mekanika bahan -...

09/02/2014

Dr. Endah Wahyuni 1

MEKANIKA BAHAN(Mechanics of Materials)

Prerequisite :

3 CREDITS

Prerequisite :

Statically Determinate Mechanics

11

Lecturers:Lecturers:

Until Until EETSTS

Endah Endah WahyuniWahyuni, ST (ITS), , ST (ITS), MScMSc (UMIST), PhD ((UMIST), PhD (UoMUoM))

[email protected]@gmail.com @end222@end222

ETSETS -- EASEASETS ETS -- EASEAS

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

22

09/02/2014

Dr. Endah Wahyuni 2

BILINGUAL CLASSBILINGUAL CLASS

Module in English, Class in Indonesian; or Module in English, Class in Indonesian; or iivice versa.vice versa.

Delivery of contents in 2 languages Delivery of contents in 2 languages (Indonesian & English).(Indonesian & English).

Technical terms in EnglishTechnical terms in English

Students???Students??? Students???Students???

33

MaterialsMaterials Books: Books: 1.1. E.P. Popov, 1978, Mechanics of Materials E.P. Popov, 1978, Mechanics of Materials 22 GereGere & Timoshenko& Timoshenko 1997 Mechanics of1997 Mechanics of2.2. GereGere & Timoshenko& Timoshenko, 1997, Mechanics of , 1997, Mechanics of

Materials Materials 3.3. R.C. Hibbeler, 1997, Mechanics of MaterialsR.C. Hibbeler, 1997, Mechanics of Materials4.4. Any related books, with topic: Mechanics of Any related books, with topic: Mechanics of

MaterialMaterial55 OnlineOnline5.5. OnlineOnline

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

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

09/02/2014

Dr. Endah Wahyuni 3

E.P. Popov, 1978, Mechanics of E.P. Popov, 1978, Mechanics of Materials, 2Materials, 2ndnd edition edition

55

GereGere & Timoshenko& Timoshenko, 2008, Mechanics , 2008, Mechanics of Materials, 7of Materials, 7thth edition edition

66

09/02/2014

Dr. Endah Wahyuni 4

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

77

Other books: Mechanics of MaterialOther books: Mechanics of Material

88

09/02/2014

Dr. Endah Wahyuni 5

Learning MethodsLearning Methods

ClassClassStudents are Students are requiredrequired to read the course to read the course material to be provided in the existing class material to be provided in the existing class scheduleschedule

ResponsivenessResponsivenessExercises in class with guidanceExercises in class with guidance

QuizQuizII l t i til t i tiIInn--class exam at any given timeclass exam at any given time

Home workHome workStudents do the work to be done Students do the work to be done at home at home with with the 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

09/02/2014

Dr. Endah Wahyuni 6

Notes:Notes: 20 minutes late20 minutes late, not permitted to enter the class, not permitted to enter the class.. Disturbing class Disturbing 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 Pembelajaran Bobot Nilai

dan Evaluasi %

1 1 Ketepatan penjelaskan Kuliah lihat UTStentang tegangan, rergangan, modulus elastisitas serta modulus geser

No

Dapat menjelaskan tentang tegangan, a. pendahuluanregangan, modulus elastisitas serta modulus b. pengertian tegangan, regangangeser c. pengertian modulus elastisitas

Minggu ke Kompetensi Indikator Kompetensi Materi Pembelajaran

g

2 2 & 3 Ketepatan perhitungan tegangan pada Kuliah lihat UTSbalok yang menerima beban lentur murni Responsi

PR 1 2

3 4 & 5 Ketepatan perhitungan tegangan geser Kuliah lihat UTSpada balok akibat beban lentur Responsi

PR 2 2

4 6 Ketepatan perhitungan tegangan dan Kuliah lihat UTSd kib t b b t i R i

Dapat menghitung tegangan dan regangan a. pengertian torsid kib t b b t i b t t i

penampang. lenturc. shear centerd. geser pada profil berdinding tipis

Dapat menghitung tegangan geser pada balok a. hubungan momen dan gayayang disebabkan oleh beban lentur, lintangpada balok-balok dengan berbagai bentuk b. tegangan geser akibat beban

baik semasih pada kondisi elastis maupun non elastissesudah mencapai kondisi non elastis

pada sebuah balok akibat beban lentur murni b. lentur muni pada balok denganbaik pada balok dengan bahan tunggal dua bahanmaupun pada balok dengan dua bahan, c. lentur murni pada balok

d. static test

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

g p g

1212

regangan pada poros akibat beban torsi Responsic. regangan oleh torsi PR 3 2

5 7 & 8 Ketepatan perhitungan kombinasi tegangan Kuliah lihat UTSdan ketepatan penggambaran bentuk kern Responsi

PR 4 2

6 9 Test 40UTS

berbagai bentuk penampang penampang kolomc. kern

Dapat mengkombinasikan tegangan-tegangan a. kombinasi tegangan pada balok sejenis pada penampang balok atau kolom tidak simetrisdan 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

09/02/2014

Dr. Endah Wahyuni 7

Contents1. Introduction

2. Slicing Methodg

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

- Poisson’s Ratio

- Relationship of Stress, Strain and Poisson’s Ratio

11 Shear Stress and Strain11. Shear Stress and Strain

- Shear Stress

- Shear Strain

1414

09/02/2014

Dr. Endah Wahyuni 8

12. Pure Bending on beams

13. Moment of Inertia

14. Calculating Stress on beams

15 Beams with two materials15. Beams with two materials

16. Pure bending on non-elastic beams

17. Shear-bending Stress

18. Torsion

19. Multiple Stressesp

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 prismatic and shear for prismatic barbar andand definite staticdefinite static

3.3. Definite Static Beam’s deformation Definite Static Beam’s deformation

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

Unit Load methodUnit Load method Unit Load methodUnit Load 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

09/02/2014

Dr. Endah Wahyuni 9

ReviewsReviews::Statically Determinate MechanicsStatically Determinate Mechanics

Determinate Structure Determinate Structure : : If?If?Static Equation ??Static Equation ??

11 22 33

1717

rol rolrol rol

sendi rol

sendi

sendi1818

09/02/2014

Dr. Endah Wahyuni 10

rol rolrol rol

sendi rol

sendi sendi1919

ReactionsReactions

Simply supported beamsSimply supported beams

Cantilever beamsCantilever beams

TrussesTrusses

2020

09/02/2014


LoadingsLoadings

-- PointPoint LoadLoad At At midspanmidspan, , Within certain locationWithin certain location

-- Distribution LoadsDistribution Loads Full distributed loadsFull distributed loads Partially distributed loads Partially distributed loads

-- Moment LoadsMoment Loads-- Moment LoadsMoment Loads At the end of cantileverAt the end of cantilever MidspanMidspan Within certain locationWithin certain location

2121

Modul 1Modul 1

Tegangan dan ReganganTegangan dan Regangan

Stress & StrainStress & Strain

2222

09/02/2014


IntroductionIntroduction

At a structure, each elements of a structure At a structure, each elements of a structure should be having a dimension. The elements should be having a dimension. The elements have to be calculated to resist the loading onhave to be calculated to resist the loading onhave to be calculated to resist the loading on have to be calculated to resist the loading on them or maybe applied to them. To calculate the them or maybe applied to them. To calculate the dimension of the elements, we should know the dimension of the elements, we should know the methods to analyses, which are:methods to analyses, which are: strength strength ( ( kekuatankekuatan), ), stiffness stiffness ( ( kekakuankekakuan)),, stabilitystability (( kestabilankestabilan )) stability stability ( ( kestabilankestabilan ),),

The methods will be discussed in this Mechanic of The methods will be discussed in this Mechanic of Materials. Materials.

2323

Mechanics of materials is a subject of a very old Mechanics of materials is a subject of a very old age, which generally begins with Galileo in the early 17th age, which generally begins with Galileo in the early 17th century. The first one describes the behavior of the century. The first one describes the behavior of the structure of load rationally.structure of load rationally.

2424

09/02/2014


The behavior of the structure to obtain the force depends The behavior of the structure to obtain the force depends not only on the fundamental laws of Newtonian not only on the fundamental laws of Newtonian mechanics that govern force equilibrium but also to the mechanics that govern force equilibrium but also to the physical characteristics of the structural parts, which can physical characteristics of the structural parts, which can be obtained from the laboratory, where they are given be obtained from the laboratory, where they are given h f f i i k lh f f i i k lthe force of action is known accurately.the force of action is known accurately.

Mechanics of Material is a mixed knowledge from the Mechanics of Material is a mixed knowledge from the experiment and the Newtonian principals on elastic experiment and the Newtonian principals on elastic mechanics.mechanics.

O f th i bl i h i f t i l i tO f th i bl i h i f t i l i t One of the main problems in mechanics of materials is to One of the main problems in mechanics of materials is to investigate the resistance of an object, that is the investigate the resistance of an object, that is the essence of the internal forces for balancing the external essence of the internal forces for balancing the external forces.forces.

2525

APPLICATIONS Planning of a Structure

STRUCTURAL ANALYSESSTRUCTURAL ANALYSES

MATERIALS

CONTROL PLANNING OF THE DIMENSIONS

STRENGTH / STRESS

STRUCTURES: STABLE2626

09/02/2014


EXAMPLE

TUBE TRUSSES

2727

EXAMPLEBUILDING FRAME

70/70

50/50

2828

09/02/2014


EXAMPLEP1

P2

H1 H2

B1 B2

Because of P2 > P1, thus from stress analysis, dimension will be obtained

where B2 > B1, H2 > H12929

Metode IrisanP2

P1 P2P1

S2

GAYA DALAM

S1S2

S3

S1

S2

S3

P3P4

P3P4GAYA DALAM

3030

09/02/2014


Tegangan (Stress)

TEGANGAN NORMAL TEGANGAN GESERTEGANGAN NORMAL TEGANGAN GESER

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= A 0Lim

NORMAL STRESS A= A 0 NORMAL STRESS

V= A 0Lim

ASHEAR STRESS

= Normal Stress

= Shear Stress

F

A

V

= Shear Stress

= Cross-section area

= Forces on perpendicular of cross-section

= Forces on parralel of cross-section 3232

09/02/2014


Stress (Tegangan)Stress symbols on elements related with coordinates : z

yy

z

xz

yz

zyzx

x

xxy

yx

3333

Normal Stresses

NORMAL STRESS

TensionNORMAL STRESS

Compressionp

= P/A = P/AP P

PP 3434

09/02/2014


Average Shear Stresses

SHEAR STRESSFORCES ACTING PARRALEL SECTION

CREATING

= P Cos/ ANormal

AShear

P

AShear

ANormal

= P / AShear3535

Average Shear Stress

P

P

½ P½ P

AShear

= P / Total AShear

Total Ashear = 2 x Sectional Area of Bolts

3636

09/02/2014


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 SECTIONAL

AREA

CALCULATION RESULT

DON’T UNDERSTAND STATICALLY

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

09/02/2014


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 of A concrete wall as shown in the figure, received distributed loads of 20 20 kN/mkN/m22. . Calculate the stress on Calculate the stress on 1 m 1 m above the based. The gravitation above the based. The gravitation load of the concrete isload of the concrete is 25 kN/m25 kN/m33

4040

09/02/2014


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 reaction the reaction R = W + P = 30 kNR = W + P = 30 kN

using upper part of the wall as a free thing, thus the weight using upper part of the wall as a free thing, thus the weight of the wall upper the crossof the wall upper the cross--section is section 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 on Normal stress on aa--a a is is aa = Pa/A = 14,4/(0,5x1) = 28,8 = Pa/A = 14,4/(0,5x1) = 28,8 KN/m2KN/m2The stress is a compression normal stress that worked as The 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 = 4cm2, cable BD = 7 cm2, so calculatestress that happened in ABC and BD

1.

B

D

cables.

A

B

W

C If bolt diameter = 30mm, b = 200 mm, d1 =8 mm, d2 = 12 mm, P =2000 kg, so calculate

2.P

P

d1d2

b

g,the maximum stressof each frame andshear stress of thebolt.

4242

09/02/2014


Static TestP

P LOAD INCREASE CONTINUOUSLY

FRACTURE TEST INGFRACTURE TEST ING MATERIAL

P ULTIMATE LOAD

TESTING MATERIAL

P ULTIMATE STRESSPUlt

A4343

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

4444

09/02/2014


FLEXURE TESTFLEXURE TEST

4545

STRAIN TESTING MATERIAL

STATIC TEST LOAD

P

STRAIN

P increase continuously

L

-. Pload increase continuously

- Every Pload increasing, list deformation of testing material that shows in dial gauge.P

4646

09/02/2014


Strain

P (Load) = StrainL

=

Change as every Loading changes

(Deformation)P – Diagram

4747

Stress – Strain DiagramPhysical properties of every material can be shownfrom their stress – strain diagram relationship.

P (load) (Stress)

P – Diagram – Diagram

= Strainpict. A pict. B

4848

09/02/2014


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 DIFFERENTWITH MATERIAL - 2

- THE – RELATIONSHIP OF MATERIAL - 1 ARE SIMILAR WITHMATERIAL - 2, ALTHOUGH THEY HAVE DIFFERENT CROSSSECTIONAL AREA

THEREFORE, MORE SUITABLE USING PICTURE B TO KNOW PHYSICAL PROPERTIES OF SOME

MATERIAL4949

Stress – Strain Diagram

(Stress) (Stress)

Proportional Limit

StrainStrain

STEEL MATERIAL CONCRETE MATERIAL

5050

09/02/2014


HOOKE LAW

E X= ELASTIC CONDITION

=E

(Stress)

DETERMINATION OF YIELD POINT

P ti l

OFF-SET METHOD

= STRESS

= STRAIN

E = ELASTICITY MODULUS

Strain

Proportional Limit

5151

HOOKE’s LAW

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

Static Test was done. If Pload that’s givenP

Static Test was done. If Pload that s givento this frame is 4000 kg, this frame is stillin elastic condition, and goes on 2 mm inlength, so calculate of stress and strainvalue of that frame. If modulus elasticityvalue is 2 x 106 kg/cm2 and then calculatethe cross sectional area of that frame.

L

P 5252

09/02/2014


Bar Deformation due to Axial Load

P2P3

PP P4

dx

d d

P1

PxPx Px force to dx elemen and

cause d deformation

d x + dx

d= dx E

dxP

E

dx

Ax

=

5353


example :

B = P d / A E

B

L

P = Px Px

P

dx

A

= Px / Ax . E dx

0

= Px . dx / Ax . E

A L

= P . X / Ax . E0

L

P P

PxA 0

Ax = A , Px = P

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

09/02/2014


B L


DEFORMATION DUE TO SELFWEIGHT IS :

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

A 0

= ½ . W.x2 / A . E = w . L2 / 2 . A . E = WT . L / 2 . A . E0

L

DEFORMATION DUE TO P LOAD AND SELFWEIGHT IS :

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

= L (P + ½.WT) / A.E5555

Contoh 2Contoh 2--1:1:Tentukan pergeseran relatif dari titikTentukan pergeseran relatif dari titik--titik A dan D pada titik A dan D pada batang baja yang luas penampangnya bervariasibatang baja yang luas penampangnya bervariasisepertiseperti terlihat terlihat pada pada gambar gambar di di bbawah awah bila bila diberikan diberikan empat empat gaya gaya terpusat terpusat PP11, , PP22, , PP33

dd PP A bill h E 200 10A bill h E 200 1066 kN/kN/ 22dan dan PP44.. Ambillah E = 200 x 10Ambillah E = 200 x 1066 kN/mkN/m22..

5656

09/02/2014


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 kNAntara titik C dan D, PAntara titik C dan D, Pxx = +50 kN= +50 kNDengan menggunakan persamaan:Dengan menggunakan persamaan:

Dengan memasukkan hargaDengan memasukkan harga--harga numeric dari contoh, harga numeric dari contoh, maka diperoleh:maka diperoleh:pp

5757

BAR DEFORMATION DUE TO AXIAL LOAD

Problem :

1. A

C100 cm 100 cm

If the bar diameter of ABand BC is 20 mm, = 30o

and Elasticit Mod l s is1. A

B

DE1000 kg

and Elasticity Modulus is2x106 kg/cm2, calculatedeformation of point B.

Calculate P /P then after P and P2.

P1 ½ P2

b1

b2

b3

h1

h2

Calculate P1/P2, then after P1 and P2

working, the length of both bar stillbe similar, if b1 = 50 mm, b2 = 50 mm,b3 = 25 mm, h1 = 500 mm, h2 = 500mm and thickness of both bar = 20mm.

P2

5858

09/02/2014


Poisson’s Ratio

STRAIN

AXIAL STRAIN LATERAL STRAIN

The shape is being LONGER and

SMALLER

POISSON’S RATIO ( ) =

Lateral

Axial

SMALLER

Concrete = 0.1 – 0.2

Rubber = 0.5 – 0.65959

The Relationship of Poisson’s Ratio, Stress and Strain

xz

yz

zyzx

y

z y

xxy

yx

x

6060

09/02/2014


z


z

y

y

z 6161


x EE E

x y z+ - -=

y EE E

x y z- + -=

z EE E

x y z- - +=

6262

09/02/2014


Shear Stress and Shear StrainSHEAR STRESS

zy zyA

y

z

zy zy

yz

zy

yz

AB

C

/2

/2C

B

= SHEAR STRAINO O

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

zy yz=

Fz = 0 yz left = yz right 6363

Shear Stress and Shear StrainSHEAR STRAIN:

SHAPE TRANSFORMATION THAT IS EXPRESSED

WITH ANGLE TRANSFORMATION ‘ ‘ AREWITH ANGLE TRANSFORMATION ‘ ‘ ARECALLED “SHEAR STRAIN”

HOOKE LAW for Shear stress and shear strain:

= Shear Stress

= Shear Strain

. G=

E

= Shear Modulus

= Poisson’s Ratio

GE

2 (1+ )=

The relationship between Normal Modulus Elasticity and Shear Modulus

G

6464

09/02/2014


Modul 2Modul 2

beam flexure beam flexure (pure bending)(pure bending)

6565

Pure Bending in Beam

Flexure due to MOMEN only

6666

09/02/2014



Ya

Y C

max max/2/2

Initial Length

Yb = C

Force Equilibrium:

( Y/C . max ) dA = 0

A

C Y . dA = 0

A

FX = 0

6767


MOMENT :

M = ( Y/C . max ) dA . Y = max Y 2 . dA A A

Y2 . dA = I = Inertia Moment

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

max = M . Ya / I

TOP FIBER STRESS BOTTOM FIBER STRESS

max = M . Yb / I6868

09/02/2014



GENERALLY:

max = M . Y / Imax

I / Y = W (Resistance Moment)

I / Ya = Wa

I / Yb = Wbb b

I = Y 2 . dAA

INERTIA MOMENT

6969

INERTIA MOMENTEXAMPLE :

h/2

Ix = y 2 . dAA

= Y 2 . b . dy

h/2

-h/2

1/ 3 b 1/ (1/ 1/ ) h3 bh/2

y

h/2 = 1/3 . 1/4. h3. b = 1/12 . b. h3

1/2Ix = 3.y 2 . dy + 2 y 2 . dy

-11/2 11/2

= 1/3 . y3. b = 1/3 . (1/8 + 1/8) . h3. b-h/2

b

x

y

2

2

3

11

y y-2

y y

-11/2

+ 3.y 2 . dy11/2

2x

7070

09/02/2014


INERTIA MOMENTEXAMPLE :

= 3/3 . y3

-11/2+ 2 . 1/3 . y3

11/2+ 3/3 . y3

2

3 y-2

3 y-11/2

3 y11/2

= (-11/2)3 – (-2)3 + 2/3 . (11/2)3 - 2/3 . (-11/2)3 + 23 - (11/2)3

= 13,75

CARA LAIN :CARA LAIN :

= 1/12 . 3 . 43 – 1/12 . 1 . 33 = 16 – 2,25 = 13,75

SHORTER CALCULATION

7171

STRESS CALCULATION OF THE BEAM

10 cm10.000 kg

30 cm

10 cm

30 cm

10 cm

400 cm

CROSS SECTIONAL AREA :

A = ( 2 30 10 ) + (10 30 ) = 900 cm2A = ( 2 . 30 . 10 ) + (10 . 30 ) = 900 cm2

INERTIA MOMENT:

I = 1/12 . 30 . 503 – 2 . 1/12 . 10 . 303 = 267.500 cm4

7272

09/02/2014


STRESS CALCULATION OF THE BEAM

RESISTANCE MOMENT:

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

WORKING MOMENT (Beban Hidup Diabaikan) :

MMax = ¼ . 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 Beam

Max

y1 = 20 cm

yMax

+

-1

Max

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

W1 = I / y17474

09/02/2014


EXERCISE – MOMENT INERTIA

30 cm

10 cm

Calculate Inertia Moment of its strong axis( I )

Sb Y

1

40 cm

10 cm

strong axis( Ix ) and weak axis ( Iy )

Sb X

Calculate Inertia Sb Y10 cm

8 cm2

Moment of its strong axis( Ix ) andweak axis ( Iy )

Sb X10 cm

8 cm

8 cm

10 10 10

20 cm

7575

EXERCISE – PURE BENDING

1 2

A B C

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

400 cm 200 cm1500 kg

A B C

30 cm

10 cm

- Draw its momen diagram

- Calculate Inertia Moment of Beam Section

30 cm

10 cm

10 c

m8

cm

8 cm

- Calculate edge fiber stresses of section - 1 and 2, then draw its stress diagram

- Calculate its maximum stress

30 cm

7676

09/02/2014


ASSYMETRIC FLEXUREq

qqCos

qSin

L

Moment occurs of X-axis (MX)and Y-axis (MY)

MX = 1/8 . qC . L2 MY = 1/8 . qSi . L2MX 1/8 . qCos . L MY 1/8 . qSin . L

Moment that its flexureround ‘X’-axis

Moment that its flexureround ‘Y’-axis

7777

Stress of the Section due toAssymetric Flexure

L

q

c L

qSin

a

b

c

d

oa

b

c

MX . h/2

Ix+

My . b/2

Iy= +

MX . h/2

Ix-

My . b/2

Iy= +

MX . h/2-

My . b/2= -

q

qCos c

d

Ix Iy-

MX . h/2

Ix+

My . b/2

Iy= -

MX = 1/8 . qCos . L2

MY = 1/8 . qSin . L2 Ix = 1/12 . b . h3 Iy = 1/12 . h . b37878

09/02/2014


Exercise - Stress of the Section due toAssymetric Flexure

q P

BA

L = 300 cm, q = 100 kg/m, P = 200 kg, h = 20 cm, b = 10 cm, = 30o

P i i 150 f di t

c

d

o

L

BA P is in 150 cm of distance from B

Calculate stress that occurs in the midspan a, b, c, d, e and f. Where point -e is 5 cm of distance from f

a

b

ox-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 sectionarea of the steel cableABC = 4 cm2, eaxh of BDframe = 6 x 3 cm2, socalculate stress that

1.

D

occurs in ABC cable andmaximum stress of BDframe.

Calculate the deflectionof point - b and shearstress of As B bolt Bolt

AB

C

50 cm

B

stress of As.B bolt. Boltdiameter of As.B = 20mm.

Modulus Elasticity of BDframe = 2x106 kg/cm2.

W W

8080

09/02/2014


1 2

A B C

2000 kg/m (include its selfweight)200 cm 80 cm80 cm

2.

400 cm 200 cm1000 kg

A B C

30 cm

10 cm

- Dram its moment diagram

- Calculate Inertia Moment of Beam

- Calculate edge fiber stresses of25 cm

1000 kg

20 cm

10 c

m8

cm

8 cm

Calculate edge fiber stresses of section – 1 and 2, then draw its stress diagram.

- Calculate Maximum stress that occurs in ABC beam.

8181

q P

BAc de

f

3.

L

L = 300 cm, q = 1000 kg/m, P = 2000 kg, = 30o, P is100 cm from B.

ab

e

100 cm from B.

Calculate stress that occurs in the midspan of pointa, b, c, d, e and f.

8282

09/02/2014


Composite Beam (2 Material)dx

dy

a y

x

1

xE1

a

e

yh

b1

b2

a

e

2

1

eE1

eE2

DISTRIBUTION OF ELASTIC STRESS

DISTRIBUTION OF SINGLE MATERIAL -

STRESS

8383

Composite Beam (2 Material)

b2 n2 b2

b1

b2.n2

b2/n1

b1.n1

b1/n2

b2

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

Cross Section of Frame with 1st Material

Cross Sestion of Frame with 2nd Material

8484

09/02/2014


Exercise -Composite Beam (2 Material)

1

1000 kg

A B12 cm

a

b

400 cm1Concrete

2 1200 cm36 cm

12 1210c

1

1st Material = Concrete2nd Material = Steel

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

Ste

el

Calculate stress that occured in the section 1 – 1 and in fiber ‘a’, concrete fiber ‘b’, steel fiber ‘b’ and fiber ‘c’.

Draw its stress diagram.

(Selfweight of the beam is ignored) 8585

Pure Bending ofNon-Elastic Beam

STRESS-STRAIN DIAGRAM

ELASTIC NON - ELASTIC

8686

09/02/2014


Pure Bending ofNon-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

h h/4

C

h/4T

Plastic moment that can be held = C . ½ . h = T . ½ . h

C = T = ( bh/2)C T yp ( /2)

Plastic momen of a rectangular beam is:

Mp = yp . bh/2 . h/2 = yp . bh /4

2

8888

09/02/2014



Generally can be written as:

h/2

0

h/2

yp . y2 . b = yp . bh /4

Mp = . y dA = 2 ( yp ) . y . b . dy 0

If l l t ith l ti ti

2

If calculate with elastic equation :

Myp = yp . I / (h/2) = yp . 1/12 b h3 / ( h/2 )

= yp . b . h2 / 6

8989


Mp / Myp = yp . b . h2 / 4 yp . b . h2 / 6

1 5 SHAPE FACTOR= 1,5 SHAPE FACTOR

Section that have Elastic – Plastic condition

yo

h/2

Minor Yield (Elastic-Plastic)

Major Yield (Elastic-Plastic)

All Yield (Plastic) 9090

09/02/2014


Section that have Elastic – Plastic conditionElastic-Plastic moment that can be held with stress distibution which have partial yield is:

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

yo

0+ 2 ( yp) . b . y. dy

h/2

yo

yp . y3//yo . bo

yo

= 2/3yo

+ yp . b . y2

h/2

= 2/3 yp . yo2 . b + yp . bh2 / 4 - yp . b . yo

2

= yp . bh2 / 4 – 1/3 yp . b . yo2 = Mp – 1/3 yp . b . yo

2 9191

Modul 3Modul 3

Shear Stress of BeamShear Stress of Beam

9292

09/02/2014


Shear Stress - Flexureq (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 + ½ . q . dx2 = 0½ q

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/3

Bid M NO SHEAR

M M

Bid. D

Bid. M NO SHEAR

SHEAR

M M+dM

9494

09/02/2014


Shear Stress - FlexureShear Stress due to Flexure Load

a

b

h

je g

R

d f

- MB . Y

FBFA

- MBFB =Afghj

MB . Y

IdA

MB

I= Y . dA

Afghj

=- MB . Q

IQ = Y . dA

Afghj

= Afghj . Y9595

Shear Stress - FlexureShear Stress due to Flexure Load

- MA

I= Y . dA

Aabde

FA =- MA . Q

Iabde

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

09/02/2014


Shear Stress due to Flexure LoadExample :

=50 . 200 . 25 + 50 . 200 . 150

50 . 200 + 50 . 200

= 87,5 cm200 mm

Yc

V = 30 000 kg nail strength = 7000 kg50 mm Yc

I = 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

V = 30.000 kg, nail strength = 7000 kg

200 mm

625 cm or,

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 nails capacity is 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 of spacing.

q = 3000 kg/m

600 cm

A B

100 100 200 100 100

9898

09/02/2014


Shear Stress DiagramLongitudinal Direction:

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

=V . QI . t

q

t=

Example :t = b

hj

1/8 . V. h2

I

V . Q q

h

dyf g

j

yy1

=V . Q

I . t q

t=

V

I . tY . dA

A=

9999

Shear Stress Diagram

V

I= x

Y2

2

h/2

y1

VI . b y1

h/2b . y . dy=

V

( b/2 ) 2 – y12V

2 . I=

If y1 = 0, so

V2 . I

=h2

4x = 1/8

V . h2

1/12 . b .h3

=3 . V

2 . b. h=

3 . V2 . A

100100

09/02/2014


Problem :20 cm

5 cma

bc

q = 3000 kg/m

A B

P = 1500 kg 1200 cm

5 cm

3 cm

20 cm

15 cm

c

d

e

600 cm

A B

Draw shear stress diagram of the section in support – A and of the section - 1 that is 100 cm of distance from point B.

101101

Working steps:

1. Calculate the Neutral Axis

20 . 5 . 2,5 + 20 . 5 . 15 + 15 . 3 . 26,5 Yc 12 01 cm

, ,

20 . 5 + 20 . 5 + 15 . 3=Yc = 12,01 cm

From TOP

2. Calculate Inertia Moment

I = 1/12 . 20 . 53 + 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

09/02/2014


3. Calculatie shear forces

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

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

ab1

b2

0

100100100

12.019,519,519 51

20205

0 0951951

0

In section ‘A’ with 10.000 kg of shear force

414,6414,6

20,7382,92

c

d1

d2

e 0

100

4545

35.059,51

3.50514.4914.4915.99

5

515150 0

1073,85

652.05652.05

0

468,16

284,27284,27

93,63

56,85418,951

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

b1

b2

c

d1

d2

100100100

4545

35.05

12.019,519,519,51

3.50514.4914.49

20205

5

515

0 0951951

1073,85

652.05652.05

0269,49269,49

304,30

184,774184,774

13,47453,89

60,86

36,95512,3182

e 045 14.49

15.9915150 0

652.050, ,

104104

09/02/2014


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

36,95512,318

20,7393,63

56,854

82,92

18,951

15 cm0 0

Shear 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 its SHEAR

CENTER106106

09/02/2014


Shear Center

PV=P

F1

V P

eh

F1

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

b. t. h . V . Q½ . . b . t . h

P=

=. b . t . h

2 . P I . t

V . ½ . h . b . tx =

b2 . h2 . t

4 . I107107

Problem :

e

P 10 cm

50 cm

Determine the SHEAR CENTER of this section

V=P

F1 F2

e 50 cm

10 cm

10 15 30

section.

Equation that is used:

e . P + F1 . 60 = F2 . 60

e = ( F2 . 60 – F1 . 60 ) / P

½ . . 17,5 . 10F1 = F2 = . 37,5 . 10½ . 108108

09/02/2014


I = 1/12 . 55 . 703 - 1/12 . 40 . 503 = 1.155.416,67 cm4

P . 17,5 . 10 . ½ . 60 =

V . Q

I . t=

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

Calculation :

P . 37,5 . 10 . ½ . 60

I . t 1.155.416,67 . 10

= = =V . Q

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

F1 = 0,00045 . P . 17,5 . 10½ . = 0,0394 . P

F2 = 0,00097 . P . 37,5 . 10½ . = 0,1820 . P

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

8,556 cm

In order to make frame didn’t induce torsion , so thePload must be placed in e = 8,556 cm ( see Picture)

109109

KERN / GALIH / INTIVariety of KERN :

Limited with 4 point


p

Li it d ith 4 i t

Unlimited


110110

09/02/2014


KERN / GALIH / INTIDetermine Inertia moment of sloping axis:

XY

Y

x df Cos Sin x x= + y

XY

X

Cos Sin y y= - x

2y=Ix df

Ix = y2

2

222Cos x+ Sin - 2xy Sin Cos df

= 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 df

= Ix Sin + Iy Cos + 2 SxySin Cos 2

112112

09/02/2014


KERN / GALIH / INTIExample of determining KERN limits :

Determine the Neutral axis :

2.20.1 + 8.2.6.22 cm

y

2.20.1 + 8.2.6.2

2.20 + 8.2.2==x 3,2 cm

A = 2.20 + 8.2.2 = 72 cm

Ix = 1/12.2.203 + 1/12.8.23.2

+ 8.2.92.2 = 3936 cm42

2

16x

=Wax393610

= 393,6 cm3

=Wbx393610

= 393,6 cm3

10

3,2

113113

KERN / GALIH / INTIContoh Menentukan batas – batas KERN :

Iy = 1/12.20.23 + 1/12.2.83.2

+ 20 2 (2 2)2 + 2 2 8 (2 8)2 = 628 48 cm4+ 20.2.(2,2) + 2.2.8.(2,8) = 628,48 cm

=Wkr y628,48

3,2= 196,4 cm3

=Wkn y 6,8= 92,42 cm3628,48

KWbx 393,6

KWkn y 92,42

Ka x = bx

A=

,

72= 5,46 cm

Kb x =Wax

A=

393,6

72= 5,46 cm

Kkr y =A 72

=,

= 1,28 cm

Kkny =Wkr y

A 72=

196,4

= 2,72 cm 114114

09/02/2014


KERN / GALIH / INTIPicture of KERN limits :

2,72 cm1,28 cm

2 cm

16

y

x5,46 cm

5,46 cm

2

2

10

3,2

5,46 cm

115115

Modul 4Modul 4

TorsiTorsionon

116116

09/02/2014


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 Material’s subject was limited in rounded section only.

117117

TORSION (Puntiran )

M M

Torsion Moment at both end of the barM M

MM

Torsion Moment distributed along the

M(x)g

bar

118118

09/02/2014


TORSION (Puntiran )

maxCmax

Cmax . dA . = T

St

A

C

C Stress

Area

Forces Distance

Torsion MomentOr can be written as:

max

C . dA = T

2

= IP . dA2

= Polar Inertia Moment

A

A119119

Example of Polar Inertia Moment for CIRCLE

. dA2

=A

3 d2 =0

C

2 4

.

4

.4

.0

C

= C =

32 d

4

Torsion of the CIRCLE can be determined with

2

Torsion of the CIRCLE can be determined with this equation:

maxT =

C. IP

T . C

TORSION MOMENT

max = T . C

IPTORSION STRESS

120120

09/02/2014


For Circle For Circle –– Hollow SectionHollow Section::

121121

TWIST ANGLE OF CIRCULAR BARTWIST ANGLE OF CIRCULAR BAR

122122

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

09/02/2014


If :If :

Then:Then:

So general statement of the twist angle of a section from So 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 EXERCISE PROBLEM EXERCISE -- 11

See a tiered bar that shown in this following picture, it’s 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

09/02/2014


Polar Inertia MomentPolar Inertia Moment::

BarBar AB = BCAB = BC

BBarar CD = DECD = DE

Considering its left section, torsion moment in every part will be:Considering its left section, torsion moment in every part will be: Considering its left section, torsion moment in every part will be:Considering its left section, torsion moment in every part 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 every To get rotation of edge A, can be done with add up every integration limit:integration limit:

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

126126

09/02/2014


EXERCISE EXERCISE --11

Calculate maximum torsion shear stress of AC Calculate maximum torsion shear stress of AC –– bar (as bar (as seen in AC bar seen in AC bar –– exercise 1)exercise 1). . Assume that bar diameter Assume that bar diameter from Afrom A –– C is 10 mm.C is 10 mm.from A from A C is 10 mm.C is 10 mm.

AnswerAnswer::

127127

ExercisesExercises

Soal 4.1Soal 4.1S b h b iS b h b iSebuah poros berongga mempunyai Sebuah poros berongga mempunyai diameter luar 100 mm dan diameter dalam diameter luar 100 mm dan diameter dalam 80 mm. Bila tegangan geser ijin adalah 55 80 mm. Bila tegangan geser ijin adalah 55 MPa, berapakah besar momen puntir yang MPa, berapakah besar momen puntir yang bisa diteruskan ? Berapakah tegangan bisa diteruskan ? Berapakah tegangan pada mukaan poros sebelah dalam bila pada mukaan poros sebelah dalam bila diberikan momen puntir ijin?diberikan momen puntir ijin?

128128

09/02/2014


129129

Sebuah poros inti berongga berdiameter Sebuah poros inti berongga berdiameter 200 di l h d l b i200 di l h d l b i200 mm diperoleh dengan melubangi 200 mm diperoleh dengan melubangi poros melingkar padat berdiameter 300 poros melingkar padat berdiameter 300 mm hingga membentuk lubang aksial mm hingga membentuk lubang aksial berdiameter 100 mm. berdiameter 100 mm. Berapakah Berapakah persentase kekuatan puntiran yang hilang persentase kekuatan puntiran yang hilang oleh operasi ini ?oleh operasi ini ?

130130

09/02/2014


131131

Poros padat berbentuk silinder dengan ukuran yang Poros padat berbentuk silinder dengan ukuran yang bervariasi yang terlihat dalam gambar digerakkan olehbervariasi yang terlihat dalam gambar digerakkan olehbervariasi yang terlihat dalam gambar digerakkan oleh bervariasi yang terlihat dalam gambar digerakkan oleh momenmomen--momen puntirmomen puntir seperti ditunjukkanseperti ditunjukkan dalam dalam gambargambar tersebut. tersebut. Berapakah tegangan puntir Berapakah tegangan puntir maksimum dalam poros tersebut, dan diantara kedua maksimum dalam poros tersebut, dan diantara kedua katrol yang ada ?katrol yang ada ?

132132

09/02/2014


133133

a.a. Tentukanlah tegangan geser maksimum dalam poros Tentukanlah tegangan geser maksimum dalam poros yang dihadapkan pada momenyang dihadapkan pada momen--momen puntir, yangmomen puntir, yangyang dihadapkan pada momenyang dihadapkan pada momen momen puntir, yang momen puntir, yang diperlihatkan dalam gambar.diperlihatkan dalam gambar.

b.b. b. Hitunglah dalam derajat sudut pelintir antara kedua b. Hitunglah dalam derajat sudut pelintir antara kedua ujungnya. Ambillah G = 84.000 MN/m².ujungnya. Ambillah G = 84.000 MN/m².

134134

09/02/2014


135135

ModulModul 55

STRESS COMBINATIONSTRESS COMBINATION

136136

09/02/2014


Equation Equation that have learned before about linier elastic that have learned before about linier elastic material, can be simplified as:material, can be simplified as:

Normal Normal StressStress::D tD t i l l di l l da. a. Due to Due to axial loadaxial load

b. b. Due to Due to flexureflexureA

P

My

137137

I

My

Shear StressShear Stress::a. a. Due to Due to torsiontorsion

pI

T

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

p

tI

VQ

Superposition of the stress, only considered in Superposition of the stress, only considered in elastic problem when deformation that elastic problem when deformation that happened is small.happened is small.

138138

09/02/2014


EXERCISE:EXERCISE: A bar A bar 50x75 mm 50x75 mm that is that is 1.5 1.5 metermeter of length, selfweight is of length, selfweight is

not considered, was loaded as seen in this following not considered, was loaded as seen in this following picture.picture. (a). (a). Determine maximum tension and Determine maximum tension and compression stress that work pependicularly of beam compression stress that work pependicularly of beam p p p yp p p ysection, 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 two so it can be solved in two

stepssteps. . In Picture In Picture (b)(b), it shows that the bar only take axial , it shows that the bar only take axial load only. Then In Picture load only. Then In Picture ((cc)), it shows that the bar only , it shows that the bar only take transversal load onlytake transversal load only

Axial LoadAxial Load normal stress that the bar have along its lengthnormal stress that the bar have along its length

140140

Axial LoadAxial Load, , normal stress that the bar have along its length normal stress that the bar have along its length is:is:

09/02/2014


Normal stress due to tranversal load depends on flexure Normal stress due to tranversal load depends on flexure moment value and the maximum flexure moment is in moment value and the maximum flexure moment is in force that use:force that use:

Stress superposition woks perpendicularly of beam Stress superposition woks perpendicularly of beam section and linearly decreased to the neutral axis as section and linearly decreased to the neutral axis as seen in picture (g)seen in picture (g)

141141

142142

09/02/2014


STRESS COMBINATION ON COLUMNSTRESS COMBINATION ON COLUMN

Similar equation can be done to assymetric Similar equation can be done to assymetric section:section:

MMP

WhenWhen::Flexure Moment Flexure Moment MyyMyy = +P z= +P z00 that works of ythat works of y--axisaxisFlexure MomentFlexure Moment MzzMzz == --P yP y00 that works of zthat works of z--axisaxis

yy

yy

zz

zzx I

zM

I

yM

A

P

Flexure MomentFlexure Moment MzzMzz P yP y00 that works of zthat works of z axis axis A A is cross section area of frameis cross section area of frameIzzIzz andand IyyIyy isis inertia moment of the section to each their inertia moment of the section to each their principal axisprincipal axisPositive symbol Positive symbol (+) (+) is tension stress, and is tension stress, and NegatiNegative ve 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. Beam’s weight is not considered.

144144

09/02/2014


AnswerAnswer::

Forces that work in Forces that work in ABCDABCD sectionsection,, on the picture on the picture (c), (c), isis

P = P = --64 64 kNkN, ,

MM == 640 (0 15) =640 (0 15) = 9 69 6 kN mkN m andandMMyyyy = = --640 (0.15) = 640 (0.15) = --9,6 9,6 kN.mkN.m,, andand

MMzzzz = = --64 (0.075 + 0.075) = 64 (0.075 + 0.075) = --9,6 9,6 kN.mkN.m..Cross section area of the beamCross section area of the beam A = (0.15)(0.3) = 0,045 m²,A = (0.15)(0.3) = 0,045 m²,

And its Inertia moment isAnd its Inertia moment is::

145145

JadiJadi dengandengan menggunakanmenggunakan hubunganhubungan yang yang setarasetara dapatdapatdiperolehdiperoleh tegangantegangan normal normal majemukmajemuk untukuntuk elemenelemen--elemenelemen sudutsudut ::

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

146146

09/02/2014


147147

THE END

148148

mekanika bahan -...

Documents