# Confused about forces acting in this simple statics problem

I'm trying to solve this statics problem to find the internal tension that results, in each member:

It shows 3 idealised weightless rigid members hinged in a vertical plane, in an isoceles triangle. Two of the members carry identical loads at their midpoint (W = Mg). The bottom member is constrained in such a way that the internal tension exactly balances the "spreading" forces developed by the two members it supports.

But when I try and calculate it (which should be simple!) I get an indeterminate solution. it looks like there should be another constraint, beyond the usual ones, but I can't see any constraint that I'm missing.

As a thought experiment, it should have a single, static, unique, solution. If I built a rigid triangle like this, from 3 lengths of wood, dangled weights from 2 of them, and somehow balanced it vertically with one bottom corner on a brick and the other bottom corner level with it on a roller skate (or just stood it up in an ice-rink), that would match the setup. The resulting concoction would clearly have a static equilibrium (ignoring any bending or overbalancing!) and therefore a unique + static set of tensions, for any given reasonable lengths of wood and magnitudes of weight.

# My attempt at solving

By symmetry, I should be able to analyse just one of the diagonal members, to solve for all tensions. So I just draw one diagonal strut, and the reactions/forces at its ends:

For simplicity I'm using resolved reaction forces F1 and F2 for the support at the apex from the other diagonal member, and F2 and F3 for the horizontal internal tension and roller support at the bottom right corner (the horizontal components must be equal and opposite):

1. Vertically:
2W = F1 + F3
=> F3 = 2W - F1
2. Horizontally: zero net force (only 2 forces, equal and opposite)
3. Moments round top-left end:
F3.(2d) = (2W).d + F2.(2h)
=> F3d = Wd + F2h
Substituting from (1):
( 2W - F1 ).d = Wd + F2h
=> Wd = F1d + F2h
4. Moments round bottom-right end:
(2W).d = F1.(2d) + F2.(2h)
=> Wd = F1d + F2h
Same equation as (3).

So I now have 2 equations in 3 unknowns. I'm guessing there should be an extra constraint, but I can't see it.

• Was this problem given in a statics class analyzing trusses? – Bob D Mar 17 at 16:51
• No, homework and classes are many years in the past :) Its a problem/situation I noticed, arising casually in "real life", in which the scale is metres and kN's, and I got curious about the strength of the bottom restraints that was required to stop the top from pancaking. I can see why you wondered if it's a truss though. It seemed a solution should be easy, but for some reason it works out indeterminate, and I can't think what I'm missing. – Stilez Mar 17 at 17:31
• I don't think this truss possess all the properties required of a statically determinate truss. But it's been a while for me too. – Bob D Mar 17 at 17:37
• But what's missing? If built, it would be pretty determinate - it would settle at a specific position with specific tensions. I must be missing some extra constraint. Maybe something about the angles that the forces apply at, that some or other of them are in a constrained ratio (angle of force they apply) or some component is zero. I'm not sure, but that's my best guess. I assumed no constraint on that, since reaction forces aren't always parallel/perpendicular to the objects they apply to. Maybe I'm wrong? – Stilez Mar 17 at 17:44
• See my answer to follow – Bob D Mar 17 at 17:45

There is no horizontal reaction force at the support on the bottom right. Rollers do not resist horizontal forces. $$F_2$$ is the tensile force in the bottom member.

Look at the structure as a whole.

There are only two externally applied forces, each 2W downward.

There are only two reaction forces, the vertical reaction force at the pin and the vertical reaction force at the roller, call them $$R_A$$ and $$R_B$$.

Therefore for equilibrium,

$$+R_{A}-2W-2W+R_{B}=0$$

$$R_{A}+R_{B}=4W$$

The vertical loading on the structure is symmetrical with respect to the two supports. Therefore

$$R_{A}=R_{B}=2W$$

You can prove this by summing the moments of the loads and reaction at the roller about the pin and set to zero.

And by the way, this is not a simple statics truss problem in terms of determining the stresses in its members. A statically determinate truss possesses all of the following properties:

1. Members of the truss lie in the same plane (planar truss)
2. Members are connected by frictionless pins
3. All external loads lie in the plane of the truss and applied at the joints (pins) only
4. Truss reactions can be determined by equations for equilibrium (sum of forces and moments zero)
5. All forces on the truss are concurrent and meet at the joints
6. All member forces are axial and either in tension or compression.

I have bold faced those aspects of the properties that do not seem to be met by your truss. The main issue is that the two external 2W loads are not applied at the pins. The bottom member is surely in axial tension. But the other two other members seem to be subjected to positive bending stress, meaning the top fibers of each member are in compression and the bottom fibers tension. So it is neither pure tension or compression.

UPDATE:

The following is per your request. I have to caution you, however, that it’s been a while since I did these problems and I am not an ME, so you might want to check with others who are more familiar with them. Maybe @npojo can add something here..

The stresses in the side members are complex because they involve shear stress and bending moments due to the loads in addition to compressive force component in the member due to the externally applied load.

If we can assume negligible deformation (bending) of the side members then I think we can resolve the two 2W loads into components parallel to and normal to the members as shown in the diagram below. The normal component causes vertical shear force and bending moments in each member The parallel compressive component of the force (axial force in the member) is then what is transmitted to the bottom member based on summing the forces at pin A.

Hope that helps.

• It does. But I'm not sure how to extend it to get the tension in the bottom member, and especially, the forces applying to each end of the bottom member and the angles they are applied at (I understand your point about the mixed tension/compression in the diagonals, very helpful). My concern is that they might have a line of action that's neither horizontal nor vertical, so I'd need to calculate the H + V components of the forces pulling that member at each end... You've explained how to get F(3), but can you update your answer a bit, to explain how to get the transverse tensions F(2)? – Stilez Mar 17 at 20:00
• @Stilez I'll give it a try – Bob D Mar 17 at 20:19
• Also fixed question to correct description of F(2) as internal tension in member. The main question, I realise, is the magnitude and direction of the internal tension in the lower member. I figured that I could ignore the compression/tension issue (much as it's ignored in the classic exercise "ladder with man on it is propped against a smooth wall, what is the friction force supporting it" from school days, where we assume a rigid beam and simply resolve forces + moments. I don't understand why this hasn't worked for me here, as the question seems identical - find the horizontal force). – Stilez Mar 17 at 20:39
• @Stilez See my update. You can see that in order for the sum of the horizontal forces at pin A to be zero, that the force in the lower member, which I call AC, will be negative (arrow in opposite direction) which shows that the member is in tension. – Bob D Mar 17 at 20:55

$$F_1=0$$. Reasoning: forces on left diagonal must be symmetric to right diagonal. Hence $$F'_1$$ on the left diagonal equals $$F_1$$. But $$F_1$$ and $$F'_1$$ are action and reaction forces. Hence are opposites.

• That's helpful reasoning. I also got F1=0 (net force on structure as a whole gives 4W = 2.F(3), but also 2W = F(1)+F(3) for each diagonal considered separately), but I couldn't justify it to myself on any intuitive basis. Your point about equal and opposite as they are action/reaction, and by symmetry F1=0, provides the step of intuitive justification that I couldn't get. Not a solution to the overall problem of tensions in the (especially bottom) member, but very appreciated. – Stilez Mar 17 at 20:29