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Homework1

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mohamed Eraky
mohamed Eraky

This document contains the solutions to three problems from a homework assignment on aircraft design. Problem 1 uses MATLAB code to derive equations for stresses as functions of strains in an isotropic material. Problem 2 calculates the forces and deflections in a beam under two concentrated loads. Problem 3 determines the force and deflection at a joint between two beams with different thermal expansion coefficients.

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Fourth Year Aircraft Design Report: Solution to Homework1 Report No: 1 Date: 18/3/2013 Submitted to: Dr. Mohammad Tawfik Name  Mohammad Tawfik Eraky ‫عراقي‬ ‫أحمد‬ ‫توفيق‬ ‫محمد‬ 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 − 𝜖 𝑥 + 𝜐(𝛼. 𝑇 + 𝜖 𝑥 − 𝜖 𝑦 − 𝜖 𝑧) (2𝜐 − 1)(𝜐 + 1) 𝛼. 𝑇 − 𝜖 𝑦 + 𝜐(𝛼. 𝑇 + 𝜖 𝑦 − 𝜖 𝑥 − 𝜖 𝑧) (2𝜐 − 1)(𝜐 + 1) 𝛼. 𝑇 − 𝜖 𝑧 + 𝜐(𝛼. 𝑇 + 𝜖 𝑧 − 𝜖 𝑥 − 𝜖 𝑦) (2𝜐 − 1)(𝜐 + 1) ]
Pb.2 Static equilibrium ∑𝐹𝑋 = 0 → 𝐹𝐴 + 𝐹𝐵 = 0 𝛿 𝑐1 + 𝛿 𝑐2 = 0 → 𝐼 𝛿 𝑐1 = −𝜎.𝑙 𝐸 = −𝑃.𝐿 𝐸𝐴 = −16∗300 200∗ 𝜋 4 ∗152 = −0.135 (𝑚𝑚) ⟶ 1 𝛿 𝑐2 = 𝜎.𝑙 𝐸 = 𝐹 𝐶.𝑙 𝐸𝐴 = 𝐹 𝐶∗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) 𝛿 𝐵1 = −0.135 𝑚𝑚 𝛿 𝐵2 = 4.8 ∗ 300 200 ∗ 𝜋 4 ∗ 152 = 0.04 𝑚𝑚 𝛿 𝐵 = 𝛿 𝐵1 + 𝛿 𝐵2 𝛿 𝐵 = 0.094 𝑚𝑚
Pb.3 𝛿 𝑇 + 𝛿 𝑅 = 0 → 𝐼 𝛿𝑡 = 𝛼 𝑆. 𝑙 𝑆. 𝑇𝑆 + 𝛼 𝐵. 𝑙 𝐵. 𝑇𝐵 𝛿𝑡 = 12 ∗ 10−6 ∗ 0.3 ∗ 10 + 21 ∗ 10−6 ∗ 0.3 ∗ 10 → 1 𝛿𝑡 = 0.099 (𝑚𝑚) 𝛿 𝑅 = ∑ −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

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Homework1

  • 1. Fourth Year Aircraft Design Report: Solution to Homework1 Report No: 1 Date: 18/3/2013 Submitted to: Dr. Mohammad Tawfik Name  Mohammad Tawfik Eraky ‫عراقي‬ ‫أحمد‬ ‫توفيق‬ ‫محمد‬ 2013/2014
  • 2. 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 − 𝜖 𝑥 + 𝜐(𝛼. 𝑇 + 𝜖 𝑥 − 𝜖 𝑦 − 𝜖 𝑧) (2𝜐 − 1)(𝜐 + 1) 𝛼. 𝑇 − 𝜖 𝑦 + 𝜐(𝛼. 𝑇 + 𝜖 𝑦 − 𝜖 𝑥 − 𝜖 𝑧) (2𝜐 − 1)(𝜐 + 1) 𝛼. 𝑇 − 𝜖 𝑧 + 𝜐(𝛼. 𝑇 + 𝜖 𝑧 − 𝜖 𝑥 − 𝜖 𝑦) (2𝜐 − 1)(𝜐 + 1) ]
  • 3. Pb.2 Static equilibrium ∑𝐹𝑋 = 0 → 𝐹𝐴 + 𝐹𝐵 = 0 𝛿 𝑐1 + 𝛿 𝑐2 = 0 → 𝐼 𝛿 𝑐1 = −𝜎.𝑙 𝐸 = −𝑃.𝐿 𝐸𝐴 = −16∗300 200∗ 𝜋 4 ∗152 = −0.135 (𝑚𝑚) ⟶ 1 𝛿 𝑐2 = 𝜎.𝑙 𝐸 = 𝐹 𝐶.𝑙 𝐸𝐴 = 𝐹 𝐶∗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) 𝛿 𝐵1 = −0.135 𝑚𝑚 𝛿 𝐵2 = 4.8 ∗ 300 200 ∗ 𝜋 4 ∗ 152 = 0.04 𝑚𝑚 𝛿 𝐵 = 𝛿 𝐵1 + 𝛿 𝐵2 𝛿 𝐵 = 0.094 𝑚𝑚
  • 4. Pb.3 𝛿 𝑇 + 𝛿 𝑅 = 0 → 𝐼 𝛿𝑡 = 𝛼 𝑆. 𝑙 𝑆. 𝑇𝑆 + 𝛼 𝐵. 𝑙 𝐵. 𝑇𝐵 𝛿𝑡 = 12 ∗ 10−6 ∗ 0.3 ∗ 10 + 21 ∗ 10−6 ∗ 0.3 ∗ 10 → 1 𝛿𝑡 = 0.099 (𝑚𝑚) 𝛿 𝑅 = ∑ −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