   # A floor system in a small building consists of wood planks supported by 2-in. (nominal width) joists spaced at distance s and measured from center to center (see figure). The span length L of each joist is 12 ft, the spacing s of the joists is 16 in., and the allowable bending stress in the wood is 1250 psi. The uniform floor load is 120 lb/ft", which includes an allowance for the weight of the floor system itself. Calculate the required section modulus S for the joists, and then select a suitable joist size (surfaced lumber) from Appendix G, assuming that each joist may be represented as a simple beam carrying a uniform load. What is the maximum floor load that can be applied to your final beam selection in part (a)? ### Mechanics of Materials (MindTap Co...

9th Edition
Barry J. Goodno + 1 other
Publisher: Cengage Learning
ISBN: 9781337093347
Chapter 5, Problem 5.6.9P
Textbook Problem
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## A floor system in a small building consists of wood planks supported by 2-in. (nominal width) joists spaced at distance s and measured from center to center (see figure). The span length L of each joist is 12 ft, the spacing s of the joists is 16 in., and the allowable bending stress in the wood is 1250 psi. The uniform floor load is 120 lb/ft", which includes an allowance for the weight of the floor system itself. Calculate the required section modulus S for the joists, and then select a suitable joist size (surfaced lumber) from Appendix G, assuming that each joist may be represented as a simple beam carrying a uniform load. What is the maximum floor load that can be applied to your final beam selection in part (a)? (a)

To determine

The required section modulus S .

### Explanation of Solution

Given information:

The span length of each joist is 12ft , the spacing of the joists is 16in , the permissible bending stress is 1250psi , the total floor load is 120lb/ft2 .

The following figure gives the free body diagram of the beam:

Figure-(1)

Write the expression for the uniformly distributed load.

q=FLL2×s.....(I)

Here, the uniformly distributed load is q , the total floor load is FL , the span length is L , the spacing between the joists is s .

Write the expression for the maximum bending moment.

Mmax=qL28.....(II)

Here, the maximum bending moment is Mmax .

Write the expression for the section modulus.

S=Mmaxσallow.....(I)

Here, the permissible stress is σallow .

Calculation:

Substitute 120lb/ft2 for FL , 12in for L and 16in for s in Equation (II)

q=(120lb 122 in2)×(16in)=13

(b)

To determine

The maximum floor load applied on the section modulus.

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