midspan deflection:
δ=5WL4384EI=5π4sinx384EI=1.27⋅sinxEI\delta=\frac{5WL^4}{384EI}=\frac{5\pi^4sinx}{384EI}=1.27\cdot\frac{sinx}{EI}δ=384EI5WL4=384EI5π4sinx=1.27⋅EIsinx
bending moment at any point x:
Mx=WLx/2−Wx2/2=(π/2−x/2)xsinxM_x = WLx/2 -Wx2 /2=(\pi/2-x/2)xsinxMx=WLx/2−Wx2/2=(π/2−x/2)xsinx
shear:
F=Mx/x=(π/2−x/2)sinxF=M_x/x=(\pi/2-x/2)sinxF=Mx/x=(π/2−x/2)sinx
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment