σ τ τ τ σ τ τ τ σ Review Chapter Four States of Stress Part Three Review Review

Transcription

1 Chapter Four States of Stress Part Three When makng your choce n lfe, do not neglect to lve. Samuel Johnson Revew When we use matrx notaton to show the stresses on an element The rows represent the axs whch the face s parallel to σ τ τ τ σ τ τ τ σ yx y yz zx zy z Meetng Twenty One - States of Stress III X-face Y-face Z-face Revew The columns represent the axs whch the stress s drected along or parallel to X-axs Y-axs Z-axs σ τ τ τ σ τ τ τ σ yx y yz zx zy z Revew If we look at a element n the shape of a cube we can draw the stresses on that element σ x τxy τ xz τ yx σ y τ yz τ zx τzy σ z Meetng Twenty One - States of Stress III 3 Meetng Twenty One - States of Stress III 4 1

2 Revew If all the stresses act n one plane, we have plane stress For the xy plane we would have σx τxy τ yx σ y If we return to our orgnal defnton of an element that has stresses on all faces of the cube we have σ x τxy τ xz τ yx σ y τ yz τ zx τzy σ z Meetng Twenty One - States of Stress III 5 Meetng Twenty One - States of Stress III 6 In our prevous dscussons we rotated the orentaton of the axs so that we were able to locate the maxmum stresses on each face σ x τxy τ xz τ yx σ y τ yz τ zx τzy σ z In three dmensons, we have to rotate our axs system s a more complex fashon than the case wth plane stress σ x τxy τ xz τ yx σ y τ yz τ zx τzy σ z Meetng Twenty One - States of Stress III 7 Meetng Twenty One - States of Stress III 8

3 If we rotate our axs system to some new prme poston, the stresses on the new σ τ τ faces wll also be transformed τ σ τ yz yx y τ τ σ z zx zy There exsts some transformaton to a new coordnate system (x, y, z ) such that all σ τ τ the shear stresses wll be equal to τ σ τ yz yx y τ τ σ z zx zy Meetng Twenty One - States of Stress III 9 Meetng Twenty One - States of Stress III 1 In ths case we wll have the stuaton where The axs (x, y, and z ) are call the prncpal axs σ ' τ ' τ ' σ1 τ ' σ ' τ ' σ yx y yz = τ ' τ ' σ ' zx zy z σ3 σ ' τ ' τ ' σ1 τ ' σ ' τ ' σ yx y yz = τ ' τ ' σ ' zx zy z σ3 Meetng Twenty One - States of Stress III 11 Meetng Twenty One - States of Stress III 1 3

4 The axal stresses σ 1, σ and σ 3 are the prncpal stresses σ ' τ ' τ ' σ1 τ ' σ ' τ ' σ yx y yz = τ ' τ ' σ ' zx zy z σ3 In order to solve for the prncpal stresses we must solve for the roots of a 3 rd order equaton σ ' τ ' τ' σ1 τ' σ ' τ' σ yx y yz = τ' τ ' σ ' σ zx zy z 3 σ I σ + I σ I = Meetng Twenty One - States of Stress III 13 Meetng Twenty One - States of Stress III 14 The coeffcents, I s, n the expresson are known as stress nvarants and do not depend on the orentaton of the transformed coordnate system σ ' τ ' τ' σ1 τ' yx σ ' y τ' yz = σ τ' τ ' σ ' σ zx zy z 3 σ I σ + I σ I = There are multple possble orentatons that can gve the prncpal stresses on alternate faces σ ' τ ' τ' σ1 τ' yx σ ' y τ' yz = σ τ' τ ' σ ' σ zx zy z 3 σ I σ + I σ I = Meetng Twenty One - States of Stress III 15 Meetng Twenty One - States of Stress III 16 4

5 The three stress nvarants are gven by σ ' x τ ' xy τ' xz σ1 τ' yx σ ' y τ' yz = σ τ' τ ' σ ' zx zy z σ3 σ I σ + I σ I = I = σ + σ + σ 1 3 x y z I = σσ + σσ + σσ τ τ τ x y y z z x xy yz zx x y z x yz y zx z xy xy yz zx I = σσσ στ στ στ + τ τ τ Meetng Twenty One - States of Stress III 17 Problem Gven the followng state of stress (all unts are ks) σ x := 3 σ y := σ z := τ xy := 15 τ yz := 1 τ zx := 1 Meetng Twenty One - States of Stress III 18 Problem We can use our expressons to calculate the stress nvarants σ x := 3 σ y := σ z := τ xy := 15 τ yz := 1 τ zx := 1 I 1 := σ x + σ y + σ z I := σ x σ y + σ y σ z + σ z σ x τ xy ( ) I 3 := σ x σ y σ z σ x τ yz ( ) ( τ yz ) ( ) ( ) σ y τ zx σ z τ xy ( τ zx ) + τ xy τ yz τ zx Problem Substtutng we have σ x := 3 σ y := σ z := τ xy := 15 τ yz := 1 τ zx := 1 I 1 := σ x + σ y + σ z I := σ x σ y + σ y σ z + σ z σ x τ xy ( ) I 3 := σ x σ y σ z σ x τ yz ( ) ( τ yz ) ( ) ( ) σ y τ zx σ z τ xy ( τ zx ) + τ xy τ yz τ zx I 1 = 3 I = I 3 = Meetng Twenty One - States of Stress III 19 Meetng Twenty One - States of Stress III 5

6 Problem We then solve for the values of σ that solve the expresson I 1 = 3 I = I 3 = σ σ σ 3 I1 + I I3 = I 3 I Vector := σ := polyroots ( Vector) I 1 1 Problem So the prncpal stresses on the element are ks ks 49.5 ks I 3 I Vector := σ := polyroots ( Vector) I 1 1 σ 1 = σ = σ3 = σ 1 = σ = σ3 = Meetng Twenty One - States of Stress III 1 Meetng Twenty One - States of Stress III Problem The maxmum shear stress on the element can then be found by σ 1 = σ = σ3 = τ max max σ 1 σ σ σ 3 σ 3 σ 1 :=,, τ max = Traxal Stress Traxal stress s a state of stress that can be defned as σ x σ y σ z Meetng Twenty One - States of Stress III 3 Meetng Twenty One - States of Stress III 4 6

7 Traxal Stress These stresses are by defnton also the prncpal stresses Traxal Stress So we can defne the maxmum shear stress on an element n traxal stress by σ x σ y σ z Maxmum σ σ σ σ σ σ,, x y y z z x Meetng Twenty One - States of Stress III 5 Meetng Twenty One - States of Stress III 6 Traxal Stress Traxal stress becomes mportant when we consder pressure vessels wth crcular cross sectons Traxal Stress Traxal stress becomes mportant when we consder pressure vessels wth crcular cross sectons Meetng Twenty One - States of Stress III 7 Meetng Twenty One - States of Stress III 8 7

8 We can start by assumng we have a sphercal pressure vessel wth a wall thckness, t, and a radus, R We must make the assumpton that the thckness, t, s much smaller than the radus of the vessel, R Meetng Twenty One - States of Stress III 9 Meetng Twenty One - States of Stress III 3 The confned gas wthn the pressure vessel s at some pressure p whle the external pressure on the vessel s p o If we cut through the center of the vessel, we can look at the forces developed on the vessel Remember we are cuttng a sphere n half Meetng Twenty One - States of Stress III 31 Meetng Twenty One - States of Stress III 3 8

9 The force to the rght exerted by the external pressure, p o, s equal to the product of the pressure tmes the area projected Thnk of the projected area as what you would see s you were lookng along an axs straght at the half of the ball. Meetng Twenty One - States of Stress III 33 Meetng Twenty One - States of Stress III 34 You would see a crcular area wth a radus of R So the force exerted by the external pressure would be equal to By the same dea, the force exerted by the nternal pressure wll be equal to ( π ( ) ) F = p R t p p ( ) F = p R π p ( ) F = p R π Meetng Twenty One - States of Stress III 35 Meetng Twenty One - States of Stress III 36 9

10 But snce R s >> t, we can make a good approxmaton to the force as ( π ) F = p R p So the net force from the pressure dfference would be ( ) ( ) F F = p πr p πr p p p ( ) F = p R π Meetng Twenty One - States of Stress III 37 Meetng Twenty One - States of Stress III 38 In order for the system to be n equlbrum, the dfference between the outward force F p and the nward force F po must be made up by the walls of the vessel Ths force s developed as the stress n the wall tmes the area The area over whch the stress acts s A= πr π R t ( ) ( ) ( ) F = π π p Fp p R p R Wednesday, October 16, Meetng Twenty One - States of Stress III 39 ( ) ( ) F = π π p Fp p R p R Wednesday, October 16, Meetng Twenty One - States of Stress III 4 1

11 Expandng the second term we have ( ) A = πr π R Rt t Snce t s very small, we can gnore the t term and we have A= πr π R Rt A= π Rt ( ) ( ) ( ) F = π π p Fp p R p R Wednesday, October 16, Meetng Twenty One - States of Stress III 41 ( ) ( ) F = π π p Fp p R p R Wednesday, October 16, Meetng Twenty One - States of Stress III 4 So the force requred to generate the confnng stress σ s equal to Fconfnng = σπ Rt So the sum of the forces s equal to ( ) ( ) = p πr p πr + σπ Rt ( ) ( ) F = π π p Fp p R p R Wednesday, October 16, Meetng Twenty One - States of Stress III 43 Meetng Twenty One - States of Stress III 44 11

12 Isolatng the stress we have p πr + p πr ( ) ( ) π Rt = σ And reducng ( p ) R p t = σ Meetng Twenty One - States of Stress III 45 Meetng Twenty One - States of Stress III 46 The stress σ s the same at every pont on the vessel R( p p ) = σ t The state of stress s dfferent on the nner and outer wall of the vessel R ( p p ) t = σ Meetng Twenty One - States of Stress III 47 Meetng Twenty One - States of Stress III 48 1

13 If we look at the outer wall, we have the pressures actng as a axal stress along one axs and the stress σ actng along the other two axs σ σ p o And on the nner wall we have σ σ p Meetng Twenty One - States of Stress III 49 Meetng Twenty One - States of Stress III 5 Homework Read Secton on Cylndrcal Vessel and Allowable Stress Problem Problem 5-5. Problem Meetng Twenty One - States of Stress III 51 13