(MPa) compute (a) The traction vector acting on an internal material plane with normal n ( e1 e

Transcription

1 EN10: Continuum Mechanics Homework : Kinetics Due 1:00 noon Friday February 4th School of Engineering Brown University 1. For the Cauchy stress tensor with components (MPa) compute (a) The traction vector acting on an internal material plane with normal n ( e1e ) / The traction follows as t n σ (b) The principal stresses One principal is 00 by inspection. MATLAB gives the others -105MPa and 405MPa. (c) The hydrostatic stress The hydrostatic stress is one third of the trace of the stress tensor: 00 MPa (d) The deviatoric stress tensor D The deviatoric stress is ij ij kkij / (e) The Von-Mises equivalent stress D D The V-M stress is 466 e ij ij MPa

2 . The figure shows two designs for a glue joint. The glue will fail if the stress acting normal to the joint exceeds 60 MPa, or if the shear stress acting parallel to the plane of the joint exceeds 00 MPa. (a) Calculate the normal and shear stress acting on each joint, in terms of the applied stress (b) Hence, calculate the value of that will cause each joint to fail. Glue joint Taking e 1 to be horizontal, the normal vectors to the two joints are n1 cos(0) e1 sin(0) e n e 1. The stress tensor is σ e1e 1. The tractions follow as t cos(0) e t e, and hence the normal and tangential components of traction are n t1 n1 t t1 nn1 e1 e1 e t cos (0) t t cos(0) cos(0)(cos(0) sin(0) sin(0) cos(0) For the second joint t t 0. n t. An internal surface plane that makes equal angles with each of the three principal stress directions is known as the octahedral plane. Show that the normal component of stress acting on this plane is I 1 /, and that the magnitude of the shear traction acting on the plane is 1 ( 1 ) ( 1 ) ( ) I / where ( 1,, ) are the three principal stresses, and I is the second invariant of the deviatoric stress tensor S / ij ij kk ij Denote the principal stresses and directions by { 1,, },{ n1, n, n } - we can then write σ 1n1 n1 n n n n If the normal to the octahedral plane makes an equal angle to { n1, n, n }, then m ( n1 n n ) / The traction on this plane follows as t nσ n n n 1 1 / The normal traction is tn tm ( 1 ) /. This is trace( σ ) /. The shear traction follows as t t t t nm 1n1 n n / ( 1 )( n1 n n) / ( 1 ) n1 / ( 1 ) n / ( 1 ) n / ( 1 ) ( 1 ) ( ) / 7 Finally note also that I ( ij kkij / )( ij nnij / ) kk nn / ijij. Substituting the principal stresses into this expression gives the required result.

3 4. In this problem we consider further the beam bending calculation discussed in HW. Suppose that the beam is made from a material in which the Material Stress tensor is related to the Lagrange strain tensor by E ij (this can be regarded as representing an elastic material with zero Poisson s ratio and shear modulus ) (a) Calculate the distribution of material stress in the bar, expressing your answer as components in the { e1, e, e } ij e H e 1 H e m R m 1 e 1 From HW, we have the Lagrange Strain follows as T 1 ( x ( x 0 E ( F F I ) /. The stress ( x ( x 0 Σ (b) Calculate the distribution of nominal stress in the bar expressing your answer as components in the { e1, e, e } The nominal stress is Recall that S ΣF T (1 x cos( x1 sin( x1 F (1 x sin( x cos( x, and so 1 1 ( x ( x 0(1 x cos( x1 (1 x sin( x1 S sin( x1 cos( x1 ( x ( x (1 x cos( x1 ( x ( x (1 x sin( x1 (c) Calculate the distribution of Cauchy stress in the bar expressing your answer as components in the { e1, e, e } The Cauchy stress is 1 (1 x cos( x1 sin( x1 cos( x1 sin( x1 σ FS ( x ( x J (1 x / R )sin( x1 / R ) cos( x1 / R ) cos ( x1 sin( x1 cos( x1 ( x ( x (1 x sin( x1 cos( x1 sin ( x1 (d) Repeat (a)-(c) but express the stresses as components in the { m1, m, m }

4 The change matrix is cos( x1 sin( x1 Q sin( x1 cos( x1 ( m) T 1 0 Thus Q Q ( x ( x and similarly cos ( x1 0 S ( x ( x (1 x sin( x1cos( x1 0 cos ( x1 sin( x1 cos( x1 Σ ( x ( x sin( x1 cos( x1 sin ( x1 (e) Calculate the distribution of traction on a surface in the beam that has normal e 1 in the undeformed beam. Give expressions for the tractions in both { e1, e, e } and { m1, m, m } The traction can be computed from the nominal stress as cos( x1 sin( x1 t n S ( x ( x (1 x 1 0 ( x ( x (1 x [cos( x sin( x ] 1 1 Note that this is in the { e1, e, e }. In the { m1, m, m } the answer is simply t ( x ( x (1 x [1 0] (f) Show that the surfaces of the beam that have positions x h/ in the undeformed beam are traction free after deformation The traction on a surface with normal in the e direction is cos( x1 sin( x1 t n S ( x ( x (1 x 0 1 [] (g) Calculate the resultant moment (per unit out of plane distance) acting on the ends of the beam. The resultant moment is H M ( r r ) t ( x ( x (1 x x ( m m ) dx H H 15 R 0 1 H H 10R H

5 5. A solid is subjected to some loading that induces a Cauchy stress ij at some point in the solid. The solid and the loading frame are then rotated together so that the entire solid (as well as the loading frame) is subjected to a rigid rotation R. This causes the components of the Cauchy stress tensor to change to new values ij. The goal of this problem is to calculate a formula relating ij, ij and R ij. ij a. Let n i be a unit vector normal to an internal material plane in the solid before rotation. After rotation, this vector (which rotates with the solid) is n i. Write down the formula relating n i and n i ni Rijn j b. Let T i be the internal traction vector that acts on a material plane with normal the solid before application of the rigid rotation. Let i same material plane after rotation. Write down the formula relating Ti RijTj n i in T be the traction acting on the T i and T i c. Finally, using the definition of Cauchy stress, find the relationship between ij, and R ij. Ti n j ji RijTj Rkl nl ki RinRijTj Tn nl Rkl ki Rin ln kl ki in ki kl ln in Hence R R R R 6. Repeat problem 5, but instead, calculate a relationship between the components of Nominal stress S ij and S ij before and after the rigid rotation. For nominal stress, letting ln kl ki in ki kl ln in Thus R R R R N i denote the normal to the reference configuration, we have Ti N js ji RijTj N js ji RinRijTj Tn N js ji Rin ij 7. Repeat problem 5, but instead, calculate a relationship between the components of material stress ij and ij before and after the rigid rotation. The material stress is unaffected by the rigid rotation you can see this from the physical interpretation of material stress, or else use the formula relating material stress to Cauchy stress, and note that the

6 deformation gradient for stretch followed by rotation is related to that associated with just the stretch by 1 1 F RF. Substituting this and the results of (5) into the formula ij JFik kl Fjl and simplifying the result shows that ij ij 8. One constitutive model for metallic glass (Anand and Su, J. Mech Phys. Solids 5 16 (005)) assumes that plastic flow in the glass takes place by shearing on planes that are oriented at an angle to the principal stress directions, calculated as follows. Let e1, e, e be unit vectors parallel to the three principal stresses (with 1 ) and suppose that shearing takes place on a plane with normal m, with shearing direction (tangent to the plane) s. let mσs and pn m σm denote the resolved shear stress and (compressive) normal stress acting on the shear plane. The constitutive model assumes that shearing in the { e1, e } plane occurs on the plane for which f p ( ) ( ) is a maximum with respect to. Here is a material property known as the internal friction coefficient. Their paper (eq 71) states that it is easily shown that f ( ) is a maximum for Derive this result. 1 tan 4 The stress can be expressed as 1 0 [cos,sin ] [ sin, cos ] 0 s m The normal pressure and shear traction follow as ( )sin cos p n 1 1sin cos pn ( 1)sin cos 1sin cos Differentiation leads to the following equation ( )(cos sin ) 0 1 This can be rearranged into the form stated.