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Stress Analysis & Design Example

Stress Analysis & Design Example 17
MECHANICS OF CHAPTER MATERIALS 1 Introduction – Concept of StressMECHANICS OF MATERIALS Contents Stress in Two Force Members Normal Stress Stress on an Oblique Plane Shearing Stress Maximum Stresses Bearing Stress Stress Under General Loadings State of Stress Factor of Safety Stress Analysis Design Example 1 2 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Concept of Stress • The main objective of the study of 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 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Axial Loading: Normal Stress • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. • The force intensity on that section is defined as the normal stress. F P  lim  ave A A A0 • The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy P  A dF  dA ave A • The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics alone. 1 4 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Centric Eccentric Loading • A uniform distribution of stress in a section implies that the line of action of the resultant of the internal forces passes through the centroid of the section. • A uniform distribution of stress is only possible if the concentrated loads on the end sections of twoforce members are applied at the section centroids. This is referred to as centric loading. • If a twoforce member is eccentrically loaded, then the resultant of the stress distribution in a section must yield an axial force and a moment. • The stress distributions in eccentrically loaded members cannot be uniform or symmetric. 1 5 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Shearing Stress • Forces P and P’ are applied transversely to the member AB. • 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 P. • The corresponding average shear stress is, P  ave A • Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value. • The shear stress distribution cannot be assumed to be uniform. 1 6 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Shearing Stress Examples Single Shear Double Shear P F P F   ave ave A A A 2A 1 7 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS 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 exerted on the pin. • Corresponding average force intensity is called the bearing stress, P P  b A t d 1 8 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Analysis and Design example • 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 • Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports 1 9 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Structure FreeBody Diagram • Structure is detached from supports and the loads and reaction forces are indicated • Conditions for static equilibrium: M 0 A0.6m30kN0.8m  C x A 40kN x F 0A C  x x x CA40kN x x F 0 A C 30kN 0  y y y A C 30kN y y • A and C can not be determined from y y these equations 1 10 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Component FreeBody Diagram • In addition to the complete structure, each component must satisfy the conditions for static equilibrium • Consider a freebody diagram for the boom: M 0A0.8m  B y A 0 y substitute into the structure equilibrium equation C 30kN y • Results: A 40kN C 40kN C 30kN x y Reaction forces are directed along boom and rod 1 11 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Method of Joints • The boom and rod are 2force 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 axis between the force application points, equal in magnitude, and in opposite directions • Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle:  F 0  B F F 30kN AB BC  4 5 3 F 40kN F 50kN AB BC 1 12 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Stress Analysis Can the structure safely support the 30 kN load • From a statics analysis F = 40 kN (compression) AB F = 50 kN (tension) BC • At any section through member BC, the internal force is 50 kN with a force intensity or stress of 3 P 5010 N d = 20 mm BC  159 MPa BC 6 2 A 31410 m • From the material properties for steel, the allowable stress is  165 MPa all • Conclusion: the strength of member BC is adequate 1 13 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS 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 s = 100 MPa). What is an all appropriate choice for the rod diameter 3 P P 5010 N 6 2  A 50010 m all 6 A  10010 Pa all 2 d A  4 6 2 4A 450010 m 2 d 2.5210 m 25.2mm   • An aluminum rod 26 mm or more in diameter is adequate 1 14 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Stress Analysis Design Example • Would like to determine the stresses in the members and connections of the structure shown. • From a statics analysis: F = 40 kN (compression) AB F = 50 kN (tension) BC • Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection 1 15 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Rod Boom Normal Stresses • The rod is in tension with an axial force of 50 kN. • At the rod center, the average normal stress in the 6 2 circular crosssection (A = 314x10 m ) is s = +159 BC MPa. • At the flattened rod ends, the smallest crosssectional area occurs at the pin centerline, 6 2 A20mm40mm 25mm 30010 m 3 P 5010 N  167MPa BC,end 6 2 A 30010 m • The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa. • Since AB in compression, minimum section remain unstressed at A in the boom. Only bearing stress takes place. 1 16 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Pin Shearing Stresses • The crosssectional area for pins at A, B, and C, 2 25mm  26 2 A  r  49110 m  2  • The force on the pin at C is equal to the force exerted by the rod BC, 3 P 5010 N  102MPa C,ave 6 2 A 49110 m • The pin at A is in double shear with a total force equal to the force exerted by the boom AB, P 20kN  40.7 MPa A,ave 6 2 A 49110 m 1 17 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Pin Shearing Stresses • Divide the pin at B into sections to determine the section with the largest shear force, P 15kN E 50 kN P 25kN (largest) G • Evaluate the corresponding average shearing stress, P 25kN G  50.9MPa B,ave 6 2 A 49110 m 1 18 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Pin Bearing Stresses • To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm, P 40kN  53.3MPa b td30mm25mm • To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm, P 40kN  32.0MPa b td50mm25mm 1 19 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Example 2 Strut Gusset Base Anchor Bolts Plate (A) Bearing stress between strut and pin (C) Bearing stress between pin and gusset P P  b2  b1 2t d G pin 2td pin (B) Shear Stress in pin (D) Shear stress in Anchor bolts o P cos 40 P   pin pin 2 2 4d 4 2d 4 bolt pin o P cos 40 (E) Bearing stress between anchor bolts and base plate  bolt 4t d B bolt © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Stress in Two Force Members • We saw that axial forces on a two force member result in only normal stresses on a plane cut perpendicular to the member axis. • We saw that transverse forces on bolts and pins result in only shear stresses on the plane perpendicular to bolt or pin axis. • Will show that either axial or transverse forces may produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member axis. 1 21 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Stress on an Oblique Plane • Pass a section through the member forming an angle q with the normal plane. • From equilibrium conditions, the distributed forces (stresses) on the plane must be equivalent to the force P. • Resolve P into components normal and tangential to the oblique section, F P cosq V Psinq • The average normal and shear stresses on the oblique plane are F Pcosq P 2  cos q A A A 0 q 0 cosq V Psinq P  sinq cosq A A A 0 q 0 cosq 1 22 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Maximum Stresses • Normal and shearing stresses on an oblique plane P P 2  cos q  sin q cosq A A 0 0 • The maximum normal stress occurs when the reference plane is perpendicular to the member axis, P    0 m A 0 • The maximum shear stress occurs for a plane at o + 45 with respect to the axis, P P   sin 45 cos45  m A 2A 0 0 1 23 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS 2   cos q q x  sinq cosq q x Graph of normal stresses  and shear stress  versus angle q of the qq inclined section o   max x Maximum Normal Stress occurs when θ=0 and o   / 2 max x Maximum Shear Stress occurs when θ=±45 and © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Maximum shear stress plane Maximum normal stress plane Normal and shear stresses acting on stress elements oriented at (b) q = 0° and (c) q = 45° for a bar in tension © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Failure of Aluminum in Tensile Test o Failure at 45 plane from a simple tensile test © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Formation of persistent slip bands at yielding results in Slip bands in a polished the failure of ductile steel specimen loaded in materials in shear almost at tension o an angle 45 w.r.t the load under uniaxial loading, even though the maximum o shear stress at 45 plane is half of the maximum normal stress induced © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Stress Under General Loadings • A 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, x F  lim x A A0 x x V V y z   lim lim xy xz AA A0A0 • For equilibrium, an equal and opposite internal force and stress distribution must be exerted on the other segment of the member. 1 28 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS State of Stress • Stress components are defined for the planes cut parallel to the x, y and z axes. For equilibrium, equal and opposite stresses are exerted on the hidden planes. • The combination of forces generated by the stresses must satisfy the conditions for equilibrium: F F F 0  x y z M M M 0  x y z • Consider the moments about the z axis: M 0AaAa  z xy yx   xy yx similarly,   and   yz zy yz zy • It follows that only 6 components of stress are required to define the complete state of stress 1 29 © IIT Bombay , CE Dept, Lecture Notes for CE201 MECHANICS OF MATERIALS Factor of Safety Structural members or machines Factor of safety considerations: must be designed such that the • uncertainty in material properties working stresses are less than the • uncertainty of loadings ultimate strength of the material. • uncertainty of analyses • number of loading cycles FS Factor of safety • types of failure  ultimate stress u FS • maintenance requirements and  allowable stress all deterioration effects • importance of member to structures integrity • risk to life and property • influence on machine function 1 30 © IIT Bombay , CE Dept, Lecture Notes for CE201
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