Question 7 of 9: Fixed-end moments of a non-prismatic beam by the flexibility method (24 marks)

Question 7: Fixed-end moments 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 symmetric, non-prismatic, built-in beam under three equal point loads.

Find. The fixed-end moments at both supports.

Approach. Release both end moments to leave a simply supported primary beam, exploit symmetry so the two redundants collapse into one, and enforce zero end rotation with a single flexibility equation.

1. Choose the primary structure and the redundant. Cut both built-in moments; the primary structure is a simply supported beam of span 8 m. Call the released end moments $X_{A}$ and $X_{B}$ (hogging positive). The beam, its supports and its loads are all symmetric about mid-span, so $$X_{A}=X_{B}=X$$ and only one compatibility equation is needed.
2. Primary bending moment diagram. The simply supported reactions are $R=\tfrac{3}{2}(10)=15.0$ kN, so $$M_{0}(x)=15x\ \ (0\le x\le 2),\qquad M_{0}(x)=5x+20\ \ (2\le x\le 4),$$ with $M_{0}$ mirrored on the right half. The key ordinates are $M_{0}(2)=30.0$, $M_{0}(4)=40.0$ kN·m.
3. Unit-redundant diagram. A unit couple applied at $A$ on the simply supported beam produces the linear field $$\bar{m}(x)=1-\frac{x}{8},$$ and because the pair $X_{A}=X_{B}=X$ is applied at both ends the redundant contributes a constant $-X$ to the moment field: $$M(x)=M_{0}(x)-X\!\left(1-\frac{x}{8}\right)-X\frac{x}{8}=M_{0}(x)-X.$$ That single observation is what symmetry buys.
4. Write the compatibility equation. The rotation at $A$ must vanish, $$\delta_{10}-X f_{11}=0,\qquad \delta_{10}=\int_{0}^{8}\frac{M_{0}\bar{m}}{EI(x)}\,dx,\qquad f_{11}=\int_{0}^{8}\frac{\bar{m}}{EI(x)}\,dx.$$ Note that $EI(x)$ is inside both integrals — that is the only place the non-prismatic geometry enters, because $M_{0}$ and $\bar{m}$ come from equilibrium alone and do not care how the stiffness varies.
5. Evaluate the flexibility coefficients. Carrying out the integrals piecewise with $EI(x)=3EI$ on $0\!-\!2$ and $6\!-\!8$ and $EI$ on $2\!-\!6$, $$\delta_{10}=\frac{80.0}{EI},\qquad f_{11}=\frac{8}{3EI}=\frac{2.6667}{EI},$$ so $$X=\frac{\delta_{10}}{f_{11}}=\frac{80.0}{2.6667}=\boxed{30.0\ \text{kN}\cdot\text{m}}$$
6. Cross-check with the half-span form. For a symmetric built-in beam the tangent is horizontal at both the support and mid-span, so the area of the $M/EI$ diagram over half the span must vanish: $$\int_{0}^{4}\frac{M_{0}(x)-X}{EI(x)}\,dx=0\;\Rightarrow\;\bigl(10.0+\tfrac{2}{3}X\bigr)+\bigl(70.0+2X\bigr)\cdot(-1)^{0}=0,$$ i.e. $80.0+\tfrac{8}{3}X=0$ for the sagging sign of $X$, giving the same magnitude 30.0 kN·m. Two independent routes, one answer.
7. Assemble the final moment field. With $M(x)=M_{0}(x)-30.0$, $$M(0)=M(8)=\boxed{-30.0\ \text{kN}\cdot\text{m (hogging)}},\qquad M(2)=M(6)=0,\qquad M(4)=\boxed{+10.0\ \text{kN}\cdot\text{m}}$$ The points of contraflexure land exactly on the two outer loads, which is a coincidence of these particular stiffness ratios and worth reporting. The end reactions are $R_{A}=R_{B}=15.0$ kN by symmetry.
8. Interpret the stiffening. A prismatic beam under the same three loads would give $$\sum \frac{Pab^{2}}{L^{2}}=11.25+10.00+3.75=25.0\ \text{kN}\cdot\text{m}.$$ Tripling $EI$ over the outer quarters raises the end moment to 30.0 kN·m, a 20 % increase, and drops the mid-span moment from 15.0 to 10.0 kN·m. Stiffness attracts moment — the haunch pulls the diagram towards the supports.