东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A10 Shearing and Bearing Stress

Shearing and Bearing Stressmi@seeo.cn

Shearing and Bearing Stress mi@seu.edu.cn

Contents·Introduction to Pin Shearing and Bearing（连接件中的剪切和挤压简介）·Shearing Stress（剪切应力）·Bearing Stress（挤压应力）·StressAnalysis&DesignExample（应力分析与设计示例)vRod&BoomNormal Stresses（杆件拉压正应力）vPin Shearing Stresses（销钉剪切应力）vPinBearingStresses（销钉挤压应力）·More Examples（更多示例）2

Contents • Introduction to Pin Shearing and Bearing（连接件中的 剪切和挤压简介） • Shearing Stress（剪切应力） • Bearing Stress（挤压应力） • Stress Analysis & Design Example（应力分析与设计示 例）  Rod & Boom Normal Stresses（杆件拉压正应力）  Pin Shearing Stresses（销钉剪切应力）  Pin Bearing Stresses（销钉挤压应力） • More Examples（更多示例） 2

Introduction to Pin Shearing and BearingLoadLoadFailure of a boltin single shear3

Failure of a bolt in single shear Introduction to Pin Shearing and Bearing 3

Shearing Stress1. Forces P and P'are applied transversely to themember AB.BA: Corresponding internal forces act in the planeof section C and are called shearing forcesTheresultantoftheinternal shearforcedistribution is defined as the shear of the sectionand is equal to the load P.The corresponding average shearing stress is,BAPTaveA? Shearing stress distribution varies from zero atthemembersurfacestomaximumvaluesthatAmay be much larger than the average value: The shearing stress distribution cannot be assumedto be uniform.4

Shearing Stress • Forces P and P’ are applied transversely to the member AB. A P  ave  • The corresponding average shearing stress is, • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • Corresponding internal forces act in the plane of section C and are called shearing forces. • Shearing stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value. • The shearing stress distribution cannot be assumed to be uniform. 4

Shearing StressSingle ShearDouble ShearHEPKKEBAFBNLLDDGHFcKF5-PFPFTave2ave2AAAA5

Shearing Stress A F A P  ave   Single Shear A F A P 2  ave   Double Shear 5

Bearing Stress. Bolts, rivets, and pinscreate stresses on thepoints of contact orbearing surfaces of thedmembers they connect.. The resultant of the forcedistribution on the surfaceis equal and opposite to theforce exerted on the pin: Corresponding averageforce intensity is calledthe bearing stress,PPohtdA6

Bearing Stress • Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect. t d P A P  b   • Corresponding average force intensity is called the bearing stress, • The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin. 6

Stress Analysis & Design Exampled=25mm: Would like to determinethe stresses in the20mmFlat endTOPVIEWOFRODBCmembers and connectionsof the structure shown40 mmd=20mm: From a statics analysis:d=20mm25mmFAB= 40 kN600mmFRONTVIEW(compression)FBc = 50 kN (tension)Flat end50mm. Must consider maximumnormal stresses in AB and800mmQ=30kNQ=30kNBC, and the shearingENDVIEW125mm120mmstress and bearing stress30mmT20mm25mmat each pinned connectionBTOPVIEWOFBOOMAB7d=25mm

• Would like to determine the stresses in the members and connections of the structure shown. Stress Analysis & Design Example • Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection • From a statics analysis: FAB = 40 kN (compression) FBC = 50 kN (tension) 7

Rod & Boom Normal Stresses: The rod is in tension with an axial force of 50 kN.25mm? At the rod center, the average normal stress in the20mmcircular cross-section (A = 314x10-6m2) is Bc =Flat endTC+159 MPa.40mmAt the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,A=(20mm)(40mm-25mm)= 300x10-6m250×103NP= 167MPa BC,end300×10-6m2A: The boom is in compression with an axial force of40 kN and average normal stress of -26.7 MPa.: The minimum area sections at the boom ends areunstressed since the boom is in compression.8

Rod & Boom Normal Stresses • The rod is in tension with an axial force of 50 kN. • The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa. • The minimum area sections at the boom ends are unstressed since the boom is in compression.    167MPa 300 1 0 m 5 0 1 0 2 0mm 4 0mm 2 5mm 300 1 0 m 6 2 3 , 6 2            N A P A  B C end • At the flattened rod ends, the smallest cross￾sectional area occurs at the pin centerline, • At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2 ) is BC = +159 MPa. 8

Pin Shearing Stresses. The cross-sectional area for pins atA, B, and C,25mm50kN= 491x10-6m2A=元r2(a)-d=25mm. The force on the pin at C is equal50kN50kMto the force exerted by the rod BC50×103NP491x10-0m2=102MPatC,aveA0k. The pin at A is in double shearwith a total force egual to thead=25mmforce exerted by the boom AB.P20kNFD10kN= 40.7MPaT A,ave40kN491x10-℃m2AE9

Pin Shearing Stresses • The cross-sectional area for pins at A, B, and C, 6 2 2 2 491 10 m 2 25mm          A   r   102MPa 491 10 m 50 10 N 6 2 3 ,      A  P C ave  • The force on the pin at C is equal to the force exerted by the rod BC, • The pin at A is in double shear with a total force equal to the force exerted by the boom AB, 40.7MPa 491 10 m 20kN 6 2 ,     A  P A ave  9

Pin Shearing Stresses: Divide the pin at B into sections to determineFAB=20kNthe section with the largest shear forceFAB=20kNQ=15kNPe =15kNGPin BE0FBc=50kNQ=15kNPG = 25kN (largest)(a)PEEvaluatethecorresponding averageshearing stress,PG25kN=50.9MPaTB,ave6m2A491x10-mQ=15kN(b)1FAB=20kNIQ=15kN10

• Divide the pin at B into sections to determine the section with the largest shear force, 25kN (largest) 15kN   G E P P 50.9MPa 491 10 m 25kN 6 2 ,     A  PG B ave  • Evaluate the corresponding average shearing stress, Pin Shearing Stresses FBC = 50kN 10