Chapter 12, Problem 27AP

The lintel of prestressed reinforced concrete in Figure P12.27 is 1.50 m long. The concrete encloses one steel reinforcing rod with cross-sectional area 1.50 cm². The rod joins two strong end plates. The cross-sectional area of the concrete perpendicular to the rod is 50.0 cm². Young’s modulus for the concrete is 30.0 × 10⁹ N/m². After the concrete cures and the original tension T₁ in the rod is released, the concrete is to be under compressive stress 8.00 × 10⁶ N/m². (a) By what distance will the rod compress the concrete when the original tension in the rod is released? (b) What is the new tension T₂ in the rod? (c) The rod will then be how much longer than its unstressed length? (d) When the concrete was poured, the rod should have been stretched by what extension distance from its unstressed length? (e) Find the required original tension T₁ in the rod.

Figure P12.27

To determine

The compressed length of the rod when the original tension in the rod is released.

Answer to Problem 27AP

The compressed length of the rod when the original tension in the rod is released is $0.400\text{mm}$.

Explanation of Solution

The length of the concrete is $1.50\text{}\text{m}$, cross-sectional area of the rod is $1.50\text{}{\text{cm}}^2$, cross-sectional area of the concrete is $50.0\text{}{\text{cm}}^2$, Young’s modulus of the concrete is $30.0\times {10}^9\text{N}/{\text{m}}^2$, and the compressive stress on concrete is $8.0\times {10}^6\text{N}/{\text{m}}^2$.

Formula to calculate the compressive strain is,

    $\text{Compressive}\text{strain}=\frac{\Delta l}{l}$

Here, $\Delta l$ is the compressive length and $l$ is the original length.

Formula to calculate the modulus of rigidity of the concrete is,

    $Y=\frac{\text{Compressive}\text{stress}}{\text{Compressive}\text{strain}}$

Substitute $\frac{\Delta l}{l}$ in the above equation to find $Y$.

    $Y=\frac{\text{Compressive}\text{stress}}{\frac{\Delta l}{l}}$

Substitute $30.0\times {10}^9\text{N}/{\text{m}}^2$ for $Y$, $8.0\times {10}^6\text{N}/{\text{m}}^2$ for compressive stress and $1.50\text{}\text{m}$ for $l$ in the above equation to find $\Delta l$.

    $\begin{array}{c}\left(30.0\times {10}^9\text{N}/{\text{m}}^2\right)=\frac{\left(8.0\times {10}^6\text{N}/{\text{m}}^2\right)}{\frac{\Delta l}{\left(1.50\text{}\text{m}\right)}}\\ \Delta l=\left(0.0004\text{m}\left(\frac{1000\text{mm}}{1\text{m}}\right)\right)\\ =0.4\text{mm}\end{array}$

Conclusion:

Therefore, the compressed length of the rod when the original tension in the rod is released is $0.400\text{mm}$.

To determine

The magnitude of the new tension in the rod.

Answer to Problem 27AP

The magnitude of the new tension in the rod is $40.0\text{kN}$.

Explanation of Solution

Formula to calculate the compressive stress is,

    $\text{Compressive}\text{stress}=\frac{T_2}{A_{\text{C}}}$

Here, $T_2$ is the new tension on the concrete and $A_C$ is the cross-sectional of the concrete.

Substitute $8.0\times {10}^6\text{N}/{\text{m}}^2$ for compressive stress, and $50.0\text{}{\text{cm}}^2$ for $A_{\text{C}}$ in the above equation to find $T_2$.

    $\begin{array}{c}\left(8.0\times {10}^6\text{N}/{\text{m}}^2\right)=\frac{T_2}{\left(50.0\text{}{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)}\\ T_2=\left(4000\text{N}\right)\left(\frac{1\text{kN}}{{10}^3\text{N}}\right)\\ =40.0\text{kN}\end{array}$

Conclusion:

Therefore, the magnitude of the new tension in the rod is $40.0\text{kN}$.

To determine

The increase in length of the rod due to tension.

Answer to Problem 27AP

The increase in length of the rod due to tension is $2.00\text{mm}$.

Explanation of Solution

Formula to calculate the tensile stress on the rod is,

    $\text{Tensile}\text{stress}=\frac{T_2}{A_{\text{r}}}$

Here, $A_{\text{r}}$ is the cross-sectional area of the rod.

Formula to calculate the tensile strain is,

    $\text{Tensile}\text{strain}=\frac{\Delta l_{\text{t}}}{l}$

Here, $\Delta l_{\text{t}}$ is the increase in length.

Formula to calculate Young’s modulus is,

    $Y=\frac{\text{Tensile}\text{stress}}{\text{Tensile}\text{strain}}$

Substitute $\frac{T_2}{A_{\text{r}}}$ for tensile stress and $\frac{\Delta l_{\text{t}}}{l}$ for tensile strain in the above equation to find $Y$.

    $Y=\frac{\frac{T_2}{A_{\text{r}}}}{\frac{\Delta l}{l}}$

Substitute $40\text{kN}$ for $T_2$, $1.5\text{}{\text{cm}}^2$ for $A_{\text{r}}$, $30.0\times {10}^9\text{N}/{\text{m}}^2$ for $Y$ and $1.5\text{m}$ for $l$ in the above equation to find $\Delta l$.

    $\begin{array}{c}\left(30.0\times {10}^9\text{N}/{\text{m}}^2\right)=\frac{\frac{\left(40\text{kN}\right)}{\left(1.5\text{}{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)}}{\frac{\Delta l}{1.5\text{m}}}\\ \Delta l=\left(0.00133\text{m}\left(\frac{1000\text{mm}}{1\text{m}}\right)\right)\\ \approx 2.00\text{mm}\end{array}$

Conclusion:

Therefore, the increase in length of the rod due to tension is $2.00\text{mm}$.

To determine

The required extension of the rod while concrete was poured.

Answer to Problem 27AP

The required extension of the rod while concrete was poured is $2.40\text{mm}$.

Explanation of Solution

The length of the concrete is $1.50\text{}\text{m}$, cross-sectional area of the rod is $1.50\text{}{\text{cm}}^2$, cross-sectional area of the concrete is $50.0\text{}{\text{cm}}^2$, Young’s modulus of the concrete is $30.0\times {10}^9\text{N}/{\text{m}}^2$, and the compressive stress on concrete is $8.0\times {10}^6\text{N}/{\text{m}}^2$.

Formula to calculate the required extension length of the concrete is,

    $\Delta l_{\text{required}}=\Delta l+\Delta l_{\text{t}}$

$\Delta l_{\text{required}}$ is the required extension length.

Substitute $0.400\text{mm}$ for $\Delta l$ and $2.0\text{mm}$ for $\Delta l_{\text{t}}$ in the above equation to find $L_{\text{required}}$.

    $\begin{array}{c}\Delta l_{\text{required}}=\left(0.400\text{mm}\right)+\left(2.0\text{mm}\right)\\ =2.40\text{mm}\end{array}$

Conclusion:

Therefore, the required extension of the rod while concrete was poured is $2.40\text{mm}$.

To determine

The required original tension on the rod.

Answer to Problem 27AP

The required original tension on the rod is $48.0\text{kN}$.

Explanation of Solution

Formula to calculate the stress due to original tension is,

    $\text{Tensile}\text{stress}=\frac{T_1}{A_{\text{r}}}$

Here, $T_1$ is the original tension on the rod.

Formula to calculate the tensile strain is,

    $\text{Tensile}\text{strain}=\frac{\Delta l_{\text{required}}}{l}$

Formula to calculate Young’s modulus of the steel rod is,

    $Y_{\text{s}}=\frac{\text{Tensile}\text{stress}}{\text{Tensile}\text{strain}}$

Here, $Y_{\text{s}}$ is the Young’s modulus of the steel rod.

Substitute $\frac{T_1}{A_{\text{r}}}$ for tensile stress and $\frac{\Delta l_{\text{required}}}{l}$ for tensile strain in the above equation to find $T_1$.

`    $\begin{array}{c}{Y}_{\text{s}}=\frac{\left(\frac{T_1}{A_{\text{r}}}\right)}{\left(\frac{\Delta l_{\text{required}}}{l}\right)}\\ T_1=\left(\frac{\Delta l_{\text{required}}}{l}\right)A_r{Y}_{\text{s}}\end{array}$

Substitute $2.40\text{mm}$ for $\Delta l_{\text{required}}$, $1.5\text{}{\text{cm}}^2$ for $A_{\text{r}}$, $20.0\times {10}^{10}\text{N}/{\text{m}}^2$ for $Y_{\text{s}}$ and $1.5\text{m}$ for $l$ in the above equation to find $T_1$.

    $\begin{array}{c}{T}_1=\left(\frac{\Delta l_{\text{required}}}{l}\right)A_r{Y}_{\text{s}}\\ =\frac{\left(2.40\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{1.5\text{m}}\left(1.5{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)\left(20.0\times {10}^{10}\text{N}/{\text{m}}^2\right)\\ =\left(48000\text{N}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ \approx 48.0\text{kN}\end{array}$

Conclusion:

Therefore, the required original tension on the rod is $48.0\text{kN}$.