ARC 341 Structural Analysis II. Lecture 10: MM1.3 MM1.13


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1 ARC241 Structural Analysis I Lecture 10: MM1.3 MM1.13 MM1.4) Analysis and Design MM1.5) Axial Loading; Normal Stress MM1.6) Shearing Stress MM1.7) Bearing Stress in Connections MM1.9) Method of Problem Solution MM1.10) Analysis & Design of Simple Structures MM1.11) Stresses on an Oblique Plane MM1.12) Stress Under General Loading Conditions
2 MM1.4) Analysis and Design The study of Mechanics of Materials aims towards: 1) Analysis of Structures: determination of stresses and deformations in structural members. 2) Design of Structures: sizing of structural members. Stress, σ, is the amount of force per unit area distributed over a crosssection (intensity of force). It can be classified as a) normal stress, b) shearing stress, and c) bearing stress. 1
3 MM1.5) Axial Loading; Normal Stress When the loading on a member is directed along its axis, we say that the member is under axial loading. Axial loading causes a stress that is perpendicular (normal) to the plane of the sections normal stress. Normal stress can be calculated by dividing the force P by the crosssectional area A. σ = In general, this formula gives the average stress over a crosssection. Why? P A 2
4 σ varies across the crosssection of a members, where it is maximum near the loading and minimum far away from the loading, but this variation is small and can be ignored at points far away from the point of loading along the axis of the member. To define σ at a given point Q of the crosssection, we should consider a small are A subjected to differential load F. The stress at point Q then is: F σ = (lim A > 0) A 3
5 MM1.6) Shearing Stress When transverse forces are applied to a members, internal forces are developed in the plane of the section to resist these forces. These internal forces are represented by shearing stress: τ P A ave = = Unlike normal stress, shearing stress cannot be assumed uniform. Why? F A 4
6 MM1.7) Bearing Stress Pins and bolts can create stresses on the members they connect along the bearing surface. The distribution of these stresses are quite complicated, so an average nominal stress called bearing stress can be computed as an approximation. Bearing stress is calculated by dividing the load by the area of the rectangle representing the projection of the pin or bolt on the plate section. Thus: σ P A b = = P td 5
7 Problem MM1.20 A 40kN axial load is applied to a short wooden post that is supported by a concrete footing resting on undisturbed soil. Determine (a) the maximum bearing stress on the concrete footing, (b) the size of footing for which the average bearing stress in the soil is 145 kpa. 6
8 MM1.9) Method of Problem Solution Steps to solving a Mechanics Problem: 1) Clearly understand the statement of the problem. 2) Draw necessary FBD s. 3) Write equilibrium equations. 4) Solve for unknown forces. 5) Calculate stresses and deformations. 7
9 MM1.9) Numerical Accuracy The accuracy of a solution cannot exceed the (a) accuracy of given data, (b) accuracy of computations. Example: If P = 300 kn with possible error of, then the relative error is 0.13%. Thus, the solution accuracy cannot be greater than 0.13% regardless of the accuracy of computations. In engineering problems, the data are seldom known with an accuracy greater than 0.2%. 8
10 MM1.11) Stresses on an Oblique Plane Thus far, axial loads were found to cause normal stresses. And, transverse loads were found to cause shearing stresses. Why? Because stresses were determined in planes perpendicular to the axis of the member. What will happen if stresses were determined on an oblique plane? 9
11 Consider the same member subjected to an axial load P and P, but this time the stress is to be determined on an oblique plane with an angle θ. FBD shows that distributed forces on oblique section must equal P. P can be resolved to F (normal force) and V (tangential force) where: F = P cosθ V = Psinθ Divide these forces by the area of the oblique section to get normal and shearing stresses: σ = F A θ = P cosθ = A / cosθ 0 P A 0 cos 2 θ τ = V A θ = Psinθ = A / cosθ P 0 A 0 sinθ cosθ 10
12 What can we learn from the equations of stresses on an oblique plane in member under axial loading? P 2 P σ = cos θ τ = sinθ cosθ A 0 1) When θ = 0, σ = σ max and τ = 0. 2) When θ = 45 o, τ = τ max and σ = τ. The stress value in this case is P 2A 0 A 0 11
13 MM1.12) Stress Under General Loading Conditions Consider a body subjected to several loads P 1, P 2, etc. Pass a section at point Q using a plane parallel to the yz plane. To maintain equilibrium, the sliced section must be subjected to normal force F x and shear force V x. The portion of these forces at point Q are F x and V x. V x can be further decomposed to V xy and V xz. Dividing by the area A, stresses can be defined as: σ x lim = A 0 F x A τ xy lim = A 0 V x y A τ xz lim = A 0 z A V x 12
14 Note: First subscript in σ x, τ xy, τ xz is used to indicated that the stresses under consideration are excreted on a surface perpendicular to the xaxis. The second subscript in τ τ xy, τ xz identifies the direction of the component. The normal stress σ x is positive if the corresponding arrow points in the positive x direction. τ xy, τ xz are positive if the corresponding arrows point in the positive y and z directions. If the passing section faces the negative x directions, then positive stresses will point in the negative direction. 13
15 Stress Components in 3D: If passing planes are taken parallel to all the planes around point Q, a small cube with all the stress components can be visualized. Let each side of the cube has length a, and each face of the cube has area A. By inspection, the three equilibrium equations for the summation of forces can be satisfied. Considering one plane, the three equilibrium equations for the summation of moments can be satisfied. From this equilibrium, one concludes that τ xy = τ yx, τ yz = τ zy, τ zx = τ xz. 14
16 Other Conclusions: Only 6 components are required to define the condition of stress at a given point Q. Namely, σ x, σ y, σ z, τ xy, τ yz, τ zx. Shear can t take place in one plane only. An equal shearing stress must be excreted on another plane perpendicular to the first one. The same load condition may lead to different interpretation of the stress situation at a given point, depending on upon the orientation of the element considered. 15
17 Problem MM1.42 Members AB and BC of the truss shown are made of the same alloy. It is known that a 20mm square bar of the same alloy has tested to failure and that an ultimate load of 120 kn was recorded. If bar AB as a crosssectional are of 225 mm 2, determine: a) the factor of safety for bar AB b) the crosssectional are of bar AC of it is to have the same factor of safety as bar AB. 16
18 Problem MM1.46 Two wooden members of 90 x 140 mm uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 520 kpa, determine the largest axial load P that can be safely applied. 140 mm 90 mm 17
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