find the ratio of d/b such that the largest stress in the beam will be minimum(adsbygoogle = window.adsbygoogle || []).push({});

i know that the maximum stress is

σ_{max}=Y_{max}[tex]\frac{M}{I}[/tex]

and i know that for a rectangle I=bh^{3}/12

now in the question i am asked to find the ration of the (diameter of the log)/(the width of the rectangle) such that σ_{max}is minimal

since the rectangle is contained in the circle

d^{2}=b^{2}+h^{2}

where h is the height of the rectangle

σ_{max}=Y_{max}[tex]\frac{M}{I}[/tex]

σ_{max}=(h/2)*(12M/(bh^{2})

σ_{max}=6M/(b*h^{2})

σ_{max}=6M/(b*(d^{2}-b^{2})

basically from here i need to find the ratio d/b so that (b*(d^{2}-b^{2}) is maximum,

but how can i do this??

d/b=K

(b*(d^{2}-b^{2})

=(bd^{2}-b^{3})

=d/b*(d*b_{2}-b^{4}/d)

but i cant get to the ratio, i feel i am so close but just not getting it

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# Homework Help: Beams and bending

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