Question

 E = 240 GPa and I = 65.0(10⁶) mm4. (Figure 1) 

Part A Determine the slope of end A of the cantilevered beam. 

Part B Determine the deflection of end A of the cantilevered beam.  

Answer 1

A) Slope at A = 0.0048 radian.

B) Deflection at A = 8.645 mm

Given, E = 240 GPa , I = 65 (10⁶) mm^{​​​​​4}​

The slope and deflection will be find out by superposition principle ,

I.e the slope at A will be sum of slope at A due to load at A and slope at A due to moment at A .

Similarly, for deflection.

The slope at A is given as ,

$\theta_A = \frac{PL^2}{2EI}+\frac{ML}{EI}$

As, from figure , P = 10 KN ,. M = 10 KN m .

L = 3 m .

Putting all velues in Newton and meter .

$\theta_A = \frac{10*10^3*3^2}{2*240*10^9*65*10^{-6}}+\frac{10*10^3*3}{240*10^9*65*10^{-6}}$

Slope = 0.00288 + 0.00192 = 0.0048 radian.

Hence, slope at A = 0.0048 radian .

B)

Now, Deflection at A =

$y_A = \frac{PL^3}{3EI}+\frac{ML^2}{2EI}$

$y_A = \frac{10*10^3*3^3}{3*240*10^9*65*10^{-6}}+\frac{10*10^3*3^2}{2*240*10^9*65*10^{-9}}$

Deflection = 0.00576 + 0.00288 = 0.008645 m

Hence, deflection at A = 8.645 mm