1 2 3 4
5 6
1 2 3
4 5 6
7 8
9
4 m
3@4 m = 12 m Problem.3.2.
PROBLEM 3.3
The.lower.chord.members.1,.2,.and.3.of.the.truss.shown.next.are.hav- ing. a. 20°C. increase. in. temperature.. Find. the. horizontal. displacement.
of.node.5,.given.E =.10.GPa.and.A =.100.cm2.for.all.bars.and.the.linear.
thermal.expansion.coefficient.is.α =.5(10–6)/°C.
1 2 3 4
5 6
1 2 3
4 5 6
7 8
9
4 m
3@4 m = 12 m Problem.3.3
3.2 Indeterminate Truss Problems: Method
We.notice.that.if.the.vertical.reaction.at.the.central.support.is.known,.then.
the.number.of.force.unknowns.becomes.18.and.the.problem.can.be.solved.
by.the.18.equilibrium.equations.from.the.nine.nodes..The.key.to.the.solution.
is.then.to.find.the.central.support.reaction,.which.is.called.the.redundant.
force..Denoting.the.vertical.reaction.of.the.central.support.by.Rc,.the.original.
problem.is.equivalent.to.the.problem.shown.next.as.far.as.force.equilibrium.
is.concerned.
P Rc
c
Statically.equivalent.problem.with.the.redundant.force.Rc.as.unknown.
The.truss.above,.with.the.central.support.removed,.is.called.the.primary structure..Note.that.the.primary.structure.is.statically.determinate..The.mag- nitude.of.Rc.is.determined.by.the.condition.that.the.vertical.displacement.
of.node.c.of.the.primary.structure,.due.to.(1).the.applied.load.P.and.(2).the.
redundant.force.Rc,.is.zero..This.condition.is.consistent.with.the.geometric.
constraint.imposed.by.the.central.support.on.the.original.structure..The.ver- tical.displacement.at.node.c.due.to.the.applied.load.P.can.be.determined.by.
solving.the.problem.associated.with.the.primary.structure.as.shown.next.
c
P ∆c
Displacement.of.node.c.of.the.primary.structure.due.to.the.applied.load.
The.displacement.of.node.c.due.to.the.redundant.force.Rc.cannot.be.com- puted.directly.because.Rc.itself.is.unknown..We.can.compute,.however,.the.
displacement.of.node.c.of.the.primary.structure.due.to.a.unit.load.in.the.direc- tion.of.Rc..This.displacement.is.denoted.by.δcc,.the.double.subscript.cc.signifies.
displacement.at.c.(first.subscript).due.to.a.unit.load.at.c.(second.subscript).
c 1 kN δcc
Displacement.at.c due.to.a.unit.load.at.c.
The.vertical.displacement.at.c.due.to.the.redundant.force.Rc.is.then.Rcδcc,.as.
shown.in.the.following.figure.
c Rc Rcδcc
Displacement.at.c.due.to.the.redundant.force.Rc.
The.condition.that.the.total.vertical.displacement.at.node.c,.Δc,.be.zero.is.
expressed.as
Δc.=.Δ ′c.+.Rcδcc.=.0. (3.11) This.is.the.additional.equation.needed.to.solve.for.the.redundant.force.Rc..
Once.Rc.is.obtained,.the.rest.of.the.force.unknowns.can.be.computed.from.
the.regular.joint.equilibrium.equations..Equation.3.11.is.called.the.condition of compatibility.
We.may.summarize.the.concept.behind.the.aforementioned.procedures.
by.pointing.out.that.the.original.problem.is.solved.by.replacing.the.inde- terminate.truss.with.a.determinate.primary.structure.and.superposing.the.
solutions.of.two.problems,.each.determinate,.as.shown.next.
c
P ∆c
c Rc Rcδcc
+
The.superposition.of.two.solutions.
And,.the.key.equation.is.the.condition.that.the.total.vertical.displacement.at.
node.c.must.be.zero,.consistent.with.the.support.condition.at.node.c.in.the.
original.problem..This.method.of.analysis.for.statically.indeterminate.struc- tures.is.called.the.method of consistent deformations.
EXAMPLE 3.5
Find the force in bar 6 of the truss shown next, given E = 10 GPa and A = 100 cm2 for all bars.
1 kN 0.5 kN
1 4
2 3
4 m
3 m 6 3 1
2
4 5
Example of an indeterminate truss with one redundant force.
Solution
The primary structure is obtained by introducing a cut at bar 6 as shown in the left panel of the following figure. The original problem is replaced by that of the left panel and that of the middle panel.
1 kN
1 4
2 3
1 2
3
4 5
1 4
2 3
1 2
3
4 5 F6
F6δ
∆
1 kN 0.5 kN
1 4
2 3
1 2
3
4
5 6
0.5 kN
+ =
Superposition of two solutions.
The condition of compatibility in this case requires that the total relative displacement across the cut obtained from the superposition of the two solutions be zero:
Δ = Δ′ + F6 δ = 0
where Δ′ is the overlap length (opposite of a gap) at the cut due to the applied load and δ is, as defined in the following figure, the overlap length across the cut due to a pair of unit loads applied at the cut.
1 4 3 2
1
2
3
4 5
1 1
Overlap displacement at the cut due to the unit-force pair.
The computation needed to find Δ′ and δ is tabulated next.
Computing.for.Δ′.and.δ Member
Real Load For Δ′ For δ
Fi EA/L Vi fi fi Vi vi fi vi
(kN) (kN/m) (mm) (kN/kN) (mm) (mm/kN) (mm/kN)
1 –0.33 25,000 –0.013 –0.8 0.010 –0.032 0.026
2 0 33,333 0 –0.6 0 –0.018 0.011
3 0 25,000 0 –0.8 0 –0.032 0.026
4 0.50 33,333 0.015 –0.6 –0.009 –0.018 0.011
5 –0.83 20,000 –0.042 1.0 –0.042 .0.050 0.050
6 0 20,000 0 1.0 0 .0.050 0.050
Σ –0.040 0.174
Note:. Fi =.ith.member.force.due.to.the.real.applied.load;.Vi =.Fi/(EA/L)i.=.ith.member.
elongation.due.to.the.real.applied.load;.fi =.ith.member.force.due.to.the.virtual.
unit.load.pair.at.the.cut;.vi.=.fi/(EA/L)i.is.the.ith.member.elongation.due.to.the.
virtual.unit.load.pair.at.the.cut;.Δ′.=.−0.040.mm;.δ.=.0.174.mm/kN.
0.040 mm, 0.174 mm/kN
= − δ =
From the condition of compatibility:
∆′ + F6 δ = 0 F6 = – 0.174
–0.040 = 0.23 kN
EXAMPLE 3.6
Formulate the conditions of compatibility for the truss problem shown.
P
c d
Statically indeterminate truss with two degrees of redundancy.
Solution
The primary structure can be obtained by removing the supports at node c and node d. Denoting the reaction at node c and node d as Rc and Rd, respectively, the original problem is equivalent to the superposition of the three problems as shown in the following figure.
c
P ∆c ∆d
c Rc Rcδcc
c
Rd Rdδcd
d
Rcδdc d
d Rdδdd
+
+
Superposition of three determinate problems.
In the figure:
Δ′c: vertical displacement at node c due to the real applied load Δ′d: vertical displacement at node d due to the real applied load δcc: vertical displacement at node c due to a unit load at c δcd: vertical displacement at node c due to a unit load at d δdc: vertical displacement at node d due to a unit load at c δdd: vertical displacement at node d due to a unit load at d
The conditions of compatibility are that the vertical displacements at nodes c and d are zero:
Δc = Δ′c + Rcδcc + Rd δcd = 0
Δd = Δ′d + Rc δdc + Rd δdd = 0 (3.12)
Equation 3.12 can be solved for the two redundant forces Rc and Rd. Denote Vi: ith member elongation due to the real applied load
fic: ith member force due to the unit load at c vic: ith member elongation due to the unit load at c fid: ith member force due to the unit load at d vid: ith member elongation due to the unit load at d
We can express the displacements according to the unit load method as Δ′c = Σ fic (Vι )
Δ′d = Σ fid (Vι ) δcc = Σ fic (vιc ) δdc = Σ fic (vιd ) δcd = Σ fid (vιc ) δdd = Σ fid (vιd )
The member elongation quantities in the equations are related to the member forces through
.
V FL
i E Ai i i i
=
. v f L
ic E Aic i i i
=
. v f L
id E Aid i i i
=
Thus, we need to find only member forces Fi, fic, and fid, corresponding to the real load, a unit load at node c, and a unit load at node d, respectively, from the primary structure.