---
title: "Truss ideal assumptions and member forces"
subject: "Applied Mechanics"
chapter: 7
key: "semester_1/applied_mechanics:truss-assumptions-member-forces-40kn-30kn"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-assumptions-member-forces-40kn-30kn
site: BCT Engineering
---

# Truss ideal assumptions and member forces

Applied Mechanics · Chapter 7: Analysis of Plane Trusses

- Asked in: 2073 Shrawan, 2+6 marks; 2072 Kartik, 2+8 marks (asked 2 times)
- Video: [Truss Analysis- Assumptions and Procedure](https://www.youtube.com/watch?v=SXxhBBkXpZs) — Mechanical Engineering with Dr. Sanei
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/truss-assumptions-member-forces-40kn-30kn
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-7

## Question

Write down the ideal assumptions of Truss. Calculate the member forces in all members of the truss loaded as shown in the figure below by using a suitable method. [Figure: truss with pin support at A (left) and roller support at C (right); horizontal span 10 m + 10 m; apex B with 40 kN downward and 30 kN horizontal load; heights 3 m and 3 m marked at the right; internal joint D.]

## Short answer

Truss assumptions: straight members, frictionless pins, loads at joints, negligible self-weight, two-force members, rigid planar geometry. Using method of joints, reactions: Ax=30 kN left, Ay=11 kN up, Cy=29 kN up. Member forces: AB 77.75 kN compression, BC 112.73 kN compression, AD 100.92 kN tension, DC 100.92 kN tension, BD 58.00 kN tension.

## Answer

#### Ideal assumptions of a truss
1. All members are straight and joined at their ends by smooth (frictionless) pins.
2. Loads and support reactions act only at the joints.
3. The self-weight of members is negligible (or is lumped at the joints).
4. Every member is a two-force member, so it carries only axial force (tension or compression), with no bending or shear.
5. Members are rigid enough that the truss geometry does not change appreciably under load, and all members lie in one plane.

#### Data
Take A(0,0) pin, C(20,0) roller, apex B(10,6) and internal joint D(10,3) (the 3 m + 3 m heights in the figure). Members: AB, BC, AD, DC, BD (m = 5, j = 4, r = 3, so m + r = 2j and the truss is determinate and stable). Loads at B: 40 kN ↓ and 30 kN →. Lengths: AB = BC = √(10² + 6²) = 11.66 m, AD = DC = √(10² + 3²) = 10.44 m, BD = 3 m.

```text
          30kN→ B  (40 kN ↓)
              / | \
            /   |   \
          /     D     \
        /  .    |   .    \
      A(pin) ------------- C(roller)
```

#### Support reactions
- ΣM_A = 0: C_y(20) − 40(10) − 30(6) = 0, so **C_y = 29 kN ↑**.
- ΣFy = 0: A_y = 40 − 29 = **11 kN ↑**.
- ΣFx = 0: A_x + 30 = 0, so **A_x = 30 kN ←**.

#### Method of joints (tension +, compression −)
**Joint C** (cos = 10/11.66 = 0.8575, sin = 6/11.66 = 0.5145 for CB; 0.9578 and 0.2874 for CD):
- ΣFy: 29 + 0.5145F_CB + 0.2874F_CD = 0.
- ΣFx: −0.8575F_CB − 0.9578F_CD = 0, so F_CD = −0.8953F_CB.
- Substituting: 29 + 0.5145F_CB − 0.2573F_CB = 0, so **F_CB = −112.73 kN (C)** and **F_CD = +100.92 kN (T)**.

**Joint D** (members DC, DA, DB; no load):
- ΣFx: 0.9578(100.92) − 0.9578F_DA = 0, so **F_DA = 100.92 kN (T)**.
- ΣFy: −0.2874(100.92) − 0.2874(100.92) + F_DB = 0, so **F_DB = 58.00 kN (T)**.

**Joint A**: AD pulls A with (96.67, 29.00); AB gives (0.8575, 0.5145)F_AB. ΣFx: −30 + 96.67 + 0.8575F_AB = 0, so **F_AB = −77.75 kN (C)**. Check ΣFy: 11 + 29.00 − 0.5145(77.75) = 11 + 29.00 − 40.00 = 0 ✓.

#### Result
| Member | Force (kN) | Nature |
|---|---|---|
| AB | 77.75 | compression |
| BC | 112.73 | compression |
| AD | 100.92 | tension |
| DC | 100.92 | tension |
| BD | 58.00 | tension |

Joint B closes the check: ΣFy = −40 + 0.5145(77.75) + 0.5145(112.73) − 58 = −40 + 40 + 58 − 58 = 0 ✓, and ΣFx = 30 + 0.8575(77.75) − 0.8575(112.73) = 30 + 66.67 − 96.67 = 0 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Under the ideal truss assumptions, each member carries:
   A. Bending and shear
   B. Torsion only
   C. Only axial force, because loads act only at the joints and joints are smooth pins
   D. Axial force plus bending from its self-weight
   Answer: C. With frictionless pins and joint loads only, every member is a two-force member carrying only tension or compression. (Key point: Truss assumptions)

2. Why is this truss determinate?
   A. m = 5, j = 4, r = 3, so m + r = 2j
   B. It has only three joints
   C. It has a roller at C
   D. It has no loads at D
   Answer: A. m + r = 5 + 3 = 8 = 2j, so the number of unknowns equals the number of equations. (Key point: Determinacy check)

3. How is Cy = 29 kN found?
   A. Sum of horizontal forces
   B. Moments about A: 20Cy = 40(10) + 30(6)
   C. Moments about C ignoring the 30 kN load
   D. Joint method at B
   Answer: B. Moments about A include both the 40 kN vertical load (10 m) and 30 kN horizontal load (6 m height), giving Cy = 580/20 = 29 kN. (Key point: Vertical reaction at C)

4. What is the force in member BC?
   A. 112.73 kN tension
   B. 58 kN tension
   C. 100.92 kN compression
   D. 112.73 kN compression
   Answer: D. From joint C the equations give F_CB = -112.73 kN, a negative value meaning compression. (Key point: Nature of BC)

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