Question

Problem 2: Cantilever beam In class we derived the spring constant of a beam of length L and constant bending stiffness El clamped at x=0 and subject to a vertical force Fat x= L. Let's study a few different variations of that problem. Let's replace the vertical force F by a counter clockwise bending moement Mo applied at x = L. Recom- pute the equivalent spring constant. Note that in the class we computed spring constant with force and displacement. Here you should compute the torsional spring constant with moment and slopes. (2a) (2b) Now let's consider a composite cantilever beam shown in the figure below. Two beams are welded to each other end to end. Let's derive the equivalent spring constant for this system. Hint: Remember, the EL E212 L/2 L/2 rods are welded end to end. So the displacement and slope of rod 1 at the right end, is the same as the displacement and the slope of the rod 2 at the left end.

Answer 1

o [ apple betren trapent at a 4B] = 1 max Since the beam is dampw at end Mo 1x=0 = 0 TX - J X-6 Mo ا آج - [x]o - MOL MOL Mo MOL = Moment Arple of terpent at XL Torsimal spring = on thes EI unit of moment rad, or dyre

Er d'z - M(*)' = AL-X ay = 1 SL - x) dx For x 20 to 42 dy & [Lx 144 FL At x=0, dy so fyzo co & C 20 3 4 = 18 [ 2x2 - 4 ] Deflection at 4:2 is For x2 4 to L region a 1 - 2 y = 11 2 [2x - 27 3] + D, R+ P2 Shop s y fh It - tel. 0 [Lx=*]+D, 165,34 5,5, 18] av x= 42 Y dy F C++ ] +, - Day [-] [ - ] + [at, talz +d2 - - - 1] + en Po-so) De er to HD)

-y [x-13 / ]- [- ] | + in tv - [最后」 - Deffect in a X L 4 FLE] = =|| |- - - - 241 will chan we & h 26 & I = I2 I --
Hope calculations are ok. Let me know if you find something odd somewhere.