| MECHANICS - THEORY 
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    |  |  | Introduction 
 
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    |  The  internal structure of racing cars are complex and require detailed
 knowledge of mechanics.
 |  | Mechanics of Materials presents  topics
        that build on the concepts learned in your Statics course. In general,
        this course introduces stress and strain (both normal and shear) for basic
        structural members (beams, shafts, columns and axial rods), deflection
        of beams, column buckling and material failure. Similar to all engineering
        courses, there are also many sub-topics.  This course is critical to structural engineers, product reliability
        engineers, and equipment design engineers. Without a good knowledge of
        stress and strain, and how they can be calculated, no structure can be
        well designed and premature failures can and probably will occur. | 
  
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    |  |  | Static Equilibrium 
 
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    |  |  | Mechanics of Materials starts with the basic static equilibrium equations, Each object in any static structure must be in both force and moment
      equilibrium. These equations are used in almost every problem in mechanics
        to calculate needed information, such as support reactions,
      internal forces, equivalent loads and shear.  Another tool used extensively in mechanics is the Free Body Diagram
        (FBD). These diagrams isolate a single member or structure with all boundary
        conditions
      converted to forces and moments. A FBD helps to identify all loads, forces
      and moments (both known and unknown) on a object which can then be used
        to solve for unknowns. | 
  
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    |  |  | Problem Solving 
 
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    |  Steps in Solving Problems
 |  | In engineering, following a set procedure in solving
        problems saves time and decreases the number of errors. Generally, the
        following five
        steps will help solve problems in mechanics. 
        Read and understand the question(s).a) Identify all given data.
 b) Identify exactly what is being asked.
Make sketches and draw free-body diagrams.Set up the problem and apply all relevant            equations.
Perform the calculations.Check and review the answer. | 
  
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    |  |  | Units 
 
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            | Basic Conversion Ratios |  
            |  | 1 ft | = | 0.3048 m |  |  
            |  | 1 slug | = | 14.59 kg |  |  
            |  | 1 lb | = | 4.448 N |  |  
            |  | 1 psi | = | 6.895 kPa |  |  |  | Whether we like it or not, engineers must be able to work in both US
      and SI (metric) units. Conversion ratios relate one set of units to another
      set and are commonly used to quickly convert between units. Some basic
      ratios are shown at the left. The
		Units Appendix
	has a large selection of conversion ratios for both SI and US units. | 
  
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    |  |  | Normal Stress 
 
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    |  Normal Stress
 |  | The most common type of stress in mechanics is a normal stress. A stress is defined as the load divided by the area. For a normal stress, it is all loads perpendicular to the surface. (non-perpendicular loads are addressed in later sections). This relationship can be summarized as Another way to express the same concept is to define the total perpendicular
        load P as        where A is the area of the surface. The symbol σ is used for normal
          stress. Later, other symbol, such as τ, are used for other types of
          stress | 
  
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    |  |  | Stress on an Inclined Plane 
 
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    |  Stress on an Inclined Plane
 
 |  | In many cases, the stress needs to be determined on a plane that is
        not perpendicular to the load direction. This can obtained by using trigonometry
        relations, but remember, both the area and load needs to be rotated to
      the inclined plane.       N = P cosθ V = P sinθ
 Ap = A/cosθ
 Using the definition of normal stress, the stress perpendicular to the inclined
        plane is The stress parallel to the plane (referred to as shear stress) is | 
  
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    |  |  | Saint-Venant Principle 
 
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    |  Stress Distribution in a Tensile Test
 |  | The Saint-Venant Principle simply states that localized
        stress concentrations disappear a short distance from the concentration
        (after Barre de Saint-Venant,
        1864). This means that the stress is uniform in a uniaxial loaded bar
        away from the ends. Generally, stress concentrations dissipate quickly. This is illustrated
      at the left for a square bar that is pulled using friction grips at both
      ends. The grips introduce the tensile load on the surface of the bar, but
        quickly, the stress is evenly distributed over the bar. The relationship,
        σ = P/A, is only valid away from boundary conditions
        that introduce stress concentrations. | 
  
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    |  | Factor of Safety 
 
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    |  |  | In all design situations, a factor of safety needs to be considered
        to help insure the structure does not fail due to unforeseen conditions
        such as excessive loading, errors,
        and poor craftsmanship. Of course, including a factor of safety does
        not
guarantee that the structure will not fail, but it does decrease the possibility
        of failure. Basically, a "factor of safety" (FS) is a multiplication constant that
        is applied to the calculated load to give a higher load.  In all cases, the FS is greater than 1. In most cases, the factor of
        safety will be 2 or 3. While a high factor of safety sounds good, the penalty
        will be an increase in weight. Another way
      to consider the FS of a structure is  where the failure stress is the stress level when the material will
        fail. Failure or ultimate stress levels for various materials can be
        found in handbooks (see 
		Materials Properties appendix). The allowable stress is the stress
        found in the design calculations for the actual applied load. In addition to FS, some design codes talk about a "margin of safety"
        (MS). This is simply a FS less 1, or      MS = FS - 1.0 | 
  
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