan bahan tunggal dua bahan

maupun pada balok dengan dua bahan, c. lentur murni pada balok

d. static test

Dapat menghitung tegangan yang terjadi a. lentur muni pada balok elastis

1212

regangan pada poros akibat beban torsi Responsi

c. regangan oleh torsi PR 3 2

5 7 & 8 Ketepatan perhitungan kombinasi tegangan Kuliah lihat UTS

dan ketepatan penggambaran bentuk kern Responsi

PR 4 2

6 9 Test 40UTS

berbagai bentuk penampang penampang kolom

c. kern

Dapat mengkombinasikan tegangan-tegangan a. kombinasi tegangan pada balok

sejenis pada penampang balok atau kolom tidak simetris

dan dapat menggambar bentuk kern dari b. kombinasi tegangan pada

d. tegangan oleh torsi pada poros

non elastis

pada poros akibat beban torsi b. tegangan geser torsi


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Contents1. Introduction

2. Slicin Method

3. Understanding of Stress

4. Normal Stress

5. Average Shear Stress

6. Determine of and

7. STATIC TEST

8. Allowed Stress

9. Strain1313

10. Diagram, Normal Stress - Strain

- HOOKE law

- Yield Point

- Deformation of bars from Axial loads

- Poissons Ratio

- Relationship of Stress, Strain and Poissons Ratio

.

- Shear Stress

- Shear Strain

1414


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12. Pure Bending on beams

13. Moment of Inertia

14. Calculating Stress on beams

.

16. Pure bending on non-elastic beams

17. Shear-bending Stress

18. Tors ion

19. Multi le Stresses

20. Combination of stresses on Columns

21. KERN

22. ..etc ETS1515

After midsemester evaluation:After midsemester evaluation:

1.1. Plane stress analysisPlane stress analysis

Maximum and minimum stressMaximum and minimum stress

MohrMohr CircleCircle

2.2. Bar design based on stressBar design based on stress

Based onBased on axial stressaxial stress,, flexureflexure and shear for prismaticand shear for prismatic

barbar andand definite staticdefinite static3.3. Definite Static Beams deformationDefinite Static Beams deformation

Equation of elastic line deformation method.Equation of elastic line deformation method.

Area moment methodArea moment method

4.4. Stability of Compression BarStability of Compression Bar

Centric Load and Shear Force.Centric Load and Shear Force.

1616


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ReviewsReviews::Statically Determinate MechanicsStatically Determinate Mechanics

Determinate StructureDeterminate Structure :: If?If?

Static Equation ??Static Equation ??

11

22

33

1717

sendi rol

sendi

sendi1818


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sendi rol

sendi sendi1919

ReactionsReactions

Simply supported beamsSimply supported beams

Cantilever beamsCantilever beams

TrussesTrusses

2020


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LoadingsLoadings-- PointPoint LoadLoadAtAt midspanmidspan,,

Within certain locationWithin certain location

-- Distribution LoadsDistribution Loads Full distributed loadsFull distributed loads

Partially distributed loadsPartially distributed loads

-- At the end of cantileverAt the end of cantilever

MidspanMidspan

Within certain locationWithin certain location2121

Modul 1Modul 1

Tegangan dan ReganganTegangan dan Regangan

Stress & StrainStress & Strain

2222


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IntroductionIntroduction At a structure, each elements of a structureAt a structure, each elements of a structure

should be having a dimension. The elementsshould be having a dimension. The elements

them or maybe applied to them. To calculate thethem or maybe applied to them. To calculate thedimension of the elements, we should know thedimension of the elements, we should know themethods to analyses, which are:methods to analyses, which are:

strengthstrength (( kekuatankekuatan),),

stiffnessstiffness (( kekakuankekakuan)),,

,,The methods will be discussed in this Mechanic ofThe methods will be discussed in this Mechanic of

Materials.Materials.

2323

Mechanics of materials is a subject of a very oldMechanics of materials is a subject of a very old

age, which generally begins with Galileo in the early 17thage, which generally begins with Galileo in the early 17th

century. The first one describes the behavior of thecentury. The first one describes the behavior of the

structure of load rationally.structure of load rationally.

2424


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The behavior of the structure to obtain the force dependsThe behavior of the structure to obtain the force dependsnot only on the fundamental laws of Newtoniannot only on the fundamental laws of Newtonianmechanics that govern force equilibrium but also to themechanics that govern force equilibrium but also to thephysical characteristics of the structural parts, which canphysical characteristics of the structural parts, which canbe obtained from the laboratory, where they are givenbe obtained from the laboratory, where they are givent e orce o act on s nown accurate y.t e orce o act on s nown accurate y.

Mechanics of Material is a mixed knowledge from theMechanics of Material is a mixed knowledge from theexperiment and the Newtonian principals on elasticexperiment and the Newtonian principals on elasticmechanics.mechanics.

ne o e ma n pro ems n mec an cs o ma er a s s one o e ma n pro ems n mec an cs o ma er a s s oinvestigate the resistance of an object, that is theinvestigate the resistance of an object, that is theessence of the internal forces for balancing the externalessence of the internal forces for balancing the externalforces.forces.

2525

APPLICATIONS Planning of a Structure

MATERIALS

CONTROL

PLANNING OF THE

DIMENSIONS

STRENGTH /

STRESS

STRUCTURES: STABLE2626


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EXAMPLE

TUBE TRUSSES

2727

EXAMPLEBUILDING FRAME

70/70

50/50

2828


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EXAMPLE

P1P2

H1 H2

B1 B2

Because of P2 > P1, thus from stress

analysis, dimension wil l be obtained

where B2 > B1, H2 > H12929

Metode Irisan

P2P1 P2

P1

GAYA DALAM

S1S2

S3

S1S3

P3P4

P3P4GAYA DALAM

3030


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Tegangan (Stress)

Tegak Lurus

Bidang Potongan

Sejajar Bidang

Potongan

DEFINISI :

TEGANGAN ADALAH GAYA DALAM YANG

BEKERJA PADA SUATU LUASAN KECIL

TAK BERHINGGA DARI SUATU

POTONGAN 3131

Stress (Tegangan)MATHEMATICS EQUATIONS

F=

LimA

V=

A 0Lim

A

SHEAR STRESS

= Normal Stress

F

A

V

= Cross-section area

= Forces on perpendicular of cross-section

= Forces on parralel of cross-section3232


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Stress (Tegangan)

Stress symbols on elements related with

coordinates : z

yy

z

xz

yz

zyzx

x

xxy yx

3333

Normal Stresses

NORMAL STRESS

Tension

NORMAL STRESS

Compression

= P/A = P/AP P

PP 3434


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Average Shear Stresses

SHEAR STRESSFORCES ACTING

PARRALEL SECTIONCREATING

= P Cos / ANormal

AShear

P

AShear

ANormal

= P / AShear3535

Average Shear Stress

P

P

AShear

= P / Total AShear

Total Ashear=

2 x Sectional Area of Bolts3636


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Calculation of

STRESS

Determine

and

NEED TO UNDERSTAND

THE PURPOSE AND THE GOAL

CALCULATION

DETERMINATION OF FORCE

CHOOSE THE EQUATION

or

WILL BE PROBLEM IF

AND CROSS SECTIONALAREA

CALCULATION RESULT

DONT UNDERSTANDSTATICALLY

DETERMINATED

ENGINEERING MECHANIC

3737

DETERMINE FORCE VALUEUSE STATIC EQUATION:

FX = 0 MX = 0

FY = 0 MY = 0

FZ = 0 MZ = 0

Define Cross Sectional Area

Choose the smallest AreaTo get

The Maximum Stress

3838


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Determine Cross Sectional Areaexample :

The smallest cross

sectional area that was

choosen to get the

maximum stress value

3939

Example 1Example 1::

A concrete wall as shown in the figure, received distributed loads ofA concrete wall as shown in the figure, received distributed loads of 2020

kN/mkN/m22.. Calculate the stress onCalculate the stress on 1 m1 m above the based. The gravitationabove the based. The gravitation

load of the concrete isload of the concrete is 25 kN/m25 kN/m33

4040


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Answer:Answer:

Self weight of concrete wallSelf weight of concrete wall::W = [(0,5 + 1,5)/2] (0,5) (2) (25) = 25 kNW = [(0,5 + 1,5)/2] (0,5) (2) (25) = 25 kN

Total loadTotal load:: P = 20 0,5 0,5 = 5 kNP = 20 0,5 0,5 = 5 kN

FromFrom FFyy = 0,= 0, the reactionthe reaction R = W + P = 30 kNR = W + P = 30 kN

using upper part of the wall as a free thing, thus the weightusing upper part of the wall as a free thing, thus the weight

of the wall upper the crossof the wall upper the cross--section issection is WW11 = (0,5 + 1) (0,5)= (0,5 + 1) (0,5)

(25/2) = 9,4 kN(25/2) = 9,4 kN

FromFrom FFyy = 0,= 0, the Load on sectionthe Load on section :: FFaa = P + W= P + W11 = 14,4 kN= 14,4 kNNormal stress onNormal stress on aa--aa isis aa = Pa/A = 14,4/(0,5x1) = 28,8= Pa/A = 14,4/(0,5x1) = 28,8

KN/m2KN/m2

The stress is a compression normal stress that worked asThe stress is a compression normal stress that worked as

FaFa on the section.on the section.4141

StressTASK :

If W = 10 Ton, a = 30o and cross

sectional area of steel cable ABC = 4

cm2, cable BD = 7 cm2, so calculate

stress that happened in ABC and BD1.

D

cables.

A W

C If bolt diameter = 30mm, b = 200 mm, d1 =

8 mm, d2 = 1 2 m m , P =

2000 k so calculate

2.

P

P

d1d2

b

the maximum stress

of each frame and

shear stress of the

bolt.

4242


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Static Test

P

P LOAD INCREASE

CONTINUOUSLY

MATERIAL

P ULTIMATE LOAD

TESTING MATERIAL

P ULTIMATE STRESSPUlt

A4343

Universal Test Machine (UTM)Universal Test Machine (UTM)

4444


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FLEXURE TESTFLEXURE TEST

4545

STRAIN TESTING MATERIAL

STATIC TEST LOAD

P

STRAINL

-. load

- Every Pload increasing, list deformation

of testing material that shows in dial

gauge.P

4646


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Strain

P (Load)

= StrainL

=

Change as every

Loading changes

(Deformation)P Diagram

4747

Stress Strain Diagram

Physical properties of every mater ial can be shownfrom their stress strain diagram relationship.

P (load) (Stress)

P Diagram Diagram = Strainpict. A pict. B

4848


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STRESS STRAIN DIAGRAM

- MATERIAL 1 AND MATERIAL - 2, BOTH ARE IDENTICAL

MATERIAL

- THE CROSS SECTIONAL AREA OF MATERIAL - 2 < MATERIAL - 1

- THE P

RELATIONSHIP OF MATERIAL - 1 ARE DIFFERENT

WITH MATERIAL - 2

- THE

RELATIONSHIP OF MATERIAL - 1 ARE SIMILAR WITH

MATERIAL - 2, ALTHOUGH THEY HAVE DIFFERENT CROSS

SECTIONAL AREA

THEREFORE, MORE SUITABLE USING PICTURE B

TO KNOW PHYSICAL PROPERTIES OF SOME

MATERIAL

4949

Stress Strain Diagram

(Stress)(Stress)

Proportional

Limit

StrainStrain

STEEL MATERIAL CONCRETE MATERIAL

5050


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HOOKE LAW

E X= ELASTICCONDITION

=E

(Stress)

DETERMINATION OF YIELD POINT

OFF-SET METHOD

= STRESS

= STRAIN

E = ELASTICITY MODULUS

Strain

ropor onaLimit

5151

HOOKEs LAW

problem

: In some frame with L =100 cm in length,

Static Test was done. If P thats ivenP

to this f rame is 4000 kg, this f rame is st il l

in elastic condit ion, and goes on 2 mm in

length, so calculate of s tress and strainvalue of that frame. If modulus elastici ty

value is 2 x 106 kg/cm2 and then calculate

the cross sectional area of that frame.L

P 5252


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Bar Deformation due to

Axial Load

P2P3

4

dx

1

PxPx Px force to dx elemen and

cause d deformation

x + x

d = dx

Edx

P

E

dxAx

=

5353


Axial Loadexample :

B

B

L

P = Px Px

dx = Px / Ax . E dx

0

x . x x .

A L

= P . X / Ax . E

L

P P

x

Ax = A , Px = P

= P . L / E . ADeformation due to P load,selfweight was ignored 5454


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B L


Axial LoadDEFORMATION DUE TO SELFWEIGHT IS :

= Px . dx / Ax . E = 1 / A . E w . X . dx

A 0

= . W.x2 / A . E = w . L2 / 2 . A . E = WT . L / 2 . A . E

0

L

DEFORMATION DUE TO P LOAD AND SELFWEIGHT IS :

= P.L / A.E + WT.L / 2.A.E =

= L (P + .WT) / A.E 5555

Contoh 2Contoh 2--1:1:

Tentukan pergeseran relatif dari titikTentukan pergeseran relatif dari titik--titik A dan D padatitik A dan D pada

batang baja yang luas penampangnya bervariasibatang baja yang luas penampangnya bervariasi

sepertiseperti terlihatterlihat padapada gambargambar didi

bbawahawah bilabila diberikandiberikan empatempat gayagaya terpusatterpusat PP11,, PP22,, PP33anan 44.. m a = xm a = x mm ..

5656


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Gaya dalam batang adalah :Gaya dalam batang adalah :

Antara titik A dan B, PAntara titik A dan B, Pxx = +100 kN= +100 kNAntara titik B dan C, PAntara titik B dan C, Pxx == --150 kN150 kN

Antara titik C dan D, PAntara titik C dan D, Pxx = +50 kN= +50 kN

Dengan menggunakan persamaan:Dengan menggunakan persamaan:

Dengan memasukkan hargaDengan memasukkan harga--harga numeric dari contoh,harga numeric dari contoh,

maka diperoleh:maka diperoleh:

5757

BAR DEFORMATION DUE TOAXIAL LOAD

Problem :

C100 cm 100 cm

If the bar diameter of AB

and BC is 20 mm, = 30o.

B

DE1000 kg

y u u

2x106 kg/cm2, calculate

deformation of point B.

2.

P1 P2

b1

b2

b3

h1

h2

1 2, 1 2working, the length of both bar still

be similar, i f b1 = 50 mm, b2 = 50 mm,

b3 = 25 mm, h1 = 500 mm, h2 = 500

mm and thickness of both bar = 20

mm.

P2

5858


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Poissons Ratio

STRAIN

AXIAL STRAIN LATERAL STRAIN

The shape is being

LONGER and

SMALLER

POISSONS RATIO ( ) =

Lateral

Axial

Concrete = 0.1 0.2

Rubber = 0.5 0.65959

The Relationship of Poissons

Ratio, Stress and Strain

xzyz

zy

zxy

zy

xxy yx

x

6060


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y

y

z 6161



x EE E

x y z+ - -=

y EE E

x y z- + -=

z EE E

x y z- - +=

6262


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Shear Stress and Shear Strain

SHEAR STRESS

A

yz yz

zy

yz

AB

C

/2

/2

C

B

= SHEAR STRAIN

O O

MO = 0 zy(dy.dx).dz - (dx.dz.).dy = 0yz

zy yz=

Fz = 0 yz left = yz right 6363

Shear Stress and Shear Strain

SHEAR STRAIN:

SHAPE TRANSFORMATION THAT IS EXPRESSED

CALLED SHEAR STRAIN

HOOKE LAW for Shear stress and shear strain:= Shear Stress

= Shear Strain

. G=

= Shear Modulus

= Poissons Ratio

G2 (1+ )

=

The relationship between Normal Modulus Elasticity and

Shear Modulus

G

6464


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Modul 2Modul 2

beam flexurebeam flexure

(pure bending)(pure bending)

6565

Pure Bending in Beam

Flexure due to

MOMEN only

6666


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Ya

max max

/2/2

Initial Length

b =

Force Equilibrium:

( Y/C . max ) dA = 0

A

C Y . dA = 0

A

FX = 0

6767


MOMENT :

M = ( Y/C . max ) dA . Y = max Y2 . dA

A A

Y2 . dA = = Inertia Moment

M = ( max / C ) . max = M . C /

A

max = M . Ya /

TOP FIBER STRESS BOTTOM FIBER STRESS

max = M . Yb /

6868


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GENERALLY:

= M . Y /

/ Y = W (Resis tance Moment)

/ Ya = Wa

/ Y = W

= Y 2 . dA

A

INERTIA MOMENT

6969

INERTIA MOMENTEXAMPLE :

h/2

x = y 2 . dAA

= Y 2 . b . dy

h/2

-h/2

h/2

y

h/2= 1/

3. 1/

4. h3. b = 1/

12. b. h3

1/2x = 3. 2 . d + 2 y 2 . dy

-11/2 11/2

= 3 . y . = 3 . 8 + 8 . .-h/2

b

x

y

2

2

3

11

-2 -11/2

+ 3.y 2 . dy11/2

2x

7070


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INERTIA MOMENTEXAMPLE :

= 3/3 . y3

-11/2

+ 2 . 1/3 . y3

11/2

+ 3/3 . y3

2

-2 -11/2 11/2

= (-11/2)3 (-2)3 + 2/3 . (11/2)

3 - 2/3 . (-11/2)

3 + 23 - (11/2)3

= 13,75

= 1/12 . 3 . 43 1/12 . 1 . 3

3 = 16 2,25 = 13,75

SHORTER CALCULATION

7171

STRESS CALCULATION

OF THE BEAM

10 cm10.000 kg

30 cm

10 cm30 cm

10 cm

400 cm

CROSS SECTIONAL AREA :

. . .

INERTIA MOMENT:

= 1/12 . 30 . 503 2 . 1/12 . 10 . 30

3 = 267.500 cm4

7272


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STRESS CALCULATIONOF THE BEAM

RESISTANCE MOMENT:

Wa = Wb = I/y = 267.500 / 25 = 10.700 cm3

WORKING MOMENT (Beban Hidup Diabaikan) :

MMax = . 10.000 . 400 = 1.000.000 kgcm.

MAXIMUM STRESS OCCURED:

Max = MMax / W = 1.000.000 / 10.700 = 93,46 kg/cm2

7373

Stress Calculation

of BeamMax

y1 = 20 cmyMax

+

-1

Max

= M / W1 = 1.000.000 . 20 / 267.500 = 74.77 kg/cm21

W1 = / y17474


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EXERCISE MOMENT INERTIA

30 cm

10 cm

Calculate Inertia

Moment of i ts

Sb Y

1

40 cm

10 cm

xand weak axis ( Iy )

Sb X

Calculate InertiaSb Y10 cm2

Moment of itsstrong axis( Ix ) and

weak axis ( Iy )Sb X

10 cm

8 cm

10 10 10

20 cm

7575

EXERCISE PURE BENDING

1 2

100 kg/m (include its selfweight)200 cm 80 cm

400 cm 200 cm

1500 kg

30 cm

10 cm

- Draw its momen diagram

- Calculate Inertia Moment of Beam

Section

10 cm

10cm

8cm

8cm

- Calculate edge fiber stresses of

section - 1 and 2, then draw its

stress diagram

- Calculate its maximum stress

7676


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ASSYMETRIC FLEXURE

q

q

qCos

qSin

L

Moment occurs of X-axis (MX)

and Y-axis (MY)

= 2 = 2 . os

. . n

.

Moment that its flexure

round X-axis

Moment that its flexure

round Y-axis7777

Stress of the Section due to

Assymetric Flexure q

qSin

a

b

d

o a

b

MX . h/2

x

+My . b/2

y

= +

MX . h/2

x

-My . b/2

y

= +

MX . h/2 -My . b/2= -

q

qCos

d

x y

MX . h/2

x

+My . b/2

y

= -

MX = 1/8 . qCos

. L2

MY = 1/8 . qSin

. L2 x = 1/12 . b . h

3y = 1/12 . h . b

37878


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Exercise - Stress of the Section due to

Assymetric Flexureq

P

L = 300 cm, q = 100 kg/m, P= 200 kg, h = 20 cm, b = 10

cm, = 30o

c

d

L

s n cm o s ance

f rom B

Calculate stress that

occurs in the midspan a, b,

c, d, e and f. Where point -

e is 5 cm of distance fromf

a

bx-axis and 3 cm from y-axis.

Point - f is 6 cm of distance

from x-axis and 4 cm from

y-axis

e

7979

Problem - Iassume W = 8 Ton, =

90o and cross section

area of the steel cable

ABC = 4 cm2, eaxh of BD

frame = 6 x 3 cm2, so

calculate stress that

1.

D

occurs in ABC cable and

maximum stress of BD

frame.Calcu late the def lect ion

of point - b and shear

AB

C

50 cm

B

. .

diameter of As.B = 20

mm.

Modulus Elastici ty of BD

frame = 2x106 kg/cm2.

W W

8080


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1 2

2000 kg/m (include its selfweight)200 cm 80 cm

80 cm

2.

400 cm 200 cm

1000 kg

30 cm

10 cm

- Dram i ts moment diagram

- Calculate Inertia Moment of Beam

- Calculate ed e fiber stresses of25 cm

1000 kg

20 cm

10cm

8cm

8cm

section 1 and 2, then draw itsstress diagram.

- Calculate Maximum stress that

occurs in ABC beam.8181

qP

BAc df

3.

L

L = 300 cm, q = 1000 kg/m, P = 2000 kg, = 30o, P is

ab

.

Calculate stress that occurs in the midspan of poin t

a, b, c, d, e and f.

8282


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Composite Beam (2 Material)dx

dy

x

1

xE1

e

h

b1

b2

a

e

2

1

eE1

eE2

DISTRIBUTION OF

ELASTIC STRESS

DISTRIBUTION OF

SINGLE MATERIAL -

STRESS

8383

Composite Beam (2 Material)

b1

. 2

b2/n1

b1.n1

b1/n2

E1 > E2, n1 = E1 / E2, n2 = E2 / E1

Cross Section of

Frame with 1st Material

Cross Sestion of

Frame with 2nd Material

8484


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Exercise -Composite Beam (2 Material)

1

1000 kg

A B12 cm

a

b

400 cm1Concrete

2 1200 cm36 cm

12 1210c

1

1st Material = Concrete

2nd Material = Steel

E concrete = 200.000 kg / cm2 ; E stel = 2.000.000 kg /cm2

Steel

Calculate stress that occured in the section 1 1 and infibera , concrete fiber b , steel f iberb and fiber c .

Draw its stress diagram.

(Selfweight of the beam is ignored) 8585

Pure Bending of

Non-Elastic Beam

STRESS-STRAIN DIAGRAM

ELASTIC NON - ELASTIC

8686


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Pure Bending of

Non-Elastic Beam

Strain

distribution

Elastic Strain

distribution

Non Elastic Strain

distribution

a

bc

d

o

If effect of D aob andcod are small

8787

Rectangular Beam that have Full Plastic Condition

hh/4

C

h/4T

Plastic moment that can be held = C . . h = T . . h

= = bh yp

Plastic momen of a rectangular beam is:

Mp = yp .bh/2 .

h/2 = yp .bh /4

2

8888


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Rectangular Beam that have Full Plastic ConditionGenerally can be written as:

h/2

0

h/2

yp . y2 . b = yp .

bh /4

Mp = . y dA = 2 ( yp ) . y . b . dy0

2

ca cu a e w e as c equa on :Myp = yp . / (

h/2) = yp .1/12 b h

3 / ( h/2 )

= yp . b . h2 / 6

8989

Rectangular Beam that have Full Plastic Condition

Mp / Myp = yp . b . h2 / 4

yp . b . h2 / 6

= ,

Section that have Elastic Plastic condition

yoh/2

Minor Yield

(Elastic-Plastic)

Major Yield

(Elastic-Plastic)

Al l Yield

(Plastic) 9090


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Section that have Elastic Plastic condition

Elastic-Plastic moment that can be held with stressdistibution which have partial yield is:

M = . y dA = 2 ( yp ) . y/yo . b . y. dy 0

+ 2 ( yp) . b . y. dy2

yo

yp . y3//yo . b

o

yo

= 2/3

yo

+ yp . b . y2

h/2

= 2/3 yp . yo2 . b + yp . bh

2 / 4 - yp . b . yo2

= yp . bh2 / 4 1/3 yp . b . yo

2 = Mp 1/3 yp . b . yo2

9191

Modul 3Modul 3

Shear Stress of BeamShear Stress of Beam

9292


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Shear Stress - Flexure

q (x)

V V+dV

dxM+dM

dx

M x

S MA = 0

(M + dM) M (V + dV) . dx + q . dx . dx/2 = 0

M + dM M V . dx + dV . dx + . . dx2 = 0

dM V . dx = 0

dM = V . dx

small small

OR

dM / dX = V

9393

dM / dx = V

Shear Stress - FlexureThis equation is giving

explanation that :

IF THERE IS FLEXURE MOMENT

DIFFERENCE AT SIDE BY SIDE

SECTION, THERE WILL BE A SHEAR.

Example :

L/3 L/3 L/3M M

Bid. D

.

SHEAR

M M+dM

9494


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Shear Stress - Flexure

Shear Stress due to Flexure Loada

b

h

je g

R

d f

-

FBFA

-FB =

Afghj

.

IdA

I= Y . dA

Afghj

=- MB . Q

IQ = Y . dA

Afghj

= Afghj . Y9595

Shear Stress - FlexureShear Stress due to Flexure Load

- MA

I= Y . dA

A

FA =- MA . Q

I

FB FA = R Held up by shear connector

=- MB . Q

I-

- MA . Q

I= dF

( MA + dM ) . Q MA . Q dM . Q

Troughout dx

=I

=I

dF/dx = q = SHEAR FLOW

q = dM . Q / dx . I = V . Q / I9696


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Shear Stress due to Flexure Load

Example : = 50 . 200 . 25 + 50 . 200 . 15050 . 200 + 50 . 200

= 87,5 cm200 mm

Yc

50 mm YcI = 200 . 503 / 12 + 50 . 200 . 62,52

= 50 . 2003 / 12 + 50 . 200 . 62,52

= 113.500.000 mm4 = 11.350 cm4

Q = 50 . 200 ( 87,5 25 ) = 625.000 mm3

= 625 cm3 or

Y1

. ,

200 mm

Y1 = 200 Yc - 200 / 2 = 62,5 mm

Q = 50 . 200 . 62,5 = 625.000 mm3 = 625 cm350 mm

q = V . Q / I = 30.000 x 625 / 11.350 = 1.651 kg / cm

Nail spacing = 7000 / 1651 = 4,24 cm 9797

200 mm

50 mm

Problem : Assume that top nai ls capacityis 7000 kg and bottom nails is

5000 kg. Then calculate spacing

of top and bottom nail, from A

until B, so the section strength

50 mm

30 mm

200 mm

150 mm

enough to carried on q load.

Spacing of top and bottom nails

was made in 3 different type ofspacing.

q = 3000 kg/m

600 cm

A B

100 100 200 100 100

9898


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Shear Stress Diagram

Longitud inal Direction:

= dF / t .dx = ( dM / dx ) . ( A . Y / I . t ) = V . A . Y / I . t

=.

I . t

t=

Example :t = b

h

1/8 . V. h2

I

h

dyf g yy1

=.

I . t

t=

V

I . tY . dA

A

=

9999

Shear Stress Diagram

V

I= x

Y2

2

h/2

y1

V

I . by1

h/2

b . y . dy=

( b/2 ) 2 y12

2 . I=

If y1 = 0, so

V

2 . I=

h2

4x = 1/8

V . h2

1/12 . b .h3

=3 . V

2 . b. h=

3 . V

2 . A

100100


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Problem :

20 cm

5 cma

bq = 3000 kg/m

P = 1500 kg 1200 cm

5 cm

3 cm

20 cm

15 cm

d

e

600 cm

Draw shear stress diagram of the section in support A

and of the section - 1 that is 100 cm of dis tance from

point B.

101101

Working steps:

1. Calculate the Neutral Axis

20 . 5 . 2,5 + 20 . 5 . 15 + 15 . 3 . 26,5

20 . 5 + 20 . 5 + 15 . 3= = ,

From TOP

2. Calculate Inertia Moment

I =1/12 . 20 . 5

3 + 20 . 5 . 9,512 + 1/12 . 5 . 203

+ 20 . 5 . 2,952 + 1/12 . 15 . 33 + 15 . 3 . 14,492

= 208,33 + 9044,01 + 3333,33 + 870,25 +

33,75 + 9448,20

= 22937,88 cm4

102102


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3. Calculatie shear forces

Ra = 3000 . 6/2 + 2/3 . 1500 = 10.000 kgRb = 3000 . 6 + 1500 - 10.000 kg = 9.500 kgVa = 10.000 kg ; V1 = - 9.500 + 3000 . 1= - 6.500 kg

Position A y Q q = V.Q / I = q / tt

a

b1b2

0

100

100

12.01

9,51

9,51

20

20

5

0 0

951951

0

In section A with 10.000 kg of shear force

414,6

414,6

20,73

82,92

c

d1d2e 0

45

45

35.05 ,3.505

14.49

14.49

15.99

5

5

15

150 0

1073,85

652.05

652.05

0

468,16

284,27

284,27

93,63

56,854

18,951

103103

Posisi A y Q q = V.Q / I = q / tt

a 0 12.01

200 00

In Section 1 with 6.500 kg of shear force

b1b2

c

d1d

100

100

100

45

35.05

9,51

9,51

9,513.505

14.49

14.49

20

5

5

5

15

951951

1073,85

652.05

652.05

269,49

269,49

304,30

184,774

184 774

13,474

53,89

60,86

36,955

12 318

e 0 15.99 150 00

104104


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20 cma 0 0

Shear Stress Diagram:

5 cm

3 cm

20 cm

5 cm b

c

d

e

13,474

0 0

53,89

60,68

36,955

12,318

20,73

93,63

56,854

82,92

18,951

15 cmShear Force

10.000 kg

Shear Force

6.500 kg

105105

Shear Flow Variation

Shear flow variation is used to determine the SHEAR

CENTER, so that vertical loading that works will not

induce torsion to the section, if works in i ts SHEAR

CENTER106106


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Shear Center

P=

F1

eh

F1

e = F1 . h / P =2 . P . I . t

b. t. h . V . Q .

. b . t . h

P=

=. b . t . h

2 . P I . t

V . . h . b . tx =

b2 . h2 . t

4 . I107107

Problem :

P 10 cm Determine the SHEAR

CENTER of thisV=P

F1 F2

10 cm

10 15 30

.

Equation that is used:

e . P + F1 . 60 = F2 . 60

e = ( F2 . 60 F1 . 60 ) / P

. . 17,5 . 10F1 = F2 = . 37,5 . 10 . 108108


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I = 1/12 . 55 . 703 - 1/12 . 40 . 50

3 = 1.155.416,67 cm4

P . 17,5 . 10 . . 60=

V . Q

I . t=

1.155.416 67 . 10= 0,00045 . P kg/cm2

Calculation :

P . 37,5 . 10 . . 60= = =

V . Q

I . t 1.155.416,67 . 100,00097 . P kg/cm2

F1 = 0,00045 . P . 17,5 . 10 . = 0,0394 . P

F2 = 0,00097 . P . 37,5 . 10 . = 0,1820 . P

e = 0,0394 . P. 60 -0,182 . P . 60 =:

P

8,556 cm

In order to make frame didnt induce torsion , so the

Pload must be placed in e = 8,556 cm ( see Picture)109109

KERN / GALIH / INTIVariety of KERN :

Limited with 4 oint

Limited with 6 point

Unlimited

m e w po n

110110


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KERN / GALIH / INTIDetermine Inertia moment of s loping axis:

Y

x df Cos Sin x

x= + y

X

Cos Sin y

y= - x

2y

=Ix

df

Ix

= y2

2

222Cos x+ Sin - 2xy Sin Cos d

f

= Ix Cos + Iy Sin -2 Sxy Sin Cos 2

111111

2x

=Iy

df

KERN / GALIH / INTIDetermine Inertia Moment of Sloping axis:

= x2

2

222Cos y+ Sin +2xy Sin Cos d

f

= Ix Sin + Iy Cos + 2 Sxy Sin Cos 2

112112


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KERN / GALIH / INTIExample of determining KERN limits :

Determine the Neutral axis :

+2 cm

y

. . . . .

2.20 + 8.2.2==x 3,2 cm

A = 2.20 + 8.2.2 = 72 cm

Ix = 1/12.2.203 + 1/12.8.2

3.2

+ 8.2.92.2 = 3936 cm42

2

16x

=Wax3936

10= 393,6 cm3

=Wbx3936

10= 393,6 cm3

10

3,2

113113

KERN / GALIH / INTIContoh Menentukan batas batas KERN :

Iy = 1/12.20.23 + 1/12.2.8

3.2

2 2 = 4 . . , . . . , ,

=Wkr y628,48

3,2= 196,4 cm3

=Wkn y 6,8= 92,42 cm3

628,48

W 393,6 Wkn y 92 42a x =

A=

72

= 5,46 cm

Kb x =Wax

A=

393,6

72

= 5,46 cm

kr y = A 72=

= 1,28 cm

Kkny =Wkr y

A 72=

196,4

= 2,72 cm114114


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KERN / GALIH / INTIPicture of KERN limits :

2,72 cm1,28 cm

2 cm

16

y

x

5,46 cm

2

2

10

3,2

,

115115

Modul 4Modul 4

TorsiTorsionon

116116


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TORSION (Puntiran )

30 N-m

30 N-mSection Plane

20 N-m

10 N-m

10 N-m

INNER TORSION MOMENT equal with OUTTER TORSION MOMENT

Torsion that is learned in this Mechanics of Materials

subject was limited in rounded section only.

117117

TORSION (Puntiran )

M

Torsion Moment at

both end of the bar

MM

Torsion Moment

distr ibuted along theM(x) bar

118118


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TORSION (Puntiran )

max

Cmax

Cmax . dA . = T

A

C

ress

Area

Forces Distance

Torsion MomentOr can be writ ten as:

max

C . dA = T

2

= IP. dA2

= Polar Inertia Moment

A

A119119

Example of Polar Inertia Moment forCIRCLE

. dA2

=

A

3 d2 =

0

C

2

4

.

4

.4

.

0

C

= C

=32

d4

2

this equation:

maxT =

C. IP

TORSION MOMENT

max =.

IPTORSION STRESS

120120


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For CircleFor Circle Hollow SectionHollow Section::

121121

TWIST ANGLE OF CIRCULAR BARTWIST ANGLE OF CIRCULAR BAR

122122

With determine small angle of DAB in this followingpicture. The maximum stress of its geometry is:


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If :If :

Then:Then:

So general statement of the twist angle of a section fromSo general statement of the twist angle of a section from

the bar with linier elastic material is:the bar with linier elastic material is:

123123

PROBLEM EXERCISEPROBLEM EXERCISE -- 11

See a tiered bar that shown in this following picture, its outboard in

the wall (point E), determine rotain of point A if torsion moment in B

and D was given. Assume that the shear modulus (G) is 80 x 109

N/m2.

124124


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Polar Inertia MomentPolar Inertia Moment::

BarBar AB = BCAB = BC

BBarar CD = DECD = DE

Considerin its left section torsion moment in ever art will be:Considerin its left section torsion moment in ever art will be:

TTABAB = 0, T= 0, TBDBD = T= TBCBC = T= TCDCD = 150= 150 N.mN.m, T, TDEDE = 1150= 1150 N.mN.m

125125

To get rotation of edge A, can be done with add up everyTo get rotation of edge A, can be done with add up every

integration limit:integration limit:

Value ofValue of TT andand IIpp areare constant,constant, so the equation will beso the equation will be::

126126


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EXERCISEEXERCISE --11

Calculate maximum torsion shear stress of ACCalculate maximum torsion shear stress of AC bar (asbar (as

seen in AC barseen in AC bar exercise 1)exercise 1).. Assume that bar diameterAssume that bar diameter

from Afrom A C is 10 mm.C is 10 mm.

AnswerAnswer::

127127

ExercisesExercises

Soal 4.1Soal 4.1

e ua poros erongga mempunyae ua poros erongga mempunya

diameter luar 100 mm dan diameter dalamdiameter luar 100 mm dan diameter dalam

80 mm. Bila tegangan geser ijin adalah 5580 mm. Bila tegangan geser ijin adalah 55MPa, berapakah besar momen puntir yangMPa, berapakah besar momen puntir yang

bisa diteruskan ? Berapakah teganganbisa diteruskan ? Berapakah tegangan

pada mukaan poros sebelah dalam bilapada mukaan poros sebelah dalam bila

diberikan momen puntir ijin?diberikan momen puntir ijin?

128128


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129129

Sebuah poros inti berongga berdiameterSebuah poros inti berongga berdiameter

mm pero e engan me u angmm pero e engan me u ang

poros melingkar padat berdiameter 300poros melingkar padat berdiameter 300

mm hingga membentuk lubang aksialmm hingga membentuk lubang aksialberdiameter 100 mm.berdiameter 100 mm. BerapakahBerapakah

persentase kekuatan puntiran yang hilangpersentase kekuatan puntiran yang hilang

oleh operasi ini ?oleh operasi ini ?

130130


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131131

Poros padat berbentuk silinder dengan ukuran yangPoros padat berbentuk silinder dengan ukuran yang

bervariasi an terlihat dalam ambar di erakkan olehbervariasi an terlihat dalam ambar di erakkan oleh

momenmomen--momen puntirmomen puntir seperti ditunjukkanseperti ditunjukkan dalamdalam

gambargambar tersebut.tersebut. Berapakah tegangan puntirBerapakah tegangan puntir

maksimum dalam poros tersebut, dan diantara keduamaksimum dalam poros tersebut, dan diantara keduakatrol yang ada ?katrol yang ada ?

132132


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133133

a.a. Tentukanlah tegangan geser maksimum dalam porosTentukanlah tegangan geser maksimum dalam poros

an dihada kan ada momenan dihada kan ada momen--momen untir anmomen untir an

diperlihatkan dalam gambar.diperlihatkan dalam gambar.

b.b. b. Hitunglah dalam derajat sudut pelintir antara keduab. Hitunglah dalam derajat sudut pelintir antara kedua

ujungnya. Ambillah G = 84.000 MN/m.ujungnya. Ambillah G = 84.000 MN/m.

134134


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135135

ModulModul 55

STRESS COMBINATIONSTRESS COMBINATION

136136


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EquationEquation that have learned before about linier elasticthat have learned before about linier elasticmaterial, can be simplified as:material, can be simplified as:

NormalNormal StressStress::

a.a. ue oue o axa oaaxa oa

b.b. Due toDue to flexureflexure

A

P

137137

I

Shear StressShear Stress::

a.a. Due toDue to torsiontorsion

I

T

b.b. Due toDue to shear forceshear force of beamof beam

tI

VQ

Superposition of the stress, only considered inSuperposition of the stress, only considered in

elastic problem when deformation thatelastic problem when deformation that

happened is small.happened is small.

138138


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EXERCISE:EXERCISE: A barA bar 50x75 mm50x75 mm that isthat is 1.51.5 metermeter of length, selfweight isof length, selfweight is

not considered, was loaded as seen in this followingnot considered, was loaded as seen in this following

picture.picture. (a).(a). Determine maximum tension andDetermine maximum tension and

com ression stress that work e endicularl of beamcom ression stress that work e endicularl of beam

section, assume that it is an elastic materialsection, assume that it is an elastic material..

139139

ANSWERANSWER Using superposition method,Using superposition method, so it can be solved in twoso it can be solved in two

stepssteps.. In PictureIn Picture (b)(b), it shows that the bar only take axial, it shows that the bar only take axial

load only. Then In Pictureload only. Then In Picture ((cc)), it shows that the bar only, it shows that the bar only

take transversal load onlytake transversal load only

140140

,,

is:is:


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Normal stress due to tranversal load depends on flexureNormal stress due to tranversal load depends on flexure

moment value and the maximum flexure moment is inmoment value and the maximum flexure moment is in

force that use:force that use:

Stress superposition woks perpendicularly of beamStress superposition woks perpendicularly of beam

section and linearly decreased to the neutral axis assection and linearly decreased to the neutral axis asseen in picture (g)seen in picture (g)

141141

142142


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STRESS COMBINATION ON COLUMNSTRESS COMBINATION ON COLUMN

Similar equation can be done to assymetricSimilar equation can be done to assymetric

section:section:

WhenWhen::

Flexure MomentFlexure Moment MyyMyy = +P z= +P z00 that works of ythat works of y--axisaxis

== -- --

yy

yy

zz

zzx

I

z

I

y

A

AA is cross section area of frameis cross section area of frame

IzzIzz andand IyyIyy isis inertia moment of the section to each theirinertia moment of the section to each their

principal axisprincipal axis

Positive symbolPositive symbol (+)(+) is tension stress, andis tension stress, and NegatiNegativeve

symbolsymbol ((--)) isis compression stress.compression stress. 143143

ExampleExampleDetermine stress distribution of ABCD section of the

beam as seen on this following picture. if P = 64 kN.

Beams weight is not considered.

144144


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AnswerAnswer::Forces that work inForces that work in ABCDABCD sectionsection,, on the pictureon the picture (c),(c), isis

P =P = --6464 kNkN,,

yyyy -- .. -- ,, .. ,,

MMzzzz == --64 (0.075 + 0.075) =64 (0.075 + 0.075) = --9,69,6 kN.mkN.m..

Cross section area of the beamCross section area of the beam A = (0.15)(0.3) = 0,045 m,A = (0.15)(0.3) = 0,045 m,

And its Inertia moment isAnd its Inertia moment is::

145145

JadiJadi dengandengan menggunakanmenggunakan hubunganhubungan yangyang setarasetara dapatdapat

diperolehdiperoleh tegangantegangan normalnormal majemukmajemuk untukuntuk elemenelemen--

elemenelemen sudutsudut ::

BilaBila tandatanda hurufhuruf tegangantegangan menandakanmenandakan letaknyaletaknya,, makamakategangantegangan normalnormal sudutsudut adalahadalah ::

146146


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147147

THE END

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