Answer:
[tex]\mathbf{\Delta_{max}=0.78 \ mm}[/tex]
Explanation:
From the given information;
A simply supported beam (E = 12 GPa)
load q = 125 N/m
point load P = 200 N
the rectangular cross section
b = 75 mm
h = 200 mm
length = 3.6 m
The objective is to calculate the maximum deflection of the beam;
Using the formula;
[tex]I = \dfrac{1}{E}*bh^3[/tex] about the z-axis that goes through the central
[tex]I = \dfrac{1}{12}*(75 \ mm)*(200 \ mm)^3[/tex]
[tex]I = 5*10^7 \ mm^4[/tex]
The length L = 3.6 m = 3600 mm
The maximum deflection of the beam can be calculated by using the formula:
[tex]\Delta _{max} = \dfrac{5}{384}* \dfrac{qL^4}{EI}+\dfrac{PL^3}{48EI}[/tex]
[tex]\Delta _{max} = \dfrac{1}{12*10^3 \frac{N}{mm^2}*5*10^7 \ mm^4 }[ \dfrac{5*\frac{125 \ N}{100 \ mm}*3600 \ mm^4}{384}+\dfrac{200*(3600 \ mm )^3}{48}][/tex]
Thus; the maximum deflection of the beam is [tex]\mathbf{\Delta_{max}=0.78 \ mm}[/tex]
