Trusses

External forces are only applied to the internal hinges (nodes).

In each rod, the only internal action is the normal load $N$

• If $N>0$, then the beam is loaded by traction
• If $N<0$, then the beam is loaded by compression

Simplified degree of static determinacy

$$h=\text{Number of Rods}+\text{Number of External Reactions}-2\cdot \text{Number of Internal Hunges}$$

Solution Strategy

1. Compute $h$ (Check if isostatic).
2. External Reactions: Treat the whole truss as a single rigid body first. Ignore the insides. Solve for the ground supports ($H_A,V_A,etc.$).
3. Internal Forces ($N$):
   • Method of Nodes: Solves every rod (good for full analysis).
   • Method of Sections (Ritter): Solves specific rods (good for finding just one or two forces).

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Method 1: Equilibrium at Nodes

The Logic

Isolate a specific node (hinge) and treat it as a particle. Since the node is in equilibrium, the sum of forces acting on it must be zero.

Procedure:

1. Isolate a Node: Start with a node where you have at least 1 known force and no more than 2 unknown rods (since you only have 2 equations: $\sum F_x,\sum F_y$).
2. Draw Vectors:
   • Draw known external forces.
   • Draw unknown rod forces ($N$) pointing AWAY from the node.
   • Note: We always assume Traction (+) initially.
3. Solve:
   • $\sum F_x=0$
   • $\sum F_y=0$
4. Interpret signs:
   • Result $>0$: Assumption correct (Traction).
   • Result $<0$: Rod is in Compression.
5. Repeat: Move to the next connected node and use the values you just found.

Spotting "Zero Force" Members

Sometimes you can spot rods with $N=0$ just by looking.

• Example: Look at Node B. A vertical force $F$ balances the vertical rod ($N_3$). However, nothing opposes the horizontal rod ($N_2$). Therefore, to maintain equilibrium, $N_2$ must be zero.

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Method 2: Method of Sections (Ritter)

The Logic

Instead of solving node-by-node, you “cut” the truss into two separate rigid bodies. The internal forces of the cut rods become external forces on the new sections.

The Rules of the Cut:

1. Cut Max 3 Rods: You only have 3 equilibrium equations ($\sum F_x,\sum F_y,\sum M$), so you cannot solve for more than 3 unknowns.
2. No Common Node: The three cut rods cannot all meet at the same node (otherwise the moment equation becomes useless).

The “Ritter” Strategy: To find the force in a specific rod (e.g., Rod A) without doing extra maths:

1. Identify the other two rods you cut (Rods B and C).
2. Find their intersection point (Point P).
3. Sum Moments about Point P ($\sum M_P=0$).
   • Since Rods B and C pass through P, their moment is zero.
   • You are left with an equation containing only Rod A.