For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
Generally, the equation for the is mathematically given as
[tex]A=\frac{PL^2}{2EI}+\frac{ML}{EI}[/tex]
Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
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The value of slope of end a of the cantilevered beam when E = 200 gpa and i = 65. 0(106) mm4 is 0.0048 rads.
The formula used to find the slope at the end of the cantilevered beam is,
[tex]A=\dfrac{PL^2}{2EI}+\dfrac{ML}{EI}[/tex]
Here, (M) is the momentum, (EI) is flexural rigidity, (P) is concentrated load and (L) is the length of the beam.
Concentrated load is 10 KN and uniform load is 3 kN/m. E = 200 gpa and i = 65. 0(106) mm4.
[tex]A=\dfrac{10+10^2+3^2}{2\times240\times10^9\times65\times10^6}+\dfrac{10+10^3\times3}{240\times10^9\times65\times10^{-6}}\\A=0.00288+0.00192A=0.0048 \rm\; rads[/tex]
Thus, the value of slope of end a of the cantilevered beam when E = 200 gpa and i = 65. 0(106) mm4 is 0.0048 rads.
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