- #1
Dell
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find the ratio of d/b such that the largest stress in the beam will be minimum
i know that the maximum stress is
σmax=Ymax[tex]\frac{M}{I}[/tex]
and i know that for a rectangle I=bh3/12
now in the question i am asked to find the ration of the (diameter of the log)/(the width of the rectangle) such that σmax is minimal
since the rectangle is contained in the circle
d2=b2+h2
where h is the height of the rectangle
σmax=Ymax[tex]\frac{M}{I}[/tex]
σmax=(h/2)*(12M/(bh2)
σmax=6M/(b*h2)
σmax=6M/(b*(d2-b2)
basically from here i need to find the ratio d/b so that (b*(d2-b2) is maximum,
but how can i do this??
d/b=K
(b*(d2-b2)
=(bd2-b3)
=d/b*(d*b2-b4/d)
but i can't get to the ratio, i feel i am so close but just not getting it
i know that the maximum stress is
σmax=Ymax[tex]\frac{M}{I}[/tex]
and i know that for a rectangle I=bh3/12
now in the question i am asked to find the ration of the (diameter of the log)/(the width of the rectangle) such that σmax is minimal
since the rectangle is contained in the circle
d2=b2+h2
where h is the height of the rectangle
σmax=Ymax[tex]\frac{M}{I}[/tex]
σmax=(h/2)*(12M/(bh2)
σmax=6M/(b*h2)
σmax=6M/(b*(d2-b2)
basically from here i need to find the ratio d/b so that (b*(d2-b2) is maximum,
but how can i do this??
d/b=K
(b*(d2-b2)
=(bd2-b3)
=d/b*(d*b2-b4/d)
but i can't get to the ratio, i feel i am so close but just not getting it
Last edited: