Transcript
ELEMEN MESIN (RI.1232)
LECTURE II
Tegangan Konsep Tegangan
Tegangan
Tarik dan Tekan
Tegangan Tegangan
lentur
geser dan puntir
Dosen: Fahmi Fah mi Mub Mubaro arok, k, ST., ST., MSc. MSc. Metallurgy Laboratory Mechanical Engineering ITS-- Sur ITS Surab abay aya a 2008
http://www.its.ac.id/personal http://www.its.ac.id/personal/material. /material.php?id=fahmi php?id=fahmi
Definition
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V
Start with internal system system of forces as shown below to get proper signs for V, N and M. - Tegangan (stress) intensitas gaya persatuan luas - Regangan (strain) deformasi (perubahan bentuk) akibat tegangan yang bekerja
σ
ε
=
=
P Ao ∆l lo
Jenis-jenis tegangan 1. Tegangan tarik dan tekan (Tensile dan compression stress). 2. Tegangan Geser (Shears (Shears stress) disi disini ni term termas asuk uk tega tegan ngan gan punt puntir ir (Torsional (Torsional Stress ). 3. Tega Tegang ngaan Bendi ending ng / len lengk gku ung ( ng ( Bending stress ). 4. Tegangan kombinasi ( Combination stress ).
Various of Average Normal Stress
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Tegangan Tarik dan Tekan
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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
Tegangan dan regangan akan memberikan nilai positif pada kondisi tegangan tarik sedang pada kondisi tegangan tekan akan memberikan nilai negatif
∆l = per perpa panj njan anga gan n
lo = panjang awal sebelum beban diberikan
Normal Stress due to Bending Moment
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Key Points: 1. Inte Internal rnal bendin bending g moment moment causes causes beam beam to deform. deform. 2. For this this case, top fiber fibers s in compression compression,, bottom bottom in tension. tension.
Normal Stress due to Bending Moment
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Internal bending moment, lb-in
Bending stress, psi
σ
=
My I
Distance from NA to point of interest, in
Moment of inertia, in4
Design of Beams
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Tegangan Geser dan Puntir
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Average Shear Stress
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• Forces P and P and P’ P’ are are applied transversely to the member AB. member AB. • Corres Correspon pondin ding g intern internal al forc forces es act act in the the plane plane of of section C and C and are called shearing called shearing forces. forces. • The resultant resultant of the the inter internal nal shear force force distr distribut ibution ion is defined as the shear the shear of of the section and is equal to the load P load P . • The corres correspon pondin ding g avera average ge shea shearr stre stress ss is, is, ave =
τ
P A
• Shear Shear stres stresss distr distribu ibutio tion n varies varies from from zero zero at at the member surfaces to maximum values that may be much larger than the average value. • The shea shearr stress stress dis distr tribu ibuti tion on canno cannott be assum assumed ed to be uniform.
Average Shear Stress
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Single Shear
ave =
τ
P F =
Double Shear
ave =
τ
P
=
F 2
Torsion
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Torque is a moment that tends to twist a member about its longitudinal axis.
J = 1 π c 4 2
J = 1 π c24 − c14 2
Shear and Bending Moment Diagrams
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• Vari Variat atio ion n of shea shearr and and bend bendin ing g moment along beam may be plotted. • Dete Determ rmin inee reac reacti tion onss at sup suppo port rts. s. • Cut beam at C and C and consider member AC member AC , V = + P 2 M = + Px 2 • Cut beam at E and E and consider member EB member EB,, V = − P 2 M = + P ( L − x ) 2 • For For a beam beam subj subjec ecte ted d to to concentrated loads, shear is constant between loading points and moment varies linearly.
Sample Problem Bending and Shear
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SOLUTION: • Trea Treati ting ng the the ent entir iree beam beam as as a rigi rigid d body, determine the reaction forces forces
For the timber beam and loading shown, draw the shear and bend-moment diagrams and determine the maximum normal stress due to bending.
• Sect Sectio ion n the the beam beam at poin points ts near near supports and load application points. Apply equilibrium analyses on resulting free-bodies to determine internal shear forces and bending couples
• Iden Identi tify fy the the max maxim imum um shea shearr and and bending-moment bending-moment from plots of their distributions. • Apply Apply the the elasti elasticc flex flexure ure form formula ulass to determine the corresponding maximum normal stress.
Sample Problem 5.1
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SOLUTION: • Treati Treating ng the the entire entire beam beam as as a rigid rigid body, body, dete determ rmine ine the the reaction forces from
∑ F y
=0=
∑ M B :
R B = 46 kN
R D = 14 kN
• Sectio Section n the bea beam m and appl apply y equili equilibri brium um anal analyses yses on on resulting free-bodies ∑ F y = 0
− 20 kN − V 1 = 0
V 1 = − 20 kN
∑ M 1 = 0
(20 kN )(0 m ) + M 1 = 0
M 1 = 0
∑ F y = 0
− 20 kN − V 2 = 0
∑ M 2 = 0
(20 kN )(2.5 m ) + M 2 V 3 = + 26 kN
V 2 = − 20 kN = 0 M 2 = − 50 kN ⋅ m
3 = − 50 kN ⋅ m
V 4 = + 26 kN M 4 = + 28 kN ⋅ m V 5 = − 14 kN
M 5 = + 28 kN ⋅ m
V 6 = − 14 kN
M 6 = 0
Sample Problem 5.1
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• Identi Identify fy the the maxi maximu mum m shea shearr and and bend bending ing-moment from plots of their distributions. distribution s. V m = 26 kN
M m = M B = 50 kN ⋅ m
• Apply Apply the the elasti elasticc flexur flexuree form formula ulass to to determine the corresponding maximum normal stress. S = 1 b h 2 = 1 (0 .080 m )(0 .250 m )2 6
6
= 833 .33 × 10
m =
σ
m =
σ
M B S
=
−6
m3
50 × 10 3 N ⋅ m 833 . 33 × 10 − 6 m 3
60 .0 × 10 6 Pa
ELEMEN MESIN (RI.1232)
LECTURE III
Sambungan -Sa Samb mbun unga gan n Ke Keli ling ng Dosen: Fahmi Fah mi Mub Mubaro arok, k, ST., ST., MSc. MSc. Metallurgy Laboratory Mechanical Engineering ITS-- Sur ITS Surab abay aya a 2008
