Displacement of an axially loaded beam

[NEGATIVE_40]

Homework Statement

The 30mm diameter A-36 steel rod (E=200x10^9 Pa) is subjected to the loading shown. Determine the displacement of end A with respect to end C. (see attached picture)

Homework Equations

\delta = \frac{PL}{EA} (eq. 1)
where delta is the deflection, P is the internal force, E is Young's modulus and A is the cross sectional area.

\delta_{A/C} = \delta_C - \delta_B (eq. 2)

The Attempt at a Solution

I first turned the two forces at B into a single force of +48kN. Since \sum F_x = 0 I determined the reaction A to be 42kN. I then took a cut at a point between A and B, and determined the displacement of B by eq.1 giving -0.1188mm (to the left). Similarly I then took a cut between B and C and determined the displacement of C and found a dispacement of -0.3819mm. Then by eq. 2 the displacement of A relative to C is -0.2632mm. This answer is incorrect though.

I spoke to my tutor about this earlier today, and he told me I could move the reaction at B to any point along the line of the beam (so any where between A and C). Doing this I found an internal force of -42kN (compression), which by equation 1, for the whole beam, gives a displacement of -0.2971mm, which is also wrong.

The stated answer in the back of the book is -0.772mm (is the textbook's answer correct?) I've been trying all sorts of combinations on this problem for days now, and am no closer to solving it. Any help would be appreciated!

 

[Mapes]

I'm not seeing where you get your eq. 2. If AB contracts and BC contracts, why isn't the changing distance between A and C just the sum of the contractions?

Even with this revision, though, my answer doesn't match the one in your book.

 

[NEGATIVE_40]

I'm not seeing where you get your eq. 2. If AB contracts and BC contracts, why isn't the changing distance between A and C just the sum of the contractions?

Even with this revision, though, my answer doesn't match the one in your book.

I just rearranged eq. 2 from another equation in the textbook, so it may not apply in this case. What you suggest makes intuitive sense. I'm still a bit confused though. You can move the reaction at B to any point along the beam, can't you?

If I do what you suggest I get -0.5007mm, which is still off. Do you think the given answer is incorrect?

(Mechanics of Materials by Hibbeler, 8e, question F4-3)

 

[Mapes]

You can move the reaction at B to any point along the beam, can't you?

No way! Move it all the way to the left and you have a uniform 90kN compressive load. All the way to the right and it's 48kN. These certainly aren't equivalent.

 

[NEGATIVE_40]

No way! Move it all the way to the left and you have a uniform 90kN compressive load. All the way to the right and it's 48kN. These certainly aren't equivalent.

So...I would have to take a cut between A and B (to get displacement of B ) and then take a cut between B and C (to get displacement of C) The displacement of A relative to C would therefore be \delta_C + \delta_B, which is what I did originally, and according to the textbook is incorrect. Is my methodology correct (when I don't move the reaction at B, that is) ?

 

[NEGATIVE_40]

Just to makes things clear, this is what I have done.

\delta_{A/C}= \sum \frac{PL}{EA}= \frac{P_{AB}L_{AB}}{EA}+ \frac{P_{BC}L_{BC}}{EA}= \frac{-42000N \cdot 0.4m}{200\times 10^9Pa \cdot (\frac{\pi}{4}\cdot 0.03^2)} + \frac{-90000N \cdot 0.6m}{200\times 10^9Pa \cdot (\frac{\pi}{4}\cdot 0.03^2)}= -0.5007mm

 

[pongo38]

Yea i get that too. Books are sometimes wrong.

 

[Mapes]

Looks good to me.