Chapter 2, Problem 6P

To determine

Find the dead loads acting on the girder AE and beam CD.

Answer to Problem 6P

The dead load acting on the beam CD is $\underset{\_}{1082.3\text{lb/ft}}$.

Dead load in Girder AE:

The dead load at C, A, and E are $\underset{\_}{13,529\text{lb}}$, $\underset{\_}{4,529\text{lb}}$, and $\underset{\_}{4,529\text{lb}}$ respectively.

The uniformly distributed load in the girder AE is $\underset{\_}{111.3\text{lb/ft}}$.

Explanation of Solution

Given information:

The thickness of the reinforced concrete slab is $4\text{in}\text{.}$

The area of cross-section of the steel floor beam is $A_{\text{steel}\text{beam}}=18.3\text{in}{\text{.}}^2$.

The area of cross-section of the steel girder is $A_{\text{steel}\text{girder}}=32.7\text{in}{\text{.}}^2$.

The length, height, and thickness of the brick wall are $l=25\text{ft},h=7\text{ft},t=6\text{in}\text{.}$.

Calculation:

Show the floor systemof the building as shown in Figure 1.

Refer Figure 1.

The tributary area of the beam CD is represented by the shaded region.

Tributary area of the beam CD:

The width of the tributary area of the beam CD is $B=12\text{ft}$.

The width of the tributary area of the beam CD is same as the length of the beam CD. Then,

The length of the tributary area of the beam CD is $25\text{ft}$.

The thickness of the concrete slab is $T=4\text{in}\text{.}$

Refer Table 2.1 “Unit Weights of Construction Materials” in the text book.

The unit weight of the reinforced concrete is $w_c=150{\text{lb/ft}}^3$.

The unit weight of the structural steel is $w_s=490{\text{lb/ft}}^3$.

The unit weight of the brick wall is $w_b=120{\text{lb/ft}}^3$.

Calculate the dead load per unit length of the beam CD as follows:

Concrete Slab:

Calculate the dead load of the concrete slab using the relation:

${\left(DL\right)}_{\text{Concrete}\text{Slab}}=L\times B\times H\times w_c$ (1)

Substitute $1\text{ft}$ for $L$, $12\text{ft}$ for $B$, $4\text{in}\text{.}$ for $T$, and $150{\text{lb/ft}}^3$ for $w_c$ in Equation (1).

$\begin{array}{c}{\left(DL\right)}_{\text{Concrete}\text{Slab}}=1\text{ft}\times 12\text{ft}\times 4\text{in}\text{.}\times \left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)\times 150{\text{lb/ft}}^3\\ =600\text{lb}\end{array}$

Steel beam:

Calculate the dead load of the steel beam using the relation:

${\left(DL\right)}_{\text{Steel}\text{Beam}}=A_{\text{steel}\text{beam}}\times L\times w_s$ (2)

Substitute $1\text{ft}$ for $L$, $18.3\text{in}{\text{.}}^2$ for $A_{\text{steel}\text{beam}}$ and $490{\text{lb/ft}}^3$ for $w_s$ in Equation (2).

$\begin{array}{c}{\left(DL\right)}_{\text{Steel}\text{Beam}}=A_{\text{steel}\text{beam}}\times L\times w_s\\ =18.3\text{in}{\text{.}}^2\times \left(\frac{1{\text{ft}}^2}{144\text{in}{\text{.}}^2}\right)\times 1\text{ft}\times 490{\text{lb/ft}}^3\\ =62.27\text{lb}\\ \approx 62.3\text{lb}\end{array}$

Calculate the dead load of the brick wall using the relation:

$\begin{array}{c}{\left(DL\right)}_{\text{Brick wall}}=A_{\text{brick wall}}\times L\times w_b\\ =1\text{ft}\times 7\text{ft}\times 6\text{in}\text{.}\times \left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)\times 120{\text{lb/ft}}^3\\ =420\text{lb}\end{array}$

Calculate the dead load of the beam CD as follows:

$\begin{array}{c}{\left(DL\right)}_{\text{Beam}CD}={\left(DL\right)}_{\text{Concrete}\text{Slab}}+{\left(DL\right)}_{\text{Steel}\text{Beam}}+{\left(DL\right)}_{\text{Brick wall}}\\ =600\text{lb}+62.3\text{lb}+420\text{lb}\\ =1082.3\text{lb}\end{array}$

The dead load of $1082.3\text{lb/ft}$ is distributed over the length of the beam CD.

Show the dead load acting on the beam as shown in Figure 2.

Refer Figure 2.

The reaction at C and D are denoted by $C_y$ and $D_y$.

The dead load on the beam is symmetrical. Then,

$\begin{array}{l}{C}_y=D_y=\left(\frac{1082.3\text{lb/ft}\times 25\text{ft}}{2}\right)\\ C_y=D_y\approx 13529\text{lb}\end{array}$

Show the dead load acting on the beam as shown in Figure 3.

Refer Figure 3.

Thus, the dead load acting on the beam CD is $\underset{\_}{1082.3\text{lb/ft}}$.

Show the floor system of the building as shown in Figure 4.

Refer Figure 4.

Tributary area of the girder AE:

The width of the tributary area of the girder AE is $\left(\frac{25}{2}\right)\text{ft}$.

The width of the tributary area of the girder AE is same as the length of the girder AE. Then,

The length of the tributary area of the girder AE is $24\text{ft}$.

The thickness of the reinforced concrete slab is $T=4\text{in}\text{.}$

Calculate the dead load per unit length of the girder AE as follows:

Steel beam:

Calculate the dead load of the girder AE using the relation:

${\left(DL\right)}_{\text{Steel}\text{Girder}}=A_{steelgirder}\times L\times w_s$ (3)

Substitute $1\text{ft}$ for $L$, $32.7\text{in}{\text{.}}^2$ for $A_{steelgirder}$ and $490{\text{lb/ft}}^3$ for $w_s$ in Equation (3).

$\begin{array}{c}{\left(DL\right)}_{\text{Steel}\text{Girder}}=32.7\text{in}{\text{.}}^2\times \left(\frac{1{\text{ft}}^2}{144\text{in}{\text{.}}^2}\right)\times 1\text{ft}\times 490{\text{lb/ft}}^3\\ =\text{111}\text{.3}\text{lb}\end{array}$

Concentrated load at A and E.

Show the tributary area of column at A and E as shown in Figure 5.

Refer Figure 5.

The tributary area of column at A and E are Equal.

The tributary area of column at A and E are $\left(\frac{25}{2}\right)\text{ft}\times 6\text{ft}$.

Calculate the concentrated load (P) at the column A and E the using the relation:

$\begin{array}{c}P=P_{\text{Concret}}+P_{\text{Steel}}\\ =\left[\left(\text{Tributary}\text{Area}\text{of}\text{}\text{column}\right)\times T\times w_C\right]+\left[A_{\text{steel}\text{beam}}\times w_S\times \left(\frac{25}{2}\right)\right]\end{array}$

Substitute $\left(\frac{25}{2}\right)\text{ft}\times 6\text{ft}$ for $\text{Tributary}\text{Area}\text{of}\text{}\text{column}$, $150{\text{lb/ft}}^3$ for $w_c$, $18.3\text{in}{\text{.}}^2$ for $A_{\text{steel}\text{beam}}$ and $490{\text{lb/ft}}^3$ for $w_s$, and $4\text{in}\text{.}$ for $T$.

$\begin{array}{c}P=\left[150\times \left(\frac{25}{2}\right)\text{ft}\times 6\text{ft}\times 4\text{in}\times \left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{+18}\text{.3}\text{in}{\text{.}}^2\times \left(\frac{1{\text{ft}}^2}{144\text{in}{\text{.}}^2}\right)\times 490\times \left(\frac{25}{2}\right)\text{ft}\right]\\ =\left[3,750\text{ft+778}\text{.38}\right]\\ \approx \text{4,529}\text{lb}\end{array}$

Refer Figure 3.

The concentrated load at the column C is $\underset{\_}{8,279\text{lb}}$.

Show the loading on the girder AE as shown in Figure 6.

Refer Figure 6.

The reaction at A and E are denoted by $A_y$ and $E_y$.

The dead load on the beam is symmetrical. Then,

$\begin{array}{l}{A}_y=E_y=\left(\frac{4529\text{lb}+13529\text{lb}+111.3\text{lb/ft}\times 24\text{ft}+4529\text{lb}}{2}\right)\\ A_y=E_y\approx 12629\text{lb}\end{array}$

Show the loading on the girder AE as shown in Figure 7.

Refer Figure 7.

The dead load at C, A, and E are $\underset{\_}{13,529\text{lb}}$, $\underset{\_}{4529\text{lb}}$, and $\underset{\_}{4529\text{lb}}$ respectively.

The uniformly distributed load in the girder AE is $\underset{\_}{111.3\text{lb/ft}}$.