Simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The beam has a length L 5 4 m and rectangular cross section with a width of 200 mm and height of 300 mm. Determine the maximum permissible value for the maximum inten- sity, 0 q , if the allowable normal stresses in tension and compression are 120 MPa.

Answer:q₀ = 350,740.2885 N/mExplanation:Given[tex]q(x)=\frac{x}{L} q_{0}[/tex]σ = 120 MPa = 120*10⁶ Pa[tex]L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\[/tex]We can see the pic shown in order to understand the question.We apply∑MB = 0  (Counterclockwise is the positive rotation direction)⇒ - Av*L + (q₀*L/2)*(L/3) = 0⇒ Av = q₀*L/6   (↑)Then, we apply[tex]v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x[/tex]Then, we can get the maximum bending moment as follows[tex]M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m[/tex]then we get  [tex]M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}[/tex]We get the inertia as follows[tex]I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}[/tex]We use the formulaσ = M*y/I⇒ M = σ*I/ywhere[tex]y=\frac{h}{2} =\frac{0.3m}{2}=0.15m[/tex]If M = Mmax, we have[tex](\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}[/tex]