Question 5 of 8: Frame Analysis by Slope Deflection

Question 5: Frame Analysis by Slope Deflection (18 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 horizontal beam 15 m long carrying a rigid column at mid-length, dimensioned 8 m + 6 m + 1 m along the beam.

Find. The joint rotations, the member-end moments, the support reactions, and the shear and bending moment diagrams with maximum and minimum ordinates labelled for every member.

Approach. Establish first that the frame cannot sway, so the only unknowns are the rotations at nodes 2 and 3; write the slope-deflection equation for each member end, impose moment equilibrium at those two joints, and back-substitute for the end moments, shears and reactions.

1. Show that there is no sidesway. The members are inextensible and node 1 is fully fixed, so beam 1-2 pins node 2 horizontally; column 2-5 then pins node 2 vertically against the fixed base at node 5. Beam 2-3 carries the horizontal restraint on to node 3, whose roller removes its vertical freedom. Every joint translation is therefore suppressed and the only unknowns are $\theta_2$ and $\theta_3$. This is what the phrase “all members are inextensible” in the question is there to license.
2. Compute the fixed-end moments. For a uniformly loaded member, $\text{FEM} = w L^{2} / 12$ at each end, hogging at the left and sagging at the right in the clockwise-positive convention: $$\text{FEM}_{12} = -\frac{(12)(8)^{2}}{12} = -64, \qquad \text{FEM}_{21} = +64\ \text{kN}\cdot\text{m}$$ $$\text{FEM}_{23} = -\frac{(8)(6)^{2}}{12} = -24, \qquad \text{FEM}_{32} = +24\ \text{kN}\cdot\text{m}$$ The column carries no transverse load, so both its fixed-end moments are zero.
3. Treat the overhang as a known applied moment. The 1 m cantilever beyond the roller is determinate and has no rotational stiffness, so it must never be given a share of the joint balance. It delivers $$M_{34} = -\frac{w a^{2}}{2} = -\frac{(8)(1)^{2}}{2} = -4\ \text{kN}\cdot\text{m}$$ to node 3, together with its 8 kN of load.
4. Write the slope-deflection equations. With $M_{ij} = \dfrac{2EI}{L}\left( 2\theta_i + \theta_j \right) + \text{FEM}_{ij}$ and $\theta_1 = \theta_5 = 0$, writing $X = EI\theta_2$ and $Y = EI\theta_3$, $$M_{12} = 0.25X - 64, \qquad M_{21} = 0.5X + 64$$ $$M_{23} = \tfrac{2}{3}X + \tfrac{1}{3}Y - 24, \qquad M_{32} = \tfrac{2}{3}Y + \tfrac{1}{3}X + 24$$ $$M_{25} = X, \qquad M_{52} = 0.5X$$
5. Impose joint equilibrium and solve. At node 2 the three member ends must balance, and at node 3 the beam end must balance the overhang: $$M_{21} + M_{23} + M_{25} = 0 \quad\Rightarrow\quad 2.1667X + 0.3333Y = -40$$ $$M_{32} + M_{34} = 0 \quad\Rightarrow\quad 0.3333X + 0.6667Y = -20$$ $$EI\theta_2 = \boxed{-15.0\ \text{kN}\cdot\text{m}^{2}}, \qquad EI\theta_3 = \boxed{-22.5\ \text{kN}\cdot\text{m}^{2}}$$ Both come out as exact numbers, which is the arithmetic check that the sign of the overhang moment was entered correctly.
6. Back-substitute for the member-end moments. $$M_{12} = \boxed{-67.75}, \quad M_{21} = \boxed{+56.5}, \quad M_{23} = \boxed{-41.5}$$ $$M_{32} = \boxed{+4.0}, \quad M_{25} = \boxed{-15.0}, \quad M_{52} = \boxed{-7.5}\ \text{kN}\cdot\text{m}$$ Node 2 balances exactly, $56.5 - 41.5 - 15.0 = 0$, and node 3 balances against the overhang, $4.0 - 4.0 = 0$.
7. Convert the end moments to shears. Using $M_{\text{sag}} = +M_{ij}$ at a left end and $-M_{ji}$ at a right end, the shear at the left end of a uniformly loaded member is $V_i = \left( M_{\text{sag},j} - M_{\text{sag},i} + wL^{2}/2 \right)/L$. On beam 1-2 that gives $V = +49.41$ kN falling to $-46.59$ kN at node 2; on beam 2-3, $+30.25$ kN falling to $-17.75$ kN; on the overhang, $+8$ kN falling to zero; and on the column, a constant $5.625$ kN.
8. Locate the sagging peaks and assemble the reactions. Zero shear on 1-2 occurs at $x = 49.41/12 = 4.12$ m, where $$M = -67.75 + \frac{(49.41)^{2}}{2(12)} = \boxed{+33.96\ \text{kN}\cdot\text{m}}$$ and on 2-3 at $3.78$ m past node 2, where $$M = -41.5 + \frac{(30.25)^{2}}{2(8)} = \boxed{+15.69\ \text{kN}\cdot\text{m}}$$ The reactions follow from the shear jumps: $$R_1 = \boxed{49.41\ \text{kN}}, \qquad R_3 = 8 - (-17.75) = \boxed{25.75\ \text{kN}}, \qquad R_5 = \boxed{76.84\ \text{kN}}$$ and these sum to $152$ kN, exactly the applied $12(8) + 8(7) = 152$ kN.