Frame with an Inclined Member and Two Built-in Bases at Different Levels: 07-Str-A4, May 2016, Q7

Question 7: Frame with an Inclined Member and Two Built-in Bases at Different Levels (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
Joint coordinates	1 (0, 0); 2 (0, 5); 3 (4, 8); 4 (4, 3) m
Member lengths	1–2 = 5 m; 2–3 = 5 m (4–3–5 triangle); 3–4 = 5 m
Supports	joints 1 and 4 fully built in
Load	42 kN at joint 3, along member 2–3 (direction 0.8, 0.6)
Components	$F_x$ = 33.6 kN; $F_y$ = 25.2 kN
Members	uniform EI, inextensible

Find. The member end moments, shears, axial forces and support reactions, and the bending moment diagram with its extreme ordinates.

Question 7 — both column bases are fully built in and the two bases are at different levels; the 42 kN load acts at joint 3 along the axis of member 2–3. All members inextensible with the same EI.

Approach. Prove that the sway pattern has a single parameter, write the three slope-deflection equations, close the system with the virtual-work sway equation for a unit horizontal translation, and finish with statics. All three members happen to be 5 m long, which makes the algebra unusually clean.

Resolve the load. Member 2–3 has a 4 m run and a 3 m rise, so its unit vector is $(0.8,\,0.6)$ and$$F_x = 42(0.8) = 33.6\ \text{kN},\qquad F_y = 42(0.6) = 25.2\ \text{kN}$$both positive, i.e. up and to the right.
Establish the sway pattern from inextensibility alone. Both columns are vertical and inextensible, so $v_2 = v_1 = 0$ and $v_3 = v_4 = 0$. The inclined member is then inextensible only if$$(u_3-u_2)(0.8) + (v_3-v_2)(0.6) = 0 \;\Longrightarrow\; u_3 = u_2 = \Delta$$so both joints translate horizontally by the same amount and the inclined member has zero chord rotation:$$\psi_{12} = \psi_{34} = \frac{\Delta}{5},\qquad \psi_{23} = 0$$That single observation removes the usual difficulty with inclined members and leaves three unknowns, $\theta_2,\ \theta_3,\ \Delta$.
Slope-deflection equations. With $k = EI/5$ for every member, $\theta_1 = \theta_4 = 0$, and no span loads (the 42 kN acts at a joint):$$M_{12}=2k(\theta_2-3\psi),\quad M_{21}=2k(2\theta_2-3\psi),\quad M_{23}=2k(2\theta_2+\theta_3)$$$$M_{32}=2k(2\theta_3+\theta_2),\quad M_{34}=2k(2\theta_3-3\psi),\quad M_{43}=2k(\theta_3-3\psi)$$
Joint equilibrium. Balancing joints 2 and 3,$$4\theta_2 + \theta_3 - 3\psi = 0,\qquad 4\theta_3 + \theta_2 - 3\psi = 0$$Subtracting shows $\theta_2 = \theta_3 = \theta$, whence $5\theta = 3\psi$, i.e. $\theta = 0.6\psi$. The equality of the two joint rotations is a consequence of the equal member lengths and of the load acting only at a joint.
Sway equation by virtual work. Give the frame a virtual translation $\Delta^{*} = 1$ to the right; the two columns take $\psi^{*} = 1/5$, the inclined member takes $\psi^{*} = 0$, and joint 3 moves one unit horizontally so the applied load does work $F_x$. Then$$\frac{M_{12}+M_{21}}{5} + \frac{M_{34}+M_{43}}{5} + F_x = 0$$$$\Longrightarrow\; 4k\left(3\theta - 6\psi\right) = -5(33.6) \;\Longrightarrow\; k\psi = \boxed{10.0},\quad k\theta = 6.0$$so $EI\theta = 30.0$ and $EI\Delta = 5EI\psi = 250$, a sway to the right of $250/EI$ metres — the same sense as the horizontal component of the load, as it must be.
Member end moments. Substituting $k\theta = 6.0$ and $k\psi = 10.0$,$$M_{12} = 2(6-30) = -48.0,\qquad M_{21} = 2(12-30) = -36.0$$$$M_{23} = M_{32} = 2(3\times 6) = +36.0,\qquad M_{34} = -36.0,\qquad M_{43} = -48.0\ \text{kN}\cdot\text{m}$$Both joints balance to zero. In the sagging convention the base moments are$$\boxed{M_1 = M_4 = 48.0\ \text{kN}\cdot\text{m}},\qquad \boxed{M_2 = M_3 = 36.0\ \text{kN}\cdot\text{m}}$$the frame being point-symmetric in its response even though its geometry is not mirror-symmetric.

Question 7 — bending moment diagram drawn normal to each member (ordinates in kN·m). Every member has a straight moment line because none carries a span load; the contraflexure point in each column is where the diagram crosses the axis.

Shears and the storey check. With no span load each member has a constant transverse shear:$$V_{1\text{-}2} = \frac{48.0+36.0}{5} = 16.8\ \text{kN},\qquad V_{3\text{-}4} = \frac{48.0+36.0}{5} = 16.8\ \text{kN}$$$$V_{2\text{-}3} = \frac{36.0+36.0}{5} = 14.4\ \text{kN}$$The two column shears sum to $33.6$ kN, exactly the applied horizontal component — the decisive arithmetic check on the sway equation.
Axial forces and reactions. Global equilibrium, using the two base fixing moments of 48.0 kN·m, gives$$H_1 = H_4 = -16.8\ \text{kN},\qquad V_1 = -30.6\ \text{kN},\qquad V_4 = +5.4\ \text{kN}$$so the left base is a hold-down of 30.6 kN. Resolving at the joints, member 1–2 carries 30.6 kN tension, member 2–3 carries 31.8 kN tension and member 3–4 carries 5.4 kN compression. Moments about joint 4 close the check: $-168 + 72 + 96 = 0$.
Reading the diagrams. Every member has a straight moment line, so each has exactly one point of contraflexure. In column 1–2 it lies $48.0/(48.0+36.0) \times 5 = 2.857$ m above the base; in the inclined member it is exactly at mid-length, because the two end moments are equal in magnitude and opposite in sagging sense; in column 3–4 it lies 2.857 m above joint 4. The largest moment anywhere in the frame is 48.0 kN·m at each built-in base.

Quantity	Value
Load components	33.6 kN horizontal, 25.2 kN vertical (up)
Joint rotations	$EI\theta_2 = EI\theta_3 = 30.0$ (clockwise positive)
Sway	$\Delta = 250/EI$ to the right
Base moments $M_1,\ M_4$	48.0 kN·m each (maximum ordinates)
Joint moments $M_2,\ M_3$	36.0 kN·m each
Member shears	16.8 / 14.4 / 16.8 kN in members 1–2, 2–3, 3–4
Axial forces	30.6 kN T, 31.8 kN T, 5.4 kN C
Reactions	$H_1 = H_4 = 16.8$ kN (leftward); $V_1 = 30.6$ kN down; $V_4 = 5.4$ kN up
Points of contraflexure	2.857 m from each base; mid-length of member 2–3

Topic. Slope-deflection analysis of a sway frame containing an inclined member, solved by establishing the sway mechanism from inextensibility before writing any stiffness.

Key relations. Inextensibility $(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e} = 0$; $\psi_{ij} = [(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_{\perp}]/L$; virtual-work sway equation $\sum (M_{ij}+M_{ji})\psi^{*}_{ij} + \sum F\cdot d^{*} = 0$; $V = -(M_{ij}+M_{ji})/L$.

Why this works. Inclined members frighten candidates because the sway pattern normally couples horizontal and vertical joint movement. Here the two columns are vertical, so their inextensibility pins both joints to their own levels, and the inclined member can then only survive if the two joints move horizontally by the same amount. Its chord rotation is consequently zero and it contributes nothing to the sway equation, leaving a system no harder than an ordinary portal. Writing the sway equation by virtual work rather than by free bodies also avoids having to resolve the inclined member’s shear into global directions, and it automatically picks up the work done by the applied load.

Common pitfall. The two recurring errors are assuming the inclined member must have a chord rotation (it does not, once the columns are shown to be vertical and inextensible) and using the total 42 kN in the sway equation instead of its horizontal component of 33.6 kN. The storey-shear check — the two column shears must add to the applied horizontal force — catches both immediately.

Source. Hibbeler, Structural Analysis, 10th ed., §11.5 (frames with sidesway, inclined members); Kassimali, 6th ed., §16.6; McCormac, 4th ed., Ch. 16.

Question 7 of 8: Frame with an Inclined Member and Two Built-in Bases at Different Levels

Question 7: Frame with an Inclined Member and Two Built-in Bases at Different Levels (21 marks)