Question

USING MATLAB 1) Before the internet, engineers used tables to find coefficients for materials such as steel. Structural Engineers must know the bending stiffness (H) for materials such as steel to design and construct an efficient system of support. There can be no bending or torsion in a building or else it will eventually fail and collapse. The equation below relates bending stiffness to moment of inertia (1y) and deformation coefficient (E). Use the information below to calculate (no conversions needed) the bending stiffness for steel and then use fprintf to display the both the deformation and bending stiffness values. Also, display a title and titles for the columns above the values like a table in a textbook that an engineer can use. Hzz=E I For steel Iy = Ask the user E = 103, 104, 106, 1.2 x 107, and 2.56 x 107 Pa

Answer 1

Matlab Script

clear all; close all; clc Iy = input('Enter the value of moment of inertia, Iy: '); $to input the value of Iy E = [1043, 10^4, 10^6, 1.2*10^7, 2.56*10*7]; Deformation coefficient UNE EcoJon WN IIIIIIIIII Hzz = E*Iy; Computes Bending stiffness m = [E' Hzz')'; fprintf("\nBending stiffness values for steels (Iy = $.4f) \n', Iy); fprintf('$25s $25s \n', 'Deformation coefficient (E), Pa', 'Bending Stiffness (Hz)'); fprintf("$20.5e $28.7e \n', m);

Sample output for Iy = 6666.67

Command Window Enter the value of moment of inertia, Iy: 6666.667 Bending stiffness values for steels (Iy = 6666.6670) De formation coefficient (E), Pa Bending Stiffness (Hzz) 1.00000e+03 6.6666670e+06 1.00000e+04 6.6666670e+07 1.00000e+06 6.6666670e+09 1.20000e+07 8.0000004e+10 2.56000e+07 1.7066668e+11