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## Q. 1.1

The floor beam in Fig. 1–8 is used to support the 6-ft width of a lightweight plain concrete slab having a thickness of 4 in. The slab serves as a portion of the ceiling for the floor below, and therefore its bottom is coated with plaster. Furthermore, an 8-ft-high, 12-in.-thick lightweight solid concrete block wall is directly over the top flange of the beam. Determine the loading on the beam measured per foot of length of the beam.

## Verified Solution

Using the data in Tables 1–2 and 1–3, we have

 TABLE 1-2 Minimum Densities for Design Loads from Materials ${lb}/{ft^{3} }$ ${kN}/{m^{3}}$ Aluminum 170 26.7 Concrete, plain cinder 108 17.0 Concrete, plain stone 144 22.6 Concrete, reinforced cinder 111 17.4 Concrete, reinforced stone 150 23.6 Clay, dry 63 9.9 Clay, damp 110 17.3 Sand and gravel, dry, loose 100 15.7 Sand and gravel, wet 120 18.9 Masonry, lightweight solid concrete 105 16.5 Masonry, normal weight 135 21.2 Plywood 36 5.7 Steel, cold-drawn 492 77.3 Wood, Douglas Fir 34 5.3 Wood, Southern Pine 37 5.8 Wood, spruce 29 4.5
 TABLE 1-3 Minimum Design Dead Loads Walls psf ${kN}/{m^{2}}$ 4-in. (102 mm) clay brick 39 1.87 8-in. (203 mm) clay brick 79 3.78 12-in. (305 mm) clay brick 115 5.51 Frame Partitions and Walls Exterior stud walls with brick veneer 48 2.30 Windows, glass, frame and sash 8 0.38 Wood studs $2\times 4in.$,$\left(51\times 102 mm\right)$  unplastered 4 0.19 Wood studs $2\times 4in.$,$\left(51\times 102 mm\right)$plastered one side 12 0.57 Wood studs $2\times 4in.$,$\left(51\times 102 mm\right)$plastered two sides 20 0.96 Floor Fill Cinder concrete, per inch (mm) 9 0.017 Lightweight concrete, plain, per inch (mm) 8 0.015 Stone concrete, per inch (mm) 12 0.023 Ceilings Acoustical fiberboard 1 0.05 Plaster on tile or concrete 5 0.24 Suspended metal lath and gypsum plaster 10 0.48 Asphalt shingles 2 0.1 Fiberboard,$\frac{1}{2} in.$ (13 mm) 0.75 0.04

Concrete slab: $\left[{8 lb}/{\left(ft^{2}\cdot in. \right) }\right] \left(4 in.\right) \left(6 ft\right) =192{lb}/{ft}$

Plaster ceiling: $\left(5{lb}/{ft^{2} }\right) \left(6 ft\right)=30 {lb}/{ft}$

Block wall:$\left(105{lb}/{ft^{3} }\right) \left(8 ft\right)\left(1 ft\right)={840 {lb}/{ft}}$

Total load:                                              ${1062 {lb}/{ft}} =1.06 { {k}/{ft}}$

Here the unit k stands for “kip,” which symbolizes kilopounds. Hence, $1 k= 1000 lb.$