1 MECHANICS OF MATERIALS Prepared by Engr. John Paul Timola
2 Mechanics of materials branch of mechanics that studies the internal effects of stress and strain in a solid body. stress is associated with the strength of the material from which the body is made, while strain is a measure of the deformation of the body
3 vital importance for the design of any machine or structure many of the formulas and rules of design cited in engineering codes are based upon the principles of this subject. dates back to the beginning of the seventeenth century, when Galileo Galilei performed experiments to study the effects of loads on rods and beams made of various materials.
4 LOADS Surface loads act on a small area of contact reported by concentrated forces Distributed loading act over a larger surface area of the body. When coplanar, there is a resultant force equal to the area under the distributed loading diagram resultant acts through the geometric center or centroid of this area.
5 Body force developed when one body exerts a force on another body without direct physical contact between the bodies. examples include the effects caused by the earth s gravitation or its electromagnetic field. normally represented by a single concentrated force acting on the body. In the case of gravitation, this force is called the weight W of the body and acts through the body s center of gravity.
6 Support Reactions
7 Equations of Equilibrium Equilibrium of a body requires both a balance of forces, to prevent the body from translating or having accelerated motion along a straight or curved path, and a balance of moments, to prevent the body from rotating. F 0 M O 0
8 F 0 F 0 F 0 x y z M 0 M 0 M 0 x y z
9 Internal Resultant Loadings statics is primarily used to determine the resultant loadings that act within a body using the method of sections. Normal force, N. acts perpendicular to the area. developed whenever the external loads tend to push or pull on the two segments of the body. Shear force, V. lies in the plane of the area, the body to slide over one another.
10 Internal Resultant Loadings Torsional moment or torque, T. developed when the external loads tend to twist one segment of the body with respect to the other about an axis perpendicular to the area. Bending moment, M. caused by the external loads that tend to bend the body about an axis lying within the plane of the area.
11 Important Points 1. The study of the relationship between the external loads applied to a body is. 2. Internal loads within the body cause stress and. 3. produce a resultant force having a magnitude equal to the area under the load diagram, and having a location that passes through the centroid of this area. 4. The mathematical equations in order to prevent a body from translating with accelerated motion and from rotating are. 5. The method used to determine the internal resultant loadings acting on the surface of a sectioned body is. 6. When attached to a member, it prevents translation of the member in that direction, and it produces a couple moment to prevents rotation. 7. Three types of external forces that can be applied to a body.
12 Example 1 Determine the resultant internal loadings acting on the cross section at C of the cantilevered beam.
13 Example 2 The 500-kg engine is suspended from the crane boom. Determine the resultant internal loadings acting on the cross section of the boom at point E.
14 Example 3 Determine the resultant internal loadings acting on the cross section at C of the beam.
15 STRESS quotient that describes the intensity of the internal force acting on a specific plane (area) passing through a point
16 Normal Stress, σ (sigma) intensity of the force acting normal to A z lim A 0 F z A If the normal force or stress pulls on A as shown, it is tensile stress, whereas if it pushes on A, it is compressive stress.
17 Shear Stress, (tau) intensity of the force acting tangent to A zx lim A 0 F x A zy lim A 0 F y A
18 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/m 2 ) or a Pascal (1 Pa = 1 N/m 2 ), In the Foot-Pound Second system of units, usually express stress in pounds per square inch (psi) or kilopounds per square inch (ksi), where 1 kilopound (kip) = 1000 lb.
19 AVERAGE NORMAL STRESS IN AN AXIALLY LOADED BAR Provided the material of the bar is both homogeneous and isotropic, then when the load P is applied to the bar through the centroid of its cross-sectional area, the bar will deform uniformly throughout the central region of its length.
20 Average Normal Stress Distribution If we pass a section through the bar, and separate it into two parts, then equilibrium requires the resultant normal force N at the section to be equal to P.
21 Because the material undergoes a uniform deformation, it is necessary that the cross section be subjected to a constant normal stress distribution. Each small area A on the cross section is subjected to a force N = σ A, 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.
22 N A where σ = average normal stress at any point on the cross-sectional area N = internal resultant normal force, which acts through the centroid of the cross-sectional area. A = cross-sectional area of the bar where s is determined
23 Maximum Average Normal Stress If the bar may be subjected to several external axial loads, or a change in its cross-sectional area may occur, the normal stress within the bar may be different from one section to the next. The maximum average normal stress is to be determined, where the ratio N/A is a maximum.
24 Important Points 1. When a body subjected to external loads is sectioned, there is a distribution of force acting over the sectioned area which holds each segment of the body in equilibrium called. 2. Stress is the limiting value of per unit area, as the area approaches zero. 3. The of the stress components at a point depends upon the type of loading acting on the body, and the orientation of the element at the point. 4. When a prismatic bar is made of homogeneous and material, and is subjected to an axial force acting through the centroid of the cross-sectional area, then the center region of the bar will deform. 5. True or False. Stress can determine if the material is continuous and cohesive.
25 Example 1 The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading.
26 Example 2 The 80-kg lamp is supported by two rods AB and BC. If AB has a diameter of 10 mm and BC has a diameter of 8 mm, determine the average normal stress in each rod.
27 Activity The cylinder is made of steel having a specific weight of 490 lb/ft 3. Determine the average compressive stress acting at points A and B.
28 AVERAGE SHEAR STRESS Shear stress is the component that acts in the plane of the sectioned area.
29 How it develops Consider the effect of applying a force F to the bar. If F is large enough, it can 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.
30 The average shear stress distributed over each sectioned area that develops this shear force is defined by V A avg = average shear stress at the section V = internal resultant shear force on the section determined from the equations of equilibrium A = area of the section
31 Procedure for Analysis 1. Section the member at the point where the average shear stress is to be determined. 2. Draw the necessary free-body diagram, and calculate the internal shear force V acting at the section that is necessary to hold the part in equilibrium.
32 Example 1 Determine the average shear stress in the 20-mmdiameter pin at A and the 30-mm-diameter pin at B that support the beam.
33 Example 2 If the wood joint has a thickness of 150 mm, determine the average shear stress along shear planes a a and b b of the connected member. For each plane, represent the state of stress on an element of the material.
34 Activity Determine the largest internal shear force resisted by the bolt with a diameter of 180 mm then solve for the average shear stress.
35 Example 3 The inclined member is subjected to a compressive force of 600 lb. Determine the average compressive stress along the smooth areas of contact at AB and BC, and the average shear stress along the horizontal plane DB.
36 ALLOWABLE STRESS DESIGN To ensure the safety of a structural or mechanical member, it is necessary to restrict the applied load to one that is less than the load the member can fully support.
37 Reasons 1. The intended measurements of a structure or machine may not be exact, due to errors in fabrication or in the assembly of its component parts. 2. Unknown vibrations, impact, or accidental loadings can occur that may not be accounted for in the design. 3. Atmospheric corrosion, decay, or weathering tend to cause materials to deteriorate during service. 4. Some materials, such as wood, concrete, or fiber-reinforced composites, can show high variability in mechanical properties.
38 Factor of Safety One method of specifying the allowable load for a member is to use a number Ratio of the failure load F fail to the allowable load F allow F.S. F F fail allow
39 Factor of Safety If the load applied to the member is linearly related to the stress developed within the member, F.S. fail fail allow allow
40 Required Area at Section N A or A= allow V allow
41 Procedure for Analysis First find the section over which the critical stress is acting Section the member through the area and draw a free-body diagram of a segment of the member. Determine the internal resultant force at the section using the equations of equilibrium. Solve the required area needed to sustain the calculated load or factored load at the section provided either the allowable stress or the load and resistance factors are known
42 Example 1 Determine the largest load P that can be applied to the bars of the lap joint. The bolt has a diameter of 10 mm and an allowable shear stress of 80 MPa. Each plate has an allowable tensile stress of 50 MPa, an allowable bearing stress of 80 MPa, and an allowable shear stress of 30 MPa.
43 Activity If each of the three nails has a diameter of 4 mm and can withstand an average shear stress of 60 MPa, determine the maximum allowable force P that can be applied to the board.
44 Example 2 The control arm is subjected to the loading. Determine to the nearest 1/4 inch the required diameters of the steel pins at A and C if the factor of safety for shear is F.S. = 1.5 and the failure shear stress is 12 ksi.
45 Example 3 The suspender rod is supported at its end by a fixed-connected circular disk. If the rod passes through a 40- mm-diameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20-kN load. The allowable normal stress for the rod is σ allow = 60 MPa, and the allowable shear stress for the disk is τ allow = 35 MPa.
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