Chapter 8.3, Problem 48P

Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

Fig. P8.47

8.48 Solve Prob. 8.47, assuming that the 750-N force is directed vertically upward.

To determine

The normal and shearing stress at point a.

Answer to Problem 48P

The normal stress at point a is $\underset{\_}{-7.98\text{MPa}}$.

The shear stress at point a is $\underset{\_}{0.391\text{MPa}}$.

Explanation of Solution

Given information:

Assume $750\text{N}$ directly vertically upward.

Calculation:

At point A:

Find the area of cross section $\left(A\right)$ using the equation:

$A=bh$ (1)

Here, b is the width of the bar and h is the height of the bar.

Substitute $60\text{mm}$ for b and $32\text{mm}$ for h in Equation (1).

$\begin{array}{c}A=60\times 32\\ =1.920{\text{mm}}^2{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =1920\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the moment of inertia $\left(I_z\right)$ of section using the relation:

$I_z=\frac{b{h}^3}{12}$ (2)

Substitute $60\text{mm}$ for b and $32\text{mm}$ for h in Equation (2).

$\begin{array}{c}{I}_z=\frac{60\times {32}^3}{12}\\ =\frac{1,966,080}{12}\\ =163.84\times {10}^3{\text{mm}}^{\text{4}}\left(\frac{{10}^{-12}{\text{m}}^{\text{4}}}{1{\text{mm}}^{\text{4}}}\right)\\ =163.84\times {10}^{-9}{\text{m}}^{\text{4}}\end{array}$

Find the moment of inertia $\left(I_y\right)$ of section using the relation:

$I_y=\frac{b{h}^3}{12}$ (3)

Substitute $32\text{mm}$ for b and $60\text{mm}$ for h in Equation (3).

$\begin{array}{c}{I}_y=\frac{32{\left(60\right)}^3}{12}\\ =\frac{6,912,000}{12}\\ =576\times {10}^3{\text{mm}}^{\text{4}}\left(\frac{{10}^{-12}{\text{m}}^{\text{4}}}{1{\text{mm}}^4}\right)\\ =576\times {10}^{-9}{\text{m}}^{\text{4}}\end{array}$

Sketch the side view of bar as shown in Figure 1.

At the section containing point a, b, and c.

Refer to Figure 1.

$\begin{array}{l}P=10\text{kN}\\ V_y=750\text{N}\\ V_z=500\text{N}\end{array}$

Find the moment about z axis as follows:

$\begin{array}{c}{M}_Z=\left(180\times {10}^{-3}\right)\left(-750\right)\\ =-135\text{N}\cdot \text{m}\end{array}$

Sketch the side view of bar as shown in Figure 2.

Find the moment about y axis as follows:

$\begin{array}{c}{M}_y=\left(220\times {10}^{-3}\right)\left(500\right)\\ =110\text{N}\cdot \text{m}\end{array}$

$T=0$

Find the normal stress $\left(\sigma \right)$ at point a using the formula as follows:

$\sigma =\frac{P}{A}+\frac{M_{z}y}{I_z}-\frac{M_{y}z}{I_y}$ (4)

Here, P is the centric force, A is the area of rectangular cross section, $I_y$ moment of inertia about y axis, $I_z$ is moment of inertia about z axis, y is the distance from point a, $M_y$ and $M_z$ is bending couples, and z is distance from z axis.

Substitute $10\text{kN}$ for P, $1920\times {10}^{-6}{\text{m}}^{\text{2}}$ for A, $110\text{N}\cdot \text{m}$ for $M_y$, $-135\text{N}\cdot \text{m}$ for $M_z$, $576\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_y$, $163.84\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_z$, $16\text{mm}$ for y, and $0$ for z in Equation (4).

$\begin{array}{c}\sigma =\frac{10\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)}{1920\times {10}^{-6}}+\frac{-135\times 16\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{163.84\times {10}^{-9}}+\frac{110\times 0}{576\times {10}^{-9}}\\ =\frac{10\times {10}^3}{1920\times {10}^{-6}}-\frac{135\times 0.016}{163.84\times {10}^{-9}}\\ =5,208,333.333-13,183,593.75+0\\ =-7,975,260.42\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\sigma =-7.98\text{MPa}$

Thus, the normal stress at point a is $\underset{\_}{-7.98\text{MPa}}$.

Sketch the cross section at point a as shown in figure 3.

Determine the first moment area (Q) as follows:

$Q=A^{\prime }\overline{z}$ (5)

Here, $A^{\prime }$ is the cross section area and $\overline{z}$ is the distance from neutral axis.

Refer to Figure 2.

Substitute $\left(32\times 30\right){\text{mm}}^2$ for $A^{\prime }$ and $15\text{mm}$ for $\overline{y}$ in Equation (5).

$\begin{array}{c}Q=\left(32\times 30\right)\times 15\\ =14,400{\text{mm}}^{\text{3}}\end{array}$

Find the shear stress $\left(\tau \right)$ at point a using the formula as follows:

$\tau =\frac{V_{z}Q}{I_{y}t}$ (6)

Here, $V_z$ is shear about z axis, Q is the first moment area, and t is the thickness of the bar.

Substitute $500\text{N}$ for $V_z$, $14,400{\text{mm}}^{\text{3}}$ for Q, $576\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_y$, and $32\text{mm}$ for t in Equation (6).

$\begin{array}{c}\tau =\frac{500\times 14.4\times {10}^3{\text{mm}}^{\text{3}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^3}{576\times {10}^{-9}\times 32\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =\frac{7.2\times {10}^{-3}}{1.8432\times {10}^{-9}}\\ =390,625\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\tau =0.391\text{MPa}$

Thus, the shear stress at point a is $\underset{\_}{0.391\text{MPa}}$.

To determine

The normal and shearing stresses at point b.

Answer to Problem 48P

The normal stress at point b is $\underset{\_}{-5.11\text{MPa}}$.

The shear stress at point b is $\underset{\_}{0.293\text{MPa}}$.

Explanation of Solution

Calculation:

At point b:

Find the normal stress $\left(\sigma \right)$ at point b using the formula as follows:

$\sigma =\frac{P}{A}+\frac{M_{z}y}{I_z}-\frac{M_{y}z}{I_y}$ (7)

Substitute $10\text{kN}$ for P, $1920\times {10}^{-6}{\text{m}}^{\text{2}}$ for A, $110\text{N}\cdot \text{m}$ for $M_y$, $-135\text{N}\cdot \text{m}$ for $M_z$, $576\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_y$, $163.84\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_z$, $16\text{mm}$ for y, and $-15\text{mm}$ for z in Equation (7).

$\begin{array}{c}\sigma =\frac{10\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)}{1920\times {10}^{-6}}-\frac{135\times 16\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{163.84\times {10}^{-9}}-\frac{110\times \left(-15\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}{576\times {10}^{-9}}\\ =\frac{10\times {10}^3}{1920\times {10}^{-6}}-\frac{135\times 0.016}{163.84\times {10}^{-9}}+2,864,583.33\\ =5,208,333.333-13,183,593.75+2,864,583.33\\ =-5,110,677.087\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\sigma =-5.11\text{MPa}$

Thus, the normal stress at point b is $\underset{\_}{-5.11\text{MPa}}$.

Sketch the cross section at point b as shown in figure 4.

Determine the first moment area (Q) as follows:

$Q=A^{\prime }\overline{z}$ (8)

Here, $A^{\prime }$ is the cross section area and $\overline{z}$ is the distance from neutral axis.

Refer to Figure 2.

Substitute $\left(32\times 15\right){\text{mm}}^2$ for $A^{\prime }$ and $22.5\text{mm}$ for $\overline{y}$ in Equation (8).

$\begin{array}{c}Q=\left(32\times 15\right)22.5\\ =10,800{\text{mm}}^{\text{3}}\end{array}$

Find the shear stress $\left(\tau \right)$ at point b using the formula as follows:

$\tau =\frac{V_{z}Q}{I_{y}t}$ (9)

Here, $V_z$ is shear about z axis, Q is the first moment area, and t is the thickness of the bar.

Substitute $500\text{N}$ for $V_z$,  $10,800{\text{mm}}^{\text{3}}$ for Q, $576\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_y$, and $32\text{mm}$ for t in Equation (9).

$\begin{array}{c}\tau =\frac{500\times 10.8\times {10}^3{\text{mm}}^{\text{3}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^3}{576\times {10}^{-9}\times 32\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}\\ =\frac{5.4\times {10}^{-3}}{1.8432\times {10}^{-9}}\\ =292,968.75\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\tau =0.293\text{MPa}$

Thus, the shear stress at point b is $\underset{\_}{0.293\text{MPa}}$.

To determine

The normal and shearing stresses at point c.

Answer to Problem 48P

The normal stress at point c is $\underset{\_}{-2.25\text{MPa}}$.

The shear stress at point c is $\underset{\_}{0}$.

Explanation of Solution

Calculation:

Find the normal stress $\left(\sigma \right)$ at point b using the formula as follows:

$\sigma =\frac{P}{A}+\frac{M_{z}y}{I_z}-\frac{M_{y}z}{I_y}$ (10)

Substitute $10\text{kN}$ for P, $1920\times {10}^{-6}{\text{m}}^{\text{2}}$ for A, $110\text{N}\cdot \text{m}$ for $M_y$, $-135\text{N}\cdot \text{m}$ for $M_z$, $576\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_y$, $163.84\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_z$, $16\text{mm}$ for y, and $-30\text{mm}$ for z in Equation (10).

$\begin{array}{c}\sigma =\frac{10\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)}{1920\times {10}^{-6}}-\frac{135\times 16\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{163.84\times {10}^{-9}}-\frac{110\times \left(-30\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}{576\times {10}^{-9}}\\ =\frac{10\times {10}^3}{1920\times {10}^{-6}}-\frac{135\times 0.016}{163.84\times {10}^{-9}}+5,729,166.67\\ =5,208,333.333-13,183,593.75+5,729,166.67\\ =-2,246,093.75\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\sigma =-2.25\text{MPa}$

Thus, the normal stress at point c is $\underset{\_}{24.1\text{MPa}}$.

Find the shear stress $\left(\tau \right)$ at point c using the formula as follows:

$\tau =\frac{V_{z}Q}{I_{y}t}$ (11)

The point c is edge on the cross section. Since Q is zero.

Substitute $500\text{N}$ for $V_z$, 0 for Q, $576\times {10}^{-9}{\text{m}}^{\text{4}}$ for $I_y$, and $32\text{mm}$ for t in Equation (11).

$\begin{array}{c}\tau =\frac{500\times 0}{576\times {10}^{-9}\times 32}\\ =0\end{array}$

Thus, the shear stress at point c is $\underset{\_}{0}$.