Read Chapter 1 Basic Elasticity - Equilibrium Equations - Plane Stresses - Principal Stresses - Mohr’s Circle - Stress Strain Relationships

Stress Equilibrium under external forces Continuous and deformable material Forces are transmitted throughout its volume

Stress Resultant force at point O  δP Force must be in equilibrium: Equal and opposite force δP on the particle Divide the particle along plane nn containing O δP can be considered as uniformly distributed over a small area δA

Stress Normal or direct stress Shear stress

Equilibrium Equations First two terms Taylor Series Expansion Plane Stresses Body Forces per unit mass times density

Equilibrium Equations Taking sum of the moments about an axis through the centre line of the element parallel to the z axis it would be found that:

Determination of stresses on an incline plane Body forces ignored (second order term) δx and δy are small  stress distribution is assumed uniform

Determination of stresses on an incline plane Sum of the Forces perpendicular to ED: Sum of the Forces parallel to ED:

Example Cantilever beam of solid cross-section Compressive load of 50 KN, 1.5 mm below horizontal diameter plane Torque of 1200 Nm Calculate direct and shear stresses on a plane 600 to the axis of the cantilever beam on a point located the lower edge of the vertical plane of symmetry

Example cont’d

Example cont’d θ = 300 σy = 0

Example cont’d

Principal Stresses Principal stresses determine the maximum or minimum value given a loading stress, σn Starting from the plane stress equation, derivating with respect to θ and equating to zero, we can obtain an expression of the maximum and minimum stresses: Two solutions are obtained: θ and θ + π/2 These planes correspond to those on which there is no shear stresses The direct stresses on these planes (principal planes) are called principal stresses

Principal Stresses Derivation on page 16 Megson. Maximum principal stress Minimum principal stress If negative (compressive) could be numerically larger that σI

Principal Stresses Similarly, we can find the maximum and minimum shear stresses (principal): Two solutions are obtained: 450 inclined from principal planes These planes correspond to those on which there is no normal stresses The shear stresses on these planes (principal planes) are called principal shear stresses

Principal Stresses Occur 450 inclination from principal planes Principal stresses Summary:

Mohr’s Circle Graphical Representation of the stresses at a point in a deformable body Recall Stresses on an incline plane: The stress equation might be rewritten as (using trig relations):

Mohr’s Circle Squaring: Adding: Equation of Mohr’s Circle:

Mohr’s Circle Circle of radius:

Mohr’s Circle Example: Direct stress of 160 N/mm2 (tension, x direction) 120 N/mm2 (compression, y direction) Applied to elastic material on two mutually perpendicular planes Principal stresses are limited to 200 N/mm2 CALCULATE THE ALLOWABLE VALUE OF SHEAR STRESS ON THE GIVEN PLANES DETERMINE ALSO THE VALUE OF THE OTHER PRINCIPAL STRESSES AND THE MAXIMUM SHEAR STRESS AT THAT POINT

Mohr’s Circle

Strain Let’s Look at εx:

Strain Higher order powers of are ignored Similarly

Compatibility Equations Since the six strains are defined in terms of three displacement functions then they must bear some relationship to each other Derivation of these equations is described in Megson 1.10.

Plain Strain We have a 3D compatibility equation and expressions for strain We shall concern ourselves with 2D problems (displacement on only one plane, xy) Then εz, γxz, γyz become zero Compatibility equation plain strain

Principal Strains

Stress-Strain Relationships So far, we have three equations of equilibrium for 3D deformable body & Six strain-displacement relationships TOTAL: 9 independent equations towards the solution of the 3D stress problem The total of unknowns is 15: - 6 stresses - 6 trains - 3 displacements

Stress-Strain Relationships So far we have not made any assumptions regarding the force-displacement relations (i.e. stress-strain relations) For isotropic materials (homogenous) experiments show that: Modulus of Elasticity, or Young’s Modulus

Stress-Strain Relationships For isotropic materials

Material Constitutive Equations Sample problem: E1, A1, L, α1 E2, A2, L, α2 Rigid Rigid What are the stresses on the bars when exposed to a change in temperature equal toT0? E1, A1, L, α1

Sample Problem (Plane Stresses) FBD F1 F1 F2 F2 F3 F3 Sum of the Forces: 2F1 + F2 = 0 Solve for and

Sample Problem FOS Given the information below, determine the limit and ultimate loads of the structure as well as the corresponding margins of safety.