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- 麻辣Lv 76 years agoFavorite Answer
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) AreasA1=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) Centroidyc=(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 InertiaIy=[(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 weightq=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 weightdw=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 midpointdp=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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