兆緯 asked in 科學工程學 · 7 years ago

結構力學 求解析

http://ocw.mit.edu/courses/aeronautics-and-astrona...

麻煩高手幫忙解析pdf的第8題, 謝謝

2 Answers

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  • 麻辣
    Lv 7
    7 years ago
    Favorite 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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  • 6 years ago

    參考下面的網址看看

    http://phi008780520.pixnet.net/blog

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