8.L=9'=108" long simply-supported beam has an unsymmetric cross-section as shown below. A F=220# person walks across the beam. Assume the weight is applied through the shear center of the beam. The beam is made of steel with a modulus E=30 Msi and a Poisson ratio u=0.3.
L Type CROSS-SECTION:
Big rectangle: z=0.2"*y=2"
Small rectangle: z=0.1"*y=1.5"
Determine:
(a) Areas
A1=Big rectangle
=2*1.5
=3(in^2)
A2=Small rectangle
=1.9*1.3
=2.47(in^2)
A=L-type
=A1-A2
=0.53(in^2)
(b) Centroid
yc=(A1*1/2-A2*1.9/2)/A
=(3/2-2.47*0.38)/0.53
=1.233"
zc=(A1*1.5/2-A2*1.3/2)/A
=(3*0.75-2.47*0.65)/0.53
=1.216"
(c) Momentum of Inertia
Iy=[(2*1.5^3)/12-A1*(1.5-zc)^2]-[1.9*1.3^2/12-A2*(1.3/2-zc)^2]
=6.355(in^4)
(d) Beam weight
q=Density
=7850(kg/m^3)*2.2(#/kg)*10^(-6)m^3/cm^3
=2.512*10^(-2)(#/cm^3)*2.54^3(cm^3/in)
=0.411643(#/in^3)
w=q*A
=0.411643*0.53
=0.218(#/in)
(e) deflection due to weight
dw=w*L^4/384E*Iy.....from Material Mechanics
=0.218*108^4/(384*30*10^6*6.355)
=4.05*10(-4)"
(f) deflection due to Person on midpoint
dp=F*L^3/48E*Iy.....from Material Mechanics
=202*108^3/(48*30*10^6*6.355)
=2.78*10^(-2)"
dz=dw+dp
=(2.78+0.0405)/100
=0.0282".....ans
(g) Magnitude of that deflection, divided into y, z components.
Without any load applys to y-direction => dy=0
(h) Max stress σx, in the beam for this condition and its location.
Z=Sectional modulus
=Iy/zc
=6.355/1.216
=5.226(in^3)
R=Reaction
=(F+w*L)/2
=(220+0.218*108)/2
=122#
Mmax=R*L/2.....at midpoint
=122*108/2
=6576(#.in)
σx=Mmax/Z
=6567/5.226
=1258(psi).....ans
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