Chapter 6.5, Problem 60P

(a)

To determine

Show that the magnitude of the horizontal shearing force H exerted on the lower face of the portion of the beam ACKJ is $H=\frac{1}{2}b{\sigma }_Y\left(2c-y_Y-\frac{y^2}{y_Y}\right)$.

Answer to Problem 60P

The magnitude of the horizontal shearing force H exerted on the lower face of the portion of the beam ACKJ is $H=\frac{1}{2}b{\sigma }_Y\left(2c-y_Y-\frac{y^2}{y_Y}\right)$.

Explanation of Solution

Given information:

K is a point at a distance $y<y_Y$ above the neutral axis.

${\sigma }_x={\sigma }_Y$ between C and E.

${\sigma }_x=\left(\frac{{\sigma }_Y}{y_Y}\right)y$ between E and K.

Calculation:

The point K is located a distance y above the neutral axis.

Provide the stress distribution as shown below.

$\sigma ={\sigma }_Y\frac{y}{y_Y}$ for $0\le y<y_Y$

$\sigma ={\sigma }_Y$ for $y_Y\le y\le c$.

Sketch the stress distribution for $\sigma ={\sigma }_Y\frac{y}{y_Y}$ as shown in Figure 1.

Sketch the stress distribution for $\sigma ={\sigma }_Y$ as shown in Figure 1.

Calculate the horizontal forces acting on ACKJ as shown below.

$H={\displaystyle \int \sigma dA}$

Substitute ${\sigma }_Y\frac{y}{y_Y}$ for $\sigma$ and apply the limits.

$\begin{array}{c}H={\displaystyle {\int }_y^{y_Y}{\sigma }_Y\frac{yb}{y_Y}dy}+{\displaystyle {\int }_{y_Y}^c{\sigma }_{Y}bdy}\\ =\frac{{\sigma }_{Y}b}{y_Y}{\left(\frac{y^2}{2}\right)}_y^{y_Y}+{\sigma }_{Y}b{\left(y\right)}_{y_Y}^c\\ =\frac{{\sigma }_{Y}b}{y_Y}\left(\frac{y_Y^2-y^2}{2}\right)+{\sigma }_{Y}b\left(c-y_Y\right)\\ =\frac{{\sigma }_{Y}b{y}_Y}{2}-\frac{{\sigma }_{Y}b{y}^2}{2y_Y}+{\sigma }_{Y}bc-{\sigma }_{Y}b{y}_Y\end{array}$

$=\frac{{\sigma }_{Y}b}{2}\left(2c-y_Y-\frac{y^2}{y_Y}\right)$

Therefore, the magnitude of the horizontal shearing force H exerted on the lower face of the portion of the beam ACKJ is $H=\frac{1}{2}b{\sigma }_Y\left(2c-y_Y-\frac{y^2}{y_Y}\right)$.

(b)

To determine

The shearing stress at K.

Answer to Problem 60P

The shearing stress at K is ${\tau }_{xy}=\frac{3P}{4b{y}_Y}\left(1-\frac{y^2}{y_Y^2}\right)$.

Explanation of Solution

Given information:

K is a point at a distance $y<y_Y$ above the neutral axis.

${\sigma }_x={\sigma }_Y$ between C and E.

${\sigma }_x=\left(\frac{{\sigma }_Y}{y_Y}\right)y$ between E and K.

$y_Y$ is a function of x.

Calculation:

Refer to part (a).

The horizontal shearing force is $H=\frac{1}{2}b{\sigma }_Y\left(2c-y_Y-\frac{y^2}{y_Y}\right)$`

Calculate the shear stress as shown below.

${\tau }_{xy}=\frac{1}{b}\frac{\partial H}{\partial x}$

Substitute $\frac{1}{2}b{\sigma }_Y\left(2c-y_Y-\frac{y^2}{y_Y}\right)$ for H.

$\begin{array}{c}{\tau }_{xy}=\frac{1}{b}\frac{\partial }{\partial x}\left(\frac{1}{2}b{\sigma }_Y\left(2c-y_Y-\frac{y^2}{y_Y}\right)\right)\\ =\frac{{\sigma }_Y}{2}\frac{\partial }{\partial x}\left(2c-y_Y-\frac{y^2}{y_Y}\right)\\ =\frac{{\sigma }_Y}{2}\left(-1-y^2\left(-\frac{1}{y_Y^2}\right)\right)\frac{d{y}_Y}{dx}\\ =\frac{{\sigma }_Y}{2}\left(\frac{y^2}{y_Y^2}-1\right)\frac{d{y}_Y}{dx}\end{array}$ (1)

Provide the relation of moment as shown below.

$M=Px=\frac{3}{2}{M}_Y\left(1-\frac{1}{3}\frac{y_Y^2}{c^2}\right)$

Differentiate both sides of the Equation as shown below.

$\begin{array}{c}\frac{dM}{dx}=\frac{d}{dx}\left(Px\right)=\frac{d}{dx}\left(\frac{3}{2}{M}_y\left(1-\frac{1}{3}\frac{y_Y^2}{c^2}\right)\right)\\ =P=\frac{3}{2}{M}_y\left(-\frac{2}{3c^2}{y}_Y\right)\frac{d{y}_Y}{dx}\end{array}$

$\begin{array}{l}P=-\frac{M_y{y}_Y}{c^2}\frac{d{y}_Y}{dx}\\ \frac{d{y}_Y}{dx}=-\frac{P{c}^2}{M_y{y}_Y}\end{array}$

Substitute $\frac{2}{3}{\sigma }_{Y}b{c}^2$ for $M_y$.

$\begin{array}{c}\frac{d{y}_Y}{dx}=-\frac{P{c}^2}{\frac{2}{3}{\sigma }_{Y}b{c}^2{y}_Y}\\ =-\frac{3P}{2{\sigma }_{Y}b{y}_Y}\end{array}$

Substitute $-\frac{3P}{2{\sigma }_{Y}b{y}_Y}$ for $\frac{d{y}_Y}{dx}$ in Equation (1).

$\begin{array}{c}{\tau }_{xy}=\frac{{\sigma }_Y}{2}\left(\frac{y^2}{y_Y^2}-1\right)\left(-\frac{3P}{2{\sigma }_{Y}b{y}_Y}\right)\\ =\frac{3P}{4b{y}_Y}\left(1-\frac{y^2}{y_Y^2}\right)\end{array}$

Therefore, the shearing stress at K is ${\tau }_{xy}=\frac{3P}{4b{y}_Y}\left(1-\frac{y^2}{y_Y^2}\right)$.