Indeterminate Stress SystemsMechanics of Materials 材料力学材料力学Professor Shibin WANG （）Indeterminate Stress SystemsMechanics of Materials关于超静定的基本概念关于超静定的基本概念静定问题与静定结构静定问题与静定结构未知力（内力或外力）个数未知力（内力或外力）个数未知力（内力或外力）个数未知力（内力或外力）个数 等于独立的平衡方程数等于独立的平衡方程数等于独立的平衡方程数等于独立的平衡方程数超静定问题与超静定结构超静定问题与超静定结构未知力个数多于独立未知力个数多于独立未知力个数多于独立未知力个数多于独立 的平衡方程数的平衡方程数的平衡方程数的平衡方程数超静定次数超静定次数未知力个数与独立平衡方程数之差未知力个数与独立平衡方程数之差未知力个数与独立平衡方程数之差未知力个数与独立平衡方程数之差多余约束多余约束保持结构静定保持结构静定保持结构静定保持结构静定多余的约束多余的约束多余的约束多余的约束简单的超静定问题简单的超静定问题Ch.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of Materials静定与静定与超超静定的辩证关系静定的辩证关系多余约束的两种作用：多余约束的两种作用：多余约束的两种作用：多余约束的两种作用： 增加了未知力个数，同时增加对变形增加了未知力个数，同时增加对变形增加了未知力个数，同时增加对变形增加了未知力个数，同时增加对变形 限制与约束，前者使问题变为不可解，限制与约束，前者使问题变为不可解，限制与约束，前者使问题变为不可解，限制与约束，前者使问题变为不可解， 后者使问题变为可解。后者使问题变为可解。后者使问题变为可解。后者使问题变为可解。求解超静定问题的基本方法求解超静定问题的基本方法平衡、变形协调、平衡、变形协调、平衡、变形协调、平衡、变形协调、 物性关系。现在的物性关系体现为力与物性关系。现在的物性关系体现为力与物性关系。现在的物性关系体现为力与物性关系。现在的物性关系体现为力与 变形关系。变形关系。变形关系。变形关系。 求解超静定问题的基本方法求解超静定问题的基本方法简单的超静定问题简单的超静定问题Ch.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of Materials 拉压超静定问题拉压超静定问题E2A2 l2E3A3 l3=E2A2 l2E1A1 l1yxABCD例题例题5 5FPFPFN3FN2FN1ACh.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of Materials平衡方程平衡方程平衡方程平衡方程超静定次数：超静定次数：3-2=1yxFPFN3FN2FN1Ch.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of Materials l1 l3 变形协调方程变形协调方程变形协调方程变形协调方程: : : :各各各各杆变形的几何关系杆变形的几何关系杆变形的几何关系杆变形的几何关系E2A2 l2E3A3 l3=E2A2 l2E1A1 l1BCDAFP l2Ch.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of Materials平衡方程平衡方程平衡方程平衡方程: :变形协调方程：变形协调方程：变形协调方程：变形协调方程：物性关系物性关系物性关系物性关系: : : :Ch.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of Materials结果：结果：结果：结果：由平衡方程、由平衡方程、变形协调方程、变形协调方程、物性关系物性关系联立解出联立解出例题例题5 5Ch.4 Indeterminate Stress SystemIndeterminate Stress SystemsMechanics of MaterialsxyCh.4 Indeterminate Stress SystemsBig Question What is an Indeterminate System?Big Answer One in which unknown forces can not be determined from equilibrium alone.This is a DETERMINATE systemFor example:P1P2RxSince:fromWe have ONE unknown, and ONE available equilibrium relationship.Indeterminate Stress SystemsMechanics of MaterialsxyThis is an INDETERMINATE systemPRAxSince RAx & RBx CAN NOT be found from equilibrium:1 equation, 2 unknownsRBxTherefore, we need MORE INFORMATION!4.1 Consider the following problem:Indeterminate Stress SystemsMechanics of MaterialsExamples of Statically Determinate Systems:Beams:Trusses:Other:10 kNVAVBHAVAHAMAzVAVBHAVAHAMANo more than 2 unknowns at a jointIndeterminate Stress SystemsMechanics of MaterialsExamples of Statically Indeterminate Systems:Beams:Trusses:VBVAHAMAOther:How much load carried by concrete, how much by the steel?10 kN- Too many reactions- Too many unknowns at a jointVAHAMAMBHBVBIndeterminate Stress SystemsMechanics of Materialsxye.g.We need more tools!4.2 The Direct Method of AnalysisEquilibrium of ForcesGeometric Compatibility of DeformationsHookes LawPRAxRBxABCuab12E, AEquilibrium:1Indeterminate Stress SystemsMechanics of MaterialsCompatibility:2i.e. Contraction of part = Extension of part12Hookes Law:3xye.g.PRAxRBxABCuab12E, AIndeterminate Stress SystemsMechanics of MaterialsRAxRBxabxyPPRAxRAxs sx1s sx1s sx1s sx2s sx2Fu1Fu2Indeterminate Stress SystemsMechanics of Materials1(From )From Compatibility :2Indeterminate Stress SystemsMechanics of MaterialsabxyPs sx1s sx2E, A0u-ve+veCompressionTensionNOTE:And from :2Indeterminate Stress SystemsMechanics of MaterialsConsider some numbersP=20 kNABCu200 mm100 mm12E=70 GPa, A=100 mm2Indeterminate Stress SystemsMechanics of MaterialsA Scottish engineer, Rankine made observations about the expansion and contraction of materials due to changes in temperature.4.3 Thermal Strains (William Rankine, 1870)xya a=Coefficient of Linear Expansion (A material property)He noted that these deflections were proportional to the change in temperature the material experienced .Indeterminate Stress SystemsMechanics of MaterialsFrom Hookes Law:(Due to Forces and Temperature Changes)Mechanical StrainThermal StrainCoef Th expanax 10-6/oCMild Steel12Aluminium23Concrete10.8Wood -Nylon0.9Rubber130-200Indeterminate Stress SystemsMechanics of MaterialsxyExample: Consider a bar constrained between two walls.From Hookes Law:RxRxE, n n, a aL (fixed)bdu=0But,Indeterminate Stress SystemsMechanics of Materialsi.e.And,Indeterminate Stress SystemsMechanics of MaterialsIf a steel bar has a maximum axial stress of 300 MPa, what is the greatest allowable temperature increase?E, n n, a a, s sMaxE=200 GPan n=0.25a a=12x10-6 1/oC+ve (D DT) i.e. HOTCompression-ve (D DT) i.e. COLDTensionIndeterminate Stress SystemsMechanics of MaterialsReduce an indeterminate system into a series of determinate parts, and sum the parts in a way which would satisfy (i) Equilibrium & (ii) Compatibility.4.4 The Method of SuperpositionRAxRBxuPABCabxye.g.12RAxPABCuuRAx”ABu”RBx=P= -RBxIndeterminate Stress SystemsMechanics of MaterialsSuperposition:(i) At end A,RAx= RAx + RAx”(ii) At end B,u + u” = 0From (ii), u + u” = 0Indeterminate Stress SystemsMechanics of MaterialsFrom (i), RAx= RAx + RAx”(Since RAx = P & RAx”= -RBx)EtcSee (a) The Direct Method of Analysis for the normal force diagram, s sx1, s sx2 and u.Indeterminate Stress SystemsMechanics of MaterialsExample: Composite Column.A short composite column is constructed by filling a round steel pipe with concrete as shown. If the column is placed between two rigid plates and loaded in compression, we would like to find the load-deflection relation for the column. Also, we wish to find the maximum allowable load if the allowable stresses for the steel and concrete are 100 MPa and 8 MPa.Take L=2m, d=0.5m, t=13mm, ES=200 GPa, and EC=14 GPa. Neglect the weight of the materials. Equilibrium Compatibility Hookes Law? u vs P, and PMax ?Indeterminate Stress SystemsMechanics of MaterialsDraw FBDs of the steel pipe, concrete column, and an end plate. Equilibrium Compatibility Hookes LawFS+FCPFSFSuSFCFCuCFrom the FBD of the top plate:12Indeterminate Stress SystemsMechanics of MaterialsConsider FC and FS:3into :31Inserting numbers:L=2 m, AS=20.95x10-3 m2, AC=196.3x10-3 m2i.e. Stiffness!Indeterminate Stress SystemsMechanics of MaterialsMaximum Load?Assume Max Stress reached in steel first:Check:OK!So finally:Indeterminate Stress SystemsMechanics of MaterialsExample: Suspended Bar.A rigid bar DC is attached to two elastic wires AD and BC, having diameters d and 2d, modulus of elasticity E, and length L1. A load P is applied at point H. We wish to find the specific distance x from D at which the load P should be applied so that the rigid bar remains horizontal after the load is applied. Neglect the weight of the rigid bar and wires. Equilibrium Compatibility Hookes Law? x ?Indeterminate Stress SystemsMechanics of MaterialsDCHDraw FBDs of the cables, and rigid bar.FADFBCuCuDPxL-xFADFADFBCFBCL1uDuC EquilibriumFrom bar:12 Compatibility3 Hookes Lawor,xy+veIndeterminate Stress SystemsMechanics of MaterialsConsidering both cables:4Subbing into :34(NOTE: ABC=4AAD)5Subbing into :25Indeterminate Stress SystemsMechanics of Materials4.5 SummaryStatically indeterminate problems require more than equilibrium conditions to solve them. The following tools are required to determine unknown forces and deformations. Equilibrium Compatibility (of geometry) Hookes Law (stress vs strain behaviour)Indeterminate Stress SystemsMechanics of MaterialsExampleIndeterminate Stress SystemsMechanics of MaterialsExampleTwo aluminum bars (EAl = 10.0 x 106 psi) are attached to a rigid support at the left and a cross-bar on the right. An iron bar (EFe = 28.5 x 106 psi)is attached to the rigid support and at the left end there is a gap of b = 0.02 in. The cross sectional area of each bar is 0.5 in2 and the length is 10 in. If the iron bar is stretched until it contacts the cross bar and welded to it what are the normal stresses in the bars?Indeterminate Stress SystemsMechanics of MaterialsSolutionIndeterminate Stress SystemsMechanics of MaterialsSolutionIndeterminate Stress SystemsMechanics of MaterialsIndeterminate Stress SystemsMechanics of MaterialsCh.3 ExperimentThank You