Displacement of an axially loaded beam

In summary: So the given answer is incorrect?In summary, the 30mm diameter A-36 steel rod is subjected to loading, and the displacement of end A with respect to end C is determined using equations 1 and 2. The attempt at a solution involves converting the forces at B into a single force and finding the reaction at A. However, after discussion with a tutor, it is discovered that the reaction at B can be moved to any point along the beam, resulting in a different displacement. The given answer in the textbook is incorrect.
  • #1
NEGATIVE_40
23
0

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



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

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


The Attempt at a Solution



I first turned the two forces at B into a single force of +48kN. Since [tex] \sum F_x = 0 [/tex] 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!
 

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  • #2
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.
 
  • #3
Mapes said:
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)
 
Last edited:
  • #4
NEGATIVE_40 said:
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.
 
  • #5
Mapes said:
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 [tex] \delta_C + \delta_B [/tex], 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) ?
:confused:
 
  • #6
Just to makes things clear, this is what I have done.

[tex] \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 [/tex]
 
  • #7
Yea i get that too. Books are sometimes wrong.
 
  • #8
Looks good to me.
 

1. What is the definition of displacement in relation to an axially loaded beam?

Displacement refers to the change in position or movement of a point on an axially loaded beam from its original position.

2. How is displacement calculated for an axially loaded beam?

Displacement can be calculated by using the formula: δ = FL/AE, where δ is the displacement, F is the applied force, L is the length of the beam, A is the cross-sectional area, and E is the modulus of elasticity.

3. What factors affect the displacement of an axially loaded beam?

The displacement of an axially loaded beam can be affected by the applied force, length of the beam, cross-sectional area, modulus of elasticity, and the material properties of the beam.

4. How does the direction of the applied force affect the displacement of an axially loaded beam?

The direction of the applied force can affect the displacement of an axially loaded beam. If the force is applied along the axis of the beam, it will cause elongation and increase the displacement. If the force is applied perpendicular to the axis of the beam, it will cause bending and decrease the displacement.

5. How can the displacement of an axially loaded beam be minimized?

The displacement of an axially loaded beam can be minimized by using a stronger material with a higher modulus of elasticity, increasing the cross-sectional area of the beam, or reducing the length of the beam. Additionally, proper support and bracing can also help minimize displacement.

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