104L Lab 5.docx - Subject Laboratory 5 Simply Supported...

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Subject: Laboratory 5. Simply Supported Wide Flange Beam: Purpose: The purpose of this lab was to calculate experimental and theoretical values of bending and shearing stress in the web and flange of a simply supported beam. Procedure and Theory: We applied 0, 5,000, 10,000, and 15,000-pound loads to the beam. For each loading, we recorded the strain readings from each of the strain gauges for the web and the flange. The experimental values for bending stress and shear stress were calculated using equations from lab two. The orientation of strain gauges simplifies the equations to find the corresponding stress. See lab two in the manual for those equations. For the theoretical values, σ y for the web region should be zero and σ z for the flange region should be zero. The bending stresses in the web and flange, σ x and σ z , can be calculated from My I = ( P 2 ) a y I . The shear stresses in the web, τ xy , can be calculated from VQ It = P 2 ´ y A It . The shear stress in the web, τ xz , can be calculated from VQ It = P 2 ( 1.065 c ) It . Sample Calculations:
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For the sample calculations, Youngs Modulus is 30*10^6 and Poisson’s Ratio is 0.3. The following sample calculations are from “Area One” of the web region with a load of 5,000 lbs. See Table One for other variables. ε y = ε + 45 + ε 45 ε 0 = 4 144 + 82 =− 58 microns ε x = ε 0 =− 82 microns ϒ xy = ε + 45 ε 45 = 4 + 144 = 148 microns Experimental Values for Stress: σ x = E ( ε x + υε y ) ( 1 υ 2 ) = 30 10 6 (− 82 10 6 + 0.3 ∗− 58 10 6 ) 1 0.3 2 =− 3277 psi σ y = E ( ε y + υ ε x ) ( 1 υ 2 ) = 30 10 6 (− 58 10 6 + 0.3 ∗− 82 10 6 ) 1 0.3 2 =− 2723 psi 1 0.3 2 ¿ 2 ¿ τ xy = E ϒ xy 2 ( 1 + υ ) = 30 10 6 144 10 6 ¿ Theoretical Values for Stress: 26 ¿ ¿ 1.5 ¿¿ −( 5000 lbs 2 ) ¿ σ x My I = ( P 2 ) a y I . = ¿
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0.26 ¿ ¿ ( 41.4 i n 4 ) ¿ τ xy = VQ It = P 2 ´ y A It = ( 5000 lbs 2 )( 7.16 i n 3 ) ¿ Sample Calculations for Flange Region (excluding experimental calculations as they are the same as web’s): ( 5000 lbs 2 ) 26 ¿ 3.1 ¿ 41.4 in 4 =− 4867 psi σ x My I = ( P 2 ) a y I =− ¿ 0.365 ¿ ¿ ( 41.4 in 4 )
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