Civil Engineer Equation Descriptions

**Stress:**

σ= axial normal or shear stress, F= force, A= cross sectional area

**Strain:**

ε= strain, Δ*L*= change in length, L_{0}= initial length

Strain is the occasion of material moving out of its equilibrium position within a body as a result of an applied force, known as the stress. Any physical object, when a force is applied, will deform. The image below provides a dramatic example: the stress imparted by the club face at the moment of impact causes a strain within the golf-ball.

Golf ball compression. Harold Edgerton, 1935 *© 2010 MIT. Courtesy of MIT Museum. More information and images may be found at http://edgerton-digital-collections.org/*

Strain is likewise a measurement of how much the part deforms. By convention, if the part extends, the strain will be positive, and if the part contracts, the strain will be negative. In one dimension it takes the form:

where L_{f} is the final length of the part, Δ*L* the change in length, and L_{0} the equilibrium length. It should be noted that in this, and all dimensionalities, the strain has units of pure number. The sign convention for normal stress holds that tension is positive, and compression is negative.

Stress is the amount of force giving rise to a particular strain, either in compression or tension. It is defined, in the simplest case, as

where σ is the one-dimensional stress, F is force and A the cross-sectional area normal to the applied force. Its SI is a Pascal (Pa), or one Newton per meter-squared.

The relationship between stress and strain is Hooke's Law:

*E* is Young’s Modulus (or Modulus of Elasticity). It is a material property. Some illustrative values would be rubber, 0.01-0.1 GPa; oak, ~11 GPa; ASTM-A36 steel, 180 GPa; tungsten carbide, 450-650 GPa. (Data from The Engineering Toolbox.) The higher the modulus, the more stiff the material. Furthermore, the Young’s modulus has quite a range, informs how various materials will react in the day to-day environment, and suggests to which applications they might be put. The equation has its limitations, however: eventually the response of the material will become non-linear, and the equation will fail. Understanding where this will occur is a matter of experiment and plotting stress versus strain---for this work, however, it is reasonable to remain within the linear region.

Cross-sectional area is also important when it comes to predicting failure due to strain. For example, if you have a machine that pulls on both ends of a pencil with 100 pounds of force, the machine will probably pull the pencil apart and break it. If you put a wooden baseball bat in that same machine and pulled on both ends with 100 pounds of force, it would likely be minimally affected. Though both are wooden, the base-ball bat can distribute the same force over a much larger area. Similarly, spider-silk has an exceptionally high tensile strength, but breaks easily due to its slender aspect.

It should also be noted that this treatment has been very explicitly one-dimensional, for simplicity. This is an acceptable treatment when one dimension of the system clearly dominates the others and the material is isotropic. A taught piano-wire would be well-handled by such an approximation, or a square cross-sectioned bar, in which the bar length is much greater than the cross-section length. A more thorough treatment might consider an anisotropic material, being pulled in one direction, which would be described as a tensor equation. Though such is beyond the scope of this material, the depth and power of the tensor formulation make it advisable to all scientists and engineers.

As the Young's Modulus describes resilience of materials under tension, so the bulk modulus describes their resillience under compressive loading. Noteworthy, however, is the formulation that associates these two quantities. A material under compressive stress, after all, tends to bulge out at the sides---a stress ball, for example. This puts part of the material under tensile load, so that the Young's Modulus comes into play. The degree of this expansion is expressed by Poisson's ratio.

*G*=Shear Modulus

The **Shear modulus** relationship is very similar to Young's modulus, but it is applied to a different loading condition. It is fit to examine all of the components individually.

Shear stress is a result of loading the specimen so that the load acts perpendicular to the cross sectional area. Shear stress, like normal stress, has SI units of or a Pascal (Pa). Because shear and normal stresses affect the part differently, they cannot be combined for a total stress. Materials typically behave very differently when stressed normally versus in shear. The same equation is used to calculated the amount of shear stress that is present, but it is written as .

Shear strain, like normal strain, measures the deformation due to stress. Shear strain measures a change in shape of the specimen, and it is measured as an angle given in radians. This concept can be more clearly understood from the visual below. Shear stress and shear strain are related through a parameter known as the shear modulus. This parameter is material specific, and it works much like Young's modulus. The shear modulus is typically determined through experimental testing, but if the shear stress and shear strain are known, it can be determined from the shear modulus equation.

Example 1:

If the shear stress is 800 Pa and the shear strain is 3 μ radians, determine the shear modulus.

α_{L} *= coefficient of thermal expansion*

**Thermal expansion** is another way, other than stress, that a specimen can deform. Many materials are sensitive to a change in temperature as this equation describes. As temperature increases or decreases, the particles of the material become more or less active respectively. Excitation produces more space between molecules, thus elongating the entire specimen. The coefficient of thermal expansion () is a dimensionless material property that describes how much a material is affected by temperature change. Higher values for the coefficient means the material is more affected by temperature change. The value of these coefficients is determined experimentally, and they can be looked up for specific materials. Because this equation deals with a change or difference in temperatures, it is important to remember to match the temperature scale with the correct coefficient.

As the equation is currently written, we are actually solving for the strain that results from the change in temperature. We can rearrange the equation so that it looks like:

This way, we can solve for the expected change in length as long as we are given initial length, material, and the change in temperature.

**Example**: Given a material with a length of 2 meters, a coefficient of thermal expansion of 23 , an initial temperature of 10°C, and a final temperature of 35°C, find the expected change in length.

**Beam bending** creates a normal stress in a part, but it involves a different loading scenario than the axial normal stress. In order to understand beam bending, we must first examine its individual components.

Bending is produced when a specimen is subject to a bending moment. A moment occurs when a force is applied to an object at some distance away from the **point of interest**. The POI is any point we want to analyze. In other words, when a moment is applied, it causes rotation about the point of interest. Because a moment is a force multiplied by a distance, the SI unit is a Newton meter (Nm). In the case of a beam in bending, we must calculate the internal moment of the beam. While external loads are applied to the beam, the beam must also exert an opposing moment in order to maintain equilibrium.

Now consider the “*y”* term in our beam bending equation. This term is typically defined as the distance between the **neutral axis** of the beam and the point of interest. The neutral axis is an interesting concept in beam bending. Although the beam is experiencing stress, there is no stress present at the neutral axis of the beam. In a beam that is experiencing bending so that it looks like a smile, the beam is in compression above the neutral axis and tension below the neutral axis. The neutral axis will typically be the center of the beam height. As the distance away from the neutral axis increases, the bending stress becomes larger.

Finally, we must evaluate the second moment (*I)*. This term represents how much the beam resists bending based on its geometry. The SI unit for the second moment of area is m^{4}---an area integrated by the square of distance. Formally, this term is defined as

Fortunately, it can be simplified to *I = bh*^{3}/12 if the cross section of the beam is symmetric about the bending axis.

The beam bending equation is also commonly written as . The negative sign helps account for the sign convention for stress mathematically. In the previous example, a positive moment, positive distance to the point of interest, and positive second moment of area combined with this negative sign, show that the point of interest is actually experiencing compression. The negative sign can be left off of the equation, and you just apply the negative sign to the final stress when the point of interest is in compression.

*v*= velocity of fluid in a pipe, *q*= volumetric flow rate, *D*= pipe diameter

**Pipe flow** is used to determine the velocity at which a fluid will move through a section of piping. There are several factors that can change the velocity of the moving fluid that we must account for, and this equation assumes negligible friction in the pipe. The volumetric flow rate is defined as the amount or volume of a fluid which passes through a given area per unit time. The SI unit for volumetric flow rate is cubic meters per second.

The pipe flow equation, at first glance, seems to be counterintuitive because the diameter of the pipe is on the bottom of the equation. It seems that the fluid would be able to move faster through a larger pipe, but the opposite is actually true. As the size of the pipe decreases for the same volumetric flow rate, the velocity of the fluid will increase. Consider using a water hose to wash your car. The water is flows out of the hose at a steady rate if there is no attachment on it. If you were to place your finger over the end of the hose, the water sprays out at a much greater velocity than it was before. In this scenario, the volumetric flow rate through the hose was constant, but we decreased the area of the hose opening with our thumb, thus accelerating the flow of the water.

The only other way to increase the speed of the fluid flowing through a pipe is to increase the volumetric flow rate, or in more simple terms, to push more fluid through the pipe at one time. The water hose connected to your home water has a steady state velocity. If we were to hypothetically connect that same water hose to a fire hydrant, the velocity of the fluid in the hose would be much greater than from the home source, but the volumetric flow rate would be significantly increased.

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