Download - Tegangan Geser, Lengkung Dan Puntir
ELEMEN MESIN(RI.1232)
Dosen:
Fahmi Mubarok, ST., MSc.
Metallurgy Laboratory
Mechanical Engineering
ITS- Surabaya
2008
Tegangan� Konsep Tegangan
� Tegangan Tarik dan Tekan
� Tegangan lentur
� Tegangan geser dan puntir
LECTURE IILECTURE II
http://www.its.ac.id/personal/material.php?id=fahmi
Fahmi Mubarok
I ���� 2Mech. Eng. Dept. ITS Surabaya
Definition
- Tegangan (stress)
intensitas gaya persatuan luas
- Regangan (strain)
deformasi (perubahan bentuk) akibat tegangan yang bekerja
oA
P=σ
ol
l∆=ε
Jenis-jenis tegangan
1. Tegangan tarik dan tekan (Tensile dan compression stress).
2. Tegangan Geser (Shears stress) disini termasuk tegangan puntir (Torsional Stress ).
3. Tegangan Bending / lengkung ( Bending stress ).
4. Tegangan kombinasi ( Combination stress ).
V
Start with internal system of forces as shown below to get proper signs for V, N and M.
Fahmi Mubarok
I ���� 3Mech. Eng. Dept. ITS Surabaya
Various of Average Normal Stress
Fahmi Mubarok
I ���� 4Mech. Eng. Dept. ITS Surabaya
Tegangan Tarik dan Tekan
Untuk membandingkan spesimen
dengan berbagai ukuran, maka
digunakan konsep tegangan
teknik
F = beban yang diberikan tegak
lurus terhadap penampang
spesimen
Ao = luas penampang awal
sebelum beban diberikan
∆l = perpanjanganlo = panjang awal sebelum beban
diberikan
Tegangan dan regangan akan memberikan
nilai positif pada kondisi tegangan tarik
sedang pada kondisi tegangan tekan akan
memberikan nilai negatif
Fahmi Mubarok
I ���� 5Mech. Eng. Dept. ITS Surabaya
Normal Stress due to Bending Moment
Key Points:
1. Internal bending moment causes beam to deform.
2. For this case, top fibers in compression, bottom in tension.
Fahmi Mubarok
I ���� 6Mech. Eng. Dept. ITS Surabaya
Normal Stress due to Bending Moment
I
My=σ
Bending stress, psi
Internal bending moment, lb-in
Distance from NA to
point of interest, in
Moment of inertia, in4
Fahmi Mubarok
I ���� 7Mech. Eng. Dept. ITS Surabaya
Design of Beams
Fahmi Mubarok
I ���� 8Mech. Eng. Dept. ITS Surabaya
Tegangan Geser dan Puntir
Fahmi Mubarok
I ���� 9Mech. Eng. Dept. ITS Surabaya
Average Shear Stress
• Forces P and P’ are applied transversely to the
member AB.
A
P=aveτ
• The corresponding average shear stress is,
• The resultant of the internal shear force distribution is
defined as the shear of the section and is equal to the
load P.
• Corresponding internal forces act in the plane of
section C and are called shearing forces.
• Shear stress distribution varies from zero at the
member surfaces to maximum values that may be
much larger than the average value.
• The shear stress distribution cannot be assumed to be
uniform.
Fahmi Mubarok
I ���� 10Mech. Eng. Dept. ITS Surabaya
Average Shear Stress
A
F
A
P==aveτ
Single Shear
A
F
A
P
2ave ==τ
Double Shear
Fahmi Mubarok
I ���� 11Mech. Eng. Dept. ITS Surabaya
Torsion
Torque is a moment that tends to twist a
member about its longitudinal axis.
4
21 cJ π=
( )41422
1 ccJ −= π
Fahmi Mubarok
I ���� 12Mech. Eng. Dept. ITS Surabaya
Shear and Bending Moment Diagrams
• Variation of shear and bending
moment along beam may be
plotted.
• Determine reactions at supports.
• Cut beam at C and consider
member AC,
22 PxMPV +=+=
• Cut beam at E and consider
member EB,
( ) 22 xLPMPV −+=−=
• For a beam subjected to
concentrated loads, shear is
constant between loading points
and moment varies linearly.
Fahmi Mubarok
I ���� 13Mech. Eng. Dept. ITS Surabaya
Sample Problem Bending and Shear
For the timber beam and loading shown,
draw the shear and bend-moment
diagrams and determine the maximum
normal stress due to bending.
SOLUTION:
• Treating the entire beam as a rigid
body, determine the reaction forces
• Identify the maximum shear and
bending-moment from plots of their
distributions.
• Apply the elastic flexure formulas to
determine the corresponding maximum
normal stress.
• Section the beam at points near
supports and load application points.
Apply equilibrium analyses on resulting
free-bodies to determine internal shear
forces and bending couples
Fahmi Mubarok
I ���� 14Mech. Eng. Dept. ITS Surabaya
Sample Problem 5.1
SOLUTION:
• Treating the entire beam as a rigid body, determine the
reaction forces
∑ ∑ ==== kN14kN46:0 from DBBy RRMF
• Section the beam and apply equilibrium analyses on
resulting free-bodies
( )( ) 00m0kN200
kN200kN200
111
11
==+∑ =
−==−−∑ =
MMM
VVFy
( )( ) mkN500m5.2kN200
kN200kN200
222
22
⋅−==+∑ =
−==−−∑ =
MMM
VVFy
0kN14
mkN28kN14
mkN28kN26
mkN50kN26
66
55
44
33
=−=
⋅+=−=
⋅+=+=
⋅−=+=
MV
MV
MV
MV
Fahmi Mubarok
I ���� 15Mech. Eng. Dept. ITS Surabaya
Sample Problem 5.1
• Identify the maximum shear and bending-
moment from plots of their distributions.
mkN50kN26 ⋅=== Bmm MMV
• Apply the elastic flexure formulas to
determine the corresponding maximum
normal stress.
( )( )
36
3
36
2
612
61
m1033.833
mN1050
m1033.833
m250.0m080.0
−
−
×
⋅×==
×=
==
S
M
hbS
Bmσ
Pa100.60 6×=mσ
ELEMEN MESIN(RI.1232)
Dosen:
Fahmi Mubarok, ST., MSc.
Metallurgy Laboratory
Mechanical Engineering
ITS- Surabaya
2008
LECTURE IIILECTURE III
Sambungan-Sambungan Keling
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