Unit M. (All About) Stress Readings: CDL 4., 4.3, 4.4 16.001/00 -- Unified Engineering Department of Aeronautics and Astronautics Massachusetts Institute of Technology
LEARNING OBJECTIVES FOR UNIT M. Through participation in the lectures, recitations, and work associated with Unit M., it is intended that you will be able to.explain the concept and types of stress and how such is manifested in materials and structures.use the various ways of describing states of stress.apply the concept of equilibrium to the state of stress Unit M. - p.
Thus far we ve talked about loads, both external and internal. But when we talk about how a structure carries a load internally, we need to consider not just how this is carried in an aggregate sense, but how it is carried point to point. For this we need to introduce.. The Concept of Stress Definition: Stress is the measure of the intensity of force acting at a point --> Think about this as follows: Figure M.-1 Representation of a bar with a colinear force acting in the x 1 direction x 1 F 1 F 1 Unit M. - p. 3
Now divide the x 1 - force surface into a n x m grid Figure M.- Representation of bar cross-section divided into n x m grid n Δ F 1 = Δ A (1) = The stress is the intensity, thus let n and m go to infinity (so ) Δ A 0 σ 1 = lim n m F 1 n m A (1) n m m on (1) surface A = total area Δ = individual block area ΔF 1 ΔA (1) Units: Force length ( ) Unit M. - p. 4
σ 1 Where is the stress at a point on the x 1 - face - magnitude σ 1 - direction i 1 --> Consider a more generalized case Figure M.-3 Cross-section of general body subjected to internal force x 3 x F 3 F x 1 F i 1 F 1 Force F is acting on x 1 - face (i 1 is normal to plane of face) Resolve force into three components: F = F 1 i 1 + F i + F 3 i 3 Then take the limit as the force on the face is carried by a smaller set of areas (bigger grid) until we have: Unit M. - p. 5
lim ΔA 0 Δ F Δ A (1) = σ 1 This gives: σ 1 = σ 11 i 1 + σ 1 i + σ 13 i 3 = σ 1 = Δ F 1 Δ F Δ F 3 Δ A 1 Δ A 1 Δ A 1 where is the stress vector and the σ mn are the components acting in the three orthogonal directions We can do the same on the x and x 3 faces and get: σ = σ 1 i 1 + σ i + σ 3 i 3 (on i - face) σ 3 = σ 31 i 1 + σ 3 i + σ 33 i 3 (on i 3 - face) Note that using tensor format we can write this as: = σ m = σ mn i n Unit M. - p. 6
This leads us to consider the.. Stress Tensor and Stress Types σ mn is the stress tensor and this has particular meaning. Figure M.-4 Infinitesimal element (a cube) representing a very small piece of a body x 3 σ 33 σ 31 σ 3 σ 13 σ σ 1 σ 1 σ 3 x (known as Differential Element) x 1 σ 11 Unit M. - p. 7
σ mn tells the face and direction of the stress stress acts in n-direction stress acts on face with normal vector in the m-direction Note: one important convention If face has positive normal, positive stress is in positive direction If face has negative normal, positive stress is in negative direction --> Also note that there are two types of stress: Acts normal to the face = Normal/extensional stress Acts in-plane of face = Shear/stress What do these terms mean? Normal/extensional -- extends element and Shear -- caused angular changes (think of deck of cards) Unit M. - p. 8
So there are (at first) 9 components of the stress tensor σ 11 σ 1 σ 1 σ σ 3 σ 3 σ 33 σ 31 σ 13 extensional shear But not all of these are independent due to the Symmetry of Stress Tensor One needs to consider the Moment Equilibrium of the differential element --> Note important notation/convention: Unit M. - p. 9
Figure M.-5 Consider x faces and stresses acting on such x 3 x σ σ σ + dx x dx - σ acts on one side - length of element in x direction is dx - on other side, stress is: base value σ + σ dx x σ + σ x dx infinitesimal length over which change is occurring rate of change of σ with respect to x Unit M. - p. 10
σ is field variable: σ ( x 1, x, x 3 ) Do the same with all stresses (and all directions!) Let s, using this concept, consider moment equilibrium about the x 1 axis (Thus, any stresses with a subscript of 1 do not contribute since they act parallel to x 1 or have no moment arm) Figure M.-6 Differential stress element under gradient stress field x 3 x 1 x σ σ 3 + 3 dx x 3 3 σ σ 3 + 3 dx σ x σ σ + σ 3 dx x σ 3 σ σ 33 + 33 dx x 3 3 dx 3 dx 1 σ 33 dx Unit M. - p. 11
Take moments about center of cube: M = 0 + + --> Note that σ, σ 33 and associated stresses act through m (no moment arm,) so they don t contribute. This leaves us with: σ 3 ( dx 1 dx 3 ) dx σ 3 ( dx 1 dx ) dx 3 + σ 3 + σ 3 dx x dx 1 dx 3 ( ) dx σ 3 + σ 3 dx 3 dx 1 dx x 3 (Note: Form is (stress) (area) (moment arm) ) Canceling out common dx 1 dx dx 3 and 1/ leaves: ( ) dx 3 = 0 σ 3 σ 3 + σ 3 + σ 3 x dx σ 3 + σ 3 x 3 dx 3 = 0 Unit M. - p. 1
The two derivatives are higher order terms (HOT s) which we can disregard to first order. Why? (differential) (differential) = very small So we re left with: σ 3 σ 3 + σ 3 σ 3 = 0 σ 3 = σ 3! symmetry! Similar moment equilibrium about the other two axes yields: (about x ) σ 13 = σ 31 (about x 3 ) σ 1 = σ 1 Thus, in general: σ mn = σ nm Stress tensor is symmetric Unit M. - p. 13
This leaves us with 6 independent components of the stress tensor σ 11 σ σ 33 σ 3 = σ 3 σ 13 = σ 31 σ 1 = σ 1 rotation about x 1 rotation about x rotation about x 3 Extensional Shear Thus far we ve just defined the stress tensor and we ve used moment equilibrium to show the symmetry of the stress tensor. But there are other relations between these stress components. As in most structural problems, we always apply equilibrium. This will give us the. Stress Equations of Equilibrium We can still apply the three equations of force equilibrium. This will give us three relations among the stress components Let s do this with the x 1 - direction. Thus, we must consider all stresses with a second subscript of 1 (implies acts in x 1 - direction): Unit M. - p. 14
Figure M.-7 Infinitesimal stress element and all stresses acting in x 1 - direction x 3 σ 11 σ σ 31 + 31 dx x 3 3 σ 1 σ31 σ σ 1 + 1 dx x x dx 1 dx 3 x 1 σ σ 11 + 11 dx x 1 1 dx Now, F 1 = 0 + Unit M. - p. 15
form is (stress) (area) 1 i 1 face: σ 11 + σ 11 dx 1 x dx dx 3 1 i face: σ 1 + σ 1 dx x dx 1 dx 3 3 i 3 face: σ 31 + σ 31 dx x 3 dx 1 dx 3 + f 1 dx 1 dx dx 3 = 0 ( ) ( σ 11 ) ( dx dx 3 ) ( ) ( σ 1 ) ( dx 1 dx 3 ) ( ) ( σ 31 ) ( dx 1 dx ) 1 3 body force (acts over entire volume) Canceling out terms and dividing through by the common dx 1 dx dx 3 gives: Unit M. - p. 16
σ 11 + σ 1 + σ 31 + f x 1 - direction x 1 x x 1 = 0 3 σ σ σ Similar work in the other two directions yields: σ 1 1 3 + σ + σ 3 x 1 x x 3 + f = 0 σ 13 + σ 3 + σ 33 x 1 x x 3 + f 3 = 0 These are the three equations of stress equilibrium. These can be summarized in tensor form as: σ mn x m + f n = 0 x - direction x 3 - direction These are the key items about stress, but we also need to talk a bit more about Unit M. - p. 17
Thus far we ve used tensor notation but other notations are used and one must be able to converse in all of these Although these are a member of different notations, the most important in engineering is: --> Engineering Notation (More) Stress Notation Here, the subscripts are the directions x, y, z rather than x 1, x, x 3 Tensor x 1 Engineering x x 3, z x x 3 y z x 1, x x, y In using the subscripts, only one subscript is used on the extensional stresses. Thus: Unit M. - p. 18
Tensor Engineering σ 11 σ x σ σ y σ 33 σ z σ 3 σ 13 σ 1 σ yz σ xz σ xy = τ yz = τ xz = τ xy sometimes used for shear stresses Finally, there is a --> Matrix Notation it is sometimes convenient to represent the stress tensor in matrix form: σ mn = ~ σ 11 σ 1 σ 13 σ 1 σ σ 3 σ 31 σ 3 σ 33 symmetric matrix Unit M. - p. 19
Finally we want to consider the case of Two-Dimensional Stress In many cases, the stresses of importance are in-plane (i.e., two dimensional). There is a reduction in equations, stress components and considerations. The assumption is to get the out-of-plane stresses to zero (they may be nonzero, but they are negligible). By convention, the out-of-plane stresses are in the x 3 - direction. Thus: σ 33 = 0 σ 13 = 0 σ 3 = 0 This leaves only σ 11, σ and σ 1 as nonzero. This is known as Plane Stress (also add the condition that = 0) x 3 Unit M. - p. 0
Many useful structural configurations can be modeled this way. more later (and especially in 16.0) Now that we know all about stress, we need to turn to the companion concept of strain. Unit M. - p. 1