Method of Joints

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Introduction

Trusses are a number of straight members connected to each other by their ends by pinned connection to form a stable structure [1]. Trusses are most notably presented in roofs and bridges structures [2]. A truss’ members are most frequently made of wood or metal ( mostly Steel ).

When a force or load is applied on a structure, the load is transferred to its truss members. Sometimes, it is required to find the forces in every member in order to determine the durability and have safe designed trusses. The method of joints is a particularly simple and useful way of determining the forces transferred through each member, however, this method cannot be applied unless the external structure is in equilibrium [2] . When it is determined that whole structure is externally in equilibrium, we can perfectly assume that every internal joint in that structure is also in equilibrium. This is to say that the forces at each joint should sum up to zero. [2]


Method of joints can be done by solving the reaction forces of the whole truss. After that, the structure will be divided into joints and each joint will have its own equilibrium equations. For example, in Figure 1 there are four joints (A, B, C & D), each joint will be solved separately. Joints with less unknowns are easier to solved, and solved forced can be used as a known force in the next join that has a member in contact. e,g, joints A and B or A and C , but not A and D. .[2]




Procedure For Analysis

  1. Draw a free body diagram. [2]
  2. Check for Determinacy, Indeterminacy and Stability. [1]
  3. Find the reactions at the supports. [2]
  4. Determine the zero force members:
    1. If a joint has no external loads acting on it, or no reactions applied to it, and it has only two non co-linear members, then both of those members have internal forces equal to zero.[1]
    2. If a joint has no external loads, or reactions applied to it, and it has three members, two of which are co-linear, then the non co-linear member is equal to zero.[1]
  5. Determine the slope, or angles of the members.[1]
  6. Select a joint that has at most two unknown forces. [1]
  7. Unknown forces shall be assumed in Tension ( if the result is a negative number, then assumption was wrong and that member is in Compression). [2]
  8. Calculate the unknown forces by solving ($ \sum Fx=0 $ , $ \sum Fy=0 $ ). [2]
  9. Move to another joint that has two or less unknowns. [2]
  10. Continue step 6. to 9. until all of the members in the structure are solved.




Example

Figure 1.0: Truss ABCD


Check Determinacy

$ m+r=2j+e $

Where,
m = number of members in the structure = 5
r = number of reactions in the structure = 3
j = number of joints in the structure = 4
e = number of release conditions (such as a hinge in the middle of a member) = 0;

$ 5+3=2(4)+0 $
$ 8=8 $


Therefore this truss is determinate and stable




Find the Zero Force Members

since $ F_{AC} $ and $ F_{CD} $ are Collinear, $ F_{CB} $ is not, and there are no external loads applied to Joint C, by the rule of Zero Force Members:
$ F_{CB}=0 $
Figure 2.0: Zero Force Member of Joint C




Determine the Angles

$ \alpha=\tan^{-1}(\frac{2m}{2m})=45^{o} $
$ \sin45^{o}=\cos45^{o}=0.707 $


Figure 3.0: Reactions of Truss ABCD




Determine the Reactions

Assume that the positive axis are vertically up and horizontally right


$ \sum M_{b}=0 $
$ 50kN(4m)-A_{y}(4m) = 0 $
$ A_{y} = 50 kN $
$ \sum F_{x}=0 $
$ -50kN +B_{x} = 0 $
$ B_{x}=50 kN $
$ \sum F_{y}=0 $
$ -75 kN+50kN +B_{y} = 0 $
$ B_{y}=75-50=25 kN $






Start with Joint A

Figure 4.0: Solved Reactions of Joint A

Usually, all members are assumed to be in Tension (with the orientation arrows pointing away from the joint). After solving, if the answer turns out to be a negative number, then our initial assumption is false, and the member is in compression (with the orientation arrows pointing towards the joint). Upon solving for all the unknown internal reactions, and orienting the directions of the forces, the joint diagram will resemble that of Figure 4.0 (right).


$ \sum F_{y}=0 $
$ 50kN+ F_{AC}\sin45^{o}=0 $
$ 50 kN+F_{AC}*(0.707)= 0 $
$ F_{AC}= - 70.71 kN $ (Since negative, the member is in compression.)
$ \sum F_{x}=0 $
$ F_{AB} + F_{AC}\cos45^{o}=0 $
$ F_{AB} + (-70.71kN)*(0.707)= 0 $
$ F_{AB} = 50 kN $ (Positive, so member is in tension)

Note: After drawing the initial diagram with all the unknowns forces assumed to be in tension, no corrections are made to the drawing until another joint is to be analyzed. For example, $ F_A_C $ is found to be in compression when analyzing the forces in the Y-axis, but the direction of it was not changed while analyzing the X-axis. $ F_A_C $ was still assumed to be in Tension, since we hadn't finished analyzing the entire joint.





Continue to Joint B

Figure 5.0: Solved Reactions of Joint B
$ \sum F_{x}=0 $
$ 50kN- 50kN=0 $ (Tension)
$ \sum F_{y}=0 $
$ 25kN+ F_{BD}=0 $
$ F_{BD}= - 25 kN $ (Compression)



Continue to Joint C

$ \sum F_{O'}=0 $ (Let O' be an imaginary axis that is along the both members CA and CD)
$ F_{CA} + F_{CD}=0 $
$ 70.71 kN + F_{CD}=0 $
$ F_{CD}= -70.71 $ (Compression)

Note: Referring back to figure 2.0, $ F_A_C $ was found previously to be 70.71 kN in compression. When drawing $ F_A_C $ on Joint C, the member must be drawn in compression. Also, note that when analyzing joint C, $ F_A_C $ was taken as a positive value since it's direction was already changed when it was noted in compression, instead of tension.





Finally Solve for Joint D

Figure 6.0: Solved Reactions of Joint D

Use this as a "Check"


$ \sum F_{x}=0 $
$ -50kN+70.71\sin(45)kN=0 $
$ -50kN+ 70.71(0.7071)kN=0 $
$ -50kN+ 50kN=0 $
$ \sum F_{y}=0 $
$ 25kN+ 70.71\cos(45)-75kN=0 $
$ 25kN+ 70.71(0.7071)-75kN=0 $
$ 25kN+50kN-75kN=0 $




Final Diagram

Re-Assemble all of the joints to complete the original truss, with the internal forces noted in the correct direction.


Figure 7.0: Fully Solved Truss Using Method of Joints


$ F_{AC}=F_{CD}=70.71 kN (C), F_{BC}=0, F_{AB}=50 kN(T),F_{BD}=25 kN (C) $




References

  1. 1.0 1.1 1.2 1.3 1.4 1.5 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
  2. 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 Hibbeler, R. C. (2009). Engineering Mechanics (Statics): (12th ed.). London, UK: Pearson.