A project that involves designing, building, and analyzing how an aluminum and brass beam deflects due to certain applied loads.

1. Mechanics of Materials: Beam Deflections Engineered by Group 20 Christopher Webb, Anthony Williams, Carlaton Wong, Jonathan Wong 03/03/2010

2. The Concept Scenario: How much will a beam deflect with an applied load? Calculating the beam’s deflection will reveal the beams curvature relationships This project is designed to exhibit the mechanics of a bending moment on a single cantilever beam. Ultimately leading us to our goal.

3. Goals & Objectives Experimental Goals How much will aluminum and brass deflect with various applied loads? Determine the Young’s Modulus based on the deflection measured Does our calculated Young’s Modulus equal tabulated values?

4. Beam Failures

5. What is Beam Deflection? When loads are applied to a beam, the beam’s axis that was once straight will become curved. The displacements from the initial axis are called bending or flexural deflections.

7. At the end of the beam there will be an attachment to serve as the applied load.

9. Cantilever beams are interchangeable allowing for supplemental testing of differing materials & structural make-ups .

10. Negligible factors:

11. Load’s tiny distance from free end of member

13. Our Cantilever Beam

14. Superposition Table

15. MatLab Code hold on; grid on plot(x/in_m,v2*100,'g-*',x/in_m,v1*100,'ro-',x/in_m,v1*100+v2*100,'b*-') title('Beam Deflection of Brass w/ 0.3lb Load') xlabel('Beam Length (in)') ylabel('Deflection (cm)') legend('DistributedLoad','ConcentratedLoad','Total Deflection',3) V_max=(q*L+F)/(b*h);%Max shear stress sigma_max=(.5*q*L^2+F*L)*(h*0.5)/I; v_max_tot=(v1(end)+v2(end))*100; %Analysis of Aluminum Beam close all;clearall;clc %Conversion Factors in_m=0.0254; %inch to meter lb_N=0.22481;%Pound force to Newton %Material Properties E=69*10^9;%Pa Aluminum %Geometric Properties(in SI units) %Al bar data h=(1/8)*in_m;%Cross section height b=(3/4)*in_m;%Cross sect width L=15*in_m;%Length of Beam (m) I=(b*h^3)/12; %second momemnt of inertia(m^4) %Analysis of Brass beam close all;clearall;clc %Conversion Factors in_m=0.0254; %inch to meter lb_N=0.22481;%Pound force to Newton %Material Properties E=100*10^9;%Pa Brass %Geometric Properties(in SI units) %Brass Bar data h=(.064)*in_m;%Cross section height b=(3/4)*in_m;%Cross sect width L=9*in_m;%Length of Beam (m) I=(b*h^3)/12; %second momemnt of inertia(m^4) %Applied force F=(.3)/lb_N; %(N) %Calculating deflection x=linspace(0,L,50); v1=(F*x.^3/6 - F*L*x.^2/2)/(E*I); %Deflection due to concentrated load q=79.7*10^-3*9.81/(12*.0254); v2=zeros(1,length(x)); for i=1:length(x) v2(i)=-q*x(i)^2/(24*E*I)*(6*L^2-4*L*x(i) + x(i)^2); %Deflection due to weight of bar end %Applied force F=(.3)/lb_N; %Calculating deflection %Concetrated load x=linspace(0,L,50); v1=(F*x.^3/6 - F*L*x.^2/2)/(E*I);%Deflection in (m) %Distributed weight q=74*10^-3*9.81/(18.5*.0254); v2=zeros(1,length(x)); for i=1:length(x) v2(i)=-q*x(i)^2/(24*E*I)*(6*L^2-4*L*x(i) + x(i)^2); end hold on;grid on; plot(x/in_m,v1*100,'ro',x/in_m,v2*100,'g*',x/in_m,v1*100+v2*100,'b*') title('Beam Deflection of Aluminum w/ 0.3lb Load') xlabel('Beam Length (in)') ylabel('Deflection (cm)') legend('DistributedLoad','ConcentratedLoad','Total Deflection',3) V_max=(q*L+F)/(b*h)%Max shear stress sigma_max=(.5*q*L^2+F*L)*(h*0.5)/I; v_max_tot=(v1(end)+v2(end))*100;

18. Calculated Linear Weight Density by weighing the beams on a scale.

20. Results – Brass Bar .3 Lbs Force .4 Lbs Force .5 Lbs Force

21. Results – Aluminum Bar .3 Lbs Force .4 Lbs Force .5 Lbs Force

22. Experimental Results Aluminum : Expected Young’s Modulus: 69 GPa Brass: Expected Young’s Modulus: 100 GPa

23. Expected Results Aluminum: Brass: All measured deflections are higher than calculated. Why?

24. Max Stress in Beam Using the Flexural formula and observing that the Mmax is located at the origin we can derive an expression for Pmax

26. Aluminum = 35MPa

28. Mechanics of Materials:Beam Deflections Group 20 Christopher Webb, Anthony Williams, Carlaton Wong, & Jonathan Wong Thank you!