Chapter 3, Problem 34P

For each section illustrated, find the second moment of area, the location of the neutral axis, and the distances from the neutral axis to the top and bottom surfaces. Consider that the section is transmitting a positive bending moment about the z axis. M_z, where M_z = 10 kip · in if the dimensions of the section are given in ips units, or M_z = 1.13 kN · m if the dimensions are in SI units. Determine the resulting stresses at the top and bottom surfaces and at every abrupt change in the cross section.

Problem 3–34

(a)

(b)

(c) Dimensions in mm

(d)

To determine

The second moment of area of the section.

The distance from the neutral axis to the top surfaces.

The distance from the neutral axis to the bottom surfaces.

The resulting stress at the top surface.

The resulting stress at the bottom surface.

The resulting stress at the abrupt change in cross-section.

Answer to Problem 34P

The second moment of area of the section is $1317708.34{\text{mm}}^4$.

The distance from the neutral axis to the top surface is $37.5\text{mm}$.

The distance from the neutral axis to the bottom surface is $37.5\text{mm}$.

The resulting stress at the top surface is $32.16\text{MPa}$.

The resulting stress at the bottom surface is $32.16\text{MPa}$.

The resulting stress at the abrupt change in cross-section is $10.72\text{MPa}$.

Explanation of Solution

Figure-(1) shows two different sections divided of the in the same diagram.

Figure-(1)

Calculate the second moment of area of the section.

  $I=I_a-2I_b$ (I)

Here, the second moment area of the area $a$ is $I_a$ and the second moment of area of the area $b$ is $I_b$.

Write the moment of inertia of the rectangular section.

    $I=\frac{B{H}^3}{12}$ (II)

Calculate the location of the neutral axis.

    $\overline{y}=\frac{H}{2}$ (III)

Calculate the location of the neutral axis.

    $\overline{y}=\frac{{\displaystyle \sum A_i{y}_i}}{{\displaystyle \sum A_i}}$ (IV)

Here, the area of the each section is $A_i$ and the distance from the bottom axis to centroidal axis of each axis is $y_i$.

Write the resulting bending stress on beam.

  $\sigma =\frac{M_{z}c}{I}$ (V)

Here, the distance from the neutral axis to the top or bottom surface is $c$ and the bending moment about $z$-axis.

Conclusion:

Substitute $40\text{mm}$ for $B$ and $75\text{mm}$ for $H$ in Equation (II).

    $\begin{array}{c}{I}_1=\frac{\left(40\text{mm}\right){\left(75\text{mm}\right)}^3}{12}\\ =\frac{\left(40\times {75}^3\right)}{12}{\left(\text{mm}\right)}^4\\ =1406250{\text{mm}}^4\end{array}$

Substitute $34\text{mm}$ for $B$ and $25\text{mm}$ for $H$ in Equation (II).

    $\begin{array}{c}{I}_2=\frac{\left(34\text{mm}\right){\left(25\text{mm}\right)}^3}{12}\\ =\frac{34\times {25}^4}{12}{\left(\text{mm}\right)}^4\\ =44270.83{\text{mm}}^4\end{array}$

Substitute $1406250{\text{mm}}^4$ for $I_1$ and $44270.83{\text{mm}}^4$ for $I_2$ in Equation (I).

    $\begin{array}{c}I=1406250{\text{mm}}^4-2\times 44270.83{\text{mm}}^4\\ =\left[\left(1406250\right)\times \left(2\times 44270.83\right)\right]{\left(\text{mm}\right)}^4\\ =1317708.34{\text{mm}}^4\end{array}$

Thus, the second moment of area of the section is $1317708.34{\text{mm}}^4$.

Substitute $75\text{mm}$ for $H$ in Equation (III).

    $\begin{array}{c}\overline{y}=\frac{75\text{mm}}{2}\\ =\frac{75}{2}\left(\text{mm}\right)\\ =37.5\text{mm}\end{array}$

Since the section is symmetrical about $z$-axis so the distance from the neutral axis to the top and bottom surface is equal.

Thus, the distance from the neutral axis to the top surface is $37.5\text{mm}$.

Thus, the distance from the neutral axis to the bottom surface is $37.5\text{mm}$.

Substitute $1.13\text{kN}\cdot \text{m}$ for $M_z$, $37.5\text{mm}$ for $c$ and $1317708.34{\text{mm}}^4$ for $I$ in Equation (V).

    $\begin{array}{c}\sigma =\frac{\left(1.13\text{kN}\cdot \text{m}\right)\times \left(37.5\text{mm}\right)}{1317708.34{\text{mm}}^4}\\ =\frac{\left(1.13\text{kN}\cdot \text{m}\times \frac{{10}^6\text{N}\cdot \text{mm}}{1\text{kN}\cdot \text{m}}\right)\times \left(37.5\text{mm}\right)}{1317708.34{\text{mm}}^4}\\ =32.16\text{MPa}\end{array}$

Since top and bottom surfaces are at the same distance from the neutral axis so the resulting stresses are same.

Thus, the resulting stress at the top surface is $32.16\text{MPa}$.

Thus, the resulting stress at the bottom surface is $32.16\text{MPa}$.

Substitute $1.13\text{kN}\cdot \text{m}$ for $M_z$, $12.5\text{mm}$ for $c$ and $1317708.34{\text{mm}}^4$ for $I$ in Equation (V).

    $\begin{array}{c}\sigma =\frac{\left(1.13\text{kN}\cdot \text{m}\right)\times \left(12.5\text{mm}\right)}{1317708.34{\text{mm}}^4}\\ =\frac{\left(1.13\text{kN}\cdot \text{m}\times \frac{{10}^6\text{N}\cdot \text{mm}}{1\text{kN}\cdot \text{m}}\right)\times \left(12.5\text{mm}\right)}{1317708.34{\text{mm}}^4}\\ =10.72\text{MPa}\end{array}$

Thus the resulting stress at the abrupt change in cross-section is $10.72\text{MPa}$.

To determine

The second moment of area of the section.

The distance from the neutral axis to the top surface.

The distance from the neutral axis to the bottom surface.

The resulting stress at the point A.

The resulting stress at the point D.

The resulting stress at the abrupt change in cross-section at a point B.

The resulting stress at the abrupt change in cross-section at a point C.

Answer to Problem 34P

The second moment of area of the section is $0.44776{\text{in}}^4$.

The distance from the neutral axis to the top surface is $0.858\text{in}$.

The distance from the neutral axis to the bottom surface is $0.858\text{in}$.

The resulting stress at the point A is $19.15\text{ksi}$.

The resulting stress at the point C is $22.7\text{ksi}$.

The resulting stress at the abrupt change in cross-section at a point B is $7.99\text{ksi}$.

The resulting stress at the abrupt change in cross-section at a point C is $4.564\text{ksi}$.

Explanation of Solution

Figure-(2) shows two different sections divided of the in the same diagram.

Figure-(2)

Write the expression for the area of section 1.

    $A_1=b_1{h}_1$ (VI)

Here, the width of the section 1 is $b_1$, and the height of the section 1 is $h_1$.

Write the expression for the area of section 2.

    $A_2=b_2{h}_2$ (VII)

Here, the width of the section 2 is $b_2$, and the height of the section 2 is $h_2$.

Calculate the location of the neutral axis.

    $\overline{y}=\frac{H}{2}$ (VIII)

Calculate the location of the neutral axis.

    $\overline{y}=\frac{{\displaystyle \sum A_i{y}_i}}{{\displaystyle \sum A_i}}$ (IX)

Here, the area of the each section is $A_i$ and the distance from the bottom axis to centroidal axis of each axis is $y_i$.

Write the expression for the moment of inertia of section 1.

    $I_1=\frac{b_1{h}_1^3}{12}$ (X)

Write the expression for the moment of inertia of section 2.

    $I_2=\frac{b_2{h}_2^3}{12}$ (XI)

Write the expression for the total moment of inertia.

    $I=2\left(I_1+A_1{d}_1^2\right)+\left(I_2+A_2{d}_2^2\right)$ (XII)

Here, the distance of the neutral axis from the centroid of section 1 is $d_1$, and the distance of the neutral axis from the centroid of section 2 is $d_2$.

Write the resulting bending stress on beam.

  $\sigma =\frac{M_{z}c}{I}$ (XIII)

Here, the distance from the neutral axis to the top or bottom surface is $c$ and the bending moment about $z$-axis.

Conclusion:

Substitute $0.375\text{in}$ for $b_1$ and $1.875\text{in}$ for $h_1$ in Equation (VI).

    $\begin{array}{c}{A}_1=\left(0.375\text{in}\right)\times \left(1.875\text{in}\right)\\ =\left(0.375\times 1.875\right){\text{in}}^{\text{2}}\\ =0.703125{\text{in}}^{\text{2}}\end{array}$

Substitute $0.375\text{in}$ for $b_2$ and $1.75\text{in}$ for $h_2$ in Equation (VI).

    $\begin{array}{c}{A}_2=\left(0.375\text{in}\right)\times \left(1.75\text{in}\right)\\ =\left(0.375\times 1.75\right){\text{in}}^{\text{2}}\\ =0.65625{\text{in}}^{\text{2}}\end{array}$

Substitute $1.875\text{in}$ for $H$ in Equation (VIII).

    $\begin{array}{c}{\overline{y}}_1=\frac{1.875\text{in}}{2}\\ =\frac{1.875}{2}\left(\text{in}\right)\\ =0.9375\text{in}\end{array}$

Substitute $1.375\text{in}$ for $H$ in Equation (VIII).

    $\begin{array}{c}{\overline{y}}_2=\frac{1.375\text{in}}{2}\\ =\frac{1.375}{2}\left(\text{in}\right)\\ =0.6875\text{in}\end{array}$

Substitute $0.6875\text{in}$ for ${\overline{y}}_2$, $0.9375\text{in}$ for ${\overline{y}}_1$, $0.65625{\text{in}}^{\text{2}}$ for $A_2$ and $0.703125{\text{in}}^{\text{2}}$ for $A_1$ in Equation (IX).

    $\begin{array}{c}\overline{y}=\frac{2\left(0.703125{\text{in}}^2\times 0.9375\text{in}\right)+0.65625{\text{in}}^2\times 0.6875\text{in}}{2\times 0.703125{\text{in}}^2+0.65625{\text{in}}^2}\\ =\frac{\left(1.318359\right)+\left(0.65625\times 0.6875\right)}{2\times 0.703125+0.65625}\left(\text{in}\right)\\ =0.858\text{in}\end{array}$

Since the section is symmetrical about $y$-axis so the distance from the neutral axis to the right and left surface is equal.

Thus, the distance from the neutral axis to the top surface is $0.858\text{in}$.

Thus, the distance from the neutral axis to the bottom surface is $0.858\text{in}$.

Substitute $0.375\text{in}$ for $B$ and $1.875\text{in}$ for $H$ in Equation (X).

    $\begin{array}{c}{I}_1=\frac{\left(0.375\text{in}\right){\left(1.875\text{in}\right)}^3}{12}\\ =\frac{0.375\times {\left(1.875\right)}^3}{12}{\left(\text{in}\right)}^4\\ =0.206{\text{in}}^4\end{array}$

Substitute $1.75\text{in}$ for $B$ and $0.375\text{in}$ for $H$ in Equation (XI).

    $\begin{array}{c}{I}_2=\frac{\left(1.75\text{in}\right){\left(0.375\text{in}\right)}^3}{12}\\ =\frac{1.75\times {\left(0.375\right)}^3}{12}{\left(\text{in}\right)}^4\\ =0.00769{\text{in}}^4\end{array}$

Distance of the neutral axis from the two different centroids is $-0.0795\text{in}$ for $d_1$ and $0.1705\text{in}$ for $d_2$.

Substitute $0.206{\text{in}}^4$ for $I_1$, $0.65625{\text{in}}^{\text{2}}$ for $A_2$, $0.703125{\text{in}}^{\text{2}}$ for $A_1$, $-0.0795\text{in}$ for $d_1$, $0.1705\text{in}$ for $d_2$ and $0.00769{\text{in}}^4$ for $I_2$ in Equation (XII).

    $\begin{array}{c}I=\left[\begin{array}{l}2\left(0.206{\text{in}}^4+\left(0.703125{\text{in}}^{\text{2}}\right){\left(-0.0795\text{in}\right)}^2\right)\\ +\left(0.00769{\text{in}}^4+0.65625{\text{in}}^{\text{2}}{\left(0.1705\text{in}\right)}^2\right)\end{array}\right]\\ =\left(0.420{\text{in}}^4\right)+\left(0.02776{\text{in}}^4\right)\\ =0.44776{\text{in}}^4\end{array}$

Thus, the second moment of area of the section is $0.44776{\text{in}}^4$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $0.858\text{in}$ for $c$ and $0.44776{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_A=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(0.858\text{in}\right)}{0.44776{\text{in}}^4}\\ =\frac{8.58\text{kip}\cdot {\text{in}}^2}{0.44776{\text{in}}^4}\\ =19.15\text{ksi}\end{array}$

Thus, the resulting stress at the bottom surface at a point A is $19.15\text{ksi}$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $0.358\text{in}$ for $c$ and $0.44776{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_B=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(0.358\text{in}\right)}{0.44776{\text{in}}^4}\\ =\frac{3.58\text{kip}\cdot {\text{in}}^2}{0.44776{\text{in}}^4}\\ =7.99\text{ksi}\end{array}$

Thus, the resulting stress at the point B is $7.99\text{ksi}$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $0.2045\text{in}$ for $c$ and $0.44776{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_C=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(0.2045\text{in}\right)}{0.44776{\text{in}}^4}\\ =\frac{2.045\text{kip}\cdot {\text{in}}^2}{0.44776{\text{in}}^4}\\ =4.564\text{ksi}\end{array}$

Thus, the resulting stress at the a point C is $4.564\text{ksi}$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $1.017\text{in}$ for $c$ and $0.44776{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_B=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(1.017\text{in}\right)}{0.44676{\text{in}}^4}\\ =\frac{10.17\text{kip}\cdot {\text{in}}^2}{0.44676{\text{in}}^4}\\ =22.7\text{ksi}\end{array}$

Thus, the resulting stress at the a point D is $22.7\text{ksi}$.

To determine

The second moment of area of the section.

The distance from the neutral axis to the top surface.

The distance from the neutral axis to the bottom surface.

The resulting stress at the point A.

The resulting stress at the point D.

The resulting stress at the abrupt change in cross-section at a point B.

The resulting stress at the abrupt change in cross-section at a point C.

Answer to Problem 34P

The second moment of area of the section is $4289227{\text{mm}}^4$.

The distance from the neutral axis to the top surface is $57.29\text{mm}$.

The distance from the neutral axis to the bottom surface is $57.29\text{mm}$.

The resulting stress at the point A is $15.09\text{MPa}$.

The resulting stress at the point D is $11.25\text{MPa}$.

The resulting stress at the abrupt change in cross-section at a point B is $11.79\text{MPa}$.

The resulting stress at the abrupt change in cross-section at a point C is $7.95\text{MPa}$.

Explanation of Solution

Figure-(3) shows three different sections divided of the section in the same diagram.

Figure-(3)

Write the expression for the area of section 1.

    $A_1=b_1{h}_1$ (XIV)

Here, the width of the section 1 is $b_1$, and the height of the section 1 is $h_1$.

Write the expression for the area of section 2.

    $A_2=b_2{h}_2$ (XV)

Here, the width of the section 2 is $b_2$, and the height of the section 2 is $h_2$.

Write the expression for the area of section 3.

    $A_3=b_3{h}_3$ (XVI)

Here, the width of the section 3 is $b_3$, and the height of the section 3 is $h_3$.

Calculate the location of the neutral axis.

    $\overline{y}=\frac{H}{2}$ (XVII)

Calculate the location of the neutral axis.

    $\overline{y}=\frac{{\displaystyle \sum A_i{y}_i}}{{\displaystyle \sum A_i}}$ (XVIII)

Here, the area of the each section is $A_i$ and the distance from the bottom axis to centroidal axis of each axis is $y_i$.

Write the expression for the moment of inertia of section 1.

    $I_1=\frac{b_1{h}_1^3}{12}$ (XIX)

Write the expression for the moment of inertia of section 2.

    $I_2=\frac{b_2{h}_2^3}{12}$ (XX)

Write the expression for the moment of inertia of section 3.

    $I_3=\frac{b_3{h}_3^3}{12}$ (XXI)

Write the expression for the total moment of inertia.

    $I=\left[\left(I_3+A_3{d}_3{}^2\right)-\left(I_1+A_1{d}_1^2\right)-\left(I_2+A_2{d}_2^2\right)\right]$ (XXII)

Here, the distance of the neutral axis from the centroid of section 1 is $d_1$, and the distance of the neutral axis from the centroid of section 2 is $d_2$.

Write the resulting bending stress on beam.

  $\sigma =\frac{M_{z}c}{I}$ (XXIII)

Here, the distance from the neutral axis to the top or bottom surface is $c$ and the bending moment about $z$-axis.

Conclusion:

Substitute $12.5\text{mm}$ for $b_1$ and $50\text{mm}$ for $h_1$ in Equation (XIV).

    $\begin{array}{c}{A}_1=\left(12.5\text{mm}\right)\times \left(50\text{mm}\right)\\ =\left(12.5\times 50\right){\text{mm}}^{\text{2}}\\ =625{\text{mm}}^{\text{2}}\end{array}$

Substitute $75\text{mm}$ for $b_2$ and $75\text{mm}$ for $h_2$ in Equation (XV).

    $\begin{array}{c}{A}_2=\left(75\text{mm}\right)\times \left(75\text{mm}\right)\\ =\left(75\times 75\right){\text{mm}}^{\text{2}}\\ =5625{\text{mm}}^{\text{2}}\end{array}$

Substitute $100\text{mm}$ for $b_2$ and $100\text{mm}$ for $h_2$ in Equation (XVI).

    $\begin{array}{c}{A}_3=\left(100\text{mm}\right)\times \left(100\text{mm}\right)\\ =\left(100\times 100\right){\text{mm}}^{\text{2}}\\ =10000{\text{mm}}^{\text{2}}\end{array}$

Substitute $12.5\text{mm}$ for $H$ in Equation (XVII).

    $\begin{array}{c}{\overline{y}}_1=\frac{12.5\text{mm}}{2}\\ =\frac{12.5}{2}\left(\text{mm}\right)\\ =6.25\text{mm}\end{array}$

Substitute $100\text{mm}$ for $H$ in Equation (XVII).

    $\begin{array}{c}{\overline{y}}_2=\frac{100\text{mm}}{2}\\ =\frac{100}{2}\left(\text{mm}\right)\\ =50\text{mm}\end{array}$

Substitute $100\text{mm}$ for $H$ in Equation (XVII).

    $\begin{array}{c}{\overline{y}}_3=\frac{100\text{mm}}{2}\\ =\frac{100}{2}\left(\text{mm}\right)\\ =50\text{mm}\end{array}$

Substitute $50\text{mm}$ for ${\overline{y}}_2$, $6.25\text{mm}$ for ${\overline{y}}_1$, $50\text{mm}$ for ${\overline{y}}_3$, $10000{\text{mm}}^{\text{2}}$ for $A_3$, $5625{\text{mm}}^{\text{2}}$ for $A_2$ and $625{\text{mm}}^{\text{2}}$ for $A_1$ in Equation (XVIII).

    $\begin{array}{c}\overline{y}=\frac{\left(10000{\text{mm}}^{\text{2}}\times 50\text{mm}\right)-\left(5625{\text{mm}}^{\text{2}}\times 50\text{mm}\right)-\left(625{\text{mm}}^{\text{2}}\times 6.25\text{mm}\right)}{10000{\text{mm}}^{\text{2}}-5625{\text{mm}}^{\text{2}}-625{\text{mm}}^{\text{2}}}\\ =\frac{\left(500000\right)-\left(281250\right)-\left(3906.25\right)}{3750}\left(\text{mm}\right)\\ =57.29\text{mm}\end{array}$

Since the section is symmetrical about $y$-axis so the distance from the neutral axis to the right and left surface is equal.

Thus, the distance from the neutral axis to the top surface is $57.29\text{mm}$.

Thus, the distance from the neutral axis to the bottom surface is $57.29\text{mm}$.

Substitute $50\text{mm}$ for $B$ and $12.5\text{mm}$ for $H$ in Equation (XIX).

    $\begin{array}{c}{I}_1=\frac{\left(50\text{mm}\right){\left(12.5\text{mm}\right)}^3}{12}\\ =\frac{50\times {\left(12.5\right)}^3}{12}{\left(\text{mm}\right)}^4\\ =8138{\text{mm}}^4\end{array}$

Substitute $75\text{mm}$ for $B$ and $75\text{mm}$ for $H$ in Equation (XX).

    $\begin{array}{c}{I}_2=\frac{\left(75\text{mm}\right){\left(75\text{mm}\right)}^3}{12}\\ =\frac{75\times {\left(75\right)}^3}{12}{\left(\text{mm}\right)}^4\\ =2.637\times {10}^6{\text{mm}}^4\end{array}$

Substitute $100\text{mm}$ for $B$ and $100\text{mm}$ for $H$ in Equation (XXI).

    $\begin{array}{c}{I}_3=\frac{\left(100\text{mm}\right){\left(100\text{mm}\right)}^3}{12}\\ =\frac{100\times {\left(100\right)}^3}{12}{\left(\text{mm}\right)}^4\\ =8.333\times {10}^6{\text{mm}}^4\end{array}$

Distance of the neutral axis from the three different centroids of section is $51.04\text{mm}$ for $d_1$, $7.29\text{mm}$ for $d_3$ and $7.29\text{mm}$ for $d_2$.

Substitute $8138{\text{mm}}^4$ for $I_1$, $2.637\times {10}^6{\text{mm}}^4$ for $I_2$, $5625{\text{mm}}^{\text{2}}$ for $A_2$, $10000{\text{mm}}^{\text{2}}$ for $A_3$, $625{\text{mm}}^{\text{2}}$ for $A_1$, $51.04\text{mm}$ for $d_1$, $7.29\text{mm}$ for $d_3$, $7.29\text{mm}$ for $d_2$.and $8.333\times {10}^6{\text{mm}}^4$ for $I_3$ in Equation (XXII).

    $\begin{array}{c}I=\left[\begin{array}{l}\left(8.333\times {10}^6{\text{mm}}^4+10000{\text{mm}}^{\text{2}}{\left(7.29\text{mm}\right)}^2\right)\\ -\left(8138{\text{mm}}^4+625{\text{mm}}^{\text{2}}{\left(7.29\text{mm}\right)}^2\right)\\ -\left(2.637\times {10}^6{\text{mm}}^4+5625{\text{mm}}^{\text{2}}\left(51.04{\text{mm}}^2\right)\right)\end{array}\right]\\ =\left[\left(8864441{\text{mm}}^4\right)-\left(41353.06{\text{mm}}^4\right)-\left(2924100{\text{mm}}^4\right)\right]\\ =4289227{\text{mm}}^4\end{array}$

Thus, the second moment of area of the section is $4289227{\text{mm}}^4$.

Substitute $1.13\text{kN}\cdot \text{m}$ for $M_z$, $57.29\text{mm}$ for $c$ and $4289227{\text{mm}}^4$ for $I$ in Equation (XXIII).

    $\begin{array}{c}{\sigma }_A=\frac{\left(1.13\text{kN}\cdot \text{m}\right)\times \left(57.29\text{mm}\right)}{4289227{\text{mm}}^4}\\ =\frac{\left(1.13\text{kN}\cdot \text{m}\times \frac{{10}^6\text{N}\cdot \text{mm}}{1\text{kN}\cdot \text{m}}\right)\times \left(57.29\text{mm}\right)}{4289227{\text{mm}}^4}\\ =15.09\text{MPa}\end{array}$

Thus, the resulting stress at the bottom surface at a point A is $15.09\text{MPa}$.

Substitute $1.13\text{kN}\cdot \text{m}$ for $M_z$, $44.79\text{mm}$ for $c$ and $4289227{\text{mm}}^4$ for $I$ in Equation (XXIII).

    $\begin{array}{c}{\sigma }_B=\frac{\left(1.13\text{kN}\cdot \text{m}\right)\times \left(44.79\text{mm}\right)}{4289227{\text{mm}}^4}\\ =\frac{\left(1.13\text{kN}\cdot \text{m}\times \frac{{10}^6\text{N}\cdot \text{mm}}{1\text{kN}\cdot \text{m}}\right)\times \left(44.79\text{mm}\right)}{4289227{\text{mm}}^4}\\ =11.79\text{MPa}\end{array}$

Thus, the resulting stress at the point B is $11.79\text{MPa}$.

Substitute $1.13\text{kN}\cdot \text{m}$ for $M_z$, $30.21\text{mm}$ for $c$ and $4289227{\text{mm}}^4$ for $I$ in Equation (XXIII).

    $\begin{array}{c}{\sigma }_C=\frac{\left(1.13\text{kN}\cdot \text{m}\right)\times \left(30.21\text{mm}\right)}{4289227{\text{mm}}^4}\\ =\frac{\left(1.13\text{kN}\cdot \text{m}\times \frac{{10}^6\text{N}\cdot \text{mm}}{1\text{kN}\cdot \text{m}}\right)\times \left(30.21\text{mm}\right)}{4289227{\text{mm}}^4}\\ =7.95\text{MPa}\end{array}$

Thus, the resulting stress at the a point C is $7.95\text{MPa}$.

Substitute $1.13\text{kN}\cdot \text{m}$ for $M_z$, $42.71\text{mm}$ for $c$ and $4289227{\text{mm}}^4$ for $I$ in Equation (XXIII).

    $\begin{array}{c}{\sigma }_D=\frac{\left(1.13\text{kN}\cdot \text{m}\right)\times \left(42.71\text{mm}\right)}{4289227{\text{mm}}^4}\\ =\frac{\left(1.13\text{kN}\cdot \text{m}\times \frac{{10}^6\text{N}\cdot \text{mm}}{1\text{kN}\cdot \text{m}}\right)\times \left(42.71\text{mm}\right)}{4289227{\text{mm}}^4}\\ =11.25\text{MPa}\end{array}$

Thus, the resulting stress at the a point D is $11.25\text{ksi}$.

To determine

The second moment of area of the section.

The distance from the neutral axis to the top surface.

The distance from the neutral axis to the bottom surface.

The resulting stress at the point A.

The resulting stress at the point D.

The resulting stress at the abrupt change in cross-section at a point B.

Answer to Problem 34P

The second moment of area of the section is $5.195{\text{in}}^4$.

The distance from the neutral axis to the top surface is $2.288\text{in}$.

The distance from the neutral axis to the bottom surface is $2.288\text{in}$.

The resulting stress at the point A is $4.404\text{ksi}$.

The resulting stress at the point C is $0.408\text{ksi}$.

The resulting stress at the abrupt change in cross-section at a point B is $2.092\text{ksi}$.

Explanation of Solution

Figure-(4) shows two different sections divided of the section shown in the same diagram.

Figure-(4)

Write the expression for the area of section 1.

    $A_1=b_1{h}_1$ (XXIV)

Here, the width of the section 1 is $b_1$, and the height of the section 1 is $h_1$.

Write the expression for the area of section 2.

    $A_2=b_2{h}_2$ (XXV)

Here, the width of the section 2 is $b_2$, and the height of the section 2 is $h_2$.

Calculate the location of the neutral axis.

    $\overline{y}=\frac{H}{2}$ (XXVI)

Calculate the location of the neutral axis.

    $\overline{y}=\frac{{\displaystyle \sum A_i{y}_i}}{{\displaystyle \sum A_i}}$ (XXVII)

Here, the area of the each section is $A_i$ and the distance from the bottom axis to centroidal axis of each axis is $y_i$.

Write the expression for the moment of inertia of section 1.

    $I_1=\frac{b_1{h}_1^3}{12}$ (XVIII)

Write the expression for the moment of inertia of section 2.

    $I_2=\frac{b_2{h}_2^3}{12}$ (XXIX)

Write the expression for the total moment of inertia.

    $I=2\left(I_1+A_1{d}_1^2\right)+\left(I_2+A_2{d}_2^2\right)$ (XXX)

Here, the distance of the neutral axis from the centroid of section 1 is $d_1$, and the distance of the neutral axis from the centroid of section 2 is $d_2$.

Write the resulting bending stress on beam.

  $\sigma =\frac{M_{z}c}{I}$ (XXXI)

Here, the distance from the neutral axis to the top or bottom surface is $c$ and the bending moment about $z$-axis.

Conclusion:

Substitute $4\text{in}$ for $b_1$ and $0.875\text{in}$ for $h_1$ in Equation (XXIV).

    $\begin{array}{c}{A}_1=\left(4\text{in}\right)\times \left(0.875\text{in}\right)\\ =\left(4\times 0.875\right){\text{in}}^{\text{2}}\\ =3.5{\text{in}}^{\text{2}}\end{array}$

Substitute $2.5\text{in}$ for $b_2$ and $0.875\text{in}$ for $h_2$ in Equation (XXV).

    $\begin{array}{c}{A}_2=\left(2.5\text{in}\right)\times \left(0.875\text{in}\right)\\ =\left(2.5\times 0.875\right){\text{in}}^{\text{2}}\\ =2.1875{\text{in}}^{\text{2}}\end{array}$

Substitute $2.9375\text{in}$ for $H$ in Equation (VIII).

    $\begin{array}{c}{\overline{y}}_1=\frac{1.875\text{in}}{2}\\ =\frac{1.875}{2}\left(\text{in}\right)\\ =0.9375\text{in}\end{array}$

Substitute $2.5\text{in}$ for $H$ in Equation (VIII).

    $\begin{array}{c}{\overline{y}}_2=\frac{2.5\text{in}}{2}\\ =\frac{2.5}{2}\left(\text{in}\right)\\ =1.25\text{in}\end{array}$

Substitute $1.25\text{in}$ for ${\overline{y}}_2$, $0.9375\text{in}$ for ${\overline{y}}_1$, $2.1875{\text{in}}^{\text{2}}$ for $A_2$ and $3.5{\text{in}}^{\text{2}}$ for $A_1$ in Equation (IX).

    $\begin{array}{c}\overline{y}=\frac{\left(3.5{\text{in}}^{\text{2}}\times 0.9375\text{in}\right)+\left(2.1875{\text{in}}^{\text{2}}\times 1.25\text{in}\right)}{3.5{\text{in}}^{\text{2}}+2.1875{\text{in}}^{\text{2}}}\\ =\frac{\left(3.28125\right)+\left(2.7343\right)}{3.5{\text{in}}^{\text{2}}+2.1875{\text{in}}^{\text{2}}}\left(\text{in}\right)\\ =2.288\text{in}\end{array}$

Since the section is symmetrical about $y$-axis so the distance from the neutral axis to the right and left surface is equal.

Thus, the distance from the neutral axis to the top surface is $2.288\text{in}$.

Thus, the distance from the neutral axis to the bottom surface is $2.288\text{in}$.

Substitute $4\text{in}$ for $B$ and $0.875\text{in}$ for $H$ in Equation (X).

    $\begin{array}{c}{I}_1=\frac{\left(4\text{in}\right){\left(0.875\text{in}\right)}^3}{12}\\ =\frac{4\times {\left(0.875\right)}^3}{12}{\left(\text{in}\right)}^4\\ =0.2233{\text{in}}^4\end{array}$

Substitute $0.875\text{in}$ for $B$ and $2.5\text{in}$ for $H$ in Equation (XI).

    $\begin{array}{c}{I}_2=\frac{\left(0.875\text{in}\right){\left(2.5\text{in}\right)}^3}{12}\\ =\frac{0.875\times {\left(2.5\right)}^3}{12}{\left(\text{in}\right)}^4\\ =1.139{\text{in}}^4\end{array}$

Distance of the neutral axis from the two different centroids is $0.6495\text{in}$ for $d_1$ and $1.038\text{in}$ for $d_2$.

Substitute $0.2233{\text{in}}^4$ for $I_1$, $2.1875{\text{in}}^{\text{2}}$ for $A_2$, $3.5{\text{in}}^{\text{2}}$ for $A_1$, $0.6495\text{in}$ for $d_1$, $1.038\text{in}$ for $d_2$ and $1.139{\text{in}}^4$ for $I_2$ in Equation (XII).

    $\begin{array}{c}I=\left[\begin{array}{l}\left(0.2233{\text{in}}^4+\left(3.5{\text{in}}^{\text{2}}\right){\left(0.6495\text{in}\right)}^2\right)\\ +\left(1.139{\text{in}}^4+2.1875{\text{in}}^{\text{2}}{\left(1.038\text{in}\right)}^2\right)\end{array}\right]\\ =\left(1.699{\text{in}}^4\right)+\left(3.495{\text{in}}^4\right)\\ =5.195{\text{in}}^4\end{array}$

Thus, the second moment of area of the section is $5.195{\text{in}}^4$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $2.288\text{in}$ for $c$ and $5.195{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_A=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(2.288\text{in}\right)}{5.195{\text{in}}^4}\\ =\frac{22.88\text{kip}\cdot {\text{in}}^2}{5.195{\text{in}}^4}\\ =4.404\text{ksi}\end{array}$

Thus, the resulting stress at the bottom surface at a point A is $4.404\text{ksi}$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $0.212\text{in}$ for $c$ and $5.195{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_B=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(0.212\text{in}\right)}{5.195{\text{in}}^4}\\ =\frac{2.12\text{kip}\cdot {\text{in}}^2}{5.195{\text{in}}^4}\\ =0.408\text{ksi}\end{array}$

Thus, the resulting stress at the point B is $0.408\text{ksi}$.

Substitute $10\text{kip}\cdot \text{in}$ for $M_z$, $1.087\text{in}$ for $c$ and $5.195{\text{in}}^4$ for $I$ in Equation (XIII).

    $\begin{array}{c}{\sigma }_C=\frac{\left(10\text{kip}\cdot \text{in}\right)\times \left(1.087\text{in}\right)}{5.195{\text{in}}^4}\\ =\frac{10.87\text{kip}\cdot {\text{in}}^2}{5.195{\text{in}}^4}\\ =2.092\text{ksi}\end{array}$

Thus, the resulting stress at the a point C is $2.092\text{ksi}$.