Deformation
of a Circular Bar (Statically Indeterminate)
|
Example: (continued) |
|
From equilibrium, FA – FD =
P (1) From deflection relations, For A to B: F1 = FA, For B to C: F2 = FA
+ P, For C to D:
F3 = FA - P The relative deflections
for each section are given by Fi
Li / AE ∆B/A = FA (L/2) / AE ∆C/B = (FA + P)(L/4) /AE ∆D/C = (FA
– P)(L/4) / AE The sum of the relative
displacement must add to zero. (Geometric
condition) ∆B/A + ∆C/B + ∆D/C =
0 (Cancel
L, A, and E to simplify.) FA /2 + FA
/4 +
P/4 + FA /4 – P/4
= 0 So FA =
0 and by (1)
FD = – P (result for support reactions) So the axial stresses in
each section are σ1 = 0 , σ2 =
P/A (tension) , σ3 = –
P/A (compression) Thus the maximum axial
stress equals P/A
and occurs between B and C. (result) Click here to solve this
problem again using superposition this time. |
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