The.development.of.the.method.of.joints.and.the.method.of.sections.predates.
the.advent.of.the.electronic.computer..Although.both.methods.are.easy.to.
apply,.it.is.not.practical.for.trusses.with.many.members.or.nodes,.especially.
when.all.member.forces.are.needed..It.is,.however,.easy.to.develop.a.matrix.
formulation.of.the.method.of.joints..Instead.of.manually.establishing.all.the.
equilibrium.equations.from.each.joint.or.from.the.whole.structure.and.then.
putting.the.resulting.equations.in.a.matrix.form,.there.is.an.automated.way.
of.assembling.the.equilibrium.equations.as.shown.herein.
Assuming.there.are.N.nodes.and.M.member.force.unknowns.and.R.reac- tion.force.unknowns.and.2N =.M + R.for.a.given.truss,.we.know.there.will.be.
2N.equilibrium.equations,.two.from.each.joint..We.shall.number.the.joints.or.
nodes.from.one.to.N..At.each.joint,.there.are.two.equilibrium.equations..We.shall.
define.a.global.x-y.coordinate.system.that.is.common.to.all.joints..We.note,.how- ever,.it.is.not.necessary.for.every.node.to.have.the.same.coordinate.system,.but.it.
is.convenient.to.do.so..The.first.equilibrium.equation.at.a.node.will.be.the.equi- librium.of.forces.in.the.x-direction.and.the.second.will.be.for.the.y-direction..
These.equations.are.numbered.from.one.to.2N.in.such.a.way.that.the.x-direction.
equilibrium.equation.from.the.ith.node.will.be.the.(2i – 1)th.equation.and.the.
y-direction.equilibrium.equation.from.the.same.node.will.be.the.(2i)th.equation..
In.each.equation,.there.will.be.terms.coming.from.the.contribution.of.mem- ber.forces,.externally.applied.forces,.or.reaction.forces..We.shall.discuss.each.of.
these.forces.and.develop.an.automated.way.of.establishing.the.terms.in.each.
equilibrium.equation.
Contribution from member forces..A.typical.member,.k,.having.a.starting.node,.
i,.and.an.ending.node,.j,.is.oriented.with.an.angle.θ from.the.x-axis.as.shown.
next.
k
i
j
θ
θ
θ θ x
y
k
i
j
Fk
Fk Fk
Fk i
j
Member.orientation.and.the.member.force.acting.on.member-end.and.nodes.
The.member.force,.assumed.to.be.tensile,.pointing.away.from.the.member.
at.both.ends.and.in.opposite.direction.when.acting.on.the.nodes,.contributes.
to.four.nodal.equilibrium.equations.at.the.two.end.nodes.(we.designate.the.
right-hand. side. [RHS]. of. an. equilibrium. equation. as. positive. and. put. the.
internal.nodal.forces.to.the.left-hand.side.[LHS]):
. (2i –.1)th.equation.(x-direction):.(–Cosθ)Fk.to.the.LHS . (2i)th.equation.(y-direction):.(–Sinθ)Fk.to.the.LHS . (2j – 1)th.equation.(x-direction):.(Cosθ)Fk.to.the.LHS . (2j)th.equation.(y-direction):.(Sinθ)Fk.to.the.LHS
Contribution from externally applied forces.. An. externally. applied. force,.
applying.at.node.i.with.a.magnitude.of.Pn.making.an.angle.α from.the.x-axis,.
contributes.to:
i
Pn α
Externally.applied.force.acting.at.a.node.
. (2i –.1)th.equation.(x-direction):.(Cosα)Pn.to.the.RHS . (2i)th.equation.(y-direction):.(Sinα)Pn.to.the.RHS
Contribution from reaction forces..A.reaction.force.at.node.i.with.a.magnitude.
of.Rn.making.an.angle.β from.the.x-axis,.contributes.to:
i
β Rn
Reaction.force.acting.at.a.node.
. (2i –.1)th.equation.(x-direction):.(–Cosβ)Rn.to.the.LHS . (2i)th.equation.(y-direction):.(–Sinβ)Rn.to.the.LHS
Input and solution procedures..From.the.aforementioned.definition.of.forces,.
we.can.develop.the.following.solution.procedures.
. 1..Designate.member.number,.global.node.number,.global.nodal.coor- dinates,.and.member.starting.and.end.node.numbers..From.these.
input,.we.can.compute.member.length,.L,.and.other.data.for.each.
member.with.starting.node.i.and.end.node.j:
. x x x y y L x y x
L
y
; ; ( ) ( ) ; Cos ; Sin L
j i i 2 2
= − = = + θ = θ =
. 2..Define. reaction. forces,. including. where. the. reaction. is. at. and. the.
orientation.of.the.reaction,.one.at.a.time..The.cosine.and.sine.of.the.
orientation.of.the.reaction.force.should.be.input.directly.
. 3..Define.externally.applied.forces,.including.where.the.force.is.applied.
and.the.magnitude.and.orientation,.defined.by.the.cosine.and.sine.of.
the.orientation.angle.
. 4..Compute. the. contribution. of. member. forces,. reaction. forces,. and.
externally. applied. forces. to. the. equilibrium. equation. and. place.
them. to. the. matrix. equation.. The. force. unknowns. are. sequenced.
with.the.member.forces.first,.F1,.F2,.…,.FM,.followed.by.reaction.force.
unknowns,.FM+1,.FM+2,.…,.FM+R.
. 5..Use.a.linear.simultaneous.algebraic.equation.solver.to.solve.for.the.
unknown.forces.
Example 2.11
Find all support reactions and member forces of the loaded truss shown next.
x y
1
2
4 m
3 m
3 3 m
1 2
3 1.0 kN
0.5 kN
A truss problem to be solved by the matrix method of joint.
Solution
We shall provide a step-by-step solution.
1. Designate member number, global node number, global nodal coordinates, and member starting and end node numbers, and compute member length, L, and other data for each member.
Nodal.Input.Data
Node x-coordinate y-coordinate
1 0.0 0.0
2 3.0 4.0
3 6.0 0.0
Member.Input.and.Computed.Data Member
Start Node
End
Node Δx Δy L Cosθ Sinθ
1 1 2 3.0 4.0 5.0 0.6 0.8
2 2 3 3.0 –4.0 5.0 0.6 –0.8
3 1 3 6.0 0.0 6.0 1.0 0.0
2. Define reaction forces.
Reaction.Force.Data
Reaction At Node Cosβ Sinβ
1 1 1.0 0.0
2 1 0.0 1.0
3 3 0.0 1.0
3. Define externally applied forces.
Externally.Applied.Force.Data
Force At Node Magnitude Cosα Sinα
1 2 0.5 1.0 0.0
2 2 1.0 0.0 –1.0
4. Compute the contribution of member forces, reaction forces, and externally applied forces to the equilibrium equations and set up the matrix equation.
Contribution.of.Member.Forces Member
Number
Force Number
Equation Number and Value of Entry
2i – 1 Coeff. 2i Coeff. 2j – 1 Coeff. 2j Coeff.
1 1 1 –0.6 2 –0.8 3 0.6 4 0.8
2 2 3 –0.6 4 0.8 5 0.6 6 –0.8
3 3 1 –1.0 2 0.0 5 1.0 6 –0.0
Contribution.of.Reaction.Forces Reaction
Number
Force Number
Equation Number and Value of Entry 2i – 1 Coeff. 2i Coeff.
1 4 1 –1.0 2 0.0
2 5 1 0.0 2 –1.0
3 6 5 0.0 6 –1.0
Contribution.of.Externally.Applied.Forces
Equation Number and Value of Entry Applied Force 2i – 1 Coeff. 2i Coeff.
1 1 1.0 2 0.0
2 1 0.0 2 1.0
3 5 0.0 6 1.0
Using the aforementioned data, we obtain the equilibrium equa- tion in matrix form:
− − −
− −
−
− −
0.6 0 1.0 1.0 0.0 0
0.8 0 0.0 0.0 1.0 0
0.6 0.6 0 0 0 0
0.8 0.8 0 0 0 0
0 0.6 1.0 0 0 0.0
0 0.8 0.0 0 0 1.0
F F F F F F
0 0 0.5
1.0 0 0
1
2
3
4
5
6
= −
5. Solve for the unknown forces. An equation solver produces the following solutions, where the units are added by the user:
F1 = –0.21 kN; F2 = –1.04 kN; F3 = 0.62 kN;
F4 = –0.50 kN; F5 = 0.17 kN; F6 = 0.83 kN
PROBLEM 2.3
The.loaded.truss.shown.next.is.different.from.that.in.Example.2.11.only.
in. the. externally. applied. loads.. Modify. the. results. of. Example. 2.11. to.
establish.the.matrix.equilibrium.equation.for.this.problem.
x y
1
2
4 m
3 m
3 3 m
1 2
3 1.0 kN
0.5 kN
Problem 2.3
PROBLEM 2.4
Establish. the. matrix. equilibrium. equation. for. the. loaded. truss. shown.
next.
x y
1
2
4 m
3 m
3 3 m
1 2
1.0 kN 0.5 kN
Problem.2.4
Force transfer matrix..Consider.the.same.three-bar.truss.as.in.the.previous.
example.problems..If.we.apply.a.unit.force.one.at.a.time.at.one.of.the.six.
possible.positions,.that.is,.x-.and.y-directions.at.each.of.the.three.nodes,.we.
have.six.separate.problems.as.shown.in.the.following.figure.
x y 1
2
4 m
3 m
3 3 m
1 2
3 x
y 1
2
4 m
3 m
3 3 m
1 2
3
x y 1
2
4 m
3 m
3 3 m
1 2
3 x
y 1
2
4 m
3 m
3 3 m
1 2
3
x y 1
2
4 m
3 m
3 3 m
1 2
3 x
y 1
2
4 m
3 m
3 3 m
1 2
3
1 kN 1 kN
1 kN 1 kN
1 kN
1 kN
Truss.with.unit.loads.
The.matrix.equilibrium.equation.for.the.first.problem.appears.in.the.fol- lowing.form:
.
− − −
− −
−
− −
=
F F F F F F
0.6 0 1.0 1.0 0.0 0
0.8 0 0.0 0.0 1.0 0
0.6 0.6 0 0 0 0
0.8 0.8 0 0 0 0
0 0.6 1.0 0 0 0.0
0 0.8 0.0 0 0 1.0
1 0 0 0 0 0
1 2 3 4 5 6
. . (2.2)
The.RHS.of.the.equation.is.a.unit.vector..For.the.other.five.problems.the.
same.matrix.equation.will.be.obtained.only.with.the.RHS.changed.to.unit.
vectors.with.the.unit.load.at.different.locations..If.we.compile.the.six.RHS.
vectors.into.a.matrix,.it.becomes.an.identity.matrix:
.
I
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
=
.
(2.3)
The. six. matrix. equations. for. the. six. problems. can. be. put. into. a. single.
matrix.equation.if.we.define.the.six-by-six.matrix.at.the.LHS.of.Equation.2.2.
as.matrix.A,
.
− − −
− −
−
− −
=A
0.6 0 1.0 1.0 0.0 0
0.8 0 0.0 0.0 1.0 0
0.6 0.6 0 0 0 0
0.8 0.8 0 0 0 0
0 0.6 1.0 0 0 0.0
0 0.8 0.0 0 0 1.0
. (2.4)
and.the.six.force.unknown.vectors.as.a.single.six-by-six.matrix.F:
A6×6 F6×6.=.I6×6. (2.5)
The.solution.to.the.six.problems,.obtained.by.solving.the.six.problems.one.
at.a.time,.can.be.compiled.into.the.single.matrix.F,
.
=
−
−
− − −
− − −
− −
F×
0.0 0.0 0.83 0.63 0.0 0.0
0.0 0.0 0.83 0.63 0.0 0.0
0.0 0.0 0.5 0.38 1.0 0.0
1.0 0.0 1.0 0.0 1.0 0.0
0.0 1.0 0.67 0.5 0.0 0.0
0.0 0.0 0.67 0.5 0.0 1.0
6 6
. . (2.6)
where.each.column.of.the.matrix.F.is.a.solution.to.a.unit.load.problem..Matrix.
F.is.called.the.force.transfer.matrix..It.transfers.a.unit.load.into.the.member.
force.and.reaction.force.unknowns..It.is.also.the.“inverse”.of.the.matrix.A,.as.
apparent.from.Equation.2.5.
We. can. conclude. that. the. nodal. equilibrium. conditions. are. completely.
characterized.by.the.matrix.A..The.inverse.of.A,.matrix.F,.is.the.force.transfer.
matrix,.which.transfers.any.unit.load.into.member.and.reaction.forces.
If. the. force. transfer. matrix. is. known,. either. by. solving. the. unit. load.
problems.one.at.a.time.or.by.solving.the.matrix.equation,.Equation.2.5,.
with. an. equation. solver,. then. the. solution. to. any. other. loads. can. be.
obtained. by. a. linear. combination. of. the. force. transfer. matrix.. Thus. the.
force.transfer.matrix.also.characterizes.completely.the.nodal.equilibrium.
conditions.of.the.truss..The.force.transfer.matrix.is.particularly.useful.if.
there.are.many.different.loading.conditions.that.one.wants.to.solve.for..
Instead. of. solving. for. each. load. separately,. one. can. solve. for. the. force.
transfer.matrix,.then.solve.for.any.other.load.by.a.linear.combination.as.
shown.in.the.following.example.
Example 2.12
Find all support reactions and member forces of the loaded truss shown next, knowing that the force transfer matrix is given in Equation 2.6.
x y
1
2
4 m
3 m
3 3 m
1 2
3 1.0 kN
0.5 kN
A truss problem to be solved with the force transfer matrix.
Solution
The given loads can be cast into a load vector, which can be easily computed as the combination of the third and fourth unit load vectors as shown next.
.
− = + −
0 0 0.5
1.0 0 0
(0.5) 0 0 1.0
0 0 0
( 1.0) 0 0 0 1.0
0 0
.
. (2.7)
The solution is then the same linear combination of the third and fourth vectors of the force transfer matrix:
F F F F F F
(0.5)
0.83 0.83 0.5 1.0 0.67 0.67
( 1.0)
0.63 0.63 0.38 0.0 0.5 0.5
0.21 1.04 0.62 0.50 0.17 0.83
kN
1
2
3
4
5
6
=
−
−
−
+ −
−
−
−
=
−
−
−
.
PROBLEM 2.5
The.loaded.truss.shown.next.is.different.from.that.in.Example.2.11.only.
in.the.externally.applied.loads..Use.the.force.transfer.matrix.of.Equation.
2.6.to.find.the.solution.
x y 1
2
4 m
3 m
3 3 m
1 2
3 1.0 kN
0.5 kN
Problem.2.5
PROBLEM 2.6
The.following.loaded.truss.is.different.from.that.in.Example.2.11.only.in.
the.externally.applied.loads..Use.the.force.transfer.matrix.of.Equation.
2.6.to.find.the.solution.
x y
1
2
4 m
3 m
3 3 m
1 2
3 1.0 kN
0.5 kN
1.0 kN
Problem.2.6
67