Flexibility (force) method analysis of an L-frame: 16-Civ-B1 Advanced Structural Analysis, May 2013, Q7

Question 7: Flexibility (force) method analysis of an L-frame (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. An L-shaped frame: member 1–2 is horizontal, 6 m long, built in at joint 1 and carrying a uniformly distributed load of 3 kN/m; member 2–3 is vertical, 3 m long, rigidly connected to the beam at the corner joint 2 and pinned to the foundation at joint 3. Both members have the same $EI$ and are inextensible.

Find. The redundant reactions at the pin, the complete set of member end actions, and the shear and bending moment diagrams with the maximum and minimum ordinates in each member.

Q7 — the L-frame and its bending moment diagram. The frame is twice statically indeterminate; the two components of the pin reaction at joint 3 are the natural redundants.

Approach. Release the pin at joint 3 to leave a statically determinate cantilever, compute the released displacements $\delta_{i0}$ and the flexibility coefficients $f_{ij}$ by the unit-load method, and solve the two compatibility equations $\delta_{i0}+\sum_j f_{ij}X_j = 0$.

Count the redundants. The built-in support supplies three reaction components and the pin supplies two, giving five against three equations of equilibrium: the frame is twice statically indeterminate. Take $X_1=H_3$ (positive to the right) and $X_2=V_3$ (positive upward) as the redundants, so the released structure is a cantilever fixed at joint 1 with a free end at joint 3.
Moments in the released structure. Measuring $x$ from joint 1 along the beam and $t$ upward from joint 3 along the column, and taking the free body on the side remote from the fixed support, $$M_0 = -\frac{w(6-x)^{2}}{2}\ \text{on the beam},\qquad M_0=0\ \text{on the column},$$ $$m_1 = 3\ \text{on the beam},\qquad m_1 = 3-t\ \text{on the column},$$ $$m_2 = 6-x\ \text{on the beam},\qquad m_2 = 0\ \text{on the column}.$$ The unit horizontal force at joint 3 produces a constant moment along the beam equal to the column height, which is the geometric feature that makes $f_{11}$ large.
Evaluate the released displacements. With $EI$ cancelling from every term, $$\delta_{10}=\int_0^{6} M_0\,m_1\,dx = -\frac{w\,b\,a^{3}}{6} = -\frac{3(3)(216)}{6} = -324 ,$$ $$\delta_{20}=\int_0^{6} M_0\,m_2\,dx = -\frac{w\,a^{4}}{8} = -\frac{3(1296)}{8} = -486 ,$$ where $a=6\text{ m}$ is the beam length and $b=3\text{ m}$ the column height.
Evaluate the flexibility coefficients. $$f_{11}=b^{2}a+\frac{b^{3}}{3}=9(6)+9=63,\qquad f_{12}=f_{21}=\frac{b\,a^{2}}{2}=54,\qquad f_{22}=\frac{a^{3}}{3}=72 .$$ The matrix is symmetric, as Maxwell's reciprocal theorem requires.
Solve the compatibility equations. $$\begin{aligned} 63X_1+54X_2 &= 324\\ 54X_1+72X_2 &= 486 \end{aligned} \qquad\Longrightarrow\qquad \boxed{X_1 = H_3 = -1.80\ \text{kN},\qquad X_2 = V_3 = +8.10\ \text{kN}.}$$ The negative sign means the pin pushes the frame 1.80 kN to the left.
Complete the reactions. Vertical equilibrium gives $V_1 = wa - V_3 = 18.0-8.10 = 9.90\ \text{kN}$ upward and horizontal equilibrium gives $H_1 = +1.80\ \text{kN}$ to the right. The fixed-end moment follows from the superposition $M = M_0 + X_1m_1 + X_2m_2$ evaluated at $x=0$: $$M_1 = -54 + (-1.80)(3) + 8.10(6) = -10.80\ \text{kN}\cdot\text{m}\ \text{(hogging).}$$
Build the beam diagrams. Writing $u=6-x$ (measured back from the corner), $$M(u) = -1.5u^{2}+8.10u-5.40 .$$ At the corner ($u=0$) this gives $-5.40\ \text{kN}\cdot\text{m}$ and at the fixed end ($u=6$) it gives $-10.80\ \text{kN}\cdot\text{m}$, both hogging. Differentiating, $dM/du = -3u+8.10 = 0$ at $u = 2.70\text{ m}$, i.e. 3.30 m from the fixed support, where $$\boxed{M_{\max} = +5.535\ \text{kN}\cdot\text{m}\ \text{(sagging).}}$$ The two points of contraflexure are at 1.379 m and 5.221 m from joint 1. The beam shear runs linearly from $+9.90\text{ kN}$ at the fixed end to $-8.10\text{ kN}$ at the corner, crossing zero at 3.30 m as it must.
Column diagrams and the joint check. The column carries no transverse load, so its shear is the constant $1.80\text{ kN}$ and its moment varies linearly from zero at the pin to $-5.40\ \text{kN}\cdot\text{m}$ at the corner — identical in magnitude to the beam moment there, which is the equilibrium condition for a two-member rigid joint. The column also carries an axial compression of $8.10\text{ kN}$.

Final results — Question 7
Quantity	Member 1–2 (beam)	Member 2–3 (column)
Maximum shear	+9.90 kN at joint 1	1.80 kN (constant)
Minimum shear	−8.10 kN at joint 2	1.80 kN (constant)
Maximum bending moment	+5.535 kN·m sagging, 3.30 m from joint 1	0 at the pin
Minimum bending moment	−10.80 kN·m at joint 1	−5.40 kN·m at joint 2
Axial force	1.80 kN	8.10 kN compression
Reactions	$H_1=1.80\ \text{kN}\rightarrow$, $V_1=9.90$ kN, $M_1=10.80$ kN·m; $H_3=1.80\ \text{kN}\leftarrow$, $V_3=8.10$ kN

Topic. The flexibility (force) method with two redundants, and the unit-load evaluation of flexibility coefficients for a frame.

Key relations. $\delta_{i0}+\sum_j f_{ij}X_j = 0$; $\delta_{i0}=\displaystyle\int\frac{M_0 m_i}{EI}\,ds$ and $f_{ij}=\displaystyle\int\frac{m_i m_j}{EI}\,ds$; Maxwell's reciprocal theorem $f_{ij}=f_{ji}$; superposition $M = M_0 + \sum_j X_j m_j$.

Why this works. Releasing enough restraints to leave a determinate structure lets every internal action be written as a linear combination of the applied load and the unknown redundants. Compatibility — the fact that the released displacements must vanish at the real restraints — then supplies exactly as many equations as there are redundants. Because both members share the same $EI$, that rigidity cancels from every integral: the redundants of an indeterminate frame depend only on the relative stiffnesses of its members, never on their absolute value. That is why the question can be answered without a numerical $EI$.

Common pitfall. Two errors dominate. The first is inconsistent sign conventions between $M_0$, $m_1$ and $m_2$ — every one of the three must be computed from the same free body with the same positive sense, or the off-diagonal coefficient $f_{12}$ comes out with the wrong sign and the solution silently reverses. The second is forgetting the column contribution to $f_{11}$: the vertical member does carry moment under the unit horizontal force, and dropping its $b^{3}/3 = 9$ from the total of 63 changes $H_3$ from 1.80 kN to 3.00 kN and $V_3$ from 8.10 kN to 9.00 kN. Checking that $f_{ij}$ is symmetric catches most bookkeeping slips.

Source. Hibbeler, Structural Analysis, 10th ed., §10.1–10.5 (force method for beams and frames, Maxwell's theorem) and §9.4 (virtual work for frames); Kassimali, Structural Analysis, 6th ed., §13.1–13.4 of the method of consistent deformations; Ghali, Neville and Brown, Ch. 4.

Question 7 of 9: Flexibility (force) method analysis of an L-frame

Question 7: Flexibility (force) method analysis of an L-frame (24 marks)