Frame by Slope Deflection — Shear and Moment Diagrams: 07-Str-A1, May 2017, Q7

Question 7: Frame by Slope Deflection — Shear and Moment Diagrams (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.

Item	Value
Joint coordinates (m)	$1(0,2)$, $2(4,5)$, $3(14,5)$, $4(14,0)$, $5(4,0)$
Member 1–2	inclined, 4 m by 3 m so $L = 5$ m, stiffness $EI$; joint 1 is a pin support
Member 2–3	horizontal, $L = 10$ m, stiffness $2EI$, carrying 9 kN/m downward
Member 3–4	vertical, $L = 5$ m, stiffness $1.5EI$; joint 4 is a fixed support
Member 2–5	vertical, $L = 5$ m, stiffness $EI$; joint 5 is a pin support
Assumption stated on the paper	all members inextensible

Find. All member-end moments, and the shear force and bending moment diagrams for every member with their maximum and minimum ordinates.

Question 7: the frame, and the shear and bending moment diagrams for the loaded member 2–3. The unloaded members carry linear moment diagrams and constant shears, tabulated below.

Approach. Establish first that the frame cannot sway, so the only unknowns are the rotations $\theta_2$ and $\theta_3$. Then write the slope-deflection equations, using the modified stiffness $3EI/L$ for the two members whose far ends are pinned, solve the two joint-equilibrium equations, and back-substitute.

Prove there is no sidesway. Member 2–5 is vertical, inextensible and pinned to the ground at 5, so joint 2 cannot move vertically: $v_2 = 0$. Member 1–2 is inextensible with unit vector $(0.8,\,0.6)$ and joint 1 is a pin support, so $0.8u_2 + 0.6v_2 = 0$, giving $u_2 = 0$. The horizontal member then carries $u_3 = u_2 = 0$, and the vertical column 3–4, inextensible and fixed at 4, gives $v_3 = 0$. Both joints are held, so the only kinematic unknowns are the two rotations. This is why the paper states that the members are inextensible — without it the sway degree of freedom would have to be carried and the question would be far longer.
Fixed-end moments. Only member 2–3 is loaded: $$\text{FEM}_{23} = -\frac{wL^2}{12} = -\frac{9(10)^2}{12} = -75.0\ \text{kN}\,\text{m}, \qquad \text{FEM}_{32} = +75.0\ \text{kN}\,\text{m}$$
Slope-deflection equations. With $\psi = 0$ throughout and the modified form $M = (3EI/L)\theta$ for a member whose far end is a pin, $$M_{21} = \tfrac{3EI}{5}\theta_2 = 0.6EI\theta_2, \qquad M_{25} = \tfrac{3EI}{5}\theta_2 = 0.6EI\theta_2$$ $$M_{23} = \tfrac{2(2EI)}{10}(2\theta_2 + \theta_3) - 75.0 = 0.8EI\theta_2 + 0.4EI\theta_3 - 75.0$$ $$M_{32} = 0.8EI\theta_3 + 0.4EI\theta_2 + 75.0, \qquad M_{34} = \tfrac{2(1.5EI)}{5}(2\theta_3) = 1.2EI\theta_3, \qquad M_{43} = 0.6EI\theta_3$$ with $M_{12} = M_{52} = 0$ at the two pinned bases.
Joint equilibrium. The moments meeting at each rigid joint must sum to zero: $$\text{Joint 2:}\quad M_{21} + M_{25} + M_{23} = 0 \;\Rightarrow\; 2.0EI\theta_2 + 0.4EI\theta_3 = 75.0$$ $$\text{Joint 3:}\quad M_{32} + M_{34} = 0 \;\Rightarrow\; 0.4EI\theta_2 + 2.0EI\theta_3 = -75.0$$
Solve. The system is antisymmetric, so $EI\theta_3 = -EI\theta_2$ and $(2.0-0.4)EI\theta_2 = 75.0$: $$\boxed{EI\theta_2 = 46.875\ \text{kN}\,\text{m}^2, \qquad EI\theta_3 = -46.875\ \text{kN}\,\text{m}^2}$$
Back-substitute for the end moments. $$M_{21} = M_{25} = 0.6(46.875) = 28.125\ \text{kN}\,\text{m}$$ $$M_{23} = 0.8(46.875) + 0.4(-46.875) - 75.0 = -56.25\ \text{kN}\,\text{m}, \qquad M_{32} = +56.25\ \text{kN}\,\text{m}$$ $$M_{34} = 1.2(-46.875) = -56.25\ \text{kN}\,\text{m}, \qquad M_{43} = 0.6(-46.875) = -28.125\ \text{kN}\,\text{m}$$ Both joint checks close exactly: $28.125 + 28.125 - 56.25 = 0$ and $56.25 - 56.25 = 0$.
Member 2–3: shear and moment. The end moments are equal and opposite in the clockwise convention, so they contribute nothing to the shear and the reactions are those of a simple span: $$V_2 = \frac{wL}{2} + \frac{M_{23}+M_{32}}{L} = 45.0 + 0 = 45.0\ \text{kN}, \qquad V_3 = -45.0\ \text{kN}$$ Zero shear at mid-span, $x = 5$ m, and the sagging moment there is $$M(5) = M_{23} + V_2(5) - \frac{9(5)^2}{2} = -56.25 + 225.0 - 112.5 = \boxed{+56.25\ \text{kN}\,\text{m}}$$ with hogging moments of $-56.25$ kN·m at both ends. The two points of contraflexure sit at $x = 1.46$ m and $x = 8.54$ m, from $-56.25 + 45x - 4.5x^2 = 0$.
Unloaded members. Each carries a linear moment diagram and a constant shear $V = -(M_{ij}+M_{ji})/L$: member 3–4 runs from $-56.25$ kN·m at joint 3 to $+28.125$ kN·m at the fixed base (a point of contraflexure 3.33 m below joint 3) with a constant shear of 16.875 kN; members 1–2 and 2–5 each run from zero at their pinned end to $28.125$ kN·m at joint 2, with a constant shear of 5.625 kN.

Member	End moments (kN·m)	Shear (kN)	Extreme moment ordinates (kN·m)
1–2 (EI, 5 m, pinned at 1)	$M_{12}=0$, $M_{21}=+28.125$	5.625 constant	max 0, min −28.125 at joint 2
2–3 (2EI, 10 m, 9 kN/m)	$M_{23}=-56.25$, $M_{32}=+56.25$	+45.0 at 2, −45.0 at 3	max +56.25 at mid-span, min −56.25 at both ends
3–4 (1.5EI, 5 m, fixed at 4)	$M_{34}=-56.25$, $M_{43}=-28.125$	16.875 constant	max +28.125 at joint 4, min −56.25 at joint 3
2–5 (EI, 5 m, pinned at 5)	$M_{25}=+28.125$, $M_{52}=0$	5.625 constant	max +28.125 at joint 2, min 0 at joint 5
Rotations	$EI\theta_2 = +46.875$, $EI\theta_3 = -46.875\ \text{kN}\,\text{m}^2$

Topic. Slope-deflection analysis of a non-sway rigid frame.

Key relations. $M_{ij} = \dfrac{2EI}{L}\left(2\theta_i + \theta_j - 3\psi\right) + \text{FEM}_{ij}$, with the modified form $M_{ij} = \dfrac{3EI}{L}\left(\theta_i - \psi\right) + \text{FEM}^{\text{mod}}_{ij}$ and $M_{ji}=0$ when the far end is a pin. For a UDL, $\text{FEM} = \mp wL^2/12$. Joint equilibrium $\sum M = 0$ at each rigid joint supplies one equation per unknown rotation. Span shear $V_i = wL/2 + (M_{ij}+M_{ji})/L$ and the peak span moment is $M_i + V_i^2/2w$.

Why this works. Slope deflection is the displacement method: express every end moment in terms of the joint displacements, then impose equilibrium. The number of unknowns is the kinematic indeterminacy, which here is two, even though the frame is statically indeterminate to the fourth degree — that inversion is the whole point of the method. Using $3EI/L$ for a pin-ended member is not an approximation; it is the exact result of eliminating the pinned end's rotation in advance, and it halves the work whenever a base is pinned.

Common pitfall. Assuming sidesway must be included because the frame has an inclined member. The two inextensible members meeting at joint 2 lock it in both directions, so $\psi = 0$ everywhere; carrying a sway unknown would add a storey-shear equation and a great deal of arithmetic for a zero answer. The other frequent slip is using $4EI/L$ for members 1–2 and 2–5 and then forgetting to carry $M_{12}$ and $M_{52}$ back to zero.

Source. Hibbeler, Structural Analysis 10th ed., Ch. 11 (slope deflection: §11.3 beams, §11.4 frames without sidesway) and Ch. 12 (moment distribution); Kassimali 6th ed., Ch. 16.

Question 7 of 8: Frame by Slope Deflection — Shear and Moment Diagrams

Question 7: Frame by Slope Deflection — Shear and Moment Diagrams (22 marks)