Question

A stepped cantilever beam, made of a material having Young's modulus \(E = 200\) GPa, is shown in the figure below.

The length and the moment of inertia of the beam from point A to B are \(L_1 = 100\) mm and \(I_1 = 100\) mm\(^4\), respectively. The length and the moment of inertia of the beam from point B to C are \(L_2 = 100\) mm and \(I_2 = 700\) mm\(^4\), respectively. A shear force \(P = 30\) N is applied at point A of the beam. The magnitude of the deflection of the beam at point A is _______ mm (rounded off to 1 decimal place).

Hint:

Split the beam at the step. Use the unit load (virtual work) method, or superpose the tip deflection of segment AB with the deflection and slope carried over from segment BC.

Answer 1

Correct Answer: 1

Step 1: Treat segment BC as a cantilever loaded at its free end B.
Cut the beam at B. Everything that segment AB does to segment BC can be replaced by the force $P$ and the moment it creates at B, transferred from A:
\[ M_B = P \times L_1 = 30 \times 100 = 3000 \text{ N-mm} \]
So segment BC (length $L_2$, stiffness $EI_2$, fixed at C) is a cantilever carrying an end force $P$ and an end moment $M_B$ at its tip B.

Step 2: Find the deflection and slope this produces at B.
Using standard cantilever tip formulas,
\[ \delta_B = \frac{PL_2^3}{3EI_2} + \frac{M_B L_2^2}{2EI_2}, \qquad \theta_B = \frac{PL_2^2}{2EI_2} + \frac{M_B L_2}{EI_2} \]
With $E=200{,}000$ N/mm$^2$, $I_2=700$ mm$^4$, $L_2=100$ mm:
\[ \delta_B = \frac{30(100)^3}{3(200000)(700)} + \frac{3000(100)^2}{2(200000)(700)} = 0.0714 + 0.1071 = 0.1786 \text{ mm} \]
\[ \theta_B = \frac{30(100)^2}{2(200000)(700)} + \frac{3000(100)}{(200000)(700)} = 0.001071 + 0.002143 = 0.003214 \text{ rad} \]

Step 3: Add the contribution of segment AB, treating B as its (moving) support.
Segment AB is itself a cantilever of length $L_1$, stiffness $EI_1$, carrying $P$ at its own free tip A. Its own bending adds a deflection of A relative to B,
\[ \delta_{AB} = \frac{PL_1^3}{3EI_1} = \frac{30(100)^3}{3(200000)(100)} = 0.5 \text{ mm} \]
Also, because B itself rotates by $\theta_B$, the rigid segment AB tips over by that same angle, carrying point A sideways by an extra
\[ \theta_B \times L_1 = 0.003214 \times 100 = 0.3214 \text{ mm} \]

Step 4: Add up every contribution at A.
\[ \delta_A = \delta_B + \theta_B L_1 + \delta_{AB} = 0.1786 + 0.3214 + 0.5 \]

Final Answer:
\[ \boxed{\delta_A = 1.0 \text{ mm}} \]