Contents

Aaron Klapheck

% Notes from 10-20-08. Solving example 4-10 on page 195.
clear, clc, home
fprintf('The date and time: %s \n \n', datestr(now))

The date and time: 29-Dec-2008 10:38:46

Solving problem 4-10 Using Round Steel Tubes

fprintf('%%%%%% Answers for Round Steel Tubes %%%%%% \n')

% Given
thickness = .5; % in. Thickness of the hollow tube.
SF = 4; % factor of safety.
% For D_o innitially used the range [1:1:30] which narrowed it down to a
% value arround 12 so then used the range below.
D_o = [10:0.0001:13]; % in. Outer diameter values.
D_i = D_o - 2*thickness; % in. Inner diameter.


% 1st get length of member (l) and force applied to member (F).
F = 200000; % lb
l = 10*12; % in


% 2nd get area (A), second moment of area (I), and radious of gyration (k).
A = pi/4.*(D_o.^2-D_i.^2); % in^2
I = pi./64.*(D_o.^4-D_i.^4); % in^4
k = (I./A).^(1/2); % in.


% 3rd get the slenderness ratio (S_r).
S_r = 2.1.*l./k; % unitless. Using AISC value


% 4th Use Steel material properties
S_yc = 60000; % psi
E = 30000000; % psi
S_r_d = pi.*(2.*E./S_yc).^(1/2); % unitless


% 5th Find P_cr
leng = length(D_o);
P_cr = zeros(1,leng);
for x = 1:leng
    if (S_r(x) <= S_r_d);
        P_cr(x) = A(x).*(S_yc - 1./E.*((S_yc.*S_r(x))/(2.*pi)).^2);
    else
        P_cr(x) = A(x).*pi^2.*E./S_r(x).^2;
    end
end


% 6th get P_allow
P_allow = P_cr./SF; % lb


% 7th get actual outer diameter (D_outer)
for x = 1:leng
   if (P_allow(x) >= F)
       D_outer = D_o(x) % in
       Slenderness_ratio = S_r(x) % unitless
       Allowable_Force = P_allow(x) % lb
       Critical_Load = P_cr(x) % lb
       Radius_of_Gyration = k(x) % in
       break
   end
end

%%% Answers for Round Steel Tubes %%% 

D_outer =

   11.3544


Slenderness_ratio =

   65.5963


Allowable_Force =

  2.0000e+005


Critical_Load =

  8.0000e+005


Radius_of_Gyration =

    3.8417

Solving problem 4-10 Using Square Steel Tubes

fprintf('%%%%%% Answers for Square Steel Tubes %%%%%% \n')

% Given
thickness = .5; % in. Thickness of the hollow tube
SF = 4; % factor of safety.
% For D_o innitially used the range [1:1:30] which narrowed it down to a
% value arround 10 so then used the range below.
S_o = [9:0.0001:11]; % in. Outer length values.
S_i = S_o - 2*thickness; % in. Inner length.


% 1st get length of member (l) and force applied to member (F).
F = 200000; % lb
l = 10*12; % in


% 2nd get area (A), second moment of area (I), and radious of gyration (k).
A = S_o.^2-S_i.^2; % in^2
I = 1/12.*(S_o.^4-S_i.^4); % in^4
k = (I./A).^(1/2); % in.


% 3rd get the slenderness ratio (S_r).
S_r = 2.1.*l./k; % unitless. Using AISC value


% 4th Use Steel material properties
S_yc = 60000; % psi
E = 30000000; % psi
S_r_d = pi.*(2.*E./S_yc).^(1/2); % unitless


% 5th Find P_cr
leng = length(S_o);
P_cr = zeros(1,leng);
for x = 1:leng
    if (S_r(x) <= S_r_d);
        P_cr(x) = A(x).*(S_yc - 1./E.*((S_yc.*S_r(x))/(2.*pi)).^2);
    else
        P_cr(x) = A(x).*pi^2.*E./S_r(x).^2;
    end
end


% 6th get P_allow
P_allow = P_cr./SF; % lb


% 7th get actual outer length (S_outer)
for x = 1:leng
   if (P_allow(x) >= F)
       S_length = S_o(x) % in
       Slenderness_ratio = S_r(x) % unitless
       Allowable_Force = P_allow(x) % lb
       Critical_Load = P_cr(x) % lb
       Radius_of_Gyration = k(x) % in
       break
   end
end

%%% Answers for Square Steel Tubes %%% 

S_length =

    9.3427


Slenderness_ratio =

   69.6944


Allowable_Force =

  2.0000e+005


Critical_Load =

  8.0001e+005


Radius_of_Gyration =

    3.6158


Published with MATLAB® 7.5