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: Set up the moment expression.
The beam is fixed at C and carries a single transverse force \(P\) at the free end A. For this determinate cantilever, the bending moment at any section is simply \(P\) times the distance from that section to A, regardless of how \(I\) varies along the beam. Measure \(x\) from the free end A, so at a section a distance \(x\) from A, \(M(x) = Px\).

Step 2: Apply the unit-load (virtual work) formula for tip deflection.
For a cantilever with a real load \(P\) at the tip, the deflection at that same tip is
\[ \delta_A = \int_0^{L_{total}} \frac{M(x)\,m(x)}{E\,I(x)}\,dx \]
where \(m(x)\) is the moment from a unit virtual load at A, which has the same shape as \(M(x)/P\), i.e. \(m(x) = x\). So
\[ \delta_A = \frac{P}{E}\int_0^{L_{total}} \frac{x^2}{I(x)}\,dx \]
Near A (from \(x=0\) to \(x=L_1=100\) mm) the section has \(I_1 = 100\) mm\(^4\). From \(x=L_1\) to \(x=L_1+L_2=200\) mm, the section is the BC part, with \(I_2 = 700\) mm\(^4\).

Step 3: Evaluate the two pieces of the integral.
\[ \int_0^{100} \frac{x^2}{I_1}\,dx = \frac{1}{100}\left[\frac{x^3}{3}\right]_0^{100} = \frac{1{,}000{,}000/3}{100} = 3333.33 \text{ mm}^{-1} \]
\[ \int_{100}^{200} \frac{x^2}{I_2}\,dx = \frac{1}{700}\left[\frac{x^3}{3}\right]_{100}^{200} = \frac{(8{,}000{,}000-1{,}000{,}000)/3}{700} = \frac{2{,}333{,}333.3}{700} = 3333.33 \text{ mm}^{-1} \]
Adding, the total integral is \(3333.33 + 3333.33 = 6666.67\) mm\(^{-1}\).

Step 4: Substitute E and P.
\(E = 200\) GPa \(= 200{,}000\) N/mm\(^2\) (consistent with mm and N units used above).
\[ \delta_A = \frac{30}{200{,}000}(6666.67) = (1.5\times10^{-4})(6666.67) \]

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