Question 6 of 8: Slope-Deflection Analysis of a Propped Frame

Question 6: Slope-Deflection Analysis of a Propped Frame (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.

Find. The end moments by slope deflection, the reactions, and the shear force and bending moment diagrams of every member with maximum and minimum ordinates labelled.

Approach. Show first that the frame cannot sway, so the only unknowns are the joint rotations at ③ and ④; write the slope-deflection equations with the modified stiffness $3EI/L$ for the pin-based column, take the overhang moment into joint ④ as a known applied moment, and solve two simultaneous equations.

1. Establish that there is no sidesway. Node ② is built in, and member ②–③ is inextensible, so ③ cannot move horizontally; member ③–④ then fixes ④ horizontally as well. Vertically, the inextensible column pinned at ① holds ③ and the roller holds ④. Every joint translation is therefore prevented and the chord rotations $\psi$ are all zero — which is exactly what the phrase “members are inextensible” in the question is there to license. The unknowns are the two rotations $\theta_3$ and $\theta_4$.
2. Fixed-end moments and member stiffnesses. Only span ③–④ is loaded, giving $\mathrm{FEM}_{34} = -wL^2/12 = -12(8)^2/12 = -64$ kN.m and $\mathrm{FEM}_{43} = +64$ kN.m. The stiffness coefficients are $2E(3I)/6 = EI$ for the 6 m beam and $2E(4I)/8 = EI$ for the 8 m beam, so both spans happen to share the same coefficient. The column is pinned at ①, so use the modified form $M_{31} = (3EI/L)\theta_3 = EI\theta_3$ with $M_{13} = 0$ and no carry-over.
3. Write the slope-deflection equations. With $\theta_2 = 0$ at the built-in end, $$M_{23} = EI\theta_3, \qquad M_{32} = 2EI\theta_3$$ $$M_{34} = EI(2\theta_3 + \theta_4) - 64, \qquad M_{43} = EI(2\theta_4 + \theta_3) + 64$$ and $M_{31} = EI\theta_3$ for the column.
4. The overhang delivers a known moment to joint ④. The 1 m tip carries 30 kN, so the sagging moment at ④ taken from the right is $-30(1) = -30$ kN.m. Since ④ is the left end of member ④–⑤, that sagging value is $M_{45}$, so joint equilibrium reads $M_{43} + M_{45} = 0$, that is $M_{43} = +30$ kN.m. The overhang has no stiffness and must never be given a share of any balancing moment; it enters only through this fixed value.
5. Solve the two joint equations. Joint ③ requires $M_{32} + M_{34} + M_{31} = 0$, which gives $EI(5\theta_3 + \theta_4) = 64$; joint ④ requires $M_{43} = 30$, which gives $EI(\theta_3 + 2\theta_4) = -34$. Solving, $$EI\theta_3 = 18, \qquad EI\theta_4 = -26$$ Exact integers, which is the confirmation that the sign of the overhang moment was taken correctly — the opposite sign gives ragged decimals and nothing else in the solution lands on a round number.
6. Back-substitute for the end moments. $$\boxed{M_{23} = +18,\quad M_{32} = +36,\quad M_{34} = -54,\quad M_{43} = +30,\quad M_{31} = +18\ \text{kN.m}}$$ Joint ③ checks at once: $36 - 54 + 18 = 0$.
7. Member shears and reactions. Span ②–③ carries no load, so its shear is constant at $(-36 - 18)/6 = -9$ kN. On span ③–④, writing $M(x) = -54 + V_3x - 6x^2$ and forcing $M(8) = -30$ gives $V_3 = +51$ kN, falling to $-45$ kN at ④. The overhang carries a constant $+30$ kN. The column moment runs from $+18$ kN.m at ③ to zero at the pin over 3 m, so its shear is 6 kN. The reactions follow: at the built-in end ②, $9$ kN downward together with $6$ kN horizontal and a fixing moment of $18$ kN.m; at the column pin ①, $60$ kN upward and $6$ kN horizontal the other way; at the roller ④, $75$ kN upward. Vertical equilibrium closes: $-9 + 60 + 75 = 126 = 12(8) + 30$ kN.
8. Extreme ordinates for the diagrams. The shear is $-9$ kN throughout ②–③, rises to $+51$ kN at ③ and crosses zero $51/12 = 4.25$ m along the loaded span before reaching $-45$ kN at the roller. The peak sagging moment there is $$M_{\max} = -54 + \frac{51^2}{2(12)} = \boxed{+54.375\ \text{kN.m}}$$ The beam moment is $+18$ kN.m at the wall, falls through zero at 1.33 m, reaches $-36$ kN.m just left of ③ and jumps to $-54$ kN.m just right of it — the jump being exactly the 18 kN.m the column delivers — then swings positive over most of the loaded span and closes at $-30$ kN.m over the roller before running out to zero at the tip.