Chapter 2; Problem 40 From Mechanics of Materials 6th Edition Ferdinand P Beer.

Question

Chapter 2;  Problem 40 From Mechanics of Materials 6th Edition Ferdinand P Beer.

For those that do not have the book. her is the relevant information:

There is a vertical cylindrical rod that is 25 inches long and 1.25 inches in diameter.
This ^^^ is stacked on top of another vertical cylindrical rod of the same material that is 15 inches tall ( long) and 2 inches in diameter.
two 6 Kip forces are exerted vertically downward on the 2 inch diameter rod. The Top and bottom ends are fixed in place. so we know the total deformation must = 0.
E=.45*10^6 PSI

I came to an answer by assuming that the points where the two rods connect must deflect by equal and opposite amounts ( Ie the deflection of the top rod is equal and negative to the deflection of the bottom rod). I found that the Reaction force at A =3492 Lb and the reaction force at C = 15,492 lb

Then i decided to solve this problem the traditional way using super possition. I found that Rc = 1526 lb and RA = 10472 Lb

are either of these answers correct?
Most importantly!!!!! What is the correct way to solve this problem?

Explanation / Answer

D

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.