Chapter 8.3, Problem 54P

Three steel plates, each 13 mm thick, are welded together to form a cantilever beam. For the loading shown, determine the normal and shearing stresses at points d and e.

Fig. P8.53 and P8.54

To determine

The normal and shearing stress at point a and b.

Answer to Problem 54P

The normal stress at point d is $\underset{\_}{42.2\text{MPa}}$.

The shear stress at point d is $\underset{\_}{2.83\text{MPa}}$.

The normal stress at point e is $\underset{\_}{12.74\text{MPa}}$.

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

Explanation of Solution

Given information:

The thickness (t) of the steel plate is $13\text{mm}$.

Calculation:

Refer to Figure P8.51 in the textbook.

Forces at H are as follows:

The force in x direction, $F_x=9\text{kN}$.

The force in y direction, $F_y=-13\text{kN}$

The force in z direction, $F_z=0$

Moments at H are as follows:

Find the moment about x axis as follows:

$\begin{array}{c}{M}_x=\left(0.400\times 13\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\right)\\ =0.400\times 13,000\\ =5,200\text{N}\cdot \text{m}\end{array}$

Find the moment about y axis as follows:

$\begin{array}{c}{M}_y=\left(0.400\times 9\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\right)\\ =3,600\text{N}\cdot \text{m}\end{array}$

$M_z=0$

At point A:

Sketch the I section as shown in Figure 1.

Refer to Figure 1.

Find the centroid section $\overline{x}$ at point a as follows:

$\begin{array}{c}\overline{x}=\frac{75}{2}\\ =37.5\text{mm}\end{array}$

Find the centroid section $\overline{y}$ at point a as follows:

$\begin{array}{c}\overline{y}=\frac{13}{2}\\ =6.5\text{mm}\end{array}$

Find the area of the section using the relation:

$A=2{\left(b\times d\right)}_{\text{flange}}+{\left(b\times d\right)}_{\text{web}}$ (1)

Here, b is the width and d is the depth of the flange and web respectively.

Refer to Figure 1.

Substitute $150\text{mm}$ for $b_f$, $13\text{mm}$ for $d_f$, $13\text{mm}$ for $b_w$, and $49\text{mm}$ for $d_w$ in Equation (1).

$\begin{array}{c}A=2\times 150\left(13\right)+13\times 49\\ =3,900+637\\ =4,357{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =4537\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the moment of inertia $\left(I_x\right)$ about x axis as follows:

$I_x=2\left[\frac{b_f{d}_f^3}{12}+b_f{d}_f{\left(\overline{x}-\overline{y}\right)}^2\right]+\frac{b_w{d}_w^3}{12}$ (2)

Substitute $150\text{mm}$ for $b_f$, $13\text{mm}$ for $d_f$, $13\text{mm}$ for $b_w$, $37.5\text{mm}$ for $\overline{x}$, $6.5\text{mm}$ for $\overline{y}$ and $49\text{mm}$ for $d_w$ in Equation (2).

$\begin{array}{c}{I}_x=2\left[\frac{150\times {13}^3}{12}+150\times 13{\left(37.5-6.5\right)}^2\right]+\frac{13\times {49}^3}{12}\\ =2\left[27,462.5+1,873,950\right]+127,453.08\\ =3,930,278.08{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =3.9303\times {10}^{-6}{\text{m}}^{\text{4}}\end{array}$

Find the moment of inertia $\left(I_y\right)$ about y axis as follows:

$I_y=2\left[\frac{d_f{b}_f^3}{12}\right]+\frac{b_w^3{d}_w}{12}$ (3)

Substitute $150\text{mm}$ for $b_f$, $13\text{mm}$ for $d_f$, $13\text{mm}$ for $b_w$, and $49\text{mm}$ for $d_w$ in Equation (3).

$\begin{array}{c}{I}_y=2\left[\frac{13\times {150}^3}{12}\right]+\frac{49\times {13}^3}{12}\\ =7,312,500+8,971.08\\ =7,321,471.083{\text{mm}}^{\text{4}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^4\\ =7.3215\times {10}^{-6}{\text{m}}^{\text{4}}\end{array}$

At point d:

Sketch the I section for point b as shown in Figure 2.

Refer to Figure 2.

Find the area of section (A) as follows:

$A=bd$ (4)

Substitute $60\text{mm}$ for b and $13\text{mm}$ for d in Equation (4).

$\begin{array}{c}A=60\times 13\\ =780{\text{mm}}^{\text{2}}\end{array}$

Refer to Figure 2.

The centroid of the $\overline{x}$ is $45\text{mm}$.

The centroid of the $\overline{y}$ is $31\text{mm}$.

Find the first moment area $\left(Q_x\right)$ at point b as follows:

$Q_x=A\overline{y}$ (5)

Substitute $780{\text{mm}}^{\text{2}}$ for A and $31\text{mm}$ for $\overline{y}$ in Equation (5).

$\begin{array}{c}{Q}_x=780\times 31\\ =24,180{\text{mm}}^{\text{3}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^3\\ =24.18\times {10}^{-6}{\text{m}}^{\text{3}}\end{array}$

Find the first moment area $\left(Q_y\right)$ at point b as follows:

$Q_y=A\overline{x}$ (6)

Substitute $780{\text{mm}}^{\text{2}}$ for A and $45\text{mm}$ for $\overline{y}$ in Equation (6).

$\begin{array}{c}{Q}_y=780\times \left(45\right)\\ =35,100{\text{mm}}^{\text{3}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^3\\ =35.1\times {10}^{-6}{\text{m}}^{\text{3}}\end{array}$

Find the first moment area $\left(Q_x\right)$ and $\left(Q_y\right)$ at point e as follows:

The point e is located at edge. Since $\left(Q_x\right)$ is zero.

The point e is located at edge. Since $\left(Q_y\right)$ is zero.

Determine the normal stress at point d using the relation:

$\sigma =\frac{M_{x}y}{I_x}-\frac{M_{y}x}{I_y}$ (7)

Here, $M_x$ and $M_y$ is couple moments, $I_y$ is moment of inertia about y axis, $I_z$ is moment of inertia about z axis, and x is the distance.

Substitute $5,200\text{N}\cdot \text{m}$ for $M_x$, $37.5\text{mm}$ for y, $7.3215\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I_y$, $3.9303\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I_x$, $3,600\text{N}\cdot \text{m}$ for $M_y$, and $\text{15}\text{mm}$ for x in Equation (7).

$\begin{array}{c}\sigma =\frac{5,200\times 37.5\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{3.9303\times {10}^{-6}}-\frac{3,600\times 15\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{7.3215\times {10}^{-6}}\\ =\frac{5,200\times 0.0375}{3.9303\times {10}^{-6}}-\frac{3,600\times \left(0.015\right)}{7.3215\times {10}^{-6}}\\ =49,614,533.24-7,375,537.8\\ =42,238,995\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\sigma =42.2\text{MPa}$

Thus, the normal stress at point d is $\underset{\_}{42.2\text{MPa}}$.

Determine the shear stress at point a due to $V_x$ using the relation:

$\tau =\frac{\left|V_x\right|\left|Q_y\right|}{I_{y}t}$ (8)

Substitute $9\text{kN}$ for $V_x$, $35.1\times {10}^{-6}{\text{m}}^{\text{3}}$ for $Q_y$, and $7.3215\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I_y$, and $13\text{mm}$ for t in Equation (8).

$\begin{array}{c}\tau =\frac{9\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)35.1\times {10}^{-6}}{7.3215\times {10}^{-6}\left(13\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\frac{9000\times 35.1\times {10}^{-6}}{7.3215\times {10}^{-6}\times 0.01}\\ =33,318,992.01\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =3.32\text{MPa}\end{array}$

Sketch the horizontal direction of shear stress as shown in figure 3.

Determine the shear stress at point a due to $V_y$ using the relation:

$\tau =\frac{\left|V_y\right|\left|Q_x\right|}{I_{x}t}$ (8)

Substitute $13\text{kN}$ for $V_x$, $24.18\times {10}^{-6}{\text{m}}^{\text{3}}$ for $Q_x$, and $3.9303\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I_x$, and $13\text{mm}$ for t in Equation (8).

$\begin{array}{c}\tau =\frac{13\text{kN}\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)24.18\times {10}^{-6}}{3.9303\times {10}^{-6}\left(13\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}\\ =\frac{13000\times 24.18\times {10}^{-6}}{3.9303\times {10}^{-6}\left(0.013\right)}\\ =6,152,202.12\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =6.15\text{MPa}\end{array}$

Sketch the vertical direction of shear stress as shown in figure 4.

Find the shear stress at point d using superposition method.

$\begin{array}{c}\tau =6.15-3.32\\ =2.83\text{MPa}\end{array}$

Thus, the shear stress at point d is $\underset{\_}{2.83\text{MPa}}$.

Determine the normal stress at point e using the relation:

$\sigma =\frac{M_{x}y}{I_x}-\frac{M_{y}x}{I_y}$ (8)

Substitute $5,200\text{N}\cdot \text{m}$ for $M_x$, $37.5\text{mm}$ for y, $7.3215\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I_y$, $3.9303\times {10}^{-6}{\text{m}}^{\text{4}}$ for $I_x$, $3,600\text{N}\cdot \text{m}$ for $M_y$, and $75\text{mm}$ for x in Equation (8).

$\begin{array}{c}\sigma =\frac{5,200\times 37.5\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{3.9303\times {10}^{-6}}-\frac{3,600\times 75\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{7.3215\times {10}^{-6}}\\ =\frac{5,200\times 0.0375}{3.9303\times {10}^{-6}}-\frac{3,600\times \left(0.075\right)}{7.3215\times {10}^{-6}}\\ =49,614,533.24-36,877,689\\ =12,736,844\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\end{array}$

$\sigma =12.74\text{MPa}$

Thus, the normal stress at point b is $\underset{\_}{12.74\text{MPa}}$.

Determine the shear stress at point e using the relation:

The point e is located on edge.

The shear stress at point e is zero.