See the theoretical introduction to the Cremona method!
See STATEMENTS ABOUT ZERO MEMBERS - definition with examples!

See the solution for this truss using the method of joint equilibrium.
See the solution for this truss using Ritter's method.

We will start with a summary of forces in the members calculated using the method of joint equilibrium, so we know what results to expect.
The graphical method of Cremona often serves as a check for analytical calculations done by other methods.

Summary of forces in the members [kN]

\begin{aligned} \begin{array}{llll} \mathrm{N}_1=20 \ \mathrm{kN} & \mathrm{N}_6=20 \ \mathrm{kN} & \mathrm{N}_{11}=0 \ \mathrm{kN} & \mathrm{N}_{16}=0 \ \mathrm{kN} \\ \mathrm{N}_2=60 \ \mathrm{kN} & \mathrm{N}_7=-28.284 \ \mathrm{kN} & \mathrm{N}_{12}=-21.213 \ \mathrm{kN} & \mathrm{N}_{17}=-20 \ \mathrm{kN} \\ \mathrm{N}_3=60 \ \mathrm{kN} & \mathrm{N}_8=40 \ \mathrm{kN} & \mathrm{N}_{13}=15 \ \mathrm{kN} & \mathrm{N}_{18}=-60 \ \mathrm{kN} \\ \mathrm{N}_4=60 \ \mathrm{kN} & \mathrm{N}_9=-56.569 \ \mathrm{kN} & \mathrm{N}_{14}=-63.64 \ \mathrm{kN} & \mathrm{N}_{19}=-45 \ \mathrm{kN} \\ \mathrm{N}_5=45 \ \mathrm{kN} & \mathrm{N}_{10}=0 \ \mathrm{kN} & \mathrm{N}_{15}=0 \ \mathrm{kN} & \mathrm{N}_{20}=0 \ \mathrm{kN} \end{array} \end{aligned}

Cremona method

To begin the solution using the Cremona method, we must analytically calculate the support reactions:

We label the joints, number the members, and proceed to calculations

We notice that members number 10, 11, 15, 20 are zero
If we do not notice this, nothing will happen,
but it is always a check of our calculations.
The angle of inclination of the diagonal members: \( \alpha=45^{\circ}

We cut the truss through members 3 and 18 and calculate the reactions separately for the left and right sides (the horizontal reaction is calculated before we cut the system).

Reactions

\[ \Sigma \mathrm{x}=0 \quad \mathrm{H}_{\mathrm{IX}}=0 \\ \]

right side:

\begin{array}{ll} \Sigma \mathrm{M}_{\mathrm{XI}}=0 & 15 \cdot 3+45 \cdot 3-\mathrm{N}_3 \cdot 3=0 \quad \mathrm{~N}_3=\frac{15 \cdot 3+45 \cdot 3}{3}=60 \cdot \mathrm{kN} \\ \Sigma \mathrm{x}=0 & \mathrm{~N}_{18}=-\mathrm{N}_3=-60 \cdot \mathrm{kN} \\ \Sigma \mathrm{y}=0 & 15-45+\mathrm{V}_{\mathrm{XI}}=0 \quad \mathrm{~V}_{\mathrm{XI}}=30 \mathrm{kN} \\ \end{array}

left side:

\begin{array}{lll} \Sigma \mathrm{M}_{\mathrm{IX}}=0 & \mathrm{~N}_3 \cdot 3-20 \cdot 3+\mathrm{V}_{\mathrm{VII}} \cdot 6=0 & \mathrm{~V}_{\mathrm{VII}}=\frac{\mathrm{N}_3 \cdot 3-20 \cdot 3}{-6}=-20 \cdot \mathrm{kN} \\ \Sigma \mathrm{y}=0 & \mathrm{~V}_{\mathrm{VII}}-20+\mathrm{V}_{\mathrm{IX}}=0 & \mathrm{~V}_{\mathrm{IX}}=20-\mathrm{V}_{\mathrm{VII}}=40 \cdot \mathrm{kN} \end{array}