**Note:**One may ask at this point as to why as we need the horizontal rod (13). It is because point 3 will otherwise keep moving to the right making the whole structure unstable. Rod (13) has two forces acting on it: one vertical force due to the wheel and the other at end 2. However these two forces cannot be collinear so without the rod (13) the system will not be in equilibrium. Generally, in a truss each joint must be connected to at least three rods or two rods and one external support. Let us now analyze forces in the structure that just formed. For simplicity I take the lengths of all rods to be equal. To get the forces I look at all the forces on each pin and find conditions under which the pins are in equilibrium. The first thing we note that each rod in equilibrium under the influence of two forces applied by the pins at their ends. As I discussed in the previous lecture, in this situation the forces have to be collinear and therefore along the rods only. Thus each rod is under a tensile or compressive force. Thus rods (12), (23) and (13) experience forces as shown in figure 3.

_{12}due to rod (12) and F

_{23}due to rod (23). Further, it is pulled down by the weight W. Thus forces acting on pin 2 look like shown in figure 4.

_{23}, normal reaction N and a horizontal force F

_{13}. Applying equilibrium condition

_{13}is coming out to be negative, the direction should be opposite to that assumed. Balance of forces in the vertical direction gives

*j*and the number of members (rods)

*m*are related as follows:

*m = 2j - 3 *

*m + 3*. We solve for these unknowns by writing equilibrium conditions for each pin; there will be

*2j*such equations. For the system to be determinate we should have

*m + 3 = 2j*, which is the condition given above. If we add any more members, these are redundant. On the other hand, less number of members will make the truss unstable and it will collapse when loaded. This will happen because the truss will not be able to provide the required number of forces for all equilibrium conditions to be satisfied. Statically determinate trusses are known as simple trusses.

**Exercise 1:**Shown in figure 5 are three commonly used trusses on the sides of bridges. Show that all three of them are simple trusses.

- If the middle line of the members of a truss meet at a point that point is taken as a pin joint. This is a very god assumption because as we have seen earlier while introducing a truss (triangle with pin joint), the load is transferred on to other member of the trusses so that forces remain essentially collinear with the member.
- All external loads are applied on pin connections.
- All members' weight is equally divided on connecting pins.

**Truss Analysis - Method of joints: **

In method of joints, we look at the equilibrium of the pin at the joints. Since the forces are concurrent at the pin, there is no moment equation and only two equations for equilibrium viz. **Example 1:**As the first example, I take truss ABCDEF as shown in figure 6 and load it at point E by 5000N. The length of small members of the truss is 4m and that of the diagonal members is

*Nx*at point A is zero because there is no external horizontal force on the system. To find

*N*I take moment about A to get

_{2}^{rd}law). Therefore force on member AB is compressive (pushes pin A away) whereas that on AF is tensile (pulls A towards itself). Next I consider joint F where force AF is known and two forces BF and FE are unknown. For pint F

_{BE}to be compressive, it is actually tensile.

_{CE}to be tensile whereas F

_{CD}to be compressive

**Truss Analysis - Method of sections: **

As the name suggests in method of sections we make sections through a truss and then calculate the force in the members of the truss though which the cut is made. For example, if I take the problem we just solved in the method of joints and make a section S_{1}, S

_{2}(see figure 9), we will be able to determine the forces in members BC, BE and FE by considering the equilibrium of the portion to the left or the right of the section.

*N*at D and

*N*at A. Now let us consider the section of the truss on the left (see figure 10).

*N*reaction at A. This clearly tells us that

*F BE*is tensile. Similarly, to counter the torque about B generated by

*N*force at A, the force on FE should also be from F to E. Thus this force is also tensile. If we next consider the balance of torque about A,

_{FE}do not give any torque about A. So to counter torque generated by F

_{BE}, the force on BC must act towards B, thereby making the force compressive. Let us now calculate individual forces. F

_{FE}is easiest to calculate. For this we take the moment about B. This gives 4 Ã—

*N*Next we calculate F

_{BE}. For this, we use the equation

_{BC}, we can use either the equation