% Bending/Shear stress example close all; length = 20; % in force = 10000; %lbs %eye-beam dimensions max_width = 4; % in min_width = 1; % in y_max = 4; % in center_y = 2; % in %max bending moment at the root... M = force*length; I = min_width*(2*center_y)^3/12 + 2*( max_width*(y_max-center_y)^3/12 + ... max_width*(y_max-center_y)*((y_max+center_y)/2)^2); sigma_max = M * y_max / I; % solve for shear stress distribution % V / (I * t) * int(y*da) % Point 1: evaluated at location just before thickness changes from 4 to 1 in tempCoeff = force / (I * max_width); int_y_da = ((y_max+center_y)/2) * max_width*(y_max-center_y); shear_1 = tempCoeff*int_y_da; % Point 2: evaluated at location just after thickness changes from 4 to 1 in tempCoeff = force / (I * min_width); shear_2 = tempCoeff*int_y_da; % Point 3: evaluated at center of beam tempCoeff = force / (I * min_width); int_y_da = (center_y/2) * min_width*center_y; shear_3 = shear_2+tempCoeff*int_y_da; %evaluating continous integral for width of 4.. int_y_da_4 = force / (I * max_width)*4*(y_max^2/2 - (center_y:.1:y_max).^2/2); %evaluating continous integral for width of 1.. int_y_da_1 = shear_2 + force / (I * min_width)*1*(center_y^2/2 - (0:.1:center_y).^2/2); figure; grid on; hold on;set(gcf,'color',[1 1 1]); plot(int_y_da_4,center_y:.1:y_max,'linewidth',2) plot(int_y_da_1,0:.1:center_y,'linewidth',2) plot(int_y_da_1,0:-.1:-center_y,'linewidth',2) plot(int_y_da_4,-center_y:-.1:-y_max,'linewidth',2) plot([shear_1 shear_2],[center_y center_y],'linewidth',2) plot([shear_1 shear_2],[-center_y -center_y],'linewidth',2) plot(shear_1,center_y,'o') plot(shear_2,center_y,'o') plot(shear_3,0,'o') plot(shear_2,-center_y,'o') plot(shear_1,-center_y,'o') xlabel('shear stress (lb/in^2)','fontsize',16,'fontweight','bold');ylabel('Distance from Center (in)','fontsize',16,'fontweight','bold') set(gca,'FontSize',16,'fontweight','bold'); figure; grid on; hold on;set(gcf,'color',[1 1 1]); plot([0 4 4 2.5 2.5 4 4 0 0 1.5 1.5 0 0],[4 4 2 2 -2 -2 -4 -4 -2 -2 2 2 4],'linewidth',2)