Four L-shaped members: Mechanical Analysis Problem

[jojosg]

TL;DR: This determine the support reactions at A & B

I tried calculate moment about a which is 2a(By + Bx + Ey + Ex +F) + a (Gy + Gx + Cy + Cx). But I don't think I'm correct, any help would be appreciated, this was a question on a final exam paper in my Mechanics course.

 

[wrobel]

Try to write down the equations of equilibrium for each L-like rigid body. You have three scalar static equations for a rigid body in a planar problem.
There are 2 unknown components of a reaction in each hinge and 12 static equations for 4 rigid bodies.
The number of equations and the number of unknowns matches.

 

[jojosg]

Try to write down the equations of equilibrium for each L-like rigid body. You have three scalar static equations for a rigid body in a planar problem.
There are 2 unknown components of a reaction in each hinge and 12 static equations for 4 rigid bodies.
The number of equations and the number of unknowns matches.

Like this?

 

[Lnewqban]

Why are Bx and Ex being multiplied by 2a?

 

[jojosg]

Why are Bx and Ex being multiplied by 2a?

I took moment about A and the distance from A for Ex & Bx are 2a.

 

[Lnewqban]

I took moment about A and the distance from A for Ex & Bx are 2a.

What matters is the perpendicular distance between the pivot and the force.
That distance is zero for force Bx.
For force Ex is ...

 

[jojosg]

What matters is the perpendicular distance between the pivot and the force.
That distance is zero for force Bx.
For force Ex is ...

2a(By + Ey +F) + a (Gy + Gx + Cy + Cx + Ex) like this?

 

[Lnewqban]

2a(By + Ey +F) + a (Gy + Gx + Cy + Cx + Ex) like this?

Reconsider Gx.

 

[haruspex]

Like this?

You are missing a force at D on ADC. Likewise at E on CEB.

In your moment sums, you need to be clear on the system you are considering. Your expressions seem to be taking the whole system of the four L pieces, in which case:
- the forces at D, E, C and G are internal, so should be ignored,
- the perpendicular distance from Bx to A is zero,
- assuming By acts upwards on CEB, the moments from By and F act in opposite directions around A.

Listing Ax, Ay, Bx, … etc., you have 12 unknowns, so to find them all you need 12 equations. Horizontal and vertical force balance on each L piece gives you 8 equations, so you will need four more from moment balance. Apply moment balance to some individual L pieces.

Since you are only asked for forces at A and B, you may be able to avoid finding all 12 by careful choice of which balance equations to use.

 

[jojosg]

Reconsider Gx.

on second maybe this 2a(By + Ey - F - Gx) + a(Gy + Cy - Ex - Cx) = 0.

 

[jojosg]

You are missing a force at D on ADC. Likewise at E on CEB.

In your moment sums, you need to be clear on the system you are considering. Your expressions seem to be taking the whole system of the four L pieces, in which case:
- the forces at D, E, C and G are internal, so should be ignored,
- the perpendicular distance from Bx to A is zero,
- assuming By acts upwards on CEB, the moments from By and F act in opposite directions around A.

Listing Ax, Ay, Bx, … etc., you have 12 unknowns, so to find them all you need 12 equations. Horizontal and vertical force balance on each L piece gives you 8 equations, so you will need four more from moment balance. Apply moment balance to some individual L pieces.

Since you are only asked for forces at A and B, you may be able to avoid finding all 12 by careful choice of which balance equations to use.

so how should I draw the FBD if I ignore the internal forces? I tried just the L for A and C and took Cx & Cy into account. then ACDG as one body and got F=0;

 

[haruspex]

so how should I draw the FBD if I ignore the internal forces? I tried just the L for A and C and took Cx & Cy into account. then ACDG as one body and got F=0;

In the diagram, F is applied to GE, so would not feature in any equation specific to ACDG.
Please post your equations.

 

[jojosg]

Member BE:

\sum F_x = 0 \quad \Rightarrow \quad \frac{F_G}{\sqrt{2}} = F + E_x

\sum F_y = 0 \quad \Rightarrow \quad E_y = \frac{F_G}{\sqrt{2}}

\Rightarrow F_G = \sqrt{2}E_y

From equilibrium: E_y = F + E_x



Member BC:

\sum F_x = 0 \quad \Rightarrow \quad B_x = E_x + \frac{F_C}{\sqrt{2}}

\sum F_y = 0 \quad \Rightarrow \quad B_y = E_y + \frac{F_C}{\sqrt{2}}

With substitution:

B_y = F + E_x + \frac{F_C}{\sqrt{2}}, \quad B_x = E_x + \frac{F_C}{\sqrt{2}}

Substituting E_y = F + E_x:

B_y = F + E_x + \frac{F_C}{\sqrt{2}}
B_x = E_x + \frac{F_C}{\sqrt{2}}



Joint B/Whole Structure:

\sum F_y = 0 \quad \Rightarrow \quad A_y + \frac{F_D}{2} + \frac{F_C}{2} = 0

\sum F_x = 0 \quad \Rightarrow \quad A_x + \frac{F_C}{2} + \frac{F_D}{2} = 0

Given A_y = A_x = 0:

\frac{F_G}{\sqrt{2}} - \frac{F_D}{\sqrt{2}} = 0, \quad \frac{F_G}{\sqrt{2}} = \frac{F_D}{\sqrt{2}}



Moment about B:

\sum M_B = 0 \quad \Rightarrow \quad -F(2a) = 0 \quad \Rightarrow \quad F = 0



Thus:

B_x = E_x + \frac{F_C}{\sqrt{2}}
B_y = E_x + \frac{F_C}{\sqrt{2}} \quad \text{(since } F = 0\text{)}

 

[haruspex]

I'll fix up your LaTeX by replacing each occurrence of [ tex ] and [ \ tex ] with a double #.
It is clearly not valid to conclude that F=0. Such a structure could be created and a force F applied and yet the structure resist movement.

I am having trouble guessing what your labels for forces mean. Please state for each, what it acts on, at which point, and in what direction as positive.

Member BE:
$\sum F_x = 0 \quad \Rightarrow \quad \frac{F_G}{\sqrt{2}} = F + E_x$

You've lost me already. By "member BE" do you mean the L shape from B to C via E? It'll be clearer to call that BC, since there is no label on the angle in DG or EG.
But if you do mean BC, F does not act directly on it.
What exactly is $F_G$? And why divide it by $\sqrt 2$?

$\sum F_y = 0 \quad \Rightarrow \quad E_y = \frac{F_G}{\sqrt{2}}$

$\Rightarrow F_G = \sqrt{2}E_y$

From this, it looks like you are defining $F_G$ as the total force DG exerts on EG and assuming, without justification, that it acts at 45° to the horizontal.

How about this notation for the forces:
At a junction J between two L shapes, $J_x$ and $-J_x$ are the horizontal forces they exert on each other, to the right being positive; $J_y$ and $-J_y$ are the vertical forces, up being positive. Adopt the convention that $J_x$ is the force the lower (lefthand) shape exerts on the upper (right hand) shape.
E.g. at E, BC exerts $E_x$ and $E_y$ on EG. It doesn’t matter if it turns out that BC pulls down on EG: that will simply produce a negative value for $E_y$.

I see no point in reading further till we get these issues sorted.

From equilibrium: $E_y = F + E_x$

Member BC:

$\sum F_x = 0 \quad \Rightarrow \quad B_x = E_x + \frac{F_C}{\sqrt{2}}$

$\sum F_y = 0 \quad \Rightarrow \quad B_y = E_y + \frac{F_C}{\sqrt{2}}$


With substitution:


$B_y = F + E_x + \frac{F_C}{\sqrt{2}}, \quad B_x = E_x + \frac{F_C}{\sqrt{2}}$


Substituting $E_y = F + E_x$:


$B_y = F + E_x + \frac{F_C}{\sqrt{2}}$


$B_x = E_x + \frac{F_C}{\sqrt{2}}$


Joint B/Whole Structure:


$\sum F_y = 0 \quad \Rightarrow \quad A_y + \frac{F_D}{2} + \frac{F_C}{2} = 0$


$\sum F_x = 0 \quad \Rightarrow \quad A_x + \frac{F_C}{2} + \frac{F_D}{2} = 0$


Given $A_y = A_x = 0$:


$\frac{F_G}{\sqrt{2}} - \frac{F_D}{\sqrt{2}} = 0, \quad \frac{F_G}{\sqrt{2}} = \frac{F_D}{\sqrt{2}}$


Moment about B:


$\sum M_B = 0 \quad \Rightarrow \quad -F(2a) = 0 \quad \Rightarrow \quad F = 0$


Thus:


$B_x = E_x + \frac{F_C}{\sqrt{2}}$


$B_y = E_x + \frac{F_C}{\sqrt{2}} \quad \text{(since } F = 0\text{)}$








Member BE:

$\sum F_x = 0 \quad \Rightarrow \quad \frac{F_G}{\sqrt{2}} = F + E_x$


$\sum F_y = 0 \quad \Rightarrow \quad E_y = \frac{F_G}{\sqrt{2}}$


$\Rightarrow F_G = \sqrt{2}E_y$


From equilibrium: $E_y = F + E_x$


Member BC:


$\sum F_x = 0 \quad \Rightarrow \quad B_x = E_x + \frac{F_C}{\sqrt{2}}$


$\sum F_y = 0 \quad \Rightarrow \quad B_y = E_y + \frac{F_C}{\sqrt{2}}$


With substitution:


$B_y = F + E_x + \frac{F_C}{\sqrt{2}}, \quad B_x = E_x + \frac{F_C}{\sqrt{2}}$


Substituting $E_y = F + E_x$:


$B_y = F + E_x + \frac{F_C}{\sqrt{2}}$


$B_x = E_x + \frac{F_C}{\sqrt{2}}$


Joint B/Whole Structure:


$\sum F_y = 0 \quad \Rightarrow \quad A_y + \frac{F_D}{2} + \frac{F_C}{2} = 0$


$\sum F_x = 0 \quad \Rightarrow \quad A_x + \frac{F_C}{2} + \frac{F_D}{2} = 0$


Given $A_y = A_x = 0$:


$\frac{F_G}{\sqrt{2}} - \frac{F_D}{\sqrt{2}} = 0, \quad \frac{F_G}{\sqrt{2}} = \frac{F_D}{\sqrt{2}}$


Moment about B:


$\sum M_B = 0 \quad \Rightarrow \quad -F(2a) = 0 \quad \Rightarrow \quad F = 0$


Thus:


$B_x = E_x + \frac{F_C}{\sqrt{2}}$


$B_y = E_x + \frac{F_C}{\sqrt{2}} \quad \text{(since } F = 0\text{)}$

 

[Lnewqban]

Member BE:
...
Member BC:
...
Joint B/Whole Structure:
...
Moment about B:
...

@jojosg
I would like to clarify that each of these L-shaped members work very differently than the members of a truss (like the one in your other thread).
These are rigid links which are able to transfer moments, besides compressing and tensioning forces.
For that reason, the pin connections resist forces that are that are induced by moments, besides forces that try to separate or crush them.
Those pin connections are unable to resist any moment (like any door's hinge).

The force F introduces a horizontal shear force to the whole structure.
An isolated GE member would simultaneously move to the right and rotate clockwise.
That movement is not occurring because the resistance to it from members BC and DG, via their pin connectors.
Members BC and DG are not moving or rotating either due to the resisting help of member AC and the solid ground.

External force F is reaching the ground while creating internal forces and moments in each of the members of this structure.
In the case of a truss, external forces acting on it are reaching the ground creating only internal compression and tension forces in its members and pin or roller connectors.

 

[tech99]

My ideas are as follows:-
If we are to ignore weights, the structure cannot be in equilibrium. So we must assign a weight W to the total structure and we have a centre of mass located on the vertical centre line. The structure acts exactly the same a a block of wood; you do not need to analyse twelve members, only the points at which forces are applied. Let the reactions at the supports be Av, Ah, Bv and Bh. The horizontal reaction is split equally between Ah and Bh. We cannot determine the split, so call it H.
Now analyse vertically, horizontally and then take moments about A (any point will work). This gives you 3 equations with 3 unknowns, which are Av, Bv and H.
When you have found H we have to assume an equal split of the horizontal reaction between Ah and Bh.

 

[haruspex]

My ideas are as follows:-
If we are to ignore weights, the structure cannot be in equilibrium.

Wrong.

 

[tech99]

Wrong.

I was assuming that the supports were knife edges so that the structure could lift off.

 

[brainpushups]

Not sure if this will confuse things, but perhaps approach the problem using the principle of virtual work? Was that taught as part of the course?

 

[haruspex]

I was assuming that the supports were knife edges so that the structure could lift off.

It says they are pin connected together. I agree the diagram is unclear though.

 

[Steve4Physics]

Isn’t the system statically indeterminate?
[EDIT - No it isn't! See later posts.]

Considering only external forces, it is trivial to show that $A_y = - F, B_y = F$ and $A_x + B_x = -F$:

$A_x$ and $B_x$ cannot be uniquely determined because they have the same line of action. Consideration of the internal structure won't help. To illustrate this consider two extreme cases:

1. The pinned-support at B is replaced by a simple horizontal frictionless support; to provide the required upwards reaction. In this case the structure is in equilibrium with $B_x=0$ and $A_x = -F$.

2. Alternatively, the pinned-support at A is replaced by a rope pulling vertically downwards. In this case the structure is in equilibrium with $A_x=0$ and $B_x = -F$.

Or maybe I'm missing something?
EDIT: Yes -looks like I missed something! See Posts #22 and #24.

 

[Lnewqban]

Or maybe I'm missing something?

Happy New Year, Steve!
It seems to me that, due to joints C and G, if any of the pinned-supports is replaced by a simple horizontal frictionless support, the structure will not be able to retain its shape..

 

[Steve4Physics]

Happy New Year, Steve!
It seems to me that, due to joints C and G, if any of the pinned-supports is replaced by a simple horizontal frictionless support, the structure will not be able to retain its shape..

Happy New Year @Lnewqban!

Yes, I see what you mean. So, for example, with A remaining pinned and with B free to move horizontally, the frame could be deformed into something like this

 

[haruspex]

Happy New Year @Lnewqban!

Yes, I see what you mean. So, for example, with A remaining pinned and with B free to move horizontally, the frame could be deformed into something like this

View attachment 368516

Ah, yes, I missed that. But not if B is pinned and A is free, right?, because F acts to the right. So does this mean the two horizontal forces there can be determined? It rules out the horizontal force at B being zero.

 

[Steve4Physics]

But not if B is pinned and A is free, right?, because F acts to the right.

Happy 2026 @haruspex!

If B is pinned and A is on a horizontal track, I suspect this would happen:

It satisfies the (intuitive) requirement that the system's geometrical centre moves right.

So does this mean the two horizontal forces there can be determined? It rules out the horizontal force at B being zero.

Yes. I believe so (i.e. that $A_x$ and $B_x$ are non-zero and can be determined). In fact I've just done it (hopefully correctly) using the method attempted by the OP in Post #13.

It worked out quite simply without having to use many symbols. I used the values found from the analysis of the external forces (Post #21).

Started by marking forces on the FBD of the top left piece (DG):

and continued this for the other pieces (in fact only 3 pieces needed yo be considered but using the 4th piece acts as a check).

I only needed to use symbols $a, F$ and $A_x$ to mark all the forces on every piece. It was then straightforward to express $a$ in terms of $F$ and then to find $A_x$ (and hence $B_x$).

Edit. Wording improved.

 

[Steve4Physics]

The Post #1 question seems to have only 20 minutes (exam’) working time advised, so a quick method is needed.

There is a fairly quick approach which uses (almost) no equations – essentially just diagrams.

I hope I’m not giving the OP too much assistance but they seem to have made reasonable effort but seem stuck. Click SPOLER to see.


This may be a bit long-winded but I want to make the method clear/explicit.

1. Consider the balance of external forces and moments (about A and B) for the whole system. We can directly construct/draw the following:

I don’t think any working, apart from the above diagram, is needed.
_________________________

2. Consider the top-left section (DG). Balancing forces and moments, we see that all the xy components of the pin reactions must have equal magnitudes. For simplicity/neatness call this value $a$. No equations are needed, we can just draw the diagram:

_________________________

3. Consider the top-right section, (GE). The force components at G are the Newton 3rd law partners to the ones at G for section DG. There is also an external force $\vec F$:

Balancing xy forces immediately gives the components at E if required (though this is not a necessary step).

Taking moments about E:
$-a -a-F =0 \implies a = -\frac F2$
_________________________

4. Consider the bottom-left section (AC). We know some force components from steps 1 and 2 above:

Balancing forces gives the components at C if required (though this is not a necessary step).

Taking moments about C and using the result $a = -\frac F2$ allows $A_x$ (and hence $B_x$) to be expressed in terms of $F$ .

There’s no need to analyse the bottom-right section (CB) but this can be done for a consistency check, exam’ time permitting.

Happy New Year to everyone I haven’t yet greeted!

Fairly minor edits.
Additional edit : first diagram corrected (thanks to @haruspex for spotting mistake (Post #27, below)).

 

[haruspex]

The Post #1 question seems to have only 20 minutes (exam’) working time advised, so a quick method is needed.

There is a fairly quick approach which uses (almost) no equations – essentially just diagrams.

I hope I’m not giving the OP too much assistance but they seem to have made reasonable effort but seem stuck. Click SPOLER to see.


This may be a bit long-winded but I want to make the method clear/explicit.

1. Consider the balance of external forces and moments (about A and B) for the whole system. We can directly construct/draw the following:
View attachment 368556
I don’t think any working, apart from the above diagram, is needed.
_________________________

2. Consider the top-left section (DG). Balancing forces and moments, we see that all the xy components of the pin reactions must have equal magnitudes. For simplicity/neatness call this value $a$. No equations are needed, we can just draw the diagram:
View attachment 368557
_________________________

3. Consider the top right section, (GE). The force components at G are the Newton 3rd law partners to the ones at G for section DG. There is also an external force $\vec F$.
View attachment 368558
Balancing xy forces immediately gives the components at E if required (hough this is not a necessary step).

Taking moments about E:
$-a -a-F =0 \implies a = -\frac F2$
_________________________

4. Consider the bottom left section (AC). We know some force components from steps 1 and 2 above:
View attachment 368560
Balancing forces gives the components at C if required (though this is not a necessary step).

Taking moments about C and using the result $a = -\frac F2$ allows $A_x$ (and hence $B_x$) to be expressed in terms of $F$ .

There’s no need to analyse the bottom right section (CB) but this can be done for a consistency check, exam’ time permitting.

Happy New Year to everyone I haven’t yet greeted!

Fairly minor edits.

Very neat. Small typo: in your diagram under (1), you have -F at top right instead of F.

 

[jojosg]

Ok I think I got it.

$\sum F_x = 0 : \quad A_x - B_x = -F \tag{1}$
$\sum F_y = 0 : \quad A_y + B_y = 0 \tag{2}$
$\sum M_A = 0 : \quad B_y(2a) - F(2a) = 0 \tag{3}$

$\sum M_E = 0:$
$-\frac{\sqrt{2}}{2}F_{DG}(a) - \frac{\sqrt{2}}{2}F_{DG}(a) - F(a) = 0 \tag{1}$
$-\sqrt{2}F_{DG} - F = 0 \tag{2}$
$F_{DG} = -\frac{\sqrt{2}}{2}F \tag{3}$

$\sum M_C = 0 : \quad F(a) - \frac{1}{2}F(a) - B_x(a) = 0$
$\frac{1}{2}F - B_x = 0 \quad \Rightarrow \quad B_x = \frac{1}{2}F$

$A_x - \frac{1}{2}F = -F \quad \Rightarrow \quad A_x = -\frac{1}{2}F$



$A_x = \frac{1}{2}F \, (\longleftarrow), \quad A_y = F \, (\downarrow)$

$B_x = \frac{1}{2}F \, (\longleftarrow), \quad B_y = F \, (\uparrow)$

 

[Steve4Physics]

Ok I think I got it.
.
..$A_x = \frac{1}{2}F \, (\longleftarrow), \quad A_y = F \, (\downarrow)$

$B_x = \frac{1}{2}F \, (\longleftarrow), \quad B_y = F \, (\uparrow)$

I agree. And well done for using LaTeX!

A couple of points:

1. You have assumed, without justification, that some reactions are at 45 degrees to the horizontal. It happens to be correct in this case but do you see why? (Using the separate xy components is safer.)

2. The question asks for the support reactions at points A and B. Giving each reaction’s components might be sufficient - but you might be expected to give each reaction’s magnitude and direction.

 

[jojosg]

I agree. And well done for using LaTeX!

A couple of points:

1. You have assumed, without justification, that some reactions are at 45 degrees to the horizontal. It happens to be correct in this case but do you see why? (Using the separate xy components is safer.)

2. The question asks for the support reactions at points A and B. Giving each reaction’s components might be sufficient - but you might be expected to give each reaction’s magnitude and direction.

I used basic trigo and drew a triangle with adjacent and opposite sides both a to get 45 degrees