Sample Solution from
Structural Analysis (MindTap Course List)
5th Edition
ISBN: 9781133943891
Chapter 2
Problem 1P
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To determine

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

Expert Solution

Answer

The uniformly distributed load acting on the floor beam BE is 3.84kN/m_.

The load acting at A, B, and C on the girder AC are PA=5.76kN_, PB=11.52kN_, and PC=5.76kN_.

Explanation of Solution

Given information:

The building is a single-story building.

The building is subjected to uniformly distributed load of wUDL=0.96kPa.

Calculation:

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

images

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.

images

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 (A1) as follows:

A1=4m×6m=24m2

The length of the floor beam BE is L=6m.

Calculate the uniformly distributed load (W1) acting on the floor beam BE as follows:

W1=wUDL×A1L

Substitute 0.96kPa for wUDL, 6m for L and 24m2 for A1.

W1=0.96kPa×24m26m=3.84kN/m

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

images

Refer Figure 3.

The reactions at B and E are denoted by RB and RE respectively.

The loading on the floor beam BE is symmetrical.

Calculate the value of RB and RE using the relation:

RB=RE=3.84kN/m×6m2RB=RE=11.52kN

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

images

Refer Figure 4.

Thus, the uniformly distributed load acting on the floor beam BE is 3.84kN/m_.

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

images

Refer Figure 5.

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

Calculate the tributary area of the girder AC (A2) as follows:

A2=8m×3m=24m2

Calculate the load (W2) acting on the girder AC as follows:

W2=wUDL×A2

Substitute 0.96kPa for wUDL and 24m2 for A2.

W2=0.96kPa×24m2=23.04kN

The total load acting on the tributary area of the girder AC is 23.04kN.

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

The load acting at B is 11.52kN.

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

The load acting at A is 5.76kN and the load acting at C is 5.76kN.

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

images

Refer Figure 6.

The reactions at A and C are denoted by RA and RC respectively.

The loading on the girder AC is symmetrical.

Calculate the value of RA and RC using the relation:

RA=RC=(5.76+11.52+5.762)kNRA=RC=11.52kN

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

images

Refer Figure 7.

Thus, the load acting at A, B, and C on the girder AC are PA=5.76kN_, PB=11.52kN_, and PC=5.76kN_.

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