Strength of Materials

Preface

This book is a first course in the analysis of structures. Although most of the material should be accessible to all students who have had a mechanics course, a previous exposure to Engineering Mechanics would be useful. There are no mathematical prerequisites, though some elementary calculus would be useful in certain sections which can be skipped without affecting the flow of the book.

[edit] Introduction

Solid Mechanics is the study of load carrying members in terms of forces, deformations, and stability. In this book, we take the continuum mechanics approach, where we take the material properties to be the same even when we consider infinitesimal areas and volumes. The alternative approach is to build up material properties from basic equations relating atomic forces and interactions, and extending it to larger sets of such entities (e.g., molecular dynamics). Other approaches include meso-scale theories like dislocations to explain behavior of materials, especially for plasticity. However, the approach in this book is applicable to a large variety of practical problems, where we deal with structural members, usually under elastic deformation.

The above images show a body which is acted upon by external forces F₁, F₂, etc. The supports provide reaction forces (R₁, R₂) so that the body is stationary.

In addition to external forces, which act at points on the surface, we can also have forces which are body forces, i.e., they act on each point in the body. Gravity is an example of a body force.

Consider a body which has several external forces acting on it. If we take an imaginary section of the body, each of the parts should also be in equilibrium. Now, consider an elemental area dA in that section. Let the force acting on the elemental area so that the whole body is in equilibrium be dF. We can resolve this force in three mutually perpendicular directions. Let the component of the force normal to the surface be dF_z. Let the components in the plane of dA be dF_x and dF_y respectively.

[edit] Stress

Stress is defined on the average as the force divided by the area of the body over which the force acts. More precisely, we can talk about a stress at a point, or simply a stress, in the limiting case where the area (an infinitesimal area dA) $I n s e r t f o r m u l a h e r e$ go to zero. The normal stress occurs due to the infinitesimal force normal to the infinitesimal area, while shear stresses occur due to the infinitesimal force in the plane of the infinitesimal area. (Note: strictly speaking, stress does not exist. Elongation ( $ε$ ) and stiffness (E) do. Stress is a convenient mathematical construct that allows easy manipulation of formulae)

Consider a force, dF, acting on an elemental area, dA, at any arbitrary angle. τ_xx = dF_x/dA is the normal stress, also denoted as σ_x. The shear stresses are given by τ_xy = dF_y/dA and τ_xz = dF_z/dA, where dF_y and dF_z are the y and z components of force dF respectively. These shear stresses are in the plane y-z.

Now, instead of an infinitesimal area, consider an infinitesimal volume at the point in question. Let this volume be a parallelepiped with the sides dx, dy, and dz. In this case there are, in general, nine non-zero stresses. They form a stress tensor represented by the following 3×3 matrix.
$\left( \begin{matrix} \sigma_x & \tau_{xy} & \tau_{xz}\\ \tau_{yx} & \sigma_y & \tau_{yz}\\ \tau_{zx} & \tau_{zy} & \sigma_z \end{matrix} \right)$
Accordingly, in the plane stress case (taking, for instance, the plane x-y) we will have a four component stress tensor. Further, in view of the shear stress symmetry (τ_xy = τ_yx = τ), the above 3×3 matrix reduces to the 2×2 symmetric matrix representing a symmetric plane stress tensor.
$\left( \begin{matrix} \sigma_x & \tau \\ \tau & \sigma_y \end{matrix} \right)$
The shear stress symmetry holds in the three-dimensional case. This makes the above 3×3 matrix–and hence the corresponding stress tensor–symmetric: only three of its six shear stress components are independent ones. Consequently, the stress tensor has in general six different components: three normal stresses and three shear stresses. As already shown above, in the two-dimensional case the stress tensor has only three different components: two normal stresses and one shear stress.

Equilibrium in y-direction
Suppose the body is acted upon by the forces f_x and f_y (per unit volume) in the x and y directions respectively. It then can be shown that the stress equilibrium equations, in Cartesian coordinates, has the form:
$\frac{\partial \sigma_x}{\partial x} + \frac{\partial \tau}{\partial y} = f_x$
$\frac{\partial \tau}{\partial x} + \frac{\partial \sigma_y}{\partial y} = f_y$
Thus, in the plane stress case, we have two equations involving three unknowns.

Now turn to an example of the simplest one-dimensional loading of a body. Consider a rod, pulled at each end, along the longitudinal axis.

If we take a section perpendicular to the axis, it is easy to see σ_x = P/A where P is the load and A is the cross sectional area of the section.

Now consider the case where the cross section is angled θ with respect to the longitudinal axis. In this case, we can resolve the force P along and perpendicular to the section. The force components are, P cosθ and P sinθ respectively. The area of the cross section is now A / cosθ. The normal stress is σ_θ = P cosθ/(A/ cosθ) = (P/A) cos²θ. Similarly, the shear stress is τ_θ = (P/A) sinθ cosθ.

[edit] Strain

The strain in a material is defined as the fractional change in length, and is a dimensionless quantity. It is denoted by the symbol ε, where ε = (L − L₀)/L₀. Here L₀ is called the gauge length.

The strain defined above is called the engineering strain or the nominal strain, and is different from natural strain or true strain, which is defined as

$\epsilon_n = \int_{L_0}^L \frac{dL'}{L'} = \ln \left( \frac{L}{L_0} \right) = \ln (1 + \epsilon)$

Note that the engineering strain and natural strain are the same for small values of ε, and for the most part for the kinds of loads and displacements in this book, the differences are not relevant. Also, it is not uncommon to use the symbol e to represent engineering strain and the symbol ε to represent true strain.

In the infinitesimal case, we have, strain ε_x = du/dx, where du is the change in length for the segment dx. Thus, the total change in length is given by Δ = ∫₀^L du = ∫₀^L ε_x dx.

[edit] Hooke’s Law

Hooke’s law states that the stress is linearly related to strain for some materials. This is an empirical law by Robert Hooke, who observed this behavior in springs. Thus, for 1 dimension we have,

$σ = E ε$

where E is the constant of proportionality called Young's modulus.

Note that we have considered this value of E to be the same in all directions. Materials with whose properties don’t have a directional variation are called isotropic materials. Materials which have different properties in different directions are called anisotropic. The most common cause of anisotropy is the crystalline nature of materials. However, most common structural materials do not have the same crystal orientation over large ranges. Another cause for anisotropy is the kind of processing done on a material. Some processes like drawing tend to create stresses in a particular direction.

Applying Hooke’s law to the definition of length change, we have,

$\Delta = \int_0^L\frac{\sigma}{E}dx = \int_0^L\frac{P}{AE}dx$

Or,

$\Delta = \frac{PL}{AE}$

From mechanics we know that a spring has linear variation of extension with force, the constant of proportionality usually denoted by k. Thus, the spring constant for a beam under axial load is AE/L. This can be extended to components of different shapes, so that a structure is an assembly of springs arranged in a complicated manner.

[edit] Poisson’s Ratio

The longitudinal strain is linearly related to lateral strain in many materials. The ratio of the lateral to the longitudinal strain is called Poisson’s ratio and is denoted by ν.
$\nu = \left| \frac{\epsilon_{lat}}{\epsilon_{long}} \right|$
The value of Poisson’s ratio varies from material to material, and is about 0.30 for most common metals.

Recently,the discovery of re-entrant materials has excited much interest. These are unique honeycombs and foams which exhibit negative Poisson’s ratios. The allowable range of Poisson’s ratio in three dimensional isotropic solids is from -1 to 1/2 (though some polymer foams have been found to exhibit a value of 1.0).

[edit] Temperature Effects

Materials change length due to changes in temperature. To quantify this value, the length and temperature must be referenced to a determined temperature. The rate of change is quantified by the coefficient of linear thermal expansion, denoted by variables, CoF & α, By definition of strain, we have, ε = α ΔT, where ΔT is the change in temperature (compared to a given reference point). If we have a structural member between rigid supports, a strain of ΔT develops within the member. There is also an associated stress, which is refered to as a thermal stress. Structural members whose cross-section is composed of materials with differing CoFs, present a much more complex problem.

[edit] Energy Stored due to Deformation

When a rod is compressed, the work done on it is stored as energy. In the above figure, the energy stored is shown as the shaded area. Now, we know that the energy stored in a spring is 1/2 k x², for a spring constant of k and extension x. For a rod, whose constant is AE/L and extension is PL/AE, the energy stored is 1/2 P²L / AE. From the above equation, we see that it is better to use long bolts rather than short ones as they will have lower peak load P for the same diameter and material, without doing any dynamical analysis.

[edit] Compatibility Conditions

In an Engineering Mechanics course, we use force equilibrium to solve for forces in individual members. However, there were cases where it was not possible and such problems were considered statically indeterminate problems. They require some constitutive equations to solve completely, and Hooke’s law is such an equation.

For instance, consider a column fixed at both ends, and a load P applied at the center. Now, we know that reaction forces R₁ and R₂ act at either ends. Further, using force equilibrium, we have R₁ + R₂ = P. Here we have two unknowns, R₁ and R₂, but only one equation. We can apply Hooke’s law, with displacements Δ₁ and Δ₂, where Δ₁ = R₁L/2AE and Δ₂ = R₂L/2AE. The second equation is Δ₁ = Δ₂, since the supports are fixed. Thus, we can solve the problem.

[edit] Finite Element Method

Consider a rod with different cross sections A_i and lengths L_i, so that each section has an equivalent spring constant of A_iE/L_i. If the forces at each node (where the area changes) are F_i, then we can write equations in terms of displacements of each node and the spring constant of each section for the force F_i. If the beam is fixed at both ends, then the forces can only act on the other nodes.

In the general case, if you have loads being applied to connected thin rods, you can use this method to solve for forces and deformations. The whole structure can be discretized, and the nodes numbered so that we get the matrix [k_ij].

This procedure of breaking down a complicated problem into a large number of simple ones is called the Finite Element Method.

The individual pieces are called elements, and in the above case we have used the line element. For a general 2-D problem, we would need to use triangular elements at least. The matrix for a typical problem will contain hundreds of such elements and the size of the matrix be same as the number of elements. However, as most of the members are zero, the matrix is called a sparse matrix, and it can be solved much faster than the typical matrix you would encounter, say, in a mathematics course.

[edit] Saint-Venant’s Principle

So far we have been dealing with point loads acting at the ends of beams. We have assumed that the stress is the same as σ_av = P/A. However, this is not true close to the point of application of the load. Saint-Venant’s principle (by the French elastician Saint-Venant) states that the stresses remote from the point of application of the load are not affected by the precise behaviour of the structure close to the point of application of the load. In the drawing shown, this means that although the stress field at A-A might be hard to calculate, at B-B the stress can be approximated as P/A. S-V’s principle is consistent with many years of experiment and analysis, but has not been proven.

[edit] Stress Concentration Factors

So far we have assumed that the stress at all points in a cross section is the same, and by and large that is correct. However, there is an increase in stress near the point loads and discontinuities. The peak loads are a function of the average stress calculated by methods stated previously. The correct peak loads can only be obtained by methods like finite element analysis. However, charts have been published for several typical scenarios. As mentioned earlier, the peak stresses σ_p is a function of the average stress σ_av. Now, we have σ_p = K×σ_av, where K is called the stress concentration factor and can be looked up from tables for common scenarios. The study of stress concentration factors is also a design guideline, and indicates the types of features that you should avoid in a real world situation. For instance, the stress concentration factor for a fillet radius increases rapidly beyond r < 0.1 d, so you should specify large fillet radii for your shafts or eliminate the change in diameter.

The above schematic shows the possible values of the stress concentration ratios K for a plate with a hole in it. The stress at the narrowest section is calculated based on d = D − 2r, where r is the radius of the hole and D is the width of the slab. As can be seen, the stress concentration is not significant till about r/d = 0.3.

Note that local stresses might not be as high as that implied by the stress concentration factors due to plastic flow of the material, since the stress concentration analysis is done for elastic flow.

[edit] Elastic and Plastic Deformation

So far, we have considered members which deform elastically when a load is applied. Some of them obey Hooke’s law, so that the relationship between stress and strain is linear. Elastic deformation refers to the ability of the material to regain its original shape after the external load is removed. However, we know that for large loads, the material deformation is permanent, and this is called plastic deformation. Metals can undergo last plastic deformation, and metals with such a property are called ductile. Materials which cannot undergo plastic deformation are called brittle.

Materials for which the elastic region is very small are called plastic materials. The above image shows the idealization for such a material. The material flows when a certain stress state is reached.

Materials which are elastic for large deformations are called elastic materials. The above image shows an ideal elastic material–in this case, it also obey’s Hooke’s law, and is called a linearly elastic material. Note that a substance can be elastic and still not obey Hooke’s law (non-linearly elastic).

Most metals undergo elastic deformation for small loads and deform plastically for larger loads. There is certain amount of increase in stress when the material deforms plastically for most metals. However, the elastic plastic approximation shown above is good enough simple analysis.

Note that material which exhibit such behavior will show what is known as elastic recovery, where the sample will show some tendency to the earlier state. In the above image, the amount of recovery is Δ after the load has been removed. Note that there is a residual deformation in the material in the unloaded state.

Some materials show increased stress during plastic flow, with a phenomenon called strain hardening. In this case they can be approximated with a stress strain curve with two slopes, but this kind of analysis is rarely used in practice as most structural deformation is in the elastic zone, and plastic flow will most likely be considered a failure mode.

[edit] Shear Strain and Modulus

Earlier we introduced stress-strain relationship for materials under an axial load, where we found that the stress σ was proportional to the strain ε. If we plot the shear stress τ against the shear strain γ, we get a straight line for material which obey Hooke’s law. Thus, Hooke’s law can be extended to shear stress using the following equation: τ = Gγ, where G is the shear modulus, and γ is the shear strain.

[edit] Generalized Hooke’s Law

Poisson’s ratio shows the amount of change in dimension that occurs in another direction when a certain change occurs in one direction. The strain in each direction is the superposition of all the strains–the direct strains and the ones due to Poisson’s ratio. We can therefore state the generalized Hooke’s law as follows:

$\epsilon_x = {\sigma_x \over E} - {\nu\sigma_y \over E} - {\nu\sigma_z \over E}$

$\epsilon_y = {\nu\sigma_x \over E} - {\sigma_y \over E} - {\nu\sigma_z \over E}$

$\epsilon_z = {\nu\sigma_x \over E} - {\nu\sigma_y \over E} - {\sigma_z \over E}$

$\gamma_{xy} = {\tau_{xy} \over G}$

$\gamma_{yz} = {\tau_{yz} \over G}$

$\gamma_{zx} = {\tau_{zx} \over G}$

Note that these equations apply only to isotropic materials (steel, aluminum, copper, etc.) whose elastic properties are identical in all directions.

[edit] Relationships Between Moduli

Consider an element which has stresses σ₁ and σ₂ acting on it along the x and y directions respectively. Now consider further that σ₁ = σ and σ₂ = −σ.

The shear modulus can be expressed in terms of the Young’s modulus:

$G = \frac{E}{2 \left( 1 + \nu \right)}$

Consider a body undergoing compression, so that the strains are ε_x, ε_y, and ε_z respectively. The fractional change in volume is thus ε_x + ε_y + ε_z. Applying the generalized version of Hooke’s law, we have, the bulk modulus k is given by

$k = \frac{E}{3 \left( 1 - 2\nu \right)}$

[edit] Pressure Vessels

Vessels, tanks, and pipelines that carry, store or receive fluids, at a pressure different than its surroundings, are called pressure vessels. Pressure vessels usually are thin walled structures containing fluids at high pressure. The design decision is to find the value for the thickness t of the pressure vessel for a given maximum pressure p. Since the thickness is much smaller than the radius of the vessel r, the inner and outer radius are taken to be equal. For the longitudinal direction, we have the force due to pressure πr²p. This is balanced by the stress developed in the walls, acting on the circumference of the circle 2πrtσ. Thus the stress in the longitudinal direction is pr/2t. For the tangential direction, the pressure force is π2rl and the stress force is 2tlσ. The stress is pr/t.

[edit] Torsion

Torsion acting on a long bar tends to twist it in the direction of the torque. The analysis is performed on a section perpendicular to the axis, and the sum of the internal resisting torque is set equal to the external torque acting on the system. The following assumptions will be made for the bars studied in this chapter:

Plane sections remain plain after the torque is applied.
The shear strain γ varies linearly in the radial direction.
The material is linearly elastic, so that Hooke’s law applies.

[edit] Torsion Formula

We want to find the maximum shear stress τ_max which occurs in a circular shaft of radius c due to the application of a torque T. Using the assumptions above, we have, at any point r inside the shaft, the shear stress is τ_r = r/c τ_max.

∫τ_rdA r = T

∫ r²/c τ_max dA = T

τ_max/c∫r² dA = T

Now, we know,

J = ∫ r² dA

is the polar moment of intertia of the cross sectional area..

Thus, the maximum shear stress

τ_max = Tc/J

The above equation is called the torsion formula.

Now, for a solid circular shaft, we have,

J = πc⁴/2

Further, for any point at distance r from the center of the shaft, we have, the shear stress τ is given by

τ = Tr/J

We only consider the torsional loading of simple circular shafts in this analysis, ie cylinders or non-eccentric tubes without splits. Circular shafts are most commonly used as torque carrying members in machinery with rotating parts (like drive shafts of motors). This is fortuitous, as the analysis of non circular members under torsion is not simple to perform analytically.

[edit] Angle of Twist

The above image shows the twist of a shafted acted upon by a torque T at one end. We know that the shear angle γ is given by

γ = τ/G

For a shaft of radius c, we have

φ c = γ L

where L is the length of the shaft. Now, τ is given by

τ = Tc/J

so that

φ = TL/GJ

The angle of twist φ for a circular shaft acted upon by a torque T_x at a point x along its axis is given by:

$\phi = \phi_B - \phi_A = \int_A^B\frac{T_x}{J_xG}dx$

where J_x is the moment at section x. Note that you can use the torsion formula for shafts with slowly varying area as long as they are circular.

[edit] Loading of Beams

One of the areas where solid mechanics as discussed in this book is most effective is in the case beam loading. The loads on a beam can be point loads, distributed loads, or varying loads. There can also be point moments on the beam. The beam itself is supported at one or more points. The conditions at the support depend on the kind of support used. If the support is a roller, it can only have a reaction perpendicular to the motion of the roller. If the support is a pin, it cannot carry a moment. If the support is fixed, then it can have a reaction in any direction and support a moment as well.

In the above image, a simple beam is loaded at the center by a load P. It has a pinned contact at one end, and a rolling contact at the other end.

The above image shows an ideal moment acting at the center of a beam. An ideal moment is one which is not associated with a force.

The general method is analyzing beam problems is to find the loads, reactions, and moments, and come up with the values for the loads and moments at each section. This, in general will be a piecewise function of the distance along the beam.

For loads, we set up axes and this decides the sign of the force. For moments, we set the convention that the clockwise moment is positive.

[edit] Examples

1. Consider the case of the beam discussed earlier shown above.

The beam has reactions R₁ and R₂ acting on each of the supports.

R₁ + R₂ = P

Applying symmetry, we have,

R₁ = R₂

So that,

R₁ = R₂ = P/2

The above image shows the shear diagram for this problem. Note that the positive and negative directions are conventions, but it is important to choose one direction for positive shear and stick to it. As can be seen, the shear value changes at the point of application of the load.

The above image shows the moment diagram for the beam. The moment varies linearly from the supports to the middle. (The value at the middle should be PL/4). Note that the moment is maximum at the middle and zero at the ends, so that the effect of the moment is maximum at the center. In later chapters we will see that the stresses and strains due to moments are the most important ones for beams.

[edit] Exercises

1. In the above beam, find the reactions in the supports and the shear force at a position x. Also, find the moment at that point.

2. The above beam shows loading by two separate point loads. Find the shear and moment at location x from one end of the beam.

3. Find the shear and moment at each point along the beam with a point load and moment acting at two different points. How can a point moment occur in practice?

[edit] Calculus for Solving Beam Problems

Methods of calculus can be used to deal with continuous load functions. However these methods can be extended to point loads and moments by the use of the Dirac delta function.

The above shows a beam with uniform load per unit length w. Such loads are used to model the self weight of the beam where it acts uniformly throughout its length. This can also be used for the loading of, say, a bridge due to all the vehicles on it. The individual point loads acting through the tires can be modeled as a continuous load if the number of vehicles is large.

Consider a beam with load per unit length q(x) for any point x along its length. Consider an elemental length dx of the beam. Applying force equilibrium for this segment, we have,

V + dV = q dx + V

where V is the shear at the point x. Thus, we have the relation

dV/dx = q

Applying moment equilibrium to the same segment, we have,

M + dM − V dx − M − q dx dx/2 = 0

Neglecting the second order terms in dx, we have,

dM/dx = V

Or,

d²M/dx² = q

The above differential equations can be integrated with appropriate boundary conditions to get the shear and moment at each point.

[edit] Examples

Consider a beam of length L supported at its ends by two pins, with a uniform load per unit length of w.

The total load on the beam is thus wL. Thus, the reaction at each support is wL/2.

We have, the general relation for shear

$V = \int_0^x q dx + C_1$

The shear at origin is just the reaction at that point (=wL/2). If we take the vertical direction as the positive direction, we have, the shear at the origin is wL/2. Also, q = −w. Thus, we have, V = −wx + wL/2 for any point x.

The Moment at the centre should be wL^2/8

Similarly, we have the moment M

$M = \int_0^x V dx + C_2$

It is easy to see that the moment at the origin is zero. Thus, we have M = w x (L − x)/2.

[edit] Calculus for Point Loads

The above works well for well for continuous, smooth load functions. But in real world situations, we have to deal with point loads and moments. Thus we use the Dirac functions for a point load P at location a as

$P\left\langle x - a \right\rangle^{-1}$

and for a point moment M at location a as

$M \left\langle x - a \right\rangle^{-2}$

Now the previously stated equations can be used under the rules for the Dirac function. The function itself is defined as

$\left\langle x - a \right\rangle^n = \left\{ \begin{matrix} 0 & x < a\\ (x - a)^n & x > a \end{matrix} \right.$

Consider a beam of length L with supports at both ends and a point load P at the center. From the Dirac definition, we have,

$q = -P\left\langle x - L/2 \right\rangle^{-1}$

The shear at one end is just the reaction P/2. The shear at any point is given by

$V = \int_0^x q dx = \int_0^x -P\left\langle x - L/2 \right\rangle^{-1} dx = -P \left\langle x - L/2 \right\rangle^{0} + P/2$

The moment at one end is zero. The moment at any point is given by

$M = \int_0^x V dx = -P \left\langle x - L/2 \right\rangle + \frac{Px}{2}$

Applying the definition of the Dirac function, we have, the shear

$V = \left\{ \begin{matrix} P/2 & x < L/2\\ -P/2 & x > L/2 \end{matrix} \right.$
$M = \left\{ \begin{matrix} \frac{Px}{2} & x < L/2\\ \frac{P(L - x)}{2} & x > L/2 \end{matrix} \right.$

Note that the above is a special case of the general method called Laplace Transforms.

[edit] Exercises

1. Find the shear and moment in the above beam using Dirac methods.

2. The beam shown above has two loads which can be modeled as shown. Find the shear and moment at any point along the beam.

3. Consider the beam shown above with an overhang. Find the shear and moment at points along the axis.

[edit] Stresses and Strains in Beams

Given the loads and moments at each cross section, we can calculate the stress and strain at each location.

Consider the bending of a slender beam (one for which the cross section is much smaller than the length). A moment acting on the beam causes a deformation called flexure. Let the radius of the osculating circle of the beam be ρ. Consider an elemental length ds in the neutral plane (for which the deformation is zero). This element subtends an angle θ at the center of curvature, so that

ds/dθ = ρ

If we move a distance y along the radius, we have the length of the arc subtended would be (ρ − y) dθ. Thus the elemental extension would be y dθ. For a small enough curvature, we have the distance along the neutral plane would be the same as the initial undeformed length, dx. This gives us the axial strain at any point x as

ε_x = y/ρ

Also, using the Hooke’s law, we have,

σ_x = E ε_x = Ey/ρ

Now, we know that a pure moment M acts on the beam, so that there is no axial force. Or,

∑ F_x = 0

∫ σ_x dA = 0

∫ y dA = 0

The above equation gives us the location of the neutral plane.

Further, applying the moment conservation part of the equilibrium relations, we have,

M − ∫ σ_x dA y = 0

Now, we know,

σ_x = Ey/ρ

and

I = ∫ y² dA

where I is the moment of inertia, so that

M = E/ρ I

E/ρ = M/I

Thus, we have the expression for the stress due to a pure moment as

σ_x = My/I

Suppose the maximum value of y is c (the distance from the neutral plane). Then, we have,

σ_max = Mc/I

The quantity I/c is called the section moment of a beam and it denoted by S.

σ_max = M/S

[edit] Beams with Arbitrary Moments

So far we have considered beams which were acted upon by moments in the M_z. Suppose there is another moment in the M_y direction. Note that a moment in the M_x direction would simply be a torsional effect as the axis is in the x direction. Now, it is easy to see that the combination of moments is, in fact, equivalent to a moment acting on a beam of arbitrary cross section. It is fairly straightforward to show that, in this case, the stress at any point (y, z) on the face of the section for the beam is a function of I_z, I_y, and I_yz.

[edit] Shear in Beams

Earlier, we saw the methods to calculate the shear forces on a beam. Now we can analyze the stresses due shear forces, like we did for stresses due to bending moments. We showed that for the shear force V is given by V = dM/dx, where M is the moment acting at the point x.

[edit] The General State of Stress

[edit] Principal Stresses

The general state of stress can be represented by a symmetric 3 x 3 matrix.

It is always possible to choose a coordinate system such that all shear stresses are zero. The 3 x 3 matrix is then diagonalised, with the three principal stresses on the diagonal, and all other components equal to zero. The three principal stresses are conventionally labelled σ₁, σ₂ and σ₃.
σ₁ is the maximum (most tensile) principal stress, σ₃ is the minimum (most compressive) principal stress, and σ₂ is the intermediate principal stress.

[edit] Stress Invariants

$I 1 = σ 3 + σ 2 + σ 1$

$I 2 = σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1$

$I 3 = σ 1 σ 2 σ 3$

[edit] Hydrostatic Stress

The so-called hydrostatic stress, σ_h, is given by:

$\sigma_{h} = \left . \frac {\sigma_{1} + \sigma_{2} + \sigma_{3}} {3} \right .$

[edit] Deviatoric Stresses

[edit] Mohr’s Circle

Consider the two dimensional stress condition where the stresses are σ_x, σ_y, and τ_xy. We have, for another set of orthogonal axes x’-y’ at angle θ with x-y, the stresses are

$\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x + \sigma_y}{2}\cos 2\theta + \tau\sin 2\theta$

$\tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta$

From the above equations, we can see that for any stress states given by σ_x, σ_y, and τ_xy, we can find a value of θ such that the value of σ_x’ is maximum. This value is called the principal stress σ₁ (for maximum) or σ₂ (for minimum).

The principal stresses are given by

$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}$

and the maximum shear stress is given by

τ_max = (σ₁ − σ₂)/2

From the definitions for σ_x’ and τ_x’y’, we have

$\left( \sigma_{x'} - \frac{\sigma_x + \sigma_y}{2} \right)^2 + \tau_{x'y'}^2 = \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2$

In the σ-τ graph, this is a circle with center on the x axis, and the distance of the center from the origin is given by (σ_x + σ_y)/2 and a radius given by

$R = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}$

This circle is known as Mohr’s circle, and is useful for visualizing the stress state at a point.

The above figure shows Mohr’s circle for a stress state (σ, τ). The center and radius of the circle are obtained from equations stated above. The other stress σ_y can be read off by the point diametrically opposite the (σ, τ) point. The stress at any plane can be found using simple geometrical constructs.

[edit] Mohr’s Circle for Common Cases

The above shear stress in this case is σ₁/2.

A liquid is one which by definition cannot take a shear. Thus the Mohr-circle diagram is just a point.

For pure shear, the Mohr circle is centered at the origin.

[edit] Failure Criteria

In the case of isotropic materials, the state of stress at any point of the body is completely defined by the triad of the principal stresses. Now that we are able to transform stresses to get the principal stresses, we can use these stresses to consider some of the criteria (theories) postulated for the failure of materials in two- and three-axial states of stress usually based on experiments on yielding and fracture of materials in the uniaxial state of stress. According to such experiments, the kind of failure depends on the type of material. Failure of ductile materials (most of metals) occurs when the elastic limit is reached and yielding commences. Failure of non-ductile materials (e.g., cast iron, concrete) occurs by brittle fracture.

[edit] Maximum Shear Stress Criterion

For ductile materials, one failure theory is that of maximum shear stress. We know that the maximum shear stress is given by τ_max = (σ₁ − σ₂)/2. The yield stress, σ_y can be determined by uniaxial tensile tests. Thus, if the maximum shear stress theory is valid, failure occurs when the maximum shear stress reaches σ_y/2.

In the above image, the material will fail if the stress state is outside the shaded region.

[edit] Maximum Distortion Energy Criterion

If the maximum shear stress theory is valid, failure occurs when the maximum shear stress reaches ${\frac {\sqrt{2} \sigma_y} {3}}$ .

[edit] Failure of Materials

The actual failure mode of each material is unique, though certain criteria can be applied to classes of materials. mode of failure . fatigue failure ,

http://en.wikibooks.org/wiki/Strength_of_Materials

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30

Mission must be fulfilled