## Introduction to Mechanics of Deformable Solids |

### From inside the book

Results 1-3 of 35

Page 300

3 : 17 ) dp - jo ? + 4 = [ 3 + ( * ) * ] - * [ 3 + ( ) ' ( 139 ] AO ( 13 . 3 : 18 ) Toa ) ] Tae ( r

/ b ) ? ( b¢ / 8 ) 2

explicit expression for Tac in terms of r / b , bö / L , and bø / .

3 : 17 ) dp - jo ? + 4 = [ 3 + ( * ) * ] - * [ 3 + ( ) ' ( 139 ] AO ( 13 . 3 : 18 ) Toa ) ] Tae ( r

/ b ) ? ( b¢ / 8 ) 2

**Substitution**of Eq . ( 13 . 3 : 18 ) in ( 13 . 3 : 14 ) would give anexplicit expression for Tac in terms of r / b , bö / L , and bø / .

**Substitution**of Eq ...Page 304

4 : 1 ) , after

- dy2 – 372 do * deg2 de maione ( 13 . 4 : 7 ) Oa1 where each set of stress

components 0 1 , T1 and 0 a2 , T2 satisfies the yield condition J2 = JZY [ ( 13 .

4 : 1 ) , after

**substitution**of r = b for the outer tube and r = a for the inner , b dų dyz- dy2 – 372 do * deg2 de maione ( 13 . 4 : 7 ) Oa1 where each set of stress

components 0 1 , T1 and 0 a2 , T2 satisfies the yield condition J2 = JZY [ ( 13 .

Page 352

... Combination of Eqs . ( 14 . 5 : 3 ) and ( 14 . 5 : 4 ) gives RA® M = Co2 M = 4M

RA = V4M6qo and the inequality ( 14 . 5 : 6 ) becomes ( 14 . 5 : 7 ) In = 2M or Mo

2 gol ? ( 14 . 5 : 8 ) Vqo 64 The plastic - limit load qo can be found by

...

... Combination of Eqs . ( 14 . 5 : 3 ) and ( 14 . 5 : 4 ) gives RA® M = Co2 M = 4M

RA = V4M6qo and the inequality ( 14 . 5 : 6 ) becomes ( 14 . 5 : 7 ) In = 2M or Mo

2 gol ? ( 14 . 5 : 8 ) Vqo 64 The plastic - limit load qo can be found by

**substitution**...

### What people are saying - Write a review

We haven't found any reviews in the usual places.

### Other editions - View all

### Common terms and phrases

actual addition angle answer applied assemblage axes axial axis beam behavior bending circle circular column combined compatibility components compression compressive stress Consider constant creep cross section curve cylinder deflection deformation determined diameter direction displacement effect elastic equal equation equilibrium example Figure Find force given gives homogeneous idealization increase initial interior isotropic length limit linear linear-elastic load material maximum Maxwell modulus moment needed nonlinear normal obtained outer plane plastic positive pressure principal Prob problem produced pure radius range ratio relation replaced requires residual response result rotation shaft shear stress shell shown shows simple sketch solid solution steel strain stress-strain relations structural Suppose surface symmetry temperature tensile tension thickness thin-walled torsion tube twisting uniform unloading versus viscous yield zero