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# Solve Problem 2.5 if a 6 in. thick brick wall, which is 7 ft high and 25 ft long, hears directly on the top of beam CD. See Fig. P2.5. 2.5 The floor system of an apartment building consists of a 4 in. thick reinforced concrete slab resting on three steel floor beams, which in turn are supported by two steel girders, as shown in Fig. P2.5. The areas of cross section of the floor beams and the girders are 18.3 in. 2 and 32.7 in. 2 , respectively. Determine the dead loads acting on the beam CD and the girder AE . FIG. P2.5, P2.6, P2.9

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### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

#### Solutions

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Section
BuyFindarrow_forward

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 2, Problem 6P
Textbook Problem
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## Solve Problem 2.5 if a 6 in. thick brick wall, which is 7 ft high and 25 ft long, hears directly on the top of beam CD. See Fig. P2.5.2.5 The floor system of an apartment building consists of a 4 in. thick reinforced concrete slab resting on three steel floor beams, which in turn are supported by two steel girders, as shown in Fig. P2.5. The areas of cross section of the floor beams and the girders are 18.3 in.2 and 32.7 in.2, respectively. Determine the dead loads acting on the beam CD and the girder AE.FIG. P2.5, P2.6, P2.9

To determine

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

### Explanation of Solution

Given information:

The thickness of the reinforced concrete slab is 4in.

The area of cross-section of the steel floor beam is Asteelbeam=18.3in.2.

The area of cross-section of the steel girder is Asteelgirder=32.7in.2.

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

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=12ft.

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 25ft.

The thickness of the concrete slab is T=4in.

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

The unit weight of the reinforced concrete is wc=150lb/ft3.

The unit weight of the structural steel is ws=490lb/ft3.

The unit weight of the brick wall is wb=120lb/ft3.

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:

(DL)ConcreteSlab=L×B×H×wc (1)

Substitute 1ft for L, 12ft for B, 4in. for T, and 150lb/ft3 for wc in Equation (1).

(DL)ConcreteSlab=1ft×12ft×4in.×(1ft12in.)×150lb/ft3=600lb

Steel beam:

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

(DL)SteelBeam=Asteelbeam×L×ws (2)

Substitute 1ft for L, 18.3in.2 for Asteelbeam and 490lb/ft3 for ws in Equation (2).

(DL)SteelBeam=Asteelbeam×L×ws=18.3in.2×(1ft2144in.2)×1ft×490lb/ft3=62.27lb62.3lb

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

(DL)Brick wall=Abrick wall×L×wb=1ft×7ft×6in.×(1ft12in.)×120lb/ft3=420lb

Calculate the dead load of the beam CD as follows:

(DL)BeamCD=(DL)ConcreteSlab+(DL)SteelBeam+(DL)Brick wall=600lb+62.3lb+420lb=1082.3lb

The dead load of 1082.3lb/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 Cy and Dy.

The dead load on the beam is symmetrical. Then,

Cy=Dy=(1082.3lb/ft×25ft2)Cy=Dy13529lb

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

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