Fifth SI Edition CHTER 1 MECHNICS OF MTERILS Ferdinand. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Introduction Concept of Stress Lecture Notes: J. Walt Oler Teas Tech University
Contents Concept of Stress Review of Statics Structure Free-Body Diagram Component Free-Body Diagram Method of Joints Stress nalysis Design ial Loading: Normal Stress Centric & Eccentric Loading Shearing Stress Shearing Stress Eamples Bearing Stress in Connections Stress nalysis & Design Eample Rod & Boom Normal Stresses in Shearing Stresses in Bearing Stresses Stress in Two Force Members Stress on an Oblique lane Maimum Stresses Stress Under General Loadings State of Stress Factor of Safety 1-
Concept of Stress The main objective of the study of the mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures. Both the analysis and design of a given structure involve the determination of stresses and deformations. This chapter is devoted to the concept of stress. 1-3
Review of Statics The structure is designed to support a 30 kn load The structure consists of a boom and rod joined by pins (zero moment connections) at the junctions and supports erform a static analysis to determine the internal force in each structural member and the reaction forces at the supports 1-4
Structure Free-Body Diagram Structure is detached from supports and the loads and reaction forces are indicated Conditions for static equilibrium: M 0 ( 0.6m) ( 30kN)( 0.8m) F C F y y C 40kN 0 + C y 0 30kN y and C y can not be determined from these equations + C 40kN y + C y 30kN 0 1-5
Component Free-Body Diagram In addition to the complete structure, each component must satisfy the conditions for static equilibrium Consider a free-body diagram for the boom: M 0 ( 0.8m) y C y B 0 30kN y substitute into the structure equilibrium equation Results: 40kN C 40kN Cy 30kN Reaction forces are directed along boom and rod 1-6
Method of Joints The boom and rod are -force members, i.e., the members are subjected to only two forces which are applied at member ends For equilibrium, the forces must be parallel to to an ais between the force application points, equal in magnitude, and in opposite directions Joints must satisfy the conditions for static equilibrium which may be epressed in the form of a force triangle: F 4 F F B B B 0 F 5 BC 40kN 30kN 3 F BC 50kN 1-7
Stress nalysis Can the structure safely support the 30 kn load? From a statics analysis F B 40 kn (compression) F BC 50 kn (tension) d BC 0 mm t any section through member BC, the internal force is 50 kn with a force intensity or stress of σ BC 3 50 10 314 10-6 N m 159 Ma From the material properties for steel, the allowable stress is σ all 165 Ma Conclusion: the strength of member BC is adequate 1-8
Design Design of new structures requires selection of appropriate materials and component dimensions to meet performance requirements For reasons based on cost, weight, availability, etc., the choice is made to construct the rod from aluminum (σ all 100 Ma). What is an appropriate choice for the rod diameter? σ d π 4 d all 4 π σ 6 ( 10 m ).5 10 m 5.mm 4 500 all π 50 10 100 10 3 6 N 500 10 a n aluminum rod 6 mm or more in diameter is adequate 6 m 1-9
ial Loading: Normal Stress The resultant of the internal forces for an aially loaded member is normal to a section cut perpendicular to the member ais. The force intensity on that section is defined as the normal stress. σ lim 0 F σ ave The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy σ ave df The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics alone. σ d 1-10
Centric & Eccentric Loading uniform distribution of stress in a section infers that the line of action for the resultant of the internal forces passes through the centroid of the section. uniform distribution of stress is only possible if the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as centric loading. If a two-force member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an aial force and a moment. The stress distributions in eccentrically loaded members cannot be uniform or symmetric. 1-11
Shearing Stress Forces and are applied transversely to the member B. Corresponding internal forces act in the plane of section C and are called shearing forces. The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load. The corresponding average shear stress is, τ ave Shear stress distribution varies from zero at the member surfaces to maimum values that may be much larger than the average value. The shear stress distribution cannot be assumed to be uniform. 1-1
Shearing Stress Eamples Single Shear Double Shear τ ave F τ ave F 1-13
Bearing Stress in Connections Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect. The resultant of the force distribution on the surface is equal and opposite to the force eerted on the pin. Corresponding average force intensity is called the bearing stress, σ b t d 1-14
Stress nalysis & Design Eample Would like to determine the stresses in the members and connections of the structure shown. From a statics analysis: F B 40 kn (compression) F BC 50 kn (tension) Must consider maimum normal stresses in B and BC, and the shearing stress and bearing stress at each pinned connection 1-15
Rod & Boom Normal Stresses The rod is in tension with an aial force of 50 kn. t the rod center, the average normal stress in the circular cross-section ( 31410-6 m ) is σ BC +159 Ma. t the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline, 6 ( 0mm)( 40mm 5mm) 300 10 m σ BC, end N m 167 Ma The boom is in compression with an aial force of 40 kn and average normal stress of 6.7 Ma. The minimum area sections at the boom ends are unstressed since the boom is in compression. 3 50 10 300 10 6 1-16
in Shearing Stresses The cross-sectional area for pins at, B, and C, 5mm π r π 491 10 6 m The force on the pin at C is equal to the force eerted by the rod BC, 3 50 10 N τ C, ave 6 491 10 m 10Ma The pin at is in double shear with a total force equal to the force eerted by the boom B, 0kN 6 491 10 m τ, ave 40.7 Ma 1-17
in Shearing Stresses Divide the pin at B into sections to determine the section with the largest shear force, E G 15kN 5kN (largest) Evaluate the corresponding average shearing stress, 5kN τ, G B ave 6 491 10 m 50.9Ma 1-18
in Bearing Stresses To determine the bearing stress at in the boom B, we have t 30 mm and d 5 mm, σ b td 40kN ( 30mm)( 5mm) 53.3Ma To determine the bearing stress at in the bracket, we have t (5 mm) 50 mm and d 5 mm, σ b td 40kN ( 50mm)( 5mm) 3.0Ma 1-19
Stress in Two Force Members ial forces on a two force member result in only normal stresses on a plane cut perpendicular to the member ais. Transverse forces on bolts and pins result in only shear stresses on the plane perpendicular to bolt or pin ais. Will show that either aial or transverse forces may produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member ais. 1-0
Stress on an Oblique lane ass a section through the member forming an angle θ with the normal plane. From equilibrium conditions, the distributed forces (stresses) on the plane must be equivalent to the force. Resolve into components normal and tangential to the oblique section, F cosθ V sinθ The average normal and shear stresses on the oblique plane are σ τ F V θ θ cosθ 0 cosθ sinθ 0 cosθ 0 0 cos θ sinθ cosθ 1-1
Maimum Stresses Normal and shearing stresses on an oblique plane σ The maimum normal stress occurs when the reference plane is perpendicular to the member ais, σ cos θ τ 0 0 τ m 0 0 The maimum shear stress occurs for a plane at + 45 o with respect to the ais, τ m sin 45 cos 45 σ 0 sinθ cosθ 0 1-
Stress Under General Loadings member subjected to a general combination of loads is cut into two segments by a plane passing through Q The distribution of internal stress components may be defined as, σ τ y lim 0 lim 0 F V y τ z lim 0 Vz For equilibrium, an equal and opposite internal force and stress distribution must be eerted on the other segment of the member. 1-3
State of Stress Stress components are defined for the planes cut parallel to the, y and z aes. For equilibrium, equal and opposite stresses are eerted on the hidden planes. The combination of forces generated by the stresses must satisfy the conditions for equilibrium: F F F 0 y M z y M 0 τ τ y M y z M 0 Consider the moments about the z ais: similarly, ( τ ) a ( τ ) yz y zy It follows that only 6 components of stress are required to define the complete state of stress z y τ τ and τ τ a yz zy 1-4
Factor of Safety Structural members or machines must be designed such that the working stresses are less than the ultimate strength of the material. FS Factor of safety FS σ σ u all ultimate stress allowable stress Factor of safety considerations: uncertainty in material properties uncertainty of loadings uncertainty of analyses number of loading cycles types of failure maintenance requirements and deterioration effects importance of member to integrity of whole structure risk to life and property influence on machine function 1-5