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Chris Fox's Engineering Section

Direct Stress and Strain

"For a spring undergoing elastic [reversible] deformation, the extension is directly proportional to the applied force".

Hooke's Law

Stress (σ) may be defined as the internal effect of an applied force on a body. That force may be tensile, causing the body to expand in the direction of the force, or compressive, causing it to contract in that direction. The formula for stress is σ=F/A, where A is the cross-sectional area of the specimen (The area perpendicular to the force's line of action). Because area is measured in m² (square metres), the units of stress are Nm-2.

The deformation of a body due to an applied force is termed the strain. Strain (ε) is given by the formula ε=x/l; x denotes the extension and l the original length, in both cases in the direction of the force's line of action. Because it is a ratio, strain has no units.

The ratio of stress to strain is termed the Young's Modulus (E). Young's Modulus is given by E=σ/ε or E=Fl/Ax. Its units are Nm-2.

As a specimen expands in the direction of the applied force, it contracts perpendicular to that direction. If the strains parallel to and perpendicular to the force's line of action are denoted by εx and εy respectively, then εy=-νεx; ν denotes Poisson's Ratio, whose value is normally 1/3 for elastic deformation. It follows from this that the cross-sectional area changes when a force is applied, which in turn means that the exact value of the stress can only be obtained if the change in cross-sectional area can be calculated. However, the engineering stress, which is obtained by dividing the force by the original cross-sectional area, may be used as an approximation.

The force acting on a specimen cannot be increased indefinitely. If the stress reaches the yield stress σy, the material undergoes plastic (irreversible) deformation. If the applied force is increased further, the stress will eventually reach the ultimate stress σu, at which point the material breaks. These effects can be represented on a stress-strain graph. Starting at the origin, the graph is linear until the stress reaches σy, after which the gradient gradually decreases; the line ends when the stress reaches σu. If the force is removed after the specimen has yielded, the specimen does not fully return to its original length; this is represented on the graph by a straight line running parallel to the first straight line and ending at the ε-axis.

Because of the potential for failure, a safety factor is introduced into stress calculations. The value of this number can range from 3 if failure would be an inconvenience, to 10 if failure would cause death. The yield stress is divided by the safety factor to give σs, which presumably stands for "safe stress". σs is then portrayed as the maximum stress that can be withstood.

Consider a wheel of radius r rotating with angular velocity ω. The rim has cross-sectional area A. Let us examine a small section of rim of angular length δθ. The mass of that section is given by m=ρArδθ, where ρ denotes density, the mass per unit volume. The centrifugal force P acting on the section is given by P=mrω²; substituting for m gives P=ρAr²δθω².

At each end of the section, a tangential force F acts along the rim. The line of action of this force forms an angle δθ/2 with the tangent to the midpoint of the section. Balancing forces radially therefore gives P = 2F sin (δθ/2). Because δθ is very small, sin (δθ/2) ≈ δθ/2, in which case P=Fδθ. Substituting for F gives P=σAδθ. Combining this with the other equation for P gives σAδθ=ρAr²ω²δθ, and simplifying this gives σ=ρr²ω².

Strain energy (u) is defined as the potential energy stored in a spring or other object due to the work done on it. If one end of a bar is attached to the ceiling and a longitudinal force P applied at the other end, then u=0.5Px. Substituting for P gives u=0.5σAx, and replacing x with σl/E gives u=σ²V/2E.

Strain energy can be used to examine cases of shock loading. Consider a bar of length l, with one end attached to the ceiling and a base attached to the other end. A mass m drops through a height h to the base, causing extension x in the bar. The strain energy in the bar equals the potential energy lost by the mass. Therefore, 0.5Px=mg(x+h), where P denotes the force in the bar at maximum extension. Replacing x with Pl/AE, multiplying both sides by 2AE/l and rearranging gives P²-2mgP-2mghAE/l=0. Solving this gives P={2mg±√(4m²g²+8mghAE/l)}/2. And factorising and simplifying gives P=mg{1+√(1+2hAE/mgl)}. If h>>x, as usual, then mgh≈0.5Px and mgh=0.5P²l/AE. Substituting for P and rearranging gives the formula for the maximum stress as σmax=√(2mghE/Al).

If h=0, there is no free fall, but weight is still suddenly applied. In this case the formula for P becomes P=mg{1+√(1+0)}. This simplifies to P=2mg. The maximum force in the bar, and in turn the maximum stress, is therefore twice that at equilibrium.

Now let us turn to stress in non-uniform bars. A bar with a single step effectively consists of two bars, which we shall denote with 1 and 2. The overall extension xTOT=x1+x2. Substituting for the x values gives xTOT=Fl1/A1E+Fl2/A2E. In general, for a stepped bar xTOT = (F/E) Σ(ln/An). If the cross-section changes continuously, A may be a function of y, the distance from one end. If δ denotes the overall extension, then for such a bar δ = (F/E) ∫0l (1/A) dy.

An example of a continuously changing cross-section is that of a truncated cone. If d1 and d2 denote the diameters at the ends, and y denotes the distance from end 1, then because dd/dy is constant, (dy-d1)/y=(d2-d1)/l. Rearranging this gives dy=d1+(d2-d1)y/l. The area of a cross-section is given by Ay=πry², or Ay=π(dy/2)². Substituting for dy gives Ay=0.25π{d1+(d2-d1)y/l}².

Stress analysis can be used to solve statically indeterminate problems - problems that cannot be solved solely using force equilibrium equations. An example is one where three struts are attached to the ceiling at points B, C and D, with their other ends joined together at point A. Angles ∠BAC and ∠DAC are equal and shall be denoted by α. The original length of each of AB and AD is l / cos α, where l is the original length of AC. If a downward force F is applied at A, causing it to move down by a distance δ, the resulting extensions in AB and AD both equal δ cos α.

If the resulting force in AC is Fx, and that in each of AB and AD is Fy, then F = Fx + 2Fy cos α. Because F=EAx/l, Fx=ExAxδ/l and Fy = EyAyδ cos²α / l. Therefore, Fy/Fx = EyAy cos²α / ExAx. Substituting for Fy in the force equation and rearranging gives Fx=F/{1+2(EyAy/ExAx) cos³α}. If all three struts are of the same material and have equal cross-sectional areas, then Fx=F/(1+2 cos³α).

Two bars, which we shall denote with A and B, are joined together to form a single column which is built in at both ends; A is on top. A downward force P is applied at the joint, causing a reaction R1 at the bottom of the column and R2 at the top. Let us assume both the reactions are tensile; compressive forces and strains will then be negative.

Because the forces must balance, P=R2-R1. And because the overall length of the column is fixed, xA+xB=0. The latter equation can be rewritten as R2lA/EAAA+R1lB/EBAB=0. We now have a pair of simultaneous equations that can be solved to obtain the reactions. If the bars are of the same material and have the same cross-sectional area, R2lA+R1lB=0.

We will now consider stress and strain in compound bars. Bar 2 is built into the floor, and is surrounded by bar 1 which is attached to the floor. A single plate is attached to the tops of both bars, and a downward force P exerted on the plate.

Because of the plate, x1=x2; this leads to σ1l1/E12l2/E2. Also, the forces in the bars must equal the applied force, i.e. P=F1+F2; this can be rewritten as P=σ1A12A2. We therefore have a pair of simultaneous equations that can be solved to give the stresses.

The dimensions of a body can be changed by heat as well as by mechanical forces. The formula for extension due to a change in temperature is x=lαΔT, where α is the linear thermal expansion coefficient, measured in K-1. This applies to any direction.

Imagine that the aforementioned compound bar is heated. Both bars have a tendency to expand, but will have different natural (unstressed) extensions. Because of the plate, however, they are forced to take a common intermediate position, which we shall call the equilibrium position. Only the extensions from the natural positions to the equilibrium position, which we shall call x1 and x2, cause stress.

Equating extensions gives l1α1ΔT+x1=l2α2ΔT+x2; replacing x with σl/E and rearranging gives σ1l1/E12l2/E2=ΔT(l2α2-l1α1). Because there is no applied force, F1+F2=0; this leads to σ1A12A2=0. We therefore have a pair of simultaneous equations that can be solved to obtain the stresses.