mekanika bahan -...
TRANSCRIPT
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MEKANIKA BAHAN(Mechanics of Materials)
Prerequisite :
3 CREDITS
Prerequisite :
Statically Determinate Mechanics
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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
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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???
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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
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E.P. Popov, 1978, Mechanics of E.P. Popov, 1978, Mechanics of Materials, 2Materials, 2ndnd edition edition
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GereGere & Timoshenko& Timoshenko, 2008, Mechanics , 2008, Mechanics of Materials, 7of Materials, 7thth edition edition
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R.C. Hibbeler, R.C. Hibbeler, 20102010, Mechanics of , Mechanics of MaterialsMaterials, 8, 8thth editionedition
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Other books: Mechanics of MaterialOther books: Mechanics of Material
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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..
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EvaluationsEvaluationsUTS (30%) UAS (30%)
Quiz1 (10%) Quiz2 (10%)
PR1 (10%) PR2 (10%)
*Prosentase bisa diubah sesuai yang menguntungkan mahasiswa
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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!
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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
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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
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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
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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
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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
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rol rolrol rol
sendi rol
sendi sendi1919
ReactionsReactions
Simply supported beamsSimply supported beams
Cantilever beamsCantilever beams
TrussesTrusses
2020
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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
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Modul 1Modul 1
Tegangan dan ReganganTegangan dan Regangan
Stress & StrainStress & Strain
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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.
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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.
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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.
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APPLICATIONS Planning of a Structure
STRUCTURAL ANALYSESSTRUCTURAL ANALYSES
MATERIALS
CONTROL PLANNING OF THE DIMENSIONS
STRENGTH / STRESS
STRUCTURES: STABLE2626
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EXAMPLE
TUBE TRUSSES
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EXAMPLEBUILDING FRAME
70/70
50/50
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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
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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
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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
TensionNORMAL STRESS
Compressionp
= P/A = P/AP P
PP 3434
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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
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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 SECTIONAL
AREA
CALCULATION RESULT
DON’T UNDERSTAND STATICALLY
DETERMINATED ENGINEERING MECHANIC
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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
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Determine Cross Sectional Areaexample :
The smallest cross sectional area that was
choosen to get the maximum stress value
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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
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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 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.
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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.
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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)
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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
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Strain
P (Load) = StrainL
=
Change as every Loading changes
(Deformation)P – Diagram
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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
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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 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
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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
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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
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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
Bar Deformation due to Axial Load
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
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B L
Bar Deformation due to Axial Load
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..
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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 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
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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
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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
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z
The Relationship of Poisson’s Ratio, Stress and Strain
z
y
y
z 6161
The Relationship of Poisson’s Ratio, Stress and Strain
x EE E
x y z+ - -=
y EE E
x y z- + -=
z EE E
x y z- - +=
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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
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Modul 2Modul 2
beam flexure beam flexure (pure bending)(pure bending)
6565
Pure Bending in Beam
Flexure due to MOMEN only
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Pure Bending in Beam
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
Pure Bending in Beam
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
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Pure Bending in Beam
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
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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
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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
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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
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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
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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
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Dr. Endah Wahyuni 41
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.
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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
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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 = 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
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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
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Rectangular Beam that have Full Plastic Condition
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
Rectangular Beam that have Full Plastic Condition
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
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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
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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
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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
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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
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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
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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
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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 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, ,
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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
36,95512,318
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
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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
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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
Limited with 6 point
p
Li it d ith 4 i t
Unlimited
Limited with 4 point
110110
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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
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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
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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
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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 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
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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
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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:
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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.
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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::
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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?
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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 ?
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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 ?
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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.
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135135
ModulModul 55
STRESS COMBINATIONSTRESS COMBINATION
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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.
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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..
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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
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Axial LoadAxial Load, , normal stress that the bar have along its length normal stress that the bar have along its length is:is:
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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)
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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.
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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::
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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 ::
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THE END
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