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Given a three member truss structure shown here: Truss structure

How do I find the vertical and horizontal displacements at M? Assume all members are elastic and E,A, I and constant.

I start with finding the reaction forces with $F_x=2P-H_B=0, H_B=2P$ (rightwards) and $M_B=2l*V_A-2Pl=0, V_A=P$ (upwards)

Hence $N_{AB}=2P$ and here is where I get stuck. I cannot find the forces that put joint A into equilibrium.

Edit: Additionally, show all member forces

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TL;DR This analysis is long-winded and I realise at the end that it's incorrect.


Your reactions are right. You are incorrect to state that $N_{AB}=2P$, as you are ignoring the shear in member BC.

After getting the reactions, you should calculate the member forces at A:

  • Resolving vertically: $N_{AC}*cos(45)=P$
  • Resolving horizontally: $N_{AC}*sin(45) + N_{AB}=0$

Given that $cos(45)=sin (45)$, we find that $N_{AB}=P$.

Similarly resolving at C, we'll find that the shear in member BC is $P$, and the horizontal component of member AC is also $P$.

So basically member BC will act as a simply supported member spanning from B to C, with a point load at midspan of $2P$, and an axial compression of $P$.

Ignoring second order effects (which is probably oversimplistic), the horizontal displacement from $2P$ on a simply supported beam is given by the standard solution $WL^3 / (48EI)$, or in this case $2P(2l)^3 / (48EI)$; and the vertical displacement from the standard solution $FL/EA$, or in this case $Pl/EA$.

Conclusion:

  • Horizontal displacement $2P(2l)^3 / (48EI)$
  • Vertical displacement $Pl/EA$

Note: I've ignored shear in members AB and AC; I think these may be zero but I've not proved that. I've also ignored that the axial tension in AB will cause position A to move, dragging member AC and hence C with it, plus the axial compression in AC will move point C even further.

On second thoughts this analysis presented here is inadequate, certainly for the displacements and possibly for the forces. Perhaps building a stiffness matrix is required.

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    $\begingroup$ On the drawing, I would interpret the circles as pin joints that can't transfer bending moments. This would mean that there won't be any shear in AB and AC, only in BC. As for your calculation of the displacements: Shouldn't you be using l instead of 2l for the vertical displacement at M? $\endgroup$ – ingenørd Jun 4 '19 at 12:20
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    $\begingroup$ Sure the joints can transfer shear, but members AB and AC can't. Without any transversal loads as on BC, if the shear is non-zero the bending moment can't be zero in both ends. You were on the right path. $\endgroup$ – ingenørd Jun 4 '19 at 15:38
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    $\begingroup$ I agree with @ingenørd, this solution is correct. I even created a model on Ftool (that's the normal force diagram) to check, and it's right. As ingenørd says, there's no shear on the other members because there'd be no way for them to balance it out. The rest of the math works out just fine, too. $\endgroup$ – Wasabi Jun 5 '19 at 0:32
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    $\begingroup$ The only detail which probably should be accounted for (though the effect is really small) is the fact that the other members deflect, which moves the position of C, such that the undeformed shape of BC is a slight diagonal. So you need to add that incline to the total deflection at M. (You're currently effectively assuming that C is fixed in space) $\endgroup$ – Wasabi Jun 5 '19 at 0:36
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    $\begingroup$ I think after adding the displacement for joint members, your answer is correct. It has been most helpful. Thank you. $\endgroup$ – Thomas Jun 6 '19 at 9:43
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As the OP and others Have done we set the moment about B = 0 to get reactions.

$$ \Sigma M_B=0,\ 2lA_v-2Pl=0\ A_v=P $$ $$\Sigma F_v=0\ A_v+B_v=0\ B_v=-P\ and\ C_v=P\ BC\ under\ tension$$

$$ F_{AC}=\sqrt(P^2+P^2=\sqrt2P^2 =P\sqrt2\\ and\ F_{AB}=P \ tension $$

$$ \Sigma F_x=0,\quad B_h=2P$$

Horizontal displacements of M as has been mentioned in @AndyT, but we need to add to that the horizontal component of movement of point C multiplied by 1/2, which is $1/2 \sqrt2*(P*l/AI )\sqrt2/2= Pl/2AI \ \text{and add to that the horizontal expansion of AB/2 which is }=lP/EA $

$$ \therefore H_{displacement\ of\ M}= 2P(2l)^3 / (48EI) + Pl/2EA+Pl/EA $$

On vertical displacement we note that under the tension point M located at hlf height of BC is moving up by $Pl/EA$ .

please check my arithmetic.

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  • $\begingroup$ Ah, of course, displacement of A is solely due to FL/EA in member AB, horiz displacement of C is this plus horizontal component of FL/EA in member AC, and horiz displacement of M from this effect is half of horiz displacement of C. Well done, I missed that. $\endgroup$ – AndyT Jun 5 '19 at 8:51
  • $\begingroup$ I think you've missed the vertical displacement of C from the vertical component of FL/EA in member AC though. And I think member BC is in compression, so M will move downwards, plus the compression in member AC will move C downwards, and hence move M further downwards. $\endgroup$ – AndyT Jun 5 '19 at 8:53
  • $\begingroup$ As to the axial shortening of AC: I have axial force $N_{AC} = P / cos(45)$, giving axial shortening of $P/cos(45) * 2*sqrt(2)*l / EA$. Multiply by cos(45) for horizontal component and I get $2Pl*sqrt(2)/EA$. So I think your $PL/2EA$ term should be $sqrt(2)*PL/EA$. $\endgroup$ – AndyT Jun 5 '19 at 8:58
  • $\begingroup$ But +1 for working out the steps I'd missed. And I'm willing for my last comment to be corrected, I'm not confident in it! $\endgroup$ – AndyT Jun 5 '19 at 8:59
  • $\begingroup$ @AbdyT, indeed the displacement of C will change the geometry of the truss, but it's effect will be insignificant as compared to other deflections. $\endgroup$ – kamran Jun 6 '19 at 1:29
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After some discussion with my classmates, I analyzed the $BC$ member as a simple beam. and horizontal displacement $\delta_{mh} = \delta_{bending} + \delta_{truss}$ Considering load $2P$ on $l$, we have bending load $Pl$ on point $M$. The displacement from this displacement load can be calculated via whichever method of choice. Here I used the principle of virtual work: $$\delta_{bending} = \frac{2}{EI}\int_0^l Px \cdot \frac{x}{2} dx = \frac{Pl^3}{3EI}$$ Next, the displacement of the truss structure is also determined through virtual work. Here is where I am unsure of the member forces. Positive: right, upwards, tension We have reaction forces $V_A=-P,V_B=P,H_A=-2P$ from equilibrium conditions. Point load $2P$ at $M$ is treated as two loads $P$ on $C,B$. Hence we have member forces $N_{AB}=-\sqrt2P , N_{BC}=P, N_{CA}=P$. Placing virtual horizontal load at $C$, we compute $$ 2\delta_{truss}=\frac{Pl}{EA}(2+2+4\sqrt2) \\ \delta_{truss}=\frac{Pl(2+2\sqrt2)}{EA}$$ and hopefully I have the correct displacement here.

Note: I believe vertical displacement is a simple virtual load on C. i.e. $$\delta_{CV} = \delta_{MV} = \frac{2Pl}{EA}$$

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