Deformation
of a Circular Bar
Example: (continued) Now for each segment of
the bar the axial deformation is
δi = PiLi/AiEi and the sum of all
deformations must add to zero. δ1 +
δ2 + δ3 = 0 which gives P1L1/AE + P2L2/AE + P3L3/AE =
0 or FABL1/AE + FBCL2/AE + FCDL3/AE =
0 (24 -F1)(1.0)/AE +
24(0.4)/AE + 9(0.6)L3/AE =
0 F1 = 24 + 24(0.4) + 9(0.6) = 39
kN
(result) Note that the force in the
section of the bar between A and B is FAB = 24 – 39
= - 15 kN. So that portion of the bar
is in compression. The other two sections
(between B and C and between
C and D ) are in tension. FBC= 24 kN and
FCD = 9 kN Click here for an
alternate solution to this example |
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