Fourth Year

Aircraft Design

Report: Solution to Homework1

Report No: 1 Date: 18/3/2013

Submitted to: Dr. Mohammad Tawfik

Name

Mohammad Tawfik Eraky

محمد توفيق أحمد عراقي

2013/2014

Pb.1

Using MATLAB code to get the stresses as functions of strains

Script:

clc;clear;close all; syms v alpha_t eps_x eps_y eps_z I A=[1 -v -v; -v 1 -v ;-v -v 1 ]; I=inv(A); Sigma=I*[eps_x-alpha_t ;eps_y-alpha_t;eps_z-alpha_t]; disp(Sigma);

[

𝜎𝑥

𝜎𝑦

𝜎𝑧

] = 𝐸.

[ 𝛼. 𝑇 − 𝜖𝑥 +

𝜐(𝛼. 𝑇 + 𝜖𝑥 − 𝜖𝑦 − 𝜖𝑧)

(2𝜐 − 1)(𝜐 + 1)

𝛼. 𝑇 − 𝜖𝑦 +𝜐(𝛼. 𝑇 + 𝜖𝑦 − 𝜖𝑥 − 𝜖𝑧)

(2𝜐 − 1)(𝜐 + 1)

𝛼. 𝑇 − 𝜖𝑧 +𝜐(𝛼. 𝑇 + 𝜖𝑧 − 𝜖𝑥 − 𝜖𝑦)

(2𝜐 − 1)(𝜐 + 1) ]

Pb.2

Static equilibrium ∑𝐹𝑋 = 0 → 𝐹𝐴 + 𝐹𝐵 = 0

𝛿𝑐1 + 𝛿𝑐2 = 0 → 𝐼

𝛿𝑐1 =−𝜎.𝑙

𝐸=

−𝑃.𝐿

𝐸𝐴=

−16∗300

200∗𝜋

4∗152

= −0.135 (𝑚𝑚) ⟶ 1

𝛿𝑐2 =𝜎.𝑙

𝐸=

𝐹𝐶.𝑙

𝐸𝐴=

𝐹𝐶∗1000

200∗𝜋

4∗152

→ 2

By equaling first and second equation we got

𝐹𝐶 = 4.8 𝐾𝑁 , 𝐹𝐴 = 11.2 𝐾𝑁

Deflection at point (B)

𝛿𝐵1 = −0.135 𝑚𝑚

𝛿𝐵2 =4.8 ∗ 300

200 ∗𝜋4

∗ 152= 0.04 𝑚𝑚

𝛿𝐵 = 𝛿𝐵1 + 𝛿𝐵2

𝛿𝐵 = 0.094 𝑚𝑚

Pb.3

𝛿𝑇 + 𝛿𝑅 = 0 → 𝐼

𝛿𝑡 = 𝛼𝑆. 𝑙𝑆. 𝑇𝑆 + 𝛼𝐵. 𝑙𝐵. 𝑇𝐵

𝛿𝑡 = 12 ∗ 10−6 ∗ 0.3 ∗ 10 + 21 ∗ 10−6 ∗ 0.3 ∗ 10 → 1

𝛿𝑡 = 0.099 (𝑚𝑚)

𝛿𝑅 = ∑−𝐹. 𝑙

𝐸. 𝐴=

−𝐹 ∗ 0.3

200 ∗ 10−6 ∗ 200 ∗ 109+

−𝐹 ∗ 0.3

200 ∗ 10−6 ∗ 200 ∗ 109→ 2

Substituting equations 1 , 2 into equation I

F = 6.988 KN

Deflection at joint

𝛿𝐽 = 21 ∗ 10−6 ∗ 10 ∗ 0.3 −7 ∗ 103 ∗ 0.3

200 ∗ 10−6 ∗ 200 ∗ 109

𝛿𝐽 = 0.01 𝑚𝑚