Question

Find the forces in all the members of the truss shown in Fig. (a). Tabulate the results.

Answer 1

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 [Ref. Fig.(b)] ∑ = 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° = 40√2 x 1/√2 = 40 kN

Note: If the assumed direction of 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 analysed to indicate the forces on the joint. At the other end, mark the arrows in the reverse direction. 

In the present case, near the joint C, the arrows are marked on the members 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 analyse that joint. 

In this case, there are only two unknown forces at the joint D as shown in Fig.(c).

∑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 Fig. (d).

∑V = 0, gives

F_BE sin 45° – 40 – 40√2 × sin 45° = 0

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 [Ref. Fig. (e)]. In this case,