Activity 2.1 6 Step By Step Truss System Answers

Activity 2.1 – 6‑Step‑by‑Step Truss System Answers

Designing and analysing a truss is a classic exercise in statics that helps students grasp how forces travel through structures. Activity 2.1 in many engineering curricula asks you to solve a truss problem by following a six‑step procedure. Below is a complete, step‑by‑step guide that not only provides the correct answers but also explains the reasoning behind each move, so you can apply the method to any similar truss later on.

Introduction

A truss is a framework of straight members connected at joints, usually assumed to be pin‑connected and to carry only axial forces (tension or compression). In Activity 2.1 you are typically given:

1. A planar truss diagram (often a simple Pratt or Warren configuration).
2. External loads (point loads, distributed loads, or reactions).
3. Support conditions (usually a pinned support and a roller).

The objective is to determine the internal force in each member and identify whether it is in tension or compression. The six‑step method is a systematic approach that eliminates guesswork and ensures every equilibrium condition is satisfied No workaround needed..

Step 1 – Draw a Clear Free‑Body Diagram (FBD)

What to do:

• Sketch the entire truss as a single free‑body.
• Indicate all external loads, including their magnitudes and directions.
• Mark the support reactions: a vertical and horizontal reaction at the pinned support (usually A) and a vertical reaction at the roller (usually B).

Why it matters:

The FBD provides the foundation for the equilibrium equations that will be used in Step 2. Any missing load or reaction will produce incorrect member forces later on Worth keeping that in mind..

Common pitfalls:

• Forgetting the horizontal reaction at the pinned support.
• Misreading the direction of a load (e.g., a downward load shown as upward).

Step 2 – Calculate the Support Reactions

Because the truss is a statically determinate structure, the three global equilibrium equations are sufficient:

[ \sum F_x = 0,\qquad \sum F_y = 0,\qquad \sum M = 0 ]

Procedure:

1. Sum of horizontal forces – set the horizontal reaction (A_x) equal to any horizontal external loads (often zero).
2. Sum of vertical forces – solve for the vertical reactions (A_y) and (B_y).
3. Sum of moments – take moments about either support to eliminate one unknown reaction and solve for the other.

Example calculation (typical data):

• Load (P = 10;{\rm kN}) acts downward at joint C, 4 m from support A.
• Span (AB = 8;{\rm m}).

[ \sum M_A = 0 \Rightarrow B_y \times 8;{\rm m} - P \times 4;{\rm m}=0 \Longrightarrow B_y = \frac{P \times 4}{8}=5;{\rm kN} ]

[ \sum F_y = 0 \Rightarrow A_y + B_y - P = 0 \Longrightarrow A_y = P - B_y = 5;{\rm kN} ]

[ \sum F_x = 0 \Rightarrow A_x = 0;{\rm kN} ]

These values become the known forces for the next steps Worth keeping that in mind..

Step 3 – Identify Zero‑Force Members

Zero‑force members can be eliminated early, reducing the amount of calculation. Use the two classic rules:

1. Two‑member joint with no external load or support reaction – both members are zero‑force.
2. Three‑member joint with only two members collinear and no external load – the non‑collinear member is zero‑force.

Application to Activity 2.1:

• Joint D has only members DC and DG meeting, no external load → DC and DG are zero‑force.
• Joint E is a three‑member joint (members EA, EB, EF) with EA and EB collinear and no load → EF is zero‑force.

Mark these members on the diagram; they will be omitted from subsequent calculations.

Step 4 – Choose a Method of Sections

Two common techniques are used to find member forces:

• Method of Joints – solve each joint sequentially, using the fact that the sum of forces in the x‑ and y‑directions must be zero.
• Method of Sections – cut through the truss, isolate a portion, and apply equilibrium equations to that segment.

Why the method of sections is often preferred for Activity 2.1:

• The problem usually asks for the force in a specific member (e.g., the diagonal BC).
• Cutting through a maximum of three members allows you to solve directly for the unknowns using only (\sum F_x = 0), (\sum F_y = 0), and (\sum M = 0).

How to cut:

1. Draw a straight line that passes through no more than three members whose forces are unknown.
2. Separate the truss into left and right sections; keep the section containing the known reactions for easier moment calculations.

Example cut:

Cut through members AB, BC, and CD to find the force in BC Most people skip this — try not to..

Step 5 – Apply Equilibrium to the Chosen Section

With the cut defined, isolate the left side of the truss (including support A). Write the three equilibrium equations for that segment The details matter here..

1. Sum of moments about a convenient point – choose a point that eliminates two of the three unknown member forces. In our example, taking moments about joint A removes the unknown forces in AB and AD, leaving only BC.

[ \sum M_A = 0 \Rightarrow (F_{BC}) \times (distance;from;A;to;BC) - (B_y) \times (horizontal;distance;AB) = 0 ]

2. Solve for the remaining unknown – rearrange to isolate (F_{BC}).

[ F_{BC} = \frac{B_y \times \text{AB}}{\text{lever arm of }BC} ]

Using the numbers from Step 2 ( (B_y = 5;{\rm kN}) ), a typical geometry (AB = 4 m, lever arm = 3 m), we get

[ F_{BC}= \frac{5;{\rm kN}\times 4;{\rm m}}{3;{\rm m}} = 6.67;{\rm kN} ]

3. Determine the nature of the force – if the calculated value is positive when assuming tension, the member is indeed in tension; if negative, it is in compression. In the example, the positive result indicates BC is in tension Practical, not theoretical..

4. Complete the analysis – use (\sum F_x = 0) and (\sum F_y = 0) on the same section to find the remaining two unknown member forces (AB and CD) It's one of those things that adds up..

Sample results:

• (F_{AB}= -5.0;{\rm kN}) → compression
• (F_{CD}= 2.5;{\rm kN}) → tension

Step 6 – Verify Results and Summarize Findings

The final step is a sanity check:

• Equilibrium check – make sure at every joint, the vector sum of member forces equals zero.
• Consistency with zero‑force members – confirm that the members previously identified as zero‑force indeed have zero axial force in the calculations.
• Physical intuition – compression members are usually the top chords in a simply supported truss, while tension members appear in the bottom chords and diagonals under downward loads.

Summary table for Activity 2.1

Some disagree here. Fair enough.

(Only the most relevant members are shown; all other members can be filled in using the same procedure.)

Scientific Explanation Behind the Truss Behavior

Axial Load Distribution

In a pin‑joined truss, each member experiences only axial forces because the joints cannot resist moments. This simplification allows the use of statics alone to predict internal forces. The principle of superposition also applies: the response to multiple loads equals the sum of the responses to each load applied individually But it adds up..

Material Considerations

The calculated axial forces are used to check against allowable stress limits:

[ \sigma = \frac{F}{A} ]

where ( \sigma ) is stress, ( F ) is axial force, and ( A ) is the cross‑sectional area. For steel trusses, a common allowable stress is about 250 MPa. If a member’s stress exceeds this value, the designer must increase the area or select a stronger material.

Easier said than done, but still worth knowing.

Stability and Determinacy

A planar truss is statically determinate when

[ m = 2j - 3 ]

where ( m ) is the number of members and ( j ) the number of joints. Practically speaking, activity 2. 1 typically satisfies this condition, guaranteeing that the three global equilibrium equations are sufficient to solve for all reactions and internal forces That's the whole idea..

Frequently Asked Questions (FAQ)

Q1: What if the truss has more than three unknown members in the cut section?
A: The method of sections is limited to three unknowns because only three equilibrium equations exist in a plane. Redraw the cut to pass through a different set of members, or revert to the method of joints for a systematic solution.

Q2: Can I assume all members are either tension or compression without calculation?
A: Not safely. While intuition (top chords compress, bottom chords tension) helps, complex loading or geometry can reverse the expected nature. Always verify with equilibrium equations And that's really what it comes down to. That alone is useful..

Q3: How do I handle distributed loads on a truss?
A: Convert the distributed load into an equivalent point load acting at the centroid of the distribution, then treat it as a standard external load in the FBD.

Q4: Why are zero‑force members useful?
A: Identifying them early reduces the number of equations you must solve, saves time, and prevents unnecessary calculations that could introduce rounding errors Not complicated — just consistent..

Q5: What software can I use to check my hand calculations?
A: Free tools like Free Body Diagram apps, or professional packages such as SAP2000, STAAD.Pro, and RISA‑3D can model trusses and output member forces for verification Simple, but easy to overlook..

Conclusion

Activity 2.1’s six‑step truss analysis provides a clear roadmap from a raw diagram to a complete set of member forces. By drawing an accurate free‑body diagram, calculating support reactions, identifying zero‑force members, selecting an appropriate method of sections, applying equilibrium equations, and finally verifying the results, you develop both a deep conceptual understanding and a reliable computational technique It's one of those things that adds up..

Mastering this process equips you to tackle more advanced structures—such as three‑dimensional space frames or trusses with redundant supports—while retaining the confidence that every force balance is satisfied. Keep the steps handy, practice with different truss configurations, and soon the six‑step method will become second nature, enabling you to solve real‑world engineering challenges efficiently and accurately.