Statically Indeterminate Bar with alternating area and length

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SUMMARY

The discussion focuses on solving a statically indeterminate bar problem with alternating area and length. The user calculated the reactions RA and RD, arriving at RA = 10.5 kN to the left and RD = 2 kN to the right. The user struggled with the algebra involving the total change in length equation, specifically the contributions from three sections of the bar. A suggestion was made to utilize a Free Body Diagram to determine the force in the middle segment, which is not simply P.

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hboyd3
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So first I did a force balance (dFx=0)

P=PB-PC=8.5kN

RD= RA - P
RA = RD + P

I know the total change in length is equal to 0, so

dAB + dBC + dCD = 0

==>

(RA*L1)/(E*A1) + (P*L2)/(E*A2) + ((RA-P)*L1)/(E*A1) = 0

but when I do the algebra I can never get it to work. my numbers never match the answer no matter which way I approach it. I can do problems with only 2 different sections but this 3 sectioned one has me baffled.

the answers are
RA= 10.5kN to the left
RD= 2kN to the right

anyone have any insight?
 
Last edited by a moderator:
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You are assuming that the force in the middle segment is P. It is not. Cut a section (Free Body Diagram) thru the middle bar and determine the force in that bar as a function of the known and unknown variables.
 

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