Chapter 2-4 Lecture Notes
The Chapter 2 Objectives are to:
1. Explain and apply elongation of axially loaded members principles under uniform and non-uniform
conditions,
2. Explain and apply statically indeterminate principles,
3. Explain and apply thermal, misfit, and pre-strain principles,
4. Explain and apply stresses on inclined plane principles,
5. Demonstrate positive teamwork, and
6. Demonstrate problem solving ability.
Chapter 2-4
Statically Indeterminate Structures
Statically indeterminate structures have more supports than are necessary for static equilibrium, that is,
the equations ΣF=0 and ΣM=0 aren’t sufficient to determine the reactions and internal forces in the
structures. There are more variables than equations, thus, a unique solutions can’t be obtained. The
principles are illustrated below.
ΣF=0=R-P1-P2
R= P1+P2
A unique solution.
ΣF=0=RA+ RB-P=0
RA= P-RB
A non-unique solution, RA, depends on
the unknown value of RB. The second
equation is the compatibility equation
(or boundary condition) that δAB=0
because the bar cannot elongate
because of the boundary constraints.
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Now look at the next example. More supports than required for equilibrium.
The force, P, is stretching AC and
compressing CB.
For section AB, using the notation of the previous section (See Fig. 2-9):
N1 = RA
ΣF=RA-P+RB=0
δAB= (RA-P+RB)*(a+b)/EA = 0
The above equations aren’t independent, but we see that δAB = 0 as the boundary conditions require, so
different equations are needed.
N2
For section AC,
ΣF=N2 – P + RB=0
δAC=N2/EA= (P - RB)*a/EA
For section CB,
ΣF=N3 + RB = 0
δCB=N3/EA= (-RB)*b/EA
N3
The boundary condition is that the total elongation, δ= δAC+ δCB =0
(-RB+P)*b/EA - RB*a/EA =0 and solving,
RB*(a+b)=P*a and L=a+b so
RB=P*a/L
Returning to the force balance and solving,
RA-P+RB=0
RA=P-RB=P-P*a/L
=P(1-a/L)
2
=P(L-a)/L
=P(a+b-a) so
RA=P*b/L
Note: Eq 2-12, p 143, has an error. What is it?
In summary, the reactions and internal forces for statically determinate structures can be calculated by the
equilibrium equations:
ΣF=0 and ΣM=0
The reactions and internal forces for statically indeterminate structures can be determined by the
equilibrium equations because there are more unknowns than there are equations.
UNIQUE SOLUTIONS REQUIRE AS MANY
INDEPENDENT EQUATION AS UNKNOWNS
Consequently, the set of independent equations must come from:
1. Equilibrium equations ΣF=0 and ΣM=0
2. Compatibility equations (or boundary condition equations such as δ=0)
3. Force displacement equations such as δ=PL/EA or σ=Eε.
Note: Equations of items 1 and 2 are independent of the materials whereas Equations of item 3 depend
on the material properties such as modulus of elasticity or the shear modulus of elasticity.
Go to http://web.mst.edu/~mecmovie/index.html ch 5.5 for an interesting animation.
See Ex 2-5 in a separate file to illustrate the principles.
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