Stiffness Matrix of a Straight Non-Prismatic Beam: 07-Str-A4, December 2016, Q6

Question 6: Stiffness Matrix of a Straight Non-Prismatic Beam (22 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 straight, symmetric, stepped beam of total length $L=12$ m, made of three prismatic segments.

Given data
Segment	Extent (m)	Flexural rigidity
A to the first step	$0\le x\le 3$	$2EI$
Central length	$3\le x\le 9$	$EI$
Second step to B	$9\le x\le 12$	$2EI$
Total length	$L=12$ m

Find. The four terms of the member stiffness matrix $[K]$ relating the end moments to the end slopes, $\{M_A,\,M_B\}^{\mathsf T}=EI[K]\{\theta_A,\,\theta_B\}^{\mathsf T}$.

Question 6: the stepped beam. The stiffened end lengths are 2EI over 3 m at each end, with EI over the central 6 m.

Approach. Build the $2\times2$ flexibility matrix of the simply supported beam by integrating the unit-moment diagrams against $1/EI(x)$, then invert it. Inversion is cheap for a $2\times2$ matrix and avoids any need to integrate a variable-stiffness differential equation.

Set up the flexibility problem. Support the member as a simple beam and apply the end moments $M_A$ and $M_B$. With the sign convention $M_{\text{sag}}(A)=-M_A$ and $M_{\text{sag}}(B)=+M_B$, the sagging moment is linear: $$m(x)=-M_A\left(1-\frac{x}{L}\right)+M_B\left(\frac{x}{L}\right).$$ The unit-moment fields conjugate to $\theta_A$ and $\theta_B$ are therefore $\bar m_A=-(1-x/L)$ and $\bar m_B=x/L$.
Integrate through the three segments. The flexibility coefficients are $f_{ij}=\int_0^{L}\bar m_i\bar m_j\,\mathrm{d}x/EI(x)$. Splitting each integral at the two steps and dividing the end segments by 2, $$f_{AA}=f_{BB}=\frac{1}{EI}\left(\frac{2.3125}{2}+1.625+\frac{0.0625}{2} \right)=\frac{2.8125}{EI}=\frac{45}{16EI},$$ $$f_{AB}=-\frac{1}{EI}\left(\frac{0.3125}{2}+1.375+\frac{0.3125}{2}\right) =-\frac{1.6875}{EI}=-\frac{27}{16EI}.$$ Symmetry of the beam is what makes $f_{AA}=f_{BB}$; if it were not symmetric the two diagonal terms would differ.
Invert to get the stiffness matrix. With $\det[f]=\left(\tfrac{45}{16}\right)^{2}-\left(\tfrac{27}{16}\right)^{2} =\tfrac{1296}{256}$ (all divided by $EI^{2}$), $$[K]=[f]^{-1}=\frac{EI}{\det}\begin{bmatrix}45/16 & 27/16\\ 27/16 & 45/16\end{bmatrix} \;\Longrightarrow\; \boxed{\begin{Bmatrix}M_A\\ M_B\end{Bmatrix} =EI\begin{bmatrix}\dfrac{5}{9} & \dfrac{1}{3}\\[6pt] \dfrac{1}{3} & \dfrac{5}{9}\end{bmatrix} \begin{Bmatrix}\theta_A\\ \theta_B\end{Bmatrix}.}$$ Numerically $k_{AA}=k_{BB}=0.5556\,EI$ and $k_{AB}=k_{BA}=0.3333\,EI$.
Interpret the two numbers. A prismatic beam of the same length would give $k_{AA}=4EI/L=0.3333\,EI$ and $k_{AB}=2EI/L=0.1667\,EI$, so the haunched member is $\boxed{1.667}$ times stiffer. More telling is the carry-over factor $$\mathrm{COF}=\frac{k_{AB}}{k_{AA}}=\frac{1/3}{5/9}=\boxed{0.600},$$ against $0.500$ for a prismatic member. Stiffening the ends drives a larger fraction of any applied end moment across to the far end, which is precisely why haunched members change the outcome of a moment distribution and why non-uniform members need their own stiffness and carry-over tables.
Check the result independently. Impose $\theta_A=1$ with $\theta_B=0$ on a three-element stiffness model of the same beam; the support reactions return $M_A=0.5556\,EI$ and $M_B=0.3333\,EI$, matching the inverted flexibility matrix. Symmetry of $[K]$ (guaranteed by Maxwell–Betti) and positive definiteness ($k_{AA}\gt|k_{AB}|$) are the two structural checks that should be made on any hand-assembled member matrix.

Question 6 — results
Quantity	Value
$f_{AA}=f_{BB}$	$45/(16EI)=2.8125/EI$
$f_{AB}=f_{BA}$	$-27/(16EI)=-1.6875/EI$
$k_{AA}=k_{BB}$	$\tfrac{5}{9}EI=0.5556\,EI$
$k_{AB}=k_{BA}$	$\tfrac{1}{3}EI=0.3333\,EI$
Carry-over factor	0.600 (prismatic: 0.500)
Rotational stiffness relative to a prismatic beam	1.667

Topic. Member stiffness of a non-prismatic (haunched) beam, obtained by inverting the end-rotation flexibility matrix.

Key relations. $f_{ij}=\displaystyle\int_0^{L}\frac{\bar m_i\bar m_j}{EI(x)}\mathrm{d}x$ with $\bar m_A=-(1-x/L)$, $\bar m_B=x/L$; $[K]=[f]^{-1}$; carry-over factor $\mathrm{COF}=k_{AB}/k_{AA}$; prismatic limits $k_{AA}=4EI/L$, $k_{AB}=2EI/L$, $\mathrm{COF}=0.5$.

Why this works. The stiffness matrix relating end moments to end rotations is the inverse of the flexibility matrix relating end rotations to end moments, provided both are written for the same pair of conjugate quantities on the same simply supported release. That equivalence is what lets a variable-$EI$ member be handled with nothing more than two integrals and a $2\times2$ inversion. The unit-moment diagrams stay linear no matter how $EI$ varies, because they come from equilibrium alone; all the geometry of the haunch enters through the weighting $1/EI(x)$ inside the integral. Maxwell–Betti guarantees $f_{AB}=f_{BA}$ and hence a symmetric $[K]$, which is the single most useful check on the arithmetic.

Common pitfall. The frequent error is to integrate $\bar m_i\bar m_j$ over the whole span and only afterwards divide by some average $EI$. That is not the same integral: the segments must be weighted individually, and here the difference is a factor of $5/3$ in the stiffness and a shift of the carry-over factor from 0.5 to 0.6. A second error is to report $[K]$ without stating the sign convention for $M_A$ and $M_B$; the numerical terms depend on it, and a mixed convention produces an unsymmetric matrix that should be recognised immediately as wrong.

Source. Hibbeler, Structural Analysis, 10th ed., §12.9 (non-prismatic members, stiffness and carry-over factors) and Ch. 15; Ghali, Neville & Brown, 7th ed., §6.9; Kassimali, 6th ed., §16.2.

Question 6 of 9: Stiffness Matrix of a Straight Non-Prismatic Beam

Question 6: Stiffness Matrix of a Straight Non-Prismatic Beam (22 marks)