Application of Derivatives

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Question

A beam of length l is supported at one end. If W is the uniform load per unit length, the bending moment M at a distance x from the end is given by $straight M space equals 1 half space straight l space straight x space space minus space 1 half space straight W space straight x squared.$ maximum value. Find the point on the beam at which the bending moment has th

Solution

straight M space equals 1 half space straight l space straight x space minus space 1 half space straight w space straight x squared space space space space space space space rightwards double arrow space space space space dM over dx space equals space 1 half straight l space minus space wx

Now,

dM over dx space equals space 0

give us

1 half straight l space minus space wx space equals space 0 comma space space space space space rightwards double arrow space space space space straight x space equals space space fraction numerator straight l over denominator 2 straight w end fraction

fraction numerator straight d squared straight M over denominator dx squared end fraction space equals negative straight w

When

straight x equals fraction numerator straight l over denominator 2 straight w end fraction comma space space fraction numerator straight d squared straight M over denominator dx squared end fraction space equals space minus straight w less than 0

therefore space space space space space straight M space is space maximum space when space straight x space equals space fraction numerator straight l over denominator 2 space straight w end fraction therefore space space space space space space required space point space is space at space space straight a space distance space fraction numerator straight l over denominator 2 straight w end fraction space from space the space supporting space end.