Flexibility method — fixed-end moment of a non-prismatic beam (24 marks): 07-Str-A4, December 2017, Q7

Question 7: Flexibility method — fixed-end moment of a non-prismatic beam (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 beam simply supported at joint 1 and built in at joint 2. The left 6 m has flexural rigidity $EI$ and the right 6 m has $3EI$; a 10 kN/m UDL covers the stiffer right-hand half only.

Question 7 — data
Quantity	Value
Left segment, joint 1 to mid-point	6 m at $EI$
Right segment, mid-point to joint 2	6 m at $3EI$
Support at joint 1	simple (roller)
Support at joint 2	built in
UDL	10 kN/m over the right-hand 6 m only

Find. The fixed-end moment at the right-hand support.

Q7: non-prismatic beam, simply supported at joint 1 and built in at joint 2; the 10 kN/m UDL covers the stiffer right-hand half only.

Approach. Release the vertical reaction at joint 1, so the primary structure is a cantilever built in at joint 2; then enforce zero deflection at joint 1 with flexibility coefficients that carry the change of section inside the integrals.

Confirm the degree of redundancy. A roller plus a built-in end gives four reaction components against three equations, so the beam is once statically indeterminate and one release is enough.
Write the primary moment fields. Measuring $x$ from joint 1 and taking sagging positive, the released structure carries only the UDL, and the unit redundant is an upward force at joint 1: $$M_{0}(x)=\begin{cases}0, & 0\le x\le 6\\[2pt] -5(x-6)^{2}, & 6\le x\le 12\end{cases}\qquad m_{1}(x)=x$$
Integrate the flexibility coefficient. The $1/EI(x)$ weighting is what makes the beam non-prismatic: $$f_{11}=\int_{0}^{6}\frac{x^{2}}{EI}dx+\int_{6}^{12}\frac{x^{2}}{3EI}dx=\frac{72}{EI}+\frac{168}{EI}=\frac{240}{EI}$$
Integrate the load term. Only the loaded half contributes, and it is the stiffer half, which is why the answer is smaller than the prismatic value: $$\Delta_{10}=\int_{6}^{12}\frac{-5(x-6)^{2}\,x}{3EI}dx=-\frac{5}{3EI}\int_{0}^{6}u^{2}(u+6)\,du=-\frac{5(756)}{3EI}=-\frac{1260}{EI}$$
Enforce compatibility. The support at joint 1 does not move, so $\Delta_{10}+R_{1}f_{11}=0$: $$R_{1}=\frac{1260}{240}=\boxed{5.25\ \text{kN upward}}$$
Take moments about the built-in end. With the redundant known the beam is determinate: $$M_{2}=R_{1}(12)-\frac{w(6)^{2}}{2}=63-180=\boxed{-117\ \text{kN}\cdot\text{m}}$$ that is, 117 kN·m hogging, with the top fibre in tension.
Check the remaining actions. Vertical equilibrium gives $R_{2}=60-5.25=54.75$ kN, and the moment where the section changes is $M(6)=5.25(6)=31.5$ kN·m sagging — the whole left-hand half carries a straight line, because it has no load on it.

It is worth comparing the answer with the prismatic case. A uniform propped cantilever loaded over its whole length would carry $wL^{2}/8=180$ kN·m at the built-in end; here only half the span is loaded and the stiff half sits next to the fixity, so the reaction at the simple support drops to 5.25 kN and the fixed-end moment to 117 kN·m. Stiffening the region next to a fixed end always pulls moment into that end, while loading only the far half pushes it the other way; the two effects partly cancel here, and only the integrals settle which wins.

Q7: bending moment diagram (sagging plotted downward). The reaction 5.25 kN at the simple support fixes the whole diagram; the built-in end carries 117 kN.m hogging.

Question 7 — results
Quantity	Value
Flexibility coefficient $f_{11}$	$240/EI$
Primary displacement $\Delta_{10}$	$-1260/EI$
Redundant reaction at joint 1	5.25 kN upward
Reaction at the built-in end	54.75 kN upward
Moment at the change of section	31.5 kN·m sagging
Fixed-end moment at the right end	117 kN·m hogging

Topic. The force (flexibility) method on a beam whose stiffness varies along its length.

Key relations. $\Delta_{10}+X_{1}f_{11}=0$ with $f_{11}=\int m_{1}^{2}\,dx/EI(x)$ and $\Delta_{10}=\int M_{0}m_{1}\,dx/EI(x)$; the moment diagrams $M_{0}$ and $m_{1}$ come from equilibrium alone and are therefore independent of how $EI$ varies.

Why this works. Because $m_{1}(x)=x$ and $M_{0}(x)$ are fixed by statics, the only place the change of section can enter is as the weighting $1/EI(x)$ inside the two integrals. Splitting each integral at the step is all that non-prismatic really means here.

Common pitfall. Averaging the two rigidities and using $2EI$ throughout. That gives $f_{11}=288/EI$ and $\Delta_{10}=-1890/EI$, hence $R_{1}=6.56$ kN and a fixed-end moment of only 101 kN·m — about 13 per cent light, because the averaged beam is too flexible next to the fixity and too stiff away from it. Integrate segment by segment.

Source. Hibbeler, Structural Analysis, 10th ed., §10.1–10.4 (force method, beams); Kassimali, 6th ed., §9.1–9.4.

Question 7 of 9: Flexibility method — fixed-end moment of a non-prismatic beam (24 marks)

Question 7: Flexibility method — fixed-end moment of a non-prismatic beam (24 marks)