Continuous frame by moment distribution: 16-Civ-A1 Elementary Structural Analysis, May 2018, Q6

Question 6: Continuous frame by moment distribution (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. Four $4\text{ m}$ columns, pinned at their bases (1,4,5,8) and rigidly framed into the continuous top beam at 2,3,6,7. Two internal hinges $1\text{ m}$ either side of the interior columns leave a $6\text{ m}$ suspended span at mid-length; end spans $2\!-\!3$ and $6\!-\!7$ are $8\text{ m}$; UDL $30\text{ kN/m}$ throughout. Symmetric about mid-span. Find. reactions, SFD and BMD.

Q6 — symmetric portal-and-suspended-span frame.

Suspended span. The $6\text{ m}$ span between the two hinges is simply supported: hinge reactions $=30(6)/2=90\text{ kN}$, mid-span sag $=30(6)^2/8=+135\text{ kN}\cdot\text{m}$.
Moment distribution on each half. Column stiffnesses are $3EI/L$ (pinned far end); distributing the fixed-end moments of the loaded end span into the columns and iterating (or exploiting symmetry) yields the joint moments $$\boxed{M_{@2}=M_{@7}=106.9\text{ kN}\cdot\text{m},\qquad M_{@3}=M_{@6}=159.4\text{ kN}\cdot\text{m (beam, hogging)}}$$ with interior columns carrying $54.4\text{ kN}\cdot\text{m}$ at their tops.
Reactions. Vertical support forces $$\boxed{V_1=V_8=113.4\text{ kN},\qquad V_4=V_5=246.6\text{ kN}}\quad(\Sigma V=720\text{ kN}=30\times24).$$
Diagram extremes. End-span beam: hogging $106.9$ at the exterior joint, hogging $159.4$ at the interior joint, sagging $+107.6\text{ kN}\cdot\text{m}$ near mid-span (shear zero at $3.78\text{ m}$). Columns are straight with tip moments $106.9$ (exterior) and $54.4\text{ kN}\cdot\text{m}$ (interior) and horizontal shears $26.7$ and $13.6\text{ kN}$ that cancel in pairs by symmetry.

Quantity	Value
Exterior column base reaction $V_1,V_8$	$113.4\text{ kN}$
Interior column base reaction $V_4,V_5$	$246.6\text{ kN}$
Beam hog at interior joint (3,6)	$159.4\text{ kN}\cdot\text{m}$
Beam hog at exterior joint (2,7)	$106.9\text{ kN}\cdot\text{m}$
End-span max sag	$+107.6\text{ kN}\cdot\text{m}$
Suspended-span max sag	$+135\text{ kN}\cdot\text{m}$
Interior column top moment	$54.4\text{ kN}\cdot\text{m}$

Topic. Moment distribution for an indeterminate frame with internal hinges and symmetry.

Key relations. Distribution factor $DF=\dfrac{k_i}{\sum k}$ with $k=4EI/L$ (fixed far end) or $3EI/L$ (pinned far end); fixed-end moment for a UDL $=wL^2/12$; carry-over $=\tfrac12$.

Why this works. Moment distribution is Hardy Cross’s iterative relaxation of the slope-deflection equations: joints are locked, fixed-end moments applied, then released one at a time until unbalanced moments vanish. The two internal hinges detach a determinate suspended span, and symmetry lets you analyse a single half, halving the work. Because every member shares one $EI$, the distribution factors are pure length ratios.

Common pitfall. Using $4EI/L$ for a column whose base is pinned (it is $3EI/L$), or loading the frame with the suspended span still attached instead of replacing it by its $90\text{ kN}$ tip reactions.

Source. Hibbeler 10e, Ch. 12 (moment distribution); Kassimali 6e, Ch. 16.

Question 6 of 8: Continuous frame by moment distribution

Question 6: Continuous frame by moment distribution (22 marks)