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stress - Mechanics of Materials

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- 1. Stress The Material is considered as continuous and cohesive Stress is the quotient of the force and area. It describes the intensity of the internal force on a specific plane (area) passing through a point.
- 2. Normal stress: The intensity of force, or force per unit area, acting normal to ΔA is defined as the normal stress, σ (sigma), since ΔFz is normal to the area then: A Fz z lim If the normal force or stress “pulls” on the area element ΔA, it is referred to as tensile stress, whereas if it “pushes” on ΔA it is called compressive stress. Shear stress: The intensity of force, or force per unit area, acting tangent to ΔA is called shear, .Here we have shear stress components: x y z zx zy z A Fx A zx 0 lim A Fy A zy 0 lim
- 3. Units: In the International Standard or SI system, the magnitudes of both normal and shear stress are specified in the basic units of Newtons per square meter (N/m2). This unit, called a Pascal (1Pa=1N/m2) rather small, and in engineering work prefixes such as kilo (103), symbolized by k, mega (106), symbolized by M, or Giga (109), symbolized by G, are used to represent larger, more realistic values of stress.
- 4. Average normal stress distribution Provided the bar is subjected to a constant uniform deformation as noted, then this deformation is the result of a constant normal stress σ. As a result, each area ΔA on the cross section is subjected to a force ΔF=σ ΔA, and the sum of these forces acting over the entire cross-sectional area must be equivalent to the internal resultant force P at the section. dAA and thereforeIf we let dFF Then, recognizing σ is constant, we have A dAdF . ; P=σ.A, A P σ: average normal stress at any point on the cross-sectional area. P: Internal resultant normal force, which is applied through the centroid of the cross sectional area. P is determined using the method of sections and the equations of equilibrium. A: Cross sectional area of the bar.
- 5. Equilibrium ΔA σ σ’ It should be apparent that only a normal stress exists on any volume element of material located at each point on the cross section of an axially loaded bar. If we consider vertical equilibrium of the element, then applying the equation of force equilibrium. 0zF , σΔA-σ’ΔA=0, σ=σ’ In the other words, the two normal stress components on the element must be equal in magnitude but opposite in direction. This is referred to as uniaxial stress.
- 6. Average shear stress In order to show how this stress can develop, we will consider the effect of applying a force F to the bar. If the supports are considered rigid, and F is large enough, it will cause the material of the bar to deform and fail along the planes identified by AB and CD. A free body diagram of the unsupported center segment of the bar indicates that the shear force V=F/2 must be applied at each section to hold the segment in equilibrium. The average shear stress distributed over each sectioned area that develops this shear force is defined by: A V avg
- 7. A V avg Here: avg : average shear stress at the section, which is assumed to be the same at each point located on the section. V: internal resultant shear force at the section determined from the equations of equilibrium. A: Area at the section. Notice that average shear is in the same direction as V, since the shear stress must create associated forces all of which contribute to the internal resultant force V at the section.
- 8. Single shear The steel and wood joints are examples of single-shear connections and are often referred to as lap joints. V=F Double shear V=F/2
- 9. Allowable Stress An engineer in charge of the design of a structural member or mechanical element must restrict the stress in the material to a level that will be safe. To ensure safety, it is necessary to choose an allowable stress that restricts the applied load to one that is less than the load the member can fully support. One method of specifying the allowable load for the design or analysis of a member is to use a number called the factor of safety. The factor of safety (F.S.) is a ratio of the failure load Ffail divided by the allowable load, Fallow. allow fail F F SF .. If the load applied to the member is linearly related to the stress developed within the member, as in the case of using σ = P/A and τavg =V/A, then we can express the factor of safety as a ratio of the failure stress σfail (or τfail) to the allowable stress σallow (or τallow);
- 10. allow fail SF .. Or allow fail SF .. In any of these equations, the factor of safety is chosen to be greater than 1 in order to avoid the potential for failure.

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