3.4 hooke’s law - kocwcontents.kocw.net/kocw/document/2015/hanyang/hanseog... · 2016-09-09 ·...
TRANSCRIPT
3.4 HOOKE’S LAW • Modulus of Elasticity (Hooke’s Law)
• Strain Hardening
E
3.5 STRAIN ENERGY
• Modulus of Resilience
• Modulus of ToughnessEntire area under the stress-strain diagram
Eu pl
plplr
2
21
21
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 13.1
EXAMPLE 13.1
EXAMPLE 3.2 The stress–strain diagram for an aluminum alloy that is used for making aircraft parts is shown in Fig. 3–19. If a specimen of this material is stressed to 600 MPa, determine the perma-nent strain that remains in the specimen when the load is re-leased. Also, find the modulus of resilience both before and after the load application.
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.2 (cont)
• When the specimen is subjected to the load, the strain is approximately0.023 mm/mm.
• The slope of line OA is the modulus of elasticity,
• From triangle CBD,
Solutions
mm/mm008.0
100.7510600 96
CDCDCD
BDE
GPa 0.75006.0
450E
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.2 (cont)
• This strain represents the amount of recovered elastic strain.
• The permanent strain is
• Computing the modulus of resilience,
• Note that the SI system of units is measured in joules, where 1 J = 1 N • m.
Solutions
(Ans) MJ/m 40.2008.060021
21
(Ans) MJ/m 35.1006.045021
21
3
3
plplfinalr
plplinitialr
u
u
(Ans) mm/mm 0150.0008.0023.0 OC
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 13.1
EXAMPLE 13.1
3.6 POISSON’S RATIO
long
latv
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.4
A bar made of A-36 steel has the dimensions shown in Fig. 3–22. If an axial force of P = 80kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. The material behaves elastically.
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.4 (cont)
• The normal stress in the bar is
• From the table for A-36 steel, Est = 200 GPa
• The axial elongation of the bar is therefore
Solutions
mm/mm 108010200100.16 6
6
6
st
zz E
Pa 100.16
05.01.01080 6
3
AP
z
(Ans) m1205.11080 6z
zz L
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.4 (cont)
• The contraction strains in both the x and y directions are
• The changes in the dimensions of the cross section are
Solutions
m/m 6.25108032.0 6 zstyx v
(Ans) m28.105.0106.25
(Ans) m56.21.0106.256
6
yyy
xxx
L
L
Copyright ©2014 Pearson Education, All Rights Reserved
3.7 SHEAR STRESS-STRAIN DIAGRAM• Strength parameter G
Shear modulus of elasticity or the modules of rigidity• G is related to the modulus of elasticity E and Poisson’s
ratio .• Only two material constants are required for isotropic
materials.
G
vEG
12
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.5 A specimen of titanium alloy is tested in torsion and the shear stress– strain diagram is shown in Fig. 3–25a. Determine the shear modulus G, the proportional limit, and the ultimate shear stress. Also, determine the maximum distance d that the top of a block of this material, shown in Fig. 3–25b, could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement?
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.5 (cont)
• By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is
• This value represents the maximum shear stress, point B. Thus the ultimate stress is
• Since the angle is small, the top of the will be displaced horizontally by
Solutions
(Ans) MPa 504u
(Ans) MPa 360pl
mm 4.0mm 50
008.0rad 008.0tan dd
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 3.5 (cont)
• The shear force V needed to cause the displacement is
Solutions
(Ans) kN 270010075
MPa 360 ; VVAV
avg
Copyright ©2014 Pearson Education, All Rights Reserved
EXAMPLE 13.1
EXAMPLE 13.1
3.8 FAILURE OF MATERIALS DUE TO CREEP AND FATIGUE
• Creep: Change of time-dependent deformation or stressdue to time under the constant temperature
- Creep: change of time-dependent deformation under the constant stress and temperature
Copyright ©2014 Pearson Education, All Rights Reserved
Stage I: transient creepStage II: steady creepStage III: accelerate creep
3.8 FAILURE OF MATERIALS DUE TO CREEP AND FATIGUE
- Creep LimitThe maximum stress among the stresses having steady creep rate 0 at a specified temperature.
- Creep StrengthThe maximum stress a given material can withstand in a given time without exceeding a specified quantity of creep (ex.: 0.1%/1000h).
Copyright ©2014 Pearson Education, All Rights Reserved
3.8 FAILURE OF MATERIALS DUE TO CREEP AND FATIGUE
- Stress relaxation: change of time-dependent stress under the constant strain and temperature
Copyright ©2014 Pearson Education, All Rights Reserved
3.8 FAILURE OF MATERIALS DUE TO CREEP AND FATIGUE
- Fatigue Test: Rotation bending fatigue tester
Copyright ©2014 Pearson Education, All Rights Reserved
weight
specimen motor
3.8 FAILURE OF MATERIALS DUE TO CREEP AND FATIGUE
- Fatigue: S-N curve
Copyright ©2014 Pearson Education, All Rights Reserved
elS : Endurance limitor fatigue limit
: Fatigue strengthat N cycles
hS