Instability due to improper support..A.beam.or.frame.is.kinematically.unsta- ble.if.the.support.conditions.are.such.that.the.whole.structure.is.allowed.to.
move.as.a.mechanism..Examples.of.improper.support.and.insufficient.sup- port.are.shown.next.
Improper.or.insufficient.support.conditions.
Instability due to improper connection.. A. beam. or. frame. is. kinematically.
unstable. if. the. internal. connection. conditions. are. such. that. part. of. or. the.
whole.structure.is.allowed.to.move.as.a.mechanism..Examples.of.improper.
connections.are.shown.next.
Improper.internal.connections.
Statical determinacy..A.stable.beam.or.frame.is.statically.indeterminate.if.
the.number.of.force.unknowns.is.greater.than.the.number.of.equilibrium.
equations..The.difference.between.the.two.numbers.is.the.degree.of.indeter- minacy..The.number.of.force.unknowns.is.the.sum.of.the.number.of.reac- tion.forces.and.the.number.of.internal.member.force.unknowns..For.reaction.
forces,.a.roller.has.one.reaction,.a.hinge.has.two.reactions,.and.a.clamp.has.
three.reactions,.as.shown.next.
Reaction.forces.for.different.supports.
To. count. internal. member. force. unknowns,. first. we. need. to. count. how.
many.members.are.in.a.frame..A.frame.member.is.defined.by.two.end.nodes..
At.any.section.of.a.member.there.are.three.internal.unknown.forces:.T,.V,.
and.M.. The. state. of. force. in. the. member. is. completely. defined. by. the. six.
nodal.forces,.three.at.each.end.node,.because.the.three.internal.forces.at.any.
section.can.be.determined.from.the.three.equilibrium.equations.taken.from.
a.free-body.diagram.(FBD).cutting.through.the.section.as.shown.below,.if.
the.nodal.forces.are.known.
x
T V
M
Internal.section.forces.are.functions.of.the.nodal.forces.of.a.member.
Thus,. each. member. has. six. nodal. forces. as. unknowns.. Denoting. the.
number.of.members.by.M.and.the.number.of.reaction.forces.at.each.sup- port.as.R,.the.total.number.of.force.unknowns.in.a.frame.is.then.6M.+ ΣR..
On.the.other.hand,.each.member.generates.three.equilibrium.equations.
and. each. node. also. generates. three. equilibrium. equations.. Denoting.
the.number.of.nodes.by.N,.the.total.number.of.equilibrium.equations.is.
3M + 3N.
Nodal
Equilibrium Member
Equilibrium
Member Equilibrium
FBDs.of.a.node.and.two.members.
Because. the. number. of. members,.M,. appears. both. in. the. count. for.
unknowns.and.the.count.for.equations,.we.can.simplify.the.expression.for.
counting.unknowns.as.shown.next.
Number of Unknowns = 6M + ΣR Number of Equations = 3M + 3N
Number of Unknowns = 3M + ΣR Number of Equations = 3N Counting.unknowns.against.available.equations.
This. is. equivalent. to. considering. each. member. having. only. three. force.
unknowns..The.other.three.nodal.forces.can.be.computed.using.these.three.
nodal.forces.and.the.three.member.equilibrium.equations..Thus,.a.frame.is.
statically.determinate.if.3M + ΣR.=.3N.
If.one.or.more.hinges.are.present.in.a.frame,.we.need.to.consider.the.con- ditions.generated.by.the.hinge.presence..As.shown.in.the.upcoming.figure,.
the.presence.of.a.hinge.within.a.member.introduces.one.more.equation,.
which.can.be.called.the.condition.of.construction..A.hinge.at.the.junction.
of. three. members. introduces. two. conditions. of. construction.. The. other.
moment.at.a.hinge.is.automatically.zero.because.the.sum.of.all.moments.at.
the.hinge.(or.any.other.point).must.be.zero..We.generalize.to.state.that.the.
conditions.of.construction,.C,.is.equal.to.the.number.of.joining.members.
at.a.hinge,.m,.minus.one,.C =.m – 1..The.conditions.of.construction.at.more.
than.one.hinge.is.ΣC.
Since. the. conditions. of. construction. provide. additional. equations,. the.
available. equation. becomes. 3N.+ C.. Thus,. in. the. presence. of. one. or. more.
internal.hinges,.a.frame.is.statically.determinate.if.3M + ΣR =.3N + ΣC.
MT V
M = 0 TV
MT V
MT V
MT V
MT V
M = 0 TV
M = 0 TV MT
V
MT V
= 0
= 0
Presence.of.hinge.introduces.additional.equations.
Example 4.1
Discuss the determinacy of the beams and frames shown.
Solution
The computation is shown with the figures.
R = 3
R = 2
Number of unknowns = 3M + R = 7 Number of equations = 3N + ΣC = 7 Statically determinate.
R = 1 R = 3
Number of unknowns 3M + ΣR = 15 Number of equations = 3N + C = 13 Indeterminate to the 2nd degree.
Number of unknowns = 3M + ΣR = 21 Number of equations = 3N + C = 20 Indeterminate to the 1st degree.
R = 3 R = 3 R = 2 R = 1
R = 3
R = 1 Number of unknowns = 3M + ΣR = 10 Number of equations = 3N + C = 10 Statically determinate.
Number of unknowns = 3M + ΣR = 12 Number of equations = 3N + ΣC = 11 Indeterminate to the 1st degree.
R = 1 R = 3
R = 2 M = 1, N = 2, C = 0 M = 1, N = 2, C = 1
M = 3, N = 4, C = 1 M = 5, N = 6, C = 2 M = 2, N = 3, C = 1
M = 2, N = 3, C = 2
R = 3 Number of unknowns = 3M + ΣR = 8
Number of equations = 3N + ΣC = 6 Indeterminate to the 2nddegree.
Counting internal force unknowns, reactions, and available equations.
For. frames. with. many. stories. and. bays,. a. simpler. way. of. counting.
unknowns. and. equations. can. be. developed. by. cutting. through. members.
to.produce.separate.“trees”.of.frames;.each.is.stable.and.determinate..The.
number.of.unknowns.at.the.cuts.is.the.number.of.degrees.of.indeterminacy,.
as.shown.in.Example.4.2.
Example 4.2
Discuss the determinacy of the frame shown.
Multistory, multibay indeterminate frame.
Solution
We make nine cuts that separate the original frame into four “trees” of frames as shown.
1 2 3
4 5 6
7 8 9
Nine cuts pointing to 27 degrees of indeterminacy.
We can easily verify that each of the stand-alone trees is stable and stati- cally determinate, that is, the number of unknowns is equal to the number of equations in each of the tree problems. At each of the nine cuts, three internal forces are present before the cut. All together, we have removed 27 internal forces in order to have equal numbers of unknowns and equations.
If we put back the cuts, we introduce 27 more unknowns, which is the degrees of indeterminacy of the original uncut frame.
This simple way of counting can be extended to multistory, multibay frames with hinges: simply treat the conditions of construction of each hinge as “releases” and subtract the ∑C number from the degrees of inde- terminacy of the frame with the hinges removed. For supports other than fixed, we can replace them with fixed supports and count the releases for subtracting from the degrees of indeterminacy.
Example 4.3
Discuss the determinacy of the frame shown.
Indeterminate frame example.
Solution
Two cuts and five releases amounts to 2 × 3 – 5 = 1. The frame is indeter- minate to the first degree.
1 2
C = 1 C = 2 C = 2
Shortcut to count degrees of indeterminacy.
PROBLEM 4.1
Discuss.the.determinacy.of.the.beams.and.frames.shown.
(1) (2)
(3) (4)
(5) (6)
(7) (8)
Problem.4.1