Answer:
49.5 kN
Explanation:
From the information given:
[tex]R_D = 28 \ kN[/tex] [tex]\delta _D = 1.4; \ \ \ \delta _L = 1.6[/tex]
[tex]\sigma_n = 204 \ MPa; \ \ \ A_w = 6.45 \ cm^2 = 645 \ mm^2[/tex]
Thus ; [tex]P_n = \dfrac{\sigma_n}{\frac{1}{A}} \\ \\ = \ {\sigma_n}*{A} \\ \\ = 645 *204 \\ \\ = 131.58 \ kN[/tex]
From the given inequality; maximum live load (in addition to RD) that can be supported in shear by this beam is calculated by using the relation;
[tex]\phi P_n \geq \sum \delta_i R_i \\ \\ \geq \delta_DR_D + \delta_L R_L \\ \\ 0.9*131.58 \geqq [1.4*28+1.6*R_L ] \\ \\ 118.4 \geq 39.2+ 16 R_L \\ \\ 118.4 - 39.2 \geq 16R_L \\ \\ 79.2 \geq 16R_L\\ \\ R_L \leq 49.5 \ kN[/tex]
