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1、材料力学 Professor Shibin WANG （王世斌）,关于超静定的基本概念,静定问题与静定结构未知力（内力或外力）个数 等于独立的平衡方程数 超静定问题与超静定结构未知力个数多于独立 的平衡方程数 超静定次数未知力个数与独立平衡方程数之差 多余约束保持结构静定多余的约束,简单的超静定问题,Ch.4 Indeterminate Stress System,静定与超静定的辩证关系多余约束的两种作用： 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， 后者使问题变为可解。 求解超静定问题的基本方法平衡、变形协调、 物性关系。现在的物性关系体现为力与 变形关系。, 求解
2、超静定问题的基本方法,简单的超静定问题,Ch.4 Indeterminate Stress System, 拉压超静定问题,E2A2 l2,E3A3 l3=E2A2 l2,E1A1 l1,例题5,Ch.4 Indeterminate Stress System,平衡方程,超静定次数：3-2=1,Ch.4 Indeterminate Stress System,变形协调方程:各杆变形的几何关系,Ch.4 Indeterminate Stress System,平衡方程:,变形协调方程：,物性关系:,Ch.4 Indeterminate Stress System,结果：由平衡方程、变形协调方程、
3、物性关系联立解出,例题5,Ch.4 Indeterminate Stress System,Ch.4 Indeterminate Stress Systems,Big Question What is an Indeterminate System?,Big Answer One in which unknown forces can not be determined from equilibrium alone.,This is a DETERMINATE system,For example:,Since:,from,We have ONE unknown, and ONE availa
4、ble equilibrium relationship.,This is an INDETERMINATE system,Since RAx & RBx CAN NOT be found from equilibrium:,1 equation, 2 unknowns,Therefore, we need MORE INFORMATION!,4.1 Consider the following problem:,Examples of Statically Determinate Systems:,Beams:,Trusses:,Other:,Examples of Statically I
5、ndeterminate Systems:,Beams:,Trusses:,Other:,How much load carried by concrete, how much by the steel?,We need more tools!,4.2 The Direct Method of Analysis,1,2,3,RAx,RBx,a,b,x,y,P,a,b,x,y,P,E, A,NOTE:,Consider some numbers,P=20 kN,200 mm,100 mm,A Scottish engineer, Rankine made observations about t
6、he expansion and contraction of materials due to changes in temperature.,4.3 Thermal Strains (William Rankine, 1870),a=Coefficient of Linear Expansion (A material property),He noted that these deflections were proportional to the change in temperature the material experienced .,From Hookes Law:,(Due
7、 to Forces and Temperature Changes),Example: Consider a bar constrained between two walls.,From Hookes Law:,E, n, a,b,d,u=0,But,i.e.,And,If a steel bar has a maximum axial stress of 300 MPa, what is the greatest allowable temperature increase?,E, n, a, sMax,Reduce an indeterminate system into a seri
8、es of determinate parts, and sum the parts in a way which would satisfy (i) Equilibrium & (ii) Compatibility.,4.4 The Method of Superposition,=P,= -RBx,Superposition:,From (ii), u + u” = 0,From (i), RAx= RAx + RAx”,(Since RAx = P & RAx”= -RBx),Example: Composite Column.,A short composite column is c
9、onstructed 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
10、 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 ?,Draw FBDs of the steel pipe, concrete column, and an end plate.,Equilibrium,Compatibility,Hookes Law,From the FBD of t
11、he top plate:,1,2,Consider FC and FS:,3,Inserting numbers:,L=2 m, AS=20.95x10-3 m2, AC=196.3x10-3 m2,i.e. Stiffness!,Maximum Load?,Assume Max Stress reached in steel first:,Check:,OK!,So finally:,Example: Suspended Bar.,A rigid bar DC is attached to two elastic wires AD and BC, having diameters d an
12、d 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,Compatibil
13、ity,Hookes Law,? x ?,Draw FBDs of the cables, and rigid bar.,Equilibrium,From bar:,1,2,Compatibility,3,Hookes Law,or,Considering both cables:,4,(NOTE: ABC=4AAD),5,4.5 Summary,Statically indeterminate problems require more than equilibrium conditions to solve them. The following tools are required to
14、 determine unknown forces and deformations.,Example,Example,Two 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?,Solution,Solution,Ch.3 Experiment,Thank You,
