Hooke’s Law and Modulus of Elasticity (2.1-2.7) MAE 314 – Solid Mechanics Yun Jing Hooke's Law and Modulus of Elasticity

Introduction to Normal Strain Hooke's Law and Modulus of Elasticity

Introduction to Normal Strain Normal strain (ε) is defined as the deformation per unit length of a member under axial loading. Normal strain is dimensionless but can be expressed in several ways. Let us assume L = 100 mm and δ = 0.01 mm. ε = 0.01 mm / 100 mm = 1 x 10-4 or 100 x 10-6 ε = 100 μ (read as 100 microstrain) ε = 1 x 10-4 in/in (if using English units) ε = 1 x 10-4 * 100 = 0.01% Hooke's Law and Modulus of Elasticity

Mechanical Properties of Materials It is preferable to have a method of analysis that is characteristic of the properties of materials (σ and ε) rather than the dimensions or load (δ and P) of a particular specimen. Why? σ & ε are truly material properties P & δ are specimen properties Hooke's Law and Modulus of Elasticity

Mechanical Properties of Materials Stress and strain can be measured, so we want to develop a relationship between the two for a given material. How do we calculate the elongation of a bar due to loading? Apply force P Calculate σ = P/A Use material relation ε = f(σ) to calculate ε Calculate δ = εL Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram Material behavior is generally represented by a stress-strain diagram, which is obtained by conducting a tensile test on a specimen of material. Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram Stress-strain diagrams of various materials vary widely. Different tensile tests conducted on the same material may yield different results depending on test conditions (temperature, loading method, etc.). Divide materials into two broad categories: Ductile material - Material that undergoes large permanent strains before failure (e.g. steel, aluminum) Brittle material - Material that fails with little elongation after yield stress (e.g. glass, ceramics, concrete) Let’s examine the stress-strain diagram for a typical ductile material (low-carbon steel) region by region. Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram (Low-carbon steel) Linear region Stress-strain response is linear Slope = Modulus of Elasticity (Young’s modulus) = E E has units of force per unit area (same as stress) We get a relation between stress and strain known as Hooke’s Law. Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram (Low-carbon steel) Yielding region Begins at yield stress σY Slope rapidly decreases until it is horizontal or near horizontal Large strain increase, small stress increase Strain is permanent Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram (Low-carbon steel) Strain Hardening After undergoing large deformations, the metal has changed its crystalline structure. The material has increased resistance to applied stress (it appears to be “harder”). Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram (Low-carbon steel) Necking The maximum supported stress value is called the ultimate stress, σu. Loading beyond σu results in decreased load supported and eventually rupture. Breaking strength, σB Hooke's Law and Modulus of Elasticity

Stress-Strain Diagram (Low-carbon steel) Why does the stress appear to drop during necking? If we measure the true area, the graph looks like: The difference is in the area: true stress takes into account the decreased cross- section area. Thus, at the same stress level, the load drops. true stress x Hooke's Law and Modulus of Elasticity

Hooke's Law and Modulus of Elasticity Offset Method For some materials (e.g. aluminum) there is not a clear yield stress. We can use the offset method to determine σY. Choose the offset (0.002 is shown here). Draw a line with slope E, through the point (0.002, 0). σY is given by the intersection of this line with the stress-strain curve. Hooke's Law and Modulus of Elasticity

Hooke's Law and Modulus of Elasticity Elastic vs. Plastic A material is said to behave elastically if the strain caused by the application of load disappear when the load is removed – it returns to its original state. The largest value of stress for which the material behaves elastically is called the elastic limit (basically the same as σY in materials with a well-defined yield point). Once the yield stress has been obtained, when the load is removed, the stress and strain decrease linearly but do not return to their original state. This indicates plastic deformation. When a material does not have a well-defined yield point, the elastic limit can be closely approximated using the offset method. Hooke's Law and Modulus of Elasticity

Hooke's Law and Modulus of Elasticity Elastic vs. Plastic sY sY Reload Plastic deformation (Permanent strain) Hooke's Law and Modulus of Elasticity

MAE 314 – Solid Mechanics Yun Jing Deformations (2.8-2.10) MAE 314 – Solid Mechanics Yun Jing Deformations

Deformations Under Axial Loading Conditions Solid bars, cables, coil springs, etc. Axial tension or compression Prismatic Recall Hooke’s Law ( ) and equations for stress ( ) and strain ( ). Deformations

Deformations Under Axial Loading What about non-prismatic bars? Discrete Changes: total change in length is simply the summation of the change in length of each portion. Important: Each time the internal force, area, or material changes you need a new free-body diagram! Deformations

Deformations Under Axial Loading What about non-prismatic bars? Continuous Changes: continuously changing area (as shown) or continuously changing force (such as a rod hanging under its own weight) Deformation of an element of length dx can be expressed as: Integrating this over the length of the rod: This is an approximation since we made the assumption earlier that the stress distribution is constant over the cross-section. For small variations this is a good approximation. Deformations

Hooke's Law and Modulus of Elasticity Example problem The rigid bar BDE is supported by two links AB and CD. Link AB has a cross-sectional area of 500mm^2, E=70GPa. Link CD has a cross-sectional area of 600mm^2, E=200GPa. For the 30kN force shown, determine the deflection of B, D and E. Hooke's Law and Modulus of Elasticity

Statically Indeterminate Problems Statically determinate structure – reactions and internal forces can be determined uniquely from free-body diagram and equations of equilibrium. Statically indeterminate structure – there are more unknown reactions than equations of equilibrium. Where do the other equations needed to solve the unknown reactions come from? Equations of compatibility which are based on displacements. Here is a easy method to determine how many compatibility equations you need for any given problem: M = R – N M = number of compatibility equations needed R = number of unknown reactions (or internal stresses) N = number of equilibrium equations Deformations

Statically Indeterminate Problems R = 1 : R1 N = 1 : ΣFY = R1 – P = 0 M = 1 - 1 = 0 R = 2 : R1, R2 N = 1 : ΣFY = R1 – P – R2 = 0 M = 2 - 1 = 1 R2 Deformations Statically indeterminate structure

Temperature Changes Changes in temperature produce expansion or compression, which cause strain. α = coefficient of thermal expansion ΔT = change in temperature Sign convention: expansion is positive (+), contraction is negative (-) For a bar that is completely free to deform (one or both ends free): In this case, there is thermal strain but no thermal stress! Deformations

Temperature Changes Thermal stresses occur when the bar is constrained such that it cannot deform freely. In this case there is thermal stress but no thermal strain! Statically Determinate Structures Uniform ΔT in the members produces thermal strains but no thermal stresses. Statically Indeterminate Structures Uniform Δ T in the members produces thermal strains and/or thermal stresses. Deformations

3 Types of Displacement Problems Statically Determinate Statically Indeterminate Temperature Change Deformations

More Mechanical Properties (2.11-2.15) MAE 314 – Solid Mechanics Yun Jing More Mechanical Properties

More Mechanical Properties Poisson’s Ratio When an axial force is applied to a bar, the bar not only elongates but also shortens in the other two orthogonal directions. Poisson’s ratio (υ) is the ratio of lateral strain to axial strain. υ is a material specific property and is dimensionless. lateral strain axial strain Minus sign needed to obtain a positive value More Mechanical Properties

More Mechanical Properties Poisson’s Ratio Let’s generalize Hooke’s Law (σ=Eε). Assumptions: linear elastic material, small deformations So, for the case of a homogenous isotropic bar that is axially loaded along the x-axis (σy=0 and σz=0), we get Even though the stress in the y and z axes are zero, the strain is not! More Mechanical Properties

More Mechanical Properties Poisson’s Ratio What are the limits on υ? We know that υ > 0. Consider a cube with side lengths = 1 Apply hydrostatic pressure to the cube Can write an expression for the change in volume of the cube P P P P P P εx, εy, εz are very small, so we can neglect the terms of order ε2 or ε3 More Mechanical Properties

More Mechanical Properties Poisson’s Ratio ΔV simplifies to Plug σ=P/A=P into our generalized equations for strain. Plug these values into the expression for ΔV. More Mechanical Properties

More Mechanical Properties Poisson’s Ratio Since the cube is compressed, we know ΔV must be less than zero. P P P P P P More Mechanical Properties

More Mechanical Properties Shear Strain Recall that Normal stresses produce a change in volume of the element Shear stresses produce a change in shape of the element Shear strain (γ) is an angle measured in degrees or radians (dimensionless) Sign convention More Mechanical Properties

More Mechanical Properties Shear Strain Hooke’s Law for shear stress is defined as G = shear modulus (or modulus of rigidity) G is a material specific property with the same units as E (psi or Pa). More Mechanical Properties

More Mechanical Properties Shear Strain Are the three material properties E, υ,and G related or do we have to determine each separately through testing? There is a relationship between them which is derived in section 2.15 in the textbook. An isotropic material has two independent properties. Once you know two of them (E, G, or υ), you can find the third. More Mechanical Properties

More Mechanical Properties Example Problem A vibration isolation unit consists of two blocks of hard rubber bonded to a plate AB and to rigid supports as shown. Knowing that a force of magnitude P = 6 kips causes a deflection of δ=0.0625 in. of plate AB, determine the modulus of rigidity of the rubber used. More Mechanical Properties