How to calculate all forces in Truss with Methods of Joints?

A truss is a structure made up of slender members or beams pin-connected at ends and is capable of taking loads at joints. Truss is a rigid structure. They are used as roof trusses to support sloping roofs and as bridge trusses to support the deck. In many machines, steel trusses are used. Transmission towers are also examples of trusses. In the case of wooden trusses, the ends are connected by making suitable joints or by nailing and bolting whereas in steel trusses ends are connected by bolting or welding. The trusses are also known as ‘pin-jointed frames’. Let us discuss more details on Trusses and see how we can calculate all forces in Truss.

A truss in which all the members lie in a single plane is called a plane truss. In such trusses loads act in the plane of the truss only. Roof trusses and bridge trusses can be considered plane trusses.

If all the members of a truss do not lie in a single plane, then it is called a space truss. Tripod and transmission towers are examples of space trusses.

The analysis of plane trusses is discussed below.

What is a Perfect Truss?

A pin-jointed truss that has got just a sufficient number of members to resist the loads without undergoing appreciable deformation in shape is called a perfect truss.

A triangular truss is the simplest perfect truss and it has three joints and three members as shown in the below figure.

Perfect trusses with four and five joints are shown in Figures (a) and (b) respectively.

It may be observed that to increase one joint in a perfect truss, two more members are required.

Hence the following expression may be written down as the relationship between a number of joints j, and the number of members m, in a perfect truss.

m = 2j – 3

However, the above equation gives only a necessary, but not sufficient condition of a perfect truss. For example, the two trusses are shown in Figures (a) and (b) have the same number of members and joints.

The truss is shown in Figure (a) is perfect whereas the one shown in Figure (b) is not capable of retaining its shape if loaded at the joint marked 6. Therefore, the only necessary and sufficient condition of a perfect truss is that it should retain its shape when the load is applied at any joint in any direction.

What is a Deficient Truss?

A truss is said to be deficient if the number of members in it is less than that required for a perfect truss. Such trusses cannot retain their shape when loaded. A deficient truss is shown in the below figure.

What is a Redundant Truss?

A truss is said to be redundant if the number of members in it is more than that required in a perfect truss. Such trusses cannot be analyzed by making use of the equations of equilibrium alone.

Thus, a redundant truss is statically indeterminate. Each extra member adds one degree of indeterminacy. For the analysis of such members, the consistency of deformations is to be considered. The truss shown below is a typical redundant truss. In this truss one diagonal member in each panel is extra. Hence it is a two-degree redundant truss.

Let us only consider the perfect Truss for the analysis and let us see how we can calculate all the forces in Truss.

Assumptions for Analysis of Truss

In the theory that is going to be developed in this chapter, the following assumptions are made:

(1) The ends of the members are pin-connected (hinged)
(2) The loads act only at the joints
(3) Self-weights of the members are negligible
(4) Cross-section of the members is uniform

If at all the cross-section varies, the center of gravity of the section is assumed to be located along the same longitudinal line.

In reality, the members are connected by bolting, riveting, or by welding. No special care is taken to ensure perfect pin-connections. However, experiments have shown that assuming pin-connected ends is quite satisfactory since the members used are slender.

Nature Of Forces In Truss Members

The members of a truss are subjected to either tensile or compressive forces. A typical truss ABCDE loaded at joint E is shown in the below figure (a).

The member BC is subjected to compressive force C as shown in the above figure (b). The effect of this force on joint B (or C) is equal and opposite to the force C as shown in figure (b).

The member AE is subjected to tensile force T. Its effect on joints A and E are as shown in Figure (b) above. In the analysis of the frame, we mark the forces on the joints, instead of the forces in the members as shown in Figure below.

It may be noted that compressive force in a member is represented in a figure by two arrows going away from each other and a tensile force by two arrows coming towards each other. This is quite logical considering the fact that the markings on the members represent the internal reactive forces developed which are opposite in direction to the applied forces.

Methods Of Analysis

The following three methods are available for the analysis of pin-connected frames called Trusses.

(a) Method of joints
(b) Method of section
(c) Graphical method.

let us discuss how we can analyze the Truss with help of the method of Joints.

Calculating Forces in Truss by Method Of Joints

• At each joint the forces in the member’s meeting and the loads acting, if any, constitute a system of concurrent forces.
• Hence, two independent equations of equilibrium can be formed at each joint.
• First, a joint is selected where there are only two unknown forces.
• Many times such a joint can be identified only after finding the reaction at the support by considering the equilibrium of the entire frame.
• Then making use of the two equations of equilibrium at that joint the two unknown forces are found.
• Then, the next joint is selected for analysis where there are now only two unknown forces.
• Thus, the analysis proceeds from joints to joint to find the forces in all the members.
• It may be noted that if there are j number of joints, 2j number of the equations can be formed.
• There will be three reactions in a general determinate truss.
• The force in each member is unknown.
• Hence, if there are m number of members, the total number of unknowns will be m+3.

A problem can be analyzed if there are as many equations as there are unknowns.

Hence, a frame analysis problem is determinate if: 2j = m + 3

This equation is the same as the Equation above mentioned relation of a number of joints and number of members which was derived on the consideration of a perfect frame.

Hence, a perfect frame is determined.

• If m > 2j – 3, then the number of unknowns is more than the number of equations. Hence, a redundant frame is indeterminate.
• If m < 2j – 3, then the number of equations is more than the number of unknowns.

Since a set of solutions can satisfy such equations, it shows the instability of the structure. Hence, a deficient frame is not stable.

Let us explore this method of joints with help of example problems.

Example Problems to calculate all Forces in Truss with help of Method Of Joints

Problem Statement 1: Find the forces in all the members of the truss shown in the below figure. Please Tabulate the results.

Answer:

Step by Step Procedure to calculate all Forces in a Truss with help of Methods of Joints

Step 1: Determine the inclinations of all inclined members.

In this case tan θ = 3/3 = 1
∴ θ = 45°

Step 2: Look for a joint at which there are only two unknowns. If such a joint is not available,
determine the reactions at the supports, and then at the supports, these unknowns may reduce to only two.

Now at joints C, there are only two unknowns, i.e., forces in members CB and CD, say F_CB and F_CD.

Note: Usually in cantilever type frames, we find such joints without the need to find reactions.

Step 3: Now there are two equations of equilibrium for the forces meeting at the joint and two unknown forces. Hence, the unknown forces can be determined.

At joint C as shown in the below figure ∑V = 0 condition shows that the force F_CB should act away from
the joint C so that its vertical component balances the vertical downward load at C.

F_CB sin 45° = 40
∴ F_CB = 40 √2 kN

Now ∑H = 0 indicates that F_CD should act towards C.
F_CD – F_CB cos 45° = 0
F_CD = F_CB cos 45°
F_CD = 40 √2 × (1/√2) ×
F_CD = 40 kN

Note: If the assumed direction of the unknown force is opposite, the value will be negative. Then reverse the direction and proceed.

Step 4: On the diagram of the truss, mark arrows on the members near the joint analyzed to indicate the forces on the joint. At the other end, mark the arrows in the reverse direction.

In the present case, near joint C, the arrows are marked on the member’s CB and CD to indicate forces F_CB and F_CD directions as found in the analysis of joint C.

Then reversed directions are marked in the members CB and CD near joints B and D, respectively.

Step 5: Look for the next joint where there are only two unknown forces and analyze that joint.

In this case, there are only two unknown forces at the joint D, as shown in the below figure.

∑V = 0
F_DB = 40 kN
∑ H = 0
F_DE = 40 kN

Step 6: Repeat steps 4 and 5 till forces in all the members are found.

In the present case, after marking the forces in the members DB and DE, we find that analysis of joint B can be taken up. Referring to the below figure.

∑ V = 0, gives
F_BE sin 45° – 40 – 40 √2 × sin 45° = 0
∴ F_BE = 80 √2 kN
∑ H = 0
F_BA – F_BE cos 45° – 40√2 × cos 45° = 0
F_BA = 80 √2 × (1/√2) + 40 √2× (1/√2)
∴ F_BA = 120 kN

The directions of these forces are marked on the diagram.

Now the analysis is complete since the forces in all the members are determined.

Step 7: Determine the nature of forces in each member and tabulate the results. Note that if the arrow marks on a member are towards each other, then the member is in tension and if the arrow marks are away from each other, the member is in compression.

In this case,

Member	Magnitude of Force in kN	Nature
AB	120	Tension
BC	40√2	Tension
CD	40	Compression
DE	40	Compression
BE	80 √2	Compression
BD	40	Tension

Problem Statement 2: Determine the forces in all the members of the truss shown in the below figure and indicate the magnitude and nature of forces on the diagram of the truss. All inclined members are at 60° to horizontal and the length of each member is 2 m.

Solution:

Now, we cannot find a joint with only two unknown forces without finding reactions.

Consider the equilibrium of the entire frame.

∑ M _A = 0, gives

R_D × 4 – 40 × 1 – 60 × 2 – 50 × 3 = 0
∴ R_D = 77.5 kN
∑ H = 0, gives
∴ H_A = 0

∴ Reaction at A is vertical only

∑V = 0, gives
R_A + 77.5 = 40 + 60 + 50
∴ R_A = 72.5 kN

Joint A:

∑V = 0, gives
F_AB sin 60° = R_A = 72.5
F_AB = 83.7158 kN (Compression)

∑ H = 0, gives
F_AE – 83.7158 cos 60° = 0
F_AE = 41.8579 kN (Tension)

Joint D:

∑V = 0, gives
F_DC sin 60° = R_D = 77.5
∴ F_DC = 89.4893 kN (Compression)

∑ H = 0, gives
F_DE – 87.4893 cos 60° = 0
∴ F_DE = 44.7446 kN (Tension)

Joint B:

∑V = 0, gives
F_BE sin 60° – F_AB sin 60° + 40 = 0
F_BE = (72.5- 40)/ sin60°
∴ F_BE = 37 .5278 (Tension)

∑ H = 0, gives
F_BC – F_AB cos 60° – F_BE cos 60° = 0
F_BC = (83.7158 + 37.5274) × 0.5
F_BC = 60.6218 kN (Compression)

Joint C:

∑V = 0, gives
F_CE sin 60° + 50 – F_DC sin 60° = 0
F_CE = (77.5- 50)/ sin 60
F_CE = 31.7543 kN (Tension)

Now the forces in all the members are known. If joint E is analyzed it will give the check for
the analysis. The results are shown on the diagram of the truss in the below figure.

Problem Statement 3: Determine the forces in all the members of the truss shown in the below figure.

Solution:

AB = 6 sin 60° = 3 m;
BD = 6 cos 30° = 23.09 m

In ∆ABC, AC = AB = 3 m
∴ ∠ABC = ∠BCA = say θ

∴ 2θ = 180 – 60 = 120
θ = 60°

Taking moment about A, we get

R_D × 6 – 10 × 3 – 20 × 3 cos 60° = 0
R_D = 10 kN
∴ R_A = 20 + 10 – 10 = 20 kN
[Note: H₄ = 0]

Joint A: Referring to above figure

ΣV = 0, gives
20 – F_AB sin 60° = 0
∴ F_AB = 23.09 kN [Comp.]

ΣH = 0, gives
F_AC – F_AB cos 60° = 0
∴ F_AC = 23.09 cos 60° = 11.55 kN [Tensile]

Joint D:

ΣV = 0, gives
F_DB sin 30° = 10
F_DB = 20 kN [Compression]

ΣH = 0, gives
– F_DC + F_DB cos 30° = 0
F_DC = 20 cos 30°
F_DC = 17.32 kN [Tensile]

Joint C:

ΣV = 0, gives
F_CB sin 60° = 10
∴ F_CB = 11.55 kN

check ΣH = 0, gives
F_CB cos 60° = F_CD – F_CA = 17.32 – 11.55
∴ F_CB = 11.55, Checked

Let us mark all these reaction forces in a single diagram.

Calculating Forces in Truss by Method Of Section

• In the method of the section, after determining the reactions, a section line is drawn passing through not more than three members in which forces are not known such that the frame is cut into two separate parts. 
• Each part should be in equilibrium under the action of loads, reactions, and forces in the members that are cut by the section line.
• Equilibrium of any one of these two parts is considered and the unknown forces in the members cut by the section line are determined. 
• The system of forces acting on either part of the truss constitutes a non-concurrent force system. 
• Since there are only three independent equations of equilibrium, there should be only three unknown forces. 
• Hence, in this method, it is an essential condition that the section line should pass through not more than three members in which forces are not known and it should separate the frame into two parts.

Thus, the method of the section is the application of nonconcurrent force system analysis whereas the method of joints, described in the previous article was the application of analysis of concurrent force
system.

Under the following two conditions the method of the section is preferred over the method of joints:

(1) In a large truss in which forces in only a few members are required
(2) In the situation where the method of joints fails to start/proceed with the analysis

Read more about the method here. We also discussed 3 example problems to calculate all forces in Truss with the Method of section.

How to calculate all forces in Truss with Methods of Section?

Conclusion

We have discussed what is a perfect truss, a Redundance truss, and a deficient Truss. We have listed the assumptions that we make to analyze the Truss, also list the types of forces in truss members. We calculate all the forces in the truss given in 3 example problems successfully with help of the method of Joint. Let us know what do you think about this article in the comment section below.