Fixed-End Moment of a Non-Prismatic Beam by the Flexibility Method: 07-Str-A4, May 2016, Q3

Question 3: Fixed-End Moment of a Non-Prismatic Beam by the Flexibility Method (21 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.

Item	Value
Span A–B	9.0 m (3 + 3 + 3 m)
Support A	roller (vertical reaction only)
Support B	fully built in
Flexural rigidity, 0 ≤ x < 6 m	EI
Flexural rigidity, 6 m ≤ x ≤ 9 m	2EI
Point loads	8 kN at x = 3 m; 37 kN at x = 6 m

Find. The bending moment developed at the built-in end B.

Question 3 — non-prismatic beam on a roller at A and fully built in at B; the heavy portion adjacent to B has flexural rigidity 2EI.

Approach. Release the roller reaction $R_A$ as the single redundant, leaving a determinate cantilever fixed at B; compute the tip deflection under the loads and under a unit force, restore compatibility, then take moments for $M_B$. The step change in $EI$ enters only as a change of integrand at $x = 6$ m.

Establish the degree of indeterminacy. The unknown reaction components are $R_A$ at the roller and $R_B,\,H_B,\,M_B$ at the built-in end — four, against three equations of planar statics. The beam is therefore one degree statically indeterminate, and one release suffices.
Choose the release and the primary structure. Remove the roller at A. The primary structure is a cantilever fixed at B and free at A, which is determinate and whose bending moment is written directly from the free body to the left of any section, measuring $x$ from A:$$M_0(x) = -8\langle x-3\rangle - 37\langle x-6\rangle$$where $\langle\;\rangle$ is the Macaulay bracket (zero when negative).
Unit-load moment field. Apply a unit upward force at A on the same primary structure; the free body to the left of a section then contains only that force, so$$m_1(x) = x$$a single straight line over the whole 9 m — the step in $EI$ changes the flexibility integrals, not the statics.
Flexibility coefficients. With $\Delta_{10} = \int M_0 m_1 / EI\,\mathrm{d}x$ and $f_{11} = \int m_1^{2}/EI\,\mathrm{d}x$, split each integral at $x = 6$ m. For the displacement term,$$\int_3^6 \frac{-8(x-3)\,x}{EI}\,\mathrm{d}x = -\frac{180}{EI},\qquad \int_6^9 \frac{[-8(x-3)-37(x-6)]\,x}{2EI}\,\mathrm{d}x = -\frac{1080}{EI}$$so that$$\Delta_{10} = \boxed{-\frac{1260}{EI}}$$the minus sign meaning that A deflects downward under the loads, as it must.
Flexibility of the released structure. Likewise$$f_{11} = \int_0^6\frac{x^{2}}{EI}\mathrm{d}x + \int_6^9\frac{x^{2}}{2EI}\mathrm{d}x = \frac{72}{EI} + \frac{85.5}{EI} = \boxed{\frac{157.5}{EI}}$$Notice how little the stiff outer bay contributes relative to its lever arm: doubling $EI$ there halves the largest part of the integrand exactly where $m_1$ is greatest.
Compatibility at the roller. The real support permits no vertical movement at A, so$$\Delta_{10} + R_A f_{11} = 0 \quad\Longrightarrow\quad R_A = -\frac{\Delta_{10}}{f_{11}} = \frac{1260}{157.5} = \boxed{8.00\ \text{kN (upward)}}$$The $1/EI$ cancels, which is why a numerical value of $EI$ was never needed — only the ratio of the two rigidities matters.
Recover the fixed-end moment by statics. Taking moments about B for the whole beam,$$M_B = R_A(9) - 8(6) - 37(3) = 72 - 48 - 111 = \boxed{-87.0\ \text{kN}\cdot\text{m}}$$that is a hogging moment of magnitude 87.0 kN·m at the built-in end, with tension on the top fibre.
Check the answer independently. A direct-stiffness model of the same beam, with the $EI$/2$EI$ elements assembled and the roller and built-in end imposed, returns $R_A = 8.000$ kN and $M_B = -87.00$ kN·m. The shape of the moment diagram is also sensible: $M = +24.0$ kN·m sagging under the 37 kN load and $+24.0$ kN·m under the 8 kN load, falling to $-87.0$ at the wall.

Quantity	Value
Degree of indeterminacy	1
$\Delta_{10}$ (deflection at A, primary structure)	−1260/EI
$f_{11}$	157.5/EI
Roller reaction $R_A$	8.00 kN upward
Moment at the built-in end, $M_B$	87.0 kN·m hogging
Sagging moment under each point load	24.0 kN·m

Topic. The force (flexibility) method applied to a singly-indeterminate non-prismatic beam.

Key relations. $\Delta_{10} + R f_{11} = 0$; $\Delta_{10} = \int M_0 m_1/EI\,\mathrm{d}x$; $f_{11} = \int m_1^{2}/EI\,\mathrm{d}x$; Macaulay brackets for stepped loading.

Why this works. The force method replaces an unknown reaction by an unknown force and then insists that the deflection it was preventing is genuinely zero. Both flexibility integrals are evaluated on the same determinate primary structure, so the statics never gets harder than a cantilever; a variable $EI$ is absorbed simply by splitting the integral at each change of section, which is why the method is the natural choice for haunched and stepped members where slope-deflection would need modified stiffness and carry-over factors. Because $EI$ appears in the same power in both integrals it cancels in the ratio, so a stepped beam is solved from the rigidity ratio alone.

Common pitfall. Two errors dominate. Integrating with a single constant $EI$ and applying a correction afterwards is not equivalent, because $m_1$ is largest exactly where the section changes. And once $R_A$ is found, candidates often try to write $M_B$ from another integral instead of from simple statics, which is both slower and easier to get wrong.

Source. Hibbeler, Structural Analysis, 10th ed., §10.1–10.4 (force method, beams with variable moment of inertia); Ghali, Neville & Brown, 7th ed., Ch. 4.

Quantity	Value
Degree of indeterminacy	1
\(\Delta_{10}\) (deflection at A, primary structure)	−1260/EI
\(f_{11}\)	157.5/EI
Roller reaction \(R_A\)	8.00 kN upward
Moment at the built-in end, \(M_B\)	87.0 kN·m hogging
Sagging moment under each point load	24.0 kN·m

Question 3 of 8: Fixed-End Moment of a Non-Prismatic Beam by the Flexibility Method

Question 3: Fixed-End Moment of a Non-Prismatic Beam by the Flexibility Method (21 marks)