What makes a beam statically indeterminate

A structure is stable if its centre of gravity lies above its base. An object is unstable when its centre of gravity lies outside its base. In other words, an object is unstable if a line drawn between its centre of gravity and the centre of the Earth does not pass through its base. A beam is said to be stable if it satisfy the following conditions.

All the reactions should not be parallel to each other. There should be no concurrent force system i. Besides truss-supported beams, beams are classified in the following groups: Simply supported: A beam supported on the ends, which are free to rotate and have no moment resistance.

Fixed: A beam supported on both ends, which are fixed in place. Overhanging: A simple beam extending beyond its support on one end. A fluid, such as air, tending to become or remain turbulent is said to be statically unstable; one tending to become or remain laminar is statically stable; and one on the borderline between the two which might remain laminar or turbulent depending on its history is statically neutral.

In a statically determinate system, all reactions and internal member forces can be calculated solely from equations of equilibrium. A truss which has got just sufficient number of members to resist the loads without undergoing deformation in its shape is called a perfect truss. Triangular truss is the simplest perfect truss and it has three joints and three members. There are five basic idealized support structure types, categorized by the types of deflection they constrain: roller, pinned, fixed, hanger and simple support.

If the structure can be analyzed by use of the equation of equilibrium is known as statically determinate structure. The equation of equilibrium are,. Sum of horizontal force is zero, Sum of vertical force is zero, Sum of the moment about any point is zero. The structure which cannot be analyzed by just using the equation of equilibrium is known as a statically indeterminate structure.

At A it is hinged support and at B it has roller support. So, In the above beam, the reaction occurred at A is vertically upward and also in a horizontal direction. Because hinge support has two reactions. So, in the above beam, Totally three unknown reaction which is two reactions at support A and one unknown reaction at B.

If the number of unknown reactions is less than or equal to equilibrium reaction i. The above beam, It has two hinge support in which, A has two reactions and B has also two reactions. The image below is a common situation of a statically indeterminate beam system. I will go into detail on how to step up and solve this problem using deflection. This is where you use the right hand rule. There are no forces in the X or Z direction and therefore nothing to cause rotation on the Y axis.

Therefore, we are left with 3 equations, force in the Y direction and moments about X and Z. Unfortunately, we have 4 supports and is statically indeterminate. Or is it? This is a case where we can use deflection to find out how the loads are distributed and what the moment on the beams are. When we look at a simply supported beam with a load in the center, we can look up the beam deflection in any table and find:.

At the center where the load is applied, the deflections will be the same, causing the loads to be unequal. It is at this point, that we are able to come up with two new equations,. Combining these equations together can get us a workable set of equations able to find all of the reactions as well as the deflection of the beam. Since the load at each end of the beam is half of the load applied, R 1 is lb and R 2 is lb. Here we have to use a different load case for the green beam, but the blue beam will remain the same.

In this situation, the following equations still hold because the load is placed at where the beams cross. Where a is 24 in and b is 48 in from the dimensioned drawing above.

Using the data from Table 1, we can now come up with our 4 equations and solve them. You can see, that moving the beams off center made very little difference to the process of solving for the indeterminate beam system.

However, we still need the deflections to match where the beams cross. We will need to have the loads be equal but in opposite directions to remain in a static conditions. Thus, our equations become:. In this case, we have 2 loads acting upon the longer beam: the load applied and the load between the two members.