FE Exam Review 65: Structures 2 Vertical Deflection at Joint

FE Exam Review – Vertical Displacement at Joint C of a Truss Using the Unit Load Method

In this FE Civil Exam problem, we calculate the vertical displacement at Joint C of a steel truss when a 40-kip downward load is applied. This is one of the most common Mechanics of Materials / Structural Analysis problems on the FE Civil Exam.

You are given:

Member forces due to the 40-kip load

Member lengths

Elastic modulus E = 29,000 ksi

Area A = 4 in²

We compute:

ΔL for each member using FL / AE

f-values by applying a 1-kip upward unit load at Joint C

Contribution of each member using (f × ΔL)

Total vertical deflection at Joint C

This is a classic FE Civil problem—if you master the Unit Load Method, you’ll score easy points on the exam.

Final Answer: 4.19 inches downward

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⏱ TIMESTAMPS (Spread Across 6:30 Minutes)

00:00 – Problem Introduction
00:26 – What the FE Exam Is Actually Testing
00:48 – Overview of the Truss and Given Values
01:10 – When to Use the Unit Load Method
01:35 – Reviewing Member Forces from 40-kip Load
01:58 – Geometry and Member Lengths
02:20 – Material Properties (A and E)
02:42 – Formula Review for ΔL = FL/AE
03:05 – Calculating Deformations for All Members
03:32 – Understanding f-Values from Unit Load Case
03:58 – Applying the Formula Δ = Σ(f × ΔL)
04:22 – Computing Contributions Member by Member
04:48 – Summing All fΔL Terms
05:12 – Understanding the Sign Convention (Tension vs Compression)
05:35 – Determining Final Vertical Deflection
05:58 – Why the Deflection Is Downward
06:15 – FE Exam Tips for Unit Load Problems
06:30 – Video Wrap-Up and Next Topic

🧩 FE Style Written Solution (Copy-Friendly)

Required: Find the downward vertical displacement at Joint C when a 40-kip load acts downward at C.

Given Member Forces (40-kip load):
AB = +100 kips
BC = +98.6 kips
CD = –150 kips
AD = –60 kips
BD = –50 kips

Member Lengths (in inches):
AB = 240
BC = 473
CD = 480
AD = 288
BD = 240

A = 4 in²
E = 29,000 ksi
Method: Unit Load Method

Step 1 — Compute ΔL = FL/AE

AB: +0.2068 in
BC: +0.4016 in
CD: –0.6206 in
AD: –0.1490 in
BD: –0.1034 in

Step 2 — Multiply by f-values (unit load case)

AB: 2.500 × 0.2068 = 0.517
BC: 2.462 × 0.4016 = 0.989
CD: (–3.750) × (–0.6206) = 2.327
AD: (–1.500) × (–0.1490) = 0.224
BD: (–1.230) × (–0.1034) = 0.127

Step 3 — Sum All Contributions

Δ = 4.187 in ≈ 4.19 in

Step 4 — Direction

Load is downward → displacement is downward.

Final Answer: 4.19 in downward

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FE Civil Practice Exam (NCEES): https://amzn.to/2Ue5YgK

FE Civil Reference Handbook (FREE via NCEES login)
FE Civil Practice – PPI: https://amzn.to/2UOPCiV

FE Civil Practice Problems – PPI: https://amzn.to/2X1BY9R

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