Plane Stress — Part 1

A slab carrying a floor load is neither: the load acts normal to its plane, so it bends as a plate, a different problem. The test for plane stress is always the same two questions: is the body thin compared with its in-plane size, and do the loads act in its plane?Slab yang menanggung beban lantai bukan kedua-duanya: beban bertindak normal kepada satahnya, jadi ia melentur sebagai plate, masalah yang berbeza. Ujian untuk plane stress sentiasa dua soalan yang sama: adakah badan itu nipis berbanding saiz dalam satahnya, dan adakah beban bertindak dalam satahnya?

Static equilibrium requires the forces in x and in y to sum to zero. Forces are stresses times face areas: σ_xx acts on the faces of area dy, σ_yx on the faces of area dx (each times the unit thickness):Static equilibrium memerlukan jumlah daya dalam arah x dan y bersamaan sifar. Daya ialah stress didarab luas muka: σ_xx bertindak pada muka seluas dy, σ_yx pada muka seluas dx (setiap satu didarab ketebalan unit):

The near-face terms cancel against the first part of the far-face terms, every remaining term carries dx dy, and dividing by it leaves the first equilibrium equation:Term muka dekat terbatal dengan bahagian pertama term muka jauh, setiap term yang tinggal membawa dx dy, dan membahagi dengannya meninggalkan equilibrium equation pertama:

In y, σ_yy acts on the faces of area dx and σ_xy on the faces of area dy. The same expansion givesDalam arah y, σ_yy bertindak pada muka seluas dx dan σ_xy pada muka seluas dy. Pengembangan yang sama memberi

Collect the two equations with a 2 × 3 matrix of derivative operators acting on the stress vector {σ_xx σ_yy σ_xy}ᵀ. This is the plane-stress version of {∂}{q} in heat transfer: there a 1 × 2 row acted on a 2 × 1 flux; here a 2 × 3 matrix acts on a 3 × 1 stress.Kumpulkan dua persamaan itu dengan matrix 2 × 3 bagi derivative operator yang bertindak ke atas stress vector {σ_xx σ_yy σ_xy}ᵀ. Ini versi plane stress bagi {∂}{q} dalam heat transfer: di sana satu baris 1 × 2 bertindak ke atas flux 2 × 1; di sini matrix 2 × 3 bertindak ke atas stress 3 × 1.

or, with the 3 × 3 constitutive matrix [E],atau, dengan constitutive matrix [E] 3 × 3,

For the text’s cantilever, E = 200 × 10⁶ N m⁻² and ν = 0.33. Keep four significant figures: Part 2 builds [k] from these numbers.Untuk cantilever dalam teks, E = 200 × 10⁶ N m⁻² dan ν = 0.33. Simpan empat angka bererti: Part 2 membina [k] daripada nombor ini.

Setting ε_zz = 0 instead of σ_zz = 0 in three-dimensional Hooke’s law gives a matrix with the same shape and different constants:Menetapkan ε_zz = 0 dan bukan σ_zz = 0 dalam Hooke’s law tiga dimensi memberi matrix dengan bentuk yang sama tetapi pemalar yang berbeza:

The shear entry is G in both. The normal entries are larger in plane strain, because a slice that cannot thin resists in-plane stretching more. For ν = 0.33 the bending stiffness of the cantilever rises by the factor 1/(1 − ν²) = 1.12, so its tip deflection falls to about 0.89 of the plane-stress value; the Explorer shows 0.90 on a fine mesh. As ν approaches 0.5 (rubber, undrained clay) the plane-strain constants grow without bound: the material becomes incompressible, a topic for later courses.Entri shear ialah G dalam kedua-duanya. Entri normal lebih besar dalam plane strain, kerana kepingan yang tidak boleh menipis lebih menahan regangan dalam satah. Untuk ν = 0.33, bending stiffness cantilever meningkat dengan faktor 1/(1 − ν²) = 1.12, jadi tip deflection-nya turun kepada kira-kira 0.89 daripada nilai plane stress; Explorer menunjukkan 0.90 pada mesh halus. Apabila ν menghampiri 0.5 (getah, undrained clay), pemalar plane strain membesar tanpa had: bahan itu menjadi incompressible, topik untuk kursus kemudian.

In matrix form the 3 × 2 operator is exactly the transpose of the equilibrium operator [∂]. That is not a coincidence: equilibrium and compatibility are dual, and it is why the final matrix [k] comes out symmetric.Dalam bentuk matrix, operator 3 × 2 itu tepat transpose bagi equilibrium operator [∂]. Itu bukan kebetulan: equilibrium dan compatibility adalah dual, dan itulah sebab matrix akhir [k] keluar simetri.

Substitute the strains into Hooke’s law and Hooke’s law into equilibrium. The result is two PDEs in the two displacements, written once with the three matrices:Gantikan strain ke dalam Hooke’s law dan Hooke’s law ke dalam equilibrium. Hasilnya dua PDE dalam dua displacement, ditulis sekali dengan tiga matrix:

Read (3.17) from the inside out: [∂]ᵀ{u}ᵀ turns displacements into strains (3 × 1); [E] turns strains into stresses (3 × 1); [∂] turns stresses into the two equilibrium equations, balanced by the body force. Compare it with the heat PDE:Baca (3.17) dari dalam ke luar: [∂]ᵀ{u}ᵀ menukar displacement kepada strain (3 × 1); [E] menukar strain kepada stress (3 × 1); [∂] menukar stress kepada dua equilibrium equation, diimbangi oleh body force. Bandingkan dengan PDE heat:

On a rectangle the normals are (±1, 0) on the vertical edges and (0, ±1) on the horizontal ones, so only two stresses act as tractions on each edge:Pada segi empat tepat, normal ialah (±1, 0) pada tepi menegak dan (0, ±1) pada tepi mendatar, jadi hanya dua stress bertindak sebagai traction pada setiap tepi:

A plate can translate in x, translate in y and rotate in its plane without straining at all: for a small rotation θ, u = −θy and v = θx give ε_xx = ε_yy = 0 and ε_xy = −θ + θ = 0. Such a motion needs no force, so [K] is singular until the supports remove all three. Holding v alone along the left edge stops the y-translation but not the x-translation, nor a rotation about a point of that edge (its points move in x, not in y); the Explorer’s “vertical only” support shows the solver refusing. Holding u alone along the edge stops x-translation and rotation but leaves the plate free to slide in y.Plat boleh translate dalam arah x, translate dalam arah y dan berputar dalam satahnya tanpa sebarang strain: untuk putaran kecil θ, u = −θy dan v = θx memberi ε_xx = ε_yy = 0 dan ε_xy = −θ + θ = 0. Pergerakan sebegini tidak memerlukan daya, jadi [K] singular sehingga support menghapuskan ketiga-tiganya. Menahan v sahaja di sepanjang tepi kiri menghalang translation y tetapi tidak menghalang translation x, mahupun putaran terhadap satu titik pada tepi itu (titik-titiknya bergerak dalam arah x, bukan y); support “vertical only” dalam Explorer menunjukkan solver menolak. Menahan u sahaja di sepanjang tepi itu menghalang translation x dan putaran tetapi membiarkan plat bebas menggelongsor dalam arah y.

Both interpolations are one matrix product. Row 1 picks the u-dofs and row 2 the v-dofs; the zeros say that u does not depend on any v. The 8-node element gives a 2 × 16 matrix of the same pattern (3.29).Kedua-dua interpolation ialah satu hasil darab matrix. Baris 1 memilih dof-u dan baris 2 dof-v; sifar menunjukkan bahawa u tidak bergantung pada mana-mana v. 8-node element memberi matrix 2 × 16 dengan pola yang sama (3.29).

The point lies in element 4, whose local node 1 is global node 5 at (2, 1). Its local coordinates are (1.5, 0.3): the Activity 4 point of Heat Transfer Part 1, with the same shape functions. The nodal displacements come from the text’s solution (3.49), in local order 5, 6, 9, 8:Titik itu terletak dalam element 4, yang local node 1-nya ialah global node 5 pada (2, 1). Koordinat local-nya (1.5, 0.3): titik Aktiviti 4 dalam Heat Transfer Part 1, dengan shape function yang sama. Nodal displacement diambil daripada penyelesaian teks (3.49), dalam susunan local 5, 6, 9, 8:

Node 6, the free-end node of the element, carries half the weight and the largest v. The plate sags by 0.11 m and hardly moves along x: bending, as expected of a cantilever (the text chose a very soft E so that the displacements are visible). At the centroid every N_i = 1/4:Node 6, node hujung bebas bagi element itu, membawa separuh weight dan v yang terbesar. Plat itu melendut 0.11 m dan hampir tidak bergerak dalam arah x: lenturan, seperti yang dijangka bagi cantilever (teks memilih E yang sangat lembut supaya displacement kelihatan). Pada centroid setiap N_i = 1/4:

Differentiating u and v gives the strains. Packed into a matrix, the derivatives of [N] become [B] = [∂]ᵀ[N], a 3 × 8 matrix, and Galerkin turns (3.17) into the element stiffness matrix:Membezakan u dan v memberi strain. Apabila dibungkus dalam matrix, derivative bagi [N] menjadi [B] = [∂]ᵀ[N], matrix 3 × 8, dan Galerkin menukar (3.17) kepada element stiffness matrix: