The ratio of the length and depth of a simply supported rectangular beam which experiences maximum bending stress equal to tensile stress, due to same load at its mid span, is

let point load P act on beam at mid. Height of beam h, width b , effective depth d and length=L, in the case we consider h=d
Tensile stress= P/(bd)
Bending moment M= (PL/4), and moment on inertia I = (bd^3)/12, and distance of extreme fiber is C=d/2
Bending Stress= MC/I, then .................................

let point load P act on beam at mid. Height of beam h, width b , effective depth d and length=L, in the case we consider h=d
Tensile stress= P/(bd)
Bending moment M= (PL/4), and moment on inertia I = (bd^3)/12, and distance of extreme fiber is C=d/2
Bending Stress= MC/I, then .................................

Please any one will explain it

Bending stress(sigma) to tensile(P) ratio=?

M/I=Sigma/y

Or Sigma=(M/I)×y=M/Z

Or sigma=(wl/4)/(bd^2/6)=(3wl)/(2bd^2)


Tensile stress , P =force /area=w/bd

So..... ratio of bending stress to tensile stress ,

Sigma=P

Or (3wl)/(2bd^2)= w/bd

Or l/d =2/3

From these relations
M/I=F/Y
TENSILE STRESS= P/A
GIVEN F=TENSILE STRESS

SO L/D= 2/3

Bending Stress=M(max)/Z
=(PL/4)/(I/y)
=(PL/4)/[(bd^3/12)/(d/2)]
=(PL/4)/[bd^2/6]
=3PL/2bd^2
Now,
Tensile Stress=P/A
=P/bd.

Since, according to question,
Bending Stress=Tensile Stress
PL/2bd^2=P/bd.
====> L/d=2/3