MECHANICS OF MATERIALS: BEAM DEFLECTIONS

Engineered by Group 20

Christopher Webb, Anthony Williams, Carlaton Wong, Jonathan Wong

03/03/2010

The ConceptSCENARIO: 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.

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?

Beam Failures

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.

Apparatus Design & Operation

The beam in use will be fixed onto the standing wooden plank by securing it at the top of the apparatus with a c-clamp.

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

This experiment used strategically filled water bottles for the applied loads.

Apparatus Design & Operation (cont.)

The deflection of the beam by the applied load can be recorded at the point of loading.

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

Negligible factors: Load’s tiny distance

from free end of member

Small bend in beam after multiple loads

Cost Factors Involved

Material Cost Vendor

½” 20”x8” Plywood $0.00 Lumber Yard

4”x4” Wood $0.00 Lumber Yard

Aluminum Beam $5.00 Home Depot

Brass Beam $5.00 Home Depot

4” C-Clamp $0.00 Home Depot

Water Bottles $0.00 Home

String $0.00 Michaels

Total Cost $10.00

Our Cantilever Beam

Superposition Table

€

δTOT =L3

24EI3qL + 8P( )

€

E =L3

24IδTOT3qL + 8P( )

MatLab Code

%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

hold on; grid onplot(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 stresssigma_max=(.5*q*L^2+F*L)*(h*0.5)/I;v_max_tot=(v1(end)+v2(end))*100;     %Analysis of Aluminum Beamclose all;clearall;clc

%Conversion Factorsin_m=0.0254; %inch to meter lb_N=0.22481;%Pound force to Newton

%Material PropertiesE=69*10^9;%Pa Aluminum

%Geometric Properties(in SI units)%Al bar datah=(1/8)*in_m;%Cross section heightb=(3/4)*in_m;%Cross sect widthL=15*in_m;%Length of Beam (m)

I=(b*h^3)/12; %second momemnt of inertia(m^4)

%Applied forceF=(.3)/lb_N;

%Calculating deflection %Concetrated loadx=linspace(0,L,50); v1=(F*x.^3/6 - F*L*x.^2/2)/(E*I);%Deflection in (m)

%Distributed weightq=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 stresssigma_max=(.5*q*L^2+F*L)*(h*0.5)/I;v_max_tot=(v1(end)+v2(end))*100;

Experiment Variables Second Moment of Inertia (I) determined by measuring geometry of the beams.

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

Water bottles with a known mass used to calculate forces applied to the beam.

Results – Initial Height

Results – Brass Bar

.3 Lbs Force

.4 Lbs Force

.5 Lbs Force

Results – Aluminum Bar

.3 Lbs Force

.4 Lbs Force

.5 Lbs Force

Experimental Results

Load(lb)Measured Deflection

(cm)Calculated E (GPa)

0.3 1.00 56.28

0.4 1.30 55.80

0.5 1.60 55.40

Average 55.83

Percent Error 19.09%

Load(lb)Measured Deflection

(cm)Calculated E (GPa)

0.3 1.00 90.76

0.4 1.30 90.80

0.5 1.50 95.14

Average 92.23

Percent Error 7.77%

Aluminum :

Brass:

Expected Young’s

Modulus:

69 GPa

Expected Young’s

Modulus:

100 GPa

Expected Results

Load(lb) Measured Deflection (cm)Calculated Deflection (cm)

(Assumed E = 69 GPa)

0.3 1.00 0.82

0.4 1.30 1.05

0.5 1.60 1.29

Load(lb) Measured Deflection (cm)Calculated Deflection (cm)

(Assumed E = 100 GPa)

0.3 1.00 0.91

0.4 1.30 1.17

0.5 1.50 1.43

Aluminum:

Brass:

All measured deflections are higher than calculated. Why?

MAX STRESS IN BEAMUsing the Flexural formula and observing that the Mmax is located at the origin we

can derive an expression for Pmax

I

hM

I

yM

2

)0(maxmaxmax

PLLqM 202

1)0(

• Approximate Yield stress:– Aluminum = 35MPa– Brass = 70 MPa

Max Stress in Beam

maxfor solve and toY.S.Set Pallow

2

2 0max

Lq

hL

IP allow

lb 595.0max AlP

lb 512 .0max BrassP

Conclusion

With the proper build and successful experimentation of our proposed apparatus, deflections of both beams were recorded.

Young’s Modulus was calculated given the measured deflection.

Comparing our calculated and measured Young’s Modulus’ error was found to be 10-20% due to possible plastic deformation.

MECHANICS OF MATERIALS:

BEAM DEFLECTIONS

Group 20

Christopher Webb, Anthony Williams, Carlaton Wong, & Jonathan Wong

Thank you!