Fixed-end moments of a non-prismatic beam by the flexibility method: 07-Str-A4, May 2015, Q3

Question 3: Fixed-end moments of a non-prismatic beam by the flexibility method (18 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.

Quantity	Value
Overall span	$L=12\ \text{m}$, built in at both ends
Outer segments	$3\ \text{m}$ each, rigidity $2EI$
Central segment	$6\ \text{m}$, rigidity $EI$
Loads	$3\times 12\ \text{kN}$ at $x=3,\ 6,\ 9\ \text{m}$

Find. The fixed-end moments at A and B.

Question 3: the non-prismatic fixed-end beam, symmetric about mid-span.

Approach. Release both built-in ends to a simply supported primary beam, apply the pair of end moments as a single symmetric redundant, and enforce zero end rotation with the virtual-work compatibility equation, integrating $1/EI$ segment by segment.

Choose the released structure and the redundant. The beam is three degrees indeterminate, but the structure and the loading are both symmetric about mid-span, so the two end moments are equal and the redundant horizontal reaction is zero for transverse loads. Releasing both end moments leaves a simply supported span, and a single redundant $X$ — a pair of equal hogging end moments — remains.
Primary moment diagram. By symmetry each end reaction of the simply supported primary beam is $$R=\tfrac{3}{2}(12)=18\ \text{kN}$$ so $$M_0(3)=18(3)=54\ \text{kN}\cdot\text{m},\qquad M_0(6)=18(6)-12(3)=72\ \text{kN}\cdot\text{m}$$ with straight lines between, and the diagram symmetric about mid-span.
Virtual system. A unit pair of end moments produces a constant bending moment $m=1$ over the whole span, which makes both integrals easy.
Flexibility coefficient. $$f=\int_0^{L}\frac{m^{2}}{EI}\,dx =\frac{2(3)}{2EI}+\frac{6}{EI}=\frac{9}{EI}$$
Load term. Integrating the primary diagram with the same weighting, using the trapezoidal areas $A_{0\text{-}3}=81$ and $A_{3\text{-}6}=189\ \text{kN}\cdot\text{m}^2$, $$\Delta_0=\int_0^{L}\frac{M_0 m}{EI}\,dx =2\left(\frac{81}{2EI}+\frac{189}{EI}\right)=\frac{459}{EI}$$
Compatibility. The built-in ends cannot rotate, so $$\Delta_0-X f=0 \;\Longrightarrow\; X=\frac{459/EI}{9/EI}=\boxed{51\ \text{kN}\cdot\text{m}}$$ The fixed-end moments are therefore $51\ \text{kN}\cdot\text{m}$ hogging at each end.
Complete the diagram and check. Superimposing, $$M(3)=54-51=+3,\qquad M(6)=72-51=+21\ \text{kN}\cdot\text{m}$$ and the points of contraflexure lie at $x=51/18=2.833\ \text{m}$ from each end. As a check, a prismatic beam with the same loads would give $\sum Pab^{2}/L^{2}=20.25+18+6.75=45\ \text{kN}\cdot\text{m}$; stiffening the end regions to $2EI$ correctly attracts more moment to the supports, $51>45$.

Question 3: resulting bending moment diagram, kN.m, sagging positive.

Question 3 — results
Quantity	Value
Fixed-end moment at A and at B	$51\ \text{kN}\cdot\text{m}$ (hogging)
End reactions	$18\ \text{kN}$ each (symmetry)
Moment at the $2EI/EI$ junctions	$+3\ \text{kN}\cdot\text{m}$ (sagging)
Mid-span moment	$+21\ \text{kN}\cdot\text{m}$ (sagging)
Points of contraflexure	$2.833\ \text{m}$ from each end

Topic. Force (flexibility) method applied to a stepped, doubly symmetric beam.

Key relations. $\sum_j f_{ij}X_j=-\Delta_{i0}$ with $f_{ij}=\int m_i m_j\,dx/EI$ and $\Delta_{i0}=\int M_0 m_i\,dx/EI$; for a symmetric structure under symmetric load the antisymmetric redundants vanish identically.

Why this works. Symmetry collapses a three-redundant problem to one. The released structure is a simple span, whose primary moment diagram is elementary; the virtual diagram for a symmetric pair of end couples is the constant $m=1$, so both integrals reduce to areas divided by the rigidity of each segment. Because only the ratio of the two integrals matters, the numerical value of $EI$ never enters — only the ratio $2EI:EI$ does, which is exactly why the question can be posed without a numerical rigidity.

Common pitfall. Using the prismatic formula $\sum Pab^{2}/L^{2}$ and then “correcting” it, or integrating $M_0 m$ with a single average $EI$. The stepped rigidity must enter the integrals segment by segment; here it changes the answer by 13 per cent. The other frequent error is to assume that stiffer end regions reduce the end moments; stiffness attracts moment, and the fixed-end moment rises.

Source. Hibbeler, Structural Analysis, ch. 10 (force method, non-prismatic members); Kassimali, ch. 13.

Question 3 of 9: Fixed-end moments of a non-prismatic beam by the flexibility method

Question 3: Fixed-end moments of a non-prismatic beam by the flexibility method (18 marks)