The.above.mentioned.description.describes.the.introduction.of.the.two.fundamental.methods.of.structural.analysis,.the.method.of.displacement.and. force.method,.and.their.applications.on.two.groups.of.constructions,.sports,. and.the.limits.of.linear.and.static.structural.analysis.are.presented.at.the.end. National.Taiwan.University),.whose.love.of.education.and.structural.theory. directed.my.career,.and.to.the.late.Professor.Yuan.Yu.Hsieh,.whose.innovation.in.the.best-selling,.groundbreaking.book.Elementary Theory of Structures.
What Is a Truss?
A Truss Member
When.there.is.strength.in.a.member,.the.member.will.deform.Each.segment.of.the.member.will.be.lengthened.or.shortened.and.the.cumulative.effect. .of.
Member Stiffness Equation in Global Coordinates
Equation.1.4.is.the.transformation.strain.equation.Now.we.find.the.transformation.between.the.force.term.in.local.coordinates,.FL =.F.and.nodal. The.significance.of.each.of.the.components.of.the.matrix,.(kG)ij,.can.be.investigated.
Unconstrained Global Stiffness Equation
Note that we are using the terminology of DOF, which stands for degrees of freedom. The configuration is fully defined for the entire truss. against.the.six.
Constrained Global Stiffness Equation and Its Solution
Consider.the.same.three-member.truss.as.in.Example.1.2.but.with.a.different.numbering.system.for.members.Construct.the.restricted.stiffness.equation, Equation 1.16.
Procedures of Truss Analysis
Armed with this table, we can easily direct the member stiffness components to the correct location in the global stiffness matrix. For example, the (2,3) component of (kG)5 will be added to the (4,7) component of the global stiffness matrix.
Kinematic Stability
Summary
Introduction
Statically Determinate Plane Truss Types
Mathematically.such.a.situation.implies.that.there.will.not.be.a.unique.solution.for.any.given.load.because.the.self-equilibrium.“solution” .can Static.indeterminate. truss problems.cannot.be.solved.by.equilibrium.conditions.alone.The.compatibility.conditions.must.be.used.to.supplement.the.equilibrium.conditions. This.solution.
Method of Joints and Method of Sections
Find the force in the inclined web elements on the third panel from the left of the K-truss shown next. Discuss methods for finding the force in the vertical web members of the K-truss shown next.
Matrix Method of Joints
Contribution from externally applied forces. applying.to.node.i.a.magnitude.of.Pn.making.an.angle.α from.the.x-axis,. 1. Set.the.member.number,.the.global.node.number,.the.global.node.coordinates,.and.the.starting.member.and.the.end.node.numbers.From these.
Truss Deflection
The.work.done.with.the.load.of.the.unit.(virtual.system).on.the.real.movement.is.
Indeterminate Truss Problems: Method of Consistent Deformationsof Consistent Deformations
The.hood.above,.with.the.central.support.removed,.the.primary.structure..Take.Take.Note.the.primary.structure.is.statically.determined.The.size.is. of.Rc.is.determined.by.the.condition.that.the.vertical.displacement. The.displacement.of.node.c.due.to.the.excess.force.Rc.cannot.be.calculated.directly because.Rc.itself.is.unknown.We.can.calculate. ,.the. Displacement.at.c due.to.a.unit.load.at.c. The.vertical.displacement.at.c.due.to.the.excess.force.Rc.is.then.Rcδcc,.as. shown.in.the.following.figure.
Displacement.at.c.due.to.the.redundant.force.Rc. The.condition.that.the.total.vertical.displacement.at.node.c,.Δc.is.zero. Fi =.the.member force.due.to.the.actual.applied.load;.Vi =.Fi/(EA/L)i.=.ith.member.
Laws of Reciprocity
We thus only need to find the element forces Fi, fic and fid, corresponding to the real load, respectively a unit load at node c and a unit load at node d from the primary structure.
Concluding Remarks
What Are Beams and Frames?
As.we.can.see.at.a.typical.section.are.three.possible.actions.or.internal forces,.bending moment,.M,.shear force,.V. ,.and.axial.force.or.thrust,.T.For.
Statical Determinacy and Kinematic Stability
If.one.or.several.hinges.are.present.in.a.frame,.we.must.take.into.account.the.conditions.generated.by.the.presence.of.the.hinges.Such.as. shown.in .the.coming.digit,. We make nine cuts that divide the original frame into four “trees” of frames, as shown. This simple method of counting can be extended to multi-story, multi-bay frames with hinges: simply treat the construction conditions of each hinge as “releases” and subtract the ∑C number from the degree of indeterminacy of the frame with the hinges removed .
Shear and Moment Diagrams
Equation.4.1b.and.Equation.4.2b.to.mark.the.fact.that.it.will.suddenly. change.across.section,.when.concentrated.load.or.concentrated. moment.externally.applied.at.the.site.of.the.section. using.equation.4.1a,.we.arrive.at.
Statically Determinate Beams and Frames
The problem is solved by using the principle of superposition, which states that for a linear structure the solution of the structure under two load systems is the sum of the solutions of the structure under each force system. After trial and error, the following FBD provides a simple solution for the axial force in the two columns. Once the axial force in the two columns is known, we can proceed to define four FBDs to expose all the nodal forces at the internal hinges, as shown in the upcoming figure, and solve for any unknown nodal forces one by one by equilibrium equations of each FBD.
Within each FBD, the force values in bold are those known from previous calculations and the other three unknowns are obtained from the equilibrium equations of the FBD itself. We note that we could have used the twelve equilibrium equations from the above four FBDs to solve the twelve force unknowns without the help of the previous FBD to first find the axial force in the two columns. For shear and moment diagrams, we use the same sign convention for both beams and columns.
For vertically oriented columns, it is common to equate the "inside" of the column with the "bottom" of the beam and draw the positive and negative shear and moment diagrams accordingly. Before drawing the thrust, shear and moment diagrams, we need to find the nodal forces that are in the direction of the axial force and the shear force. When the nodal forces are correctly oriented, drawing thrust, shear and moment diagrams is accomplished effortlessly.
Deflection of Beams and Frames
Integration Methods
If.we.express.M.as.a.function.of.x.from.the.moment.diagram,.then.we.can.integrate.equation.5.5.once.to.obtain.the.rotation. . We.can.now.define.a"conjugate.beam",.as.an.elastic.load.of.magnitude.-M/EI.is.applied.We.can.draw.it. displacement and.moment.diagrams.of. 4.Draw.the.displacement.diagram.for.the.conjugate.ray;.positive.displacement.indicates.counterclockwise.rotation.of.the.original.ray.
Energy Methods
Frame Deflection
For each of the three quantities we can place the unit charge as shown in the following figure. The calculation process, including the reaction diagram, the moment diagram for m and the integration, can be tabulated as shown below. The table also includes the calculation from Example 5.11 so that readers can see how the tabulation is performed.
Knowing the rotation and displacement at key points, we can draw the displaced configuration of the frame as shown below. It is necessary to clearly specify that the rotation at b is for the end of bar b~c, because the rotation at b is different for bar a~b.
Statically Indeterminate Beams and Frames
Indeterminate Beam Analysis
Comparing the previous two figures, we note that the combined effect of the load P and the reaction Rb must be such that the total vertical displacement at b is zero, which is dictated by the cylinder support condition of the original problem. The combination of b and Rbδbb is based on the principle of superposition, which states that the displacement of a linear structure due to two loads is a superposition of the displacement due to each of the two loads. To find b and δbb we can use the conjugate ray method for each one as shown below.
Find all reactions of the same bundle as in example 6.1, but choose a different redundant force. We now choose to place the hinge at the fixed end, effectively selecting the end moment, Ma, as the excess force. The compatibility equation is established based on the condition that the total rotation at a of the primary structure due to the combined effect of the applied load and the redundant force Ma must be zero, which is required by the fixed end support.
From the conjugate beam, the rotation at point a is calculated as the shear of the conjugate beam at a. The compatibility condition is that the vertical displacement at the center beam is zero. The primary structure, deflections at the center due to loading, redundant force and so on are shown next.
Indeterminate Frame Analysis
The matrix on the left is symmetric due to Maxwell's reciprocal law. deformations.Some.of.the.formulas.are.given.in.the.next.table. Approximate methods for statically indeterminate frames. As we can see. in the.previous.examples.the.power.method.of.analysis.for.frames.is.practical. An easier way for automation is the move method, which. Vertical load..For.regular.rigid.frames.loaded.with.vertical.floor. loads.as.shown.in.the.following.figure.is.the.deflection.of.the.beams. such.that.no.moment.exists.at.a.location.about.one-tenth.of.the.
Once.we.put.a.pair.of.hinge-and-roller.at.the.location.of.zero.moment.in. the.beams,.the.resulting.frame.is.statically.determined.and.can.be.analyzed. easy.The.following.figure.illustrates.the.solving.process. Beams.and.columns.as.statically.determined.components. fit.either.the.portal.method.or.the.tilt.method.The.portal.method.is. generally.applicable.to.low-rise.frames.of.no.more.than.five.stories.
Introduction
Moment Distribution Method
For.rotation.at.the.end.of.member.ab.and.member.bc.to.happen.a.end.
