Chapter 16, Problem 16.1P

To determine

To determine the lightest weight W shape that support the load which is uniformly distributed of 2.1 kips/ft having the span of 24 ft.

Answer to Problem 16.1P

   $W16\times 36$.

Explanation of Solution

Given Information:

Uniformly distributed load is given as 2.1 kips/ft with the span that is simple which is given as, 24ft.

Calculation:

As the load is distributed evenly the equation is given as,

   $R_A+R_B$

   $=24\times 2.1\Rightarrow 50.4$

   $R_A+R_A$

   $\Rightarrow 2R_A=50.4$

   $\Rightarrow R_A=\frac{50.4}{2}$

   $=25.2kips$

Shear moment at the maximum is given as,

   $M_a=w{l}^2/8$

   $\Rightarrow \frac{(2.1\times {24}^2)}{8}$

   $\Rightarrow 151.2kft$

The section modulus of the plastic axis is given as,

   $Z_x=\frac{1.67M_a}{s_y}$

   $=\frac{(1.67\times 151.2)}{50}\Rightarrow 5.0508f{t}^3$

   $=60.6in{.}^3$

With the help of the table of the steel beam, we have $Z_{x}ofW16\times 36$, so the required section modulus of $Z_x$ is less than $Z_{x}ofW16\times 36$. So, we have,

   $60.6in{.}^3<64in{.}^3$

The lightest W shape supporting the load that is uniformly distributed is calculated as $W16\times 36$

Conclusion:

Thus,the lightest W shape supporting the load that is uniformly distributed is calculated.