Chapter 2, Problem 1P

To determine

Calculate the loads that is acting on the floor beam BE and girder AC.

Answer to Problem 1P

The uniformly distributed load acting on the floor beam BE is $\underset{\_}{3.84\text{kN/m}}$.

The load acting at A, B, and C on the girder AC are $\underset{\_}{P_A=5.76\text{kN}}$, $\underset{\_}{P_B=11.52\text{kN}}$, and $\underset{\_}{P_C=5.76\text{kN}}$.

Explanation of Solution

Given information:

The building is a single-story building.

The building is subjected to uniformly distributed load of $w_{UDL}=0.96\text{kPa}$.

Calculation:

Show the roof of the single-story storage building as shown in Figure 1.

Refer Figure 1.

The columns are denoted by A, C, D, and F.

The floor beam is denoted by BE.

The girders are denoted by AC and DF.

Show the tributary area of the floor beam BE as shown in Figure 2.

Refer Figure 2.

The tributary area of the floor beam BE is denoted by the shaded area.

Calculate the tributary area of the floor beam BE $\left(A_1\right)$ as follows:

$\begin{array}{c}{A}_1=4\text{m}\times 6\text{m}\\ =24{\text{m}}^2\end{array}$

The length of the floor beam BE is $L=6\text{m}$.

Calculate the uniformly distributed load $\left(W_1\right)$ acting on the floor beam BE as follows:

$W_1=\frac{w_{UDL}\times A_1}{L}$

Substitute $0.96\text{kPa}$ for $w_{UDL}$, $6\text{m}$ for $L$ and $24{\text{m}}^2$ for $A_1$.

$\begin{array}{c}{W}_1=\frac{0.96\text{kPa}\times 24{\text{m}}^2}{6\text{m}}\\ =3.84\text{kN/m}\end{array}$

Show the uniformly distributed load acting on the floor beam BE as shown in Figure 3.

Refer Figure 3.

The reactions at B and E are denoted by $R_B$ and $R_E$ respectively.

The loading on the floor beam BE is symmetrical.

Calculate the value of $R_B$ and $R_E$ using the relation:

$\begin{array}{l}{R}_B=R_E=\frac{3.84\text{kN/m}\times \text{6}\text{m}}{2}\\ R_B=R_E=11.52\text{kN}\end{array}$

Show the uniformly distributed load acting on the floor beam BE as shown in Figure 4.

Refer Figure 4.

Thus, the uniformly distributed load acting on the floor beam BE is $\underset{\_}{3.84\text{kN/m}}$.

Show the tributary area of the girder AC as shown in Figure 5.

Refer Figure 5.

The tributary area of the girder AC is denoted by the shaded area.

Calculate the tributary area of the girder AC $\left(A_2\right)$ as follows:

$\begin{array}{c}{A}_2=8\text{m}\times 3\text{m}\\ =24{\text{m}}^2\end{array}$

Calculate the load $\left(W_2\right)$ acting on the girder AC as follows:

$W_2=w_{UDL}\times A_2$

Substitute $0.96\text{kPa}$ for $w_{UDL}$ and $24{\text{m}}^2$ for $A_2$.

$\begin{array}{c}{W}_2=0.96\text{kPa}\times 24{\text{m}}^2\\ =23.04\text{kN}\end{array}$

The total load acting on the tributary area of the girder AC is $23.04\text{kN}$.

Almost half the load acts at the junction of the floor beam BE and the girder AC. Then,

The load acting at B is $11.52\text{kN}$.

The remaining half of the load acts equally on the column A and C. Then,

The load acting at A is $5.76\text{kN}$ and the load acting at C is $5.76\text{kN}$.

Show the load acting on the girder AC as shown in Figure 6.

Refer Figure 6.

The reactions at A and C are denoted by $R_A$ and $R_C$ respectively.

The loading on the girder AC is symmetrical.

Calculate the value of $R_A$ and $R_C$ using the relation:

$\begin{array}{l}{R}_A=R_C=\left(\frac{5.76+11.52+5.76}{2}\right)\text{kN}\\ R_A=R_C=11.52\text{kN}\end{array}$

Show the load acting on the girder AC as shown in Figure 7.

Refer Figure 7.

Thus, the load acting at A, B, and C on the girder AC are $\underset{\_}{P_A=5.76\text{kN}}$, $\underset{\_}{P_B=11.52\text{kN}}$, and $\underset{\_}{P_C=5.76\text{kN}}$.