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Mechanics of Solids

by Suhas Mondal
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Suhas Mondal
Suhas Mondal
1 Stress 1/9 1 Stress By setting up the equilibrium conditions, the inner forces of a member subjected to an external load situation can be determined. So far neither the material nor the type of cross section applied for the member are being taken into account. But both material and type of cross section obviously have an impact on the behaviour of the member subjected to load. To design the member therefore a closer look on how the internal forces act along its cross section needs to be taken. 1.1 Normal Stress – Axial Loading Within this part of the chapter the internal forces are limited to only axial forces (normal forces) acting along the centroidal axis of a member. F F plane of cut a) F A A b) ΔF F c) ΔA σ A d) F fig 1.11: axially loaded rod A suspended rod is subjected to an axial load. The free body diagram in external equilibrium is shown in fig. 1.11a. The rod is cut perpendicular to its axis at any arbitrary distance from its ends and the equations of equilibrium are applied on the part. Thus the internal force found acting normal to the cut surface (area A) is of equal amount but opposite direction of the applied external force (fig. 1.11b). Consider the normal force to equally act on any particle ΔA of the cut surface A (fig. 1.11c). F ΔF = A ΔA
1 Stress 2/9 The intensity of a normal force acting on a surface at a certain point is described as the normal stress, denoted by the Greek letter σ (fig.1.1d). ΔF ΔA→ 0 ΔA σ = lim amount of internal force unit area σ= kN cm2 Considering a uniform distribution the normal stress is defined as: σ= F A F = ∫ σ ⋅ dA and (1.1), (1.2) A conclusion: the normal stress acting along a section of a member only depends on the external load applied (e.g. a normal force F) and the geometry of its cross section A (true for statically determinant systems). example 1.1 - stress Fig 1.12 shows a typical specimen used for uniaxial tensile testing for materials like timber or plastic. question: σ1 = F ; A1 answer: At which position will the specimen break if the applied force F is increased up to failure? σ2 = F Α2 A2 < A1, hence σ2 > σ 1 linear correlation! the specimen breaks at the maximum normal stress σ2 along the plane with the minimum cross sectional area A2. F F A2 fig 1.12: specimen for tensile test subjected to axial load A1
1 Stress 3/9 1.2 Average Shearing Stress – Transverse Loading So far the discussion focussed on normal stress, oriented perpendicular to the cutting plane or in direction of the main axis of the member. Stress can also act in the cutting plane thus perpendicular to the main axis of the member. This occurs if the member is subjected to a situation of transverse loads (fig. 1.21). F F fig 1.21: transverse load situation A situation like this is very common in a bolt or rivet connection (fig 1.22). Here the forces acting in the direction of the steel plates are transmitted by the bolt. In fig 1.23 the bolt is cut along the upper two connecting surfaces of the steel plates. To meet the equilibrium conditions, the force being transported along the cutting plane through the bolt is equal to the force being applied on the upper steel plate (F). F F 2F F F fig 1.23: plane of cut fig 1.22: bolt connection Dividing the force by the cut area of the bolt, the stress in the plane of cut is determined (fig. 1.24). Assuming the stress is uniformly distributed, the stress is defined as the average shearing stress, denoted by the Greek letter τ: τ= fig 1.24: shearing stress in the plane of cut cross section through bolt F A
1 Stress 4/9 1.3 Stress Analysis and Concept of Design Every material has its individual properties. It can be ductile, flexible or brittle. It deforms under the influence of a temperature change. It may plastically deform at a certain stress (load) and break at another. Its properties according to perpendicular directions may be equal (isotropic) or different (orthotropic). To ensure a safe design, these specific material properties have to be taken into account. The essential information is collected by conducting different tests in a material testing laboratory. At the failure of the material its ultimate stress is reached. The point of plastic deformation of the material is indicated as the yield point, corresponding to the yield stress. Taking this into account, an allowable stress can be defined for each individual material to be used within the design analysis. These stresses such as further indications concerning the maximum allowable deformation (serviceability of a structure) can be found in the respective national codes. A secure design requires a certain safety clearance towards the failure of the employed material. This is ensured by applying a safety factor (in national codes usually denoted by the Greek letter γ). In the design analysis the existing stress due to the existing load increased by the factor of safety (the design stress) has to be proofed less or equal to the allowable stress. Since the applied material might be orthotropic (different properties in different directions, e.g. timber) different allowable stresses are defined for normal and shearing stresses depending on their orientation (parallel or perpendicular, σ║ or σ ┴, see example 1.4). ratio of safety: ultimate load allowable load design analysis: Fd = F ⋅ γ σd = Fd Α σ d ≤ σ allowed τd = Fd Α τ d ≤ τ allowed design load = existing load · factor of safety design normal stress, axial loaded design stress ≤ allowable stress design average shear stress design stress ≤ allowable stress

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