Class Notes (836,518)
United States (324,534)
All (9)
Lecture

# Lecture30.pdf

5 Pages
45 Views

Department
Mechanical Engineering
Course
MECHENG 382
Professor
All Professors
Semester
Winter

Description
Lecture 30 - Modeling creep C OMBINED MODELING OF CREEP ,ELASTC AND PLASTIC DEFROMATION • Total strain consists of elastic + plastic +creep component ε total elastic plastic creep n n dε creep ˙⎜σ ⎞⎟ • For uniaxial case: ε elastic/E ; εplasticAσ ; dt = ε ⎝σ ⎠ € o • Represented by a spring, dash-pot and friction element Example: Assume that an alloy with a modulus of 100 GPa exhibits steady-state power-law creep of the form εH= 7.4 ×10 −1σ Hxp −Q(RT ) s -, where Q = 160 kJ/mole, R = 8.31 J/mol.K, and σ˜ is in MPa. An applied uniaxial stress of 30 MPa is applied. H What is the strain after 10,000 hours at 600 °C? € Creep €ate with σ = 30 MPa at 600 °C: ε = 4.75× 10 −12s−1 ε = ε + ε o elasticcreep dεo dε elasticdεcreep ∴ dt = dt + dt dε d(σ/E ) ∴ o = + 7.4× 10 σ exp5 (−Q/RT ) s (σ in MPa) dt dt dε o −12 But if σ = 30 MPa & independent of time: dt = 4.75× 10 −12 ∴ ε o 4.75 ×10 t+ ε 0 ) ε(0) = 30x10 /100x10 = 3.00x10 -4 ∴ After 10,000 hours (3.6x10 s): ε =o3.00 x 10 + 1.71 x 10 = 4.7 x 10 -4 If stress removed, then permanent strain = 1.7 x 10 .4 S TRESS RELAXATION • Fixed displacement or strain ε a 1 ME382 - 7/iv/14 Lecture 30 - Modeling creep • Initial stress can be elastic (a ε ) or plasticaif ε > yield straYn (σ /E) • Stress relaxation occurs with time Creep curve for constant applied strain • Stress relaxation occurs faster as temperature increases • Important example is stress relaxation in bolts • As nut tightens it compresses bolted components • However, equilibrium requires equal and opposite tensile load in shank of bolt • If this tensile stress relaxes with time, then components will loosen • Bolts need to be inspected and tightened if creep occurs Example: Assume that an alloy with a modulus of 100 GPa exhibits steady-state power-law creep of the form ε = 7.4 ×10 −1σ exp −(/RT ) s -, where Q = 160 kJ/mole, H H ˜ R = 8.31 J/mol.K, and σ H is in MPa. Assume initial uniaxial stress is 100 MPa. Strain is held consta€t. What is the stress after 1,000 hours at 600 °C? € εo= ε elasticεcreep 1 x 10-3 ME382 - 7/iv/14 2 Lecture 30 - Modeling creep ∴ dε o= dε elas+icεcree= 0 dt dt dt dε 1 dσ dεcreep −19 5 ∴ elas=ic 3 ; = 1.95 ×10 σ s (σ in MPa) dt 100 ×10 dt dt ∴ 1 dσ = −1.95 ×10 −1σ
More Less

Related notes for MECHENG 382
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.