Chapter 8, Problem 72RP

Forces are applied at points A and B of the solid cast-iron bracket shown. Knowing that the bracket has a diameter of 0.8 in., determine the principal stresses and the maximum shearing stress at (a) point H, (b) point K.

Fig. P8.72

To determine

The principal stresses and the maximum shearing stress at point H.

Answer to Problem 72RP

The maximum principal stress at point H is $\underset{\_}{31.9\text{ksi}}$.

The minimum principal stress at H is $\underset{\_}{\text{-6}\text{.99}\text{ksi}}$.

The shear stress at point H is $\underset{\_}{19.42\text{ksi}}$.

Explanation of Solution

Given information:

The diameter (d) of the bracket is $0.8\text{in}\text{.}$

Calculation:

Sketch the free body diagram of solid cast iron as shown in Figure 1.

Refer to Figure 1.

Find the value of P at the section containing point H and K.

$P=-2,500\text{lb}$

Find the shear force about y axis as follows:

$V_y=-600\text{lb}$

Find the shear force about x axis as follows:

$V_x=0$

Find the moment about x axis as follows:

$\begin{array}{c}{M}_x=\left(3.5-1\right)\left(600\right)\\ =2.5\times 600\\ =1,500\text{lb}\cdot \text{in}\end{array}$

Find the moment about y axis as follows:

$M_y=0$

Find the moment about z axis as follows:

$\begin{array}{c}{M}_z=-2.5\times 600\\ =-1,500\text{lb}\cdot \text{in}\end{array}$

Find the value of radius (c) using the relation:

$c=\frac{d}{2}$ (1)

Here, d is the diameter of bracket.

Substitute $0.8\text{in}\text{.}$ for d in Equation (1).

$\begin{array}{c}c=\frac{0.8}{2}\\ =0.4\text{in}\text{.}\end{array}$

Find the area (A) of the circular section using the equation:

$A=\pi c^2$ (2)

Here, c is the half of the diameter.

Substitute $0.4\text{in}\text{.}$ for c in Equation (2).

$\begin{array}{c}A=\pi {\left(0.4\right)}^2\\ =0.50265{\text{in}}^{\text{2}}\end{array}$

Find the moment of inertia (I) of section using the relation:

$I=\frac{\pi }{4}{c}^4$ (3)

Substitute $0.4\text{in}\text{.}$ for c in Equation (3).

$\begin{array}{c}I=\frac{\pi }{4}{\left(0.4\right)}^4\\ =20.106\times {10}^{-3}{\text{in}}^{\text{4}}\end{array}$

Find the moment of inertia (J) of section using the relation:

$J=2I$ (4)

Substitute $20.106\times {10}^{-3}{\text{in}}^{\text{4}}$ for I in Equation (4).

$\begin{array}{c}J=2\left(20.106\times {10}^{-3}\right)\\ =40.106\times {10}^{-3}{\text{in}}^{\text{4}}\end{array}$

Find the value of Q for semicircle using the relation:

$Q=\frac{2}{3}{c}^3$ (5)

Substitute $0.4\text{in}\text{.}$ for c in Equation (5).

$\begin{array}{c}Q=\frac{2}{3}{\left(0.4\right)}^3\\ =42.667\times {10}^{-3}{\text{in}}^{\text{3}}\end{array}$

Determine the normal stress at point H using the relation:

${\sigma }_H=\frac{P}{A}+\frac{Mc}{I}$ (6)

Here, P is the centric force, A is the area of circular cross section, I is the moment of inertia, M is the moment, and c is the centroid distance.

Substitute $-2,500\text{lb}$ for P, $0.50265{\text{in}}^{\text{2}}$ for A, $20.106\times {10}^{-3}{\text{in}}^{\text{4}}$ for I, $0.4\text{in}\text{.}$ for c, and $1,500\text{lb}\cdot \text{in}$ for M in Equation (6).

$\begin{array}{c}{\sigma }_H=-\frac{2,500}{0.50265}+\frac{1,500\times 0.4}{20.106\times {10}^{-3}}\\ =-4,973.639+29,841.838\\ =-24,868.199\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\\ =24.87\text{ksi}\end{array}$

Determine the shear stress at point H using the relation:

${\tau }_H=\frac{Tc}{J}$ (7)

Here, T is the Torque and J is the polar moment of inertia.

Substitute $1,500\text{lb}$ for T, $40.106\times {10}^{-3}{\text{in}}^{\text{4}}$ for J, and $0.4\text{in}\text{.}$ for c, in Equation (7).

$\begin{array}{c}{\tau }_H=\frac{1,500\times 0.4}{40.106\times {10}^{-3}}\\ =\frac{600}{40.106\times {10}^{-3}}\\ =14,960.355\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\\ =14.92\text{ksi}\end{array}$

Sketch the stresses at point H as shown in Figure 2.

Find the average $\left({\sigma }_{ave}\right)$ normal stress at point H using the relation:

${\sigma }_{\text{ave}}=\frac{{\sigma }_H}{2}$ (8)

Here, normal stress at point H.

Substitute $24.87\text{ksi}$ for ${\sigma }_H$ in Equation (8).

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{24.87}{2}\\ =12.435\text{ksi}\end{array}$

Find the R using the relation:

$R=\sqrt{{\left(\frac{{\sigma }_H}{2}\right)}^2+{\tau }_H^2}$ (9)

Here, shear stress at point H.

Substitute $24.87\text{ksi}$ for ${\sigma }_H$ and $14.92\text{ksi}$ for ${\tau }_H$ in Equation (9).

$\begin{array}{c}R=\sqrt{{\left(\frac{24.87}{2}\right)}^2+{14.92}^2}\\ =\sqrt{154.629+222.6064}\\ =19.423\text{ksi}\end{array}$

Determine the maximum principal stress $\left({\sigma }_{\max }\right)$ using the relation:

${\sigma }_{\max }={\sigma }_{\text{ave}}+R$ (10)

Substitute $12.435\text{ksi}$ for ${\sigma }_{\text{ave}}$ and $19.423\text{ksi}$ for R in Equation (10).

$\begin{array}{c}{\sigma }_{\max }=12.435+19.423\\ =31.858\\ =31.9\text{ksi}\end{array}$

Thus, the maximum principal stress at point H is $\underset{\_}{31.9\text{ksi}}$.

Determine the minimum principal stress $\left({\sigma }_{\min }\right)$ at point H using the relation:

${\sigma }_{\min }={\sigma }_{\text{ave}}-R$ (11)

Substitute $12.435\text{ksi}$ for ${\sigma }_{\text{ave}}$ and $19.423\text{ksi}$ for R in Equation (11).

$\begin{array}{c}{\sigma }_{\min }=12.435-19.423\\ =-6.99\text{ksi}\end{array}$

Thus, the minimum principal stress at H is $\underset{\_}{-\text{6}\text{.99}\text{ksi}}$.

Determine the maximum shear stress at point H using the relation:

${\tau }_{\max }=\frac{1}{2}\left({\sigma }_{\max }-{\sigma }_{\min }\right)$ (11)

Here, ${\sigma }_{\max }$ is the principal maximum stress and ${\sigma }_{\min }$ is the principal minimum stress.

Substitute $-6.99\text{ksi}$ for ${\sigma }_{\min }$ and $31.9\text{ksi}$ for ${\sigma }_{\max }$ in Equation (11).

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(31.9-\left(-6.99\right)\right)\\ =19.44\text{ksi}\end{array}$

Thus, the shear stress at point H is $\underset{\_}{19.42\text{ksi}}$.

To determine

The principal stresses and the maximum shearing stress at point K.

Answer to Problem 72RP

The maximum principal stress at point K is $\underset{\_}{14.21\text{ksi}}$.

The minimum principal stress at K is $\underset{\_}{-\text{19}\text{.18}\text{ksi}}$.

The shear stress at point K is $\underset{\_}{16.70\text{ksi}}$.

Explanation of Solution

Calculation:

Determine the normal stress at point K using the relation:

${\sigma }_K=\frac{P}{A}$ (12)

Substitute $-2,500\text{lb}$ for P and $0.50265{\text{in}}^{\text{2}}$ for A, in Equation (12).

$\begin{array}{c}{\sigma }_K=-\frac{2,500}{0.50265}\\ =-4,973.639\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\\ =-4.974\text{ksi}\end{array}$

Determine the shear stress at point K using the relation:

${\tau }_K=\frac{Tc}{J}+\frac{VQ}{It}$ (13)

Substitute $1,500\text{lb}$ for T, $40.106\times {10}^{-3}{\text{in}}^{\text{4}}$ for J, $60\text{lb}$ for V, $42.667\times {10}^{-3}{\text{in}}^{\text{3}}$ for Q, $20.106\times {10}^{-3}{\text{in}}^{\text{4}}$ for I, $0.8\text{in}\text{.}$ for t, and $0.4\text{in}\text{.}$ for c, in Equation (13).

$\begin{array}{c}{\tau }_H=\frac{1,500\times 0.4}{40.106\times {10}^{-3}}+\frac{600\times 42.667\times {10}^{-3}}{20.106\times {10}^{-3}\times 0.8}\\ =14,960.355+1,591.577\\ =16,551.932\text{psi}\left(\frac{1\text{ksi}}{{10}^3\text{psi}}\right)\\ =16.512\text{ksi}\end{array}$

Sketch the stresses at point K as shown in Figure 3.

Find the average $\left({\sigma }_{ave}\right)$ normal stress using the relation:

${\sigma }_{\text{ave}}=\frac{{\sigma }_K}{2}$ (14)

Here, normal stress at point H.

Substitute $-4.974\text{ksi}$ for ${\sigma }_K$ in Equation (14).

$\begin{array}{c}{\sigma }_{\text{ave}}=\frac{-4.974}{2}\\ =-2.487\text{ksi}\end{array}$

Find the R using the relation:

$R=\sqrt{{\left(\frac{{\sigma }_H}{2}\right)}^2+{\tau }_H^2}$ (15)

Here, shear stress at point H.

Substitute $-4.974\text{ksi}$ for ${\sigma }_K$ and $16.512\text{ksi}$ for ${\tau }_K$ in Equation (15).

$\begin{array}{c}R=\sqrt{{\left(\frac{-4.974}{2}\right)}^2+{16.512}^2}\\ =\sqrt{6.185+272.646}\\ =16.698\text{ksi}\end{array}$

Determine the maximum principal stress $\left({\sigma }_{\max }\right)$ at point K using the relation:

${\sigma }_{\max }={\sigma }_{\text{ave}}+R$ (16)

Substitute $-2.487\text{ksi}$ for ${\sigma }_{\text{ave}}$ and $16.698\text{ksi}$ for R in Equation (16).

$\begin{array}{c}{\sigma }_{\max }=-2.487+16.698\\ =14.21\text{ksi}\end{array}$

Thus, the maximum principal stress at point K is $\underset{\_}{14.21\text{ksi}}$.

Determine the minimum principal stress $\left({\sigma }_{\min }\right)$ at point K using the relation:

${\sigma }_{\min }={\sigma }_{\text{ave}}-R$ (17)

Substitute $-2.487\text{ksi}$ for ${\sigma }_{\text{ave}}$ and $16.698\text{ksi}$ for R in Equation (17).

$\begin{array}{c}{\sigma }_{\min }=-2.487-16.698\\ =-19.18\text{ksi}\end{array}$

Thus, the minimum principal stress at K is $\underset{\_}{-\text{19}\text{.18}\text{ksi}}$.

Determine the maximum shear stress at point K using the relation:

${\tau }_{\max }=\frac{1}{2}\left({\sigma }_{\max }-{\sigma }_{\min }\right)$ (18)

Here, ${\sigma }_{\max }$ is the principal maximum stress at point K and ${\sigma }_{\min }$ is the principal minimum stress at point K.

Substitute $-19.18\text{ksi}$ for ${\sigma }_{\min }$ and $14.21\text{ksi}$ for ${\sigma }_{\max }$ in Equation (18).

$\begin{array}{c}{\tau }_{\max }=\frac{1}{2}\left(14.21-\left(-19.18\right)\right)\\ =16.70\text{ksi}\end{array}$

Thus, the shear stress at point K is $\underset{\_}{16.70\text{ksi}}$.