fundamentals of hematological analysis

... important in theunderstanding of the behavior of a structure and the development of a feel for it The force method of analysis is then extended to beam and rigid frame analysis, almost inparallel ... sequenced to refer to aparticular DOF For example, the second DOF of member 2 is the 4th DOF in the globalnodal displacement vector Conversely, the third DOF in the global DOF nodal ... stressanalysis, limits of linear and static structural analysis are presented at the end Trang 7Truss Analysis: Matrix Displacement Method1 What Is a Truss? In a plane, a truss is composed of relatively

Ngày tải lên: 06/07/2014, 11:20

... are changed by the addition of one more diagonal member (9) Concluding remarks If the number of nodes is N and the number of constrained DOF is C, then (a) the number of simultaneous equations ... Instability Internal instability happens if the total number of force unknowns is less than the number of displacement DOFs If we denote the number of member force unknowns as M and support reaction unknowns ... 5Truss Analysis: Matrix Displacement Method by S T MauKinematic instability resulting from insufficient number of supports or members. Problem 4: Discuss the kinematic stability of each of the

Ngày tải lên: 06/07/2014, 11:20

... Treatment of side-sway The end-nodes of a member may have translation displacements perpendicular to the axis of the member, creating a “rotation” like configuration of the member This kind of displacement ... force equilibrium of the whole structure: FBD of the whole structure. The two shear forces in the third equation can be expressed in terms of member-end moments, via the FBD of each member V a ... rotation at node b, there is another DOF,which is the horizontal displacement of node b or c, designated as ∆ as shown Nodal lateral displacement that creates side-sway of memebr ab. Note that node b

Ngày tải lên: 05/08/2014, 09:20

... and M d of the beam shown and find the maximum value of each for a distributed load of intensity 10 kN/m and indefinite length of coverage Problem 1 (1) (2) Construct the influence lines of VbL ... and VbR of the beam shown and find the maximum value of each for a distributed load of intensity 10 kN/m and indefinite length of coverage Problem 1 (2) (3) Construct the influence lines of VcL, ... that satisfies all constraints of the principle Sketch of influence line for M a of section a. Example 5 Place uniformly distributed loads anywhere on the second floor of the frame shown in Example

Ngày tải lên: 05/08/2014, 09:20

... used in any displacement method of analysis. A sample of the numerical values of the factors in these two equations is given in the table below for two configurations of rectangular sections. The ... direct effect of temperature change and construction or manufacturing error is the change of shape or dimension of a structural member. For a statically determinate structure, this change of shape ... is linearly distributed from the bottom of a section to the top of the section and is constant along the length of the member. The strain at any level of the section can be computed as shown:

Ngày tải lên: 05/08/2014, 09:20

... degrees of freedom because the spatial variable, x, is continuous and represents an infinite number of points along the beam We shall pursue an approximate analysis by lumping the total mass of the ... the beam at the tip of the beam This results in a single degree of freedom (SDOF) system because we need to consider dynamic equilibrium only at the tip Dynamic equilibrium of a distributed mass ... dynamic equilibrium of this SDOF system is shown in the above figure The dynamic equilibrium equation of the lumped mass is m 2 2 dt v d where m= ρL and k is the force per unit length of lateral deflection

Ngày tải lên: 05/08/2014, 09:20

... are changed by the addition of one more diagonal member (9) Concluding remarks If the number of nodes is N and the number of constrained DOF is C, then (a) the number of simultaneous equations ... global DOF number corresponding to each local DOF of each member This table is generated using the member data given in the table in subsection (2), namely the starting and end nodes data Local DOF ... given at the beginning of this section is obtained For example, the unconstrained global stiffness matrix component k34 is the superposition of (k34)1 of member and (k12)2 of member Note that the...

Ngày tải lên: 05/08/2014, 09:20

... direction of the FBD to the left or right of the cut 52 Truss Analysis: Force Method, Part I by S T Mau (2) Method of joint 16 kN Joints used to solve for vertical web member forces At each of the ... often leads to mistakes Example Find the member forces in bars 4, 5, 6, and of the loaded Fink truss shown 10 3 10 kN 3@2m=6m Fink truss to be solved by the method of joint 39 2m 11 1 Truss Analysis: ... the accuracy of the computation We need not use these three joint equations because we have already used three equations from the equilibrium of the whole structure at the beginning of the solution...

Ngày tải lên: 05/08/2014, 09:20

... Truss Analysis: Force Method, Part I by S T Mau Matrix Method of Joint The development of the method of joint and the method of section pre-dates the advent of electronic computer ... deflection is the result of displacements of some or all of the truss nodes and nodal displacements are caused by the change of length of one or more members 2 3 Elongation of member induces nodal ... displacement method of analysis Since we are developing the force method of analysis herein, we shall not explore the application of this principle further at this point (3) Principle of Virtual Force...

Ngày tải lên: 05/08/2014, 09:20

... displacement of node of the loaded truss shown, given E=10 GPa, A=100 cm2 for all bars The magnitude of the pair of loads is 141.4 kN 1 2 3 4m 3@4m=12m Problem 5-2 (3) The lower chord members 1, 2, and of ... a counterclockwise rotation of 9.83mm/4,000mm=0.0025 radian 80 Truss Analysis: Force Method, Part II by S.T.Mau Problem (1) Find the horizontal displacement of node of the loaded truss shown, ... rotation of bar 2, we note that the –9.83 computed represents a relative vertical movement between node and node of 9.83 mm in the opposite direction of what was assumed for the pair of unit loads...

Ngày tải lên: 05/08/2014, 09:20

... value of the load (4) According to Eq 2, the slope of the moment diagram is equal to the value of the shear, and the integration of the shear function gives the moment diagram 103 Beam and Frame Analysis: ... is an applied load (q≠0), the direction of change of the shear diagram follows the direction of the load and the rate of change is equal to the intensity of the load If a concentrated load is encountered, ... N=2, C=0 M=1, N=2, C=1 R=2 Number of unknowns = 3M+ΣR =8, Number of equations = 3N+ΣC = Indeterminate to the 2nd degree R=1 Number of unknowns = 3M+ΣR =7, Number of equations = 3N+ΣC = Statically...

Ngày tải lên: 05/08/2014, 09:20