Question 7 of 9: Fixed-end moment of a non-prismatic beam by the flexibility method

Question 7: Fixed-end moment of a non-prismatic beam by the flexibility method (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 12 m propped cantilever: simple support at A, built-in at B, $EI$ over the first 9 m, $3EI$ over the last 3 m, and a 118 kN point load at the change of section.

Find. The fixing moment at the built-in support B.

Approach. Release the propping reaction at A, compute the two flexibility coefficients by integrating $M\,m/EI$ over the two different rigidities, restore compatibility, then take moments about B.

1. Choose the redundant and the primary structure. The beam is one degree statically indeterminate. Release the vertical reaction at A, leaving a cantilever built in at B and free at A, and call the redundant $R_{A}$ (positive upwards). Compatibility requires the deflection at A to be zero in the real structure.
2. Write the two moment fields. With $x$ measured from A, the primary structure under the 118 kN load carries$$M_{0}(x)=0\;(0\le x\le 9),\qquad M_{0}(x)=-118\,(x-9)\;(9\le x\le 12),$$while a unit upward force at A gives $m(x)=x$ everywhere. The load produces no moment at all over the first 9 m, because in the released structure everything to the left of the load simply hangs.
3. Integrate the flexibility coefficients. The rigidity changes at $x=9$, so each integral splits:$$\delta_{10}=\int\frac{M_{0}m}{EI}\,dx=\frac{1}{3EI}\int_{0}^{3}\!\!-118\,u\,(u+9)\,du=-\frac{1947}{EI},$$with $u=x-9$, and$$\delta_{11}=\int\frac{m^{2}}{EI}\,dx=\frac{1}{EI}\!\int_{0}^{9}\!\!x^{2}dx+\frac{1}{3EI}\!\int_{9}^{12}\!\!x^{2}dx=\frac{243+111}{EI}=\frac{354}{EI}.$$Note how the factor $\tfrac{1}{3}$ suppresses the contribution of the stiff length: the outer 3 m occupies a quarter of the span but supplies less than a third of the flexibility.
4. Restore compatibility. Setting the total deflection at A to zero, $\delta_{10}+R_{A}\,\delta_{11}=0$, so$$R_{A}=\frac{1947}{354}=\boxed{5.50\ \text{kN upwards}}.$$The common $1/EI$ cancels, which is why the question can withhold a numerical rigidity and give only the ratio 3:1.
5. Take moments about B. With the redundant known the structure is determinate:$$M_{B}=R_{A}(12)-118(3)=66-354=-288\ \text{kN}\cdot\text{m},$$that is$$\left|M_{B}\right|=\boxed{288\ \text{kN}\cdot\text{m hogging}} .$$Vertical equilibrium gives $V_{B}=118-5.5=112.5$ kN, and the sagging moment under the load is $R_{A}(9)=49.5$ kN·m. A direct-stiffness model with two elements of rigidity $EI$ and $3EI$ returns the same 5.50 kN and 288 kN·m.

For contrast, a prismatic beam of the same span and loading would give $R_{A}=P b^{2}(3L-b)/2L^{3}=118(3)^{2}(36-3)/2(12)^{3}=10.1$ kN and a fixing moment of 213 kN·m. Stiffening the outer quarter therefore drags about 35 per cent more moment into the built-in end — the general lesson that stiffness attracts load.