Stress, Strain and Failure Theories

Mohr circle stress transformation

A Mohr circle is a graphical form of the plane-stress transformation equations, in which every plane through a point corresponds to one point on a circle in the \sigma–\tau plane. For normal stresses \sigma_x, \sigma_y and shear stress \tau_{xy}, the centre lies on the \sigma-axis at C=(\sigma_x+\sigma_y)/2, and the radius is R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}. Rotating a physical plane by \theta rotates the corresponding point on the circle by 2\theta in the same sense, and the principal stresses are C\pm R. The sum of the two principal stresses equals \sigma_x+\sigma_y for every rotation of axes, because it is a stress invariant. In matrix form, a plane stress tensor [\sigma] becomes [\sigma']=[Q][\sigma][Q]^T in rotated axes. Since a rotation matrix is orthogonal, [Q]^{-1}=[Q]^T, so [Q][\sigma][Q]^{-1} and ([Q]^{-1})^T[\sigma][Q]^T are equivalent to this form, whereas [Q]^{-1}[\sigma][Q] describes the opposite rotation. The major principal stress is the algebraically largest normal stress at a point, but its magnitude need not be the largest, because a compressive minor principal stress can have a larger magnitude. For a plane at 45^\circ, the normal stress reduces to \sigma_n=\frac{\sigma_x+\sigma_y}{2}+\tau_{xy}, since \cos 90^\circ=0 and \sin 90^\circ=1. A useful check is the position of the circle. In pure bending each fibre carries a single normal stress, so its circle has its centre on the \sigma-axis at half that stress and passes through the origin, tangent to the \tau-axis there. A circle whose right-hand point lies at the origin represents compression, and one whose left-hand point lies at the origin represents tension. A circle lifted off the \sigma-axis cannot represent any stress state, because its centre must lie on that axis. For example, with \sigma_x=60, \sigma_y=20 and \tau_{xy}=30 MPa, C=40 MPa and R\approx36.1 MPa, so the principal stresses are about 76.1 MPa and 3.9 MPa.

Combined stress yield check

When several load effects such as axial force, bending moment and torque act at the same section, their stresses are superposed at each point and the resulting state is compared with the yield or ultimate strength of the material. The factor of safety is the ratio of the allowable stress to the applied equivalent or principal stress. The maximum stress (Rankine) criterion requires the largest principal stress to stay below the allowable value. For a thin-walled tube under axial load and torque, the stresses are \sigma=P/(2\pi rt) and \tau=Tr/J with J\approx2\pi r^3t, and the maximum principal stress is \sigma_1=\frac{\sigma}{2}+\sqrt{\frac{\sigma^2}{4}+\tau^2}. For a pure torsion tube the principal stresses are \pm\tau, so the maximum stress criterion limits the torque through \tau\le\sigma_u, which gives T=\tau\cdot2\pi r^2t for mean radius r and thickness t. Bending from a lateral wind or eccentric load on a pole is found from the base moment M=Fh, with stress \sigma=Mc/I. For a hollow square section with outer side a and inner side b, I=(a^4-b^4)/12 and c=a/2, so setting \sigma equal to the yield stress gives the minimum inner dimension. For plane stress, the von Mises equivalent stress is \sigma_e=\sqrt{\sigma_{xx}^2-\sigma_{xx}\sigma_{yy}+\sigma_{yy}^2+3\tau_{xy}^2}, and the factor of safety against yield is FS=\sigma_Y/\sigma_e.

Thin-walled pressure vessel stresses

For a thin-walled cylinder of radius r and wall thickness t under internal pressure p, the hoop stress is \sigma_\theta=pr/t. With closed ends, the longitudinal stress is \sigma_L=pr/(2t), so the hoop stress is exactly twice the longitudinal stress. Hooke's law then gives the hoop strain as \varepsilon_\theta=(\sigma_\theta-\nu\sigma_L)/E and the axial strain as \varepsilon_L=(\sigma_L-\nu\sigma_\theta)/E. Requiring a given ratio between these two strains fixes Poisson's ratio. For a closed-end cylinder, the von Mises equivalent stress is \sigma_e=\sqrt{\sigma_\theta^2-\sigma_\theta\sigma_L+\sigma_L^2}=\sqrt{3}\,\sigma_L=\frac{\sqrt3\,pr}{2t}. Setting \sigma_e equal to the yield strength gives the failure pressure p=\frac{2\sigma_Y t}{\sqrt3\,r}, which depends only on the radius-to-thickness ratio.

Bending stress in unsymmetric sections

In pure bending the normal stress is \sigma=-My/I, where y is measured from the neutral axis, which passes through the centroid for bending about a principal centroidal axis. The magnitude is therefore largest at the fibres farthest from the neutral axis. In a T-section, the centroid lies nearer the wide flange, so the extreme fibre on the side away from the flange lies farther from the neutral axis and carries the maximum magnitude of normal stress, while the fibres near the flange carry less. A transverse load applied through the shear centre produces no torque about the beam axis, so no twisting occurs. For an unsymmetric section, the principal axes are not aligned with the load, so the neutral axis is inclined and the deflection is not parallel to the load. The beam therefore undergoes bending only, although the bending is biaxial. Thus the shear centre location determines whether torsion appears, and the section's orientation determines whether the deflection and the load are collinear.

3D isotropic Hooke's law matrix

For an isotropic linear elastic material, the three-dimensional stiffness matrix is built from the Lamé constants \lambda and \mu. The diagonal normal terms are P=\lambda+2\mu, the off-diagonal normal terms are Q=\lambda, and the shear terms are R=\mu=G when the shear strain is the engineering strain \gamma. Because P-Q=2\mu, the shear constant follows from R=(P-Q)/2. This relation is the matrix-level statement that shear and normal stiffnesses of an isotropic material are linked. The same material also obeys G=\frac{E}{2(1+\nu)}, so E=2G(1+\nu), and only two of E, G and \nu are independent. For example, a material with \nu=0.3 and E=200 GPa has G\approx77 GPa.

Shear strain from deformed geometry

The engineering shear strain \gamma_{xy} is the decrease of the right angle between two material lines that are initially along x and y. For a displacement field u(x,y) and v(x,y), it is \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}, while the normal strains are \varepsilon_{xx}=\frac{\partial u}{\partial x} and \varepsilon_{yy}=\frac{\partial v}{\partial y}. The change in the length of a diagonal gives the normal strain along that diagonal. When the diagonals stay along the coordinate axes and remain perpendicular, the shear strain is zero even though the lengths change. A linear displacement field with all strain components equal to zero describes a rigid body motion. The rotation is \omega=\frac12\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right), so a field such as u=-\omega y, v=\omega x rotates the body as a whole without changing its shape.

Brittle vs ductile failure planes

Brittle materials usually fail by fracture when the maximum normal stress reaches the fracture strength, so the crack plane is perpendicular to the direction of the maximum principal stress. In uniaxial tension this is the transverse plane, which carries no shear stress. Ductile materials yield and fail mainly by shear, so the critical plane is the plane of maximum shear stress. In uniaxial tension this plane is inclined at 45^\circ to the load axis, which is why ductile specimens show slant shear faces in a cup-and-cone fracture. The choice of the critical plane therefore depends on the material's failure mode, not only on the applied load.

Maximum shear stress plane orientation

The maximum in-plane shear stress equals the radius of Mohr's circle, \tau_{max}=R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}. The planes carrying it are oriented by \tan 2\theta_s=-\frac{\sigma_x-\sigma_y}{2\tau_{xy}}, which places them at 45^\circ to the principal planes. On these planes the normal stress equals the average (\sigma_x+\sigma_y)/2 on both faces. The sign of \tau_{xy} and the chosen sign convention determine whether \theta_s is positive or negative, so the angle must be checked against the sign of the shear stress on the element face. For example, if \sigma_x=60, \sigma_y=20 and \tau_{xy}=0, then \tau_{max}=20 MPa on the planes at 45^\circ.