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Sagot :
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
What is the slope of end a of the cantilevered beam?
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.
How to find the slope at the end of the cantilevered beam?
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.
Learn more about the cantilevered beam here;
https://brainly.com/question/19090244
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