Tweet
Login
Mathematics Crystal
You may switch between
tex
and
pdf
by changing the end of the URL.
Home
About Us
Materials
Site Map
Questions and Answers
Skills
Topic Notes
HSC
Integration
Others
Tangent
UBC
UNSW
Calculus Advanced
Challenges
Complex Numbers
Conics
Differentiation
Integration
Linear Algebra
Mathematical Induction
Motion
Others
Polynomial Functions
Probability
Sequences and Series
Trigonometry
/
QnA /
UBC /
UBC MECH326 Stress and Strain.tex
--Quick Links--
The Number Empire
Wolfram Mathematica online integrator
FooPlot
Calc Matthen
Walter Zorn
Quick Math
Lists of integrals
List of integrals of trigonometric functions
PDF
\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large MECH326 Stress and Strain} \begin{align*} \text{\bf Stress:}&\quad\text{(1-9 p.15)}\\ &\text{Stress is a state property at a specific point within a body, which is a function of load, geometry, temperature,}\\ &\text{and manufacturing processing. Stresses are discussed mostly in the context of external forces (load-induced)}\\ &\text{but other stresses (e.g. thermal stresses) are discussed in more advanced context. When the source of stress}\\ &\text{is removed, the stress will return to zero, unless it has reached a certain limit (strength).}\\ \\ &\text{Stress is a vector, and therefore directional. E.g. normal stress $\sigma$, shear stress $\tau$, and normal stress $\sigma_1$.}\\ \\ \text{\bf Streng}&\text{\bf th:}\quad\text{(1-9 p.15)}\\ &\text{Strengths are the magnitudes of stresses at which something of interest occurs (usually loss of functions), such}\\ &\text{as the proportional limit yielding, or fracture (usually when loss of functions occurs). Depending on this ``thing of}\\ &\text{interest'', there are different types of strengths: yield strength $S_y$, ultimate strength $S_u$, shear yield strength $S_{sy}$}\\ &\text{and endurance strength $S_e$ etc.}\\ \\ %&\text{Strength is a property of a material, and depends on the choice, the treatment, and the processing of the material.}\\ &\text{Strength is an inherent property of a part. The geometry of the part has an effect on the stress it can take.}\\ &\text{This follow that different point of a part may have different strength.}\\ \\ &\text{Strength is a scaler, but as it is associated with the stress vector, even at the same point, strength may be higher in}\\ &\text{one direction but lower in another.}\\ \\ \text{\bf Uncer}&\text{\bf tainty:}\quad\text{(1-10 p.16)}\\ &\text{The strength is measured \it nominally \rm in the lab with an uncertainty. The stress in the field (where the part is used)}\\ &\text{is also measured (or predicted) with an uncertainty.}\\ \\ &\text{For example, the maximum load of the passenger lift in a building may vary by $\delta_\sigma$, say 15\% (because people's weight}\\ &\text{cannot be predicted precisely). The ``maximum allowable load'' $\sigma$ as shown in the lift needs to be $\delta_\sigma$ lower than the}\\ &\text{\it real \rm maximum load that the lift can take ($\sigma_o$).}\\ % &\sigma=\sigma_o-\sigma_o\delta_\sigma,\quad\sigma_o=\sigma\times\frac{1}{1-\delta_\sigma}.\quad \text{($\sigma$ needs to be increased by the loss-of-function parameter $\frac{1}{1-\delta_\sigma}$, or $\frac{1}{0.85}$.)}\\ \\ &\text{On the other hand, the lift manufacturer has measured the strength of the lift and found it to be $S$, say 20000 N,}\\ &\text{with an uncertainty of $\delta_S$, say 20\%. So $S$ may be $\delta_S$ higher than its \it real \rm strength $S_o$.}\\ % &S=S_o+S_o\delta_S,\quad S_o=S\times\frac{1}{1+\delta_S}.\quad \text{($S$ needs to be decreased by the maximum allowable parameter $\frac{1}{1+\delta_S}$, or $\frac{1}{1.2}$.)}\\ \\ &\text{Let us define the design factor}\quad n_d\equiv\frac{S}{\sigma}.\\ &\text{The \it loss-of-function strength \rm must be stronger than the \it maximum allowable load \rm by at least $n_d$ times.}\\ \\ &\text{For safety reasons, we must assume the worst and have the \it real \rm load always under the \it real \rm strength.}\\ % &\because\quad\sigma_o
\frac{\text{loss-of-function-parameter}}{\text{maximum-allowable-parameter}}.\\ %\\ %&\text{20000 decreased by 1/1.2 and $\sigma$ (maximum allowable load) increased by 1/0.85.}\\ %&\sigma\times\frac{1}{1-\delta_\sigma}
0$ radial clearance; $r_{int}-r_{ext}>0$ radial interference;}\\ &\text{Allowance: minimum clearance or maximum interference.}\\ \\ &\text{\underline {Tensile and Compressive Strengths}}\\ &\text{Stress}\quad\sigma=\frac{P}{A}\quad\text{where $P$ is load and $A$ is cross section (perpendicular to $P$).}\\ % &\text{Within the \it proportional limit \rm,}\quad\sigma=E\epsilon,\quad \text{where $E$ is the \it Young's Modulus.\rm}\quad\ldots\text{the Hooke's Law.}\\ % &\epsilon=\frac{\sigma}{E}=\frac{P}{AE}.\quad d\delta=\epsilon~dx,\quad \delta=\int_0^L\epsilon~dx=\int_0^L\frac{P}{AE}~dx=\frac{PL}{AE}.\\ % &\text{Different points may have different stress. The stress concentration factor }K=\frac{\sigma_{max}}{\sigma_{avg}},\quad\sigma_{max}=K\frac{P}{A}.\\ \\ &\text{Strain}\quad\epsilon=\frac{l-l_0}{l_0}.\quad\text{A \it stress-strain \rm diagram has strain as the horizontal axis and stress vertical.}\\ &\text{Over a range, }d\epsilon=\frac{dl}{l},\quad\epsilon=\int_{l_0}^l\frac{dl}{l}=ln\frac{l}{l_0}.\\ \\ &\text{The material will restore before the \it elastic limit. \rm Beyond that, it is plastic (permanently set even}\\ &\text{after the load is removed).}\\ \\ &\text{Beyond the \it yield point, \rm the strain will increase rapidly without corresponding increase in stress.}\\ &\text{The \it ultimate \rm strength correspond to the maximum stress, from which the strain will increase}\\ &\text{even regardless of the stress (which is generally decreasing as it cannot hold the load anymore).}\\ \\ &\text{When a part is under vertical tensile stress, the Poisson's Ratio $\nu=\frac{-\epsilon_{trans}}{\epsilon_{axial}}$ indicates the necking effect.}\\ &\text{where $\epsilon_{trans}<0$ is the stress perpendicular to the tension, and $\epsilon_{axial}>0$ parallel.}\\ \\ &\text{\underline {Torsional Stress}}\\ &\text{The shear stress $\tau$ of a bar of length $l_0$ at a point $\rho$ from the axis is related to the twisted angle $\theta$ by:}\\ &\tau=\frac{G\rho}{l_0}\theta.\quad\text{At the outer radius $r$, the maximum shear stress $\tau_{max}=\frac{Gr}{l_0}\theta$.}\\ &\text{$G$ is the material stiffness property called \it shear modulus \rm or \it the modulus of rigidity\rm.}\\ % &\tau=\left(\frac{\rho}{r}\right)\tau_{max}.\quad \text{The unit of $\tau$ is Pa (or Nm$^{-2}$), same as G.}\\ \\ % &\text{An imaginary line drawn parallel to the axis but $\rho$ away from it will deviate by an angle $\gamma$ after twisting.}\\ &\gamma=\frac{\rho\theta}{l_0}\quad\text{when $l_0\gg \rho$}.\quad \tau=\gamma G.\qquad \gamma_{max}=\frac{r\theta}{l_0},\quad \gamma=\left(\frac{\rho}{r}\right)\gamma_{max}.\\ % &\text{Second Moment of Area (or Polar Moment of Area): }J=\int_A\rho^2 dA.\quad \text{For a circle, }J=\int_0^r\rho^2(2\pi\rho~d\rho)=\frac{\pi r^4}{2}.\\ &\text{Shear force at $\rho$: }dV=\tau dA=\frac{\rho}{r}\tau_{max}~dA.\quad \text{Torque }dT=\rho~dV,\quad T=\int_A\rho~dV=\int_A\rho\cdot\frac{\rho}{r}\tau_{max}~dA=J\frac{\tau_{max}}{r}.\\ &\boxed{\tau_{max}=\frac{Tr}{J}.}\quad \text{Generally, }\tau=\frac{T\rho}{J}.\quad \gamma=\frac{\tau}{G}=\frac{\rho\theta}{l_0},\quad \theta=\frac{\tau l_0}{\rho G}=\frac{T\rho}{J}\cdot\frac{l_0}{\rho G}.\quad \boxed{\theta=\frac{Tl_0}{JG}.}\quad \frac{T}{J}=\frac{G\theta}{l_0}=\frac{\tau}{\rho}.\\ \\ \end{align*} \end{document}