Trusses resist axial forces (member forces) only. These forces are usually tension and compression forces. A typical example of a truss member is the roof truss with members such as kingpost, rafters (in roof truss) or boom in other usage, tie members (and not tie beams), etc.

The first step in the analysis of a statically determinate structure is to determine the support reactions. This can be done either by using the equations of static equilibrium or by the funicular polygon. Afterward, the determination of the member forces can be accomplished by either of the following methods:

- Method of Joints
- Method of Section
- Method of Tension Coefficients (commonly used for trusses in three dimensions)
- Graphical Method

**Method of Joints**

In the method of joints, understanding sign convention is important for success. When a member is in tension, it is assigned ‘+ve’ because the member by undergoing tension gains length. When the member is in compression, it is assigned ‘-ve’ because the member by undergoing compression loses length. In the resolution at the joints, free body diagrams are drawn for each joint, in turn, starting with the joint at which there are not more than two unknown forces/reactions. These can be determined by the application of two equations of equilibrium:

ΣFx = 0 and ΣFy = 0

See the Post: Resolution of Trusses: Method of Joints

**Method of Sections**

This method consists in applying equations of static equilibrium to a free-body diagram. The method is very lengthy if used to determine all forces in the members of a frame but is useful if the forces in only a few members are required. In the method, a section line is passed through the members in which forces are required to be found. A part of the structure, on any side of the section line, is then treated as a free body in equilibrium under the action of external forces. The unknown forces are then found by the application of equilibrium or principle of statics i.e. ∑M = 0. When drawing the section line, care should be taken not to cut more than three members, in which the forces are known.

**Method of Tension Coefficients**

For a space frame, three equations of static equilibrium have to be satisfied at each joint, and application of these equations to each joint in turn enables the forces in the members to be found. The method adopted for this purpose is similar to that used in the method of joints but is somewhat simplified by using the tension coefficients of the member.

In a plane frame, if AB is a bar of length, L_{AB}, having a tensile force in it of T_{AB}, then the components of this force in the x- and y-directions are T_{AB}cosBAX and T_{AB}sinBAX.

If the coordinates of A and B are xA, yA and xB, yB respectively then the component of TAB in the x-direction =

the t_{AB} is known as the tension coefficient of bar AB. Similarly, the component in the y-direction =

If at a joint A in the frame, there are a number of bars, AB, AC…AN and external loads, XA, and YA acting in the x-direction and y-direction, then since the joint is in equilibrium, the sum of the components of the external and internal forces must be zero in each of these directions. Expressing these relationships symbolically gives the expression below.

A similar pair of equations can be formed for each joint in the frame, giving in all 2j equations in the case of a frame having j-joints. These equations will contain the tension coefficients as unknowns and if the frame has n-members then there are n-unknown tension coefficients. But for a plane frame, n = 2j – 3, hence there are three superfluous equations. These can be used either to determine the reactions or to check the values of the tension coefficients obtained from the previous equations. In a space frame, each joint has three coordinates and the forces have components in three directions x, y, and z. Thus, if there are j-joints in a space frame, the consideration of the equilibrium in the three directions produces 3j equations containing n-unknown tension coefficients. But n = 3j – 6, hence there are six (6) superfluous equations.

Having found the tension coefficients t_{AB}, the force in the bar is the product t_{AB}L_{AB}. The equations of equilibrium are built up by assuming that the bars are in tension. A bar that is in compression has a negative tension coefficient. The procedure for using this method is as follows:

- Take positive direction for x, y, and z.
- Assume all members are in tension.
- Write equations for each joint in the frame.
- Solve equations for t
_{AB} - Check the values of Tab from equations of static equilibrium
- Calculate T
_{AB}= L_{AB}x t_{AB}

** ****Example 1**

The space frame shown in the plan below has the pinned supports A, B, and C at the same level. DE is horizontal and at a height of 10m above the plane of the supports. Calculate the forces in the members when the frame carries 8 kN and 4 kN loads in the horizontal plane at E and D respectively.

**Solution
**In the case of a space frame having pinned supports, the number of bars for a statically-determinate frame is 3JF where JF = number of free joints. The number of equilibrium equation will be 3JF hence sufficient equations are obtained to solve for the unknown tension coefficients.

Sign nomenclature:

**Joint D**

x-direction: – t_{DA} (5) + t_{DE }(10) + t_{DC} (5) – 4 = 0

y-direction: -t_{DA} (10) + t_{DE }(0) + t_{DC} (10) = 0

z-direction: t_{DA} (10) + t_{DE }(0) + t_{DC} (10) = 0

>>

– 5t_{DA} + 10t_{DE} + 5t_{DC} – 4 = 0 (1)

-10t_{DA} + 10t_{DC} = 0 (2)

10t_{DA} + 10t_{DC} = 0 (3)

**Joint E**

x-direction: -t_{EA}(15) + t_{EB}(5) – t_{ED}(10) – t_{EC}(5) = 0

y-direction: -t_{EA}(10) + t_{ED}(0) – t_{EB}(10) + t_{EC}(10) + 8 = 0

z-direction: t_{EA}(10) + t_{EB}(10) + t_{EC}(10) + t_{ED}(0) = 0

>>

-15t_{EA} + 5t_{EB} – 10t_{ED} – 5t_{EC} = 0 (4)

-10t_{EA} – 10t_{EB} + 10t_{EC} + 8 = 0 (5)

10t_{EA} + 10t_{EB} + 10t_{EC} = 0 (6)

From Equation (2),

-10t_{DA} + 10t_{DC} = 0 >> -10t_{DA} = -10t_{DC }>> t_{DA} = t_{DC }

From Equation (3),

10t_{DA} + 10t_{DC} = 0 >> 10t_{DA} + 10t_{DA }= 0 >> 20t_{DA }>> t_{DA }= 0

From Equation (1),

– 5t_{DA} + 10t_{DE} + 5t_{DC} – 4 = 0 >> – 5(0) + 10t_{DE} + 5(0) – 4 = 0 >> 0 + 10t_{DE }+ 0 – 4 = 0 >>

10t_{DE }= 4 >> t_{DE }= = 0.4

From Equation (4);

-15t_{EA} + 5t_{EB} – 10t_{ED} – 5t_{EC} = 0 but t_{DE }or t_{ED} = 0.4

Thus, -15t_{EA} + 5t_{EB} – 10 (0.4) – 5t_{EC} = 0 >> -15t_{EA} + 5t_{EB} – 4– 5t_{EC} = 0 or

5t_{EB} – 5t_{EC }-15t_{EA} = 4 (7)

>> 5t_{EB} = 4 + 5t_{EC }+15t_{EA}

From Equation (5)

-10t_{EA} – 2 (5t_{EB}) + 10t_{EC} + 8 = 0 >> -10t_{EA} – 2 (4 + 5t_{EC }+15t_{EA}) + 10t_{EC} + 8 = 0

-10t_{EA} – 8 – 10t_{EC }+ 10t_{EC} + 8 = 0

Collect like terms

-10t_{EA }-30t_{EA} – 10t_{EC} + 10t_{EC} + 8 – 8 = 0 >> -40t_{EA }= 0 >> t_{EA }= 0

From Equation (7)

5t_{EB} – 5t_{EC }– 0 = 4 >> 5t_{EB} – 5t_{EC} = 4 (8)

From Equation (6)

0 + 10t_{EB} + 10t_{EC} = 0 >> 10t_{EB} = – 10t_{EC }>> t_{EB} = – t_{EC }(9)

Substitute Equation (9) into (8)

5(-t_{EC}) – 5t_{EC} = 4 >> – 10t_{EC} = 4 >> t_{EC} = -0.4

From Equation (9), t_{EB} = – t_{EC }= – (-0.4) = 0.4

The summary of tension coefficients for the five (5) members of the frame are as follows

t_{DA} = 0; t_{DC} = 0; t_{EA }= 0; t_{DE }= 0.4; t_{EC} = -0.4; t_{EB} = 0.4

The summary table below shows the length of the members, tension coefficients and the forces calculated from both.

**Graphical Method**

Graphical method is one of the few common methods used in the analysis of trusses. This method involves establishing a common scale for force and dimensions (in mm or m) of the truss and drawing the truss to that scale in a graph paper while representing each of the forces sequentially from one member to another both in magnitude and direction. After wards, the forces are scaled out and the scales used to convert them to the force unit (kN).

The general procedure of this methods are as follows:

- Draw a space diagram to some scale (say 1 cm = 2 m).
- Determine the reaction either graphically or analytically.
- After finding the reactions, start with the joint in which there are not more than two unknown force.
- Draw the line parallel and equal to the force it is representing.
- Continue like this until all the forces are fully represented.