MEKANIKA BAHAN

09/02/2014
MEKANIKA BAHAN
(Mechanics of Materials)
3 CREDITS
Prerequisite :
Statically Determinate
Mechanics
1
Lecturers:
Until ETS
Endah Wahyuni,
Wahyuni, ST (ITS), MSc (UMIST), PhD (UoM
(UoM))
[email protected] @end222
ETS - EAS
Prof. Ir. Priyo Suprobo, MS, PhD
2
Dr. Endah Wahyuni
1
09/02/2014
BILINGUAL CLASS
Module in English, Class in Indonesian; or
vice
i versa.
 Delivery of contents in 2 languages
(Indonesian & English).
 Technical terms in English
 Students???

3
Materials

1.
2
2.
3.
4.
5
5.


Books:
E.P. Popov, 1978, Mechanics of Materials
Gere & Timoshenko
Timoshenko,, 1997,
1997 Mechanics of
Materials
R.C. Hibbeler, 1997, Mechanics of Materials
Any related books, with topic: Mechanics of
Material
Online
http://personal.its.ac.id/dataPersonal.php?userid=
ewahyuni
http://www.structuralconcepts.org
4
Dr. Endah Wahyuni
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09/02/2014
E.P. Popov, 1978, Mechanics of
Materials, 2nd edition
5
Gere & Timoshenko,
Timoshenko, 2008, Mechanics
of Materials, 7th edition
6
Dr. Endah Wahyuni
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R.C. Hibbeler, 2010
2010,, Mechanics of
Materials,, 8th edition
Materials
7
Other books: Mechanics of Material
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Dr. Endah Wahyuni
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Learning Methods




Class
Students are required to read the course
material to be provided in the existing class
schedule
Responsiveness
Exercises in class with guidance
Quiz
In-class
l
exam att any given
i
ti
time
Home work
Students do the work to be done at home with
the responsibility, not only collects the duty.
duty.
9
Evaluations
UTS (30%)
UAS (30%)
Quiz1 (10%)
Quiz2 (10%)
PR1 (10%)
PR2 (10%)
*Prosentase bisa diubah sesuai yang menguntungkan mahasiswa
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Dr. Endah Wahyuni
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Notes:



20 minutes late,
late, not permitted to enter the class.
class.
Disturbing class  go out
Home work is collected before the class starting
Keep the spirit on!
11
Contents
1
1
Dapat menjelaskan tentang tegangan,
regangan, modulus elastisitas serta modulus
g
geser
Ketepatan penjelaskan
tentang tegangan, rergangan, modulus
geser
elastisitas serta modulus g
a. pendahuluan
b. pengertian tegangan, regangan
c. p
pengertian
g
modulus elastisitas
d. static test
Kuliah
lihat UTS
2
2&3
Dapat menghitung tegangan yang terjadi
pada sebuah balok akibat beban lentur murni
baik pada balok dengan bahan tunggal
maupun pada balok dengan dua bahan,
baik semasih pada kondisi elastis maupun
sesudah mencapai kondisi non elastis
Ketepatan perhitungan tegangan pada
balok yang menerima beban lentur murni
a. lentur muni pada balok elastis
b. lentur muni pada balok dengan
dua bahan
c. lentur murni pada balok
non elastis
Kuliah
Responsi
PR 1
lihat UTS
Dapat menghitung tegangan geser pada balok Ketepatan perhitungan tegangan geser
yang disebabkan oleh beban lentur,
pada balok akibat beban lentur
pada balok-balok dengan berbagai bentuk
penampang.
a. hubungan momen dan gaya
Kuliah
lintang
Responsi
b. tegangan geser akibat beban
PR 2
lentur
c. shear center
d. geser pada profil berdinding tipis
lihat UTS
Dapat menghitung tegangan dan regangan
d poros akibat
kib t b
beban
b ttorsii
pada
a. pengertian torsi
b ttegangan geser ttorsii
b.
c. regangan oleh torsi
d. tegangan oleh torsi pada poros
non elastis
Kuliah
R
Responsi
i
PR 3
lihat UTS
a. kombinasi tegangan pada balok
tidak simetris
b. kombinasi tegangan pada
penampang kolom
c. kern
Kuliah
Responsi
PR 4
lihat UTS
UTS
Test
4
5
6
4&5
6
7&8
9
Dapat mengkombinasikan tegangan-tegangan
sejenis pada penampang balok atau kolom
dan dapat menggambar bentuk kern dari
berbagai bentuk penampang
Indikator Kompetensi
Ketepatan perhitungan tegangan dan
regangan pada
d poros akibat
kib t b
beban
b ttorsii
Ketepatan perhitungan kombinasi tegangan
dan ketepatan penggambaran bentuk kern
Materi Pembelajaran
Bobot Nilai
%
Minggu ke
3
Kompetensi
Metode Pembelajaran
dan Evaluasi
No
2
2
2
2
40
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Dr. Endah Wahyuni
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Contents
1. Introduction
2. Slicing
g Method
3. Understanding of Stress
4. Normal Stress
5. Average Shear Stress
6. Determine of
 and 
7. STATIC TEST
8. Allowed Stress
9. Strain
13
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 Strain
11.
- Shear Stress
- Shear Strain
14
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12. Pure Bending on beams
13. Moment of Inertia
14. Calculating Stress on beams
15 Beams with two materials
15.
16. Pure bending on non-elastic beams
17. Shear-bending Stress
18. Torsion
19. Multiple
p Stresses
20. Combination of stresses on Columns
21. KERN
22. …………..etc
ETS
15
After midsemester evaluation:
1.
Plane stress analysis
Maximum and minimum stress
Mohr Circle
2.
Bar design based on stress
Based on axial stress
stress,, flexure and shear for prismatic
bar and definite static
3.
Definite Static Beam’s deformation

Equation of elastic line deformation method.

Unit Load method

Area moment method
4.
Stability of Compression Bar
Centric Load and Shear Force.
16
Dr. Endah Wahyuni
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Reviews::
Reviews
Statically Determinate Mechanics
Determinate Structure : If?
Static Equation ??
1
2
3

17
rol
Dr. Endah Wahyuni
rol
sendi
rol
send
i
sendi
18
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09/02/2014
rol
sendi
sendi
rol
rol
sendi
19
Reactions

Simply supported beams

Cantilever beams

Trusses
20
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Loadings
-
Point Load
At midspan,
midspan,
 Within certain location

-
Distribution Loads
Full distributed loads
 Partially distributed loads

-
Moment Loads
At the end of cantilever
 Midspan
 Within certain location

21
Modul 1
Tegangan dan Regangan
Stress & Strain
22
Dr. Endah Wahyuni
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Introduction
At a structure, each elements of a structure
should be having a dimension. The elements
have to be calculated to resist the loading on
them or maybe applied to them. To calculate the
dimension of the elements, we should know the
methods to analyses, which are:
 strength ( kekuatan
kekuatan),
),
 stiffness ( kekakuan)
kekakuan),
 stability ( kestabilan ),
)
The methods will be discussed in this Mechanic of
Materials.

23

Mechanics of materials is a subject of a very old
age, which generally begins with Galileo in the early 17th
century. The first one describes the behavior of the
structure of load rationally.
24
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09/02/2014

The behavior of the structure to obtain the force depends
not only on the fundamental laws of Newtonian
mechanics that govern force equilibrium but also to the
physical characteristics of the structural parts, which can
be obtained from the laboratory, where they are given
the
h fforce off action
i is
i kknown accurately.
l

Mechanics of Material is a mixed knowledge from the
experiment and the Newtonian principals on elastic
mechanics.

O off the
One
th main
i problems
bl
iin mechanics
h i off materials
t i l iis tto
investigate the resistance of an object, that is the
essence of the internal forces for balancing the external
forces.
25
APPLICATIONS
Planning of a Structure
STRUCTURAL ANALYSES
MATERIALS
PLANNING OF THE
DIMENSIONS
STRUCTURES: STABLE
Dr. Endah Wahyuni
CONTROL
STRENGTH /
STRESS
26
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EXAMPLE
TUBE
TRUSSES
27
EXAMPLE
BUILDING FRAME
70/70
50/50
28
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EXAMPLE
P2
P1
H2
H1
B1
B2
Because of P2 > P1, thus from stress
analysis, dimension will be obtained
where B2 > B1, H2 > H1
29
Metode Irisan
GAYA DALAM
P1
P1
P2
P2
S2
S1
S1
S3
S3
S2
P4
P3
GAYA DALAM
P4
P3
30
Dr. Endah Wahyuni
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09/02/2014
Tegangan (Stress)
TEGANGAN NORMAL
Tegak Lurus
Bidang Potongan
TEGANGAN GESER
Sejajar Bidang
Potongan
DEFINISI :
TEGANGAN ADALAH GAYA DALAM YANG
BEKERJA PADA SUATU LUASAN KECIL
TAK BERHINGGA DARI SUATU
POTONGAN
31
Stress (Tegangan)
MATHEMATICS EQUATIONS
=
A
Lim
=
A
Lim
0
F
A
NORMAL STRESS
0
V
A
SHEAR STRESS
 = Normal Stress
 = Shear Stress
A = Cross-section area
F = Forces on perpendicular of cross-section
V = Forces on parralel of cross-section
Dr. Endah Wahyuni
32
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09/02/2014
Stress (Tegangan)
Stress symbols on elements related with
coordinates :
z
z
zx
xz
x
zy
yz
y
xy yx
y
x
33
Normal Stresses
NORMAL STRESS
NORMAL STRESS
Tension
Compression
p
P
P
Dr. Endah Wahyuni
 = P/A
P
P
= P/A
34
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09/02/2014
Average Shear Stresses
FORCES ACTING
PARRALEL SECTION
P
CREATING
SHEAR STRESS
= P Cos/ A
Normal
AShear
ANormal
= P / A
AShear
Shear
35
Average Shear Stress
P
P
½P

AShear
= P / Total AShear
Total Ashear =
2 x Sectional Area of Bolts
36
Dr. Endah Wahyuni
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09/02/2014
Determine  and 
Calculation of
NEED TO UNDERSTAND
STRESS
THE PURPOSE AND THE GOAL
CHOOSE THE EQUATION
CALCULATION
DETERMINATION OF FORCE
AND CROSS SECTIONAL
AREA
 or 
WILL BE PROBLEM IF
DON’T UNDERSTAND
STATICALLY
DETERMINATED
ENGINEERING MECHANIC
CALCULATION RESULT
37
DETERMINE FORCE VALUE
USE STATIC EQUATION:
 FX = 0
 MX = 0
 FY = 0
 MY = 0
 FZ = 0
 MZ = 0
Define Cross Sectional Area
To get
Choose the smallest Area
The Maximum Stress
38
Dr. Endah Wahyuni
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09/02/2014
Determine Cross Sectional Area
example :
The smallest cross
sectional area that was
choosen to get the
maximum stress value
39
Example 1
1::
A concrete wall as shown in the figure, received distributed loads of 20
kN/m2. Calculate the stress on 1 m above the based. The gravitation
load of the concrete is 25 kN/m3
40
Dr. Endah Wahyuni
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09/02/2014
Answer:
Self weight of concrete wall:
wall:
W = [(0,5 + 1,5)/2] (0,5) (2) (25) = 25 kN
, ) (0,5)
( , ) = 5 kN
Total load:
load: P = 20 ((0,5)
From Fy = 0, the reaction R = W + P = 30 kN
using upper part of the wall as a free thing, thus the weight
of the wall upper the cross
cross--section is W1 = (0,5 + 1) (0,5)
(25/2) = 9,4 kN
From Fy = 0, the Load on section : Fa = P + W1 = 14,4 kN
Normal stress on a-a is a = Pa/A = 14,4/(0,5x1) = 28,8
KN/m2
The stress is a compression normal stress that worked as
Fa on the section.
41
Stress
TASK :
D
1.
B

A
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 BD
cables.
C
W
P
2.
b
P
d1
d2
Dr. Endah Wahyuni
If bolt diameter = 30
mm, b = 200 mm, d1 =
8 mm, d2 = 12 mm, P =
2000 kg,
g, so calculate
the maximum stress
of each frame and
shear stress of the
bolt.
42
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Static Test
P LOAD INCREASE
CONTINUOUSLY
P
FRACTURE TEST ING
MATERIAL
TESTING MATERIAL
P
P
PUlt
A
ULTIMATE LOAD
ULTIMATE STRESS
43
Universal Test Machine (UTM)
44
Dr. Endah Wahyuni
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09/02/2014
FLEXURE TEST
45
STRAIN
TESTING MATERIAL
P
STATIC TEST
LOAD
STRAIN
L
-. Pload increase continuously
P
- Every Pload increasing, list deformation
of testing material that shows in dial
gauge.
46
Dr. Endah Wahyuni
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09/02/2014
Strain

=
L
P (Load)
=
Strain
Change as every
Loading changes
P –  Diagram
(Deformation)
47
Stress – Strain Diagram
Physical properties of every material can be shown
from their stress – strain diagram relationship.
P (load)
pict. A
P –  Diagram
Dr. Endah Wahyuni
 (Stress)
pict. B
= Strain
 –  Diagram
48
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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
49
Stress – Strain Diagram
 (Stress)
 (Stress)
Proportional
Limit
Strain
STEEL MATERIAL
Strain
CONCRETE MATERIAL
50
Dr. Endah Wahyuni
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09/02/2014
HOOKE LAW
 = EX
E=


ELASTIC
CONDITION
DETERMINATION OF YIELD POINT
OFF-SET METHOD
 (Stress)
Proportional
P
ti
l
Limit
 = STRESS
 = STRAIN
E = ELASTICITY MODULUS
Strain
51
HOOKE’s LAW
problem
:
P
L
P
Dr. Endah Wahyuni
In some frame with L =100 cm in length,
Static Test was done. If Pload that
that’s
s given
to this frame is 4000 kg, this frame is still
in elastic condition, and goes on 2 mm in
length, so calculate of stress and strain
value of that frame. If modulus elasticity
value is 2 x 106 kg/cm2 and then calculate
the cross sectional area of that frame.
52
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09/02/2014
Bar Deformation due to
Axial Load
P3
P2
P4
P1
Px
Px
dx
dx+
d=
Px force to dx elemen and
cause d deformation
dx

 dx
E
d
dx = P x
Ax E
53
Bar Deformation due to
Axial Load
example :
B
B
=
P = Px
Px
A
dx
L
A
Px
Px . dx / Ax . E
 = Px / Ax . E dx
0
 = P . X / Ax . E
Ax = A ,
P
P
Deformation due to P load,
selfweight was ignored
Dr. Endah Wahyuni
L
L
0
Px = P
=P.L/E.A
54
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09/02/2014
Bar Deformation due to
Axial Load
DEFORMATION DUE TO SELFWEIGHT IS :
=
B
L
Px . dx / Ax . E = 1 / A . E
w . X . dx
A
= ½ . W.x2 / A . E
0
L
= w . L2 / 2 . A . E = WT . L / 2 . A . E
0
DEFORMATION DUE TO P LOAD AND SELFWEIGHT IS :
 = P.L / A.E + WT.L / 2.A.E =
 = L (P + ½.WT) / A.E
55
Contoh 22-1:
Tentukan pergeseran relatif dari titiktitik-titik A dan D pada
batang baja yang luas penampangnya bervariasi
seperti terlihat pada gambar di
bawah bila diberikan empat gaya terpusat P1, P2, P3
d
dan
P4. Ambillah
A bill h E = 200 x 106 kN/m
kN/ 2.
56
Dr. Endah Wahyuni
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09/02/2014

Gaya dalam batang adalah :
Antara titik A dan B, Px = +100 kN
Antara titik B dan C, Px = -150 kN
Antara titik C dan D, Px = +50 kN
Dengan menggunakan persamaan:


Dengan memasukkan hargaharga-harga numeric dari contoh,
maka diperoleh:
p
57
BAR DEFORMATION DUE TO AXIAL LOAD
Problem :
1. A
100 cm

100 cm
B
1000 kg
2.
P1
Dr. Endah Wahyuni
P2
If the bar diameter of AB
and BC is 20 mm,  = 30o
and Elasticity
Elasticit Modulus
Mod l s is
2x106 kg/cm2, calculate
deformation of point B.
E D
b2
b1
C
b3
½ P2
h1
h2
Calculate P1/P2, then after P1 and P2
working, the length of both bar still
be similar, if b1 = 50 mm, b2 = 50 mm,
b3 = 25 mm, h1 = 500 mm, h2 = 500
mm and thickness of both bar = 20
58
mm.
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09/02/2014
Poisson’s Ratio
STRAIN
AXIAL STRAIN
LATERAL STRAIN
The shape is being
LONGER and
SMALLER
POISSON’S RATIO (
)=
 Lateral
 Axial
Concrete = 0.1 – 0.2
Rubber = 0.5 – 0.6
59
The Relationship of Poisson’s
Ratio, Stress and Strain
z
zx
xz
x
zy
yz
y
y
xy yx
x
60
Dr. Endah Wahyuni
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09/02/2014
The Relationship of Poisson’s
Ratio, Stress and Strain
z
y
y
z
61
The Relationship of Poisson’s
Ratio, Stress and Strain
x =
+
y =
-
z =
-
x
E
x
E
x
E
y
-
E
+
-
y
E
z
-
E
y
E
z
-
E
+
z
E
62
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09/02/2014
Shear Stress and Shear Strain
SHEAR STRESS
zy
y
z
B
zy
yz
O
A
A
yz
zy
C
B
/2
C
/2
O
 = SHEAR STRAIN
zy(dy.dx).dz - yz (dx.dz.).dy = 0
zy = yz
yz left =  yz right
 MO = 0
 Fz = 0
63
Shear Stress and Shear Strain
SHEAR STRAIN:
SHAPE TRANSFORMATION THAT IS EXPRESSED
WITH ANGLE TRANSFORMATION ‘  ‘ ARE
CALLED “SHEAR STRAIN”
HOOKE LAW for Shear stress and shear strain:
 = . G
E
G=
2 (1+
 = Shear Stress

)
= Shear Strain
G = Shear Modulus
= Poisson’s Ratio
The relationship between Normal Modulus Elasticity and
64
Shear Modulus
Dr. Endah Wahyuni
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09/02/2014
Modul 2
beam flexure
(pure bending)
65
Pure Bending in Beam

Flexure due to
MOMEN only
66
Dr. Endah Wahyuni
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09/02/2014
Pure Bending in Beam
Ya
Yb = C
max
 max
/2
/2
Initial Length
Force Equilibrium:
FX = 0
( Y/C .  max ) dA = 0
A
C
Y . dA = 0
A
67
Pure Bending in Beam
MOMENT :
M=
A
( Y/C .
A
max ) dA . Y = max
Y 2 . dA
A
Y2 . dA = I = Inertia Moment
M=(
max / C ) . I
TOP FIBER STRESS
max = M . Ya / I
max = M . C / I
BOTTOM FIBER STRESS
max = M . Yb / I
68
Dr. Endah Wahyuni
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09/02/2014
Pure Bending in Beam
GENERALLY:
max
= M.Y/I
I/Y = W
(Resistance Moment)
I / Ya = Wa
I / Yb = Wb
I =
Y 2 . dA
A
INERTIA MOMENT
69
INERTIA MOMENT
EXAMPLE :
y
3
b
y
h/2
Y 2 . b . dy
Ix =
y 2 . dA =
A
-h/2
h/2
h/2
= 1/3 . y3. b
= 1/3 . (1/8 + 1/8) . h3. b
x
-h/2
h/2 = 1/ . 1/ . h3. b = 1/ . b. h3
1/
-11/2
2
Ix =
2
x
11
2
3
Dr. Endah Wahyuni
4
12
11/2
y
3.y
y 2 . dy
y + 2 y 2 . dy
-2
-11/2
2
+ 3.y 2 . dy
11/2
70
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09/02/2014
INERTIA MOMENT
EXAMPLE :
-11/2
11/2
2
3
3
1
3
3
3
= /3 . y
+ 2 . /3 . y
+ /3 . y
-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
CARA LAIN :
= 1/12 . 3 . 43 – 1/12 . 1 . 33 = 16 – 2,25 = 13,75
SHORTER CALCULATION
71
STRESS CALCULATION
OF THE BEAM
10 cm
10.000 kg
10 cm
30 cm
400 cm
10 cm
30 cm
CROSS SECTIONAL AREA :
A = ( 2 . 30 . 10 ) + (10 . 30 ) = 900 cm2
INERTIA MOMENT:
I = 1/12 . 30 . 503 – 2 . 1/12 . 10 . 303 = 267.500 cm4
72
Dr. Endah Wahyuni
36
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
73
Stress Calculation
of Beam
Max
1
-
y1 = 20 cm
yMax
+
Max
1 = M / W1 = 1.000.000 . 20 / 267.500 = 74.77 kg/cm2
W1 = I / y1
Dr. Endah Wahyuni
74
37
09/02/2014
EXERCISE – MOMENT INERTIA
Sb Y
30 cm
1
10 cm
40 cm
Sb X
Calculate Inertia
Moment of its
strong axis( Ix )
and weak axis ( Iy )
10 cm
2
Sb Y
10 cm
8 cm
20 cm
8 cm
10 cm
Sb X
Calculate Inertia
Moment of its
strong axis( Ix ) and
weak axis ( Iy )
10 10 10
75
EXERCISE – PURE BENDING
1
A
400 cm
100 kg/m (include its selfweight)
80 cm
200 cm
2
B
200 cm
C
1500 kg
30 cm
- Draw its momen diagram
10 cm
30 cm
8 cm
10 cm
8 cm
10 cm
Dr. Endah Wahyuni
- Calculate Inertia Moment of Beam
Section
- Calculate edge fiber stresses of
section - 1 and 2, then draw its
stress diagram
- Calculate its maximum stress
76
38
09/02/2014
ASSYMETRIC FLEXURE
q
qSin 
L

qCos 
Moment occurs of X-axis (MX)
and Y-axis (MY)
q
2
MX = 1/8 . qCos
C  . L
2
MY = 1/8 . qSin
Si  . L
Moment that its flexure
round ‘X’-axis
Moment that its flexure
round ‘Y’-axis
77
Stress of the Section due to
q
Assymetric Flexure
c
L
d
a
o
qSin 
b
a

qCos 
q
MX = 1/8 . qCos  . L2
MY = 1/8 . qSin  . L2
Dr. Endah Wahyuni
MX . h/2
Ix
MX . h/2
b = +
Ix
MX . h/2
c = Ix
MX . h/2
d = Ix
= +
Ix = 1/12 . b . h3
My . b/2
Iy
My . b/2
Iy
My . b/2
Iy
My . b/2
+
Iy
+
Iy = 1/12 . h . b783
39
09/02/2014
Exercise - Stress of the Section due to
Assymetric Flexure L = 300 cm, q = 100 kg/m, P
q
= 200 kg, h = 20 cm, b = 10
cm,  = 30o
P
A
B
L
Calculate stress that
occurs in the midspan a, b,
c, d, e and f. Where point e is 5 cm of distance from
x-axis and 3 cm from yaxis.
c
d
f
o
e
b
a

Point - f is 6 cm of distance
from x-axis and 4 cm from
79
y-axis
assume 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
occurs in ABC cable and
maximum stress of BD
frame.
Problem - I
1.
D
50 cm
A
P is
i in
i 150 cm off di
distance
t
from B
B
B

C
W
W
Calculate the deflection
of point - b and shear
stress of As.B
As B bolt.
bolt Bolt
diameter of As.B = 20
mm.
Modulus Elasticity of BD
frame = 2x106 kg/cm2. 80
Dr. Endah Wahyuni
40
09/02/2014
2.
80 cm
1
A
2000 kg/m (include its selfweight)
80 cm
200 cm
2
B
400 cm
C
200 cm
1000 kg
1000 kg
30 cm
- Dram its moment diagram
10 cm
- Calculate Inertia Moment of Beam
25 cm - Calculate edge fiber stresses of
20 cm
section – 1 and 2, then draw its
stress diagram.
8 cm
10 cm
8 cm
- Calculate Maximum stress that
occurs in ABC beam.
81
3.
q
f
c d
e
a
P
A
b
B
L

L = 300 cm, q = 1000 kg/m, P = 2000 kg,  = 30o, P is
100 cm from B.
Calculate stress that occurs in the midspan of point
a, b, c, d, e and f.
82
Dr. Endah Wahyuni
41
09/02/2014
Composite Beam (2 Material)
dx
x
dy
1
a
y
2
h
e
1
b1
b2
DISTRIBUTION OF
ELASTIC STRESS
xE1
a
e
eE2
eE1
DISTRIBUTION OF
SINGLE MATERIAL STRESS
83
Composite Beam (2 Material)
b2 n2
b2.n
b2
b2/n1
b1.n1
b1
b1/n2
Cross Section of
Frame with 1st Material
Cross Sestion of
Frame with 2nd Material
E1 > E2, n1 = E1 / E2, n2 = E2 / E1
Dr. Endah Wahyuni
84
42
09/02/2014
Exercise -Composite Beam (2 Material)
Concrete
Steel
1
a
12 cm
b
1000 kg
A
1
1200 cm
36 cm
2
c
12 10 12
E concrete = 200.000 kg / cm2 ;
1 400 cm
B
1st Material = Concrete
2nd Material = Steel
E stel = 2.000.000 kg /cm2
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)
85
Pure Bending of
Non-Elastic Beam



ELASTIC

NON - ELASTIC
STRESS-STRAIN DIAGRAM
86
Dr. Endah Wahyuni
43
09/02/2014
Pure Bending of
Non-Elastic Beam
Strain
Elastic Strain
distribution distribution

c
a
o
If effect of D aob and
cod are small
Non Elastic Strain
distribution

b
d


87
Rectangular Beam that have Full Plastic Condition
C
h
T
h/
4
h/
4
Plastic moment that can be held = C . ½ . h = T . ½ . h
C = T =  yp ( bh/2)
Plastic momen of a rectangular beam is:
Mp =
yp . bh/2 . h/2 = yp . bh /24
88
Dr. Endah Wahyuni
44
09/02/2014
Rectangular Beam that have Full Plastic Condition
Generally can be written as:
h/
Mp =
. y dA =
2
(
2

yp ) . y . b . dy
0
h/
2

yp . y . b
2
= yp . bh /4
2
0
If calculate
l l t with
ith elastic
l ti equation
ti :
h
Myp = 
yp . I / ( /2) =
2
= 
yp . b . h / 6
1
3

yp . /12 b h
/ ( h/2 )
89
Rectangular Beam that have Full Plastic Condition
yp . b . h2 / 4
Mp / Myp =
= 1,5
15
yp . b . h2 / 6
SHAPE FACTOR
Section that have Elastic – Plastic condition
yo
Minor Yield
(Elastic-Plastic)
Dr. Endah Wahyuni
h/2
Major Yield
(Elastic-Plastic)
All Yield
(Plastic)
90
45
09/02/2014
Section that have Elastic – Plastic condition
Elastic-Plastic moment that can be held with stress
distibution which have partial yield is:
yo
M = 
. y dA = 2 (
yp ) . y/yo . b . y. dy + 2 (
0
yo
3/
= 2/3
yp . y /yo . b
2

yp) . b . y. dy
yo
2
+ yp . b . y2
o
2
= 2/3 
yp . yo . b +
h/
h/
yo
2
2

yp . bh / 4 - 
yp . b . yo
2
2
1
1
= 
yp . bh / 4 – /3 
yp . b . yo = Mp – /3
2

yp . b . yo
91
Modul 3
Shear Stress of Beam
92
Dr. Endah Wahyuni
46
09/02/2014
Shear Stress - Flexure
q (x)
V+dV
V
dx
x
M
M+dM
S MA = 0
dx
(M + dM) – M – (V + dV) . dx + q . dx . dx/2
=0
M + dM – M – V . dx + dV . dx + ½ . q . dx2 = 0
small
small
dM – V . dx = 0
dM / dX = V
OR
dM = V . dx
93
Shear Stress - Flexure
This equation is giving
explanation that :
IF THERE IS FLEXURE MOMENT
DIFFERENCE AT SIDE BY SIDE
SECTION, THERE WILL BE A SHEAR.
dM / dx = V
Example :
L/3
L/3
L/3
NO SHEAR
Bid M
Bid.
Bid. D
M
M
M+dM
M
SHEAR
Dr. Endah Wahyuni
94
47
09/02/2014
Shear Stress - Flexure
Shear Stress due to Flexure Load
a
e
b
j
d
f
FA
FB =
- MB . Y
I
Afghj
=
dA =
- MB
I
- MB . Q
R
FB
Y . dA
Afghj
Q=
I
h
g
Y . dA = Afghj . Y
Afghj
95
Shear Stress - Flexure
Shear Stress due to Flexure Load
FA =
- MA
I
Y . dA =
Aabde
FB – FA = R
=
=
- MA . Q
I
Held up by shear connector
- MB . Q
-
- MA . Q
I
I
( MA + dM ) . Q – MA . Q
Troughout dx
= dF
dM . Q
=
I
dF/dx = q = SHEAR FLOW
q = dM . Q / dx . I = V . Q / I
Dr. Endah Wahyuni
I
96
48
09/02/2014
Shear Stress due to Flexure Load
Example :
200 mm
50 . 200 . 25 + 50 . 200 . 150
50 . 200 + 50 . 200
= 87,5 cm
V = 30.000
30 000 kg,
kg nail strength = 7000 kg
Yc =
50 mm Yc
I = 200 . 503 / 12 + 50 . 200 . 62,52
= 50 . 2003 / 12 + 50 . 200 . 62,52
200 mm
= 113.500.000 mm4 = 11.350 cm4
Q = 50 . 200 ( 87,5 – 25 ) = 625.000 mm3
= 625 cm3
or,
Y1
50 mm
Q = 50 . 200 . 62,5 = 625.000 mm3 = 625 cm3
Y1 = 200 – Yc - 200 / 2 = 62,5 mm
q = V . Q / I = 30.000 x 625 / 11.350 = 1.651 kg / cm
Nail spacing = 7000 / 1651 = 4,24 cm
Problem :
200 mm
50 mm
50 mm
200 mm
30 mm
97
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
enough to carried on q load.
Spacing of top and bottom nails
was made in 3 different type of
spacing.
150 mm
100 100
200
A
100 100
q = 3000 kg/m
B
600 cm
98
Dr. Endah Wahyuni
49
09/02/2014
Shear Stress Diagram
Longitudinal Direction:
 = dF / t.dx = ( dM / dx ) . ( A . Y / I . t ) = V . A . Y / I . t
=
V.Q
I.t
q
=
t
1/8 . V. h2
I
Example :
t=b
j
h
f
g
=
dy
y
y1
h
q
V.Q
=
I.t
t
V
=
Y . dA
I.t A
99
Shear Stress Diagram

=
=
V
I.b
V
2.I
h/
2
V
b . y . dy =
I
y1
Y2
x
2
h/
2
y1
( b/2 ) 2 – y12
If y1 = 0, so
=
=
h2
V
= 1/8
x
2.I
4
3.V
2 . b. h
=
V . h2
1/
12
. b .h3
3.V
2.A
100
Dr. Endah Wahyuni
50
09/02/2014
Problem :
20 cm
P = 1500 kg
1
200 cm
q = 3000 kg/m
a
5 cm
5 cm
b
c
20 cm
A
B
600 cm
d
3 cm
e
15 cm
Draw shear stress diagram of the section in support – A
and of the section - 1 that is 100 cm of distance from
point B.
101
Working steps:
1. Calculate the Neutral Axis
Yc =
20 . 5 . 2,5
, + 20 . 5 . 15 + 15 . 3 . 26,5
,
20 . 5 + 20 . 5 + 15 . 3
12,01
01 cm
= 12
From TOP
2. Calculate Inertia Moment
1
3
2
1
3
I = /12 . 20 . 5 + 20 . 5 . 9,51 + /12 . 5 . 20
+ 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
102
Dr. Endah Wahyuni
51
09/02/2014
3. Calculatie shear forces
Ra = 3000 . 6/2 + 2/3 . 1500 = 10.000 kg
Rb = 3000 . 6 + 1500 - 10.000 kg = 9.500 kg
Va = 10.000 kg ; V1 = - 9.500 + 3000 . 1= - 6.500 kg
In section ‘A’ with 10.000 kg of shear force
Position
a
b1
b2
c
d1
d2
e
A
0
100
100
100
35.05
45
45
0
y
Q
12.01
0
951
9,51
951
9,51
9 51
9,51
1073,85
3.505
14.49 652.05
14.49 652.05
0
15.99
q = V.Q / I
t
 =q/t
0
414,6
414,6
20
20
5
0
20,73
82,92
468,16
5
93,63
284,27
284,27
0
5
15
15
56,854
18,951
0
103
In Section ‘1’ with 6.500 kg of shear force
Posisi
a
b1
b2
c
d1
d2
e
A
0
100
100
100
35.05
45
45
0
y
Q
12.01
0
951
9,51
951
9,51
9,51
1073,85
3.505
14.49 652.05
14.49 652.05
0
15.99
q = V.Q / I
t
 =q/t
0
269,49
269,49
20
20
5
0
13,474
53,89
304,30
5
60,86
184,774
184,774
,
0
5
15
15
36,955
12,318
,
0
104
Dr. Endah Wahyuni
52
09/02/2014
Shear Stress Diagram:
20 cm
a
0
0
5 cm b
82,92
c
5 cm
93,63
20 cm
d
3 cm
e
53,89
13,474
20,73
18,951
56,854
0
60,68
12,318
36,955
0
15 cm
Shear Force
10.000 kg
Shear Force
6.500 kg
105
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
CENTER
106
Dr. Endah Wahyuni
53
09/02/2014
Shear Center

F1

P
V=P
V
P
h
e
F1
e = F1 . h / P =
=

½. .b.t.h
=
P
b. t. h . V . Q
2.P.I.t
.b.t.h
V.½.h.b.t
b2 . h2 . t
x
=
2.P
I.t
4 . I 107
Problem :
F1
F2
10 cm
P
V=P
e
50 cm
Determine the SHEAR
CENTER of this
section
section.
10 cm
10 15
30
Equation that is used:
e . P + F1 . 60 = F2 . 60
e = ( F2 . 60 – F1 . 60 ) / P

F1 = ½ .  . 17,5 . 10
Dr. Endah Wahyuni

F2 = ½ .  . 37,5 . 10
108
54
09/02/2014
Calculation :
I = 1/12 . 55 . 703 - 1/12 . 40 . 503
= 1.155.416,67 cm4
 =
V.Q
P . 17,5 . 10 . ½ . 60
=
I.t
1.155.416,67 . 10
= 0,00045 . P kg/cm2
 =
V.Q
P . 37,5 . 10 . ½ . 60
=
I.t
1.155.416,67 . 10
= 0,00097 . P kg/cm2
F1 = ½ . 0,00045 . P . 17,5 . 10
=
0,0394 . P
F2 = ½ . 0,00097 . P . 37,5 . 10
=
0,1820 . P
: = 8,556 cm
P
In order to make frame didn’t induce torsion , so the
Pload must be placed in e = 8,556 cm ( see Picture)
e=
0,182 . P . 60 - 0,0394 . P . 60
109
KERN / GALIH / INTI
Variety of KERN :
Limited with 4 p
point
Limited with 6 point
Li it d with
Limited
ith 4 point
i t
Unlimited
110
Dr. Endah Wahyuni
55
09/02/2014
KERN / GALIH / INTI
Determine Inertia moment of sloping axis:
Y
x
Y
x = x Cos  + y Sin 
df

X
y = y Cos  - x Sin 
2
Ix =

Ix =

2
y df
X
2
2
2
y Cos  + x Sin  - 2xy Sin  Cos  df
2
2
= Ix Cos  + Iy Sin  -2 Sxy Sin  Cos 
111
KERN / GALIH / INTI
Determine Inertia Moment of Sloping axis:
2
Iy =
=
x df
2
2
2
2
x Cos  + y Sin + 2xy Sin  Cos  df
2
2
= Ix Sin  + Iy Cos  + 2 Sxy Sin  Cos 
112
Dr. Endah Wahyuni
56
09/02/2014
KERN / GALIH / INTI
Example of determining KERN limits :
y
Determine the Neutral axis :
2 cm
x=
16
x
A = 2.20 + 8.2.2
Ix =
2
1/
3
12.2.20
= 3,2 cm
= 72 cm
+ 1/12.8.23.2
+ 8.2.92.2
2
10
2.20.1 + 8.2.6.2
2.20 + 8.2.2
= 3936 cm4
3936
= 393,6 cm3
10
3936
= 393,6 cm3
=
10
Wax =
3,2
Wbx
113
KERN / GALIH / INTI
Contoh Menentukan batas – batas KERN :
Iy =
1/
3
12.20.2
+ 1/12.2.83.2
+ 20
20.2.(2,2)
2 (2 2)2 + 2.2.8.(2,8)
2 2 8 (2 8)2
= 628,48
628 48 cm4
628,48
= 196,4 cm3
3,2
628,48
=
= 92,42 cm3
6,8
Wkr y =
Wkn y
Dr. Endah Wahyuni
Ka x =
Wbx
A
Kb x =
Wax
A
393,6
,
72
5,46
cm
=
393,6
=
72
= 5,46 cm
=
Kkr y =
Kkny =
Wkn y
A
Wkr y
A
92,42
,
72
= 1,28 cm
196,4
=
72
= 2,72 cm
=
114
57
09/02/2014
KERN / GALIH / INTI
Picture of KERN limits :
1,28 cm
2,72 cm
y
2 cm
16
5,46 cm
x
2
5,46 cm
2
10
3,2
115
Modul 4
Torsion
Torsi
on
116
Dr. Endah Wahyuni
58
09/02/2014
TORSION (Puntiran )
30 N-m
Section Plane
30 N-m
10 N-m
10 N-m
20 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.
117
TORSION (Puntiran )
M
Torsion Moment at
both end of the bar
M
M
M
M(x)
Torsion Moment
g the
distributed along
bar
118
Dr. Endah Wahyuni
59
09/02/2014
TORSION (Puntiran )

C
max
max

AC
max . dA .  = T
St
Stress

C
Area
Forces
Distance
Torsion Moment
Or can be written as:
max
C
 . dA = T
2
A
 . dA = IP = Polar Inertia Moment
2
119
A
Example of Polar Inertia Moment for CIRCLE
C
2 .  . d  = 2 .
 . dA =
3
2
0
A

4
4
C
4
=
0
C
2
=
d
4
32
Torsion of the CIRCLE can be determined with
this equation:
T=
max
max
C
=
. IP
T.C
IP
TORSION MOMENT
TORSION STRESS
120
Dr. Endah Wahyuni
60
09/02/2014
For Circle – Hollow Section:
Section:
121
TWIST ANGLE OF CIRCULAR BAR
With determine small angle of DAB in this following
picture. The maximum stress of its geometry is:
122
Dr. Endah Wahyuni
61
09/02/2014

If :

Then:

So general statement of the twist angle of a section from
the bar with linier elastic material is:
123
PROBLEM EXERCISE - 1
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 Moment:
Moment:

Bar AB = BC

Bar CD = DE

Considering its left section, torsion moment in every part will be:
TAB = 0, TBD = TBC = TCD = 150 N.m
N.m,, TDE = 1150 N.m

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
To get rotation of edge A, can be done with add up every
integration limit:

Value of T and Ip are constant, so the equation will be
be::
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EXERCISE -1

Calculate maximum torsion shear stress of AC – bar (as
seen in AC bar – exercise 1)
1).. Assume that bar diameter
from A – C is 10 mm.

Answer::
Answer
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Exercises

Soal 4.1
S b h poros b
Sebuah
berongga mempunyaii
diameter luar 100 mm dan diameter dalam
80 mm. Bila tegangan geser ijin adalah 55
MPa, berapakah besar momen puntir yang
bisa diteruskan ? Berapakah tegangan
pada mukaan poros sebelah dalam bila
diberikan momen puntir ijin?
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
Sebuah poros inti berongga berdiameter
200 mm di
diperoleh
l hd
dengan melubangi
l b
i
poros melingkar padat berdiameter 300
mm hingga membentuk lubang aksial
berdiameter 100 mm. Berapakah
persentase kekuatan puntiran yang hilang
oleh operasi ini ?
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
Poros padat berbentuk silinder dengan ukuran yang
bervariasi yang terlihat dalam gambar digerakkan oleh
momen--momen puntir seperti ditunjukkan dalam
momen
gambar tersebut. Berapakah tegangan puntir
maksimum dalam poros tersebut, dan diantara kedua
katrol yang ada ?
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133
a.
b.
Tentukanlah tegangan geser maksimum dalam poros
yang dihadapkan pada momen
momen--momen puntir, yang
diperlihatkan dalam gambar.
b. Hitunglah dalam derajat sudut pelintir antara kedua
ujungnya. Ambillah G = 84.000 MN/m².
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Modul 5
STRESS COMBINATION
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
Equation that have learned before about linier elastic
material, can be simplified as:
Normal Stress
Stress::
a. Due
D tto axial
i l lload
d
P
A
b. Due to flexure

 
My
I
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
Shear Stress
Stress::
a. Due to torsion

T
Ip
b. Due to shear force of beam

VQ
It
Superposition of the stress, only considered in
elastic problem when deformation that
happened is small.
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EXERCISE:

A bar 50x75 mm that is 1.5 meter of length, selfweight is
not considered, was loaded as seen in this following
picture. (a). Determine maximum tension and
compression
p
stress that work p
pependicularly
p
y of beam
section, assume that it is an elastic material.
material.
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

ANSWER
Using superposition method, so it can be solved in two
steps.. In Picture (b)
steps
(b),, it shows that the bar only take axial
load only. Then In Picture (c), it shows that the bar only
take transversal load only
Axial Load,
Load, normal stress that the bar have along its length
is:
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
Normal stress due to tranversal load depends on flexure
moment value and the maximum flexure moment is in
force that use:
Stress superposition woks perpendicularly of beam
section and linearly decreased to the neutral axis as
seen in picture (g)
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STRESS COMBINATION ON COLUMN

Similar equation can be done to assymetric
section:
x  

P M zz y M yy z


A
I zz
I yy
When:
When:
Flexure Moment Myy = +P z0 that works of yy-axis
Flexure Moment Mzz = -P y0 that works of zz-axis
A is cross section area of frame
Izz and Iyy is inertia moment of the section to each their
principal axis
Positive symbol (+) is tension stress, and Negati
Negative
ve
symbol (-) is compression stress.
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Example
Determine 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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Answer:
Answer:
Forces that work in ABCD section,
section, on the picture (c), is
P = -64 kN,
kN,
Myy = -640 (0.15)
(0 15) = -9,6
9 6 kN.m
kN m, and
kN.m,
Mzz = -64 (0.075 + 0.075) = -9,6 kN.m.
kN.m.
Cross section area of the beam A = (0.15)(0.3) = 0,045 m²,
And its Inertia moment is:
is:
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
Jadi dengan menggunakan hubungan yang setara dapat
diperoleh tegangan normal majemuk untuk elemen
elemen-elemen sudut :
Bila tanda huruf tegangan menandakan letaknya
letaknya,, maka
tegangan normal sudut adalah :
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THE END
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