Let us consider a differential length dx of the beam shown

In this section, fv will be used for shearing
stress instead of the standard symbol τ.

For the upper shaded portion of the beam, the forces acting are the total normal forces FR and FL due to the bending stresses to the left and to the right of the beam. These forces will be resisted by the shearing force fvb dx acting at the boundary surface between the shaded and the unshaded portions.

For equilibrium of the upper shaded portion
$F_L + F_V - F_R = 0$

$F_V = F_R - F_L$

Where
$F_V = f_v b \, dx$

$\displaystyle F_L = \int f_{b1} \, dA ; \,\, f_{b1} = \frac{My}{I}$

$\displaystyle F_R = \int f_{b2} \, dA ; \,\, f_{b2} = \frac{(M + dM)y}{I}$

$\displaystyle F_v = \int \frac{(M + dM)y}{I} dA - \int \frac{My}{I} dA$

$\displaystyle f_v b \, dx = \int \frac{My}{I} dA + \int \frac{dM}{I} y \, dA - \int \frac{My}{I}$

$\displaystyle f_v b \, dx = \frac{dM}{I} \int y \, dA$

$\displaystyle f_v = \frac{1}{Ib} \frac{dM}{dx} \int y \, dA$

But $\,\, \dfrac{dM}{dx} = V, \,\,$ where $\,\, V \,\,$ represents the shear at the section in $\,\, N \,\,$, and $\,\, \int y \, dA = A \bar{y} \,\,$ represents the first moment of an area of the shaded section about N.A. in mm3 which we will denote as $\,\, Q \,\,$, then

$f_v = \dfrac{VQ}{Ib}$

# Shear Flow

If the shearing stress fv is multiplied by the width b, we obtain a quantity q, known as the shear flow, which represents the longitudinal force per unit length transmitted across a section at a level y1 from the neutral axis.

$q = f_v b = \dfrac{VQ}{I}$

Application of Flexural and Shearing Stresses to Rectangular Section
For rectangular section, $\,\, I = \dfrac{bd^3}{12} \,\,$ and $\,\, c = d/2. \,\,$

The bending stress at a level $\,\,y\,\,$ from the neutral axis is

$f_b = \dfrac{12My}{bd^3}$

and the maximum bending stress in the section is

$(\,f_b\,)_{max} = \dfrac{6M}{bd^2}$

The maximum shearing stress in the section is

$(\,f_v\,)_{max} = \dfrac{3V}{2bd} = \dfrac{3V}{2A}$

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